Split (2)

train · ~239k rows (showing the first 191k)

id stringlengths 1 6	url stringlengths 16 1.82k	content stringlengths 37 9.64M
17000	https://www.unitsconverters.com/en/M3-To-Fm3/Utu-225-9189	m³ to fm³ \| Cubic Meter to fm³ \| m³ to Cubic Femtometer Units Converters AccelerationAngleAreaEnergyForceLengthPowerPressureSpeedTemperatureTimeVolumeWeight Percentage error Subtract fraction LCM of three numbers Discover more Unit Converter m³ to fm³ (Cubic Meter to Cubic Femtometer) -10%Copy+10%-10%Copy+10% = ⇄ m³ to A³ \| m³ to fL \| m³ to mm³ \| m³ to mL \| m³ to cm³ \| m³ to L \| m³ to gal (US) \| m³ to gal (UK) \| m³ to ft³ \| m³ to bbl \| m³ to yd³ \| m³ to nm³ More 👎) 👍 Converting 2X of 1 ▶1/2X of 1 ▶ 5X of 1 ▶1/5X of 1 ▶ 8X of 1 ▶1/8X of 1 ▶ Cubic Angstrom Smallest Cubic Meter Base Volume of Earth Biggest Result 1 m³ is equivalent to 1E+45 fm³ Discover more Unit Converter Home » Volume » m³ to fm³ Formula Used 1 Cubic Meter = 1E+45 Cubic Femtometer ∴ 1 Cubic Meter = 1E+45 Cubic Femtometer Other m³ Conversions m³ to Earth Volume⇄ [Cubic Meter to Volume of Earth⇄] (Biggest) m³ to EL⇄ [Cubic Meter to Exaliter⇄] m³ to PL⇄ [Cubic Meter to Petaliter⇄] m³ to mi³⇄ [Cubic Meter to Cubic Mile⇄] m³ to TL⇄ [Cubic Meter to Teraliter⇄] m³ to km³⇄ [Cubic Meter to Cubic Kilometer⇄] m³ to GL⇄ [Cubic Meter to Gigaliter⇄] m³ to Acre ft (US)⇄ [Cubic Meter to Acre-Foot (US Survey)⇄] m³ to acft⇄ [Cubic Meter to Acre-Foot⇄] m³ to ML⇄ [Cubic Meter to Megaliter⇄] m³ to acin⇄ [Cubic Meter to Acre-Inch⇄] m³ to Dekastere⇄ [Cubic Meter to Dekastere⇄] m³ to cd⇄ [Cubic Meter to Cord⇄] m³ to ccf⇄ [Cubic Meter to Centum Cubic Foot⇄] m³ to hcf⇄ [Cubic Meter to Hundred-Cubic Foot⇄] m³ to ton reg⇄ [Cubic Meter to Ton Register⇄] m³ to am³⇄ [Cubic Meter to Cubic Attometer⇄] m³ to fm³⇄ [Cubic Meter to Cubic Femtometer⇄] (You are Here) m³ to m³⇄ [Cubic Meter to Cubic Meter⇄] m³ to μm³⇄ [Cubic Meter to Cubic Micrometer⇄] m³ to nm³⇄ [Cubic Meter to Cubic Nanometer⇄] m³ to pm³⇄ [Cubic Meter to Cubic Picometer⇄] m³ to kL⇄ [Cubic Meter to Kiloliter⇄] m³ to st⇄ [Cubic Meter to Stere⇄] m³ to tu⇄ [Cubic Meter to Tun⇄] m³ to yd³⇄ [Cubic Meter to Cubic Yard⇄] m³ to Hhd⇄ [Cubic Meter to Hogshead⇄] m³ to Cor (Biblical)⇄ [Cubic Meter to Cor (Biblical)⇄] m³ to Homer (Biblical)⇄ [Cubic Meter to Homer (Biblical)⇄] m³ to bbl (UK)⇄ [Cubic Meter to Barrel (UK)⇄] m³ to bbl (oil)⇄ [Cubic Meter to Barrel (Oil)⇄] m³ to bbl⇄ [Cubic Meter to Oil Barrel⇄] m³ to bbl (US)⇄ [Cubic Meter to Barrel (US)⇄] m³ to hL⇄ [Cubic Meter to Hectoliter⇄] m³ to Decistere⇄ [Cubic Meter to Decistere⇄] m³ to ft³⇄ [Cubic Meter to Cubic Foot⇄] m³ to Bath (Biblical)⇄ [Cubic Meter to Bath (Biblical)⇄] m³ to daL⇄ [Cubic Meter to Decaliter⇄] m³ to gal (UK)⇄ [Cubic Meter to Gallon (UK)⇄] m³ to gal (US)⇄ [Cubic Meter to Gallon (US)⇄] m³ to Hin (Biblical)⇄ [Cubic Meter to Hin (Biblical)⇄] m³ to fbm⇄ [Cubic Meter to Board Foot⇄] m³ to Cab (Biblical)⇄ [Cubic Meter to Cab (Biblical)⇄] m³ to qt (UK)⇄ [Cubic Meter to Quart (UK)⇄] m³ to L⇄ [Cubic Meter to Liter⇄] m³ to dm³⇄ [Cubic Meter to Cubic Decimeter⇄] m³ to qt (US)⇄ [Cubic Meter to Quart (US)⇄] m³ to pt (UK)⇄ [Cubic Meter to Pint (UK)⇄] m³ to pt (US)⇄ [Cubic Meter to Pint (US)⇄] m³ to Log (Biblical)⇄ [Cubic Meter to Log (Biblical)⇄] m³ to Cup (UK)⇄ [Cubic Meter to Cup (UK)⇄] m³ to Cup (Metric)⇄ [Cubic Meter to Cup (Metric)⇄] m³ to Cup (US)⇄ [Cubic Meter to Cup (US)⇄] m³ to Taza⇄ [Cubic Meter to Taza (Spanish)⇄] m³ to gi (UK)⇄ [Cubic Meter to Gill (UK)⇄] m³ to gi⇄ [Cubic Meter to Gill (US)⇄] m³ to dL⇄ [Cubic Meter to Deciliter⇄] m³ to fl oz (US)⇄ [Cubic Meter to Fluid Ounce (US)⇄] m³ to fl oz (UK)⇄ [Cubic Meter to Fluid Ounce (UK)⇄] m³ to Tbsp (UK)⇄ [Cubic Meter to Tablespoon (UK)⇄] m³ to in³⇄ [Cubic Meter to Cubic Inch⇄] m³ to Tbsp (Metric)⇄ [Cubic Meter to Tablespoon (Metric)⇄] m³ to Tbsp (US)⇄ [Cubic Meter to Tablespoon (US)⇄] m³ to Dstspn (UK)⇄ [Cubic Meter to Dessertspoon (UK)⇄] m³ to cL⇄ [Cubic Meter to Centiliter⇄] m³ to Dstspn (US)⇄ [Cubic Meter to Dessertspoon (US)⇄] m³ to Tsp (UK)⇄ [Cubic Meter to Teaspoon (UK)⇄] m³ to Tsp (Metric)⇄ [Cubic Meter to Teaspoon (Metric)⇄] m³ to Tsp (US)⇄ [Cubic Meter to Teaspoon (US)⇄] m³ to dr⇄ [Cubic Meter to Dram⇄] m³ to cm³⇄ [Cubic Meter to Cubic Centimeter⇄] m³ to mL⇄ [Cubic Meter to Milliliter⇄] m³ to Minim (US)⇄ [Cubic Meter to Minim (US)⇄] m³ to Minim (UK)⇄ [Cubic Meter to Minim (UK)⇄] m³ to Drop⇄ [Cubic Meter to Drop⇄] m³ to mm³⇄ [Cubic Meter to Cubic Millimeter⇄] m³ to μL⇄ [Cubic Meter to Microliter⇄] m³ to nL⇄ [Cubic Meter to Nanoliter⇄] m³ to pL⇄ [Cubic Meter to Picoliter⇄] m³ to fL⇄ [Cubic Meter to Femtoliter⇄] m³ to aL⇄ [Cubic Meter to Attoliter⇄] m³ to A³⇄ [Cubic Meter to Cubic Angstrom⇄] (Smallest) Discover more Unit Converter m³ to fm³ Conversion The abbreviation for m³ and fm³ is cubic meter and cubic femtometer respectively. 1 m³ is 1E+45 times bigger than a fm³. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including m³ to fm³ conversion. Cubic Meter to fm³ Check our Cubic Meter to fm³ converter and click on formula to get the conversion factor. When you are converting volume from Cubic Meter to fm³, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert. m³ to Cubic Femtometer The formula used to convert m³ to Cubic Femtometer is 1 Cubic Meter = 1E+45 Cubic Femtometer. Measurement is one of the most fundamental concepts. Note that we have Volume of Earth as the biggest unit for length while Cubic Angstrom is the smallest one. Convert m³ to fm³ How to convert m³ to fm³? Now you can do m³ to fm³ conversion with the help of this tool. In the length measurement, first choose m³ from the left dropdown and fm³ from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from fm³ to m³? You can check our fm³ to m³ converter. m³ to fm³ Converter Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like volume finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like m³ to fm³ through multiplicative conversion factors. When you are converting volume, you need a Cubic Meter to Cubic Femtometer converter that is elaborate and still easy to use. Converting m³ to Cubic Femtometer is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Cubic Meter to fm³, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in m³ to fm³ conversion along with a table representing the entire conversion. Discover more Unit Converter HindiFrenchSpanishMarathiPortugueseGermanPolishDutchItalianRussianGujaratiPunjabiTurkishKorean Percentage calculatorFraction calculatorLCM HCF Calculator About Contact Disclaimer Terms of Use Privacy Policy © 2016-2025. A softUsvista venture! Let Others Know ✖ Facebook Twitter Reddit LinkedIn Email WhatsApp Copied!
17001	https://www.khanacademy.org/math/in-class-7-math-foundation/xe6a68b2010f94f8c:integers	Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails.
17002	https://cs.uwaterloo.ca/journals/JIS/VOL8/Gessel/xin.pdf	23 11 Article 05.2.3 Journal of Integer Sequences, Vol. 8 (2005), 2 3 6 1 47 A Combinatorial Interpretation of the Numbers 6 (2n)! /n! (n + 2)! Ira M. Gessel1 and Guoce Xin Department of Mathematics Brandeis University Waltham, MA 02454-9110 USA gessel@brandeis.edu guoce.xin@gmail.com Abstract It is well known that the numbers (2m)! (2n)!/m! n! (m + n)! are integers, but in general there is no known combinatorial interpretation for them. When m = 0 these numbers are the middle binomial coeﬃcients ¡2n n ¢ , and when m = 1 they are twice the Catalan numbers. In this paper, we give combinatorial interpretations for these numbers when m = 2 or 3. 1 Introduction The Catalan numbers Cn = 1 n + 1 µ2n n ¶ = (2n)! n! (n + 1)! are well-known integers that arise in many combinatorial problems. Stanley [11, pp. 219–229] gives 66 combinatorial interpretations of these numbers. In 1874 Catalan observed that the numbers (2m)! (2n)! m! n! (m + n)! (1.1) are integers, and their number-theoretic properties were studied by several authors (see Dickson [4, pp. 265–266]). For m = 0, (1.1) is the middle binomial coeﬃcient ¡2n n ¢ , and for m = 1 it is 2Cn. 1Partially supported by NSF grant DMS-0200596. 1 Except for m = n = 0, these integers are even, and it is convenient for our purposes to divide them by 2, so we consider the numbers T(m, n) = 1 2 (2m)! (2n)! m! n! (m + n)!. Some properties of these numbers are given in , where they are called “super Catalan numbers”. An intriguing problem is to ﬁnd a combinatorial interpretation to the super Catalan numbers. The following identity [5, Equation (32)], together with the symmetry property T(m, n) = T(n, m) and the initial value T(0, 0) = 1, shows that T(m, n) is a positive integer for all m and n. X n 2p−2n µ p 2n ¶ T(m, n) = T(m, m + p), p ≥0. (1.2) Formula (1.2) allows us to construct recursively a set of cardinality T(m, n) but it is diﬃcult to give a natural description of this set. Shapiro gave a combinatorial interpretation to (1.2) in the case m = 1, which is the Catalan number identity X n 2p−2n µ p 2n ¶ Cn = Cp+1. A similar interpretation works for the case m = 0 of (1.2) (when multiplied by 2), which is the identity X n 2p−2n µ p 2n ¶µ2n n ¶ = µ2p p ¶ . Another intriguing formula for T(m, n), which does not appear in , is 1 + ∞ X m,n=1 CmCnxmyn = µ 1 − ∞ X m,n=1 T(m, n)xmyn ¶−1 , (1.3) which can easily be proved using the the generating function for 2T(m, n) given in formulas (35) and (37) of . Although (1.3) suggests a combinatorial interpretation for T(m, n) based on a decomposition of pairs of objects counted by Catalan numbers, we have not found such an interpretation. In this paper, we give a combinatorial interpretation for T(2, n) = 6 (2n)! /n! (n + 2)! for n ≥1 and for T(3, n) = 60 (2n)! /n! (n + 3)! for n ≥2. The ﬁrst few values of T(m, n) for m = 2 and m = 3 are as follows: m\n 0 1 2 3 4 5 6 7 8 9 10 2 3 2 3 6 14 36 99 286 858 2652 8398 3 10 5 6 10 20 45 110 286 780 2210 6460 We show that T(2, n) counts pairs of Dyck paths of total length 2n with heights diﬀering by at most 1. We give two proofs of this result, one combinatorial and one using generating functions. The combinatorial proof is based on the easily checked formula T(2, n) = 4Cn −Cn+1 (1.4) 2 which we interpret by inclusion-exclusion. Our interpretation for T(3, n) is more complicated, and involves triples of Dyck paths with height restrictions. Although we have the formula T(3, n) = 16Cn −8Cn+1 + Cn+2 analogous to (1.4), we have not found a combinatorial interpretation to this formula, and our proof uses generating functions. Interpretations of the number T(2, n) in terms of trees, related to each other, but not, apparently, to our interpretation, have been found by Schaeﬀer , and by Pippenger and Schleich [7, pp. 34]. A combinatorial interpretation of (1.2) for m = 2, using Pippenger and Schleich’s interpretation of T(2, n), has been given by Callan . 2 The main theorem All paths in this paper have steps (1, 1) and (1, −1), which we call up steps and down steps. A step from a point u to a point v is denoted by u →v. The level of a point in a path is deﬁned to be its y-coordinate. A Dyck path of semilength n (or of length 2n) is a path that starts at (0, 0), ends at (2n, 0), and never goes below level 0. It is well-known that the number of Dyck paths of semilength n equals the Catalan number Cn. The height of a path P, denoted by h(P), is the highest level it reaches. Every nonempty Dyck path R can be factored uniquely as UPDQ, where U is an up step, D is a down step, and P and Q are Dyck paths. Thus the map R 7→(P, Q) is a bijection from nonempty Dyck paths to pairs of Dyck paths. Let Bn be the set of pairs of Dyck paths (P, Q) of total semilength n. This bijection gives \|Bn\| = Cn+1, so by (1.4), we have T(2, n) = 4Cn −\|Bn\|. Our interpretation for T(2, n) is a consequence of the following Lemma 2.1. We give two proofs of this lemma, one combinatorial and the other algebraic. The algebraic proof will be given in the next section. Lemma 2.1. For n ≥1, Cn equals the number of pairs of Dyck paths (P, Q) of total semilength n, with P nonempty and h(P) ≤h(Q) + 1. Proof. Let Dn be the set of Dyck paths of semilength n, and let En be the set of pairs of Dyck paths (P, Q) of total semilength n, with P nonempty and h(P) ≤h(Q) + 1. We ﬁrst establish a bijection from En to Dn. For a given pair (P, Q) in En, since P is nonempty, the last step of P must be a down step, say, u →v. By replacing u →v in P with an up step u →v′, we get a path F1. Now raising Q by two levels, we get a path F2. Thus F := F1F2 is a path that ends at level 2 and never goes below level 0. The point v′ belongs to both F1 and F2, but we treat it as a point only in F2, even if F2 is the empty path. The condition that h(P) ≤h(Q) + 1 yields h(F1) < h(F2), which implies that the highest point of F must belong to F2. See Figure 1 below. Now let y be the leftmost highest point of F (which is in F2), and let x →y be the step in F leading to y. Then x →y is an up step. By replacing x →y with a down step x →y′, and lowering the part of F2 after y by two levels, we get a Dyck path D ∈Dn. See Figure 2 below. 3 P u v Q F1 u F v′ F2 Figure 1: First step of the bijection u F v′ F1 x y F2 u D v′ y′ x Figure 2: Second step of the bijection With the following two key observations, it is easy to see that the above procedure gives a bijection from En to Dn. First, x in the ﬁnal Dyck path D is the rightmost highest point. Second, u in the intermediate path F is the rightmost point of level 1 in both F and F1. Theorem 2.2. For n ≥1, the number T(2, n) counts pairs of Dyck paths (P, Q) of total semilength n with \|h(P) −h(Q)\| ≤1. Proof. Let F be the set of pairs of Dyck paths (P, Q) with h(P) ≤h(Q) + 1, and let G be the set of pairs of Dyck paths (P, Q) with h(Q) ≤h(P) + 1. By symmetry, we see that \|F\| = \|G\|. Now we claim that the cardinality of F is 2Cn. This claim follows from Lemma 2.1 and the fact that if P is the empty path, then h(P) ≤h(Q) + 1 for every Q ∈Dn. Clearly we have that F ∪G = Bn, and that F ∩G is the set of pairs of Dyck paths (P, Q), with \|h(P) −h(Q)\| ≤1. The theorem then follows from the following computation: \|F ∩G\| = \|F\| + \|G\| −\|F ∪G\| = 4Cn −\|Bn\| = 4Cn −Cn+1. 4 3 An algebraic proof and further results In this section we give an algebraic proof of Lemma 2.1. Although not as simple or elegant as the proof given in section 2, this proof generalizes to a larger class of paths with bounded height, while the combinatorial proof of Lemma 2.1 does not seem to generalize easily. Let c(x) be the generating function for the Catalan numbers, so that c(x) = ∞ X n=0 1 n + 1 µ2n n ¶ xn = 1 −√1 −4x 2x . Then c(x) satisﬁes the functional equation c(x) = 1 + xc(x)2. Let C = xc(x)2 = c(x) −1 and let Gk be the generating function for Dyck paths of height at most k. Although Gk is a rational function, and explicit formulas for it as a quotient of polynomials are well-known, a formula for Gk in terms of C will be of more use to us. An equivalent formula can be found in [1, Equation (16)]. Lemma 3.1. For k ≥−1, Gk = (1 + C)1 −Ck+1 1 −Ck+2. (3.1) Proof. Let P be a path of height at most k ≥1. If P is nonempty then P can be factored as UP1DP2, where U is an up step, P1 is a Dyck path of height at most k −1 (shifted up one unit), D is a down step, and P2 is a Dyck path of height at most k. Thus Gk = 1+xGk−1Gk, so Gk = 1/(1 −xGk−1). Equation (3.1) clearly holds for k = −1 and k = 0. Now suppose that for some k ≥1, Gk−1 = (1 + C) 1 −Ck 1 −Ck+1. Then the recurrence, together with the formula x = C/(1 + C)2, gives Gk = · 1 −x(1 + C) 1 −Ck 1 −Ck+1 ¸−1 = · 1 − C 1 + C 1 −Ck 1 −Ck+1 ¸−1 = · 1 −Ck+2 (1 + C)(1 −Ck+1) ¸−1 = (1 + C)1 −Ck+1 1 −Ck+2. We can prove Lemma 2.1 by showing that P∞ n=0 Gn+1(Gn −Gn−1) = 1 + 2C; this is equivalent to the statement that the number of pairs (P, Q) of Dyck paths of semilength m > 0 with h(P) ≤h(Q) + 1 is 2Cm. Theorem 3.2. ∞ X n=0 (Gn −Gn−1)Gn+1 = 1 + 2C. 5 Proof. Let Ψk = ∞ X n=k Cn 1 −Cn. Thus if j < k then Ψj = k−1 X n=j Cn 1 −Cn + Ψk. We have GnGn+1 = (1 + C)21 −Cn+1 1 −Cn+3 and Gn−1Gn+1 = (1 + C)2 (1 −Cn)(1 −Cn+2) (1 −Cn+1)(1 −Cn+3). Let S1 = ∞ X n=0 µ1 −Cn+1 1 −Cn+3 −1 ¶ and S2 = ∞ X n=0 µ (1 −Cn)(1 −Cn+2) (1 −Cn+1)(1 −Cn+3) −1 ¶ . Then P∞ n=0(Gn −Gn−1)Gn+1 = (1 + C)2(S1 −S2). We have 1 −Cn+1 1 −Cn+3 −1 = −(1 −C2)Cn+1 1 −Cn+3 , so S1 = −(1 −C2)C−2Ψ3, and (1 −Cn)(1 −Cn+2) (1 −Cn+1)(1 −Cn+3) −1 = − (1 −C)Cn (1 + C)(1 −Cn+1) − (1 −C3)Cn+1 (1 + C)(1 −Cn+3), so S2 = −1 −C 1 + C C−1Ψ1 −1 −C3 1 + C C−2Ψ3. Therefore S1 −S2 = 1 −C 1 + C C−1Ψ1 + µ1 −C3 1 + C −(1 −C2) ¶ C−2Ψ3 = 1 −C 1 + C C−1(Ψ1 −Ψ3) = 1 −C 1 + C C−1 µ C 1 −C + C2 1 −C2 ¶ = 1 + 2C (1 + C)2. Thus (1 + C)2(S1 −S2) = 1 + 2C. 6 By similar reasoning, we could prove Theorem 2.2 directly: The generating function for pairs of paths with heights diﬀering by at most 1 is ∞ X n=0 (Gn −Gn−1)(Gn+1 −Gn−2), where we take G−1 = G−2 = 0, and a calculation like that in the proof of Theorem 3.2 shows that this is equal to 1 + 2C −C2 = 4c(x) −c(x)2 −2 = 4c(x) −c(x) −1 x −2 = 1 + ∞ X n=1 (4Cn −Cn+1)xn = 1 + ∞ X n=1 T(2, n)xn. Although the fact that the series in Theorem 3.2 telescopes may seem surprising, we shall see in Theorem 3.4 that it is a special case of a very general result on sums of generating functions for Dyck paths with restricted heights. In the following lemma, the fact that C = c(x) −1 is not used, and in fact C may be completely arbitrary, as long as the series involved converge. Lemma 3.3. Let R(z, C) be a rational function of z and C of the form zN(z, C) Qm i=1(1 −zCai), where N(z, C) is a polynomial in z of degree less than m with coeﬃcients that are rational functions of C, and the ai are distinct positive integers. Let L = −limz→∞R(z, C). Then ∞ X n=0 R(Cn, C) = Q(C) + LΨ1, where Q(C) is a rational function of C, and Ψ1 = P∞ n=1 Cn/(1 −Cn). Proof. First we show that the lemma holds for R(z, C) = z/(1 −zCa). In this case, L = −limz→∞R(z, C) = C−a and ∞ X n=0 R(Cn, C) = ∞ X n=0 Cn 1 −Cn+a = C−a ∞ X n=a Cn 1 −Cn = − a−1 X n=0 Cn−a 1 −Cn + C−aΨ1. Now we consider the general case. Since R(z, C)/z is a proper rational function of z, it has a partial fraction expansion 1 zR(z, C) = m X i=1 Ui(C) 1 −zCai 7 for some rational functions Ui(C), so R(z, C) = m X i=1 Ui(C) z 1 −zCai . The general theorem then follows by applying the special case to each summand. Theorem 3.4. Let i1, i2, . . . , im be distinct integers and let j1, j2, . . . , jm be distinct integers. Then ∞ X n=0 (Gn+i1Gn+i2 · · · Gn+im −Gn+j1Gn+j2 · · · Gn+jm) (3.2) is a rational function of C. Proof. By (3.1), Gn+i1Gn+i2 · · · Gn+im (1 + C)m −1 = R(Cn, C), where R(z, C) = 1 −zCi1+1 1 −zCi1+2 · · · 1 −zCim+1 1 −zCim+2 −1. Then by Lemma 3.3, ∞ X n=0 µGn+i1Gn+i2 · · · Gn+im (1 + C)m −1 ¶ is a rational function of C. Similarly, ∞ X n=0 µGn+j1Gn+j2 · · · Gn+jm (1 + C)m −1 ¶ is a rational function of C, and the result follows easily. 4 A combinatorial interpretation for T(3, n) It is natural to ask whether the super Catalan numbes T(m, n) for m > 2 have combinatorial interpretations similar to that of Theorem 2.2. Using the partial fraction procedure described in the proof of Lemma 3.3, it is straightforward (with the help of a computer algebra system) to evaluate as rational functions of C the sums that count k-tuples of paths with height restrictions when Theorem 3.4 applies. By rationalizing the denominator, we can express any rational function of C in the form A(x) + B(x) √ 1 −4x, (4.1) where A(x) and B(x) are rational functions. Such a formula can be related to the super Catalan numbers with the help of the following formula. 8 Lemma 4.1. (1 −4x)m−1/2 = m−1 X k=0 (−4)k µm −1/2 k ¶ xk + 2(−1)m ∞ X n=0 T(m, n)xm+n. (4.2) Proof. It is easily veriﬁed that T(m, n) = 1 2(−1)n4m+n¡m−1/2 m+n ¢ . Thus (1 −4x)m−1/2 = m−1 X k=0 (−4)k µm −1/2 k ¶ xk + ∞ X k=m (−4)k µm −1/2 k ¶ xk = m−1 X k=0 (−4)k µm −1/2 k ¶ xk + ∞ X n=0 (−4)m+n µm −1/2 m + n ¶ xm+n = m−1 X k=0 (−4)k µm −1/2 k ¶ xk + 2(−1)m ∞ X n=0 T(m, n)xm+n. Thus if in (4.1) we expand the numerator of B1(x) in powers of 1 −4x, we will get an expression involving rational functions and generating functions for super Catalan numbers. For example, the sum ∞ X n=0 (Gn −Gn−1)(Gn+2 −Gn+1)(Gn+4 −Gn+3) (4.3) counts triples of paths whose heights are n, n + 2, and n + 4 for some n. It can easily be expressed in terms of sums to which Theorem 3.4 applies, and we ﬁnd that (4.3) is equal to C6(1 + C2)(1 + 2C2 + C4 −C5 + C6 −2C7 + C8) (1 + C2)2(1 + C + C2)(1 + C + C2 + C3 + C4)2 = (1 −3x)(1 −13x + 63x2 −140x3 + 142x4 −56x5 + 6x6) 2(1 −x)2(1 −2x)2(1 −3x + x2)2 − (1 −4x)5/2 (1 −x)(1 −2x)(1 −3x + x2). (4.4) The rational functions that appear in (4.4) can be simpliﬁed by partial fraction expansion, and we can write down an explicit formula involving T(3, n) for the coeﬃcients of (4.4). What we obtain is far from a combinatorial interpretation of T(3, n), but the computation suggests that perhaps some modiﬁcation of this set of paths might lead to the desired interpretation. We note also that sums with m paths instead of three empirically give similar expressions with (1 −4x)m−1/2 instead of (1 −4x)5/2. Our strategy for ﬁnding a combinatorial interpretation for T(3, n) is to consider more general paths that give us counting formulas that generalize (3.1), in the hope that this additional generality may lead us to a combinatorial interpretation for T(3, n). In this we are partially successful: we do ﬁnd in Theorem 4.3 a set of triples of paths, not too diﬀerent 9 from the set counted by (4.3) whose generating function is P∞ n=0 T(3, n+1)xn plus a rational function, and moreover we can interpret the coeﬃcients of the rational function in terms of paths. Although the set of paths is not very natural, the result does suggest that there is some hope for this approach to give a nice combinatorial interpretation of T(3, n) and perhaps even for T(m, n). For our interpretation of T(3, n), we need to consider paths that end at levels greater than 0. Let us deﬁne a ballot path to be a path that starts at level 0 and never goes below level 0. In the previous section all our paths had an even number of steps, so it was natural to assign a path with n steps the weight xn/2. We shall continue to weight paths in this way, even though some of our paths now have odd lengths. Let G(j) k be the generating function for ballot paths of height at most k that end at level j. Lemma 4.2. For 0 ≤j ≤k + 1 we have G(j) k = Cj/2(1 + C)1 −Ck−j+1 1 −Ck+2 (4.5) Proof. The case j = 0 is Lemma 3.1. Now let W be a ballot path counted by G(j) k , where j > 0, so that W is of height at most k and W ends at level j. Then W can be factored uniquely as W1UW2, where W1 is a path of height at most k that ends at level 0 and W2 is a path from level 1 to level j that never goes above level k nor below level 1. Using √x = p C/(1 + C)2 = √ C/(1 + C), we obtain G(j) k = Gk √xG(j−1) k−1 = (1 + C)1 −Ck+1 1 −Ck+2 · √ C 1 + C · G(j−1) k−1 = √ C 1 −Ck+1 1 −Ck+2G(j−1) k−1 , and (4.5) follows by induction on j. We note an alternative formula that avoids half-integer powers of C, G(j) k = xj/2(1 + C)j+11 −Ck−j+1 1 −Ck+2 , which follow easily from (4.5) and the formula √x = √ C/(1 + C). There is a similar formula for the generating function G(i,j) k for paths of height at most k that start at level i, end at level j, and never go below level 0: for 0 ≤i ≤j ≤k + 1 we have G(i,j) k = C(j−i)/2(1 + C)(1 −Ci+1)(1 −Ck−j+1) (1 −C)(1 −Ck+2) , (4.6) with G(i,j) k = G(j,i) k for i > j. Although we will not use (4.6) in this paper, it may be helpful in further applications of this method. We have not found (4.5) or (4.6) in the literature, although they may be derived from the known rational generating function for G(i,j) k described below. 10 We note two variants of (4.6), also valid for 0 ≤i ≤j ≤k + 1: G(i,j) k = x(j−i)/2(1 + C)j−i+1(1 −Ci+1)(1 −Ck−j+1) (1 −C)(1 −Ck+2) , = x−1/2C(j−i+1)/2(1 −Ci+1)(1 −Ck−j+1) (1 −C)(1 −Ck+2) . It is well known that G(i,j) k is x(j−i)/2 times a rational function of x, and it is useful to have an explicit formula for it as a quotient of polynomials. (See Sato and Cong and Krattenthaler .) Let us deﬁne polynomials pn = pn(x) by pn(x) = X 0≤k≤n/2 (−1)k µn −k k ¶ xk. The ﬁrst few values are p0 = 1 p1 = 1 p2 = 1 −x p3 = 1 −2x p4 = 1 −3x + x2 p5 = 1 −4x + 3x2 p6 = 1 −5x + 6x2 −x3 These polynomials can be expressed in terms of the Chebyshev polynomials of the second kind Un(x) by pn(x) = xn/2Un µ 1 2√x ¶ . It is not diﬃcult to show that pn = 1 −Cn+1 (1 −C)(1 + C)n, and thus we obtain G(i,j) k = x(j−i)/2pipk−j pk+1 , for 0 ≤i ≤j ≤k, and in particular, G(j) k = xj/2pk−j/pk+1 and Gk = pk/pk+1. We can now describe our combinatorial interpretation of T(3, n): T(3, n) counts triples of ballot paths whose heights are k, k −2, and k −4 for some k, ending at levels 4, 3, and 2, together with some additional paths of height at most 5. (Note that if a path of height k −4 ends at level 2, then k must be at least 6.) More precisely, let H (j) k be the generating function for ballot paths of height k that end at level j, so that H (j) k = G(j) k −G(j) k−1. Then we have: 11 Theorem 4.3. 1 + ∞ X n=0 T(3, n + 1)xn = √x ∞ X k=6 H(4) k H(3) k−2H(2) k−4 + 2G1 + 2G2 + G3 + G5. (4.7) Proof. With the help of Lemma 4.1 we ﬁnd that −(1 −4x)5/2 2x4 −10 x + 15 x2 −5 x3 + 1 2x4 = ∞ X n=0 T(3, n + 1)xn. (4.8) Using the method described in Lemma 3.3 and Theorem 3.4, we ﬁnd, with the help of Maple, that the sum √x ∞ X k=6 H(4) k H(3) k−2H(2) k−4 is equal to −(1 −4x)5/2 2x4 −10 x + 15 x2 −5 x3 + 1 2x4 +1− 2 1 −x −2 1 −x 1 −2x − 1 −2x 1 −3x + x2 − 1 −4x + 3x2 1 −5x + 6x2 −x3. Then (4.7) follows from (4.8), the formula Gk = pk/pk+1, and the formulas for pk, k = 1, . . . , 6. References N. G. de Bruijn, D. E. Knuth, and S. O. Rice, “The average height of planted plane trees”, in Graph Theory and Computing, R. C. Read, ed., Academic Press, New York-London, 1972, pp. 15–22. D. Callan, A combinatorial interpretation for a super-Catalan recurrence, J. Integer Seq. 8 (2005), no. 1, Article 05.1.8, 7 pp. (electronic). E. Catalan, Question 1135, Nouvelles Annales de Math´ ematiques (2) 13 (1874), 207. L. E. Dickson, History of the Theory of Numbers, New York: Chelsea. Vol. 1, 1966. Originally published in 1919 by the Carnegie Institute of Washington. I. Gessel, Super ballot numbers, J. Symb. Comput. 14 (1992), 179–194. C. Krattenthaler, Permutations with restricted patterns and Dyck paths, Adv. Appl. Math. 27 (2001), 510–530. N. Pippenger and K. Schleich, Topological characteristics of random surfaces generated by cubic interactions, arXiv:physics.gr-qc/0306049. M. Sato and T. T. Cong, The number of minimal lattice paths restricted by two parallel lines, Disc. Math. 43 (1983), 249–261. 12 G. Schaeﬀer, A combinatorial interpretation of super-Catalan numbers of order two, , (2001). L. W. Shapiro, A short proof of an identity of Touchard’s concerning Catalan numbers, J. Combin. Theory Ser. A 20 (1976), 375–376. R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, 1999. 2000 Mathematics Subject Classiﬁcation: Primary 05A10; Secondary 05A15. Keywords: Dyck paths, super Catalan numbers (Concerned with sequences A007054 and A007272.) Received March 5 2005; revised version received March 16 2005. Published in Journal of Integer Sequences, April 27 2005. Return to Journal of Integer Sequences home page. 13
17003	https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_General_Chemistry/03%3A_Composition_of_Substances_and_Solutions/3.3%3A_Molarity	3.3: Molarity - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 3: Composition of Substances and Solutions UK: General Chemistry { } { "3.1:Formula_Mass_and_the_Mole_Concept" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.2:_Determining_Empirical_and_Molecular_Formulas" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.3:_Molarity" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.4:_Other_Units_for_Solution_Concentrations" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "3.E:_Composition_of_Substances_and_Solutions(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "01:_Essential_Ideas" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "02:_Atoms_Molecules_and_Ions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "03:_Composition_of_Substances_and_Solutions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "04:_Stoichiometry_of_Chemical_Reactions" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "05:_Thermochemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "06:_Electronic_Structure_and_Periodic_Properties_of_Elements" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "07:_Chemical_Bonding_and_Molecular_Geometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "08:_Advanced_Theories_of_Covalent_Bonding" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "09:_Gases" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "10:_Liquids_and_Solids" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11:_Solutions_and_Colloids" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "12:_Kinetics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "13:_Fundamental_Equilibrium_Concepts" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "14:_Acid-Base_Equilibria" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "15:_Equilibria_of_Other_Reaction_Classes" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "16:_Thermodynamics" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "17:_Electrochemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "18:_Representative_Metals_Metalloids_and_Nonmetals" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "19:_Transition_Metals_and_Coordination_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "20:_Organic_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "21:_Nuclear_Chemistry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", Appendices : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Wed, 05 Jun 2019 18:45:47 GMT 3.3: Molarity 105555 105555 admin { } Anonymous Anonymous User 2 false false [ "article:topic", "Author tag:OpenStax", "concentration", "molarity", "aqueous solution", "concentrated", "diluted", "dilution", "dissolved", "solute", "solvent", "authorname:openstax", "showtoc:no", "license:ccbyncsa", "licenseversion:40" ] [ "article:topic", "Author tag:OpenStax", "concentration", "molarity", "aqueous solution", "concentrated", "diluted", "dilution", "dissolved", "solute", "solvent", "authorname:openstax", "showtoc:no", "license:ccbyncsa", "licenseversion:40" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Expand/collapse global hierarchy 1. Home 2. Campus Bookshelves 3. University of Kentucky 4. UK: General Chemistry 5. 3: Composition of Substances and Solutions 6. 3.3: Molarity Expand/collapse global location 3.3: Molarity Last updated Jun 5, 2019 Save as PDF 3.2: Determining Empirical and Molecular Formulas 3.4: Other Units for Solution Concentrations Page ID 105555 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Solutions 2. Dilution of Solutions 3. Summary 4. Key Equations 5. Glossary 6. Contributors Skills to Develop Describe the fundamental properties of solutions Calculate solution concentrations using molarity Perform dilution calculations using the dilution equation In preceding sections, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (Figure 3.3.1). In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified. Figure 3.3.1:Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage’s sweetness. (credit: Jane Whitney) Solutions We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions. The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution. A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration). Example 3.3.1: Calculating Molar Concentrations A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage? Solution Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L: M=m⁢o⁢l s⁢o⁢l⁢u⁢t⁢e L s⁢o⁢l⁢u⁢t⁢i⁢o⁢n=0.133 m⁢o⁢l 355 m⁢L×1 L 1000 m⁢L=0.375 M Exercise 3.3.1 A teaspoon of table sugar contains about 0.01 mol sucrose. What is the molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL? Answer 0.05 M Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution: (3.3.1)M=mol solute L solution Example 3.3.2: Deriving Moles and Volumes from Molar Concentrations How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example 3.3.1? Solution In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in Example 3.4.2, 0.375 M: M=mol solute L solution \begin{align} \mathrm{mol\: solute} &= \mathrm{ M\times L\: solution} \label{3.4.4} \[5pt] \mathrm{mol\: solute} &= \mathrm{0.375\:\dfrac{mol\: sugar}{L}\times \left(10\:mL\times \dfrac{1\:L}{1000\:mL}\right)} &= \mathrm{0.004\:mol\: sugar} \label{3.4.5} \end{align} Exercise 3.3.2 What volume (mL) of the sweetened tea described in Example 3.3.1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example? Answer 80 mL Example 3.3.3: Calculating Molar Concentrations from the Mass of Solute Distilled white vinegar (Figure 3.3.2) is a solution of acetic acid, C⁢H 3⁡C⁢O 2⁢H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity? Figure 3.3.3: Distilled white vinegar is a solution of acetic acid in water. Solution As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles: (3.3.2)M=mol solute L solution=25.2⁢g⁡CH⁢A 3⁢CO⁢A 2⁢H×1 mol⁢CH⁢A 3⁢CO⁢A 2⁢H 60.052⁢g⁡CH⁢A 3⁢CO⁢A 2⁢H 0.500 L solution=0.839 M (3.3.3)M=mol solute L solution=0.839 M (3.3.4)M=0.839 mol solute 1.00 L solution Exercise 3.3.3 Calculate the molarity of 6.52 g of C⁢o⁢C⁢l 2 (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL. Answer 0.674 M Example 3.3.4: Determining the Mass of Solute in a Given Volume of Solution How many grams of NaCl are contained in 0.250 L of a 5.30-M solution? Solution The volume and molarity of the solution are specified, so the amount (mol) of solute is easily computed as demonstrated in Example 3.3.3: (3.3.5)M=mol solute L solution (3.3.6)mol solute=M×L solution (3.3.7)mol solute=5.30 mol NaCl L×0.250 L=1.325 mol NaCl Finally, this molar amount is used to derive the mass of NaCl: (3.3.8)1.325 mol NaCl×58.44⁢g⁡NaCl mol NaCl=77.4⁢g⁡NaCl Exercise 3.3.4 How many grams of C⁢a⁢C⁢l 2 (110.98 g/mol) are contained in 250.0 mL of a 0.200-M solution of calcium chloride? Answer 5.55 g C⁢a⁢C⁢l 2 When performing calculations stepwise, as in Example 3.3.3, it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In Example 3.3.4, the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g. In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (Example 3.3.5). This eliminates intermediate steps so that only the final result is rounded. Example 3.3.5: Determining the Volume of Solution In Example 3.3.3, we found the typical concentration of vinegar to be 0.839 M. What volume of vinegar contains 75.6 g of acetic acid? Solution First, use the molar mass to calculate moles of acetic acid from the given mass: (3.3.9)g⁡solute×mol solute g⁡solute=mol solute Then, use the molarity of the solution to calculate the volume of solution containing this molar amount of solute: (3.3.10)mol solute×L solution mol solute=L solution Combining these two steps into one yields: (3.3.11)g⁡solute×mol solute g⁡solute×L solution mol solute=L solution (3.3.12)75.6⁢g⁡CH⁢A 3⁢CO⁢A 2⁢H⁢(mol⁢CH⁢A 3⁢CO⁢A 2⁢H 60.05 g)⁢(L solution 0.839 mol⁢CH⁢A 3⁢CO⁢A 2⁢H)=1.50 L solution Exercise 3.3.5: What volume of a 1.50-M KBr solution contains 66.0 g KBr? Answer 0.370 L Dilution of Solutions Dilution is the process whereby the concentration of a solution is lessened by the addition of solvent. For example, we might say that a glass of iced tea becomes increasingly diluted as the ice melts. The water from the melting ice increases the volume of the solvent (water) and the overall volume of the solution (iced tea), thereby reducing the relative concentrations of the solutes that give the beverage its taste (Figure 3.3.2). Figure 3.3.2: Both solutions contain the same mass of copper nitrate. The solution on the right is more dilute because the copper nitrate is dissolved in more solvent. (credit: Mark Ott). Dilution is also a common means of preparing solutions of a desired concentration. By adding solvent to a measured portion of a more concentrated stock solution, we can achieve a particular concentration. For example, commercial pesticides are typically sold as solutions in which the active ingredients are far more concentrated than is appropriate for their application. Before they can be used on crops, the pesticides must be diluted. This is also a very common practice for the preparation of a number of common laboratory reagents (Figure 3.3.3). Figure 3.3.3: A solution of K⁢M⁢n⁢O 4 is prepared by mixing water with 4.74 g of KMnO4 in a flask. (credit: modification of work by Mark Ott) A simple mathematical relationship can be used to relate the volumes and concentrations of a solution before and after the dilution process. According to the definition of molarity, the molar amount of solute in a solution is equal to the product of the solution’s molarity and its volume in liters: (3.3.13)n=M⁢L Expressions like these may be written for a solution before and after it is diluted: (3.3.14)n 1=M 1⁢L 1 (3.3.15)n 2=M 2⁢L 2 where the subscripts “1” and “2” refer to the solution before and after the dilution, respectively. Since the dilution process does not change the amount of solute in the solution, n 1 = n 2. Thus, these two equations may be set equal to one another: (3.3.16)M 1⁢L 1=M 2⁢L 2 This relation is commonly referred to as the dilution equation. Although we derived this equation using molarity as the unit of concentration and liters as the unit of volume, other units of concentration and volume may be used, so long as the units properly cancel per the factor-label method. Reflecting this versatility, the dilution equation is often written in the more general form: (3.3.17)C 1⁢V 1=C 2⁢V 2 where C and V are concentration and volume, respectively. Example 3.3.6: Determining the Concentration of a Diluted Solution If 0.850 L of a 5.00-M solution of copper nitrate, Cu(NO 3)2, is diluted to a volume of 1.80 L by the addition of water, what is the molarity of the diluted solution? Solution We are given the volume and concentration of a stock solution, V 1 and C 1, and the volume of the resultant diluted solution, V 2. We need to find the concentration of the diluted solution, C 2. We thus rearrange the dilution equation in order to isolate C 2: (3.3.18)C 1⁢V 1=C 2⁢V 2 (3.3.19)C 2=C 1⁢V 1 V 2 Since the stock solution is being diluted by more than two-fold (volume is increased from 0.85 L to 1.80 L), we would expect the diluted solution’s concentration to be less than one-half 5 M. We will compare this ballpark estimate to the calculated result to check for any gross errors in computation (for example, such as an improper substitution of the given quantities). Substituting the given values for the terms on the right side of this equation yields: (3.3.20)C 2=0.850 L×5.00 mol L 1.80 L=2.36 M This result compares well to our ballpark estimate (it’s a bit less than one-half the stock concentration, 5 M). Exercise 3.3.6 What is the concentration of the solution that results from diluting 25.0 mL of a 2.04-M solution of CH3OH to 500.0 mL? Answer 0.102 M C⁢H 3⁡O⁢H Example 3.3.7: Volume of a Diluted Solution What volume of 0.12 M HBr can be prepared from 11 mL (0.011 L) of 0.45 M HBr? Solution We are given the volume and concentration of a stock solution, V 1 and C 1, and the concentration of the resultant diluted solution, C 2. We need to find the volume of the diluted solution, V 2. We thus rearrange the dilution equation in order to isolate V 2: (3.3.21)C 1⁢V 1=C 2⁢V 2 (3.3.22)V 2=C 1⁢V 1 C 2 Since the diluted concentration (0.12 M) is slightly more than one-fourth the original concentration (0.45 M), we would expect the volume of the diluted solution to be roughly four times the original volume, or around 44 mL. Substituting the given values and solving for the unknown volume yields: (3.3.23)V 2=(0.45 M)⁢(0.011 L)(0.12 M) (3.3.24)V 2=0.041 L The volume of the 0.12-M solution is 0.041 L (41 mL). The result is reasonable and compares well with our rough estimate. Exercise 3.3.7 A laboratory experiment calls for 0.125 M H⁡N⁢O 3. What volume of 0.125 M H⁡N⁢O 3 can be prepared from 0.250 L of 1.88 M H⁡N⁢O 3? Answer 3.76 L Example 3.3.8: Volume of a Concentrated Solution Needed for Dilution What volume of 1.59 M KOH is required to prepare 5.00 L of 0.100 M KOH? Solution We are given the concentration of a stock solution, C 1, and the volume and concentration of the resultant diluted solution, V 2 and C 2. We need to find the volume of the stock solution, V 1. We thus rearrange the dilution equation in order to isolate V 1: (3.3.25)C 1⁢V 1=C 2⁢V 2 (3.3.26)V 1=C 2⁢V 2 C 1 Since the concentration of the diluted solution 0.100 M is roughly one-sixteenth that of the stock solution (1.59 M), we would expect the volume of the stock solution to be about one-sixteenth that of the diluted solution, or around 0.3 liters. Substituting the given values and solving for the unknown volume yields: (3.3.27)V 1=(0.100 M)⁢(5.00 L)1.59 M (3.3.28)V 1=0.314 L Thus, we would need 0.314 L of the 1.59-M solution to prepare the desired solution. This result is consistent with our rough estimate. Exercise 3.3.8 What volume of a 0.575-M solution of glucose, C 6 H 12 O 6, can be prepared from 50.00 mL of a 3.00-M glucose solution? Answer 0.261 L Summary Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution. Key Equations M=mol solute L solution C 1 V 1 = C 2 V 2 Glossary aqueous solution solution for which water is the solvent concentrated qualitative term for a solution containing solute at a relatively high concentration concentration quantitative measure of the relative amounts of solute and solvent present in a solution dilute qualitative term for a solution containing solute at a relatively low concentration dilution process of adding solvent to a solution in order to lower the concentration of solutes dissolved describes the process by which solute components are dispersed in a solvent molarity (M)unit of concentration, defined as the number of moles of solute dissolved in 1 liter of solution solute solution component present in a concentration less than that of the solvent solvent solution component present in a concentration that is higher relative to other components Contributors Paul Flowers (University of North Carolina - Pembroke),Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors.Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at This page titled 3.3: Molarity is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax. Back to top 3.2: Determining Empirical and Molecular Formulas 3.4: Other Units for Solution Concentrations Was this article helpful? Yes No Recommended articles 3.3: MolaritySolutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved... 1.7: MolaritySolutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved... 6.5: Weight by Volume and MolaritySolutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved... 3.3: MolaritySolutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved... 6.3: MolaritySolutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved... Article typeSection or PageAuthorOpenStaxLicenseCC BY-NC-SALicense Version4.0Show Page TOCno on page Tags aqueous solution Author tag:OpenStax concentrated concentration diluted dilution dissolved molarity solute solvent © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status×
17004	https://www.ncbi.nlm.nih.gov/books/NBK396239/	Introduction - Screening for Dyslipidemia in Younger Adults - NCBI Bookshelf An official website of the United States government Here's how you know The .gov means it's official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you're on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. Log inShow account info Close Account Logged in as: username Dashboard Publications Account settings Log out Access keysNCBI HomepageMyNCBI HomepageMain ContentMain Navigation Bookshelf Search database Search term Search Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. Chou R, Dana T, Blazina I, et al. Screening for Dyslipidemia in Younger Adults: A Systematic Review to Update the 2008 U.S. Preventive Services Task Force Recommendation [Internet]. Rockville (MD): Agency for Healthcare Research and Quality (US); 2016 Nov. (Evidence Syntheses, No. 138.) Screening for Dyslipidemia in Younger Adults: A Systematic Review to Update the 2008 U.S. Preventive Services Task Force Recommendation [Internet]. Show details Evidence Syntheses, No. 138. Chou R, Dana T, Blazina I, et al. Rockville (MD): Agency for Healthcare Research and Quality (US); 2016 Nov. Contents Search term < PrevNext > 1 Introduction Go to: Purpose and Previous U.S. Preventive Services Task Force Recommendation The purpose of this report is to update prior reviews1-3 conducted for the U.S. Preventive Services Task Force (USPSTF) on screening for lipid disorders in adults. It will be used by the USPSTF to update its 2008 recommendation.4 In 2008, the USPSTF strongly recommended lipid screening in all men age 35 years and older and in women age 45 years and older at increased risk for coronary heart disease (CHD) (A recommendation) based on good evidence that lipid-lowering drug therapy decreases the incidence of CHD events in persons with abnormal lipid levels, resulting in substantial absolute benefits.4 The USPSTF recommended screening in men ages 20 to 35 years and women ages 20 to 45 years with risk factors for CHD (B recommendation); due to the lower incidence of CHD events in these populations, screening results in lower expected benefits. The USPSTF made no recommendation for or against lipid screening in men ages 20 to 35 years or in women age 20 years and older not at increased risk (C recommendation) due to small expected benefits. A difference between this update and prior USPSTF reviews on lipid screening is that it focuses on screening in younger adults (defined in this report as adults ages 21 to 39 years). The USPSTF restricted the scope of this update to younger adults because in older adults, lipid levels are obtained as part of routine cardiovascular risk assessment, and the decision to initiate statin therapy is often based on a global assessment of cardiovascular risk or presence of cardiovascular risk factors in addition to abnormal lipid levels. In younger adults, however, there is more uncertainty about the need to perform cardiovascular risk assessment, and lipid screening might identify those who would benefit from earlier interventions to reduce lipid levels.5 A separate evidence review has been commissioned by the USPSTF on the use of statins for the prevention of cardiovascular disease (CVD) in adults age 40 years and older.6 Go to: Condition Definition Lipid disorders refer to abnormalities of cholesterol, including low-density lipoprotein cholesterol (LDL-C), high-density lipoprotein cholesterol (HDL-C), and triglycerides. The National Cholesterol Education Project (NCEP) Adult Treatment Panel III (ATP III) defined “optimal” LDL-C levels as less than 100 mg/dL and “high” as 160 mg/dL and greater; “desirable” total cholesterol (TC) levels as less than 200 mg/dL and “high” as 240 mg/dL and greater; “low” HDL-C levels as less than 40 mg/dL; and elevated triglycerides as greater than 150 mg/dL, although thresholds for treatment varied depending on the presence of risk factors for CVD.7 Go to: Prevalence and Burden of Disease/Illness The prevalence of lipid disorders is high in the United States, with an estimated 53 percent (105.3 million) of adults affected.8 Specifically, 27 percent of Americans (53.5 million) have high LDL-C, 23 percent (46.4 million) have low HDL-C, and 30 percent (58.9 million) have high triglycerides. According to National Health and Nutrition Examination Survey data from 2003 to 2006, 13 percent of adults with high LDL-C, 22 percent of adults with high non–HDL-C, and 8 percent of adults with the combination of high LDL-C, low HDL-C, and high triglycerides were ages 20 to 34 years.8 In 2003 to 2004, 64 percent of adults ages 20 to 29 years and 57 percent of adults ages 30 to 39 years met NCEP-recommended lipid levels.9 Lipid disorders are associated with CHD, which may lead to sudden coronary death and myocardial infarction. Prevalence of CHD increases with age and is higher in men than in women of the same age.10 In 2010, the overall prevalence of CHD was 6.0 percent, and among persons ages 18 to 44 years, the age-adjusted prevalence was 1.2 percent.10 Prevalence of CHD varies by race/ethnicity, affecting 11.6 percent of American Indians/Alaska Natives, 6.5 percent of blacks, 6.1 percent of Hispanics, 5.8 percent of whites, and 3.9 percent of Asian/Pacific Islanders. For young adults ages 20 to 39 years, the prevalence of CVD, including all diseases of the circulatory system and congenital CVD, was 14.9 percent in men and 8.7 percent in women.11 CHD is the leading cause of death in the United States.12,13 In 2013, the American Heart Association (AHA) estimated that approximately 635,000 Americans would have a new myocardial infarction or CHD death and 280,000 would have a recurrent cardiovascular event, with an additional 150,000 persons having silent myocardial infarctions.14 The number of myocardial infarctions or fatal CHD events is estimated at 20,000 annually for men ages 35 to 44 years and 5,000 annually for women ages 35 to 44 years.14 In 2011, CHD caused 12 percent of deaths in persons ages 25 to 44 years.13 Estimates based on Framingham Heart Study participants from 1971 to 1996 indicate that the lifetime risks (through age 80 years) of CHD for 40-year-old men with a TC level of 200, 200 to 239, and 240 mg/dL and greater were 31, 43, and 57 percent, respectively, with 10-year cumulative risks of 3, 5, and 12 percent. For younger adults, data from the Chicago Heart Association study (from 1967 to 1973), with mortality followup in 2002, estimated 10-year CHD mortality in the highest-risk decile as 0.58 percent in those ages 18 to 29 years and 1.72 percent in those ages 30 to 39 years.15 In 2010, heart disease was associated with 972 age-adjusted potential life-years lost per 100,000 persons younger than age 75 years.16,17 In 2008, heart disease and stroke accounted for nearly $300 billion in health care costs.18 Go to: Etiology and Natural History Cholesterol is a lipid that is present in all animal cells; it is vital to cell membrane structure and acts as a precursor to vitamin D, adrenal and gonadal steroid hormones, and bile acids.19 The body is able to absorb dietary cholesterol and also synthesize it de novo. In a typical Western diet, cholesterol intake is about 300 to 450 mg per day and endogenous cholesterol amounts to 800 to 1,400 mg per day. A total of 1,000 to 2,000 mg of cholesterol can be absorbed by the small intestine. Plasma cholesterol levels depend on many factors, including diet and genetics. In the general population, there is great variability in how cholesterol is synthesized and absorbed. Plasma cholesterol levels are the sum of intestinal cholesterol absorption and hepatic cholesterol synthesis balanced by net biliary excretion and cell use. Cholesterol is transported in the body as particles of lipid and protein (lipoproteins).7 There are three classes of lipoproteins: LDL-C, HDL-C, and very low-density lipoprotein cholesterol (VLDL-C). LDL-C makes up 60 to 70 percent of total serum cholesterol, HDL-C contributes 20 to 30 percent, and VLDL-C contributes 10 to 15 percent. LDL-C is the primary atherogenic lipoprotein and is the primary target of cholesterol-lowering therapy, although some forms of VLDL-C are precursors to LDL-C and promote atherosclerosis. HDL-C is inversely related to risk for CHD. LDL-C is atherogenic when it accumulates in blood vessels, contributing to plaque formation. The fully developed plaque consists of a core of cholesterol, surrounded by a capsule of connective tissue.20 The plaque core is surrounded by foam cells, which are macrophages containing intracytoplasmic cholesterol. These cells produce procoagulant and inflammatory cell mediators. Early-stage plaque formation is not associated with structural damage to the endothelium, but later-stage plaque formation leads to endothelial erosion that exposes the underlying connective tissue and allows platelets to adhere to the site, potentially leading to plaque smooth muscle cell growth through release of growth factor. Further endothelial erosion and disruption contribute to thrombus formation. As the thrombus builds, blood flow sends clumps of platelets into the distal small arteries as emboli and the thrombus may continue to grow until it occludes the artery, resulting in myocardial infarction, cerebrovascular accident (CVA), or another ischemic event. Endothelial erosion and disruption result from enhanced inflammatory activity within the plaque produced by smooth muscle cells and macrophages. Certain plaque characteristics such as a large lipid core, high density of macrophages, and low density of smooth muscle cells in the cap are markers of plaques that are more likely to undergo thrombosis. The risk of a person with coronary artery disease having a future thrombogenic event is more associated with the presence and number of vulnerable plaques than the total number of plaques. Exposure to nonoptimal lipid levels in young adulthood is associated with atherosclerotic changes later in life. One prospective cohort study of 2,824 persons ages 18 to 30 years with nonoptimal levels of LDL-C (defined as ≥100 mg/dL) at baseline found an association between cumulative exposure to higher LDL-C or lower HDL-C levels and markers of atherosclerosis two decades later.21 Persons with familial hypercholesterolemia may have dramatically high levels of LDL-C, which can lead to accelerated atherosclerosis and, if untreated, early cardiovascular death.22-24 Familial hypercholesterolemia is caused by mutations in the LDL receptor gene, which reduce the number of LDL-C receptors or prevent LDL-C from binding to these receptors, thereby reducing LDL-C removal from the blood. Patients with two mutated copies of the LDL receptor gene have the homozygous form of familial hypercholesterolemia. This condition is rare, with a prevalence of about 1 in 1,000,000.25 The characteristic clinical presentation includes skin and tendon xanthomas, TC levels of 500 to 1,000 mg/dL, and the onset in childhood of symptomatic coronary disease as well as aortic valve and proximal root disease.26,27 The heterozygous form of familial hypercholesterolemia is more common, with a prevalence of approximately 1 in 500 in the United States and United Kingdom. TC levels in persons with heterozygous familial hypercholesterolemia are less highly elevated than in those with homozygous familial hypercholesterolemia, averaging 325 to 450 mg/dL, but patients are also at increased risk for CHD and death in young adulthood due to prolonged exposure to high lipid levels that often starts in childhood.23 The estimated proportion of persons with familial hypercholesterolemia who would have an early-onset CHD event in the absence of recognition and treatment is 5 to 15 percent in men younger than age 35 years and 10 to 15 percent in women younger than age 45 years.28,29 Many patients with severe hypercholesterolemia do not have an identifiable genetic defect.24 Evidence suggests that the clinical consequences of severe hypercholesterolemia are the same regardless of the underlying cause. Go to: Risk Factors Risk factors for dyslipidemia (high LDL-C, low HDL-C, high triglycerides) include physical inactivity, obesity, abdominal obesity, metabolic syndrome, hypertension, atherogenic diet (high in saturated fatty acids, cholesterol, and sodium), consumption of added dietary sugars, genetic factors (including family history of familial hypercholesterolemia), older age, male sex, and hypothyroidism.7,8,30,31 Elevated triglycerides are associated with overweight and obesity, physical inactivity, smoking, excess alcohol intake, high carbohydrate diet, other diseases such as diabetes and nephritic syndrome, medications such as corticosteroids or estrogens, and genetic factors.7 Hyperlipidemia is also associated with HIV infection, renal transplant, and use of antipsychotic medications and protease inhibitors.32-34 Dyslipidemia is a risk factor for CHD.7 Other modifiable risk factors for CHD include hypertension, smoking, thrombogenic/hemostatic state, diabetes, obesity, physical inactivity, and atherogenic diet. Nonmodifiable risk factors include age (≥45 years in men, ≥55 years in women), male sex, and family history of early-onset CHD. Non–HDL-C (i.e., TC minus HDL-C) is a measure that includes all potentially atherogenic lipoprotein particles (LDL, VLDL, intermediate-density lipoprotein, and lipoprotein[a]) that may be a more accurate predictor of CHD risk than LDL-C.32,34,35 Apolipoprotein B is a direct measure of the total number of atherogenic particles, although it is unclear whether it adds to HDL-C and TC as a marker of CHD risk.34,36-39 In addition, TC and HDL-C are easier and less costly to measure. Other potential risk factors for CVD include alternative measures of lipid status, such as TC-to-HDL-C ratio or other lipoprotein levels, and nonlipid factors, such as inflammatory markers (e.g., C-reactive protein,40 homocysteine) and thrombogenic factors (e.g., fibrinogen, antithrombin III, factor V Leiden).7 In 2008, the USPSTF recommended screening with a fasting or nonfasting HDL-C level and either TC or a measure of LDL-C.4 In 2009, a USPSTF evidence review of nine emerging risk factors, including C-reactive protein, leukocyte count, homocysteine levels, and lipoprotein levels, found that evidence was insufficient to support the use of these risk factors to reclassify persons at intermediate risk for CHD as high risk, although it found that evidence for C-reactive protein was promising.40 Clinical practice guidelines continue to predominantly focus on LDL-C as the primary lipid risk factor. Go to: Rationale for Screening/Screening Strategies Due to the asymptomatic nature of lipid disorders, screening is required for detection. Detection of younger adults with lipid disorders could enable implementation of management strategies such as lifestyle modification or medications that could prevent negative cardiovascular outcomes in persons at immediate risk for an event or decrease risk of future events. Screening could be of particular benefit for identification of young adults with markedly elevated lipid levels due to unrecognized familial hypercholesterolemia. Screening involves blood tests that may be obtained in a fasting or nonfasting state. Although current recommendations generally recommend testing TC and LDL-C levels, they differ on the inclusion of other lipid components, the age at which to start testing, and the frequency of screening (see the “Recommendations of Other Groups” section). Go to: Interventions/Treatment Standard treatments for lipid disorders in adults include use of medications, diet, exercise, or a combination of these interventions. Prior to 2013, treatment in the United States generally followed recommendations from the Third Report of the NCEP ATP III, which recommended global cardiovascular risk evaluation, including measurement of lipids starting at age 20 years, to guide decisions regarding use of lipid-lowering therapy.7 LDL-C thresholds for initiation of lipid-lowering therapy following lifestyle intervention efforts varied from 130 mg/dL and greater to 190 mg/dL, depending on the assessed risk category (low: 10-year CVD event risk <10%; intermediate: 10% to 20%; high: >20%). Drug options for lipid reduction included statins, bile acid sequestrants, nicotinic acid, and fibrates, although statins were designated as the initial drug of choice given its proven efficacy for reducing LDL-C levels and evidence showing improved clinical outcomes. Statin or other lipid-lowering therapy was targeted to achieve LDL-C levels varying from less than 100 mg/dL to less than 160 mg/dL, depending on the risk category. Updated guidelines issued in 2013 from the American College of Cardiology (ACC) and the AHA on lipid-lowering therapy differ from those of ATP III in a number of ways. In the new guidelines, statins are the recommended first-line lipid-lowering therapy to reduce CVD risk, as evidence on its effectiveness in improving clinical outcomes is strongest.27 Target populations for statin therapy were redefined as four groups: persons with clinical CVD, persons ages 40 to 75 years with diabetes and LDL-C levels of 70 to 189 mg/dL, persons with LDL-C levels of 190 mg/dL or greater, or persons ages 40 to 75 years with an estimated 10-year CVD risk of 7.5 percent or greater. For patients in the latter group who do not meet criteria for one of the other target populations, a clinician-patient risk discussion is recommended prior to initiation of statin therapy. Rather than managing statin therapy to achieve a target LDL-C level, the ACC/AHA recommends fixed-dose statin therapy, with the intensity (based on the dose and potency of the statin used and, thus, the expected degree of LDL-C reduction) of therapy determined by the risk profile. The updated guidelines also recommend the use of a newly developed global risk calculator to estimate risk. For patients with familial hyperlipidemia, the National Lipid Association recommends lifestyle modification, moderate- to high-potency statins as first-line drug therapy (alternative drugs or combination therapy is recommended for persons who cannot tolerate statins or do not meet a LDL-C reduction target of ≥50% from baseline), and LDL apheresis in high-risk patients who do not meet lipid targets after lifestyle modification and drug therapy.41 Go to: Current Clinical Practice in the United States A study based on 1996 to 2006 National Health and Nutrition Examination Survey data from 2,587 adults ages 20 to 45 years found overall lipid screening rates of less than 50 percent.42 Screening rates varied based on the presence of cardiovascular risk factors. Lipid testing rates were 68 percent in adults with CHD or CHD equivalents, 47 percent in those with two or more risk factors, 45 percent in those with one risk factor, and 42 percent in those with no known risk factors. The presence of CHD or a CHD equivalent was associated with increased likelihood of screening compared with presence of no risk factors (relative risk, 1.5 [95% confidence interval (CI), 1.1 to 2.2]). In addition, women were more likely to have undergone screening compared with men. Among women with CHD or a CHD equivalent, two or more risk factors, one risk factor, or no risk factors, screening rates were 69, 53, 52, and 49 percent, respectively; corresponding rates for men were 64, 38, 36, and 30 percent, respectively.4,43 A study based on a 2005 National Ambulatory Medical Care Survey found disparities in rates of lipid screening in adults age 20 years and older, with higher rates in whites (40%) versus blacks (33%) or Hispanics (39%).44 Results were not reported separately for younger adults. Healthy People 2020 has set a target screening rate of 82 percent within the last 5 years for persons older than age 18 years (an increase from 75% in 2008).45 Go to: Recommendations of Other Groups Recommendations for lipid screening in young adults without risk factors for CHD vary, with some guidelines recommending screening starting at age 20 years and others not recommending screening until ages 35 to 40 years for men or 40 to 50 years for women. In general, guidelines recommend screening younger adults with CHD, CHD equivalents, or one or more CHD risk factors. The ATP III guidelines recommend screening all persons age 20 years and older every 5 years with LDL-C, HDL-C, TC, and triglycerides.7 It recommends that clinicians perform a lipoprotein analysis and risk factor evaluation to assign risk status as part of the first patient visit for adults age 20 years and older and, depending on the results, initiate efforts to control risk factors, with re-evaluation in 1 to 5 years, or initiate lifestyle changes.7 The ACC and AHA guidelines do not specifically address lipid screening but recommend, “as reasonable,” assessment of traditional cardiovascular risk factors every 4 to 6 years starting at age 20 years (grade IIa recommendation).5 Although the Pooled Cohort Equations risk calculator developed by the ACC and AHA estimates lifetime risk of atherosclerotic CVD in persons age 20 years and older, the guidelines do not make any recommendation to apply the risk calculator to persons younger than age 40 years. The American Diabetes Association recommends lipid screening in patients with diabetes at least annually, and every 2 years for adults with low-risk lipid values (LDL-C <100 mg/dL, HDL-C >50 mg/dL, and triglycerides <150 mg/dL).46 The European Society of Cardiology and the European Atherosclerosis Society recommend risk-level–based screening depending on various risk factors (e.g., diabetes, established CVD, hypertension, smoking, body mass index >30 kg/m 2, family history of premature CVD and familial dyslipidemia, chronic inflammatory disease, chronic kidney disease). Assessment of lipid levels may be considered in men older than age 40 years and women older than age 50 years.47 The Canadian Working Group on Hypercholesterolemia and Other Dyslipidemias recommends screening in men older than age 40 years, postmenopausal women, women older than age 50 years, and patients with diabetes mellitus, hypertension, smoking, abdominal obesity, strong family history of premature CVD manifestations of hyperlipidemia (e.g., xanthelasma, xanthoma, or arcus corneae), or evidence of symptomatic or asymptomatic atherosclerosis, as well as any patient for whom “lifestyle changes are indicated.” This group further recommends use of the NCEP ATP III risk estimation algorithm, with LDL-C level and TC-to-HDL-C ratio as targets.48 The American Academy of Family Physicians concurs with current USPSTF recommendations.49 The American College of Physicians Web site refers to an inactive guideline from 1996 recommending screening in men older than age 35 years and women older than age 45 years for TC levels.50 Purpose and Previous U.S. Preventive Services Task Force Recommendation Condition Definition Prevalence and Burden of Disease/Illness Etiology and Natural History Risk Factors Rationale for Screening/Screening Strategies Interventions/Treatment Current Clinical Practice in the United States Recommendations of Other Groups Copyright Notice Bookshelf ID: NBK396239 Contents < PrevNext > Share on Facebook Share on Twitter Views PubReader Print View Cite this Page PDF version of this title (424K) In this Page Purpose and Previous U.S. Preventive Services Task Force Recommendation Condition Definition Prevalence and Burden of Disease/Illness Etiology and Natural History Risk Factors Rationale for Screening/Screening Strategies Interventions/Treatment Current Clinical Practice in the United States Recommendations of Other Groups Other titles in these collections U.S. Preventive Services Task Force Evidence Syntheses, formerly Systematic Evidence Reviews Health Services/Technology Assessment Texts (HSTAT) Recent Activity Clear)Turn Off)Turn On) Introduction - Screening for Dyslipidemia in Younger AdultsIntroduction - Screening for Dyslipidemia in Younger Adults Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... Follow NCBI Connect with NLM National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov PreferencesTurn off External link. Please review our privacy policy. Cite this Page Close Chou R, Dana T, Blazina I, et al. Screening for Dyslipidemia in Younger Adults: A Systematic Review to Update the 2008 U.S. Preventive Services Task Force Recommendation [Internet]. Rockville (MD): Agency for Healthcare Research and Quality (US); 2016 Nov. (Evidence Syntheses, No. 138.) 1, Introduction. Available from: Making content easier to read in Bookshelf Close We are experimenting with display styles that make it easier to read books and documents in Bookshelf. Our first effort uses ebook readers, which have several "ease of reading" features already built in. The content is best viewed in the iBooks reader. You may notice problems with the display of some features of books or documents in other eReaders. Cancel Download Share Share on Facebook Share on Twitter URL
17005	https://arxiv.org/pdf/2401.16106	arXiv:2401.16106v3 [math.CO] 2 Jul 2024 NEW LOWER BOUNDS FOR r3(N) ZACH HUNTER Abstract. We develop recent ideas of Elsholtz, Proske, and Sauermann to construct denser subsets of {1, . . . , N } that lack arithmetic progressions of length 3. This gives the first quasipolynomial improvement since the original construction of Behrend. Introduction We say an additive set S is 3 -AP-free , if it lacks non-trivial arithmetic progressions of length three (meaning that all solutions to 2 y = x + z are of the form x = y = z). Let r3(N) denote the maximum cardinality of a 3-AP-free subset of {1, . . . , N }.For many years, the best-known lower bound has been 1 (1.1) r3(N) ≫ log C (N)N2−2√2 log 2(N ). This is based on a clever geometric construction due to Behrend from 1946 [ 1]. Since [ 1], the only improvement has been to the value of the constant ‘ C’ (the original argument gave C = −1/4, but a refinement of Elkin [ 2] (see also [ 7]) obtained C = 1 /4). In this paper, we shall get a quasipolynomial improvement to ( 1.1 ). Theorem 1. For N ≥ 1, we have r3(N) ≫ N2−(c+o(1))2 √2 log 2(N ), where c:= √√√√log 2 (√ 32 9 ) ≈0.9567 . Remark 1.1. The constant 9 /32 is not best possible. At the very least, one can obtain 7 /24 by using the set ‘ T ’ from [ 5, Definition 3.4] (as we discuss at the start of Section 4). However, it seems likely that neither constant is optimal. Thus we chose to use a slightly simpler construction to focus on the core ideas. 1.1. Additional discussion. It is useful to recall that Behrend-type arguments give rise to a more general parametrized range of bounds. Proposition 1.2. Given any integers N, D ≥ 1, we have r3(N) ≫ √DN 2−DN−2/D . Optimizing the above by taking D = ⌈√2 log 2(N)⌉ (so that N ≤ 2D2/2 and thus N−2/D ≥ 2−D) yields the aforementioned ( 1.1 ). Inspired by very recent work [ 5], we shall establish a new range of bounds. Date : July 3, 2024. 1 See Section 2 for our asymptotic notation. 12ZACH HUNTER Proposition 1.3. Fix any ǫ > 0. Given integers N, D ≥ 1, with D even, we have r3(N) ≫ǫ 1 D2 N(√9/32 − ǫ)DN−2/D . Noting that √9/32 > 0.53 > 1/2, Proposition 1.3 does noticeably better assuming that D is not too big. Whence, taking 2 D = ⌈√2 log 2(N)⌉, we obtain r3(N) ≫ log −12 (N)N(2 /0.53) −√2 log 2(N ) which already improves the bound ( 1.1 ) for all large N since (2 /0.53) < 3.78 < 4. More carefully picking D = ⌈√2 log √32 /9(N)⌉ (so that N−2/D ≥ (√32 /9) −D) gives r3(N) ≫ N(√9/32 − o(1)) 2D = N(2 + o(1)) −c2√2 log 2(N ), with c = √ log 2(√32 /9), recovering Theorem 1.1.2. Outline of ideas. We start by recalling the work of Elsholtz, Proske, and Sauer-mann [ 5] which we will build upon. In [ 5], the authors constructed a subset S of the 2-dimesional torus T2 := ( R/Z)2 with density μ(S) = 7 /24 which had certain additive properties (cf. Proposition 3.1 ). They use this set S as a building block, along with ideas from the classic Salem-Spencer construction [ 12 ], to create denser subsets of Fnp that lack arithmetic progressions of length 3 (using standard techniques, it easy to construct sets of size ≫p,ǫ (p/ 2 − ǫ)n,while they manage to get ≫p,ǫ (√7/24 p − ǫ)n). The reason they obtain this improvement is because their 2-dimensional “building block” S has density 7 /24 > (1 /2) 2. In the standard arguments, one instead starts with a 1-dimensional building block of density 1 /2. By taking product sets of S instead of the standard 1-dimensional building block, they get “useful” D-dimensional subsets which are denser (we now shrink by a √7/24-fraction per dimension, instead of 1 /2). This is simply more efficient and leads to better bounds. Recently, such types of improvements have been starting to become more frequent (see, e.g., [ 6, 4, 8]), making delightful progress on many other additive problems. Now, the main contribution of this paper will be to take an efficient 2-dimensional building block (like the set S considered by [ 5]), and get it to work in the integer setting. In principle, there are several challenges in doing this, we mention two: • There are significant differences between the group structure of Fnp and Z; [ 5]raises the specific concern that the arguments of both [ 12 ] and [ 1] must take into account potential carrying that may occur. • Generally speaking, Salem-Spencer-type properties are a bit more flexible than those of Behrend. However, in the integer setting, Salem-Spencer arguments have a much poorer quantitative performance. So we must somehow show that the relevant building block is in fact compatible with a Behrend-type argument. 2Technically, we should take D= 2 ⌈1 2 √2 log 2(N)⌉, so that Dis even, but this hardly matters. NEW LOWER BOUNDS FOR r3(N)3 The first bullet turns out to not be a serious issue; by using a variant of Behrend’s construction introduced by Green and Wolf [ 7], one can work in the torus and not worry about carries. We also note that if N = p1 · · · pD is a square-free number, then Z/N Z ∼= ( Z/p 1Z) × · · · × (Z/p DZ), which also allows one to get addition to work coordinate-wise, side-stepping carrying-related complications. The second bullet is more worrying. Currently, in the literature, there are several constructions and tricks over finite fields using Salem-Spencer-type arguments that have not been replicated over the integers. A notable example is [ 4], which showed how to construct subsets of Fnp of size ≫p (0 .99 p)n without arithmetic progressions of length 1000 but did not lead to any improvement for the analogous problem in the integer setting (despite the building blocks in [ 4] being much more efficient than the set S from [ 5]). The author has previously convinced himself that the argument of [ 4]seems “fundamentally untranslatable” to the integer setting, although in forthcoming work [ 9] we will obtain comparably strong improvements for that setting via different techniques. So while there is no reason a priori to be able to convert Salem-Spencer-type ar-guments into Behrend-type arguments, we manage to do this here. The first step is to show our building block almost enjoys the additive properties used in standard Behrend constructions (modulo certain exceptions and technical caveats). This can be extracted from intermediate elementary lemmata of [ 5], although one must be more careful and precise with certain details. From there we construct a pseudonorm which still manages to capitalize on these weaker additive properties, crucially relying on the specific nature of our technical caveats. Lastly, we show how to run a modified Behrend argument using these inputs, obtaining Proposition 1.3 . Acknowledgements. We thank Zach Chase for his support during the writing of this note. We also thank Daniel Carter, Ben Green, and Benny Sudakov for some advice and feedback on our writing. 2. Preliminaries 2.1. General notation. We use asymptotic Vinogradov notation. Thus for functions f (n), g (n), we say f ≫ g if there is some absolute constant c > 0 so that f (n) ≥ cg (n). We also write f (n) = o(1) to denote a quantity which tends to 0 as n → ∞ .Given an integer N ≥ 1, we define [ N] := {1, . . . , N }.We define a 3-AP to be an additive set P that can be written as P = {x, y, z } where 2y = x + z.2.2. Torus notation. We define TD := ( R/Z)D.Let π : RD → TD; x 7 → x + ZD be the standard projection map. As an abuse of notation, let π−1(θ) be the unique x ∈ [0 , 1) D so that π(x) = θ.Let {·} : [0 , 1) → [0 , 1/2) be the map 3 {x} = { x for x ∈ [0 , 1/2) x − 1/2 otherwise . 3Note that our definition of {·} differs from the standard definition. 4ZACH HUNTER We extend the above definition coordinate-wise (so that {(x1, . . . , x D)} = ( {x1}, . . . , {xD})). 2.3. Extra notation. Since we will be working with 2-dimensional “building blocks”, it will be convenient to have some specialized notation defined on T2.We define the sum-map ψ : T2 → [0 , 2); θ 7 → π−1(θ)1 + π−1(θ)2. We extend this coordinate-wise, i.e. for θ = ( θ(1) , . . . , θ (D0)) ∈ (T2)D0 , we define ψ(θ) := (ψ(θ(1) ), . . . , ψ (θ(D0))).Lastly, we define the projection p : T2 → [0 , 1) by p(θ) = π−1(θ)1. We extend this coordinate-wise in the same manner as ψ.3. Main Construction 3.1. Setup. We recall a construction from [ 5] which motivates our work. Proposition 3.1. There exists S ⊂ T2 with μ(S) = 7 /24 so that: If x, y, z ∈ π−1(S) solve π(x + z) = π(2 y), then: • (y1 + y2) ≥ (x1+x2)+( z1 +z2) 2 ; • if x1 + x2 = z1 + z2, then {y1} ≥ { x1} + {z1}. Remark 3.2. Bullet 1 is proved in [ 5, Fact 3.10] and Bullet 2 is proved in [ 5, Fact 3.11]. In the next section, we shall consider truncations of (a simplified version of) their construction, to obtain quantitative analogues to Proposition 3.1 . Precisely, we will establish the following. Proposition 3.3. Fix any ǫ > 0. There exists Sǫ ⊂ T2 with μ(Sǫ) ≥ 9/32 − ǫ so that: Given any 3-AP {θ, θ + α, θ + 2 α} ⊂ Sǫ, we either have: 2ψ(θ + α) ≥ ψ(θ) + ψ(θ + 2 α) + 1 /2; or otherwise, writing d := π−1(θ+2 α)−π−1(θ) 2 , so that ψ(θ + 2 α) = ψ(θ) + 2( d1 + d2), we have (3.1) ψ(θ + α) = ψ(θ) + ( d1 + d2). Furthermore, assuming \|d1 + d2\| ≤ ǫ 1000 , we also have (3.2) (1 − { p(θ)})2 + (1 − { p(θ + 2 α)})2 − 2(1 − { p(θ + α)})2 ≥ d21. The final two conditions will tell us we morally have “Freiman homomorphism”-esque properties (in the jargon of additive combinatorics). However, since we are working with a subset of T2 with density > 1/4, a simple averaging argument considering the subgroup π({0, 1/2}2) tells us that we can’t have an actual Freiman homomorphism. Thus it is natural and somewhat expected that the above statement is slightly messy and technical. NEW LOWER BOUNDS FOR r3(N ) 5 3.2. Finishing things off. Let ~1 denote the all 1’s vector in RD0 .To exploit Proposition 3.3 , we define three weights. Namely take w1(θ) := ‖ψ(θ)‖1 = D0 ∑ i=1 \|ψ(θ(i))\|,w2,ǫ (θ) := (10 10 ǫ−2)‖ψ(θ)‖22, and w3(θ) := ‖~1 − p(θ)‖22. Roughly speaking, the first weight will make us win if Eq. 3.1 fails to hold. Otherwise, the second weight (with its large constant) will let us win if Eq. 3.2 fails to hold. Lastly, when both hold, considering w2,ǫ (θ) + w3(θ) shall get the job done. Now, by considering product sets, we can establish the below: Proposition 3.4. Fix any ǫ > 0. For D0 ≥ 1 there exists Sǫ,D 0 ⊂ (T2)D0 with μ(Sǫ,D ) > (9 /32 − ǫ)D0 so that the following holds. Given a 3-AP {θ, θ + α, θ + 2 α} ⊂ Sǫ,D 0 , either 2w1(θ + α) ≥ w1(θ) + w1(θ + 2 α) + 1 /2, or otherwise writing d := π−1(θ+2 α)−π−1(θ) 2 ∈ ([ −1/2, 1/2] 2)D0 , we have that (w2,ǫ (θ) + w3(θ)) + ( w2,ǫ (θ + 2 α) + w3(θ + 2 α)) ≥ 2( w2,ǫ (θ + α) + w3(θ + α)) + D0 ∑ i=1 \|d(i)1 + d(i)2 \|2 + ( d(i)1 )2. Proof. Let Sǫ be the set from Proposition 3.3 . We take Sǫ,D 0 := SD0 ǫ (thus we get μ(Sǫ,D 0 ) = μ(Sǫ)D0 ≥ (9 /32 − ǫ)D0 ). Note that the weights sum coordinate-wise. We first handle w1. Let I1 be the set of indices i with w1(θ(i)) + w1(( θ + 2 α)(i)) 6 =2w1(( θ + α)(i)). We have that 2w1(θ + α) − (w1(θ) + w1(θ + 2 α)) ≥ \| I1\|/2(since for i ∈ I1, the i-th coordinate contributes ≥ 1/2 to the RHS, and for i 6 ∈ I1, the contribution is zero). The first condition of our criteria fails to hold only when I1 = ∅.So we now assume I1 = ∅. Write d := π−1(θ+2 α)−π−1 (θ) 2 . By assumption, Eq. 3.1 holds for all i ∈ [D0], whence the parallelogram law gives w2,ǫ (θ) + w2,ǫ (θ + 2 α) − 2w2,ǫ (θ + α) = (10 10 ǫ−2) D0 ∑ i=1 \|d(i)1 + d(i)2 \|2. Now let I2 be the set of indices i with \|d(i)1 + d(i)2 \| ≥ ǫ 1000 . Thus, one gets w2,ǫ (θ) + w2,ǫ (θ + 2 α) − 2w2,ǫ (θ + α) ≥ D0 ∑ i=1 \|d(i)1 + d(i)2 \|2 + 100 \|I2\|.6 ZACH HUNTER At the same time (invoking Eq. 3.2 for the coordinates not in I2, and noting w3(θ) ∈ [0 , 1] for θ ∈ T2 otherwise), we have that w3(θ) + w3(θ + 2 α) − 2w3(θ + α) ≥ ∑ i6∈I2 (d(i)1 )2 − 2\|I2\| ≥ D0 ∑ i=1 (d(i)1 )2 − 3\|I2\|. Summing the last two equations gives the claim. We are nearly done. We require a small observation. Lemma 3.5. Fix α ∈ (T2)D0 . For each θ ∈ (T2)D0 , there is a choice of ξ ∈ ({− 1/2, 0, 1/2}2)D0 so that π−1(θ + α) − π−1(θ) 2 = π−1(α) + ξ. Proof. We have that π(π−1(θ + α) − π−1(θ)) = α, thus π−1(θ + α) − π−1(θ) − π−1(α) ∈ (Z2)D0 . Meanwhile, π−1(θ + α), π −1(θ), π −1(α) ∈ ([0 , 1) 2)D0 , thus this difference lies in ({− 1, 0}2)D0 ⊂ ({− 1, 0, 1}2)D0 . Proof of Proposition 1.3 . Since D is even, we may write D = 2 D0 for some D0 ≥ 1. We now consider the set Sǫ,D 0 from Proposition 3.4 .Set δ := (1 /400) N−2/D . Let B0 be the set of d ∈ (R2)D0 so that D0 ∑ i=1 \|d(i)1 + d(i)2 \|2 + ( d(i)1 )2 < δ. It is not hard to see that B0 ⊂ ([ −2√δ, 2√δ]2)D0 , thus μ(B0) ≤ (4 √δ)D.Now let B be the set of α ∈ (T2)D0 so that for some choice of ξ ∈ ({− 1/2, 0, 1/2}2)D0 ,we have that d := π−1(α) + ξ ∈ B0. We get that μ(B) ≤ 3Dμ(B0) ≤ (12 √δ)D = (12 /20) D 1 N . So, randomly picking θ0 ∼ (T2)D0 , we have that E[\|B ∩ { θ0, . . . , N · θ0}\| ] < (12 /20) D < 1(since for n = 1 , . . . , N , n · θ0 will be uniformly distributed over ( T2)D0 , and thus will lie in B with probability μ(B)). Thus, we can fix some outcome of θ0 where the aforementioned intersection is empty. By Proposition 3.4 (recalling Lemma 3.5 and the definition of B), it follows that for any 3-AP P = {θ, θ + α, θ + 2 α} ⊂ Sǫ,D 0 with common difference α ∈ { θ0, . . . , N · θ0},that either w1(P ) is not contained in an interval of length 1 /4, or ( w2,ǫ + w3)( P ) is not contained in an interval of length δ/ 2. Since w1(( T2)D0 ) ⊂ [0 , D ] and ( w2,ǫ + w3)(( T2)D0 ) ⊂ [0 , (10 10 ǫ−2 + 1) D], we may apply pigeonhole principle to fix values r1, r 2 so that μ ( Sǫ,D 0 ∩ w−11 ([ r1, r 1 + 1 /4)) ∩ (w2,ǫ + w3)−1([ r2, r 2 + δ/ 2)) ) ≥ μ(Sǫ,D 0 ) δ 10 11 ǫ−2D2 ≫ǫ D−2N−2/D (9 /32 − ǫ)D0 .NEW LOWER BOUNDS FOR r3(N ) 7 Write A0 := Sǫ,D 0 ∩ w−11 ([ r1, r 1 + 1 /4)) ∩ (w2,ǫ + w3)−1([ r2, r 2 + δ/ 2)). This set shall play the role of our ‘annulus’ in the Behrend-type construction. Finally, picking μ ∼ (T2)D0 randomly, we have that E[#( n ∈ [N] : μ + n · θ0 ∈ A 0)] = Nμ (A0)(since for each n ∈ [N] and fixed θ0, μ + n · θ0 is uniformly distributed over ( T2)D0 ,over the randomness of μ). Fix an outcome of μ where the expectation is obtained. Defining A := {n ∈ [N] : μ + n · θ0 ∈ A 0} gives a 3-AP-free set (indeed, if there was some 3-AP P = {n0, n 0 + t, n 0 + 2 t} ⊂ A with common difference t > 0, we’d have that A0 contains a 3-AP with common difference α = t·θ0 ∈ { θ0, . . . , N ·θ0}, contradiction). Whence r3(N) ≥ μ(A0)N, completing the proof. Obtaining the building block We start by recalling the construction from [ 5]. Write T1 := { (a, b ) ∈ [0 , 1/2) × [1 /2, 1) : 7 12 ≤ a + b ≤ 4 3 } ∪ { (a, b ) ∈ [0 , 1/2) 2 : 5 6 < a + b } ,T2 := { (a, b ) ∈ [1 /2, 1) × [0 , 1/2) : 7 12 ≤ a + b < 5 6 and 2 a + b < 3 2 } . To prove Proposition 3.1 , [ 5] considered T := T1 ∪ T2, S := π(T ). We note that defining S∗ ǫ := S \ ψ−1([5 /6 − ǫ, 5/6)) would give a set of measure ≥ 7/24 − 2ǫ that satisfies the conditions of Proposition 3.3 (by following the proofs of [5, Facts 3.10 and 3.11]). However, to streamline our argument we shall work with a simpler construction. Namely, throughout this section, we consider the following. Definition 1. Let U1 := { (a, b ) ∈ [0 , 1/2) × [1 /2, 1) ∪ [1 /2, 1) × [0 , 1/2) : a + b < 3 4 } ,U2 := { (a, b ) ∈ [0 , 1/2) × [0 , 1) : 3 4 + ǫ < a + b < 5 4 } ;we set U := U1 ∪ U2 and take Sǫ = π(U). It is easy to see that μ(U1) = 1 2 (1 /4) 2 + 1 2(1 /4) 2 = 1 /16 while μ(U2) = 1 2 ((3 /4 − ǫ)2 − (1 /4 − ǫ)2 − (1 /4) 2) ≥ 1 2((9 − 2) /16 − 2ǫ) ≥ 7/32 − ǫ. Whence μ(Sǫ) ≥ (9 /32) − ǫ, as desired. It remains to show that Sǫ has the desired additive properties required by Proposi-tion 3.3 .We start by giving two lemmas which we will use to establish Eq. 3.2 .8 ZACH HUNTER Lemma 4.1. Let x, y, z ∈ [0 , 1] satisfy 2y = x + z. Suppose that (1 − { x})2 + (1 − { z})2 − 2(1 − { y})2 < 2 (z − x 2 )2 . Then min( x, z ) < 1/2 ≤ y.Proof. Let d := z−x 2 . By the parallelogram law, we have that ( u − d)2 + ( u + d)2 − 2u2 =2d2 for all real u ∈ R.Thus, (taking u = 1 − y and u = 1 − (y − 1/2) respectively) one gets (1 −x)2 +(1 −z)2 −2(1 −y)2 = 2 d2 = (1 −(x−1/2)) 2 +(1 −(z −1/2)) 2 −2(1 −(y −1/2)) 2. WLOG assume that x < y < z . Now, if z < 1/2, then we are done by the LHS above; similarly, if x ≥ 1/2, then we are done by the RHS above. Meanwhile, if y < 1/2 and z ≥ 1/2, then (1 − { z})2 ≥ (1 − z)2 (while 1 − { x} = 1 − x and 1 − { y} = 1 − y), so we are again done by the LHS. The final outcome which could affect {x}, {y}, {z} is that x < 1/2 ≤ y, but here we have nothing to prove. This concludes the result. Lemma 4.2. Let x, z ∈ U1, and let y∗ = x+z 2 be their midpoint. Then (1 − { x1})2 + (1 − { z1})2 − 2(1 − { y∗ 1 })2 ≥ 2 ((z − x)1 2 )2 . Proof. WLOG, assume that x1 < z 1. If z1 < 1/2 or x1 ≥ 1/2, then we are immediately done by Lemma 4.1 . So, we may assume that x1 < 1/2 ≤ z1.As x, z ∈ U1, this implies that x1 < 3/4 −x2 ≤ 1/4 (recalling that U1 ∩[0 , 1/2) 2 = ∅). Meanwhile, we also have z1 < 3/4. Thus, x1 + z2 < 1, meaning y∗ 1 < 1/2. So Lemma 4.1 applies here too, giving the result. Proposition 4.3. Let θx, θ z ∈ Sǫ and θy ∈ T2 satisfy 2θy = θx + θz . Then, writing d := π−1(θz )−π−1(θx ) 2 , we have (1 − { p(θx)})2 + (1 − { p(θz )})2 − 2(1 − { p(θy )})2 ≥ 2d21 assuming \|d1 + d2\| ≤ ǫ/ 1000 .Proof. Write x := π−1(θx), y := π−1(θy ), z := π−1(θz ) and note that x, z ∈ U by assumption. Also write y∗ := x + d = x+z 2 to denote the midpoint of x, z . Note that π(2 y∗) = π(x) + π(x + 2 d) = θx + θz = π(2 y), thus we must have 2 y∗ − 2y ∈ Z2. Since y∗, y ∈ [0 , 1) 2, we further get that y∗ − y ∈ {− 1/2, 0, 1/2}2.Hence, because {·} is appropriately periodic, it follows that {y∗} = {y}, and thus {p(θy )} = {y1} = {y∗ 1 }. So we are left to show that (1 − { x1})2 + (1 − { z1})2 − 2(1 − { y∗ 1 })2 ≥ 2d21 assuming \|d1 + d2\| < ǫ 1000 . Note that we must either have {x, z } ⊂ U1 or {x, z } ⊂ U2,since otherwise we’d have 2\|d1 + d2\| ≥ inf v∈U2 (v1 + v2) − sup w∈U1 (w1 + w2) = (3 /4 + ǫ) − (3 /4) > 2( ǫ/ 1000) .NEW LOWER BOUNDS FOR r3(N ) 9 Now, if {x, z } ⊂ U1, then we are immediately done by Lemma 4.2 . Meanwhile, if {x, z } ⊂ U2, then {x1, z 1} ⊂ [0 , 1/2). Consequently, we get that y∗ 1 ∈ [0 , 1/2) as well, so we are similarly done, by Lemma 4.1 . Lemma 4.4. Consider x, z ∈ U2, and write y∗ := x+z 2 to denote their midpoint. We have that y∗ 6 = u + ξ for all u ∈ U, ξ ∈ { (1 /2, 1/2) , (1 /2, 0) , (0 , 1/2) }.Proof. Write Ξ := {(1 /2, 1/2) , (1 /2, 0) , (0 , 1/2) }. Case 1: u ∈ U ∩ [0 , 1/2) 2. Then we have ( u1 + u2) > (3 /4) + ǫ. Meanwhile, (y∗ 1 y∗ 2 ) = (x1 + x2) + ( z1 + z2) 2 < sup v∈U (v1 + v2) = (5 /4) which is less than 5 /4 + ǫ ≤ (u1 + u2) + ( ξ1 + ξ2) for each choice of ξ ∈ Ξ. Case 2: u ∈ U \ [0 , 1/2) 2. Noting that y∗ ∈ [0 , 1) 2, we must simply rule out that y∗ ∈ [1 /2, 1) 2 (since for ξ ∈ Ξ and u ∈ [0 , 1) 2 \ [0 , 1/2) 2, u + ξ ∈ [0 , 1) 2 happens if and only if u + ξ ∈ [1 /2, 1) 2). So now we shall prove that y∗ 6 ∈ [1 /2, 1) 2. Indeed, if {x, z } ⊂ U2 ⊂ [0 , 1/2) × [0 , 1), then y∗ 1 < 1/2; otherwise, if (say) x ∈ U1, then (y∗ 1 y∗ 2 ) < sup v∈U1 (v1 + v2) + sup w∈U2 (w1 + w2) 2 = 1 2 ((3 /4) + (5 /4)) = 1 . In either case, y∗ exhibits something which cannot be possible if y∗ ∈ [1 /2, 1) 2, so we are done. Proposition 4.5. Let θx, θ y , θ z ∈ Sǫ satisfy 2θy = θx + θz . Then, writing d := π−1(θz)−π−1(θx) 2 , we have 2ψ(θy) ≥ ψ(θx) + ψ(θz ) + 1 /2 or ψ(θy) = ψ(θx) + ( d1 + d2). Proof. Write x := π−1(θx), y := π−1(θy ), z := π−1(θz ) (and note that {x, y, z } ⊂ U by assumption). Also write y∗ := x + d = x+z 2 to denote the midpoint of x, z .As noted before (cf. Proposition 4.3 ), we have y∗ − y ∈ {− 1/2, 0, 1/2}2. Write ξ := y − y∗; by Lemma 4.4 we have that ξ 6 ∈ { (−1/2, −1/2) , (−1/2, 0) , (0 , −1/2) },whence ξ1 + ξ2 ∈ { 0, 1/2, 1}. Now, we have ψ(θy) = ( y1 + y2) = ( y∗ 1 y∗ 2 ) + ( ξ1 + ξ2). Meanwhile, we clearly have (y∗ 1 y∗ 2 ) = ψ(θx) + ( d1 + d2) and 2( y∗ 1 y∗ 2 ) = ψ(θx) + ψ(θz )by definition of y∗.Recalling ξ1 + ξ2 ∈ { 0, 1/2, 1}, we are immediately done (if ξ1 + ξ2 = 0, then ψ(θy ) = (y∗ 1 y∗ 2 ) and the latter outcome holds; otherwise the former outcome holds). Proof of Proposition 3.3 . We take Sǫ to be the set from Definition 1. The desired properties hold by combining Propositions 4.3 and 4.5 . 10 ZACH HUNTER Conclusion We now briefly discuss the limits of this method and some related problems. Given an additive set X and an integer k ≥ 3, we will write rk(X) to denote the maximum cardinality of a subset A ⊂ X which lacks non-trivial k-term arithmetic progressions (sets P of the form {x0, x 0 + d, . . . , x 0 + ( i − 1) d} with d 6 = 0). 5.1. In finite fields. We note that the arguments here can recover the bounds from , but with Behrend-type little order terms. Namely, using the construction S∗ ǫ (rather than Sǫ) alluded to in Section 4, one can show: Theorem 2. Fix any prime p ≥ 3. We have that r3(Fnp ) ≫p 1 n2 ⌈(7 /24) p2⌉n/ 2 ≫p 1 n2 (√7/24 p)n. However, this is only a very minor improvement, since [ 5] proves the above only with a worse power of ‘ 1 n ’ (the exponent is roughly quadratic in p). More interestingly, the fact that this argument for r3(N) also yields bounds for r3(Fnp ) suggests that there is a natural barrier for how far these ideas can be na¨ ıvely pushed. First, recall that by the celebrated work of Ellenberg-Gijswijt [ 3], there exists an absolute constant c > 0 so that r3(Fnp ) ≪p ((1 − c)p)n for all finite fields. Already to use a building block with “efficiency > 1/2” (where the efficiency of a set S ⊂ Tt is μ(S)1/t ), one has to distinguish between the image of [ N] embedded into TD and π({0, 1/2}D) (the image of FD 2 inside the torus). Indeed, there should be a reason why we do not prove the impossible 4 bound r3(FD 2 ) > 1. In the current paper, our reason comes from considering the fractional map {·} , which is used to morally annihilate π({0, 1/2}D) (thus our argument does not attempt to say anything about 3-APs with common difference α ∈ π({0, 1/2}D)). So similarly, if one were to achieve efficiency beyond > 1 − c, they would need a way to distinguish from π({0, 1/p, . . . , (p − 1) /p }D) for any fixed prime p. This seems to suggest that one cannot “fix” a specific building block, but must instead do things in some parametrized fashion (or otherwise introduce significantly different ideas). Hence, we claim there is a natural barrier for proving r3(N) ≫ N2−η2√2 log 2(N ) for some η < √ log 2( 1 1−c ). 5.2. Longer progressions. Next, one might wonder about rk(N) := rk([ N]). Here the best-known lower bounds are of the form rk(N) ≫ log C (N)N exp( −ck log pk (N)) for certain explicit constants ck, p k (cf. [10 ] which refines the results of [ 11 ] via the framework of [ 7]). 4Recall that we wish to avoid containing any 3-APs besides singletons. A more common non-degeneracy condition is to only worry about 3-APs with three distinct points, but this glosses over some issues that are relevant to the integer setting. NEW LOWER BOUNDS FOR r3(N)11 For experts in this area, we note that a routine adjustment to the methods [ 10 ], where we now use our building block Sǫ in the final stage of the bootstrapping procedure, allows one to show that rk(N) ≫ N exp( −(1 − η)ck log pk (N)) for some absolute constant η > 0 and all k ≥ 3. However, without an improved building block, such ideas cannot replace ‘1 − η’ by (say) ‘1 /10’ (even when k is very large). We mention that in forthcoming work, the author will give a more conceptual ar-gument to obtain these sorts of improvements for all sufficiently large k . There we will construct ‘increasingly efficient’ building blocks as k gets sufficiently large, thus the bounds obtained in [ 9] (for all large k) will outperform whatever can be obtained by just na¨ ıvely applying the work from the present paper. References F. A. Behrend, On sets of integers which contain no three in arithmetic progression, in Proceedings of the National Academy of Sciences 32 (1946), p. 331-332. M. Elkin, An improved construction of progression-free sets, in SODA 10’ (2010), p. 886-905. J. S. Ellenberg and D. Gijswijt, On large subsets of Fnq with no three-term arithmetic progression, in Annals of Mathematics 185 (2017), 339–343. C. Elsholtz, B. Klahn, and G. F. Lipnik, Large subsets of Zmn without arithmetic progressions, in Des. Codes Cryptogr. 91 (2023), 1443–1452. C. Elsholtz, L. Proske, and L. Sauermann, New lower bounds for three-term progression free sets in Fnp , preprint (January 2024), 13 pp. B. J. Green, Lower bounds for corner-free sets, in New Zealand Journal of Mathematics 51 (2021). B. J. Green and J. Wolf, A note on Elkin’s improvement of Behrend’s construction, in Additive Number Theory , p. 141–144, Springer, New York 2010. L. Hambardzumyan, T. Pitassi, S. Sherif, M. Shirley, and A. Shraibman, An improved protocol for ExactlyN with more than 3 players, preprint (September 2023), 33 pp. Z. Hunter, Improved lower bounds for Szemer´ edi’s theorem, (in preparation). K. O’Bryant, Sets of integers that do not contain long arithmetic progressions, in Electronic Journal of Combinatorics 18 (2011), 18 pp. R. A. Rankin, Sets not containing more than a given number of terms in arithmetical progression, in Proc. Roy. Soc. Edinburgh Sect. A 65 (1960), p. 332-344. R. Salem and D. Spencer, On sets of integers which contain no three in arithmetic progression, in Proc. Nat. Acad. Sci. (USA) 28 (1942), p. 561-563. (Hunter) Department of Mathematics, ETH, Z¨ urich, Switzerland. Email address : zach.hunter@math.ethz.ch
17006	https://www.ixl.com/math/grade-8/multi-step-word-problems	SKIP TO CONTENT Time to get in the zone! Your teacher would like you to focus on skills in . Let's pick a skill from these categories. B.14Multi-step word problemsEHX Share video You are watching a video preview. Become a member to get full access! You've reached the end of this video preview, but the learning doesn't have to stop! Join IXL today! Become a member Incomplete answer You did not finish the question. Do you want to go back to the question? or Bernie helped design a magician costume for a school play. He used 5.75 feet of ribbon for the magician's wand. He used 11.75 feet of the same ribbon to decorate the magician's robe. If Bernie used $10.50 of ribbon in all, how much did the ribbon cost per foot? $ ref_doc_title. Excellent! You got that right! or Jumping to level 1 of 1 Excellent! Now entering the Challenge Zone—are you ready? Teacher tools 1 in 4 students uses IXL for academic help and enrichment. Pre-K through 12th grade Sign up now
17007	https://www.jmap.org/Worksheets/A.REI.B.4.UsingtheDiscriminant2.pdf	Regents Exam Questions A.REI.B.4: Using the Discriminant 2 Name: ___ www.jmap.org 1 A.REI.B.4: Using the Discriminant 2 1 The discriminant of a quadratic equation is 24. The roots are 1) imaginary 2) real, rational, and equal 3) real, rational, and unequal 4) real, irrational, and unequal 2 The roots of the equation x2 −3x −2 = 0 are 1) real, rational, and equal 2) real, rational, and unequal 3) real, irrational, and unequal 4) imaginary 3 The roots of x2 −5x + 1 = 0 are 1) real, rational, and unequal 2) real, rational, and equal 3) real, irrational, and unequal 4) imaginary 4 The roots of the equation x 2 −10x + 25 = 0 are 1) imaginary 2) real and irrational 3) real, rational, and equal 4) real, rational, and unequal 5 The roots of the equation 2x2 −5 = 0 are 1) imaginary 2) real, rational, and equal 3) real, rational, and unequal 4) real and irrational 6 The roots of the equation 2x 2 −8x −4 = 0 are 1) imaginary 2) real, rational, and equal 3) real, irrational, and unequal 4) real, rational, and unequal 7 The roots of the equation 2x 2 + 5x −6 = 0 are 1) rational and unequal 2) rational and equal 3) irrational and unequal 4) imaginary 8 The roots of the equation 5x 2 −2x + 1 = 0 are 1) real, rational, and unequal 2) real, rational, and equal 3) real, irrational, and unequal 4) imaginary 9 The roots of the equation 9x 2 + 3x −4 = 0 are 1) imaginary 2) real, rational, and equal 3) real, rational, and unequal 4) real, irrational, and unequal 10 The roots of the equation 2x2 + 4 = 9x are 1) real, rational, and equal 2) real, rational, and unequal 3) real, irrational, and unequal 4) imaginary 11 The roots of 3x2 + x = 14 are 1) imaginary 2) real, rational, and equal 3) real, rational, and unequal 4) real, irrational, and unequal 12 The roots of the equation 2x2 −x = 4 are 1) real and irrational 2) real, rational, and equal 3) real, rational, and unequal 4) imaginary Regents Exam Questions A.REI.B.4: Using the Discriminant 2 Name: ___ www.jmap.org 2 13 The roots of the equation 4(x2 −1) = −3x are 1) imaginary 2) real, rational, equal 3) real, rational, unequal 4) real, irrational, unequal 14 Which graph shows a quadratic function with two imaginary zeros? 1) 2) 3) 4) 15 Which graph has imaginary roots? 1) 2) 3) 4) Regents Exam Questions A.REI.B.4: Using the Discriminant 2 Name: ______ www.jmap.org 3 16 If f(x) is represented by the graph below, which translation of f(x) would have imaginary roots? 1) f(x + 5) 2) f(x −5) 3) f(x) + 5 4) f(x) −5 17 In the quadratic formula, b 2 −4ac is called the discriminant. The function f(x) has a discriminant value of 8, and g(x) has a discriminant value of −16. The quadratic graphs, h(x) and j(x), are shown below. Which quadratic functions have imaginary roots? 1) g(x) and h(x) 2) g(x) and j(x) 3) f(x) and h(x) 4) f(x) and j(x) 18 Which representation of a quadratic has imaginary roots? 1) 2) 2(x + 3)2 = 64 3) 4) 2x2 + 32 = 0 19 Does the equation x2 −4x + 13 = 0 have imaginary solutions? Justify your answer. ID: A 1 A.REI.B.4: Using the Discriminant 2 Answer Section 1 ANS: 4 REF: 011323a2 2 ANS: 3 REF: 080106b 3 ANS: 3 REF: 060910b 4 ANS: 3 b 2 −4ac = (−10)2 −4(1)(25) = 100 −100 = 0 REF: 011102a2 5 ANS: 4 REF: 010614b 6 ANS: 3 REF: 010513b 7 ANS: 3 b 2 −4ac = 52 −4(2)(−6) = 73 REF: 061010b 8 ANS: 4 REF: 080814b 9 ANS: 4 b 2 −4ac = 32 −4(9)(−4) = 9 + 144 = 153 REF: 081016a2 10 ANS: 2 b 2 −4ac = (−9)2 −4(2)(4) = 81 −32 = 49 REF: 011411a2 11 ANS: 3 3x 2 + x −14 = 0 12 −4(3)(−14) = 1 + 168 = 169 = 132 REF: 061524a2 ID: A 2 12 ANS: 1 . REF: 060219b 13 ANS: 4 4x 2 + 3x −4 = 0 b 2 −4ac = 32 −4(4)(−4) = 9 + 64 = 73 REF: 011618a2 14 ANS: 2 1) 1 real, mult. 2; 3) not a quadratic; 4) not a function. REF: 012324aii 15 ANS: 2 REF: 012402aii 16 ANS: 3 REF: 062409aii 17 ANS: 2 REF: 082308aii 18 ANS: 4 (1) quadratic has two roots and both are real (−2,0) and (−0.5,0), (2) x = ± 32 −3, (3) the real root is 3, with a multiplicity of 2, (4) x = ±4i REF: 011909aii 19 ANS: b 2 −4ac = (−4)2 −4(1)(13) = 16 −52 = −36 imaginary REF: 062225aii
17008	https://www.mayoclinic.org/diseases-conditions/vulvar-cancer/diagnosis-treatment/drc-20368072	Vulvar cancer - Diagnosis and treatment - Mayo Clinic This content does not have an English version. This content does not have an Arabic version. Skip to content Request appointment Log in Search Menu Request appointment Donate Diseases & conditions Find a doctor Care at Mayo Clinic Patient-Centered Care About Mayo Clinic Request Appointment Find a Doctor Locations Clinical Trials Connect to Support Groups Patient & Visitor Guide Billing & Insurance Departments & Centers International Services Contact Us Patient & Visitor Guide Health Library Diseases & Conditions Symptoms Tests & Procedures Drugs & Supplements Healthy Lifestyle Mayo Clinic Health Letter & Books Mayo Clinic Health Letter & Books For Medical Professionals Medical Professional Resources Refer a Patient Continuing Medical Education AskMayoExpert Mayo Clinic Laboratories Video Center Journals & Publications Mayo Clinic Alumni Association Continuing Medical Education Research & Education at Mayo Clinic Research Research at Mayo Clinic Research Faculty Laboratories Core Facilities Centers & Programs Departments & Divisions Clinical Trials Institutional Review Board Postdoctoral Fellowships Training Grant Programs Education Mayo Clinic College of Medicine and Science Mayo Clinic Graduate School of Biomedical Sciences Mayo Clinic Alix School of Medicine Mayo Clinic School of Graduate Medical Education Mayo Clinic School of Health Sciences Mayo Clinic School of Continuous Professional Development Mayo Clinic College of Medicine and Science Giving to Mayo Clinic Give Now Giving to Mayo Clinic Frequently Asked Questions Contact Us to Give Make a Donation Search for a disease or condition . Type 3 or more letters to display suggested search options. When results are available, use up and down arrow keys to navigate. Press space key to fill the input with the suggestion or enter key to search with the suggestion.Search by keyword or phraseSearch Patient Care & Health Information Diseases & Conditions Vulvar cancer Request an Appointment Symptoms & causes Diagnosis & treatment Doctors & departments Care at Mayo Clinic Print Diagnosis Vulvar cancer diagnosis often begins with a physical exam and a discussion of your health history. A special magnifying device may be used to closely inspect the area. A sample of tissue may be taken for lab testing. Examining the vulva Your healthcare professional will likely conduct a physical exam of your vulva to look for anything concerning. The healthcare professional may use a special magnifying instrument to look closely at the vulva. This instrument is called a colposcope. It also may be used to look at the vagina and cervix. Biopsy A biopsy is a procedure to remove a sample of tissue for testing in a lab. For vulvar cancer, a biopsy involves removing a sample of skin. A vulvar biopsy may be done in a healthcare professional's office. Medicine is used to numb the area. The health professional may use a blade or a circular cutting tool to remove some skin. Sometimes the sample is removed in an operating room. During this kind of biopsy, you receive medicine to put you in a sleep-like state so that you're not aware during the procedure. Vulvar cancer staging If you're diagnosed with vulvar cancer, the next step is to determine the cancer's extent, called the stage. Your healthcare team uses the cancer staging test results to help create your treatment plan. Staging tests may include: Examination of your pelvic area for cancer spread. Your healthcare professional may do a more thorough examination of your pelvis for signs that the cancer has spread. Imaging tests. Images of your chest, abdomen or pelvis may show whether the cancer has spread to those areas. Tests might include X-ray, MRI, CT and positron emission tomography, which also is called a PET scan. The stages of vulvar cancer range from 1 to 4. A stage 1 vulvar cancer is small and confined to the vulva. As the cancer gets larger or spreads beyond the area where it started, the stages get higher. A stage 4 vulvar cancer has grown into the pelvic bone or spread to other parts of the body. Care at Mayo Clinic Our caring team of Mayo Clinic experts can help you with your vulvar cancer-related health concerns Start Here More Information Vulvar cancer care at Mayo Clinic Colposcopy CT scan MRI Needle biopsy Positron emission tomography scan Sentinel node biopsy Show more related information Treatment Vulvectomy Enlarge image Close Vulvectomy Vulvectomy Treatment for vulvar cancer may involve removing part of the vulva, called a partial vulvectomy. Surgery to remove the entire vulva and the underlying tissue is called a radical vulvectomy. Treatment for vulvar cancer usually begins with surgery to remove the cancer. Other treatments may include radiation therapy, chemotherapy, targeted therapy and immunotherapy. Your healthcare team considers many factors when creating a treatment plan. These factors may include your overall health, the type and stage of your cancer, and your preferences. Surgery For most vulvar cancers, surgery is the first treatment. Procedures used to treat vulvar cancer include: Removing the cancer and some healthy tissue. An excision involves cutting out the cancer and a small amount of healthy tissue that surrounds it, called a margin. Cutting out a margin of healthy-looking tissue helps ensure that all of the cancer cells have been removed. This procedure also might be called a wide local excision or a radical excision. Removing part of the vulva or the entire vulva. Vulvectomy is a surgery to remove the vulva. When part of the vulva is removed, it's called a partial vulvectomy. When the entire vulva and the underlying tissue are removed, it's called a radical vulvectomy. Vulvectomy may be an option for larger cancers. Radiation therapy and chemotherapy may be used before surgery to shrink the cancer. This may allow for a less extensive operation. Removing a few nearby lymph nodes. A sentinel node biopsy looks for signs of cancer in the nearby lymph nodes. This procedure identifies the lymph nodes most likely to contain cancer. Those lymph nodes are removed and tested. If no cancer is found, it's unlikely that the cancer has spread. For vulvar cancer, the sentinel lymph nodes may be removed from one or more areas. Removing many lymph nodes. If the cancer has spread to the lymph nodes, many lymph nodes may be removed to reduce the risk that cancer will spread to other parts of the body. Surgery carries a risk of complications. These may include infection and problems with healing around the incision. Removing lymph nodes can cause fluid retention and leg swelling, a condition called lymphedema. Radiation therapy Radiation therapy treats cancer with powerful energy beams. The energy can come from X-rays, protons or other sources. During radiation therapy, you lie on a table while a machine moves around you. The machine directs radiation to precise points on your body. Radiation therapy is sometimes used to shrink large vulvar cancers before surgery. Sometimes radiation therapy is combined with chemotherapy. Using a low dose of chemotherapy medicine during radiation treatments makes the radiation more effective. If cancer cells are found in your lymph nodes, radiation therapy may be used on the area around your lymph nodes. This treatment may kill any cancer cells that might remain after surgery. Radiation is sometimes combined with chemotherapy in these situations. Chemotherapy Chemotherapy treats cancer with strong medicines. Many chemotherapy medicines exist. Most chemotherapy medicines are given through a vein. Some come in pill form. For those with vulvar cancer that has spread to other areas of the body, chemotherapy may be an option. Chemotherapy is sometimes combined with radiation therapy to shrink large vulvar cancers before surgery. Chemotherapy also may be combined with radiation to treat cancer that has spread to the lymph nodes. Targeted therapy Targeted therapy for cancer is a treatment that uses medicines that attack specific chemicals in the cancer cells. By blocking these chemicals, targeted treatments can cause cancer cells to die. For vulvar cancer, targeted therapy may be used for treating advanced vulvar cancer. Immunotherapy Immunotherapy for cancer is a treatment with medicine that helps the body's immune system kill cancer cells. The immune system fights off diseases by attacking germs and other cells that shouldn't be in the body. Cancer cells survive by hiding from the immune system. Immunotherapy helps the immune system cells find and kill the cancer cells. For vulvar cancer, immunotherapy may be used for treating advanced vulvar cancer. Follow-up tests after treatment After completing vulvar cancer treatment, your healthcare professional will recommend periodic follow-up exams to check if the cancer has come back. Even after successful treatment, vulvar cancer can return. Your healthcare professional will determine the schedule of follow-up exams that's right for you. Exams are generally recommended 2 to 4 times each year for the first two years after vulvar cancer treatment. More Information Vulvar cancer care at Mayo Clinic Chemotherapy Radiation therapy Request an appointment Clinical trials Explore Mayo Clinic studies testing new treatments, interventions and tests as a means to prevent, detect, treat or manage this condition. Coping and support With time, you'll find what helps you cope with the uncertainty and worry of a vulvar cancer diagnosis. Until then, you may find it helps to: Learn enough about vulvar cancer to make decisions about your care Ask your healthcare team about your cancer, including your test results, treatment options and, if you like, your prognosis. As you learn more about vulvar cancer, you may become more confident in making treatment decisions. Keep friends and family close Keeping your close relationships strong can help you deal with vulvar cancer. Friends and family can provide the practical support you may need, such as helping take care of your home if you're in the hospital. And they can serve as emotional support when you feel overwhelmed by having cancer. Find someone to talk with Find someone who is willing to listen to you talk about your hopes and worries. This may be a friend or family member. The concern and understanding of a counselor, medical social worker, clergy member or cancer support group also may be helpful. Ask your healthcare team about support groups in your area. In the United States, other sources of information include the National Cancer Institute and the American Cancer Society. Don't be afraid of intimacy A natural reaction to changes in your body may be to avoid intimacy. Although it may not be easy, talk about your feelings with your partner. You also may find it helpful to talk to a therapist, either on your own or together with your partner. Remember that you can express your sexuality in many ways. Touching, holding, hugging and caressing may become far more important to you and your partner. Preparing for your appointment Make an appointment with your usual doctor, gynecologist or other healthcare professional if you have any symptoms that worry you. If you're found to have vulvar cancer, your health professional may refer you to a doctor who specializes in cancers of the female reproductive system. This doctor is called a gynecologic oncologist. It's a good idea to prepare for your appointment. Here's some information to help you get ready. What you can do Be aware of any pre-appointment restrictions. At the time you make the appointment, be sure to ask if there's anything you need to do in advance, such as restrict your diet. Write down symptoms you have, including any that may not seem related to the reason for which you scheduled the appointment. Write down important personal information, including major stresses or recent life changes. Make a list of all medicines, vitamins or supplements you're taking and the doses. Take a family member or friend along. Sometimes it can be very hard to remember all the information provided during an appointment. Someone who goes with you may remember something that you missed or forgot. Write down questions to ask your healthcare team. Your time with your healthcare team is limited, so preparing a list of questions can help you make the most of your time together. List your questions from most important to least important in case time runs out. For vulvar cancer, some basic questions to ask include: What kinds of tests will I need? Do I need to do anything to prepare for these tests? Other than vulvar cancer, are there any other possible causes for these symptoms? What type of vulvar cancer do I have? What stage is my cancer? What types of surgical options are available to me? What kind of success rates does each type of surgery have? What are the drawbacks to each type of surgery? Will I need to wear an ostomy bag? What about radiation or chemotherapy? Are those options available to me? What kind of success rates do those therapies have? What types of side effects does each treatment have? How will these treatments affect my sexuality? Will I be able to have children after treatment? How should I prepare for treatment? Which course of action do you recommend? What are the odds of recurrence? What is my prognosis? Don't hesitate to ask other questions. What to expect from your doctor Be prepared to answer questions, such as: When did your symptoms begin? Have your symptoms been continuous or occasional? How severe are your symptoms? What, if anything, seems to improve your symptoms? What, if anything, appears to worsen your symptoms? Have you ever been diagnosed with lichen sclerosus? Have you ever had a concerning Pap test result? Have you ever been diagnosed with HPV? By Mayo Clinic Staff Vulvar cancer care at Mayo Clinic Request an appointment Symptoms & causesDoctors & departments Jan. 10, 2025 Print Living with vulvar cancer? Connect with others like you for support and answers to your questions in the Gynecologic Cancers support group on Mayo Clinic Connect, a patient community. Gynecologic Cancers Discussions What were your first signs of endometrial cancer?69 Replies Tue, May 06, 2025 What Kind of Radiation Did You Get for Your Gynecological Cancer?119 Replies Mon, May 05, 2025 Complex endometrial hyperplasia with atypia34 Replies Sat, May 03, 2025 See more discussions Show references Niederhuber JE, et al., eds. Cancer of the cervix, vulva and vagina. In: Abeloff's Clinical Oncology. 6th ed. Elsevier; 2020. Accessed June 10, 2024. Vulvar cancer treatment (PDQ) — Patient version. National Cancer Institute. Accessed June 10, 2024. Vulvar cancer. National Comprehensive Cancer Network. Accessed June 10, 2024. About genital HPV infection. Centers for Disease Control and Prevention. Accessed June 10, 2024. Ramirez PT, et al., eds. Vulvar surgery and sentinel node mapping for vulvar cancer. In: Principles of Gynecologic Oncology Surgery. Elsevier; 2019. Accessed June 10, 2024. DeVita VT Jr, et al., eds. Cancer of the vagina and vulva. In: DeVita, Hellman, and Rosenberg's Cancer: Principles and Practice of Oncology. 12th ed. ProQuest Ebook Central. Wolters Kluwer; 2023. Accessed June 10, 2024. Kibbi N, et al. Evidence-based clinical practice guidelines for extramammary Paget disease. JAMA Oncology. 2022; doi:10.1001/jamaoncol.2021.7148. Fowler GC, et al., eds. Vulvar biopsy. In: Pfenninger and Fowler's Procedures for Primary Care. 4th ed. Elsevier; 2020. Accessed Aug. 27, 2024. Bakkum-Gamez JN (expert opinion). Mayo Clinic. July 23, 2024. Related Vulvar cancer Vulvectomy Associated Procedures Chemotherapy Colposcopy CT scan MRI Needle biopsy Positron emission tomography scan Radiation therapy Sentinel node biopsy Show more associated procedures Products & Services A Book: Mayo Clinic Family Health Book Newsletter: Mayo Clinic Health Letter — Digital Edition Show more products and services from Mayo Clinic Vulvar cancer Symptoms & causes Diagnosis & treatment Doctors & departments Care at Mayo Clinic Advertisement Mayo Clinic does not endorse companies or products. Advertising revenue supports our not-for-profit mission. Advertising & Sponsorship Policy Opportunities Ad Choices Mayo Clinic Press Check out these best-sellers and special offers on books and newsletters from Mayo Clinic Press. NEW: Mayo Clinic Guide to Better Sleep Listen to Health Matters Podcast Mayo Clinic on Incontinence The Essential Diabetes Book FREE Mayo Clinic Diet Assessment Mayo Clinic Health Letter - FREE book CON-20314004 Patient Care & Health Information Diseases & Conditions Vulvar cancer Have 2X the impact! Give Today Make a gift that can go twice as far to advance healthcare research. Give Today Find a doctor Explore careers Sign up for free e-newsletters About Mayo Clinic About this Site Contact Us Locations Health Information Policy Medicare Accountable Care Organization (ACO) Media Requests News Network Price Transparency Medical Professionals AskMayoExpert Clinical Trials Mayo Clinic Alumni Association Refer a Patient Businesses Executive Health Program International Business Collaborations Facilities & Real Estate Supplier Information Students Admissions Requirements Degree Programs Student & Faculty Portal Researchers Research Faculty Laboratories International Patients Appointments Financial Services International Locations & Offices Charitable Care & Financial Assistance Community Health Needs Assessment Financial Assistance Documents – Arizona Financial Assistance Documents – Florida Financial Assistance Documents – Minnesota Follow Mayo Clinic X YouTube Facebook LinkedIn Instagram Get the Mayo Clinic app Terms & Conditions Privacy Policy Notice of Privacy Practices Notice of Nondiscrimination Accessibility Statement Advertising & Sponsorship Policy Site Map Manage Cookies Terms and Conditions Privacy Policy Notice of Privacy Practices Notice of Nondiscrimination Accessibility Statement Advertising and Sponsorship Policy Site Map Manage Cookies Language:English English Español العربية 简体中文 © 1998-2025 Mayo Foundation for Medical Education and Research (MFMER). All rights reserved. Language:English English Español العربية 简体中文 Mayo Clinic Privacy Policy Mayo Clinic and our partners use technologies such as cookies to collect information from your browser to deliver relevant advertising on our site, in emails and across the Internet, personalize content and perform site analytics. Please visit our Privacy Policy for more information about our use of data and your rights. Click here to update your preferences I Agree x Customize Cookie Settings TrustedThese technologies are used in operating the site, including remembering your preferences, ensuring security and providing certain features. They do not track your activity online. Website AnalyticsThese technologies collect information to help us understand how our websites are being used and to make improvements. AdvertisingThese technologies are used to make advertising messages more relevant to you. They perform functions like preventing the same ad from continuously reappearing, ensuring that ads are properly displayed for advertisers and selecting advertisements that are based on your interests. Save Feedback
17009	https://mathresearch.utsa.edu/wiki/index.php?title=Sets:Cardinality	Sets:Cardinality From Department of Mathematics at UTSA Jump to navigation Jump to search The set of all Platonic solids has 5 elements. Thus . In mathematics, the cardinality of a set is a measure of the "number of elements" of the set. For example, the set contains 3 elements, and therefore has a cardinality of 3. Beginning in the late 19th century, this concept was generalized to infinite sets, which allows one to distinguish between the different types of infinity, and to perform arithmetic on them. There are two approaches to cardinality: one which compares sets directly using bijections and injections, and another which uses cardinal numbers. The cardinality of a set is also called its size, when no confusion with other notions of size is possible. The cardinality of a set is usually denoted , with a vertical bar on each side; this is the same notation as absolute value, and the meaning depends on context. The cardinality of a set may alternatively be denoted by , , , or . Contents 1 History 2 Comparing sets 2.1 Definition 1: \|A\| = \|B\| 2.2 Definition 2: \|A\| ≤ \|B\| 2.3 Definition 3: \|A\| < \|B\| 3 Cardinal numbers 4 Finite, countable and uncountable sets 5 Infinite sets 5.1 Cardinality of the continuum 6 Examples and properties 7 Union and intersection 8 Licensing History In the 1890s, Georg Cantor generalized the concept of cardinality to infinite sets, which allowed one to distinguish between the different types of infinity and to perform arithmetic on them. Comparing sets Bijective function from N to the set E of even numbers. Although E is a proper subset of N, both sets have the same cardinality. N does not have the same cardinality as its power set P(N): For every function f from N to P(N), the set T = {n∈N: n∉f(n)} disagrees with every set in the range of f, hence f cannot be surjective. The picture shows an example f and the corresponding T; Template:Color: n∈f(n)T, Template:Color:n∈Tf(n). While the cardinality of a finite set is just the number of its elements, extending the notion to infinite sets usually starts with defining the notion of comparison of arbitrary sets (some of which are possibly infinite). Definition 1: \|A\| = \|B\| : Two sets A and B have the same cardinality if there exists a bijection (a.k.a., one-to-one correspondence) from A to B, that is, a function from A to B that is both injective and surjective. Such sets are said to be equipotent, equipollent, or equinumerous. This relationship can also be denoted A ≈ B or A ~ B. : For example, the set E = {0, 2, 4, 6, ...} of non-negative even numbers has the same cardinality as the set N = {0, 1, 2, 3, ...} of natural numbers, since the function f(n) = 2n is a bijection from N to E (see picture). Definition 2: \|A\| ≤ \|B\| : A has cardinality less than or equal to the cardinality of B, if there exists an injective function from A into B. Definition 3: \|A\| < \|B\| : A has cardinality strictly less than the cardinality of B, if there is an injective function, but no bijective function, from A to B. : For example, the set N of all natural numbers has cardinality strictly less than its power set P(N), because g(n) = { n } is an injective function from N to P(N), and it can be shown that no function from N to P(N) can be bijective (see picture). By a similar argument, N has cardinality strictly less than the cardinality of the set R of all real numbers. For proofs, see Cantor's diagonal argument or Cantor's first uncountability proof. If \|A\| ≤ \|B\| and \|B\| ≤ \|A\|, then \|A\| = \|B\| (a fact known as Schröder–Bernstein theorem). The axiom of choice is equivalent to the statement that \|A\| ≤ \|B\| or \|B\| ≤ \|A\| for every A, B. Cardinal numbers In the above section, "cardinality" of a set was defined functionally. In other words, it was not defined as a specific object itself. However, such an object can be defined as follows. The relation of having the same cardinality is called equinumerosity, and this is an equivalence relation on the class of all sets. The equivalence class of a set A under this relation, then, consists of all those sets which have the same cardinality as A. There are two ways to define the "cardinality of a set": The cardinality of a set A is defined as its equivalence class under equinumerosity. A representative set is designated for each equivalence class. The most common choice is the initial ordinal in that class. This is usually taken as the definition of cardinal number in axiomatic set theory. Assuming the axiom of choice, the cardinalities of the infinite sets are denoted For each ordinal , is the least cardinal number greater than . The cardinality of the natural numbers is denoted aleph-null (), while the cardinality of the real numbers is denoted by "" (a lowercase fraktur script "c"), and is also referred to as the cardinality of the continuum. Cantor showed, using the diagonal argument, that . We can show that , this also being the cardinality of the set of all subsets of the natural numbers. The continuum hypothesis says that , i.e. is the smallest cardinal number bigger than , i.e. there is no set whose cardinality is strictly between that of the integers and that of the real numbers. The continuum hypothesis is independent of ZFC, a standard axiomatization of set theory; that is, it is impossible to prove the continuum hypothesis or its negation from ZFC—provided that ZFC is consistent. For more detail, see § Cardinality of the continuum below. Finite, countable and uncountable sets If the axiom of choice holds, the law of trichotomy holds for cardinality. Thus we can make the following definitions: Any set X with cardinality less than that of the natural numbers, or \| X \| < \| N \|, is said to be a finite set. Any set X that has the same cardinality as the set of the natural numbers, or \| X \| = \| N \| = , is said to be a countably infinite set. Any set X with cardinality greater than that of the natural numbers, or \| X \| > \| N \|, for example \| R \| = > \| N \|, is said to be uncountable. Infinite sets Our intuition gained from finite sets breaks down when dealing with infinite sets. In the late nineteenth century Georg Cantor, Gottlob Frege, Richard Dedekind and others rejected the view that the whole cannot be the same size as the part. One example of this is Hilbert's paradox of the Grand Hotel. Indeed, Dedekind defined an infinite set as one that can be placed into a one-to-one correspondence with a strict subset (that is, having the same size in Cantor's sense); this notion of infinity is called Dedekind infinite. Cantor introduced the cardinal numbers, and showed—according to his bijection-based definition of size—that some infinite sets are greater than others. The smallest infinite cardinality is that of the natural numbers (). Cardinality of the continuum One of Cantor's most important results was that the cardinality of the continuum () is greater than that of the natural numbers (); that is, there are more real numbers R than natural numbers N. Namely, Cantor showed that (see Beth one) satisfies: : (see Cantor's diagonal argument or Cantor's first uncountability proof). The continuum hypothesis states that there is no cardinal number between the cardinality of the reals and the cardinality of the natural numbers, that is, However, this hypothesis can neither be proved nor disproved within the widely accepted ZFC axiomatic set theory, if ZFC is consistent. Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space. These results are highly counterintuitive, because they imply that there exist proper subsets and proper supersets of an infinite set S that have the same size as S, although S contains elements that do not belong to its subsets, and the supersets of S contain elements that are not included in it. The first of these results is apparent by considering, for instance, the tangent function, which provides a one-to-one correspondence between the interval (−½π, ½π) and R (see also Hilbert's paradox of the Grand Hotel). The second result was first demonstrated by Cantor in 1878, but it became more apparent in 1890, when Giuseppe Peano introduced the space-filling curves, curved lines that twist and turn enough to fill the whole of any square, or cube, or hypercube, or finite-dimensional space. These curves are not a direct proof that a line has the same number of points as a finite-dimensional space, but they can be used to obtain such a proof. Cantor also showed that sets with cardinality strictly greater than exist (see his generalized diagonal argument and theorem). They include, for instance: : the set of all subsets of R, i.e., the power set of R, written P(R) or 2R the set RR of all functions from R to R Both have cardinality : (see Beth two). The cardinal equalities and can be demonstrated using cardinal arithmetic: Examples and properties If X = {a, b, c} and Y = {apples, oranges, peaches}, then \| X \| = \| Y \| because { (a, apples), (b, oranges), (c, peaches)} is a bijection between the sets X and Y. The cardinality of each of X and Y is 3. If \| X \| ≤ \| Y \|, then there exists Z such that \| X \| = \| Z \| and Z ⊆ Y. If \| X \| ≤ \| Y \| and \| Y \| ≤ \| X \|, then \| X \| = \| Y \|. This holds even for infinite cardinals, and is known as Cantor–Bernstein–Schroeder theorem. Sets with cardinality of the continuum include the set of all real numbers, the set of all irrational numbers and the interval . Union and intersection If A and B are disjoint sets, then From this, one can show that in general, the cardinalities of unions and intersections are related by the following equation: Licensing Content obtained and/or adapted from: Cardinality, Wikipedia under a CC BY-SA license Retrieved from " Navigation menu Personal tools Log in Namespaces Page Discussion Variants Views Read View source View history More Navigation Main page Recent changes Random page Help about MediaWiki Tools What links here Related changes Special pages Printable version Permanent link Page information Cite this page
17010	https://simple.wikipedia.org/wiki/Cylinder	Published Time: 2005-06-17T11:20:21Z Cylinder - Simple English Wikipedia, the free encyclopedia Jump to content [x] Main menu Main menu move to sidebar hide Getting around Main page Simple start Simple talk New changes Show any page Help Contact us About Wikipedia Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Give to Wikipedia Create account Log in [x] Personal tools Give to Wikipedia Contribute Create account Log in Pages for logged out editors learn more Talk [x] Toggle the table of contents Contents move to sidebar hide Beginning 1 Common use 2 Volume 3 Cylindric section 4 Other types of cylinders 5 References 6 Other websites Cylinder [x] 102 languages Afrikaans العربية Արեւմտահայերէն Asturianu Aymar aru Azərbaycanca تۆرکجه বাংলা Беларуская Беларуская (тарашкевіца) Български Bosanski Català Чӑвашла Čeština ChiShona Cymraeg Dansk Deutsch Eesti Ελληνικά English Español Esperanto Euskara فارسی Français Gaeilge Galego 한국어 Հայերեն हिन्दी Hrvatski Ido Bahasa Indonesia Interlingua Íslenska Italiano עברית Jawa ಕನ್ನಡ ქართული Қазақша Kiswahili Кыргызча Latina Latviešu Lietuvių Magyar Македонски Malagasy മലയാളം Bahasa Melayu Мокшень Монгол မြန်မာဘာသာ Nederlands 日本語 Nordfriisk Norsk bokmål Norsk nynorsk Occitan Олык марий Oromoo Oʻzbekcha / ўзбекча ਪੰਜਾਬੀ پنجابی Piemontèis Polski Português Română Runa Simi Русский Саха тыла Shqip Sicilianu සිංහල سنڌي Slovenčina Slovenščina Soomaaliga کوردی Српски / srpski Srpskohrvatski / српскохрватски Sunda Suomi Svenska தமிழ் Татарча / tatarça తెలుగు ไทย Тоҷикӣ Türkçe Українська Tiếng Việt Walon Winaray 吴语粵語中文 Руски ⵜⴰⵎⴰⵣⵉⵖⵜ ⵜⴰⵏⴰⵡⴰⵢⵜ Change links Page Talk [x] English Read Change Change source View history [x] Tools Tools move to sidebar hide Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Sandbox Expand all Edit interlanguage links Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikidata item From Simple English Wikipedia, the free encyclopedia For other uses, see Cylinder (disambiguation). A right circular cylinder A cylinder is one of the most basic curved three dimensional geometric shapes, with the surface formed by the points at a fixed distance from a given line segment, known as the axis of the cylinder. The shape can be thought of as a circularprism. Both the surface and the solid shape created inside can be called a cylinder. The surface area and the volume of a cylinder have been known since ancient times. In differential geometry, a cylinder is defined more broadly as a ruled surface which is spanned by a one-parameter family of parallel lines. A cylinder whose cross section is an ellipse, parabola, or hyperbola is called an elliptic cylinder, parabolic cylinder, or hyperbolic cylinder respectively. Common use [change \| change source] In common use a cylinder is taken to mean a finite section of a right circular cylinder, i.e., the cylinder with the generating lines perpendicular to the bases, with its ends closed to form two circular surfaces, as in the figure (right). If the cylinder has a radiusr{\displaystyle r} and length (height) h, then its volume is given by: V=π r 2 h{\displaystyle V=\pi r^{2}h} and its surface area is: the area of the top (π r 2{\displaystyle \pi r^{2}})+ the area of the bottom (π r 2{\displaystyle \pi r^{2}})+ the area of the side (2 π r h{\displaystyle 2\pi rh}). Therefore, without the top or bottom (lateral area), the surface area is: A=2 π r h{\displaystyle A=2\pi rh}. With the top and bottom, the surface area is: A=2 π r 2+2 π r h=2 π r(r+h){\displaystyle A=2\pi r^{2}+2\pi rh=2\pi r(r+h)}. For a given volume, the cylinder with the smallest surface area has h=2 r{\displaystyle h=2r}. For a given surface area, the cylinder with the largest volume has h=2 r{\displaystyle h=2r}, i.e. the cylinder fits in a cube (height = diameter). Volume [change \| change source] Having a right circular cylinder with a height h{\displaystyle h} units and a base of radius r{\displaystyle r} units with the coordinate axes chosen so that the origin is at the center of one base and the height is measured along the positive x-axis. A plane section at a distance of x{\displaystyle x} units from the origin has an area of A(x){\displaystyle A(x)} square units where A(x)=π r 2{\displaystyle A(x)=\pi r^{2}} or A(y)=π r 2{\displaystyle A(y)=\pi r^{2}} An element of volume, is a right cylinder of base area A w i{\displaystyle Aw^{i}} square units and a thickness of Δ i x{\displaystyle \Delta {i}x} units. Thus if _V cubic units is the volume of the right circular cylinder, by Riemann sums, V o l u m e o f c y l i n d e r=lim\|\|Δ→0\|\|∑i=1 n A(w i)Δ i x{\displaystyle \mathrm {Volume\;of\;cylinder} =\lim {\|\|\Delta \to 0\|\|}\sum {i=1}^{n}A(w_{i})\Delta {i}x}=∫0 h A(y)2 d y{\displaystyle =\int {0}^{h}A(y)^{2}\,dy}=∫0 h π r 2 d y{\displaystyle =\int _{0}^{h}\pi r^{2}\,dy}=π r 2 h{\displaystyle =\pi \,r^{2}\,h\,} Using cylindrical coordinates, the volume can be calculated by integration over =∫0 h∫0 2 π∫0 r s d s d ϕ d z{\displaystyle =\int {0}^{h}\int {0}^{2\pi }\int _{0}^{r}s\,\,ds\,d\phi \,dz}=π r 2 h{\displaystyle =\pi \,r^{2}\,h\,} Cylindric section [change \| change source] Cylindric sections are the intersections of cylinders with planes. For a right circular cylinder, there are four possibilities. A plane tangent to the cylinder, meets the cylinder in a single straight line. Moved while parallel to itself, the plane either does not intersect the cylinder or intersects it in two parallel lines. All other planes intersect the cylinder in an ellipse or, when they are perpendicular to the axis of the cylinder, in a circle. Other types of cylinders [change \| change source] An elliptic cylinder An elliptic cylinder, or cylindroid, is a quadric surface, with the following equation in Cartesian coordinates: (x a)2+(y b)2=1.{\displaystyle ({\frac {x}{a}})^{2}+({\frac {y}{b}})^{2}=1.} This equation is for an elliptic cylinder, a generalization of the ordinary, circular cylinder (a=b{\displaystyle a=b}). Even more general is the generalized cylinder: the cross-section can be any curve. The cylinder is a degenerate quadric because at least one of the coordinates (in this case z{\displaystyle z}) does not appear in the equation. An oblique cylinder has the top and bottom surfaces displaced from one another. There are other more unusual types of cylinders. These are the imaginary elliptic cylinders: (x a)2+(y b)2=−1{\displaystyle ({\frac {x}{a}})^{2}+({\frac {y}{b}})^{2}=-1} the hyperbolic cylinder: (x a)2−(y b)2=1{\displaystyle ({\frac {x}{a}})^{2}-({\frac {y}{b}})^{2}=1} and the parabolic cylinder: x 2+2 a y=0.{\displaystyle x^{2}+2ay=0.\,} In projective geometry, a cylinder is simply a cone whose apex is at infinity, which corresponds visually to a cylinder in perspective appearing to be a cone towards the sky. References [change \| change source] ↑"MathWorld: Cylindric section". Other websites [change \| change source] Wikimedia Commons has media related to Cylinder (geometry). Wikisource has the text of the 1911 Encyclopædia Britannica article Cylinder. Surface area of a cylinder at MATHguide Volume of a cylinder at MATHguide Spinning Cylinder at Math Is Fun Volume of a cylinder Interactive animation at Math Open Reference Cut a Cylinder Interactive demonstrations of the intersection of a plane and a cylinder Cylinder Calculator 3D Specify only two cylinder values and all others are immediately calculated, plus interactive 3D visualization in fullscreen. \| Collapse v t e Shapes \| \| triangle square circle rhombus parallelogram trapezoid rectangle pentagon heptadecagon polygon cube parallelepiped cuboid cone cylinder sphere ellipsoid pyramid prism tetrahedron octahedron torus dodecahedron icosahedron frustum antiprism bipyramid rhombohedron \| Retrieved from " Category: Three-dimensional shapes Hidden category: Commons link is on Wikidata This page was last changed on 9 July 2025, at 19:13. Text is available under the Creative Commons Attribution-ShareAlike License and the GFDL; additional terms may apply. See Terms of Use for details. Privacy policy About Wikipedia Disclaimers Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Cylinder 102 languagesAdd topic
17011	http://www.labptp.com/knowledge/show-435.html	为何需要设立阴阳实验对照？_基础知识_知识学堂_能力验证网参加者登录食品伙伴网 \| 食品检测服务中心 \| 收藏本页行业动态通知公告知识学堂能力验证提供机构能力验证专题汇总证书查询结果通知单查询全国免费咨询热线： 0535-2129197 首页行业动态通知公告国内能力验证公告国外能力验证公告快查中心能力验证快速检索能力验证提供机构知识学堂基础知识行业领域专题标准法规在线课堂（视频）平台介绍证书查询证书查询结果通知单查询联系方式热门关键词：能力验证指南认证认可微生物注意事项环境满意接种RB/T041 首页>知识学堂> 为何需要设立阴阳实验对照？为何需要设立阴阳实验对照？日期：2024-06-20 来源：网络核心提示：一、为什么设置阴阳对照？空白有的时候可以当作是一种阴性对照，空白的特别意义在于测定实验的背景噪声。阴性对照和阳性对照是针对预期结果而说的。凡是肯定出现预期结果的组，为阳性对照组。凡是肯定不会出现预期结果的组，为阴性对照组。空白对照是针对处理因素而说的。凡是不加处理因素的组，为空白对照组。所谓不加处理一、为什么设置阴阳对照？空白有的时候可以当作是一种阴性对照，空白的特别意义在于测定实验的“背景噪声”。阴性对照和阳性对照是针对“预期结果”而说的。凡是肯定出现预期结果的组，为阳性对照组。凡是肯定不会出现预期结果的组，为阴性对照组。空白对照是针对“处理因素”而说的。凡是不加处理因素的组，为空白对照组。所谓不加处理因素，可以是用生理盐水等溶剂代替所要加的溶液，也可以是给实验动物开刀但是不摘除某个器官，等等。阴性对照就是一个确知是阴性结果，如果实验结果呈阳性了，说明实验操作程序有问题。阳性对照的目的是排除实验操作导致的假阴性。阳性对照就是一个确知是阳性结果，如果实验结果是阴性，说明实验操作程序有问题。二、如何实验分组如何设计？改变自变量，观察因变量。通俗地讲，研究什么，就改变什么。改变自变量的过程就是实验设计的过程。 “研究DNA，就改变DNA”，组4可以在组3的基础上加入 DNA酶，催化DNA的水解。标签：能力验证能力验证网微生物能力验证能力验证样品能力验证提供下一篇：培养箱有消毒的必要? 上一篇：实验室布局要求以及都需要配备哪些检测设备？点击排行 7560 1《检验检测机构资质认定生态环境监测机构评审补充要求》官方解读 6605 2微生物检测实验室的质量控制措施 5689 3检验检测机构/实验室的质量监督与质量负责人工作职责 5471 4检验检测机构中有关人员社保等问题汇总 5014 5检验检测机构管理层的设置与分工 4845 6分享丨检验检测实验室样品受理注意事项 4832 7检验检测机构日常检查中的管理缺陷，快看看你遇到过吗？ 4747 8认监委和认可委的关系 4519 9微生物常用接种方法 4517 10能力验证注意事项 4435 11干货丨实验室授权签字人基础知识汇总 4281 12常用菌种保藏方法 4248 13实验室固液体药品取用原则及注意事项 4166 14实验室搬家，很多人忘了一件大事！ 4021 15实验室为什么要参加能力验证？ (c)2008-2020 食品伙伴网 All Rights Reserved 鲁ICP备2020046432号-151La 鲁公网安备 37060202000931号
17012	https://ocw.mit.edu/courses/18-783-elliptic-curves-spring-2021/pages/readings/	Readings \| Elliptic Curves \| Mathematics \| MIT OpenCourseWare Browse Course Material Syllabus Calendar Instructor Insights Teaching 18.783 in Person Readings Lecture Notes and Worksheets Assignments Related Resources Course Info Instructor Dr. Andrew Sutherland Departments Mathematics As Taught In Spring 2021 Level Undergraduate Topics Engineering Computer Science Mathematics Algebra and Number Theory Computation Learning Resource Types assignment Problem Sets notes Lecture Notes co_present Instructor Insights Editable Files Download Course menu search Give Now About OCW Help & Faqs Contact Us searchGIVE NOWabout ocwhelp & faqscontact us 18.783 \| Spring 2021 \| Undergraduate Elliptic Curves Menu More Info Syllabus Calendar Instructor Insights Teaching 18.783 in Person Readings Lecture Notes and Worksheets Assignments Related Resources Readings There is no required text, but lecture notes are provided. We make reference to material in the five books listed below. In addition, there are citations and links to other references. [Washington] = Washington, Lawrence C. Elliptic Curves: Number Theory and Cryptography. Chapman & Hall / CRC, 2008. ISBN: 9781420071467. (Errata (PDF)) [Preview with Google Books]. Online version. [Milne] = Milne, James S. Elliptic Curves. BookSurge Publishing, 2006. ISBN: 9781419652578. (Addendum / erratum (PDF)). Online version (PDF - 1.5MB). [Silverman] = Silverman, Joseph H. The Arithmetic of Elliptic Curves. Springer-Verlag, 2009. ISBN: 9780387094939. (Errata (PDF)) [Preview with Google Books].Online version. [Silverman (Advanced Topics)] = Silverman, Joseph H. Advanced Topics in the Arithmetic of Elliptic Curves. Springer-Verlag, 1994. ISBN: 9780387943251. (Errata (PDF)).Online version. [Cox] = Cox, David A. Primes of the Form (x^2 + ny^2): Fermat, Class Field Theory, and Complex Multiplication. Wiley-Interscience, 1989. ISBN: 9780471506546.(Errata (PDF)). Online version. Lecture 1: Introduction to Elliptic Curves No readings assigned Lecture 2: The Group Law and Weierstrass and Edwards Equations [Washington] Sections 2.1–3 and 2.6.3 Bernstein, Daniel, and Tanja Lange. “Faster Addition and Doubling on Elliptic Curves.”Lecture Notes in Computer Science 4833 (2007): 29–50. Bernstein, Daniel, and Tanja Lange. “A Complete Set of Addition Laws for Incomplete Edwards Curves.” (PDF) Lecture 3: Finite Field Arithmetic Gathen, Joachim von zur, and Jürgen Gerhard. Sections 3.2, 8.1–4, 9.1, 11.1, and 14.2–6 in Modern Computer Algebra. Cambridge University Press, 2003. ISBN: 9780521826464. [Preview with Google Books] Cohen, Henri, Gerhard Frey, and Roberto Avanzi. Chapter 9 in Handbook of Elliptic and Hyperelliptic Curve Cryptography. Chapman & Hall / CRC, 2005. ISBN: 9781584885184. [Preview with Google Books] Rabin, Michael O. “Probabilistic Algorithms in Finite Fields.” Society for Industrial and Applied Mathematics 9, no. 2 (1980): 273–80. Lecture 4: Isogenies [Washington] Section 2.9 [Silverman] Section III.4 Lecture 5: Isogeny Kernels and Division Polynomials [Washington] Sections 3.2 and 12.3 [Silverman] Section III.4 Lecture 6: Endomorphism Rings [Washington] Section 4.2 [Silverman] Section III.6 Lecture 7: Hasse’s Theorem and Point Counting [Washington] Section 4.3 Lecture 8: Schoof’s Algorithm [Washington] Sections 4.2 and 4.5 Schoof, Rene. “Elliptic Curves over Finite Fields and the Computation of Square Roots mod p.” (PDF - 1.1MB)Mathematics of Computation 44, no. 170 (1985): 483–94. Lecture 9: Generic Algorithms for the Discrete Logarithm Problem [Washington] Section 5.2 Pohlig, Stephen, and Martin Hellman. “An Improved Algorithm for Computing Logarithms over GF(p) and Its Cryptographic Significance.” (PDF)IEEE Transactions on Information Theory 24, no. 1 (1978): 106–10. Pollard, John M. “Monte Carlo Methods for Index Computation (mod p).” (PDF)Mathematics of Computation 32, no. 143 (1978): 918–24. Shor, Peter W. “Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer.” SIAM J.Sci.Statist.Comput. 26 (1997): 1484. Shoup, Victor. “Lower Bounds for Discrete Logarithms and Related Problems.” Lecture Notes in Computer Science 1233 (1997): 256–66. Lecture 10: Index Calculus, Smooth Numbers, and Factoring Integers [Washington] Sections 5.1 and 7.1 Granville, Andrew. “Smooth Numbers: Computational Number Theory and Beyond.” (PDF) In Algorithmic Number Theory: Lattices, Number Fields, Curves and Cryptography. Cambridge University Press, 2008. ISBN: 9780521808545. Lenstra, H. W. “Factoring Integers with Elliptic Curves.” (PDF - 1.3MB)Annals of Mathematics, Mathematical Sciences Research Institute 126 (1986): 649–73. Lecture 11: Elliptic Curve Primality Proving (ECPP) [Washington] Section 7.2 Goldwasser, Shafi, and Joe Kilian. “Almost All Primes Can Be Quickly Certified.”STOC'86 Proceedings of the 18 th Annual ACM Symposium on Theory of Computing (1986): 316–29. Pomerance, Carl. “Very Short Primality Proofs.” (PDF)Mathematics of Computation 48, no. 177 (1987): 315–22. Lecture 12: Endomorphism Algebras [Silverman] Section III.9 Lecture 13: Ordinary and Supersingular Curves [Silverman] Section III.1 and Chapter V [Washington] Sections 2.7 and 4.6 Lecture 14: Elliptic Curves over C (Part I) [Cox] Chapter 10 [Silverman] Sections VI.2–3 [Washington] Sections 9.1–2 Lecture 15: Elliptic Curves over C (Part II) [Cox] Chapters 10 and 11 [Silverman] Sections VI.4–5 [Washington] Sections 9.2–3 Lecture 16: Complex Multiplication (CM) [Cox] Chapter 11 [Silverman] Section VI.5 [Washington] Section 9.3 Lecture 17: The CM Torsor [Cox] Chapter 7 [Silverman (Advanced Topics)] Section II.1.1 Lecture 18: Riemann Surfaces and Modular Curves [Silverman (Advanced Topics)] Section I.2 [Milne] Section V.1 Lecture 19: The Modular Equation [Cox] Chapter 11 [Milne] Section V.2 [Washington] pp. 273–74 Lecture 20: The Hilbert Class Polynomial [Cox] Chapters 8 and 11 Lecture 21: Ring Class Fields and the CM Method [Cox] Chapters 8 and 11 (cont.) Lecture 22: Isogeny Volcanoes Sutherland, Andrew V. “Isogeny Volcanoes.” The Open Book Series. 1, no. 1 (2013): 507–530. Lecture 23: The Weil Pairing Miller, Victor S. “The Weil Pairing, and Its Efficient Calculation.” Journal of Cryptology: The Journal of the International Association for Cryptologic Research (IACR) 17, no. 4 (2004): 235–61. [Washington] Chapter 11 [Silverman] Section III.8 Lecture 24: Modular Forms and L-Functions [Milne] Sections V.3–4 Lecture 25: Fermat’s Last Theorem [Milne] Sections V.7–9 [Washington] Chapter 15 Cornell, Gary, Joseph H. Silverman, and Glenn Stevens. Chapter 1 in Modular Forms and Fermat’s Last Theorem. Springer, 2000. ISBN: 9780387989983. Online version. Course Info Instructor Dr. Andrew Sutherland Departments Mathematics As Taught In Spring 2021 Level Undergraduate Topics Engineering Computer Science Mathematics Algebra and Number Theory Computation Learning Resource Types assignment Problem Sets notes Lecture Notes co_present Instructor Insights Editable Files Download Course Over 2,500 courses & materials Freely sharing knowledge with learners and educators around the world. Learn more © 2001–2025 Massachusetts Institute of Technology Accessibility Creative Commons License Terms and Conditions Proud member of: © 2001–2025 Massachusetts Institute of Technology You are leaving MIT OpenCourseWare close Please be advised that external sites may have terms and conditions, including license rights, that differ from ours. MIT OCW is not responsible for any content on third party sites, nor does a link suggest an endorsement of those sites and/or their content. Stay Here Continue
17013	https://www.youtube.com/playlist?list=PL_87SXcKvr8WDFGXyijRSP0mLaZ93n0nk	Calculus 1 Full Course \| Stewart Calculus Explained Step-by-Step - YouTube Back Skip navigation Search Search with your voice Sign in Home HomeShorts ShortsSubscriptions SubscriptionsYou YouHistory History Play all Calculus 1 Full Course \| Stewart Calculus Explained Step-by-Step by Understand The Math • Playlist•17 videos•145 views Master Calculus 1 with This Complete Video Series...more Master Calculus 1 with This Complete Video Series...more...more Play all PLAY ALL Calculus 1 Full Course \| Stewart Calculus Explained Step-by-Step 17 videos 145 views Updated 2 days ago Save playlist Shuffle play Share Master Calculus 1 with This Complete Video Series Follow along with this full Calculus 1 course covering Chapters 2–6 of Stewart’s Calculus: Early Transcendentals. Each lesson includes clear explanations, detailed walkthroughs, and free guided notes to help you succeed — whether you’re taking calculus for the first time or reviewing before Calculus 2. 📖 What You’ll Learn 🔹 Chapter 2: Limits and Continuity • Understand limits and limit laws • Explore continuity and the Intermediate Value Theorem 🔹 Chapter 3: Derivatives • Learn the formal definition of the derivative • Master differentiation rules (power, product, quotient, chain) • Differentiate trigonometric, exponential, and logarithmic functions 🔹 Chapter 4: Applications of Derivatives • Curve sketching and optimization • Related rates examples • Mean Value Theorem 🔹 Chapter 5: Integrals • Compute definite & indefinite integrals • Apply the Fundamental Theorem of Calculus • Learn techniques of integration 🔹 Chapter 6: Applications of Integrals • Find area under curves • Calculate volumes of solids of revolution • Determine the average value of a function 📌 Free Guided Notes Each video comes with a downloadable PDF so you can follow along, take notes, and practice independently. 📥 Download here: understandthemath.com/calculus-1 🔍 About Understand The Math Hi, I’m Dr. Cheryl Hile, founder of Understand The Math! With over 25 years of university teaching experience, I’m passionate about helping students succeed. This series is designed to make calculus approachable, structured, and confidence-building. #Calculus1 #LearnCalculus #CalculusCourse #StewartCalculus #MathHelp #CalculusTutorial #LimitsAndContinuity #Derivatives #Integrals #CollegeMath #MathStudent #UnderstandTheMath #CalculusMadeEasy #StepByStepMath #OnlineMathCourse Show more Understand The Math Understand The Math Subscribe Play all Calculus 1 Full Course \| Stewart Calculus Explained Step-by-Step by Understand The Math Playlist•17 videos•145 views Master Calculus 1 with This Complete Video Series...more Master Calculus 1 with This Complete Video Series...more...more Play all 1 2:27 2:27 Now playing Learn Calculus 1 with Free Step-by-Step Tutorials and Guided Notes Understand The Math Understand The Math • 90 views • 1 month ago • 2 25:07 25:07 Now playing The Tangent Problem and the Velocity Problem \| Calculus 1 Introduction Understand The Math Understand The Math • 123 views • 3 weeks ago • 3 30:52 30:52 Now playing The Limit of a Function\| Intuitive, General, One-Sided, and Infinite Limits in Calculus 1 Understand The Math Understand The Math • 71 views • 3 weeks ago • 4 41:40 41:40 Now playing How to Calculate Limits Using Limit Laws \| Indeterminate Forms and Squeeze Theorem Understand The Math Understand The Math • 53 views • 3 weeks ago • 5 15:28 15:28 Now playing Precise Definition of a Limit (Epsilon–Delta) \| Calculus 1 Understand The Math Understand The Math • 101 views • 2 weeks ago • 6 49:34 49:34 Now playing Understanding Continuity in Calculus: Definitions, Types, and the Intermediate Value Theorem Understand The Math Understand The Math • 81 views • 2 weeks ago • 7 35:01 35:01 Now playing Limits at Infinity and Horizontal Asymptotes in Calculus Understand The Math Understand The Math • 57 views • 2 weeks ago • 8 38:20 38:20 Now playing Derivatives and Rates of Change in Calculus: Tangent Lines and Velocity Understand The Math Understand The Math • 48 views • 12 days ago • 9 31:15 31:15 Now playing The Derivative as a Function: Graphing, Differentiability, and Higher Derivatives Understand The Math Understand The Math • 43 views • 10 days ago • 10 37:34 37:34 Now playing Derivatives of Polynomials and Exponential Functions: Power Rule and More Understand The Math Understand The Math • 38 views • 7 days ago • 11 18:49 18:49 Now playing Product Rule and Quotient Rule in Calculus: Step-by-Step Examples Understand The Math Understand The Math • 37 views • 5 days ago • 12 30:30 30:30 Now playing Derivatives of Trigonometric Functions: Sine, Cosine, and More Understand The Math Understand The Math • 28 views • 3 days ago • 13 22:55 22:55 Now playing The Chain Rule in Calculus: How to Differentiate Composite Functions Understand The Math Understand The Math • 20 views • 16 hours ago • 14 34:57 34:57 Now playing Implicit Differentiation in Calculus: Step-by-Step Examples Understand The Math Understand The Math • 3 views • 5 days ago • 15 11:43 11:43 Now playing Derivatives of Inverse Trigonometric Functions \| Calculus 1 Understand The Math Understand The Math • 1 view • 5 days ago • 16 14:13 14:13 Now playing Derivatives of Logarithmic Functions and Logarithmic Differentiation \| Calculus 1 Understand The Math Understand The Math • 3 views • 4 days ago • 17 36:47 36:47 Now playing Exponential Growth and Decay in Calculus \| Population, Radioactivity, and Cooling Examples Understand The Math Understand The Math • 4 views • 2 days ago •
17014	https://thecorestandards.org/Math/Content/EE/	Common Core State Standards Initiative Common Core State Standards Initiative Expressions & Equations Print this page Grade 6 Apply and extend previous understandings of arithmetic to algebraic expressions. CCSS.Math.Content.6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents. CCSS.Math.Content.6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers. CCSS.Math.Content.6.EE.A.2.a Write expressions that record operations with numbers and with letters standing for numbers. For example, express the calculation "Subtract y from 5" as 5 - y. CCSS.Math.Content.6.EE.A.2.b Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity. For example, describe the expression 2 (8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms. CCSS.Math.Content.6.EE.A.2.c Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). For example, use the formulas V = s3 and A = 6 s2 to find the volume and surface area of a cube with sides of length s = 1/2. CCSS.Math.Content.6.EE.A.3 Apply the properties of operations to generate equivalent expressions. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. CCSS.Math.Content.6.EE.A.4 Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them). For example, the expressions y + y + y and 3y are equivalent because they name the same number regardless of which number y stands for.. Reason about and solve one-variable equations and inequalities. CCSS.Math.Content.6.EE.B.5 Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true. CCSS.Math.Content.6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. CCSS.Math.Content.6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers. CCSS.Math.Content.6.EE.B.8 Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams. Represent and analyze quantitative relationships between dependent and independent variables. CCSS.Math.Content.6.EE.C.9 Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. For example, in a problem involving motion at constant speed, list and graph ordered pairs of distances and times, and write the equation d = 65t to represent the relationship between distance and time. Grade 7 Use properties of operations to generate equivalent expressions. CCSS.Math.Content.7.EE.A.1 Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients. CCSS.Math.Content.7.EE.A.2 Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. For example, a + 0.05a = 1.05a means that "increase by 5%" is the same as "multiply by 1.05." Solve real-life and mathematical problems using numerical and algebraic expressions and equations. CCSS.Math.Content.7.EE.B.3 Solve multi-step real-life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies. For example: If a woman making $25 an hour gets a 10% raise, she will make an additional 1/10 of her salary an hour, or $2.50, for a new salary of $27.50. If you want to place a towel bar 9 3/4 inches long in the center of a door that is 27 1/2 inches wide, you will need to place the bar about 9 inches from each edge; this estimate can be used as a check on the exact computation. CCSS.Math.Content.7.EE.B.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. CCSS.Math.Content.7.EE.B.4.a Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width? CCSS.Math.Content.7.EE.B.4.b Solve word problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality and interpret it in the context of the problem. For example: As a salesperson, you are paid $50 per week plus $3 per sale. This week you want your pay to be at least $100. Write an inequality for the number of sales you need to make, and describe the solutions. Grade 8 Expressions and Equations Work with radicals and integer exponents. CCSS.Math.Content.8.EE.A.1 Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 × 3-5 = 3-3 = 1/33 = 1/27. CCSS.Math.Content.8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational. CCSS.Math.Content.8.EE.A.3 Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3 times 108 and the population of the world as 7 times 109, and determine that the world population is more than 20 times larger. CCSS.Math.Content.8.EE.A.4 Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology Understand the connections between proportional relationships, lines, and linear equations. CCSS.Math.Content.8.EE.B.5 Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed. CCSS.Math.Content.8.EE.B.6 Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. Analyze and solve linear equations and pairs of simultaneous linear equations. CCSS.Math.Content.8.EE.C.7 Solve linear equations in one variable. CCSS.Math.Content.8.EE.C.7.a Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). CCSS.Math.Content.8.EE.C.7.b Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. CCSS.Math.Content.8.EE.C.8 Analyze and solve pairs of simultaneous linear equations. CCSS.Math.Content.8.EE.C.8.a Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously. CCSS.Math.Content.8.EE.C.8.b Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. CCSS.Math.Content.8.EE.C.8.c Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair. Kindergarten-Grade 12 Standards for Mathematical Practice Introduction How to read the grade level standards Kindergarten Introduction Counting & Cardinality Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry Grade 1 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry Grade 2 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry Grade 3 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions¹ Measurement & Data Geometry Grade 4 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten¹ Number & Operations—Fractions¹ Measurement & Data Geometry Grade 5 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry Grade 6 Introduction Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability Grade 7 Introduction Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability Grade 8 Introduction The Number System Expressions & Equations Functions Geometry Statistics & Probability High School: Number and Quantity Introduction The Real Number System Quantities The Complex Number System Vector & Matrix Quantities High School: Algebra Introduction Seeing Structure in Expressions Arithmetic with Polynomials & Rational Expressions Creating Equations Reasoning with Equations & Inequalities High School: Functions Introduction Interpreting Functions Building Functions Linear, Quadratic, & Exponential Models Trigonometric Functions High School: Modeling High School: Geometry Introduction Congruence Similarity, Right Triangles, & Trigonometry Circles Expressing Geometric Properties with Equations Geometric Measurement & Dimension Modeling with Geometry High School: Statistics & Probability Introduction Interpreting Categorical & Quantitative Data Making Inferences & Justifying Conclusions Conditional Probability & the Rules of Probability Using Probability to Make Decisions Note on courses & transitions Courses & Transitions Mathematics Glossary Mathematics Glossary Table 1 Table 2 Table 3 Table 4 Table 5 Standards by Domain Please click here for the ADA Compliant version of the Math Standards. Close
17015	https://physics.stackexchange.com/questions/573345/how-can-i-calculate-deceleration-due-to-friction	Skip to main content How can I calculate deceleration due to friction? Ask Question Asked Modified 1 year ago Viewed 17k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I was wondering, how can I calculate the decelerations of an object due to friction - and therefore find the maximal distance it can reach? For example, if a car suddenly brakes in the middle of a road, how can I find how long it will take it to fully stop, its deceleration, the distance it will reach before stopping? Or if someone is skiing and reaches the bottom of a mountain, where the ground becomes flat, how can I know how long it will take the skier to come to a complete stop? Before I’ve used the formula (of which I’m not too sure): deceleration=−g×μstatic. But now that I think of it, I’m not sure if it’s correct anymore... newtonian-mechanics acceleration friction speed distance Share CC BY-SA 4.0 Improve this question Follow this question to receive notifications edited Aug 15, 2020 at 8:44 Bhavay 1,70122 gold badges1515 silver badges2828 bronze badges asked Aug 15, 2020 at 3:14 Sophia LilySophia Lily 5111 gold badge22 silver badges44 bronze badges 1 You can accept an answer by clicking on the tick next to the answer since you are new – imposter Commented Aug 15, 2020 at 19:49 Add a comment \| 3 Answers 3 Reset to default This answer is useful 1 Save this answer. Show activity on this post. I was wondering, how can I calculate the decelerations of an object due to friction - and therefore find the maximal distance it can reach? If you know the velocity of the object before friction begins to bring it to a stop you can calculate the stopping distance using the work-energy theorem which states that the net work done on an object equals its change in kinetic energy. If the only force acting on the object bringing it to a stop is the friction force then Wnet=μmgd=mv22 μgd=v22 d=v22μg Where d = stopping distance, v = velocity of object before encountering friction, μ = the coefficient of friction and g= acceleration due to gravity. In the case of vehicle braking distance, if the car is skidding you use the coefficient of kinetic friction between the tires and the road. If the wheels continue to roll you use the coefficient of static friction. Generally rolling resistance can be ignored. You can calculate the magnitude of the deceleration from Newtons second law a=Fm=μmgm=μg And finally you can calculate the stopping time from d=at22 t=2da−−−√ Hope this helps. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Aug 16, 2020 at 11:58 answered Aug 15, 2020 at 13:48 Bob DBob D 82.5k77 gold badges6161 silver badges160160 bronze badges Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Well g×coefficient of static friction is an incorrect way of finding the deceleration due to friction. The first obvious reason is that if the object is moving, then kinetic friction comes into play, or else rolling friction comes into play in real rolling conditions. Only in a perfectly ideal pure rolling scenario can we take static friction in our calculations. However most of the questions deal with ideal cases so this part is mostly correct. Also the other term "g" would be correct only in cases such as a car traveling on a straight road, etc. If the object was traveling on an incline, your formula would give you an incorrect value. This is because of how friction is defined. When an object is moving Friction=Coeffcient of kinetic friction × Normal force Only on a flat surface would the normal force be mg. On a plane inclined at an angle θ with the ground, it would be mgcosθ So deceleration here would have a gcos θ term instead of the g term in your formula. Hope it helped! Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Aug 15, 2020 at 7:44 answered Aug 15, 2020 at 4:39 Vamsi KrishnaVamsi Krishna 1,27311 gold badge1010 silver badges2424 bronze badges 2 1 This is incorrect, friction can be static if the wheels are in pure rolling. – Danny LeBeau Commented Aug 15, 2020 at 5:19 @Binod Thanks for pointing it out. Now I have edited it correctly. Please tell me if any other fixes are required. – Vamsi Krishna Commented Aug 15, 2020 at 7:44 Add a comment \| This answer is useful 0 Save this answer. Show activity on this post. Although the other answers are correct, they suppose that the friction coefficient (μ) is known and that it is constant. Friction is highly empirical, and the knowledge of a define μ is normally impossible. In the first example, μ varies depending on how much the pedals are pressed, and that pressure can change during a braking event. Furthermore, if the braking pressure on the wheels is too great, the car can slide on the road, changing the process of friction and dangerously extending the braking distance. That is the reason for ABS systems. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Aug 17, 2024 at 1:37 Claudio SaspinskiClaudio Saspinski 17.7k22 gold badges1515 silver badges4040 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions newtonian-mechanics acceleration friction speed distance See similar questions with these tags. Featured on Meta Upcoming initiatives on Stack Overflow and across the Stack Exchange network... Related 0 Finding deceleration and velocity using distance and time 15 How to brake 'beautifully'? 1 Is the "efficiency" of car brake affected by how much force is applied by the brake? 0 Deceleration with known drag force 2 Work done by friction when a bicycle comes to a skidding stop 2 Why do large trucks have a longer stopping distance than cars? Hot Network Questions Excel lookup from one matrix or array to another Make a square tikzcd diagram Why does bible claim Asa was perfect all his days when he didn't rely on God in the latter years complex 3-clause conditional: 'I realized that if I do..., I will...' vs. 'I realized that if I did..., I would...' How do you get rid of extra villagers in ACNH that have an account on the Nintendo Switch? Should I apply to volunteer while actively looking for a job, if it means I won't be volunteering for long? What is the principle of non-liability in the European Union wrt the US Section 230 and social media? What reasons do Mormons give for the usage of the name "Alma" for males? Why \bar with unicode-math was shifted right even with latin modern math by default? Film about alien spaceships ominously hovering over Earth, only to form a shield to protect it How do I pass a parameter to a locally store web page from the context (right click) menu of Notepad++? Is sickle cell anemia a pyrimidine to purine change or vice versa? Unknown Close Window key combination that is NOT Alt+F4 Is the Royal Review Board’s Five Diamond Award in Ocean's Thirteen fictional? Why wrap fruitcake in grease-proof paper? Internal`CopyListStructure does not produce packed array output from packed array input Do strong winds also cause a hovering creature to fall? Do I thank an ex-supervisor in my acknowledgements who was removed from my panel? Why Methyl Orange can be used in the titration of HCl and NaOH? Basic connection between probability space and random variable How was the th' in older poems pronounced? Major Revisions and Editor's Note How do you mount a fabric poster on an easel at a conference? Should theological questions be considered philosophical questions and dealt with by philosophers? Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
17016	https://www.algebra.com/algebra/homework/equations/eq.lesson	Lesson PARABOLAS Algebra -> Equations -> Lesson PARABOLAS Log On \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- \| \| \| \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| \| --- --- --- \| \| Algebra: EquationsSection \| SolversSolvers \| LessonsLessons \| Answers archiveAnswers \| \| \| \| \| \| \| \| This Lesson (PARABOLAS) was created by by Theo(13342) : View Source, ShowAbout Theo: This lesson covers a basic overview of PARABOLAS Parabolas are structures that are the result of quadratic equations. They have certain properties that will be discussed in more detail below. To see more information relating to the standard form of quadratic equations and how to solve them, see the lesson on QUADRATIC EQUATIONS. REFERENCES Paul's Online Notes Interactive Mathematics Purple Math Hot Math Math Warehouse Quadratic and Vertex Equation Forms Jack Sarfaty Wilson College Focus and Directrix of a Parabola (mathwords) Focusw and Directrix of a Parabola (algebralab) Focus and Directrix of a Parabola (intmath) Focus and Directrix of a Parabola (maricopa) STANDARD FORM FOR THE EQUATION OF A PARABOLA The standard form for the equation of a parabola is the standard form of a quadratic equation. The standard form of the quadratic equation is: a is the coefficient of the x^2 term. b is the coefficient of the x term. c is the constant term. VERTEX OF A PARABOLA The vertex of a parabola is the minimum / maximum point of the parabola. If the parabola opens up, the vertex is the minimum point on the parabola. If the parabola opens down, the parabola is the maximum point on the parabola. If the parabola opens right, the vertex is the leftmost point on the parabola. If the parabola opens left, the vertex is the rightmost point on the parabola. ROOTS OF THE PARABOLA In order to solve for the roots of the quadratic equation, you would set y = 0 and solve for x. Note that x^2 and are the same thing. The first is the text version and the second is the algebra.com formula generator version. An example of a parabola in standard form would be y = x^2 - 10x + 16 where: a = 1 b = -10 c = 16 This parabola opens up. That's because the "a" term in the equation of this parabola is positive. Please note that when the parabola opens up, then the vertex of the parabola is the lowest point on the parabola. This would also be the minimum point on the parabola because the value of y at this point is less than the value of y at any other point on the graph of this parabola. Graph of y = x^2 - 10x + 16 is shown below. To find the roots of this equation, set y = 0 and solve for x. Factors for this equation would be in the general form of: (dx+e)(fx+g) = 0 where: df = a dg + ef = b eg = c The factors for this equation of x^2 - 10x + 16 are: (x-2)(x-8) = 0 The standard form is: (dx+e)(fx+g) = 0 You can see that: df = 11 = 1 dg + ef = 1(-8) + (-2)1 = -8-2 = -10 eg = (-2)(-8) = 16 The roots for this equation of x^2 - 10x + 16 are: x = 2 and x = 8 If we make the "a" term in this equation negative, then the graph will open down. If we want to keep the roots the same so we have the equation in the same relative position on the graph only opening in the opposite direction, we multiply the whole equation by (-1) to get: -x^2 + 10x - 16 Original equation is x^2 - 10x + 16 Revised equation is -x^2 + 10x - 16 It's the same equation only now it is opening down. Please note that when the parabola opens down, then the vertex of the parabola is the highest point on the parabola. This would also be the maximum point on the parabola because the value of y at this point is greater than the value of y at any other point on the graph of this parabola. It has the same roots so it has essentially been pivoted on the x-axis. Graph of y = -x^2 + 10x - 16 is shown below. PARABOLA THAT OPENS TO THE LEFT OR TO THE RIGHT Let's take the original equation of y = x^2 - 10x + 16 which opens up. Replace y with x and x with y to get x = y^2 - 10y + 16 This graph will open to the right because the "a" term is positive (coefficient of y^2). Please note that when the parabola opens to the right, then the vertex of the parabola is the leftmost point on the parabola. This would also be the minimum point on the parabola because the value of x at this point is less than the value of x at any other point on the graph of this parabola. Graph of x = y^2 - 10y + 16 is shown below. Now let's take the original equation of y = -x^2 + 10x - 16 which opens down. Replace y with x and x with y to get x = -y^2 + 10y - 16 This graph will open to the left because the "a" term is negative (coefficient of y^2). Please note that when the parabola opens to the left, then the vertex of the parabola is the rightmost point on the parabola. Please note that this is also the maximum point of this parabola because the value of x at this point is greater than the value of x at any other point on the graph of this parabola. Graph of x = -y^2 + 10y - 16 is shown below. Note that we could not graph x = y^2 - 10y + 16 directly, nor could we graph x = -y^2 + 10y - 16 directly. This is because the graphing software is fixed so that your equation has to be in the form of y = f(x) only. It does not know how to handle x = f(y). In order to graph these equations, we have to solve for x in order to convert the equation that is in the form of x = f(y) into the form of y = f(x). I'll do one to show you how that was done since it's a good opportunity to show you how the completing the squares method can be helpful in solving equations. Take the equation of x = y^2 - 10y + 16 Solve for y as follows: Subtract 16 from both sides of this equation to get: y^2 - 10y = x - 16 Complete the squares on the left hand side of this equation by taking one half of -10 (the b term) to get -5 and then squaring it to get 25 to get: (y-5)^2 - 25 = x - 16 The -5 became part of the factor to be squared and the 25 is an adjustment of the result of (y-5)^2 to make it equal to y^2 - 10y rather than y^2 - 10y + 25. Add 25 to both sides of this equation to get: (y-5)^2 = x+9 Take the square root of both sides of this equation to get: y-5 = +/- Add 5 to both sides of this equation to get: y = 5 +/- You have now converted x = f(y) into the form of y = f(x). The two forms of this equation are equivalent. x = f(y) solves for x. y = f(x) solves for y. In a similar fashion, x = -y^2 + 10y - 16 was converted to: y = 5 +/- The equation of x = y^2 - 10x + 16 opens to the right. The equivalent equation of y = 5 +/- also opens to the right. The coefficient of y^2 in the x = f(y) version of the equation is positive and the coefficient of x in the y = f(x) version of the same equation is also positive. This will be generally true. The equation of x = -y^2 + 10x - 16 opens to the left. The equivalent equation of y = 5 +/- also opens to the left. The coefficient of y^2 in the x = f(y) version of the equation is negative and the coefficient of x in the y = f(x) version of the same equation is also negative. This will be generally true. FINDING THE VERTEX OF A PARABOLA WHEN THE EQUATION FOR THE PARABOLA IS IN STANDARD FORM AND THE PARABOLA OPENS UP OR DOWN Please note that when the parabola opens up, then the vertex of the parabola is the lowest point on the parabola. This is also called the minimum point on the parabola. It is called the minimum point on the parabola because the value of y at that point is less than at any other point on the parabola. Standard form of the equation of a parabola that opens up or down is y = ax^2 + bx + c The x value of the vertex of the parabola can be found by the equation x = -b/2a Once you find the x value, then you plug that value into the equation to find the y value. The generalized form of the vertex of the parabola is (x,y) = (-b/2a,f(-b/2a)). -b/2a is the x coordinate of the vertex. f(-b/2a) is the y coordinate of the vertex. You find the x coordinate of the vertex by using the formula x = -b/2a. You find the y coordinate of the vertex by plugging the value of x into the equation and solving for y. Since you equation is in the form of y = f(x), then you are solving for f(-b/2a) when you replace x with the value of -b/2a. An example will help clarify this. y = f(x) = x^2 - 10x + 16 is the standard form of the equation for a parabola that opens up. The standard form of this equation is y = ax^2 + bx + c. a is the coefficient of the x^2 term and is equal to 1. The parabola is opening up because a = 1 is positive. The vertex is the minimum point on the parabola. a = 1 b = -10 c = 16 the equation for the x value of the vertex of this parabola is found by using the formula of x = -b/2a and substituting 1 for the a value and -10 for the b value to get: x = -b/2a = -(-10)/2 = 10/2 = 5 The x value for the vertex of this parabola is 5. To find the y value of the vertex of this parabola, replace x with 5 in the original equation and solve for y. Note that y = f(x) = x^2 - 10x + 16 now becomes: y = f(5) = (5)^2 - 10(5) + 16 which becomes: y = f(5) = 25 - 50 + 16 = -25 + 16 = -9 We have an x value of 5 and a y value of -9 for the vertex of the parabola that equals y = f(x) = x^2 - 10x + 16. We found x = -b/2a = 5 and we found: y = f(-b/2a) = -9 the coordinate point of (x,y) = (-b/2a,f(-b/2a)) becomes (5,-9). If you look at the graph of the equation of y = x^2 - 10x + 16, you will see that the vertex of this equation is (5,-9) as shown below: I placed a horizontal line at y = -9 to help show this better. You can also see that the vertex is the lowest / minimum point on this parabola. FINDING THE VERTEX OF A PARABOLA WHEN THE EQUATION FOR THE PARABOLA IS IN STANDARD FORM AND THE PARABOLA OPENS LEFT OR RIGHT. Please note that a parabola that opens left or right will be in the standard form of: x = ay^2 + by + c while a parabola that opens up or down will be in the standard form of: y = ax^2 + bx + c In this section, we will be discussing the parabola that is in the form of: x = ay^2 + by + c Please note also that a parabola in the form of: x = f(y) can also be shown in the form of: y = f(x) but it will look a lot different. It will look something like: y = +/- Please note that if a is negative, then the term under the square root sign will also be negative. Bottom line is you can't miss it. If it's an equation of a parabola that opens to the left or to the right, and it is in the form of y = f(x), then there is a square root on the right hand side of the equation and it definitely does not look like the standard form of the equation of a quadratic equation. We'll continue to work with the x = f(y) form of the parabola that opens to the left or to the right. We will, however, have to convert it to the y = f(x) form when we graph it. Please note that if a is positive, then the parabola will open to the right, and if a is negative, then the parabola will open to the left. Standard form of the equation of a parabola that opens to the left or to the right is: x = ay^2 + by + c The y value of the vertex of the parabola can be found by the formula y = -b/2a Once you find the y value, then you plug that value into the equation to find the x value. The generalized form of the vertex of a parabola that opens left or right would be (f(-b/2a),-b/2a) An example will help clarify this. x = f(y) = y^2 - 10y + 16 is the standard form of the equation for a specific parabola that opens to the right. The general form of this equation is x = ay^2 + by + c. a is the coefficient of the y^2 term and is equal to 1. The parabola is opening to the right because a = 1 is positive. a = 1 b = -10 c = 16 the equation for the y value of the vertex of this parabola is found by using the formula of y = -b/2a and substituting -10 for the b value and 1 for the a value to get: y = -(-10)/2 = 10/2 = 5 The y value for the vertex of this parabola is 5. To find the x value of the vertex of this parabola, replace y with 5 in the original equation and solve for x. x = f(y) = y^2 - 10y + 16 now becomes: x = f(5) = (5)^2 - 10(5) + 16 which becomes: x = f(5) = 25 - 50 + 16 = -25 + 16 = -9 We have a y value of 5 and an x value of -9 for the vertex of the parabola that equals x = f(y) = y^2 - 10y + 16. We found y = -b/2a = 5 and we found x = f(-b/2a) = -9 the coordinate point of (f(-b/2a),-b/2a) becomes (-9,5). If you look at the graph of the equation of x = y^2 - 10y + 16, you will see that the vertex of this equation is (-9,5) as shown below: You will also see that the graph opens to the right because the "a" term is positive. PIVOT POINTS ON A PARABOLA THAT OPENS UP OR DOWN Before we go any further, I wish to point out something that may not be obvious. I reversed the direction of the parabola that opens up by multiplying the whole equation by (-1). ax^2 = bx + c became -ax^2 - bx - c. This essentially pivoted the equation on the roots of that equation. If you want to reverse the direction of the parabola at the vertex, then you have to do the following: Assume your original equation for a parabola that opens up is 3x^2 - 6x + 5 where: a = 3 b = -6 c = 5 Find the vertex at (-b/2a, f(-b/2a)) = (1,2) this is the point you want to pivot at if your want your reverse parabola to open down from that same point. Multiply your equation by (-1) to get the reverse direction equation. 3x^2 - 6x + 5 (-1) = -3x^2 + 6x - 5 Your reverse equation is: -3x^2 + 6x - 5 Remove the c term of -5 and replace it with the variable name of c because it is unknown at this time. Your vertex from the original equation of 3x^2 - 6x + 5 is (x,y) = (1,2). In your reverse equation of -3x^2 + 6x - c, replace x with the original x value of your vertex of 1 and set the equation equal to the original y value of your vertex of 2 and solve for c: Your reverse equation of: -3x^2 + 6x + c becomes: -3(1)^2 + 6(1) + c = 2 which becomes: -3 + 6 + c = 2 which becomes: c = -1 Your new c value is -1 making your reverse equation equal to: -3x^2 + 6x - 1 The original equation is: 3x^2 - 6x + 5 and the reverse equation is: -3x^2 + 6x - 1 Graph of both the original and reverse equation is shown below: As you can see, the reverse equation has now been pivoted at the vertex. AXIS OF SYMMETRY OF A PARABOLA If the parabola points up or down, then the axis of symmetry for the parabola is the x value of the vertex of the parabola. If the parabola points left or right, then the axis of symmetry for the parabola is the y value of the vertex of the parabola. The graph of the parabola that was pointing down with a vertex of (5,-9) is shown below. The axis of symmetry is the vertical line at x = 5. The graph of the parabola that was pointing left with a vertex of (-9,5) is shown below. The axis of symmetry is the horizontal line at y = 5 To see a picture of vertical and horizontal Axis of Symmetry, click on the following hyperlink. Axis Of Symmetry VERTEX FORM OF THE EQUATION FOR A PARABOLA THAT OPENS UP OR DOWN The vertex form of the equation for a parabola that opens up or down is y = a(x-h)^2 + k where: (h,k) is the vertex of the parabola. h is the x coordinate k is the y coordinate a is the coefficient of the (x-h)^2 term. When a is positive, the parabola opens up. When a is negative, the parabola opens down. Please note that when the parabola opens up, the vertex of the parabola is the lowest point on the parabola. It is also the minimum point on the parabola because the value of y at that point is less than at any other point on the parabola. Please note that when the parabola opens down, the vertex of the parabola is the highest point on the parabola. It is also the maximum point on the parabola because the value of y at that point is greater than at any other point on the parabola. VERTEX FORM OF THE EQUATION FOR A PARABOLA THAT OPENS LEFT OR RIGHT The vertex form of the equation for a parabola that opens left or right is x = a(y-k)^2 + h where: (h,k) is the vertex of the parabola. h is the x coordinate k is the y coordinate a is the coefficient of the (y-k)^2 term. When a is positive, the parabola opens to the right. When a is negative, the parabola opens to the left. Please note that when the parabola opens to the right, the vertex of the parabola is the leftmost point on the parabola. It is also the minimum point on the parabola because the value of x at that point is less than at any other point on the parabola. Please note that when the parabola opens to the left, the vertex of the parabola is the rightmost point on the parabola. It is also the maximum point on the parabola because the value of x at that point is greater than at any other point on the parabola. CONVERTING FROM THE STANDARD FORM OF A PARABOLA TO THE VERTEX FORM OF A PARABOLA WHEN THE PARABOLA OPENS UP OR DOWN Standard form of the equation for a parabola opening up or down is y = ax^2 + bx + c Vertex form of the equation for a parabola opening up or down is y = a(x-h)^2 + k To show you how we convert from the standard form to the vertex form, we'll use an example. Your equation in standard form is: y = 5x^2 + 30x + 60 where: a = 5 b = 30 c = 60 While this could be simplified further by dividing out by common factor of 5, we'll leave it the way it is for demonstration purposes because we want to keep the "a" term greater than 1 and we do not want to have to deal with fractions in the demonstration. We factor out the "a" term of 5 to get: We complete the square on the (x^2 + 6x) term to get: which becomes: which becomes: We remove the parentheses to get: which becomes: This is now in vector form of where: a = 5 h = (-3) k = 15 The vertex of this equation is (-3,15) The two equation of: and are equivalent. To prove it to yourself, simply choose any value for x and plug it in both equations and you will see that you get the same answer. I chose x = 7 and got 515 for the first equation and 515 for the second equation. Graph of this equation is shown below. I put both forms of the equation in the graph generator so that if there was a difference, you would see it. If you only see one equation being graphed, then the two forms are equivalent. This is another way of proving they are equivalent which is analagous to taking two objects and placing them on top of each other to see if they are equivalent. CONVERTING FROM THE VERTEX FORM OF A PARABOLA TO THE STANDARD FORM OF A PARABOLA WHEN THE PARABOLA OPENS UP OR DOWN To convert from the vertex form of the parabola to the standard form of the parabola, you simply reverse the process. Our equation in vertex form is: We square the (x+3) term and multiply it by 5 to get: which becomes: which becomes: which is the same equation in standard form. CONVERTING FROM THE STANDARD FORM OF A PARABOLA TO THE VERTEX FORM OF A PARABOLA WHEN THE PARABOLA OPENS LEFT OR RIGHT Standard form of the equation for a parabola opening left or right is x = ay^2 + by + c Vertex form of the equation for a parabola opening up or down is x = a(y-k)^2 + h To show you how we convert from the standard form to the vertex form, we'll use an example. Your equation in standard form is: x = 5y^2 + 30y + 60 where: a = 5 b = 30 c = 60 While this could be simplified further by dividing out by the common factor of 5, we'll leave it the way it is for demonstration purposes because we want to keep the "a" term greater than 1 and we do not want to have to deal with fractions in the demonstration. We factor out the "a" term of 5 to get: We complete the square on the (y^2 + 6y) term to get: which becomes: which becomes: We remove the parentheses to get: which becomes: This is now in vector form of where: a = 5 k = (-3) h = 15 The vertex of this equation is (15,-3) The two equations of: and are equivalent. To prove it to yourself, simply choose any value for y and plug it in both equations and you will see that you get the same answer. I chose y = 7 and got 515 for the first equation and 515 for the second equation. In order to graph this equation, we have to convert it from x = f(y) to y = f(x) by solving for y. The equation of: becomes: y = +/- Graph of this equation is shown below. CONVERTING FROM THE VERTEX FORM OF A PARABOLA TO THE STANDARD FORM OF A PARABOLA WHEN THE PARABOLA OPENS LEFT OR RIGHT To convert from the vertex form of the parabola to the standard form of the parabola, you simply reverse the process. Our equation in vertex form is: We square the (y+3) term and multiply it by 5 to get: which becomes: which becomes: which is the same equation in standard form. FOCUS OF A PARABOLA The focus of a parabola is a point on the axis of symmetry of the parabola that is a set distance from the vertex of the parabola. The point is said to be inside the parabola. If the parabola opens up, the point is above the vertex. If the parabola opens down, the point is below the vertex. If the parabola opens to the right, then the point is to the right of the vertex. If the parabola opens to the left, then the point is to the left of the vertex. DIRECTRIX OF A PARABOLA The directrix of a parabola is a line (not a point) that is perpendicular to the axis of symmetry of the parabola and does not intersect with the parabola. That line is said to be outside of the parabola. If the parabola opens up, the directrix is below the vertex. If the parabola opens down, the directrix is above the vertex. If the parabola opens to the right, the directrix is to the left of the vertex. If the parabola opens to the left, the directrix is to the right of the vertex. RELATIONSHIP BETWEEN THE FOCUS OF A PARABOLA AND THE DIRECTRIX OF A PARABOLA The focus is a point inside the parabola on the axis of symmetry of the parabola. The directrix is a line outside the parabola that is perpendicular to the axis of symmetry of the parabola. The distance between the Focus and the Vertex of the Parabola is the same as the distance between the Vertex and the Directrix at the point where the Directrix intersects with the Axis of Symmetry. This is a defining characteristic of the Focus and Directrix of a Parabola. They are in a fixed relationship with each other. Let F = the Focus point on the axis of symmetry of the parabola. Let V = the Vector point on the axis of symmetry of the parabola. Let D = the point of intersection of the Directrix with the axis of symmetry of the parabola. What we have is d = F-V = V-D. You can't just pick any d. There is only one d for each parabola that works. This is because of the special relationship between the Focus and the Directrix and any other point on the parabola that will be discussed shortly. In other words, there is a formula that determines what the distance between the Focus and the Vector, and the Vector and the Directrix, should be. That formula is The "a" in the formula is the same "a" that you have been dealing with all along. In the standard formula for a parabola of y = ax^2 + bx + c, that "a" is the coefficient of the x^2 term. In the vertex formula for a parabola of y = a(x-h)^2 + k, that "a" is the coefficient of the (x-h)^2 term. If we know "a", then we can find "d" because of this fixed relationship between them given by the formula . EXAMPLE OF FINDING THE FOCUS AND THE POINT OF INTERSECTION OF THE DIRECTRIX WITH THE AXIS OF SYMMETRY OF A PARABOLA THAT IS OPENING UP USING THE STANDARD FORM OF THE EQUATION FOR THE PARABOLA The formula we will be working with is in standard form of y = ax^2 + bx + c. Our formula is: y = 3x^2 + 24x + 43 where: a = 3 b = 24 c = 43 The formula for the distance between the focus and the vector is: d = 1/(4a) Since a equals 3, this means that d = 1/(34) = 1/12. The x value of the vector for this equation is -b/(2a) = -24/(23) = -24/6 = -4. The y value of the vector for this equation is found by replacing x with 3 in the equation of y =: 3x^2 + 24x + 43 to get: 3(-4)^2 + 24(-4) + 43 = 48 - 96 + 43 = -5 The vertex for this parabola is: V = (-4,-5) Since this parabola opens up ("a" term is positive), the focus is above the vertex and the directrix is below the vertex. Our parabola opens up, so the focus is above the vector. It has the same x value as the vector. It's y value is the y value of the vector plus d. In this case, the y value of the vector is -5 and d = 1/12. Add 1/12 to -5 and you get -5 + 1/12 which is equivalent to -60/12 + 1/12 which is equal to -59/12. The point of intersection of the directrix with the axis of symmetry of our parabola is below the vector and the same distance from it as the focus. It has the same x value as the vector. It's y value is the y value of the vector minus d. In this case, the y value of the vector is -5. Subtract 1/12 from -5 and you get -5 - 1/12 which is equivalent to -60/12 - 1/12 which is equal to -61/12. Our Focus is at (-4,-59/12) Our Vector is at (-4,-60/12) Our Directrix Point of Intersection with the Axis of Symmetry is at (-4,-61/12) EXAMPLE OF FINDING THE FOCUS AND THE POINT OF INTERSECTION OF THE DIRECTRIX WITH THE AXIS OF SYMMETRY OF A PARABOLA THAT IS OPENING UP USING THE VECTOR FORM OF THE EQUATION FOR THE PARABOLA The standard form of our equation is y = ax^2 = bx + c which equals: y = 3x^2 + 24x + 43 The vector form of our equation is y = a(x-h)^2 + k which equals: y = 3(x+4)^2 - 5 Our vector is (-4,-5) d = 1/(4a) = (1/12) Once we have the vector and the distance between the focus and the vector, all other calculations are the same. Since d is a distance, it will always be positive. If the parabola opens up or to the right, then F = V + d If the parabola opens down or to the left, then F = V - d If the parabola opens up or down, then the y value of F and V are changing, and the x value remains the same (on the Axis of Symmetry). If the parabola opens right or left, then the x value of F and V are changing, and the y value remains the same (on the Axis of symmetry). RELATIONSHIP BETWEEN ANY POINT ON A PARABOLA AND THE DISTANCE BETWEEN THAT POINT AND THE FOCUS OF THAT PARABOLA, AND THE DISTANCE BETWEEN THAT POINT AND THE DIRECTRIX OF THAT PARABOLA The distance between the focus and the Vertex is equal to the distance between the Vertex and the point of intersection between the Directrix and the Axis of Symmetry. In the picture of the graph below, d1a is the distance between the Focus and the Vector, and d1b is the distance between the Vector and the intersection between the Directrix and the Axis of Symmetry. The distance between any point on the Graph of the Parabola and the Focus is equal to the distance between that point and the Directrix. The distance between that point and the focus is a straight line between those 2 points. the distance between that point and the directrix is a vertical straight line between that point and the directrix where the intersection of that line with the directrix forms a right angle. In other words, that vertical straight line is prependicular to the directrix at the intersection point. In the picture of the graph below, d2a is the distance between any point on the graph of the Parabola, and d2b is the distance between that point on the Parabola and the intersecting point on the Directrix from a perpendicular line dropped from that point. The following details relating to the graph below are displayed below. They are also shown in the picture of the graph below. Standard Form of Equation of Parabola: y = .125x^2 + .125x - .375 Vertex Form of Equation of Parabola: y = .125(x+.5)^2 � .40625 Vertex of Parabola: (x,y) = (-.5,-.40625) Coefficient of x^2 term and (x+.5)^2 term: a = .125 Distance from Focus to Vertex: d1a = 1/(4a) = (1/.5) = 2 Distance from Vertex to Directrix: d1b = 1/(4a) = (1/.5) = 2 Axis of Symmetry: x = -.5 Focus of Parabola: (x,y) = (-.5, 1.59375) Directrix of Parabola: y = -2.40625 Point on Parabola: (x,y) = (3, 1.125) Distance from Point to Focus: d2a = sqrt((3.5)^2 + (-.46875)^2) = sqrt(12.46972656) Distance from Point to Directrix: d2b = sqrt(0^2 + (3.53125)^2) = sqrt(12.46972656) This is the graph of this parabola. To see the picture of this graph with comments and descriptive information, click on the following hyperlink. Focus and Directrix of a Parabola You are encouraged to visit the references to view the many examples and problems they have up there. This lesson has been accessed 128877 times. \| \|
17017	https://cc.ee.ntu.edu.tw/~jhjiang/instruction/courses/fall13-ld/unit05.pdf	1 Switching Circuits & Logic Design Jie-Hong Roland Jiang 江介宏 Department of Electrical Engineering National Taiwan University Fall 2013 2 §5 Karnaugh Maps K-map Walks and Gray Codes 3 Outline Minimum forms of switching functions Two- and three-variable Karnaugh maps Four-variable Karnaugh maps Determination of minimum expressions using essential prime implicants Five-variable Karnaugh maps Other uses of Karnaugh maps Other forms of Karnaugh maps 4 Limitations of Algebraic Simplification  Two problems of algebraic simplification 1. Not systematic 2. Difficult to check if a minimum solution is achieved  The Karnaugh map method overcomes these limitations  Typically for Boolean functions with 5 variables  The Quine-McCluskey method can deal with even larger functions (Subject of Unit 6, skipped) 5 Minimum Forms of Switching Functions Correspondence between Boolean expressions and logic circuits SOP (POS) can be implemented with two-level AND-OR (OR-AND) gate circuits Reducing the number of terms and literals of an SOP expression corresponds to reducing the number of gates and gate inputs Combine terms by XY'+XY=X Eliminate redundant terms by consensus theorem Minimum SOP is not necessarily unique An SOP may be minimal (locally) but not minimum (globally) E.g., F = a'b'c'+a'b'c+a'bc'+ab'c+abc'+abc = a'b'+b'c+bc'+ab (minimal but not minimum) = a'b'+bc'+ac (minimum) 6 Two-Variable Karnaugh Maps 2-variable K-map 1 0 1 0 F(m3) F(m1) F(m2) F(m0) A B A=0, B=0 A=0, B=1 A=1, B=0 A=1, B=1 1 1 0 0 00 01 10 11 F AB 1 0 1 0 0 1 0 1 A B F = A'B'+ A'B = A'(B'+B) = A' A'B' A'B A' minterm locations 7 Three-Variable Karnaugh Maps 3-variable K-map 5 1 01 7 3 11 10 00 1 0 6 2 4 0 A BC 0 0 01 0 1 11 10 00 1 0 1 1 1 0 A BC F = A'BC'+ A'BC+ AB'C'+ ABC' = A'B+ AC'+ BC' = A'B+ AC' 0 0 1 1 1 0 1 0 000 001 010 011 100 101 110 111 F ABC minterm locations 8 Three-Variable Karnaugh Maps 3-variable K-map (zeros omitted) 01 1 1 11 10 00 1 0 1 1 A BC 01 11 10 00 1 0 1 1 A BC 01 11 10 00 1 0 1 1 A BC B BC' AC' 9 Three-Variable Karnaugh Maps 3-variable K-map 1 1 01 1 11 10 00 1 0 1 1 1 a bc f(a,b,c) = abc'+b'c+a' 10 Three-Variable Karnaugh Maps 3-variable K-map 1 1 01 0 1 11 10 00 1 0 0 0 0 0 a bc F = m1+m3+m5 = M0M2M4M6M7 1 1 01 0 1 11 10 00 1 0 0 0 0 0 a bc simplify F = a'c+b'c 11 Three-Variable Karnaugh Maps 3-variable K-map 0 0 01 1 0 11 10 00 1 0 1 1 1 1 a bc G = (m1+m3+m5)' = (M0M2M4M6M7)' 0 0 01 1 0 11 10 00 1 0 1 1 1 1 a bc simplify G = ab+c' 12 Three-Variable Karnaugh Maps 3-variable K-map 1 01 1 1 11 10 00 1 0 1 x yz 1 01 1 1 11 10 00 1 0 1 x yz simplify xy+x'z+yz = xy+x'z (consensus theorem) 13 Three-Variable Karnaugh Maps 3-variable K-map 1 1 01 1 11 10 00 1 0 1 1 1 a bc 1 1 01 1 11 10 00 1 0 1 1 1 a bc F = a'b'+bc'+ac = a'c'+b'c+ab 14 Four-Variable Karnaugh Maps 4-variable K-map 6 7 5 4 01 14 15 13 12 11 9 1 01 11 3 11 10 00 10 00 10 2 8 0 AB CD 1 1 1 1 01 1 1 1 11 01 1 11 10 00 10 00 1 1 1 1 ab cd F = acd+a'b+d' minterm locations 15 Four-Variable Karnaugh Maps 4-variable K-map 1 1 01 1 1 11 1 01 1 11 10 00 10 00 1 ab cd 1 1 01 1 1 11 1 01 1 11 10 00 10 00 1 ab cd f1 = ab'cd'+a'b'd+bc' f1 = m(1,3,4,5,10,12,13) simplify 16 Four-Variable Karnaugh Maps 4-variable K-map 1 1 1 01 1 1 11 01 1 1 11 10 00 10 00 1 1 1 1 ab cd 1 1 1 01 1 1 11 01 1 1 11 10 00 10 00 1 1 1 1 ab cd f2 = c+b'd'+a'bd f2 = m(0,2,3,5,6,7,8,10,11,14,15) simplify 17 Four-Variable Karnaugh Maps Simplify incompletely specified function All the 1’s must be covered, but X’s are optional and are set to 1’s only if they will simplify the expression X 1 1 01 X X 11 1 1 01 1 11 10 00 10 00 ab cd X 1 1 01 X X 11 1 1 01 1 11 10 00 10 00 ab cd f = a'd+c'd f = m(1,3,5,7,9)+ d(6,12,13) simplify 18 Four-Variable Karnaugh Maps Simplify product-of-sums Circle 0’s instead of 1’s Apply De Morgan’s law converting SOP to POS 0 0 0 1 01 0 1 0 0 11 0 0 01 1 1 11 10 00 10 00 1 1 1 1 wx yz 0 0 0 1 01 0 1 0 0 11 0 0 01 1 1 11 10 00 10 00 1 1 1 1 wx yz f = x'z'+wyz+w'y'z'+x'y simplify f' = y'z + wxz'+w'xy f = (y+z')(w'+x'+z)(w+x'+y') 19 Determination of Minimum Expressions Using Essential Prime Implicants Implicant A product term of a function Any single 1 or any group of 1’s on a K-map combined together forms a product term Prime implicant A maximal implicant An implicant that cannot be combined with another term to eliminate a variable All of the prime implicants of a function can be obtained from a K-map by expanding the 1’s as much as possible in every possible way 20 Determination of Minimum Expressions Using Essential Prime Implicants Example 1 01 1 1 11 1 01 1 11 10 00 10 00 1 1 1 ab cd a'cd' a'b'c a'b'c'd' ac' ab'c' abc' a'b'c, a'cd', ac' are prime implicants a'b'c'd', abc', ab'c' are implicants (but not prime implicants) 21 Determination of Minimum Expressions Using Essential Prime Implicants Determine all prime implicants In finding prime implicants, don’t cares are treated as 1’s. However, a prime implicant composed entirely of don’t cares can never be part of the minimum solution Not all prime implicants are needed in forming the minimum SOP Example All prime implicants: a'b'd, bc', ac, a'c'd, ab, b'cd (composed entirely of don’t cares) Minimum solution: F = a'b'd+bc'+ac 1 1 01 1 1 1 1 11 1 01 X X 11 10 00 10 00 1 ab cd 22 Determination of Minimum Expressions Using Essential Prime Implicants Essential prime implicant (EPI) A prime implicant that covers some minterm not covered by any other prime implicant If a single term covers some minterm and all of its adjacent 1’s and X’s, then the term is an EPI Must be present in the minimum SOP AB CD f = CD+BD+B'C+AC 1 1 01 1 1 1 11 01 1 1 11 10 00 10 00 1 1 f = BD+B'C+AC EPIs already cover all 1’s Minimum SOP! AB CD 1 1 01 1 1 1 11 01 1 1 11 10 00 10 00 1 1 23 Determination of Minimum Expressions Using Essential Prime Implicants SOP minimization 1. Select all essential prime implicants 2. Find a minimum set of prime implicants which cover the minterms not covered by the essential prime implicants There may be freedom left after all essential prime implicants are selected (it affects optimality especially for functions with more variables) AB CD A'C' 1 1 1 01 1 11 1 01 1 11 10 00 10 00 1 1 A'B'D' ACD A'BD F = A'C'+ A'B'D'+ACD+ or BCD 24 Determination of Minimum Expressions Using Essential Prime Implicants Flowchart for determining a minimum SOP using K-map Choose a 1 which has not been covered Find all adjacent 1’s and X’s Are the chosen 1 and its adjacent 1’s and X’s covered by a single term? That term is an essential prime implicant. Loop it All uncovered 1’s checked? STOP Find a minimum set of prime implicants which cover the remaining 1’s on the map YES NO YES NO Note: All essential prime implicants have been determined at this point 25 Determination of Minimum Expressions Using Essential Prime Implicants Example AB CD 16 X7 15 14 01 X15 113 11 19 01 11 10 00 10 00 110 18 X0 Step 1: 14 checked Step 2: 15 checked Step 3: 16 checked EPI A'B selected Step 4: 18 checked Step 5: 19 checked Step 6: 110 checked EPI AB'D' selected Step 7: 113 checked (up to this point all EPIs determined) Step 8: AC'D selected to cover remaining 1’s 26 Five-Variable Karnaugh Maps 5-var K-map 01 11 01 11 10 00 10 00 BC DE A 1/0 minterm locations 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 16 17 19 18 20 21 23 22 28 29 31 30 24 25 27 26 01 11 01 11 10 00 10 00 BC DE A 1/0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 These two terms do not combine AB'DE' CDE BD' 27 Five-Variable Karnaugh Maps 5-var K-map 01 11 01 11 10 00 10 00 BC DE A 1/0 1 1 1 1 1 1 Expand to eliminate B Expand to eliminate D Expand to eliminate E Expand to eliminate A Expand to eliminate C 28 Five-Variable Karnaugh Maps 5-var K-map 01 11 01 11 10 00 10 00 BC DE A 1/0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 AB'C F = A'B'D'+ABE'+ACD+A'BCE+ or P1 P2 P3 P4 B'CD' P4 P3 P2 P1 Shaded 1’s show prime implicants 29 Five-Variable Karnaugh Maps 5-var K-map 01 11 01 11 10 00 10 00 BC DE A 1/0 1 1 1 1 1 1 1 1 1 1 1 1 1 AC'E F = B'C'D'+B'C'E+A'C'D'+A'BCD+ABDE+ or P1 P2 P3 P4 P5 C'D'E P1 P3 P5 P4 P2 30 Other Uses of Karnaugh Maps Use K-map to prove the equivalence of two Boolean expressions K-maps are canonical representations of Boolean functions, similar to truth tables Use K-map to perform Boolean operations AND, OR, NOT operations can be done over K-maps (truth tables) 31 Other Uses of Karnaugh Maps Use K-map to facilitate factoring Identify common literals among product terms 01 1 1 11 1 01 1 11 10 00 10 00 1 1 AB CD F = A'B'C'+A'B'D+ACB+ACD' = A'B'(C'+D)+AC(B+D') 32 Use K-map to guide simplification Other Uses of Karnaugh Maps F = ABCD+B'CDE+A'B'+BCE' = ABCD+B'CDE+A'B'+BCE'+ACDE = A'B'+BCE'+ACDE 01 11 01 11 10 00 10 00 BC DE A 1/0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Add this term Then these two terms can be removed 33 Other Forms of Karnaugh Maps Other conventions (Veitch diagrams) A B C C D A B 34 Other Forms of Karnaugh Maps Other conventions (5-var K-map) 1 1 01 1 1 1 11 1 01 11 10 00 10 00 1 1 BC DE 1 01 1 1 1 11 1 01 11 10 00 10 00 1 1 1 BC DE A=0 A=1 F = D'E'+B'C'D'+BCE+A'BC'E'+ACDE 35 Other Forms of Karnaugh Maps Other conventions (5-var Veitch diagram) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 C C D E A B F = D'E'+B'C'D'+BCE+A'BC'E'+ACDE
17018	https://www.expii.com/t/quadratic-equations-definition-examples-4524	Expii Quadratic Equations - Definition & Examples - Expii Any quadratic equation can be put into standard form, ax²+bx+c=0, where a, b, and c are constants. Solutions for the unknown x are called zeros or roots. Explanations (3) Catherine Yeo Text 14 Quadratics A quadratic equation is an equation of the following form: ax2+bx+c=0 where x represents an unknown variable, a, b, and c are constants, and a≠0. The left side has all of the fancy numbers and variables, while the right side is 0. Image source: By Caroline Kulczycky Because the term ax2 is raised to the second degree, it is called the quadratic term. The bx term is the linear term because it has a degree of 1. The c term is the constant term because it has a degree of 0. Quadratic equations can be rearranged to be equal to 0. Zeros or roots of a quadratic are known as the solutions to quadratic equations. a, b, and c don't have to be integers, so you can have a decimal as a constant! Therefore, it may initially seem like you need a perfect square, but you actually don't. Quadratics can be solved by factorization, completing the square, or using the quadratic formula. To figure out how many solutions a quadratic has, you use the discriminant! You can also find complex solutions with double roots, but that's a little advanced for this lesson! Now, let's look at some graphs of quadratic functions. Quadratics make a parabola on the graph, which makes a U shape. Graphed below in red is y=x2−2. We can see that the graph follows the pattern of a traditional quadratic equation, but its minimum point is not on the origin, but 2 units below the origin. Made using Desmos Graphed below in red is y=(x−2)2−2. Here, we can see that the graph has both been shifted down and also shifted to the right, compared to the original quadratic equation graph. Made using Desmos Graphed below in purple is y=(x+5)2−1. Just like the first two graphs, this purple graph has also been shifted from the original quadratic function graph! Here, the graph is shifted 5 units to the lift and 1 unit down. Made using Desmos Graphed below in black is y=−(x−5)2+3. Don't get frightened by this graph! The graph has been flipped because of the negative sign. The graph has also been shifted 5 units to the right and 3 units up. Made using Desmos Finally, graphed below in green is y=x2. This is the most basic quadratic function. The minimum point is always on the origin, (0,0). Made using Desmos Report Share 14 Like Related Lessons Write Systems of Linear Inequalities from a Graph Square Roots of Quadratic Equations - Definition & Examples Systems of Linear Inequalities, Word Problems - Examples Quadratic Functions - Definition & Examples View All Related Lessons Andrew Wang Text 4 Quadratic Equation A quadratic expression always includes 3 terms: the quadratic term, the linear term, and the constant term. The standard form of a quadratic equation is ax2+bx+c=0, with a quadratic expression, ax2+bx+c, set equal to 0. All quadratic equations can be rearranged into standard form. Quadratic Term The quadratic term should always be the first term of a quadratic expression in standard form: the term with the x2 in it. So, in ax2+bx+c, ax2 is the quadratic term. Report Share 4 Like Amy Feraco Video 1 (Video) Basic Introduction to Quadratic Equations by Math Booster This video by Math Booster covers quadratic equations. Summary Quadratic equations take the following forms: ax2+bx+c=yORax2+bx+c=0 where a,b, and c are constants, and a≠0. We use the y form for graphing and the 0 form for solving. The second form is called standard form. Remember, the standard form for quadratic equations is different from the standard form for linear equations. Made using Desmos There are a few important terms that help us describe the quadratic graph. The points (or point) where the curve crosses the x-axis can be called the solutions, x-intercepts, roots, or zeros. The point where the curve crosses the y-axis is called the y-intercept. The minimum is the lowest point on the graph. This will be at the bottom of the curve on a parabola that opens upwards (a≥1). Downwards facing parabolas do not have a minimum. The maximum is the lowest point on the graph. This will be at the top of the curve on a parabola that opens downwards (a≤−1). Upwards facing parabolas do not have a maximum. The vertex is the point that is the minimum point if the graph opens up. It is the maximum point if the graph opens downwards. The axis of symmetry is he vertical line that passes through the vertex. It splits the parabola in two equal parts. x=−b2a. There are many ways to find the solutions to the quadratic equation. We can use factoring, completing the square, graphing, and the quadratic formula. We can also find solutions to quadratic linear systems (which mean multiple equations with at least one linear equation and at least one quadratic equation) using the substitution, elimination, and graphing methods. You will learn all of these in depth in later sections. Quadratic equations are important! Report Share 1 Like You've reached the end How can we improve? General Bug Feature Send Feedback
17019	https://nesma.org/wp-content/uploads/2020/10/FPA-for-Software-Enhancement-v2.3-EN.pdf	Nesm.org FUNCTION POINT ANALYSIS FOR SOFTWARE ENHANCEMENT Guidelines Version 2.3 Professional guide of the Nesma Nesm.org © Copyright Nesma 2019 All rights reserved. The Netherlands Software Metrics Users Association (Nesma), formerly NEFPUG. Nothing in this publication may be reproduced or published in any form or by any method without the prior written permission of Nesma. As opposed to other Nesma handbooks and user’s guides, this stipulation also applies to Nesma members. After permission has been granted the title page of the document containing the reproduced or published material must include the following statement: "This publication contains material from the professional guide ‘Function Point Analysis for Software Enhancement’ published by Nesma. This publication appears with permission of Nesma". Contents FPA FOR SOFTWARE ENHANCEMENT 2.3 iii CONTENTS 1 Introduction ..................................................................................................... 5 1.1 Purpose of this Document ............................................................................... 5 1.2 Objectives of the Guidelines ............................................................................ 5 1.3 Starting Point ...................................................................................................... 5 1.4 Intended Audience ............................................................................................ 6 1.5 Scope of the professional guide ...................................................................... 6 1.6 Remarks .............................................................................................................. 6 1.7 Future versions .................................................................................................. 7 1.8 Organisation of the Guide ................................................................................ 7 2 General considerations ................................................................................... 8 2.1 Limiting Conditions............................................................................................ 8 2.2 Prerequisites ................................................................................................... 10 2.3 Summary of the Enhancement FPA Methodology..................................... 12 3 Methodology .................................................................................................. 14 3.1 Identify the functions within the scope of the enhancement project .... 14 3.2 Determine which transactional and data functions will be added ......... 14 3.3 Determine which transactional and data functions will be deleted ....... 15 3.4 Determine which data functions will be modified and determine the impact factor ............................................................................................ 15 3.5 Determine which transactional functions will be modified and determine the impact factor ......................................................................... 17 3.6 Calculate the size of the enhancement project .......................................... 19 3.7 Calculate the size of the system after enhancement ................................ 20 3.8 High level Function Point Analysis for Software Enhancements ............. 21 4 Testing in enhancement projects ................................................................ 22 5 Budgeting enhancement .............................................................................. 23 Contents FPA FOR SOFTWARE ENHANCEMENT 2.3 iv 6 Examples ........................................................................................................ 24 6.1 Example 1 - Expanding an ILF ....................................................................... 24 6.2 Example 2 – ELF becomes ILF ....................................................................... 24 6.3 Example 3 – ELF becomes ILF with modifications ...................................... 25 6.4 Example 4 – Splitting an ELF .......................................................................... 26 6.5 Example 5 – Adding an ELF............................................................................ 26 6.6 Example 6 – Modifying an ILF, 1 ................................................................... 26 6.7 Example 7 – Changing the name of a DET .................................................. 27 6.8 Example 8 – Modifying a heading in a report ............................................. 27 6.9 Example 9 – Adding / modifying / deleting DET’s ....................................... 27 6.10 Example 10 – Determining the % of modified DET’s and FTR’s ............... 27 7 Glossary .......................................................................................................... 29 8 Bibliography ................................................................................................... 32 FOREWORD FPA FOR SOFTWARE ENHANCEMENT 2.3 i FOREWORD TO VERSION 1 You are reading the guideline “FPA for Software Enhancement”, published by the Netherlands Software Metrics users Association (Nesma), and written by members of the working group “FPA for Enhancement and Maintenance”. To safeguard and control the development and operational status of information systems is a difficult task. This task becomes even more difficult as these systems become more complex. The reasons for this are closely aligned with measurements of (in terms of usable functionality):  the functionality that is offered to the user by the system;  the effort and materials needed to deliver this functionality to the user;  the effort and materials needed to keep the system operational in the enhancement and maintenance phase. Function point analysis (FPA) can be used to:  describe the scope of a system and measure its functional size independently of the technologies used by the system;  derive productivity and process performance metrics, estimate resource requirements and assist in project management;  evaluate the factors in a development environment that influence productivity and so provide a basis for improving the development and enhancement and maintenance processes; and  determine the scope and size of system enhancement and assist in managing system changes. The Nesma (Netherlands Software Metrics Users Association) first known as the NEFPUG (Netherlands Function Point Users Group) was founded in May 1989. Its main goals are to:  bring together individuals and organisations to exchange knowledge and experience in the development and application of software metrics;  promote accountability in the use of software metrics;  support the formulation and adoption of standards in software metrics;  encourage the development and application of software metrics. Nesma is working towards these goals through:  the activities of its study groups and working groups;  research undertaken by its members;  organising presentations, training, symposia, and the like;  making recommendations on the use of software metrics;  collecting and publishing literature on software metrics;  collaborating with organisations with similar interests; FOREWORD FPA FOR SOFTWARE ENHANCEMENT 2.3 ii  liaising and collaborating with the other software metrics user groups and associations worldwide, including IFPUG (USA), ASMA (Australia) and FESMA (Europe). In 1991 Nesma formed the working group ‘FPA for Enhancement and Maintenance’ to develop and publish guidelines for the application of function point analysis to software enhancement and maintenance. The working group consisted of five members:  J.T. Engelhart,  P.L. Langbroek,  A.J.E. Dekkers,  H.J.G. Peters,  P.H.J. Reijnders. The following people have also contributed to these guidelines:  F. X. Granneman,  P. J. M. Hickendorff,  J. W. ter Veld. Zeist, Netherlands, June 1998 FOREWORD TO VERSION 1 OF THE ENGLISH TRANSLATION There is broad international interest in the application of function point analysis to software enhancement although, to date, there have been few publications on the subject. People from all over the world have requested Nesma to translate this guide "Function Point Analysis for Software Enhancement" into English. This English version is an accurate translation of the Dutch guide and uses the terminology defined in the Nesma and IFPUG Counting Practices Manuals. The translation process has literally been a worldwide project. The Nesma Board thanks all people who participated in the translation of these guidelines. Firstly, Nesma thanks Oliver Hague (past President of ASMA, Australia), an expert in FPA and software metrics. Oliver wrote the first English language draft from a verbatim translation of the Dutch text and edited the revised text to produce this published version. Nesma also thanks Adri Timp (Netherlands, chair of the Nesma Counting Practices Committee and vice chair of the IFPUG Counting Practices Committee) and David Garmus (USA, IFPUG President 2000-2001). Adri Timp carried out a meticulous comparison of the translated English text and the original Dutch text, adding many improvements and clarifications. David Garmus reviewed the translation and provided valuable feedback. Finally, Oliver Hague and Adri Timp formatted the translated guide and made final adjustments to the text. FOREWORD FPA FOR SOFTWARE ENHANCEMENT 2.3 iii By offering this guide to the international functional software measurement community, Nesma expects to stimulate further analysis of the measurement of software enhancement. Nesma hopes, that these guidelines will be applied and tried out worldwide. (Please note the disclaimer with respect to this method in section 1.6). Nesma would like to hear about your experiences with this method and welcomes your suggestions. The practical experiences of users may lead to the publication of revised versions of this guide. Please forward your suggestions and comments to Nesma at office@nesma.org. Netherlands, August 2001 The Nesma Board FOREWORD TO VERSION 2.2.1 At the beginning of 2008 work was started to update the documentation to the new situation that was a result of the migration of the Nesma counting practises from version 2.0 to version 2.2. This migration was necessary because of the ISO certification of the Nesma FPA method (ISO 24570). The changes made to this guideline were carried out by members of the FPA core group at Getronics PinkRoccade:  Ton de Groot,  Rini Scholten, and  Theo Thijssen. In July of 2008 the updated document was extensively reviewed by the following members of the Nesma executive board:  Jabob Brunekreef,  Jolijn Onvlee,  Adri Timp,  Wim Visser, and  Hans Vonk. Editor-in-Chief was:  Ton Dekkers. This version has been translated by Robert Louwers (ABN/AMRO) and was reviewed by Ton Dekkers (Galorath International Limited) and Adri Timp (Equens SE). We especially call your attention to the disclaimer in section 1.6. Netherlands, September 2009 FOREWORD FPA FOR SOFTWARE ENHANCEMENT 2.3 iv FOREWORD TO VERSION 2.3 The 2019 update is initiated after the ISO certification of the Nesma FPA method (ISO 24570) as Nesma counting practises version 2.3. We are also glad we received two investigations done by parties using this guideline, therefore we could rephrase the disclaimer that was mentioned in section 1.6 of previous version to remarks. The changes made to this guideline were carried out by:  Alexander Vermeulen;  Ton Dekkers (Editor-in-Chief). The updated document was extensively reviewed by the following members of Nesma:  Wim Visser,  Hans Bernink,  Jolijn Onvlee,  Jacques van der Knaap,  Martin Jacobs. Netherlands, April 2019 Introduction FPA FOR SOFTWARE ENHANCEMENT 2.3 5 1 INTRODUCTION 1.1 Purpose of this Document The IFPUG "Function Point Counting Practices Manual" [IFPUG, 1] and the Nesma FPA counting practices manual “Definitions and Counting Guidelines for the Application of Function Point Analysis” [Nesma, 1] both define how FPA should be used to determine the functional size of an information system. Both guidelines follow the “Albrecht” method in maintenance and enhancement situations because a more detailed instruction is covered by the guideline. Now that the methodology of analysing new software projects has been researched and described, it is interesting to know whether FPA can be applied to maintenance and enhancement situations, and if so, in what way and within what constraints. Consideration of these issues led Nesma to form the working group on “FPA for Enhancement and Maintenance”. The first results of this working group were published in August 1993 and led subsequently to publication of the method described in these guidelines. Several organisations started using the method in actual practice. Their experiences resulted in further refinement of the guidelines and the inclusion of examples to illustrate the application of the method. In 2019 an updated version of this guideline was published in order to conform to Nesma 2.3. 1.2 Objectives of the Guidelines These guidelines are intended for anyone with an interest in the management of enhancements to information systems. The goal is to provide an objective and replicable method for assessing the scope and size of enhancements. The method is objective in that the results obtained are independent of the person applying the method: two different people using the same guideline will obtain the same result. The method is replicable in that a particular outcome can be determined a priori, and the same outcome can be produced on the second and subsequent applications of the method. 1.3 Starting Point These guidelines build on the FPA methodology described in the Nesma publication “Definitions and Counting Guidelines for the Application of Function Point Analysis” [Nesma, 1], and the IFPUG "Function Point Counting Practices Manual" [IFPUG, 1]. Introduction FPA FOR SOFTWARE ENHANCEMENT 2.3 6 1.4 Intended Audience These guidelines are intended for anyone performing Function Point Analysis and wanting to measure the size of enhancement projects more precisely. It is assumed that the reader is familiar with the standard ISO certified FPA methods as documented in [Nesma, 1] and [IFPUG, 1], (see section 1.3) and can be used in combination with both methods. 1.5 Scope of the professional guide Nesma considers the application of FPA to software enhancement from the perspective of the standard function point analysis method. The result of this work, embodied in these guidelines, is a method applicable to software enhancement and testing that is strongly related to the standard FPA method. The term Enhancement Function Point Analysis (EFPA) is used to differentiate the method from the standard function point analysis method. 1.6 Remarks Nesma is the custodian of Function Point Analysis according to the “Definitions and Counting Guidelines for the Application of Function Point Analysis” [Nesma, 1]. That version (in English) is certified by ISO under number 24570. The IFPUG "Function Point Counting Practices Manual" [IFPUG, 1] is certified by ISO under number 20926. In both these versions no mention is made of the method as set forth in this guideline. As such, this method is therefore neither a part of the Nesma ISO certified method nor part of the IFPUG ISO certified method. The method described in this guideline is an alternative one and has been found in practice to provide an indication of the size of enhancements. The large interest in this document illustrates its need in the enhancement domain. Over the years various examples of its application have been published. In the bibliography (Chapter 8) some references to documents are added. By offering this guide to the international functional software measurement community, Nesma wants to advance the application of function point analysis to enhancement projects and to broaden the understanding of measurement applied to software enhancement. Comments and suggestions for further improvement of this method may be sent to office@nesma.nl. This will help Nesma to improve and refine this method to measure the size of enhancement projects. Introduction FPA FOR SOFTWARE ENHANCEMENT 2.3 7 1.7 Future versions When changes to this guide prove necessary in the future, a new version will be released. The latest version may be downloaded from the Nesma web site 1.8 Organisation of the Guide After this introduction, Chapter 2 discusses conditions limiting the application of the enhancement FPA methodology (EFPA). Chapter 3 describes the enhancement FPA methodology. Chapter 4 discusses EFPA in relation to testing. Chapter 5 describes the use of the enhancement size measure for budgeting purposes. Application of the enhancement FPA methodology is illustrated by examples in chapter 6. Chapter 7 is a glossary of the more important concepts and abbreviations and Chapter 8 contains a concise Bibliography. GENERAL CONSIDERATIONS FPA FOR SOFTWARE ENHANCEMENT 2.3 8 2 GENERAL CONSIDERATIONS 2.1 Limiting Conditions In this paragraph we start by looking more closely at the terms which define the scope of the guideline. 2.1.1 Maintenance Maintenance includes all those activities necessary to sustain the operation of a system without modifying its functionality. Maintenance is carried out in the environment in which the system operates but is also applied to the technical infrastructure that can vary widely among organisations. In this guideline we have refrained from offering a budgeting method or model for maintenance. A separate publication "Budgeting the Operational Costs of Information Systems" [Nesma, 2] is available for this. 2.1.2 Conversion Conversion activities are not included in this guideline because they can be manifested in so many different forms. Conversion can be, for example, translating source code to a new or updated language, transferring a system to a totally different operating environment or changing the storage of physical data to accommodate the introduction of a new database management system. It is also often not evident which forms of conversion should be considered maintenance. When enhancements are done, it is sometimes necessary to develop a specific conversion system. This conversion system can be considered to be new development and its functional scope and size can be determined using the standard FPA method. 2.1.3 Enhancement In “Definitions and counting guidelines for the application of function point analysis” [Nesma, 1] a functional change is defined: Functional specifications of one or more functions may be adjusted on the basis of a change request. If these adjustments lead to activities (adjustment of the design, software code, testing, etc.) intended to bring the application in accordance with the changed specifications, then these functions should be considered as functions within the scope of the enhancement release. To determine the size of a function after the change, the same counting guidelines are used as for development. GENERAL CONSIDERATIONS FPA FOR SOFTWARE ENHANCEMENT 2.3 9 All functions mentioned in the change request, and all functions that should be changed as a result of the proposed changes, have to be analyzed in the function point analysis of the enhancement release. After the enhancement, the transactional function should be functionally changed from the user’s point of view. The user’s primary intent of this transactional function has not changed. A transactional function referred to in the change request, is to be considered as functionally changed if it meets at least one of the following criteria:  one or more DETs are added to the transactional function, and/or one or more DETs of the transactional function are changed (see subclause 3.7.4), and/or one or more DETs are removed from the transactional function;  one or more logical files are added to the transactional function and/or one or more logical files are removed from the transactional function;  in the enhancement release a data function is changed and at least one of the modified DETs of this data function is part of the transactional function;  the logical way of processing of the transactional function is changed in the enhancement release (for example as a result of added, modified and/or deleted validations or calculations). A data function referred to in the change request, is to be considered as functionally modified if it meets at least one of the following criteria:  the structure of the data function has changed because in the enhancement project one or more DETs are added to the data function, and/or one or more DETs of the data function are changed (see subclause 3.7.4) and/or one or more DETs are removed from the data function;  the nature of the data function has changed in the enhancement project. This is the case where as a result of functional changes to the transactional function (see subclause 3.7.2) the data function changes from ELF to ILF or vice versa. A DET changes if it meets at least one of the following criteria:  the length (number of positions) changes;  the data type changes (for example from alphanumeric to numeric);  the number of decimal places changes. In the context of this guideline the definition a functional change is adopted. The detailed part of the description refers to the complexity of the function and the related size. However, the phrase “If these adjustments lead to activities (adjustment of the design, software code, testing, etc.) intended to bring the application in accordance with the changed specifications, then these functions should be considered as functions within GENERAL CONSIDERATIONS FPA FOR SOFTWARE ENHANCEMENT 2.3 10 the scope of the enhancement release.” This means that an Enhancement release also covers cosmetic changes when they are required by the user. All implicated changed functions are included in the analysis of the Enhancement Release Other types of maintenance, such as corrective, preventive or perfective maintenance are not included in this guideline because they do typically not lead to functional changes to the information system. Because Function Point Analysis expresses the functionality of an information system in terms of transactional functions and data functions, these principles have also been applied to the FPA method for software enhancement. A prerequisite condition must be that the maintenance is applied to an existing and operational information system. Enhancements resulting only in the addition of new functions, with no changes being made to existing functions, should be treated as “added new development”. This situation can arise when a system is expanded to support the requirements of organisational processes not currently within the scope of a system and often results in the addition of a new component. Development of this type can be analysed using the standard FPA method as set forth in [Nesma, 1] and [IFPUG, 1]. An information system becomes operational only after it has been formally accepted by the customer. 2.2 Prerequisites The following are required to carry out a function point analysis of an enhancement:  a high level or detailed function point analysis of the current system affected by the enhancement (current FPA functional size);  documentation describing the affected part of the current system in order to judge the enhancement proposal;  an enhancement proposal that describes the modifications to be made;  a test plan for the affected changes. GENERAL CONSIDERATIONS FPA FOR SOFTWARE ENHANCEMENT 2.3 11 Figure 1 - Determining size of enhancement This information is required to determine the scope and size of the enhancement project; without it, the enhancement function point analysis cannot be carried out. In addition to the principles stated by “enhancement” in paragraph 2.1 “Limiting conditions”, the following considerations must be taken into account when determining the functional size of the enhancement project:  A Function Point functional size is available for the affected parts of the existing system. The method presented in this guideline takes as its starting point the function point analysis results obtained for the existing system. A high level or detailed function point analysis must exist or must be made of the current system, or at the very least, of the part that will be modified. This Function Point Analysis should at least summarize the individual functions, and for each function, the size in function points, the number of associated Data Element Types (DET’s) and logical data functions (FTR’s).  The current system is well documented. Proposed changes need to be assessed and current functions need to be compared with proposed functions in order to detect changes in business rules as well as changes to DETs, FTRs (ILFs and ELFs) and the user interface. Good functional system documentation (logical datamodel, functional design) is crucial to identify the scope and size of the enhancement. The analyst needs to consider: - the extent of changes to individual transactional and data functions; - the wider implications of the specified changes on other transactional and data functions.  An enhancement proposal is available The enhancement proposal, together with the documentation of the current system, GENERAL CONSIDERATIONS FPA FOR SOFTWARE ENHANCEMENT 2.3 12 must specify the enhancements to be carried out in sufficient detail to enable the effects on each impacted transactional and data function to be assessed. Sufficient detail is required to remove all ambiguity concerning the scope of the enhancement, the data and transactional functions affected and the extent of the impact on each function. If the enhancement proposal does not provide this level of detail, it must be further refined. If a function point analysis of the total system is unavailable, or if only partial function point analyses are available of earlier enhancements, but all other criteria have been met, then the scope of the enhancement can be determined only after the missing functions and their size have been determined.  A test plan is available The test plan must specify the transactional and data functions to be tested and define the scope of the tests to be carried out. The smallest unit of testing is often a system component or a sub-system, regardless of the extent of the enhancement. A test plan should identify both the components and the functions to be tested. 2.3 Summary of the Enhancement FPA Methodology Six steps are necessary for determining the size of the enhancement expressed in enhancement function points: 1. Identify the transactional and data functions within the scope of the enhancement project and determine their functional size. 2. Determine which transactional and data functions are to be added. 3. Determine which transactional and data functions are to be deleted. 4. Determine which data functions are to be changed and determine the impact factor. 5. Determine which transactional functions are to be changed and determine the impact factor. 6. Calculate the number of Enhancement Function Points. The analysis is primarily concerned with determining the FPA functions which are added, modified or deleted. For this part of the analysis Function Point Analysis (FPA) is used. The result is a summary of the impacted FPA functions with their functional size (∑FPBASE). During enhancement, transactional functions and data functions can be added, modified or deleted. With regard to deleted transactional functions and data functions, the number of function points before deletion is decisive; for added and modified transactional functions and data functions, the number of function points after modification is decisive. The impact of an enhancement may go beyond what is initially apparent from the GENERAL CONSIDERATIONS FPA FOR SOFTWARE ENHANCEMENT 2.3 13 enhancement proposal. For example, the change of a logical file or transaction may impact other transactions or logical files. After this, every impacted function needs to be carefully assessed to identify the extent of the impact of enhancement on the function. The impact factor (I) reflects the degree of change of each identified (data and transactional) function. Finally, the enhancement size of each affected transactional and data function is calculated by multiplying its base size (FPBASE) by its impact factor (I). The enhancement size is measured in “Enhancement Function Points” (EFP), not standard function points, which is a different measure. It is imperative to maintain the distinction between the standard function point unit used to express the size of software (FP) and the unit used to express the size of an enhancement (EFP). In chapter 3, “Methodology”, the relationship between the original and the new unit of measure will be described. METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 14 3 METHODOLOGY As outlined in the previous section, six steps are carried out to determine the scope and size (expressed in enhancement function points) of an enhancement project. 1. Identify the transactional and data functions within the scope of the enhancement project and determine their functional size. 2. Determine which transactional and data functions are to be added. 3. Determine which transactional and data functions are to be deleted. 4. Determine which data functions are to be changed and determine the impact factor. 5. Determine which transactional functions are to be changed and determine the impact factor. 6. Calculate the number of Enhancement Function Points. In this chapter these steps will be further refined. 3.1 Identify the functions within the scope of the enhancement project The enhancement proposal, the functional documentation of the current system and the function point analysis of the existing system are used to identify the transactional and data functions within the scope of the enhancement project. A function point analysis of the existing system is an essential prerequisite because all existing functions that are affected either directly or indirectly by the enhancement contribute to the function point size of the enhancement. If, for any reason, a function point analysis of the existing system is not available, one must be undertaken to identify, as a minimum, the functions affected by the enhancement. The size of the existing system, or that part impacted by the enhancement project, is expressed in standard function points, ∑FPBASE. 3.2 Determine which transactional and data functions will be added The enhancement proposal should specify the transactional and data functions to be added to the application. From the proposal it should be possible to calculate the size of the functions added by applying the standard FPA methodology. The total size of the added functionality is expressed as ∑FPADDED. The impact factor for added functions is 1.00. Hence, the number of enhancement function points for a single added function is determined as follows: EFPADDED = FPADDED METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 15 This means, for example, that 3 function points added will result in 3 enhancement function points. See also example 5 in Chapter 6. 3.3 Determine which transactional and data functions will be deleted The (data and transactional) functions that will be deleted from the existing system are identified from the enhancement proposal and the number of function points they represent is determined. The total size of the deleted functions is expressed as ∑FPDELETED. For deleted functions an impact factor of 0.40 is used. The number of enhancement function points for a single deleted function is determined as follows: EFPDELETED = FPDELETED x 0.40 This means, for example, that 6 function points deleted will result in 6 x 0.40 = 2.4 enhancement function points. 3.4 Determine which data functions will be modified and determine the impact factor A data function can be either an internal logical file (ILF) or an external logical file (ELF). Each type of data function is assessed to identify:  data functions that change internally: DETs added, deleted or changed; and  data functions that change type but do not change internally (that is, an ELF is changed into an ILF or vice versa). Determine which data functions will change and how many function points each data function represents after the change, applying the standard FPA rules. The function point size of the changed data function is expressed as FPCHANGED. For data functions that change internally an impact factor is calculated from the percentage of DETs changed. The percentage change is defined as the ratio of DETs changed divided by the original number of DETs (see also example 10 in Chapter 6). Number of DETs added/changed/deleted Percentage change = ---------------------------------------------------------- x 100 Number of DETs in original data function METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 16 The impact factor (ICHANGED) is taken from Table 1 using the percentage change in the number of DETs. Table 1 - Data Function Impact Factors If a data function changes type (for example, an external logical file becomes an internal logical file), a value of 0.40 is used for the impact factor. However, in case of a change of type one needs to check if there is also an internal change of the Logical File (change of DETs). If the number of DETs changes as well as the type, the impact factor due to the change in the number of DETs must be determined. The value of the impact factor due to the change in type is compared with that due to the change in the number of DETs and the higher value is used in the calculation of the enhancement function point size (ICHANGED) (see example 3 in Chapter 6). The number of enhancement function points for a single changed data function is determined as follows: EFPCHANGED = FPCHANGED x ICHANGED The number of enhancement function points arising from a change in data functions therefore depends on the extent of the change in the data function. See also examples 1, 2 and 3 in Chapter 6. If an ELF or an ILF is split into two (or more) data functions, one deleted data function and two (or more) added data functions are counted. See also example 4 in Chapter 6. If an ELF and an ILF are combined, two deleted data functions and one added data function are counted. METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 17 3.5 Determine which transactional functions will be modified and determine the impact factor The transactional functions that change are identified and the size of each transaction after the change (the enhancement) is determined. A transactional function is considered changed if it is altered in some way but retains the same name and purpose after enhancement as before enhancement. To determine the functional size of a transactional function after the change the same counting guidelines are used as for new-built systems, applying the standard FPA rules. The number of function points after the change for each transactional function is expressed as FPCHANGED. A transactional function may be affected by changes to data functions. All transactional functions specified in the enhancement proposal and those affected by changes to data functions are included in the scope of the analysis. This means that a transaction is counted when at least one of the following conditions is satisfied:  the transaction is identified in the enhancement proposal; or;  the transaction undergoes a function change as a consequence of other changes defined in the enhancement proposal. In general, a transaction must be counted if the user can identify that the transaction has changed. This means that at least one of the following criteria is met:  a transaction is affected by a DET that is added, changed or deleted;  a transaction is affected by a Logical Data File (ILF or ELF) that is added, changed or deleted;  the user interface is functionally changed (for example, the composition of a screen or a report);  the business logic supporting a transaction is changed (for example, edit rules or calculations performed on the transaction data);  a cosmetic change visible in the user interface is made, for example: - static data is changed or moved in a report or other media, - a heading is replaced or changed in a report or on a screen, - see also example 9 in Chapter 6. A change to the name of a DET is not regarded as a change in a transaction (see example 8 in Chapter 6). The nature of the DET does not change if the name only is changed. There are four steps to calculating the enhancement function point size of a change to a transaction: 1. Identify the DETs and FTRs used by the transaction. 2. Determine the percentages of DETs and FTRs changed as a result of the enhancement. 3. Determine the impact factor for the transaction. 4. Calculate the number of enhancement function points. These steps are explained below. METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 18 3.5.1 Identify the DETs and FTRs used by the transaction The enhancement function point size of a changed transactional function is calculated from the function point size of the function after the change and the change impact factor. The impact factor is determined by the percentage changes in the numbers of DETs and FTRs used by the transaction. Examples 6 and 7 in Chapter 6 illustrate how a changed transaction is assessed. If the change is cosmetic only, the number of changed DETs and FTRs is nil. The impact of such a change is considered minimal and the value of the impact factor (0.25) reflects a relatively low impact. However, the change will be included in the scope of the enhancement project. 3.5.2 Determine the percentage of DETs and FTRs changed as a result of the enhancement The impact factor is determined by the percentage changes to the numbers of DETs and FTRs used by the transaction compared with the original numbers of DETs and FTRs (see example 11 in Chapter 6). Number of DETs added/changed/deleted Percentage DETs = ---------------------------------------------------------- x 100 Number of DETs in original transaction Number of FTRs added/changed/deleted Percentage FTRs = ---------------------------------------------------------- x 100 Number of FTRs in original transaction Changes in excess of 100% are possible when DETs and FTRs are added to a transaction. METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 19 3.5.3 Determine the impact factor for the transaction The impact factor (I CHANGED) for a transaction is determined from the percentage changes in the numbers of DETs and FTRs from Table 2: Table 2 - Transactional Function Impact Factors If the impact factor is 1.00 or greater, you should consider whether enhancing the transaction is still meaningful. 3.5.4 Calculate the enhancement function point size The enhancement function point size of a single transactional function is calculated as follows (see also example 11 in Chapter 6): EFPCHANGED = FPCHANGED x ICHANGED 3.6 Calculate the size of the enhancement project The size of the enhancement project is the sum of the number of enhancement function points for all the affected transactional and data functions. EFPTOTAL = ∑EFPADDED + ∑EFPDELETED + ∑EFPCHANGED METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 20 3.7 Calculate the size of the system after enhancement The functional size of a system may change as a result of the enhancement. The size after enhancement can be calculated by analysing the whole application anew or by taking account of the changes from the original FPA analysis. Steps to take are: 1. Calculate the function point size of the application prior to the change (FPBASE) using the standard FPA method. 2. Identify the transactional and data functions deleted from the existing application and determine their function point size (∑FPDELETED). 3. Determine the transactional and data functions changed. Calculate the number of function points these represent before and after the enhancement (∑FPAFTER and ∑FPBEFORE), using the standard FPA method. 4. Determine the transactional and data functions added to the system and calculate how many function points these represent (∑FPADDED). 5. Calculate the size of the system after enhancement (FPNEW). The size of the system in unadjusted function points after enhancement is: FPNEW = FPBASE + (∑FPADDED) + (∑FPAFTER-CHANGE - ∑FPBEFORE-CHANGE) - (∑FPDELETED) Note: The Impact Factor does not play when determining the size of the system after enhancement. METHODOLOGY FPA FOR SOFTWARE ENHANCEMENT 2.3 21 3.8 High level Function Point Analysis for Software Enhancements In practise there is a tendency in FPA to speed up the sizing process and to apply High level Function Point Analysis. This will slightly change the approach of FPA for Software Enhancement. You still can apply the detailed approach at the base size obtained by the High level Function Point Analysis. A practical approach might be to define a more high level way of identifying the impact factor. The challenge is to keep it objective. Suggestion for default impact factors to be used for high level EFPA: Table 4 – High Level EFP Impact Factor defaults In contracts other default impact factors can be agreed upon, based on the specific situation. TESTING IN ENHANCEMENT PROJECTS FPA FOR SOFTWARE ENHANCEMENT 2.3 22 4 TESTING IN ENHANCEMENT PROJECTS The range of transactional and data functions that have to be tested can be much greater than the number of transactional and data functions within the scope of an enhancement. Not only do the functions directly impacted by the enhancement need to be tested, but also all other affected functions. The size of the functions to be tested is measured in test function points (TFPs). When determining the number of TFPs the impact factor per function is not taken into account. Also, no account is taken of whether a function was added or changed. Deleted functions are excluded from the analysis of test function point size (TFP). The number of TFPs is determined as follows: Transactional functions:  determine the FP size of each transaction directly involved in a test;  calculate the total number of function points for all transactions involved in the test. Data functions:  determine the FP size of each data function directly involved in a test;  calculate the total number of function points for all data functions involved in the test. The size of each function is derived using the standard FPA counting practices for new software development: the number of function points (FPTEST) is the sum of the sizes of the transactional and data functions involved in the tests. In general, testing is done on discrete components of the system and encompasses unchanged functions as well as changed and added functions. Each function included in the scope of a test is measured after enhancement. Deleted functions are therefore not included in the scope and consequently do not add to the total number of TFPs. The test function point size is equal to the size of the tested functions, where 1 FP results in 1 TFP. The total test function point size is: TFP = ∑FPTEST Note: The abbreviation TFP must not be confused with the term TP (Test Points) in Test Point Analysis (TPA). The approach used in this guideline differs from TPA. TPA is a methodology for measuring the functional size (based on FP) of structured testing and expressing this in test points, as defined by Tmap®. BUDGETING ENHANCEMENT FPA FOR SOFTWARE ENHANCEMENT 2.3 23 5 BUDGETING ENHANCEMENT The technique described in this guideline is based on the standard FPA method. When estimating small enhancement projects, on average, the estimations will be correct, but large deviations above and below may occur. With larger projects, there will likely also be deviations above and below, but when regarding the total enhancement project, these differences will most likely cancel each other out. Enhancement function point sizes and test function point sizes can be used to derive productivity metrics, for example, hours per EFP and hours per TFP. Values for hours per EFP and hours per TFP will, in general, differ from the hours per function point for new system development measured using the standard FPA method. The total enhancement effort, including testing, can be expressed as follows: Total enhancement effort = (number of EFPs x hours per EFP) + (number of TFPs x hours per TFP) EXAMPLES FPA FOR SOFTWARE ENHANCEMENT 2.3 24 6 EXAMPLES 6.1 Example 1 - Expanding an ILF Situation An internal logical file consisting of 1 RET and 37 data element types is to be expanded by adding 14 new data element types. Question Which impact factor should be used, and how many enhancement function points does the change generate? Answer The size of the internal logical file, after change, is 10 function points (ILF with average complexity). The change impact expressed as a percentage of DETs is 14/37 (x100%) = 0,378378 (x100%), which is between 1/3 (x100%) and 2/3 (x100%). From table 1, this change gives in an impact factor of 0.50. The enhancement function point size is 10 x 0.50 = 5 EFP. 6.2 Example 2 – ELF becomes ILF Situation Information system A uses an external logical file maintained by information system B. A decision is made that, in future, maintenance of this function will be carried out by information system A (the structure of the data function does not change). Question 1 Which impact factor should be used? Answer An external logical file (ELF) is imported into system A and converted to an internal logical file (FTR). An impact factor of 0.40 is used when an ELF is changed to an ILF or vice versa. Question 2 How many enhancement function points result from the change to the data function? Answer Assuming the data function is a low complexity ILF, its size after the change is 7 function points. The impact factor is 0.40, therefore the change results in 7 x 0.40 = 2.8 EFP. EXAMPLES FPA FOR SOFTWARE ENHANCEMENT 2.3 25 6.3 Example 3 – ELF becomes ILF with modifications Situation Information system A uses an external logical file that is maintained by system B and contains 45 data element types. A decision is made that system A will maintain the data function and as a consequence 25 data element types will be removed from the data function. Question 1 Which impact factor should be used? Answer An impact factor of 0.40 is used when an ELF is changed to an ILF or vice versa. However in this case the effect of the structural change to the data function must also be taken into account. The percentage change is: 25/45 (x100%) = 0,5555… (x100%). From Table 1, this change (between 1/3 and 2/3) gives an impact factor of 0.50. This is greater than the impact factor for a change of type (from an ELF to an ILF), so the higher value of 0.50 is used. Question 2 How many enhancement function points result from the change in the data function? Answer Assuming the file is a low complexity ILF, the size of the data function after change is 7 function points. The impact factor is 0.50 and so the change gives rise to 7 x 0.50 = 3.5 EFP. EXAMPLES FPA FOR SOFTWARE ENHANCEMENT 2.3 26 6.4 Example 4 – Splitting an ELF Situation An external logical file is divided into two separate external logical files. Question How do we account for this change? Answer Count one deleted external logical file and two added external logical files. The analysis must also take into account all the transactional functions affected by the change to the data function. 6.5 Example 5 – Adding an ELF Situation A new external logical file is to be added to a system. Question How do we account for this change? Answer Count one added external logical file (impact factor = 1) and take into account the transactions that use the data function. These transactions must be reassessed as part of the enhancement project. 6.6 Example 6 – Modifying an ILF, 1 Situation 3 DETs in a data function used by a transaction are to be changed. Two of the changed DETs will be used by the transaction. Question How do we account for the impact of the change on the transaction? Answer One changed data function and two changed DETs are counted to determine the impact factor for the changed transaction. EXAMPLES FPA FOR SOFTWARE ENHANCEMENT 2.3 27 6.7 Example 7 – Changing the name of a DET Situation The name of a DET is changed in a data function used by a transaction. The DET is used in the transaction. Question Do we need to account for this change? Answer The transactional and the data function are not counted. 6.8 Example 8 – Modifying a heading in a report Situation An external output prints a report that lists employee information. The heading of one of the columns is changed from "Name labourer" into "Name employee". There is no change in the DETs of the data function, so the change is considered cosmetic. Question Do we need to account for this change? Answer The function will change as a result of the cosmetic change, this on request of the user to meet requirements. As no DETs are changed the impact factor = 0.25. 6.9 Example 9 – Adding / modifying / deleting DET’s Situation Two DETs are added to a data function, 1 DET is deleted and 3 DETs are changed. Question How many DETs will be counted in this change? Answer The number of DETs counted is 2 + 1 + 3 = 6 DETs. 6.10 Example 10 – Determining the % of modified DET’s and FTR’s Situation A report (including a calculated total) displaying 16 DETs will have 3 new DETs added to it, 3 DETs will be changed and 2 DETs will be deleted. The number of affected DETs is 3 + EXAMPLES FPA FOR SOFTWARE ENHANCEMENT 2.3 28 3 + 2 = 8. The external output uses 2 FTRs. The new DETs belong to a new (to be added) FTR. Only one FTR is affected by the changed and deleted DETs. Question 1 How do we determine the percentage changes in the DETs and FTRs? Answer Measure the transactional function DETs changed relative to the original number of DETs for the transaction (16). The calculation is therefore 8/16 x 100 = 50%. Measure the transaction FTRs changed relative to the original number of FTRs (2). The calculation is therefore 2/2 x 100 = 100%. Question 2 What is the size of the change to the transaction? Answer After enhancement the number of DETs is 17 and there are 3 FTRs. The complexity of the function remains average and its function point size does not change (5 function points). The impact factor is taken from Table 2. The percentage changes in DETs (50% - first column) and FTRs (100% - third row) give an impact factor of 0.75. The enhancement size is: EFPCHANGED = 5 x 0.75 = 3.8 EFP. GLOSSARY FPA FOR SOFTWARE ENHANCEMENT 2.3 29 7 GLOSSARY Note: This paper assumes, that the reader is familiar with the standard FPA methodology, its concepts and terms. For more information, reference the IFPUG publication "Function Point Counting Practices Manual" [IFPUG, 1] or the Nesma publication “Definitions and Counting Guidelines for the Application of Function Point Analysis” [Nesma, 1]. Cosmetic change A change, requested by the user, only in the visible user interface presented by a transactional function, without any change in the processing logic underlying the transaction (independent of any descriptions of addition, change, or removal of DETs or FTRs). Cosmetic changes are not considered enhancement in the formal Nesma counting guidelines. Because these changes are requested by the user to fulfil the requirements, EFPA considers them as part of the enhancement. Data function An internal logical file (ILF) or an external logical file (ELF). DET Data element type. EFP Enhancement function point. ELF External logical file. Enhancement Enhancement is the work necessary to bring about a change in an operational information system or in the structure of a data store of an operational information system. The definition of enhancement is derived from that of Vollmar: “Enhancement is effecting a change in an information system or in the structure of the data store”. The definition is independent of the life cycle in which the information system may exist and allows for the addition of new functions, the removal of existing functions and changes to existing functions to be included. Enhancement function point A unit of measurement of the size of an enhancement. Enhancement proposal A formal request for enhancement of an operational information system. The proposal must be sufficiently comprehensive to enable the scope and impact of the enhancement to be determined. GLOSSARY FPA FOR SOFTWARE ENHANCEMENT 2.3 30 FTR (File Type Referenced)  An internal logical file (ILF) read or maintained by a transactional function.  An external logical file (ELF) read by a transactional function. Function An external input (EI), external output (EO), external inquiry (EQ), internal logical file (ILF) or external logical file (ELF) as defined in the standard FPA methodology (see for example [IFPUG, 1] or [Nesma, 1]). In function point analysis a function is either a transactional function or a data function. ILF Internal logical file. Impact factor A unit of measurement of the degree of change in a transactional function or data function. The value of the impact factor may vary according to the nature and extent of the change. LF (Logical file) A generic name for an internal logical file (ILF) and external logical file (ELF); synonymous with data function. Maintenance In the context of these guidelines, maintenance encompasses all the activities necessary to operate an automated information system and manage the associated technical, organisational and financial aspects. Maintenance involves performing the work necessary to ensure the continued operation of the system without altering the scope or structure of the system or its associated data stores (after Looijen). GLOSSARY FPA FOR SOFTWARE ENHANCEMENT 2.3 31 Standard Function Point Analysis A function point analysis performed using the standard FPA methodology as described in [IFPUG, 1] and [Nesma, 1]. Test function point A unit of measure of the size of the functions subject to testing. TFP Test function point. Transactional function An external input, external output or external inquiry function. BIBLIOGRAPHY FPA FOR SOFTWARE ENHANCEMENT 2.3 32 8 BIBLIOGRAPHY [IFPUG, 1] IFPUG Function Point Counting Practices Manual, release 4.3.1 This manual describes the standard FPA methodology and may be used together with the Nesma manual. For more information, see the IFPUG web site www.ifpug.org. Date of issue: January 2010 [Nesma, 1] Definitions and counting guidelines for the application of function point analysis. A practical manual, version 2.3. (English) This manual is also called Nesma Counting Practices Manual. It describes the standard FPA methodology, and many aspects related to the application of FPA. It may be used together with the IFPUG manual. It is identical to the version 2.2 in Dutch. For more information, see the Nesma web site www.nesma.org (English). Date of issue: 2018 Size: 112 pages A4-format ISBN 978-90-76258-45-4 [Nesma, 2] Begrotingsmodel voor de exploitatielasten van informatiesystemen (Budgeting the Operational Costs of Information Systems; Only available in Dutch) Date of issue: November 1998 Size: 117 pages A4-format ISBN 90-76258-11-2 [Nesma, 3] Onderhoud en Functiepuntanalyse Version 2.2 (Function Point Analysis for Software Enhancement) Date of issue: August 2008. Size: 21 pages A4-format ISBN 90-76258-13-9 BIBLIOGRAPHY FPA FOR SOFTWARE ENHANCEMENT 2.3 33 [Publication, 1] The Added Value of Enhancement Function Points An Empirical Evaluation Frank Vogelezang, Jelle de Vries IWSM 2014 [Publication, 2] Statistical Study of the Relationship of the FPA / EFP (Nesma) with the Effort Estimation of a Project LEDA Mc, 2018
17020	https://www.youtube.com/watch?v=Q_M_5vh622Q	Graph Quadratic Functions of the Form f(x) = ax ^ 2 + bx + c James Wenson 2300 subscribers 33 likes Description 2747 views Posted: 18 Aug 2020 Transcript: hello everyone and welcome to this video on graphing quadratic functions of the form f of x or y equals a times x squared plus b times x plus c or ax squared plus bx plus c where a b and c are real numbers and of course a is not zero because if a were zero then you wouldn't have a quadratic function all right so in this video we'll go over a couple examples of functions of this nature and how to graph them using transformations right so we're going to start with the most basic quadratic function y equals x squared and then through the use of transformations which i've gone over in previous videos so if you could please um if you if you haven't seen those check them out you know vertical translations vertical shifts horizontal shifts vertical stretching compressing and reflections we're going to be using these transformations to take the graph of y equals x squared and graph another function that looks like this say 2x squared plus 3x plus 1 or something okay all right so on my first page i talk a little bit about you know the graphs of a quadratic function right the graphs are these things called parabolas right parabolas so we have f of x equals ax squared plus bx plus c again where a is not zero so it's still quadratic right so what we're going to be doing in this video and what we're going to you know to make to make these graphs now you could make a table right you could do what you do with any function any equation when you're asked to graph you know you could make a table plug in a bunch of inputs you know seven or eight points and find you know find a bunch of outputs and then plot seven or eight or nine or ten you know plot a bunch of ordered pairs on the coordinate plane and hopefully you've plotted enough to where you can see a pattern forming and then you draw in the rest of the graph based on that pattern i mean you can always do that just what i'm doing here is going to be a little you know maybe a little quicker and you'll get an idea for the shape you know if you already have an idea for the shape of a parabola y you know y equals x squared then you can just simply move that around to get the graph of any quadratic function of any parabola so what we're going to be doing is taking a function equation looking like this f of x equals ax squared plus bx plus c and then through the use of completing the square and i'll show you this on my examples we're going to rewrite this in this form instead of ax squared plus bx plus c i'm going to rewrite it as a right that's the same a okay that a here is the same as this a here a times and then the the quantity x minus h squared so this thing's squared and then afterwards plus some k all right where h and k are real numbers so for instance like 3 times the quantity x minus 1 squared and then plus 4 all right something like that this h and k will give me the coordinates of what's called the vertex of a parabola the vertex is simply the point of the parabola that's either the lowest point if it's opening up or the highest point if it's opening down right it's just that peak point it's called the vertex of the parabola all right and then you got a couple cases here um if this a right if the coefficient on the x squared or this a here if it's positive if it's greater than zero then your graph is going to be a parabola that opens upward all right smiley so i have a little picture that here your vertex with the coordinates you know h comma k and this dotted line going through the vertex straight up and down this is an imaginary line it's not actually part of the parabola it's called the axis of symmetry and it's a vertical line it's going to have an equation of x equals some number in this case it's x equals h right the x coordinate of the vertex that's the line that if you were to fold the parabola above that line you know the two the two halves would match up and symmetrical about that line and the second case is if this value of a right this a if that's negative if it's less than zero then you're going to have a pic a graph of a parabola that's opening downward and again a little frowny face right opening down again vertex with the coordinates with this h and k that i'm talking about there and this dotted imaginary line vertical line called the axis of symmetry right the line of symmetry for the parabola goes through the vertex and has equation x equals h and how we're going to be doing this how are we going to make these graphs we're going to graph our equation written in this you know this vert it's called vertex form this f of x or y equals a times x minus a squared plus k this is called vertex form because you'll be able to easily tell what the coordinates of the vertex of the parabola are and what we're going to do to graph these is you know start with y equals x squared start with the most basic quadratic function and then use you know one transformation at a time all right uh a ver you know like a horizontal horizontal shift and then a vertical shift and then you know maybe a vertical stretch or compression or a reflection or whatever right we're going to use transformations to go from y equals x squared to this guy this vertex form okay so i have two examples for you right in the first one and in each of these examples i'm asked to do the same thing i'm asked to graph each quadratic function using transformations right so starting with y equals x squared and then afterwards based on the graph determine the domain and the range of each function so here i have f of x equals negative 3 times x squared minus 12x minus 11. so this is that ax squared plus bx plus c form where a is negative 3 b is negative 12 and c is negative 11. all right so what i'm going to be doing as i mentioned with the last on the last page so i'm going to take that equation that rule we're going to rewrite it in that vertex form right i'll write up the vertex form again here that's where i have f of x or y equals you know a that same a that's going to be negative 3 times some quantity x minus h to the second power and then after that plus some number k right and we're just going to have to determine what these a h and k are all right i already said a is going to be negative 3 but we're going to do this we're going to rewrite this in this vertex form through use of completing the square let me show you that and hopefully you recall what that is if you if you don't look up another video on that just type in you know completing the square and i'm sure plenty will come up all right so i'm going to start with we have f of x equals negative 3 times x squared minus 12x and then you have your minus 11 okay so let's take this away for a second i'll bring the graph paper back when we actually start making the graphs all right so remember completing the square i'll just do a quick refresher off to the side that was when i had 1 x squared 1 not negative 3 so this is an issue here we're going to have to take care of that issue in a second i have a variable variable squared times 1 plus and then some number b times the variable to the first power what completing the square was it was adding a number to these two terms of variables adding a constant term to make three terms that were a perfect square trinomial meaning could be written as the same factor twice could be factored as a square how you did that was you took the coefficient of the first power term b cut it in half so b over 2 b divided by 2 or half b and then squared it and then that's the number that would get added all right so you add half of b and then squared and once you do that this x squared plus bx plus half of b squared is a perfect square trinomial three terms that can be written as a perfect square and then how can they be written as a perfect square it would be the variable x plus whatever half of that b was so b divided by 2 or half b and if you want to double check take x plus half of b multiplied by x plus half of b and do all the foiling and combining like terms you you would get x squared plus bx plus half of b squared back all right so this was completing the square very very very quick refresher now as i mentioned in completing the square on the the terms of variables here the lead coefficient was 1 not negative 3 all right and again the only terms i am focusing on are the terms with variables all right so i'm only going to focus on these two terms right now the negative 3x squared and the negative twelve x well the coefficient on the x squared is not one it is negative three a very quick fix to that is to factor out that negative 3 factor out that lead coefficient all right so when i do this we have f of x equals and then just from these two terms i'm going to factor out negative 3. so negative 3 comes out then in parentheses right we're just reversing the distributive property factoring then the first term would be just just x squared and there's your there's your 1x squared that i want right and then if i pull out a factor of negative 3 from negative 12x that'd be positive 4x that's left over and you can double check by distributing right negative 3 times x squared would be negative 3x squared negative 3 times positive 4x would be negative 12x all right so now this x squared plus 4x in the parentheses here i can complete the square on these there is one x it's in this form x squared plus bx right one x squared plus a number times x so because i can complete the square i'm going to leave a little space in the parentheses there all right so i factored out the negative 3 from these two terms giving me this and then afterwards i still have the minus 11 all right so sorry there's not a lot of room over here there's still the minus 11 term i'll write that here all right now comes the completing the square part we can do that right here right in inside the parentheses right inside this set of parentheses now what do you do alright you take half remember you take half that b the the coefficient of the first power variable half of four is two and then square that well 2 squared is 4. that gets added again i'm doing this in the parentheses right so i'm adding 4 and now this x squared plus 4x plus 4 is what they call a perfect square trinomial i can rewrite these three terms as a perfect square like this over here all right now you got to be careful all right you can't just add 4 to an expression and then stop i have to undo this as well you know if you just add 4 look what would happen if i redistributed the negative 3 i'd have negative 3x squared great i'd have negative 3 times the 4x would be the negative 12x great you got the negative 11 at the end but then distributing that negative 3 in the parentheses here you see negative 3 times positive 4 would give me a negative 12. there's no negative 12 up here in the original expression so we have to counteract this right if i put the negative 3 back in right negative 3 times 4 would give me negative 12 so that means that adding 4 in the parentheses really is like adding negative 12 to the expression i started with and to counteract that you know if i add negative 12 or subtract 12 i would then also have to add back positive 12 to you know undo it right and i'm going to write that here this is this undoes what we added when completing the square that way i have an expression that's actually equal to the expression i started with and you have to be sure to do this all right don't just complete the square in the parentheses here and then add nothing right because that'll totally change the expression and you won't have what you started with i want to continue to have what i started with all right so finally now we have f of x equals right the negative 3 outside times and then in parentheses here x squared plus 4x plus 4 again the whole reason for completing the square is so i can rewrite it as a perfect square x squared plus 4x plus 4 is equivalent to the quantity x plus 2 squared and then after the parentheses we have negative 11 and then the plus 12 we added to counteract the completing the square negative 11 plus 12 is plus one positive one and this expression here and go ahead and multiply it out you know if i were to square the x plus two it would give me this and then multiply by negative 3 it would give me negative 3x squared minus 12x you know minus 12 and then i would add 1 to get minus 11. this expression down here is exactly the same as this expression up here go ahead and multiply it out and this is in what they call vertex form all right this is in that vertex form all right so look at this you know the the value of the a is the negative 3 right that's the the number in front of the square then the the x minus h remember this is x minus h squared x plus two is the same thing as x minus negative two so when i have x plus two squared that's really like having x minus negative two squared so the h the value of h here is negative two not positive two right and then we have plus k afterwards well plus one so the value of that k is one right so a is negative three h is negative two and k is 1. and the reason again this is called vertex form is because now i have this h and you know i have it in this particular form a number times a square plus another number i have it in this vertex form i can identify the vertex of the parabola very easily it is you know h comma k negative two for the x coordinate one for the y coordinate all right this'll you'll see this in my graph when i'm when i'm totally done graphing and additionally because a is a negative number a is negative three the parabola will open down we're going to see that as well in my graph all right but i'm going to put this thing in the box this vertex form back on my graph paper okay all right so we're asked to graph f of x equals you know negative 3 times x squared minus 12x minus 11. and i just showed you that that's the same as this f of x equals negative 3 times the quantity x plus two squared plus one these are exactly the same function exactly the same all right but i'm asked to do it using you know transformations okay i'm asked to do using transformations so bring this new paper up here what that means is i'm going to start with i'm going to start with the most basic parabola let's start with y equals x squared now i have plotted that many times in previous videos so i'm going to just plot that in red here real quickly now i have my axes drawn right my x-axis my y-axis so i'm going to have it so that every square represents one unit so y equals x squared that has the origin right 0 squared is 0 has 1 1 right 1 squared is 1. as negative 1 1 right it's got 2 4 2 squared is 4 right and it's got negative 2 4 and so on and you could you could find all sorts of other points all right but i'm going to draw this it's very i'm going to make it a little thin there's y equals x squared right this thin parabola here i don't want to take up too much space all right there's y equals x squared nice parabola opening up you know with the vertex the vertex is at the origin 0 0 and then through transformation i'm going to do one step at a time i'm going to change this rule x squared into this rule negative 3 times x plus two squared plus one all right again i went over there my previous videos all these transformations uh but i'm gonna take it well one step at a time all right so the first thing i notice is that the first thing i'm going to do is i'm going to do my reflection i notice that this is you know my square so the square here x squared is being multiplied by a negative number all right negative 3. it's actually the first thing i'll do is multiply by 3 all right so i'm going to take the rule the way it is i'm going to multiply the rule by 3. which is going to cause a what you know if you're a vertical stretch by a factor of three and all you're really doing is multiplying all the y-coordinates by three so we're looking at the graph of y equals you know again leave the rule the way it is and i multiply by three and get a new graph where it's vertically stretched by a factor of three all the all the points on the original i'm just multiplying the y coordinates by three so zero zero is still at zero zero one one is now at one three all right so this point again gets stretched away from the x axis by a factor of three negative one one is at negative one three 3 2 4 would be at 2 12 you know negative 2 4 would be at negative 2 12. so those are pretty high up there i'm not going to graph those so here we go and it's still a parabola shape still opening up this little thin guy here and there's y equals 3x squared all right and then after i'm done with all these different graphs i'm going to put up on another sheet of graph paper the final graph so it looks a lot cleaner okay all right so that gives me a 3 in front of the square like this rule has right the 3 in front of the square but it's also negative right and when you multiply by negative 1 so i'm going to take this rule now this 3x squared multiply by negative 1 and what does that do uh multiply the rule sorry by negative one and that causes a reflection all right that reflects the graph across the x-axis so reflects about the x-axis or across the x-axis again all you're doing is multi if you multiply the rule by negative one that means you're just taking the points on the graph of y equals three x here and multiplying all the y-coordinates by negative one so we have y equals negative three x squared so plotting some of these points now zero zero would still be at zero zero one three would go to one negative three negative 1 3 would go to negative 1 negative 3 right and you know 2 12 would go to 2 negative 12. and i can pop i can put that in here 2 3 4 5 six seven eight nine ten eleven twelve there's negative twelve two negative twelve and then negative two twelve on the orange graph the y equals y equals three x squared uh would go to negative two negative 12 and again you get this parabola here that's opening downward all right there's y equals negative 3 x squared okay now that takes care of the vertical translation uh that takes care of the vertical stretching and the reflection this number negative three in front of the square uh now i can do the addition parts you know notice that there's a plus one after this first term so after this first term here i'm going to throw a plus one so i'm going to leave the rule the way it is and add 1 to the rule so add 1 to the rule and again whenever you do something to the entire rule that's just what you do to the x uh the y coordinates it'll do a vertical transformation so i add one to this rule and that's going to cause a vertical shift up one unit all right so we're looking at the graph of y equals you know the the the previous rule negative three x squared and then plus one so again this is going to get pretty cluttered all right but again i'll draw the final result on the on a separate page but here's the graph i'm currently at y equals negative 3x squared and i'm gonna be adding one meaning moving everything up one so zero zero was the vertex now it's at zero one this one negative three is now at one negative two negative one negative three is now at negative 1 negative 2. you know 2 negative 12 is now at 2 negative 11. negative 2 negative 12 is now at negative 2 negative 11. every single point on this graph y equals negative 3x squared got bumped up one unit all right still a parabola opening down but there's y equals negative three x squared and then plus one all right and then finally all right lots of stuff going on here look at the rule now negative 3x squared plus 1. the only change that needs to be made is now i'm not doing anything outside i'm not doing anything to the whole rule anymore i only need to replace the x right inside the square i need to replace this x with x plus two so i'm going to replace x with the quantity x plus two remember x plus two is x minus negative two that the value of that h was negative two and if that if you replace x with x minus h and h is a negative number it moves you left all right this cut this will cause a horizontal shift uh you know two units left so then that we have our y equals negative three times and then we're pl just the x i'm replacing just the x with the quantity x plus two squared and then plus one and this is the f of x right this is the rule f of x that we're asked to graph all right so finally i'm taking this you know y equals negative three x squared plus one graph and moving everything left two oops i drew that wrong should be up at 0 1 here i apologize so 0 1 right 0 1 was the vertex of this graph that gets moved to the left 2 units okay so to negative two one there that's the vertex of this final graph and then uh one negative two moves left two to negative one negative two negative one negative two moves left two to negative three negative two and it's again this just the same graph earlier move left two units right so it's this parabola here and that's the graph of f right f of x equals negative 3 times x plus 2 squared plus 1. all right so i'm again i'm going to graph this on a separate piece of graph paper and then you can always you know double check your answer you always check your answers all right so lots of crazy transformations happening but then on a separate sheet i'll put up the final results right so we were graphing remember this is the graph we're trying to make i changed this to vertex form in the final graph i ended up with the vertex being negative 2 1 remember which is what i said it was going to be on that earlier page all right remember this page here where i did the completing the square change it to vertex form we said the vertex would be at negative 2 1 and that's where it ended up being after all these transformations alright so negative 2 1 and i'll label that in a second then we had negative one negative two that location negative three negative 2 and other points but you know it's just this parabola going through these three points here opening down and i'll again i'll label the these points this was negative two comma one negative three comma negative two negative one comma negative two and don't forget to check you know let's just check one of these because if these points truly are on the graph of this function then they should satisfy the equation you know the the input i have should correspond to the proper output so let's say this point negative 3 negative 2 here if i replace x with negative 3 in the rule do i get an output of negative 2 let's see when x is negative 3 you know negative 3 squared is 9 times negative 3 would be negative 27. so i have negative 27 for the first term negative 3 times negative 12 is 36 positive 36 so i have negative 27 plus 36 would be um you know positive 9 and then minus 11 would indeed be negative 2. i do get the output of negative 2 when the input's negative 3. and you can check that for negative 1. you know the when the m plus negative one is the output negative two it should be um when the input's negative two is the output one it should be right all these all the points that i plotted here should satisfy this equation i should should make that a true statement great so i know that's a lot right but you know and there are you'll see i have a video coming up where we graph parabolas in a little quicker way than this but this is just to get you used to the transformations again right you know when i mult when i do anything to the entire rule that causes some sort of vertical transformation you know vertical shift vertical stretch reflecting across the x-axis is an up and down movement of vertical movement and remember when i replace just the x's with something that causes some sort of horizontal transformation all right so the really the point of this video is to get you used to the transformations again and how how could i use them to graph a parabola all right so that was my first example right we ended up with this parabola you know see opening down right had a negative lead coefficient negative value of a it ended up opening down and a vertex at you know that h k which ended up being negative 2 1 here all right so one more example of this where again i'm using transformations to graph a parabola graph a quadratic function and here it is f of x equals 2x squared minus 8x plus 5. now as i did with the last example right we're going to take that 2x squared and i'll do it on this sheet i'm going to take that 2x squared minus 8x plus 5. and through use of completing the square on the x terms the terms of variables we'll put it in that vertex form right and i'll write out the vertex form once again it's a good thing to do when you have a formula or or an equation that you want to memorize right write it out as often as you can until you get it in your memory so that was y equals or f of x equals you know a well the a here is 2 positive 2 right so this is going to be a parabola opening up times the quantity x minus h squared and then plus some k where the h and the k will give me the coordinates of the vertex of this parabola all right and this time because the parabola is opening up right the vertex will be at the bottom of the parabola not like the last example where it was at the top because it was opening down all right so look at the graph paper i'll bring the graph paper back later so from these two terms and i want to complete the square on the these x terms uh i can't do it yet though remember to complete the square the coefficient on my x squared term needs to be one not two needs to be positive one so a very simple fix to that is again like i did in the last example just factor this number out of the two variable terms so i'm going to factor out 2 positive 2 from these two terms with variables so i do that pull 2 out and then in parentheses we'd have x squared as the first term and then minus 4x and now in the parentheses i can complete the square right there is a 1x squared in there so because i'm going to complete the square in a second i'll leave a little space and then cl then close the parentheses right so we'll add something here and then afterwards you know plus 5 outside the parentheses all right now comes the complete square part remember how to do that you know x squared minus 4x you take half of the negative 4 which is negative 2 and then square it well negative 2 squared is you know negative 2 times negative 2 that's positive 4. so again i'm adding 4 in here that completes the square on x squared minus 4x so this x squared minus 4x plus 4 is now a perfect square trinomial but you got to remember you can't just add 4 to an expression and think it's going to be the same as the original it's not think about you know earlier i factored out 2 what if i were to put that 2 back in you'd have 2 times x squared is 2x squared 2 times negative 4x would be the negative 8x from my original i've got the positive 5 from the original so what happens here 2 times this positive 4 would add 8 would have a positive 8 which would mean that this expression the way it looks right now is not the same as the original so i have to counteract that this 2 times 4 puts a positive 8 up here to undo positive 8 i have to throw on a term of negative eight again this this undoes what we added when completing the square please be sure you're doing this right if you don't put on that negative 8 then the expression you have isn't the same that you started with right and i want to always keep the same expression something equivalent to what i started with okay now the whole reason for doing that right the whole reason for completing the square so now i have f of x equals you know this 2 pulled out 2. and then the x squared minus 4x plus 4. right the reason i added the 4 in there so i could write this as a square this is going to be the quantity x minus 2 squared and you can double check take x minus 2 times x minus 2 that's x squared minus 4x plus 4. remember it's just the variable x and then the other term is just whatever half of that negative 4 was right whatever half of that linear coefficient was negative 2x minus 2 squared and then after that after the parenthesis you have positive 5 and negative 8 that makes negative 3 okay so the a here this this is in that vertex form now i'll put it in a box all right we're done this is a you know the vertex form is just a number times a square plus another number right i have that here a number times a square plus another number so i'm in vertex form the value of a is this positive 2 right the same a as earlier so the parabola is going to be opening up right that is a positive number the parabola is going to open up right when i'm done graphing the h remember this is the quantity x minus h squared so x minus well this is already minus so x minus 2 the 2 is the h all right and the k remember the form it's plus k plus k this is not a plus but you know remember when you subtract 3 that's the same thing as plus negative 3 right when i say minus 3 that's the same as plus negative 3. so negative 3 is the value of this k so i have this h and k those are the coordinates of the vertex right so the vertex when i when i'm done graphing after using all the transformations and whatnot the vertex will have coordinates you know x coordinate 2 y coordinate negative 3. and we're going to see that when we graph in a minute all right so i've done all that now i've used completing the square to change the quadratic into this vertex form and i'm going to write this again i'm back on my graph paper so this f of x equals 2x squared minus 8x plus 5 is equivalent to f of x equals two times the quantity x minus two squared and then minus three all right these are exactly the exact same expression they have the same value they're the same thing they just look different all right we're going to graph this using transformations right so i will write up the transformations on another piece of paper over here where again we're starting with the most basic quadratic equation y equals x squared the most basic parabola and i'll put that up on the coordinate plane here so i've got my x-axis horizontal y-axis vertical again i'm going to have every square represent one unit so again you have zero zero you have one one and i'll just plot these three points and draw the rest in you have negative one one and there's this parabola right opening up okay there's the graph of y equals x squared and you know the vertex is zero zero and you know so as one one negative one one opening up two four negative two four and so on right i didn't plot those other points though all right now what i'm going to try to do is you know do one transformation at a time make one little change at a time and see how the graph moves to change the rule for x squared into this rule you know to calculate outputs two times x minus two squared minus three so like i did with the last one i like to do the multiplication first you know this one has two times the square this is one times the square so i'm going to take this rule and multiply it by two all right so i'll write what i'm doing i'm going to multiply the rule by 2 so we have y equals you know 2 times the x squared right i left x squared the rule the way it was and just multiplying the outside by 2 and that causes a vertical change a vertical stretch actually by a factor of two okay so that means you're multiplying all the y-coordinates of the original graph by two all right all the points will be pushed away from the x-axis right stretched away from the x-axis now zero zero will still be at zero zero one one though will be at 1 2. see the stretch pushed away from the x axis negative 1 1 will be at negative 1 2 all right so here's the parabola all now there's y equals you know 2x squared all right okay still has a vertex 0 0. then that's before the square multiplied by the square then i notice added to the square afterwards is a minus 3. all right so i'm going to leave this rule the way it is and then subtract 3 from the rule all right so next next i'll subtract 3 from the rule now give me this this equation here y equals two x squared and then minus three well when you do something to the entire rule it causes a vertical change and if you're subtracting three that will cause a vertical shift you know down three units it's an a that's an a vertical all right vertical shift down three units all right i'll so i'll take my other graph you know this y equals 2x squared graph and move all the points down three so zero zero was the vertex now it's zero negative three you have the point one two move that down three units to one negative one you had negative one two move that down three units to negative one negative one and here's the same graph as that y equals 2x squared right just moved down 3 units there's y equals 2x squared minus 3. all right and then the only the last thing i need to do right because i've got the times 2 in front of the square i've got the minus 3 after the square but the thing being squared is not the same right this has x squared this has x minus 2 squared so i must replace i'll replace the x's replace the x with the quantity x minus two and when you replace x's in a rule you're causing a horizontal transformation not a vertical and when you replace x with x minus h and h is a positive right h is positive 2 here that that's going to move it to the right that many units so this is a horizontal shift right two units so we have our y equals you know 2 times the quantity x and i'm replacing the x in this rule with x minus two and then minus three and this was the f of x right this is the one we want that's this rule here so i'd be done now so i take the graph of y equals two x squared minus three and move all the points to the right two so this is zero negative three over to the right to two negative three which we said was going to be the vertex right that's two negative three back on the page where i changed it you know using completing the square vertex 2 negative 3 this 1 negative 1 to the right 2 to 3 negative 1 and then negative 1 negative 1 move right 2 that'll be at 1 negative 1 and this is again this graph moves to the right 2 units that's the same parabola and there there's the graph of f all right but like i said this is pretty cluttered so i'm going to put this on a separate sheet of paper but just know what happened you know know your transformations right if you're going to be doing this right so on a separate sheet i'll graph the final result okay so we had some uh now we have the vertex at 2 2 negative 3 there's the vertex i'll label that in a second we had the we have the point you know three negative one and the point one negative one on that last graph all right and a it's a parabola opening up and with this vertex there's a graph of f right and i'll again label some of these points here so here's the point three negative one the point two negative three and the point one negative one right and let's just double check all right let's double check you know are these actually points on the graph of this function so let's see i'll just take three negative one here when the input is three when x is three would the output truly be negative one would f of three be negative one so i put 3 in for x you know 3 squared is 9 9 times 2 is 18 so i get 18 for that first term negative 8 times 3 is negative 24. 18 plus negative 24 18 minus 24 is negative 6 and then negative six plus five is indeed negative one so this works and then try it out for you know when x is one with the output be negative one when x is two would the output be negative three it should be if this is truly done correctly and it was all right all right and now uh now that i've got the graphs up all right we're asked to find the domain and the range of these functions all right so i'm going to go back to that first problem all right so here's here is the final graph from the first one you know f of x equals negative 3x squared minus 12x minus 11. remember the domain is the set of all the x coordinates that are used well you can plug in anything you want for x right there's no there's nothing to make you divide by zero there's no square roots or even roots in there you can see on the graph that you know if you pick an x value it's going to go to some point on this graph i mean this graph is going this way forever this way forever all right so you pick any x value and there's a there's a point on the graph with that x value so the domain is all real numbers and i'm going to indicate that with interval notation right that's every number from negative infinity to infinity right all real numbers with parentheses on those and if you're unfamiliar with interval notation please look up a video on it i've i've put up one before on interval notation okay now the range range we're looking at the y-axis right we're looking from the bottom up low to high what y coordinates are used on points on this graph well can you see that above one there's nothing right above y equals one you know there's no points of the graph up here above one so we're seeing that though the y values that are used are one right one is the y coordinate of the vertex and all the values lower than one so from negative infinity you know down forever up until one so from negative infinity up until one those numbers are used as y coordinates now negative infinity is not a number so always parentheses on those to indicate that that's not a solution that's not part of the range one though is part of the range right one is actually a y-coordinate it's actually used on the vertex as a y-coordinate so you put that you know rectangular bracket right you put the rectangular bracket on that to indicate that one is part of the range it is actually a y coordinate all right and then lastly for this graph here right the f of x equals the second example two x squared minus eight x plus five the domain is again all real numbers and so from negative infinity to infinity right there's nothing making me divide by zero nothing i could plug in that would make me divide by zero or take even roots of negatives or anything like that and again this graph is going that way forever and that way forever so no matter what x value you pick there's going to be a point on this graph with that x value so the domain is all real numbers but the range is not all right the range and look at the vertex where the vertex is so that is a y coordinate of negative three and then there's nothing below that right there there aren't any points of the parabola down here below negative three so negative three is the lowest value of the range so i write negative three comma and then there is no highest value right negative three and then all the way all the values on the y axis above negative 3 are going to be the y coordinates of some points on this graph of the parabola so from negative 3 up right from negative 3 to positive infinity and again always parentheses on infinities they're not numbers you know infinity doesn't actually belong to the range but negative 3 does right negative 3 is actually a y coordinate of some point this point here the vertex so i put a rectangular bracket around that you know to indicate that negative 3 is part of the range wonderful all right so i think the thing to take away from this is one you know the graph of a quadratic function is called a parabola and if your lead coefficient is positive it'll open up if your lead coefficient is negative it'll open down you can take a quadratic function in this standard form here right the ax squared plus bx plus c and through use of completing the square you can put it in that vertex form right the a times the quantity x minus 8 squared plus k where h and k are the coordinates of the vertex another thing is that vertex right is going to be the highest or lowest point on the parabola and your transformations right if you start with them start with one graph and you do something to the entire rule you add something to the entire rule or multiply the whole rule by something that's going to cause a vertical transformation you're doing stuff to the y coordinates but if you take a rule and only replace the x's in that rule with something you only replace the inputs that's going to cause a horizontal transformation which means you're only changing the x coordinates of the original graph great and then of course domain is you know set of all the inputs x coordinates range is the set of all the y coordinates and outputs all right as i like to say at the end of these videos do the best you can to learn the material on your own before you go looking for help i found over i've found over the years that people that take the time and the effort to learn something on their own they tend to remember that material better and for longer and they're more proud of it you know boost confidence in your own abilities and all that this is a good thing so read the material of course you know read through the author's examples perhaps even try to work out the solutions to an author's examples before you look at how the author did them and then compare your work with the authors and see if you can learn from some mistakes there obviously do tons of exercises practice problems that's the only way to get better at anything practice right practice makes progress and check for solutions to those if they're if they're made available to you right so you can keep track of your progress see how you're doing and don't just give up on a problem after one or two attempts you know very often in mathematics there are several approaches that will work for any particular problem all right so if after several pr if after several approaches though you're still having trouble i would recommend that you actually read the material again maybe even a third time more slowly more deliberately you know take good notes because if you've been working on problems as you should you probably have a better idea as to what the author wants from you in that section or you have a better grasp you'll you'll know more about what the author finds important about that material because they're asking you about it so when you go back and read again you'll know better what to look out for right you'll know what's more what's considered more most important by the author and you might even catch some things that you didn't catch the first time through and it'll make working on the problems that much easier but if after that you know if after several read-throughs of the material and several attempts of the problems you know you're still having trouble you know it is it is not a sign of weakness whatsoever to go looking for help that is what other people are there for to help you so ask a teacher a tutor a friend you know someone in your class who you know knows the material well and is willing to help look for supplemental materials online there's tons of it out there look for videos online you know videos like this one or you know plenty of other ones out there that are better than this one i assure you they're out there and they're easy to find just keep at it stay persistent keep practicing don't give up and above all and i think this is the hardest thing for most people stay positive you know believe in yourself and your abilities even through times of struggle and failure and mistakes and frustration because it's in those moments of struggle and failure when you have the opportunity to learn the most and learn the most deeply okay you know because if you if you can persevere and persist and and make it through and actually learn how to fix your mistakes and push past your frustrations you'll remember that material far better and far longer than someone who got it right the first time and then breeze through and and and never gave it a second thought again alright so keep persistent keep practicing stay positive and don't give up and i'm sure you'll get it thank you very much for watching
17021	https://www.goodreads.com/book/show/260285.Vogel_s_Textbook_of_Practical_Organic_Chemistry	Jump to ratings and reviews Rate this book Vogel's Textbook of Practical Organic Chemistry A.I. Vogel, A.R. Tatchell, B.S. Furnis 53 ratings 7 reviews Rate this book Most widely used, established and respected reference manual for the organic chemistry laboratory. Incorporates new reactions and techniques now available to the organic chemist. Genres Chemistry Science 1552 pages, Hardcover First published January 1, 1956 18 people are currently reading 174 people want to read About the author A.I. Vogel 4 books1 follower Ratings & Reviews What do you think? Rate this book Friends & Following Create a free account to discover what your friends think of this book! Community Reviews 4.42 53 ratings 7 reviews 5 stars 31 (58%) 4 stars 15 (28%) 3 stars 6 (11%) 2 stars 0 (0%) 1 star 1 (1%) Displaying 1 - 7 of 7 reviews Chimed 1 review September 24, 2008 A great book, but why did they botch the binding so badly. Especially for a book to be referenced in the lab they should have done a better job. I have yet to see a decent looking copy outside of a shop. referance Wiktor Wyszatycki 55 reviews September 24, 2024 Uwielbiam do niej sięgać. Must have w laboratorium a dla mnie w losowych momentach nudy. 1 część składa się z omówienia sprzętu oraz metod oczyszczania 2 część składa się z ułożonych tematycznie reakcji związków organicznych oraz syntezami wytłumaczonymi krok po kroku (minus jest jeden, niektóre gramatury substratów są nieadekwatne do opisanych wydajności ale to jak w większości książek do preparatyki) Peter 306 reviews 105 followers March 4, 2023 Definitely the best handbook for the practicing organic chemist. I was a medicinal chemist before I retired and I used this practical textbook on almost a daily basis. The technical stuff in it is just superb. reviewed science Uleel 23 reviews 2 followers Read August 5, 2007 untunglah masa membaca buku ini dah berlalu...penuh perjuangan eui Ayman 2 reviews 1 follower May 9, 2011 I advice any chemist to take it as a referance Peter 1 review September 1, 2014 One of the best organic chemistry recourses available. chemistry Sara Waseem 1 review Want to read November 6, 2014 vogel's textbook of practical organic chemistry This entire review has been hidden because of spoilers. Displaying 1 - 7 of 7 reviews Join the discussion a quote a discussion a question Can't find what you're looking for? Get help and learn more about the design. Help center
17022	https://en.wikipedia.org/wiki/Disjoint_union_(topology)	Jump to content Disjoint union (topology) Français 日本語 Edit links From Wikipedia, the free encyclopedia Mathematical term \| \| \| --- \| \| \| This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Disjoint union" topology – news · newspapers · books · scholar · JSTOR (May 2025) (Learn how and when to remove this message) \| In general topology and related areas of mathematics, the disjoint union (also called the direct sum, free union, free sum, topological sum, or coproduct) of a family of topological spaces is a space formed by equipping the disjoint union of the underlying sets with a natural topology called the disjoint union topology. Roughly speaking, in the disjoint union the given spaces are considered as part of a single new space where each looks as it would alone and they are isolated from each other. The name coproduct originates from the fact that the disjoint union is the categorical dual of the product space construction. Definition [edit] Let {Xi : i ∈ I} be a family of topological spaces indexed by I. Let be the disjoint union of the underlying sets. For each i in I, let be the canonical injection (defined by ). The disjoint union topology on X is defined as the finest topology on X for which all the canonical injections are continuous (i.e.: it is the final topology on X induced by the canonical injections). Explicitly, the disjoint union topology can be described as follows. A subset U of X is open in X if and only if its preimage is open in Xi for each i ∈ I. Yet another formulation is that a subset V of X is open relative to X iff its intersection with Xi is open relative to Xi for each i. Properties [edit] The disjoint union space X, together with the canonical injections, can be characterized by the following universal property: If Y is a topological space, and fi : Xi → Y is a continuous map for each i ∈ I, then there exists precisely one continuous map f : X → Y such that the following set of diagrams commute: This shows that the disjoint union is the coproduct in the category of topological spaces. It follows from the above universal property that a map f : X → Y is continuous iff fi = f o φi is continuous for all i in I. In addition to being continuous, the canonical injections φi : Xi → X are open and closed maps. It follows that the injections are topological embeddings so that each Xi may be canonically thought of as a subspace of X. Examples [edit] If each Xi is homeomorphic to a fixed space A, then the disjoint union X is homeomorphic to the product space A × I where I has the discrete topology. Preservation of topological properties [edit] Every disjoint union of discrete spaces is discrete Separation Every disjoint union of T0 spaces is T0 Every disjoint union of T1 spaces is T1 Every disjoint union of Hausdorff spaces is Hausdorff Connectedness The disjoint union of two or more nonempty topological spaces is disconnected See also [edit] product topology, the dual construction subspace topology and its dual quotient topology topological union, a generalization to the case where the pieces are not disjoint References [edit] ^ "disjoint union topological space in nLab". ncatlab.org. Retrieved 5 May 2025. Retrieved from " Category: General topology Hidden categories: Articles with short description Short description is different from Wikidata Articles needing additional references from May 2025 All articles needing additional references
17023	https://sites.und.edu/timothy.prescott/apex/web/apex.Ch13.S8.html	13 Functions of Several Variables13.7 Tangent Lines, Normal Lines, and Tangent Planes13.9 Lagrange Multipliers 13.8 Extreme Values Given a function , we are often interested in points where takes on the largest or smallest values. For instance, if represents a cost function, we would likely want to know what values minimize the cost. If represents the ratio of a volume to surface area, we would likely want to know where is greatest. This leads to the following definition. Definition 13.8.1 Relative and Absolute Extrema ¶ Let be defined on a set containing the point . (a) If there is an open disk so that contains which contains and for all in , then has a relative maximum at ; if for all in , then has a relative minimum at . 2. (b) If for all in , then has an absolute maximum at ; if for all in , then has an absolute minimum at . 3. (c) If has a relative maximum or minimum at , then has a relative extremum at ; if has an absolute maximum or minimum at , then has an absolute extremum at . If has a relative or absolute maximum at , it means every curve on the surface of through will also have a relative or absolute maximum at . Recalling what we learned in Section 3.1, the slopes of the tangent lines to these curves at must be 0 or undefined. Since directional derivatives are computed using and , we are led to the following definition and theorem. Definition 13.8.2 Critical Point ¶ Let be continuous on an open set . A critical point of is a point in such that • and , or • or is undefined. Theorem 13.8.1 Critical Points and Relative Extrema ¶ Let be defined on an open set containing . If has a relative extrema at , then is a critical point of . Therefore, to find relative extrema, we find the critical points of and determine which correspond to relative maxima, relative minima, or neither. Watch the video: Local Maximum and Minimum Values / Function of Two Variables from The following examples demonstrate this process. Example 13.8.1 Finding critical points and relative extrema ¶ Let . Find the relative extrema of . SolutionWe start by computing the partial derivatives of : ††margin: (fullscreen) Figure 13.8.1: The surface in Example 13.8.1 with its absolute minimum indicated.¶ \| \| \| \| --- \| \| \| \| Each is never undefined. A critical point occurs when and are simultaneously 0, leading us to solve the following system of linear equations: \| \| \| \| --- \| \| \| \| This solution to this system is , . (Check that at , both and are 0.) The graph in Figure 13.8.1 shows along with this critical point. It is clear from the graph that this is a relative minimum; further consideration of the function shows that this is actually the absolute minimum. Example 13.8.2 Finding critical points and relative extrema ¶ Let . Find the relative extrema of . SolutionWe start by computing the partial derivatives of : \| \| \| \| --- \| \| \| \| It is clear that when & , and that when & . At , both and are not , but rather undefined. The point is still a critical point, though, because the partial derivatives are undefined. This is the only critical point of . ††margin: (fullscreen) Figure 13.8.2: The surface in Example 13.8.2 with its absolute maximum indicated.¶ The surface of is graphed in Figure 13.8.2 along with the point . The graph shows that this point is the absolute maximum of . In each of the previous two examples, we found a critical point of and then determined whether or not it was a relative (or absolute) maximum or minimum by graphing. It would be nice to be able to determine whether a critical point corresponded to a max or a min without a graph. Before we develop such a test, we do one more example that sheds more light on the issues our test needs to consider. Example 13.8.3 Finding critical points and relative extrema ¶ Let . Find the relative extrema of . SolutionOnce again we start by finding the partial derivatives of : \| \| \| \| --- \| \| \| \| Each is always defined. Setting each equal to 0 and solving for and , we find \| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| \| \| We have two critical points: and . To determine if they correspond to a relative maximum or minimum, we consider the graph of in Figure 13.8.3. ††margin: (fullscreen) Figure 13.8.3: The surface in Example 13.8.3 with both critical points marked.¶ The critical point clearly corresponds to a relative maximum. However, the critical point at is neither a maximum nor a minimum, displaying a different, interesting characteristic. If one walks parallel to the -axis towards this critical point, then this point becomes a relative maximum along this path. But if one walks towards this point parallel to the -axis, this point becomes a relative minimum along this path. A point that seems to act as both a max and a min is a saddle point. A formal definition follows. Definition 13.8.3 Saddle Point ¶ Let be in the domain of where and at . We say is a saddle point of if, for every open disk containing , there are points and in such that and . At a saddle point, the instantaneous rate of change in all directions is 0 and there are points nearby with -values both less than and greater than the -value of the saddle point. Before Example 13.8.3 we mentioned the need for a test to differentiate between relative maxima and minima. We now recognize that our test also needs to account for saddle points. To do so, we consider the second partial derivatives of . Recall that with single variable functions, such as , if , then is concave up at , and if , then has a relative minimum at . (We called this the Second Derivative Test.) Note that at a saddle point, it seems the graph is “both” concave up and concave down, depending on which direction you are considering. It would be nice if the following were true: \| \| \| \| --- \| and \| \| relative minimum \| \| and \| \| relative maximum \| \| and have opposite signs \| \| saddle point. \| However, this is not the case. Functions exist where and are both positive but a saddle point still exists. In such a case, while the concavity in the -direction is up (i.e., ) and the concavity in the -direction is also up (i.e., ), the concavity switches somewhere in between the - and -directions. To account for this, consider . Since and are equal when continuous (refer back to Theorem 13.3.1), we can rewrite this as . Then can be used to test whether the concavity at a point changes depending on direction. If , the concavity does not switch (i.e., at that point, the graph is concave up or down in all directions). If , the concavity does switch. If , our test fails to determine whether concavity switches or not. We state the use of in the following theorem. Theorem 13.8.2 Second Derivative Test ¶ Let be defined on an open set containing a critical point where all second order derivatives of are continuous at . Define \| \| \| \| --- \| \| \| \| (a) If and , then is a relative minimum of . 2. (b) If and , then is a relative maximum of . 3. (c) If , then is a saddle point of . 4. (d) If , the test is inconclusive. Proof Let be a unit vector. Then at the critical point , . This means that along the line , is a critical point that is a maximum or minimum according to the sign of . Now, \| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| because and are continuous and therefore equal. Suppose now that . Then we must have , and we can complete the square to see that \| \| \| \| --- \| \| \| \| Because we assumed , everything in the brackets is positive, and always has the same sign as . This shows parts 1 and 2. If , our task is easier because we only need to find two different that give opposite signs. If , let and we can choose \| \| \| \| --- \| \| \| \| Similarly, if , let and we can choose \| \| \| \| --- \| \| \| \| Finally, if , then has opposite signs for the vectors and . ∎ We first practice using this test with the function in the previous example, where we visually determined we had a relative maximum and a saddle point. Example 13.8.4 Using the Second Derivative Test ¶ Let as in Example 13.8.3. Determine whether the function has a relative minimum, maximum, or saddle point at each critical point. SolutionWe determined previously that the critical points of are and . To use the Second Derivative Test, we must find the second partial derivatives of : \| \| \| \| --- \| \| \| \| Thus . At : , and . By the Second Derivative Test, has a relative maximum at . At : . The Second Derivative Test states that has a saddle point at . The Second Derivative Test confirmed what we determined visually. Example 13.8.5 Using the Second Derivative Test ¶ Find the relative extrema of . SolutionWe start by finding the first and second partial derivatives of : \| \| \| \| \| \| \| --- --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| We find the critical points by finding where and are simultaneously 0 (they are both never undefined). Setting , we have: \| \| \| \| --- \| \| \| \| This implies that for , either or . Assume then consider : \| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Thus if , we have either or , giving two critical points: and . Going back to , now assume , i.e., that , then consider : \| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Thus if , giving the critical point . With , we apply the Second Derivative Test to each critical point. At , , so is a saddle point. At , , so is also a saddle point. At , and , so is a relative minimum. ††margin: (fullscreen) Figure 13.8.4: Graphing from Example 13.8.5 and its relative extrema.¶ Figure 13.8.4 shows a graph of and the three critical points. Note how this function does not vary much near the critical points — that is, visually it is difficult to determine whether a point is a saddle point or relative minimum (or even a critical point at all!). This is one reason why the Second Derivative Test is so important to have. Constrained Optimization When optimizing functions of one variable such as , we made use of Theorem 3.1.1, the Extreme Value Theorem, that said that over a closed interval , a continuous function has both a maximum and minimum value. To find these maximum and minimum values, we evaluated at all critical points in the interval, as well as at the endpoints (the “boundary”) of the interval. A similar theorem and procedure applies to functions of two variables. A continuous function over a closed set also attains a maximum and minimum value (see the following theorem). We can find these values by evaluating the function at the critical points in the set and over the boundary of the set. After formally stating this extreme value theorem, we give examples. Theorem 13.8.3 Extreme Value Theorem ¶ Let be a continuous function on a closed, bounded set . Then has an absolute maximum and an absolute minimum value on . ††margin: (fullscreen) (a) (b) Figure 13.8.5: Plotting the surface of along with the restricted domain .¶ Example 13.8.6 Finding extrema on a closed set ¶ Let and let be the triangle with vertices , and . Find the maximum and minimum values of on . SolutionIt can help to see a graph of along with the set . In Figure 13.8.5(a) the triangle defining is shown in the - plane in a dashed line. Above it is the surface of ; we are only concerned with the portion of enclosed by the “triangle” on its surface. We begin by finding the critical points of . With and , we find only one critical point, at . We now find the maximum and minimum values that attains along the boundary of , that is, along the edges of the triangle. In Figure 13.8.5(b) we see the triangle sketched in the plane with the equations of the lines forming its edges labeled. Start with the bottom edge, along the line . If is , then on the surface, we are considering points ; that is, our function reduces to . We want to maximize/minimize on the interval . To do so, we evaluate at its critical points and at the endpoints. The critical points of are found by setting its derivative equal to 0: \| \| \| \| --- \| \| \| \| so that we will need to evaluate at the points , , and . We need to do this process twice more, for the other two edges of the triangle. Along the left edge, along the line , we substitute in for in : \| \| \| \| --- \| \| \| \| We want the maximum and minimum values of on the interval , so we evaluate at its critical points and the endpoints of the interval. We find the critical points: \| \| \| \| --- \| \| \| \| so that we will need to evaluate at the points , , and . Finally, we evaluate along the right edge of the triangle, where . \| \| \| \| --- \| \| \| \| The critical points of are: \| \| \| \| --- \| \| \| \| so that we will need to evaluate at the points , , and . ††margin: (fullscreen) Figure 13.8.6: The surface of along with important points along the boundary of and the interior.¶ We now evaluate at a total of 7 different places, all shown in Figure 13.8.6. \| \| \| \| --- \| \| \| \| Of all the -values found, the maximum is , found at ; the minimum is 1, found at . This portion of the text is entitled “Constrained Optimization” because we want to optimize a function (i.e., find its maximum and/or minimum values) subject to a constraint — some limit to what values the function can attain. In the previous example, we constrained ourselves by considering a function only within the boundary of a triangle. This was largely arbitrary; the function and the boundary were chosen just as an example, with no real “meaning” behind the function or the chosen constraint. However, solving constrained optimization problems is a very important topic in applied mathematics. The techniques developed here are the basis for solving larger problems, where more than two variables are involved. We illustrate the technique once more with a classic problem. Example 13.8.7 Constrained Optimization ¶ The U.S. Postal Service states that the girth+length of Standard Post Package must not exceed 130”. Given a rectangular box, the “length” is the longest side, and the “girth” is twice the width+height. Given a rectangular box where the width and height are equal, what are the dimensions of the box that give the maximum volume subject to the constraint of the size of a Standard Post Package? SolutionLet , , and denote the width, height, and length of a rectangular box; we assume here that . The girth is then . The volume of the box is . We wish to maximize this volume subject to the constraint , or . (Common sense also indicates that , so that we don’t need to check the boundary where either is zero.) We begin by finding the critical points of . We find that and ; these are simultaneously 0 only when . These give a volume of 0, so we can ignore these critical points. We now consider the volume along the constraint Along this line, we have: \| \| \| \| --- \| \| \| \| The constraint is applicable on the -interval as indicated in the figure. Thus we want to maximize on . Finding the critical points of , we take the derivative and set it equal to 0: \| \| \| \| --- \| \| \| \| We found two critical points: when and when . We again ignore the solution; the maximum volume, subject to the constraint, comes at , This gives a volume of in3. ††margin: (fullscreen) Figure 13.8.7: Graphing the volume of a box with girth and length , subject to a size constraint.¶ The volume function is shown in Figure 13.8.7 along with the constraint . As done previously, the constraint is drawn dashed in the - plane and also along the surface of the function. The point where the volume is maximized is indicated. It is hard to overemphasize the importance of optimization. In “the real world,” we routinely seek to make something better. By expressing the something as a mathematical function, “making something better” means “optimize some function.” The techniques shown here are only the beginning of an incredibly important field. Many functions that we seek to optimize are incredibly complex, making the step of “find the gradient and set it equal to ” highly nontrivial. Mastery of the principles here are key to being able to tackle these more complicated problems. Exercises 13.8 Terms and Concepts 1. T/F: Theorem 13.8.1 states that if has a critical point at , then has a relative extrema at . 2. 2. T/F: A point is a critical point of if and are both 0 at . 3. 3. T/F: A point is a critical point of if or are undefined at . 4. 4. Explain what it means to “solve a constrained optimization” problem. Problems In Exercises 5–18., find the critical points of the given function. Use the Second Derivative Test to determine if each critical point corresponds to a relative maximum, minimum, or saddle point. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. In Exercises 19–24., find the absolute maximum and minimum of the function subject to the given constraint. 19. , constrained to the triangle with vertices , and . 2. 20. , constrained to the region bounded by and . 3. 21. , constrained to the region bounded by the circle . 4. 22. , constrained to the region bounded by the parabola and the line . 5. 23. , constrained to the square with vertices , , , and . 6. 24. , constrained to the region bounded by the circle . 25. Find the dimensions of the box without a top that has a volume of and the least possible surface area. 2. 26. Show that there are boxes without a top that have a volume of and arbitrarily large surface area.
17024	https://openstax.org/books/university-physics-volume-1/pages/10-4-moment-of-inertia-and-rotational-kinetic-energy	10.4 Moment of Inertia and Rotational Kinetic Energy - University Physics Volume 1 \| OpenStax This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising purposes. Privacy Notice Customize Reject All Accept All Customize Consent Preferences We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below. The cookies that are categorized as "Necessary" are stored on your browser as they are essential for enabling the basic functionalities of the site. ...Show more For more information on how Google's third-party cookies operate and handle your data, see:Google Privacy Policy Necessary Always Active Necessary cookies are required to enable the basic features of this site, such as providing secure log-in or adjusting your consent preferences. These cookies do not store any personally identifiable data. Cookie oxdid Duration 1 year 1 month 4 days Description OpenStax Accounts cookie for authentication Cookie campaignId Duration Never Expires Description Required to provide OpenStax services Cookie __cf_bm Duration 1 hour Description This cookie, set by Cloudflare, is used to support Cloudflare Bot Management. Cookie CookieConsentPolicy Duration 1 year Description Cookie Consent from Salesforce Cookie LSKey-c$CookieConsentPolicy Duration 1 year Description Cookie Consent from Salesforce Cookie renderCtx Duration session Description This cookie is used for tracking community context state. Cookie pctrk Duration 1 year Description Customer support Cookie _accounts_session_production Duration 1 year 1 month 4 days Description Cookies that are required for authentication and necessary OpenStax functions. Cookie nudge_study_guides_page_counter Duration 1 year 1 month 4 days Description Product analytics Cookie _dd_s Duration 15 minutes Description Zapier cookies that are used for Customer Support services. Cookie ak_bmsc Duration 2 hours Description This cookie is used by Akamai to optimize site security by distinguishing between humans and bots Cookie PHPSESSID Duration session Description This cookie is native to PHP applications. The cookie stores and identifies a user's unique session ID to manage user sessions on the website. The cookie is a session cookie and will be deleted when all the browser windows are closed. Cookie m Duration 1 year 1 month 4 days Description Stripe sets this cookie for fraud prevention purposes. It identifies the device used to access the website, allowing the website to be formatted accordingly. Cookie BrowserId Duration 1 year Description Sale Force sets this cookie to log browser sessions and visits for internal-only product analytics. Cookie ph_phc_bnZwQPxzoC7WnmjFNOUQpcKsaDVg8TwnyoNzbClpIsD_posthog Duration 1 year Description Privacy-focused platform cookie Cookie cookieyes-consent Duration 1 year Description CookieYes sets this cookie to remember users' consent preferences so that their preferences are respected on subsequent visits to this site. It does not collect or store any personal information about the site visitors. Cookie _cfuvid Duration session Description Calendly sets this cookie to track users across sessions to optimize user experience by maintaining session consistency and providing personalized services Cookie dmn_chk_ Duration Less than a minute Description This cookie is set to track user activity across the website. Cookie cookiesession1 Duration 1 year Description This cookie is set by the Fortinet firewall. This cookie is used for protecting the website from abuse. Functional [x] Functional cookies help perform certain functionalities like sharing the content of the website on social media platforms, collecting feedback, and other third-party features. Cookie session Duration session Description Salesforce session cookie. We use Salesforce to drive our support services to users. Cookie projectSessionId Duration session Description Optional AI-based customer support cookie Cookie yt-remote-device-id Duration Never Expires Description YouTube sets this cookie to store the user's video preferences using embedded YouTube videos. Cookie ytidb::LAST_RESULT_ENTRY_KEY Duration Never Expires Description The cookie ytidb::LAST_RESULT_ENTRY_KEY is used by YouTube to store the last search result entry that was clicked by the user. This information is used to improve the user experience by providing more relevant search results in the future. Cookie yt-remote-connected-devices Duration Never Expires Description YouTube sets this cookie to store the user's video preferences using embedded YouTube videos. Cookie yt-remote-session-app Duration session Description The yt-remote-session-app cookie is used by YouTube to store user preferences and information about the interface of the embedded YouTube video player. Cookie yt-remote-cast-installed Duration session Description The yt-remote-cast-installed cookie is used to store the user's video player preferences using embedded YouTube video. Cookie yt-remote-session-name Duration session Description The yt-remote-session-name cookie is used by YouTube to store the user's video player preferences using embedded YouTube video. Cookie yt-remote-fast-check-period Duration session Description The yt-remote-fast-check-period cookie is used by YouTube to store the user's video player preferences for embedded YouTube videos. Cookie yt-remote-cast-available Duration session Description The yt-remote-cast-available cookie is used to store the user's preferences regarding whether casting is available on their YouTube video player. Analytics [x] Analytical cookies are used to understand how visitors interact with the website. These cookies help provide information on metrics such as the number of visitors, bounce rate, traffic source, etc. Cookie hjSession Duration 1 hour Description Hotjar sets this cookie to ensure data from subsequent visits to the same site is attributed to the same user ID, which persists in the Hotjar User ID, which is unique to that site. Cookie visitor_id Duration 9 months 7 days Description Pardot sets this cookie to store a unique user ID. Cookie visitor_id-hash Duration 9 months 7 days Description Pardot sets this cookie to store a unique user ID. Cookie _gcl_au Duration 3 months Description Google Tag Manager sets the cookie to experiment advertisement efficiency of websites using their services. Cookie _ga Duration 1 year 1 month 4 days Description Google Analytics sets this cookie to calculate visitor, session and campaign data and track site usage for the site's analytics report. The cookie stores information anonymously and assigns a randomly generated number to recognise unique visitors. Cookie _gid Duration 1 day Description Google Analytics sets this cookie to store information on how visitors use a website while also creating an analytics report of the website's performance. Some of the collected data includes the number of visitors, their source, and the pages they visit anonymously. Cookie _fbp Duration 3 months Description Facebook sets this cookie to display advertisements when either on Facebook or on a digital platform powered by Facebook advertising after visiting the website. Cookie ga Duration 1 year 1 month 4 days Description Google Analytics sets this cookie to store and count page views. Cookie pardot Duration past Description The pardot cookie is set while the visitor is logged in as a Pardot user. The cookie indicates an active session and is not used for tracking. Cookie pi_pageview_count Duration Never Expires Description Marketing automation tracking cookie Cookie pulse_insights_udid Duration Never Expires Description User surveys Cookie pi_visit_track Duration Never Expires Description Marketing cookie Cookie pi_visit_count Duration Never Expires Description Marketing cookie Cookie cebs Duration session Description Crazyegg sets this cookie to trace the current user session internally. Cookie gat_gtag_UA Duration 1 minute Description Google Analytics sets this cookie to store a unique user ID. Cookie vuid Duration 1 year 1 month 4 days Description Vimeo installs this cookie to collect tracking information by setting a unique ID to embed videos on the website. Performance [x] Performance cookies are used to understand and analyze the key performance indexes of the website which helps in delivering a better user experience for the visitors. Cookie hjSessionUser Duration 1 year Description Hotjar sets this cookie to ensure data from subsequent visits to the same site is attributed to the same user ID, which persists in the Hotjar User ID, which is unique to that site. Advertisement [x] Advertisement cookies are used to provide visitors with customized advertisements based on the pages you visited previously and to analyze the effectiveness of the ad campaigns. Cookie test_cookie Duration 15 minutes Description doubleclick.net sets this cookie to determine if the user's browser supports cookies. Cookie YSC Duration session Description Youtube sets this cookie to track the views of embedded videos on Youtube pages. Cookie VISITOR_INFO1_LIVE Duration 6 months Description YouTube sets this cookie to measure bandwidth, determining whether the user gets the new or old player interface. Cookie VISITOR_PRIVACY_METADATA Duration 6 months Description YouTube sets this cookie to store the user's cookie consent state for the current domain. Cookie IDE Duration 1 year 24 days Description Google DoubleClick IDE cookies store information about how the user uses the website to present them with relevant ads according to the user profile. Cookie yt.innertube::requests Duration Never Expires Description YouTube sets this cookie to register a unique ID to store data on what videos from YouTube the user has seen. Cookie yt.innertube::nextId Duration Never Expires Description YouTube sets this cookie to register a unique ID to store data on what videos from YouTube the user has seen. Uncategorized [x] Other uncategorized cookies are those that are being analyzed and have not been classified into a category as yet. Cookie donation-identifier Duration 1 year Description Description is currently not available. Cookie abtest-identifier Duration 1 year Description Description is currently not available. Cookie __Secure-ROLLOUT_TOKEN Duration 6 months Description Description is currently not available. Cookie _ce.s Duration 1 year Description Description is currently not available. Cookie _ce.clock_data Duration 1 day Description Description is currently not available. Cookie cebsp_ Duration session Description Description is currently not available. Cookie lpv218812 Duration 1 hour Description Description is currently not available. Reject All Save My Preferences Accept All Skip to ContentGo to accessibility pageKeyboard shortcuts menu Log in University Physics Volume 1 10.4 Moment of Inertia and Rotational Kinetic Energy University Physics Volume 110.4 Moment of Inertia and Rotational Kinetic Energy Contents Contents Highlights Table of contents Preface Mechanics 1 Units and Measurement 2 Vectors 3 Motion Along a Straight Line 4 Motion in Two and Three Dimensions 5 Newton's Laws of Motion 6 Applications of Newton's Laws 7 Work and Kinetic Energy 8 Potential Energy and Conservation of Energy 9 Linear Momentum and Collisions 10 Fixed-Axis Rotation Introduction 10.1 Rotational Variables 10.2 Rotation with Constant Angular Acceleration 10.3 Relating Angular and Translational Quantities 10.4 Moment of Inertia and Rotational Kinetic Energy 10.5 Calculating Moments of Inertia 10.6 Torque 10.7 Newton’s Second Law for Rotation 10.8 Work and Power for Rotational Motion Chapter Review 11 Angular Momentum 12 Static Equilibrium and Elasticity 13 Gravitation 14 Fluid Mechanics Waves and Acoustics A \| Units B \| Conversion Factors C \| Fundamental Constants D \| Astronomical Data E \| Mathematical Formulas F \| Chemistry G \| The Greek Alphabet Answer Key Index Search for key terms or text. Close Learning Objectives By the end of this section, you will be able to: Describe the differences between rotational and translational kinetic energy Define the physical concept of moment of inertia in terms of the mass distribution from the rotational axis Explain how the moment of inertia of rigid bodies affects their rotational kinetic energy Use conservation of mechanical energy to analyze systems undergoing both rotation and translation Calculate the angular velocity of a rotating system when there are energy losses due to nonconservative forces So far in this chapter, we have been working with rotational kinematics: the description of motion for a rotating rigid body with a fixed axis of rotation. In this section, we define two new quantities that are helpful for analyzing properties of rotating objects: moment of inertia and rotational kinetic energy. With these properties defined, we will have two important tools we need for analyzing rotational dynamics. Rotational Kinetic Energy Any moving object has kinetic energy. We know how to calculate this for a body undergoing translational motion, but how about for a rigid body undergoing rotation? This might seem complicated because each point on the rigid body has a different velocity. However, we can make use of angular velocity—which is the same for the entire rigid body—to express the kinetic energy for a rotating object. Figure 10.17 shows an example of a very energetic rotating body: an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are generated as the grindstone does its work. This system has considerable energy, some of it in the form of heat, light, sound, and vibration. However, most of this energy is in the form of rotational kinetic energy. Figure 10.17 The rotational kinetic energy of the grindstone is converted to heat, light, sound, and vibration. (credit: Zachary David Bell, US Navy) Energy in rotational motion is not a new form of energy; rather, it is the energy associated with rotational motion, the same as kinetic energy in translational motion. However, because kinetic energy is given by K=1 2 m v 2 K=1 2 m v 2 K=1 2 m v 2, and velocity is a quantity that is different for every point on a rotating body about an axis, it makes sense to find a way to write kinetic energy in terms of the variable ω ω ω, which is the same for all points on a rigid rotating body. For a single particle rotating around a fixed axis, this is straightforward to calculate. We can relate the angular velocity to the magnitude of the translational velocity using the relation v t=ω r v t=ω r v t=ω r, where r is the distance of the particle from the axis of rotation and v t v t v t is its tangential speed. Substituting into the equation for kinetic energy, we find K=1 2 m v 2 t=1 2 m(ω r)2=1 2(m r 2)ω 2.K=1 2 m v t 2=1 2 m(ω r)2=1 2(m r 2)ω 2.K=1 2 m v t 2=1 2 m(ω r)2=1 2(m r 2)ω 2. In the case of a rigid rotating body, we can divide up any body into a large number of smaller masses, each with a mass m j m j m j and distance to the axis of rotation r j r j r j, such that the total mass of the body is equal to the sum of the individual masses: M=∑j m j M=∑j m j M=∑j m j. Each smaller mass has tangential speed v j v j v j, where we have dropped the subscript t for the moment. The total kinetic energy of the rigid rotating body is K=∑j 1 2 m j v 2 j=∑j 1 2 m j(r j ω j)2 K=∑j 1 2 m j v j 2=∑j 1 2 m j(r j ω j)2 K=∑j 1 2 m j v j 2=∑j 1 2 m j(r j ω j)2 and since ω j=ω ω j=ω ω j=ω for all masses, K=1 2(∑j m j r 2 j)ω 2.K=1 2(∑j m j r j 2)ω 2.K=1 2(∑j m j r j 2)ω 2. 10.16 The units of Equation 10.16 are joules (J). The equation in this form is complete, but awkward; we need to find a way to generalize it. Moment of Inertia If we compare Equation 10.16 to the way we wrote kinetic energy in Work and Kinetic Energy, (1 2 m v 2)(1 2 m v 2)(1 2 m v 2), this suggests we have a new rotational variable to add to our list of our relations between rotational and translational variables. The quantity ∑j m j r 2 j∑j m j r j 2∑j m j r j 2 is the counterpart for mass in the equation for rotational kinetic energy. This is an important new term for rotational motion. This quantity is called the moment of inertia I, with units of kg⋅m 2 kg·m 2 kg·m 2: I=∑j m j r 2 j.I=∑j m j r j 2.I=∑j m j r j 2. 10.17 For now, we leave the expression in summation form, representing the moment of inertia of a system of point particles rotating about a fixed axis. We note that the moment of inertia of a single point particle about a fixed axis is simply m r 2 m r 2 m r 2, with r being the distance from the point particle to the axis of rotation. In the next section, we explore the integral form of this equation, which can be used to calculate the moment of inertia of some regular-shaped rigid bodies. The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. Similarly, the greater the moment of inertia of a rigid body or system of particles, the greater is its resistance to change in angular velocity about a fixed axis of rotation. It is interesting to see how the moment of inertia varies with r, the distance to the axis of rotation of the mass particles in Equation 10.17. Rigid bodies and systems of particles with more mass concentrated at a greater distance from the axis of rotation have greater moments of inertia than bodies and systems of the same mass, but concentrated near the axis of rotation. In this way, we can see that a hollow cylinder has more rotational inertia than a solid cylinder of the same mass when rotating about an axis through the center. Substituting Equation 10.17 into Equation 10.16, the expression for the kinetic energy of a rotating rigid body becomes K=1 2 I ω 2.K=1 2 I ω 2.K=1 2 I ω 2. 10.18 We see from this equation that the kinetic energy of a rotating rigid body is directly proportional to the moment of inertia and the square of the angular velocity. This is exploited in flywheel energy-storage devices, which are designed to store large amounts of rotational kinetic energy. Many carmakers are now testing flywheel energy storage devices in their automobiles, such as the flywheel, or kinetic energy recovery system, shown in Figure 10.18. Figure 10.18 A KERS (kinetic energy recovery system) flywheel used in cars. (credit: “cmonville”/Flickr) The rotational and translational quantities for kinetic energy and inertia are summarized in Table 10.4. The relationship column is not included because a constant doesn’t exist by which we could multiply the rotational quantity to get the translational quantity, as can be done for the variables in Table 10.3. \| Rotational \| Translational \| --- \| \| I=∑j m j r 2 j I=∑j m j r j 2 I=∑j m j r j 2 \| m m m \| \| K=1 2 I ω 2 K=1 2 I ω 2 K=1 2 I ω 2 \| K=1 2 m v 2 K=1 2 m v 2 K=1 2 m v 2 \| Table 10.4 Rotational and Translational Kinetic Energies and Inertia Example 10.8 Moment of Inertia of a System of Particles Six small washers are spaced 10 cm apart on a rod of negligible mass and 0.5 m in length. The mass of each washer is 20 g. The rod rotates about an axis located at 25 cm, as shown in Figure 10.19. (a) What is the moment of inertia of the system? (b) If the two washers closest to the axis are removed, what is the moment of inertia of the remaining four washers? (c) If the system with six washers rotates at 5 rev/s, what is its rotational kinetic energy? Figure 10.19 Six washers are spaced 10 cm apart on a rod of negligible mass and rotating about a vertical axis. Strategy We use the definition for moment of inertia for a system of particles and perform the summation to evaluate this quantity. The masses are all the same so we can pull that quantity in front of the summation symbol. We do a similar calculation. We insert the result from (a) into the expression for rotational kinetic energy. Solution I=∑j m j r 2 j=(0.02 kg)(2×(0.25 m)2+2×(0.15 m)2+2×(0.05 m)2)=0.0035 kg⋅m 2 I=∑j m j r j 2=(0.02 kg)(2×(0.25 m)2+2×(0.15 m)2+2×(0.05 m)2)=0.0035 kg·m 2 I=∑j m j r j 2=(0.02 kg)(2×(0.25 m)2+2×(0.15 m)2+2×(0.05 m)2)=0.0035 kg·m 2. I=∑j m j r 2 j=(0.02 kg)(2×(0.25 m)2+2×(0.15 m)2)=0.0034 kg⋅m 2 I=∑j m j r j 2=(0.02 kg)(2×(0.25 m)2+2×(0.15 m)2)=0.0034 kg·m 2 I=∑j m j r j 2=(0.02 kg)(2×(0.25 m)2+2×(0.15 m)2)=0.0034 kg·m 2. K=1 2 I ω 2=1 2(0.0035 kg⋅m 2)(5.0×2 π rad/s)2=1.73 J K=1 2 I ω 2=1 2(0.0035 kg·m 2)(5.0×2 π rad/s)2=1.73 J K=1 2 I ω 2=1 2(0.0035 kg·m 2)(5.0×2 π rad/s)2=1.73 J. Significance We can see the individual contributions to the moment of inertia. The masses close to the axis of rotation have a very small contribution. When we removed them, it had a very small effect on the moment of inertia. In the next section, we generalize the summation equation for point particles and develop a method to calculate moments of inertia for rigid bodies. For now, though, Figure 10.20 gives values of moment of inertia for common object shapes around specified axes. Figure 10.20 Moment of inertia for common shapes of objects. Applying Rotational Kinetic Energy Now let’s apply the ideas of rotational kinetic energy and the moment of inertia table to get a feeling for the energy associated with a few rotating objects. The following examples will also help get you comfortable using these equations. First, let’s look at a general problem-solving strategy for rotational energy. Problem-Solving Strategy Rotational Energy Determine that energy or work is involved in the rotation. Determine the system of interest. A sketch usually helps. Analyze the situation to determine the types of work and energy involved. If there are no losses of energy due to friction and other nonconservative forces, mechanical energy is conserved, that is, K i+U i=K f+U f K i+U i=K f+U f K i+U i=K f+U f. If nonconservative forces are present, mechanical energy is not conserved, and other forms of energy, such as heat and light, may enter or leave the system. Determine what they are and calculate them as necessary. Eliminate terms wherever possible to simplify the algebra. Evaluate the numerical solution to see if it makes sense in the physical situation presented in the wording of the problem. Example 10.9 Calculating Helicopter Energies A typical small rescue helicopter has four blades: Each is 4.00 m long and has a mass of 50.0 kg (Figure 10.21). The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. Figure 10.21(a) Sketch of a four-blade helicopter. (b) A water rescue operation featuring a helicopter from the Auckland Westpac Rescue Helicopter Service. (credit b: modification of work by “111 Emergency”/Flickr) Strategy Rotational and translational kinetic energies can be calculated from their definitions. The wording of the problem gives all the necessary constants to evaluate the expressions for the rotational and translational kinetic energies. Solution The rotational kinetic energy is K=1 2 I ω 2.K=1 2 I ω 2.K=1 2 I ω 2. We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find K. The angular velocity ω ω ω is ω=300 rev 1.00 min 2 π rad 1 rev 1.00 min 60.0 s=31.4 rad s.ω=300 rev 1.00 min 2 π rad 1 rev 1.00 min 60.0 s=31.4 rad s.ω=300 rev 1.00 min 2 π rad 1 rev 1.00 min 60.0 s=31.4 rad s. The moment of inertia of one blade is that of a thin rod rotated about its end, listed in Figure 10.20. The total I is four times this moment of inertia because there are four blades. Thus, I=4 M L 2 3=4×(50.0 kg)(4.00 m)2 3=1067.0 kg⋅m 2.I=4 M L 2 3=4×(50.0 kg)(4.00 m)2 3=1067.0 kg·m 2.I=4 M L 2 3=4×(50.0 kg)(4.00 m)2 3=1067.0 kg·m 2. Entering ω ω ω and I into the expression for rotational kinetic energy gives K=0.5(1067 kg⋅m 2)(31.4 rad/s)2=5.26×10 5 J.K=0.5(1067 kg·m 2)(31.4 rad/s)2=5.26×10 5 J.K=0.5(1067 kg·m 2)(31.4 rad/s)2=5.26×10 5 J. Entering the given values into the equation for translational kinetic energy, we obtain K=1 2 m v 2=(0.5)(1000.0 kg)(20.0 m/s)2=2.00×10 5 J.K=1 2 m v 2=(0.5)(1000.0 kg)(20.0 m/s)2=2.00×10 5 J.K=1 2 m v 2=(0.5)(1000.0 kg)(20.0 m/s)2=2.00×10 5 J. To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is 2.00×10 5 J 5.26×10 5 J=0.380.2.00×10 5 J 5.26×10 5 J=0.380.2.00×10 5 J 5.26×10 5 J=0.380. Significance The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades. Example 10.10 Energy in a Boomerang A person hurls a boomerang into the air with a velocity of 30.0 m/s at an angle of 40.0°40.0°40.0° with respect to the horizontal (Figure 10.22). It has a mass of 1.0 kg and is rotating at 10.0 rev/s. The moment of inertia of the boomerang is given as I=1 12 m L 2 I=1 12 m L 2 I=1 12 m L 2 where L=0.7 m L=0.7 m L=0.7 m. (a) What is the total energy of the boomerang when it leaves the hand? (b) How high does the boomerang go from the elevation of the hand, neglecting air resistance? Figure 10.22 A boomerang is hurled into the air at an initial angle of 40°40°40°. Strategy We use the definitions of rotational and linear kinetic energy to find the total energy of the system. The problem states to neglect air resistance, so we don’t have to worry about energy loss. In part (b), we use conservation of mechanical energy to find the maximum height of the boomerang. Solution Moment of inertia: I=1 12 m L 2=1 12(1.0 kg)(0.7 m)2=0.041 kg⋅m 2 I=1 12 m L 2=1 12(1.0 kg)(0.7 m)2=0.041 kg·m 2 I=1 12 m L 2=1 12(1.0 kg)(0.7 m)2=0.041 kg·m 2. Angular velocity: ω=(10.0 rev/s)(2 π)=62.83 rad/s ω=(10.0 rev/s)(2 π)=62.83 rad/s ω=(10.0 rev/s)(2 π)=62.83 rad/s. The rotational kinetic energy is therefore K R=1 2(0.041 kg⋅m 2)(62.83 rad/s)2=80.93 J.K R=1 2(0.041 kg·m 2)(62.83 rad/s)2=80.93 J.K R=1 2(0.041 kg·m 2)(62.83 rad/s)2=80.93 J. The translational kinetic energy is K T=1 2 m v 2=1 2(1.0 kg)(30.0 m/s)2=450.0 J.K T=1 2 m v 2=1 2(1.0 kg)(30.0 m/s)2=450.0 J.K T=1 2 m v 2=1 2(1.0 kg)(30.0 m/s)2=450.0 J. Thus, the total energy in the boomerang is K Total=K R+K T=80.93+450.0=530.93 J.K Total=K R+K T=80.93+450.0=530.93 J.K Total=K R+K T=80.93+450.0=530.93 J. 2. We use conservation of mechanical energy. Since the boomerang is launched at an angle, we need to write the total energies of the system in terms of its linear kinetic energies using the velocity in the x- and y-directions. The total energy when the boomerang leaves the hand is E Before=1 2 m v 2 x+1 2 m v 2 y+1 2 I ω 2.E Before=1 2 m v x 2+1 2 m v y 2+1 2 I ω 2.E Before=1 2 m v x 2+1 2 m v y 2+1 2 I ω 2. The total energy at maximum height is E Final=1 2 m v 2 x+1 2 I ω 2+m g h.E Final=1 2 m v x 2+1 2 I ω 2+m g h.E Final=1 2 m v x 2+1 2 I ω 2+m g h. By conservation of mechanical energy, E Before=E Final E Before=E Final E Before=E Final so we have, after canceling like terms, 1 2 m v 2 y=m g h.1 2 m v y 2=m g h.1 2 m v y 2=m g h. Since v y=30.0 m/s(sin 40°)=19.28 m/s v y=30.0 m/s(sin 40°)=19.28 m/s v y=30.0 m/s(sin 40°)=19.28 m/s, we find h=(19.28 m/s)2 2(9.8 m/s 2)=18.97 m.h=(19.28 m/s)2 2(9.8 m/s 2)=18.97 m.h=(19.28 m/s)2 2(9.8 m/s 2)=18.97 m. Significance In part (b), the solution demonstrates how energy conservation is an alternative method to solve a problem that normally would be solved using kinematics. In the absence of air resistance, the rotational kinetic energy was not a factor in the solution for the maximum height. Check Your Understanding 10.4 A nuclear submarine propeller has a moment of inertia of 800.0 kg⋅m 2 800.0 kg·m 2 800.0 kg·m 2. If the submerged propeller has a rotation rate of 4.0 rev/s when the engine is cut, what is the rotation rate of the propeller after 5.0 s when water resistance has taken 50,000 J out of the system? PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: William Moebs, Samuel J. Ling, Jeff Sanny Publisher/website: OpenStax Book title: University Physics Volume 1 Publication date: Sep 19, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. Our mission is to improve educational access and learning for everyone. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. Give today and help us reach more students. Help Contact Us Support Center FAQ OpenStax Press Newsletter Careers Policies Accessibility Statement Terms of Use Licensing Privacy Policy Manage Cookies © 1999-2025, Rice University. Except where otherwise noted, textbooks on this site are licensed under a Creative Commons Attribution 4.0 International License. Advanced Placement® and AP® are trademarks registered and/or owned by the College Board, which is not affiliated with, and does not endorse, this site.
17025	https://www.varsitytutors.com/practice/subjects/sat-math/help/trigonometric-identities	SAT Math - Trigonometric Identities \| Practice Hub Skip to main content Practice Hub Search subjects AI TutorAI DiagnosticsAI FlashcardsAI WorksheetsAI SolverGamesProgress Sign In HomeSAT MathLearn by ConceptTrigonometric Identities Trigonometric Identities Help Questions SAT Math › Trigonometric Identities Questions 1 - 10 1 Using trigonometric identities prove whether the following is valid: True False Uncertain Only in the range of: Only in the range of: Explanation We can work with either side of the equation as we choose. We work with the right hand side of the equation since there is an obvious double angle here. We can factor the numerator to receive the following: Next we note the power reducing formula for sine so we can extract the necessary components as follows: The power reducing formula must be inverted giving: Now we can distribute and reduce: Finally recalling the basic identity for the cotangent: This proves the equivalence. 2 If and , evaluate . Explanation The easiest identity to use here is: Substituting in the given values we get: 3 According to the trigonometric identities, Explanation The trigonometric identity , is an important identity to memorize. Some other identities that are important to know are: 4 Simplify using the trigonometric power reducing formula. Explanation The power-reducing formulas state that: 5 Simplify the expression . Explanation Find a common denominator Multiply the numerators and leave the denominators factored Add numerators Pythagorean identity Combine like terms Factor numerator Reduce Reciprocal identity 6 Which of the following is the simplified version of ? Explanation To solve this problem we need to rewrite it in terms of and . Rewriting cotangent we get the following. Rewriting secant we get the following. Thus, we can substitute these identities into our original problem and simplify. 7 Use the power reducing formulas for trigonometric functions to reduce and simplify the following equation: Explanation The power reducing formulas for both sine and cosine differ in only the operation in the numerator. Applying the power reducing formulas here we get: Multiplying the binomials in the numerator and multiplying the denominators: Reducing the numerator: We again use the power reducing formula for cosine as follows: Combining the numerator by determining a common denominator: Now simply reducing the double fraction: 8 Using trigonometric identities, determine whether the following is valid: False True Uncertain Only valid in the range of: Only valid in the range of: Explanation In this case we choose to work with the side that appears to be simpler, the left hand side. We begin by using the power reducing formulas: Next we perform the multiplication on the numerator: The next step we take is to remove the double angle, since there is no double angle in the alleged solution: Finally we multiply the binomials in the numerator on the left hand side to determine if the equivalence holds: We see that the equivalence does not hold. 9 Simplify the expression: 1 Explanation The first step in solving this equation is to distribute : At this point, simplify using known Pythagorean identities. The left quantity simplifies such: and the right quantity simplifies such: Thus, we end up with: , which our Pythagorean identity tells us is equivalent to . Thus, 10 Simplify using identities: Explanation First we expand the inverse identities into fractional form: Invert the bottom fraction and distribute into the top, keeping track of the negative: Using the Pythagorean identity , our equation becomes: At this point, cross-cancel to obtain . Previous Page 1 of 2 Next Return to subject Powered by Varsity Tutors⋅ © 2025 All Rights Reserved
17026	https://www.youtube.com/watch?v=MfX6oJPXv4s	Algebra II: Review of Domain Restrictions MrsMcPhersonTHS 769 subscribers 232 likes Description 46822 views Posted: 11 Mar 2013 This video reviews how to find the domain of functions that are rational or contain even roots. 21 comments Transcript: this video is going to review domain restrictions which was part of chapter 7 it was in section 73 when we started talking about function operations doing addition subtraction multiplication division and composition in before I get into that I'll do a separate video about 73 I want to make a quick video just practicing finding domain and looking at the issues that happen whenever you have a denominator or an even root so all I'm doing on this is going through these six problems where we're going to look at each one talk about the issues that happen with the domain and then figure out what the domain is so you'll have this as like the second part of a question on your test is that you'll ask be asked to do some sort of function operation and then the next piece will be find the domain in this we're just finding a domain so let's look at the first one on the upper left we have our function square root of X+ 5 you want to think when you're doing domain restrictions to have kind of a mental checklist the first thing you're checking to see is there a denominator if there is it's not allowed to be zero and the second thing you're checking to see is if it has an even root if it does what is under the root has to be greater than or equal to zero so this particular problem does not have a denominator so no issue with that in our checklist but it does have an even root so the way I figure out the domain is I want to write that whatever is under the doain domain so in this case X plus 5 has to be greater than or equal to zero because you can't take the square root or an even root of any number that's negative subtract five so my domain is X has to be greater than or equal to5 the next one I'm just going to go across the x + 3x^2 minus 16 again in the checklist does it have a denominator yes what would make that denominator equal to zero so we said x^2 - 16 equal to zero we could either Factor it or we could add 16 to both sides and then take the square root we get plus or minus 4 that means my domain is all real numbers except X can't be plus or minus 4 no even root so no issues with keeping anything positive on this one the next one the 6 over radical X it has a denominator and an even root so now this piece that's under the radical has to be greater than zero is not allowed to equal zero because if it equal zero you would get zero in the bottom so this is kind of like the double issue problem you have a denominator and an even root so you have to be bigger than zero and that is your domain moving down to the bottom row the next one does not have a denominator it also doesn't have an even root this has a cube root there is no issue with cube root that's an odd root so for this one my domain is all real numbers and you'll have that happen if you don't have a denominator and you don't have an even root then you don't have a problem and your domain is any number that you want to plug in next one no denominator we do have an even root it's written a little differently here instead of writing square root we writing the 1/ half power that still is an even root so the value that's being taken to the even root 2x minus 1 has to be greater than or equal to zero so I'm going to add one to both sides to help me solve this so I get 2x has to be greater than or equal to 1 divide by 2 my domain is that X has to be greater than or equal to 12 and then finally my last one has a denominator no even root here so all I need to look at is the denominator and figure out what makes it equal to zero so 2x + 7 = 0 subtract 7 we get 2x = -7 divide by 2 so my domain is all real numbers except X cannot be -7 over 2 -7 halfes so this gives you just a little extra practice with domain it's quick they're not long problems but I want to make sure that when you take a test and you're asked to do different things and then finish with the domain that you don't leave that blank that you're comfortable writing the domain of any end result function
17027	https://en.wikipedia.org/wiki/Great-circle_distance	Jump to content Great-circle distance العربية Български Català Čeština Deutsch Eesti Ελληνικά Español Euskara Français Íslenska Italiano עברית Magyar 日本語 Polski Português Русский Slovenčina Slovenščina Українська 中文 Edit links From Wikipedia, the free encyclopedia Shortest distance between two points on the surface of a sphere This article is about shortest-distance on a sphere. For the shortest distance on an ellipsoid, see geodesics on an ellipsoid. The great-circle distance, orthodromic distance, or spherical distance is the distance between two points on a sphere, measured along the great-circle arc between them. This arc is the shortest path between the two points on the surface of the sphere. (By comparison, the shortest path passing through the sphere's interior is the chord between the points.) On a curved surface, the concept of straight lines is replaced by a more general concept of geodesics, curves which are locally straight with respect to the surface. Geodesics on the sphere are great circles, circles whose center coincides with the center of the sphere. Any two distinct points on a sphere that are not antipodal (diametrically opposite) both lie on a unique great circle, which the points separate into two arcs; the length of the shorter arc is the great-circle distance between the points. This arc length is proportional to the central angle between the points, which if measured in radians can be scaled up by the sphere's radius to obtain the arc length. Two antipodal points both lie on infinitely many great circles, each of which they divide into two arcs of length π times the radius. The determination of the great-circle distance is part of the more general problem of great-circle navigation, which also computes the azimuths at the end points and intermediate way-points. Because the Earth is nearly spherical, great-circle distance formulas applied to longitude and geodetic latitude of points on Earth are accurate to within about 0.5%. Formulae [edit] Let and be the geographical longitude and latitude of two points 1 and 2, and be their absolute differences; then , the central angle between them, is given by the spherical law of cosines if one of the poles is used as an auxiliary third point on the sphere: The problem is normally expressed in terms of finding the central angle . Given this angle in radians, the actual arc length d on a sphere of radius r can be trivially computed as Relation between central angle and chord length [edit] The central angle is related with the chord length of unit sphere : For short-distance approximation (), Computational formulae [edit] On computer systems with low floating point precision, the spherical law of cosines formula can have large rounding errors if the distance is small (if the two points are a kilometer apart on the surface of the Earth, the cosine of the central angle is near 0.99999999). For modern 64-bit floating-point numbers, the spherical law of cosines formula, given above, does not have serious rounding errors for distances larger than a few meters on the surface of the Earth. The haversine formula is numerically better-conditioned for small distances by using the chord-length relation: Historically, the use of this formula was simplified by the availability of tables for the haversine function: and . The following shows the equivalent formula expressing the chord length explicitly: where . Although this formula is accurate for most distances on a sphere, it too suffers from rounding errors for the special (and somewhat unusual) case of antipodal points. A formula that is accurate for all distances is the following special case of the Vincenty formula for an ellipsoid with equal major and minor axes: where ⁠⁠ is the two-argument arctangent. Using atan2 ensures that the correct quadrant is chosen. Vector version [edit] Another representation of similar formulas, but using normal vectors instead of latitude and longitude to describe the positions, is found by means of 3D vector algebra, using the dot product, cross product, or a combination: where and are the normals to the sphere at the two positions 1 and 2. Similarly to the equations above based on latitude and longitude, the expression based on arctan is the only one that is well-conditioned for all angles. The expression based on arctan requires the magnitude of the cross product over the dot product. From chord length [edit] A line through three-dimensional space between points of interest on a spherical Earth is the chord of the great circle between the points. The central angle between the two points can be determined from the chord length. The great circle distance is proportional to the central angle. The great circle chord length, , may be calculated as follows for the corresponding unit sphere, by means of Cartesian subtraction: Substituting and this formula can be algebraically manipulated to the form shown above in § Computational formulae. Radius for spherical Earth [edit] Main article: Earth radius The shape of the Earth closely resembles a flattened sphere (a spheroid) with equatorial radius of 6378.137 km; distance from the center of the spheroid to each pole is 6356.7523142 km. When calculating the length of a short north-south line at the equator, the circle that best approximates that line has a radius of (which equals the meridian's semi-latus rectum), or 6335.439 km, while the spheroid at the poles is best approximated by a sphere of radius , or 6399.594 km, a 1% difference. So long as a spherical Earth is assumed, any single formula for distance on the Earth is only guaranteed correct within 0.5% (though better accuracy is possible if the formula is only intended to apply to a limited area). Using the mean Earth radius, (for the WGS84 ellipsoid) means that in the limit of small flattening, the mean square relative error in the estimates for distance is minimized. For distances smaller than 500 kilometers and outside of the poles, a Euclidean approximation of an ellipsoidal Earth (Federal Communications Commission's (FCC)'s formula) is both simpler and more accurate (to 0.1%). See also [edit] Air navigation Angular distance Circumnavigation Elliptic geometry § Elliptic space (the 3D case) Flight planning Geodesy Geodesics on an ellipsoid Geodetic system Geographical distance Isoazimuthal Loxodromic navigation Meridian arc Rhumb line Spherical geometry Spherical trigonometry Versor References and notes [edit] ^ Admiralty Manual of Navigation, Volume 1, The Stationery Office, 1987, p. 10, ISBN 9780117728806, The errors introduced by assuming a spherical Earth based on the international nautical mile are not more than 0.5% for latitude, 0.2% for longitude. ^ Kells, Lyman M.; Kern, Willis F.; Bland, James R. (1940). Plane And Spherical Trigonometry. McGraw Hill Book Company, Inc. pp. 323-326. Retrieved July 13, 2018. ^ "Calculate distance, bearing and more between Latitude/Longitude points". Retrieved 10 Aug 2013. ^ Sinnott, Roger W. (August 1984). "Virtues of the Haversine". Sky and Telescope. 68 (2): 159. ^ Vincenty, Thaddeus (1975-04-01). "Direct and Inverse Solutions of Geodesics on the Ellipsoid with Application of Nested Equations" (PDF). Survey Review. 23 (176). Kingston Road, Tolworth, Surrey: Directorate of Overseas Surveys: 88–93. Bibcode:1975SurRv..23...88V. doi:10.1179/sre.1975.23.176.88. Retrieved 2008-07-21. ^ Gade, Kenneth (2010). "A non-singular horizontal position representation" (PDF). The Journal of Navigation. 63 (3). Cambridge University Press: 395–417. Bibcode:2010JNav...63..395G. doi:10.1017/S0373463309990415. ^ McCaw, G. T. (1932). "Long lines on the Earth". Empire Survey Review. 1 (6): 259–263. doi:10.1179/sre.1932.1.6.259. ^ Agafonkin, Vladimir (30 August 2017). "Fast geodesic approximations with Cheap Ruler". Mapbox. "mapbox/cheap-ruler". Mapbox. 10 May 2024. External links [edit] GreatCircle at MathWorld Retrieved from " Categories: Metric geometry Spherical trigonometry Distance Spherical curves Hidden categories: Articles with short description Short description is different from Wikidata
17028	https://www.palmbeachstate.edu/prepmathlw/Documents/solving%20percent%20problems.pdf	SOLVING PERCENT PROBLEMS Percent Equation In problems involving the percent equation, two parts of the equation are given, the other part is unknown. Percent • Base = Amount To write the equation, identify the given and unknown parts using the guide below: "of" translates to "multiply" "is" translates to "=" The base usually follows "of" The amount usually follows "is" "what" identifies the unknown When the percent is given, convert the percent to decimal form. EXAMPLE 1: 15 is what percent of 120? To solve the problem, identify the given and unknown parts: Given: Base = 120 Unknown: Percent = x Amount = 15 Equation: 120 • x = 15 120 • x 15 120 120 x 0.125 12.5% = = = 0.125 120 15.000 Percent Proportion Problems involving the percent equation can also be solved with the proportion: Percent Amount (is) 100 Base (of) = When the percent is given, drop the percent sign and place the percent over 100. Cross multiply to solve the proportion. Example 2: 27 is 45% of what number? Given: Percent = 45% Unknown: Base = x Amount = 27 Proportion: 45 27 100 x = PBCC Page 1 of 4 SLC Lake Worth Math Lab6/7/2005 Next, cross multiply to solve the proportion: 45 27 100 x = 45(x) 27(100) 45x 2700 45x 2700 45 45 x 60 = = = = 60 45 2700 Percent of Increase or Decrease To find the percent of increase or decrease, 1) Subtract the new amount from the original amount to find the decrease. Subtract the original amount from the new amount to find the increase. 2) Solve for the percent. The "original amount" is the "base;" the "increase" or "decrease" is the "amount." Example 3: The amount in a savings account increased from $560 to $672. Find the percent of increase. 1) Find the increase: New Amount – Original Amount = Increase $672 – $560 = $112 2) Solve for the percent: Given: Base = $560 Unknown: Percent = x Amount = $112 Proportion: x 112 100 560 = Cross-multiply to solve: x 112 100 560 560(x) 112(100) 560x 11 ,200 560x 11 ,200 560 560 x 20% = = = = = 20 560 11 ,200 The percent of increase is 20% PBCC Page 2 of 4 SLC Lake Worth Math Lab6/7/2005 Price Increases and Discounts Suppose a shirt is on sale for 40% of the original price. If the total cost, including the discount, is $27.00, what was the price of the shirt before the discount? The discount is 40% of the original price. If we let x = the original price or the shirt, then Discount = 40%(Original Price) = 40%(x) = 0.4x The total cost of the shirt, including the discount, is found by subtracting the amount of the discount from the original price of the shirt: Original Price – Discount = Total Cost x – 0.4x $27.00 To find the original price of the shirt, solve the equation for x: x 0.4x 27 0.6x 27 0.6x 27 0.6 0.6 x 45 − = = = = 45. 0.6. 27.0. The price of the shirt before the discount was $45.00. Simple Interest The amount of simple interest earned when $1000 is invested at 4.5% for 1 year is found using the formula I = Prt, where I is the interest earned, P is the principal or amount invested, r is the interest rate (percent) and t is the time: I = Prt I = 1000(0.045)(1) = $45.00 The amount in the account at the end of one year is found by adding the interest earned to the principal: Principal + Interest = Amount. $1000 + $45 = $1045 EXAMPLE 4: Money is invested at a simple interest rate of 5%. At the end of one year the amount in the account is $3,675. How much was the original investment? To find the amount of the original investment, let x = the Principal. Given r = 5% = 0.05 and t = 1, the interest earned on the principal is given by I = Prt I = (x)(0.05)(1) = 0.05x PBCC Page 3 of 4 SLC Lake Worth Math Lab6/7/2005 At the end of one year the amount in the account is found by adding the interest to the principal: Principal + Interest = Amount x + 0.05x = $3,675 To find the original investment, solve the equation for x: x 0.05x 3,675 1.05x 3,675 1.05x 3,675 1.05 1.05 x 3,500 + = = = = 3500. 1.05. 3,675.00. The amount of the original investment was $3,500. PBCC Page 4 of 4 SLC Lake Worth Math Lab6/7/2005
17029	https://www.youtube.com/watch?v=TvrgL0rHbMI	Chi Square Test of Independence and Homogenuity The Math Lane 162 subscribers 3 likes Description 403 views Posted: 18 Mar 2021 Learn how to conduct a Chi Square 2- way Test of Independence or Homogeneity. Learn the assumptions, null, alternative hypothesis, calculator work and conclusion statement. Prep for AP Statistics test. Practice on Khan Academy.org: -Expected Counts in Chi-Square Tests with Two-Way Tables -Test Statistic and P-value in Chi-Square Tests with Two- Tables -Making Conclusion in Chi-Square Tests for Two- way Tables Transcript: okay today we're going to go over the chi-square test of independence and homogeneities so remember a chi-square test are dealing with categorical data and not numerical so let's start with the chi-square test for independence it is used for categorical bivariate data from one sample and it's used to see if two categorical variables are associated i.e called dependent or not associated i.e independent so our assumptions and formula remain the same just the same as goodness of fit we need an srs and we need expected counts at least five um we do have a different button on the calculator that we're going to use since we're doing a chi-square two-way test instead of a goodness of it um so it's a little more work involved on the calculator i'll show you here an example with an example of what how we use the calculator so we're first going to have to create a matrix okay because the whole point of a two-way test is that we have a cop we have a table of values you're going to create a matrix by going to menu matrix create matrix you are then going to list the number of rows and columns three by two three by three don't include the total columns you're going to enter your values of the observed matrix into that matrix then you're going to hit the enter button and then we need to name that matrix so how you name a matrix is you're going to hit control var and if you notice right above a var on the ti inspires it says stow with an arrow it actually means store okay so we're going to store that matrix and then you can name it any letter that you want a b c whatever you want to name it and so that's going to name your matrix so once you have your matrix named we can then complete the test so go to menu statistics test two-way test and it will ask for the observed matrix and you will put whatever you need if you named it a you'll type in a okay and then from the output uh the output will give you your chi-square value your p-value degrees of freedom etc so yay assumptions um and formula are the same just a different button on the calculator so what are the hypothesis for tests of independence well your null is simply that whatever variables you're talking about are independent and your alternative is that whatever variables are dependent so remember independent has to be your null dependent has to be your alternative [Music] and again make sure to relate to the exact variables that you are talking about so let's look at example here we have a beef distributor of wishes to determine whether there is a relationship between geographic region and cut of meat preferred if there is no relationship we will say that beef preference is independent of geographic region suppose that in a random sample 500 300 are from the north 200 from the south 150 prefer cut a 275 prefer cut b and 75 prefer cut c so if beef preference is independent of geographic region how would we expect this table to be filled in so what we're going to do here is create the expected values okay so chi-square test of independence again we're going to kind of have a chart rows and columns of data so utilizing the information already given we have a random sample of 500 total so the very bottom right hand side of my table is 500 that's my overall total i know that my overall north is 300 i know that my overall south is 200 i know that my overall cut a is 150 my overall cut beat is 275 and my overall cut c is 75 so that is given in the problem so how we calculate expected is we need to use proportions so how do i figure out what value is expected of cut a meet from the north is we're going to take our row column our row total times our column total and divide it by our overall total so i'm going to take my total of my row our column 300 total of my row 150 divided by overall total and that should get 290. so how do we figure it out from the south take your column total total total of south is 200 divided by your overall total of 500 and we're multiplying that by our cut a total of 150. okay so rope total times column total divided by overall total so again if you do this accordingly for cup b and north you're going to take 300 divided by 500 times 275 and you should get 165 for cup b south you're going to take 200 divided by 500 times 275 you should get 110 for cut c and north you're going to take 300 divided by 500 times 75 to get 45 and cut c south you're going to take 200 divided by 500 times 75 to get 30. so this is your expected value matrix uh so again easy way to think of expected value again assuming h your null is true is your expected counts equals your row total times your column total divided by overall total your degrees of freedom for chi-square two-way table is going to be your row minus one times your column minus one and do not include the uh total rows and columns just the actual value data value rows and columns okay so going back to our beef distributor problem okay we have the same scenario here but now i have this extra information of now let's suppose in an actual sample of 500 the observed cuts and meat totals are given as such so it tells me that i observe that a of north meat was 100 a of south meat was 50 and so forth so we have the observed values here is there sufficient evidence to suggest that geographic regions and beef preference are not independent i.e is there a difference between the expected and observed counts okay so notice that we have categorical data because i'm just talking about north south cut a b c k that's not numerical these are just counts of that but you're also clues that you have a table anytime you have a table of values typically that's going to be your chi square independent or homogeneity which here it clearly states independent so how do we perform a full test as always we start with assumptions do i have a random sample notice over here in the problem it clearly states the word a random sample of 500 so random srs check are all expected counts greater than 5. remember on the previous slide we hand calculated these and you do have to show the expected matrix listed out to show that all accounts were greater than five but we did hand calculate this in a previous slide so expected counts check next we need our null and our alternative so our null is that geographic region and b preference are independent our alternative would be that geographic region and beef preference are dependent okay once i've wrote checked my assumptions i found my expected value matrix i've written my null and my alternative i am ready to conduct the test so i'm going to show you how to use the graphing calculator to conduct your test okay so using your ti inspire cx you're first going to go to your calculate tab okay and the first thing we need to do is we need to create a matrix of the observed matrix so to create a matrix you're going to go to menu you're going to go to matrix um and go to menu matrix go to create matrix okay now how many rows and columns not including the totals i have three rows and i have two columns three cuts of meat so three rows and i have two columns north and south again do not include the row and column totals so hit ok i'm then going to input in my data 100 150 100 150. sure there you go 50 150 125 and tab key works best 125 50 and 25. so again make sure that you're entering in the observed matrix not the expected um calculator smart calculator can actually hand calculate the expected based on the row and column totals um but for ap test standards you need to list out the expected matrix as a showing work component okay so after you create the matrix you're going to hit enter okay that's going to save it in the calculator now we need to name this matrix okay so what you're going to do is you're going to hit control the blue button control and then var so notice above vars is sto that means store so hit control vars and notice it's going to take this uh answer matrix that you just made and we need to name it something so you can name it any letter you want a b c d so forth i'm just to name it a so we're going to do control vars to store it as a hit enter so now this matrix is named a once i've named my matrix i can now conduct the test so go to menu statistics stat test and chi-square two-way test and notice it says well what is my observed matrix with the right arrow i named it a or find whatever you named it enter and notice there we have our chi-square value of 7.57 we have a p-value of 0.0226 and it also tells me my degrees of freedom okay so you got to create a matrix number one you got to name the matrix by using control store control var and then you can go to menu statistics test uh two-way test okay uh so now that we've used calculator again if you want to hand calculate this would be the formula just be aware for ap standards um there might be a multiple choice question oh with the formula um with values um inserted in so just make sure you know where that formula is on the formula chart um based on the calculator output we're going to name the test calc chi-square two-way test of independence we're going to list the chi-square value of 7.576 that we got output from the calculator we're going to list the p-value of 0.0226 we're going to list degrees of freedom of 2. and notice alpha is not stated so when it's not stated let's just use alpha equals 0.05 so make sure that you have all parts of this as the showing work but just like with all hypothesis tests we then make a conclusion based on p-value and alpha so compare p to alpha and .02 is less than 0.25 so since p is less than alpha i reject h o there is sufficient evidence to suggest that what's my alternative geographic region to be preference are dependent okay uh so very similar to a goodness of fit but notice on the key here how do we know if something is in a chi-square test of independence a it will state independent and you will have a table a two-way table so what about the other last chi-square test for homogeneity well it is used with a single categorical variable from two or more independent samples and it is used to see if the two populations are the same um so think about that root word homo means the same so we're testing to see if something is the same or is not the same so again assumptions and formula remain the same we still need to check for srs we still need to create the expected value of matrix and make sure all expected values are five we still use the same button on the calculator uh chi square two two-way table so yes you will have to create your observed matrix and name it store it and conduct your hypothesis test the only change is how we state the null and alternative so what is my null alternative for a test of homogeneity it is the null is that the two distributions are the same so we're going to assume that they're equal an alternative is that the distributions are different and of course you always need to make sure that you write it in context [Music] so let's do an example of this we have following data's on drinking behavior for independently chosen random samples of male and female does there appear to be a gender difference with respect to drinking behavior and again there they denote low drinking is one to seven a week moderate as eight to twenty four high is 25 or more so notice we have table here of values observed values for men and women based on non-low moderate or high so notice what is it asking us here is there a difference so that is your clue that this is a chi-square test of homogeneity okay so if this is our problem we always need to start with assumptions we first need to have an srs does we have a stated word random and notice it's independently chosen random samples so yes srs checks we then need to check expected counts here you need to hand calculate expected counts so again how do we do that again you're going to use your row and column totals so to find your expected value of men that don't drink at all again you're going to take your men total your column total of 981 divided by your overall total 2017 times your row total of 326 so 981 divided by 20 17 times 326 will get you 158.6 and again do this for every value in the table so how do we get a female again we're going to take the female total 1036 divided by the column total 10 36 divided by 2017 times the column total of 326 and that will get you your 167.4 so again you need to do that for each uh value to get your expected matrix and notice are all the values greater than five very much so okay so assumptions haven't met next we do no alternative our null is that is drinking behavior the same for male and female our alternative is that it's drinking behavior not the same for male and female so next we are ready to actually conduct the test and again i will show you again how to conduct the test using the builder okay again uh showing you how to use utilize the calculator to get your chi-square and p-value so first we need to create a matrix so first go to calculator you go to menu matrix and you need to create matrix so how many rows and columns do not include totals so total rows we have none low moderate high that is four columns we have men women that is two hit enter again um input your values input your values for the observed again make sure that you are using observed 140 186 478 661 300 173 63 and 16. again hit enter to stay save it into the calculator now we need to name it so you're going to hit control var which is really store and store it as um name it as you want we can name it a you can name it b name it whatever you want enter so i've named it b and once we've named it we are now ready to conduct the test so go to menu statistics stack calculation us test at the end stat test and we have a two-way table so chi-square two-way test again it wants the observed matrix i named it b find whatever you named it hit enter and voila i have my chi-square value i have my p-value now be very careful on this p-value notice it says 8.67 e to the negative 21. e to the negative 21 okay this is scientific notation this means times 10 to the negative 21. anytime you have scientific notation negative means you have a really really small number so this means it's essentially zero very important it is 90.67 okay um this is moving the decimal to the left 21 spots so essentially zero and degrees of freedom are three okay uh so going back to a showing work problem again a chi-squared value we need to write down is 96.53 again here is if you want to see how this would be found by using the formula our p-value notice is pretty much zero again you had times 10 to the negative 21. that would literally be point 20 zeros okay so be very very careful on that calculator if you see e to the negative that means you are moving the decimal to the left and probably um that p-value is going to be zero not always um but double check but uh e to the negative 21 for sure pretty much zero degrees of freedom we need a list as three and alpha is not stated therefore we will use .05 so a conclusion statement we compare p to alpha and if basically our probability of getting this value is pretty much zero that means that p is low reject h o so since p is less than alpha i reject h o there is sufficient evidence to suggest that drinking behavior is not the same for female and male again we can somewhat look at the table and maybe suspect that you know check as a reasonableness again never just do that solely based on appearance but again if you look at the numbers you can see that they're a little off if you just think about logic you know beauty of stats is that it is very logical if you think logically um men and women um differ on a variety of things um so this problem is complete again make sure that you have all parts for a chi-square two-way test of independence or homogeneity assumptions create expected value matrix have to write it down even though the calculator is smart and does it itself you have to show that to show that each value is greater than 5 you need a null alternative and utilizing calculator oh here i forgot to name the test you should have put down chi-square test of homogeneity or two-way table i believe they will count that as well make sure to have list out the chi-square value the p-value degrees of freedom and the alpha and then a conclusion statement based on comparing p to alpha so that is your chi-square two-way table test
17030	https://www2.mathematik.tu-darmstadt.de/~paffenholz/daten/preprints/20201007_Lattice_Polytopes.pdf	Lecture Notes on Lattice Polytopes (draft of October 7, 2020) October 7, 2020 Discrete Geometry III; Summer 2019 TU Berlin Christian Haase • Benjamin Nill • Andreas Paﬀenholz 1 An invitation to lattice polytopes . . . . . . . . . . . . . . . . . . . . . 5 1.1 Before we begin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2 Lattice polygons and isomorphisms . . . . . . . . . . . . . . . . . . . 6 1.3 Triangulations and Pick’s formula . . . . . . . . . . . . . . . . . . . . 10 1.4 A Classiﬁcation of Lattice Polygons . . . . . . . . . . . . . . . . . . . 12 1.5 Dilations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2 Polytopes and Lattices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.1 Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.1.1 Cones and Polytopes. . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.1.2 Convex hulls and half-spaces . . . . . . . . . . . . . . . . . . . 23 2.1.3 The face lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 2.2 Decompositions of Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.2.1 Polyhedral Complexes . . . . . . . . . . . . . . . . . . . . . . . . 30 2.2.2 Regular Subdivisions and Triangulations . . . . . . . . 31 2.3 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.3.1 Discrete Subgroup and Lattice Bases. . . . . . . . . . . . 33 2.3.2 Coordinates and Normal Forms . . . . . . . . . . . . . . . . 41 2.3.3 Metric Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 2.3.4 Hilbert Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.4 Lattice polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 2.4.1 Equivalence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 2.4.2 Examples of Lattice Polytopes and Constructions. 49 VIII 2.4.3 Volumes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.5 Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 3 Ehrhart Theory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 3.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.2 Old Motivation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.2.1 Examples of Ehrhart polynomials . . . . . . . . . . . . . . 65 3.3 Generating Functions for Lattice Points . . . . . . . . . . . . . . . 67 3.4 Ehrhart’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 3.5 Stanley’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.5.1 Half-Open Decompositions of Cones . . . . . . . . . . . . 76 3.5.2 The integer point generating function of half-open cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.5.3 Stanley’s theorem and the h∗-polynomial of a lattice polytope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 3.5.4 Where does the h∗-notation come from? . . . . . . . . . 81 3.6 Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 3.6.1 Stanley reciprocity for cones . . . . . . . . . . . . . . . . . . . 83 3.6.2 Ehrhart-Macdonald reciprocity for lattice polytopes 85 3.7 Properties of the h∗-polynomial . . . . . . . . . . . . . . . . . . . . . . 87 3.7.1 Degree and codegree of lattice polytopes . . . . . . . . 87 3.7.2 Ehrhart polynomials of lattice polygons . . . . . . . . . 90 3.7.3 Polytopes with Small Degree . . . . . . . . . . . . . . . . . . . 91 3.8 Brion’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 4 Geometry of Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 4.1 Minkowski’s Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 4.2 Coverings and Packings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 4.3 Flatness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 4.4 Finiteness of lattice polytopes with few interior lattice points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 4.4.1 Finiteness of barycentric coordinates of lattice simplices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 4.4.2 Coeﬃcient of asymmetry . . . . . . . . . . . . . . . . . . . . . . 115 4.4.3 Bounding the volume . . . . . . . . . . . . . . . . . . . . . . . . . 116 4.5 Lower Bounds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 4.6 Empty lattice simplices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 4.7 Lattice polytopes without interior lattice points . . . . . . . . 124 4.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 5 Minkowski meets Ehrhart . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 5.1 Lattice Polytopes of given h∗-Polynomial . . . . . . . . . . . . . . 133 IX 5.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 5.1.2 The pyramid theorem for lattice simplices . . . . . . . 135 5.1.3 Proof of the pyramid theorem . . . . . . . . . . . . . . . . . . 137 5.2 Lattice polytopes of small degree . . . . . . . . . . . . . . . . . . . . . 139 5.2.1 Cayley-Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 5.2.2 Small Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 5.2.3 Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 5.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 6 Short Rational Generating Functions . . . . . . . . . . . . . . . . . 149 6.1 Hermite and Smith Normal Forms . . . . . . . . . . . . . . . . . . . . 150 6.2 Computing Short Rational Generating Functions . . . . . . . 150 6.2.1 Polynommial Time Evaluation . . . . . . . . . . . . . . . . . 155 6.2.2 Integer Linear Programming via Evaluation. . . . . . 157 6.3 The Shortest Vector Problem . . . . . . . . . . . . . . . . . . . . . . . . 157 6.4 Short Lattice Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 6.4.1 Applications of LLL . . . . . . . . . . . . . . . . . . . . . . . . . . 170 6.5 Variations of Barvinok’s Algorithm . . . . . . . . . . . . . . . . . . . 170 6.6 The Closest Vector Problem . . . . . . . . . . . . . . . . . . . . . . . . . 171 6.7 Integer Programming in Fixed Dimension . . . . . . . . . . . . . 172 6.8 Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 6.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 7 Reﬂexive and Gorenstein polytopes . . . . . . . . . . . . . . . . . . . 179 7.1 Reﬂexive polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180 7.1.1 Dimension 2 and the number 12 . . . . . . . . . . . . . . . . 182 7.1.2 Dimension 3 and the number 24 . . . . . . . . . . . . . . . . 185 7.2 The combinatorics of simplicial reﬂexive polytopes . . . . . . 187 7.2.1 The maximal number of vertices . . . . . . . . . . . . . . . 187 7.2.2 The free sum construction . . . . . . . . . . . . . . . . . . . . . 188 7.2.3 The addition property . . . . . . . . . . . . . . . . . . . . . . . . 189 7.2.4 Vertices between parallel facets. . . . . . . . . . . . . . . . . 189 7.2.5 Special facets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 7.3 Gorenstein polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 7.4 Finiteness of Gorenstein polytopes of given degree . . . . . . 197 7.5 Classiﬁcation of reﬂexive polytopes . . . . . . . . . . . . . . . . . . . 198 7.5.1 Smooth reﬂexive polytopes . . . . . . . . . . . . . . . . . . . . 198 7.5.2 All reﬂexive polytopes . . . . . . . . . . . . . . . . . . . . . . . . 198 7.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 8 Unimodular Triangulations . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 8.1 Regular Triangulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 8.2 Pulling Triangulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 8.3 Compressed Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 X 8.4 Special Simplices in Gorenstein Polytopes . . . . . . . . . . . . . 210 8.5 Dilations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 8.5.1 Composite Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 8.5.2 Prime Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 8.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 A Some Convex Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 A.1 Convex Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 A.2 Ellipsoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 A.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 B Solutions to some Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 B.1 Solutions for Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 B.2 Solutions for Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Name Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 kritisch überprüfen, welche Notatio-nen/Begriﬀe/Namen man auch we-glassen könnte! R real numbers \R R>0 positive integers \Rg R≥0 nonnegative integers \Rge R<0 negative integers \Rl R≤0 nonpositive integers \Rle Q rational numbers \Q Z integers \Z Z>0 positive integers \Zg Z≥0 nonnegative integers \Zge Z<0 negative integers \Zl Z≤0 nonpositive integers \Zle (Rd)⋆, (Qd)⋆, (Zd)⋆the dual spaces \Rdual, \Qdual, \Zdual T a triangulation \triang ∆d a simplex \simplex a the vector with variable name a, all others similar \va V(P) the vertices of a polytope P \verts(P) F(P) the facets of a polytope P \facets(P) ∂P the boundary (complex) of P \boundary P Π(V ) the fundamental parallelepiped of V \fp(V) G the integer point generating function\IntPtGenF(shortcuts siehe .sty oder Text) b G the integer point generating series\IntPtGenS Lecture Notes Lattice Polytopes (draft of October 7, 2020) ⌊a⌋the largest z ∈Z with z ≤a \floor ⌈a⌉the smallest z ∈Z with z ≥a \ceil {a} the fractional part a −⌊a⌋of a \fracpart A a point/vector conﬁguration, usually P ∩Zd \configuration ehr the Ehrhart counting function \ehrcount b Ehr the Ehrhart series \ehrseries \| a ﬁeld \kk 1 the column vector with all entries equal to 1 \1 0 the column vector with all entries equal to 1 \0 C the cone over a polytope \pcone C⋆the dual cone to C \dpcone h⋆coeﬃcients of the h⋆-polynomial \hstar f f-vector entries \fvec T F P Tangent cone of a face \tangentcone{P}{F} b L Laurent series\LaurentS L Laurent polynomials\LaurentP Q Laurent quotient\LaurentQ R Laurent rational functions \LaurentR Φ The map onto the rational functions \LaurentHom F ⪯P F is a face of P \isfaceof C⋆\dual C C⋆⋆\ddual C C⋆⋆⋆\dddual C C a polyhedral complex \pcomplex S a subdivision \psubdiv pull(S; v) a pulling reﬁnement \pull{\psubdiv}{\vv} star(§; F) open star of a face F in a complex S \Star{\S}{F} star(§; F) closed star of a face F in a complex S \clStar{\S}{F} V(P) vertices of P rank A rank of A lin A linear space spanned by A aﬀA aﬃne hull of A conv A convex hull of A cone A conic hull of A P ⋆Q the join of P and Q P \join Q vol the (normalized?) volume \vol nvolZd the normalized volume \Vol lineal A lineality space of P Sw(V ) regular subdivision induced by w on V \regsubdiv{\vw}{V} lift(w) the convex hull of the lift of V by w \reglift{\vw}{V} Ψw the convex piecewise linear function \regfunction{\vw}{V} id The identity matrix The zero matrix — 2 — Haase, Nill, Paffenholz: Lattice Polytopes (draft of October 7, 2020) Λ A lattice \lattice ˆ Λ The mother lattice \lambdahat B A basis of a lattice width(K; a) width of convex body K wrt functional a ∈Λ⋆\width{K}{\va} widthΛ(K) width of convex body K wrt lattice Λ \lwidth{K}{\lattice} ϱ Λ packing radius of a lattice \packingradius{\lattice} µ Λ covering radius of a lattice \coveringradius{\lattice} ηF The unique primitive inner normal of a facet F of a polytope \innP_F uF The unique primitive inner normal of CP corresponding to the facet F of a polytope P \innC_F uP The unique lattice point of the Gorenstein cone CP satisfying ⟨uP , x ⟩= 1 for any primitive generator of CP \innC_P P ∨The Gorenstein polytope dual to P dualG␣Pﬁ ∆d The standard simplex conv(0, e1, . . . , ed) Cd The unit cube {x ∈Rd \| 0 ≤x ≤1} = [0, 1]d. T F P The tangent cone to P at F \tangentcone{P}{F} visibleP (m) The complex of faces of P visible from m \visible{P}{\vm} invitation durchlesen, fertig? exercises? B neuere Resultate zusammenstellen, klassiﬁzieren rein/exer-cise/notes/raus – Ehrhart – GoN – UT Struktur sortieren Kap. GoN vs MinkowskiEhrhart; zus"at-zliche Themen?? 2.3 Lattices streamlinen C Ch. 2 running examples A Ch. 3 Ehrhart – Todos Ch. 4 GoN – gegen Ende gro"se Baustelle (include size?) Ch. 5 Ehrhart meets Minkowski zu kurz Ch. 6 Algorithmen schreiben A Ch. 7 Gorenstein Ch. 8 UT somewhere: Cayley-korrespondenz via lattice-point-Zerlegung im dualen Kegel somehwere: Minkowski-sums and reﬁnement of normal fan, Cayley-Trick Haase, Nill, Paffenholz: Lattice Polytopes — 3 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) how do we attribute results? in parantheses after “Theorem xy”? In a separate senctence before or after? Do we give a citation, a year? How do we do this for theorems with a name (like Ehrhart’s Theorem)? — 4 — Haase, Nill, Paffenholz: Lattice Polytopes An invitation to lattice polytopes 1 Contents 1.1 Before we begin . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2 Lattice polygons and isomorphisms . . . . . . . 6 1.3 Triangulations and Pick’s formula . . . . . . . . . 10 1.4 A Classiﬁcation of Lattice Polygons . . . . . . . 12 1.5 Dilations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 In this chapter we would like to give the reader a gentle introduction to the main players of this book. For this, we will mainly stick to objects in two dimension, that is, we will consider polygons, their subdivisions into smaller pieces and the lattice of all points with integer coordinates. However, the reader will already encounter many methods and types of results studied in more detail later, among them are triangulations, lattice point counting, estimating volumes, and several classiﬁcation results. Our goal is to convey a ﬁrst impression of the rich ﬂavours of lattice polytope theory. This is a branch of convex-discrete geometry with an algebraic touch and a hint of number theory. It is surprising how many of the interesting features of lattice polytopes already appear in dimensions one, two or three. And there is no shortage of open questions for these seemingly elementary objects. Lecture Notes Lattice Polytopes (draft of October 7, 2020) 1.1 Before we begin Before reading on, we would like to invite the reader to pull out a sheet of graph paper and play with lattice polygons, making one’s ﬁrst own discoveries and building a feeling and an appreceation for the objects and the questions treated in this book. Fig. 1.1: A lattice polygon Fig. 1.2: non-lattice Fig. 1.3: non-convex When we say lattice polygon, we mean a closed convex polygon with all vertices at crossing points of your graph paper as in Figure 1.1 (and not as in Figs. 1.2 and 1.3). For such a polygon, we can record the number b of graph-paper-crossing-points on the boundary, the number i of graph-paper-crossing-points in the interior, and the enclosed area a measured in units of graph-paper-squares. For example, our ﬁrst polygon above has (b, i, a) = (6, 5, 7). Your task is now to play with these ﬁgures and ﬁnd out what triples (b, i, a) you can achieve. That is, if I say (3, 0, 1/2), you draw the picture in Figure 1.4, and if I say (9, 1, 4.5), you draw the picture in Figure 1.5. The reader is invited to try realizing the cases (5, 2, 4), (18, 0, 9), (3, 17, 17.5) and (11, 2, 6.5). Fig. 1.4: (b, i, a) = (3, 0, 1/2) Fig. 1.5: (b, i, a) = (9, 1, 4.5) 1.2 Lattice polygons and isomorphisms In this section, we formally introduce lattice polygons, which are the key player of this chapter, ﬁnd a famous connection between its lattice points and its volume, and develop the appropriate notion of when to consider two such polygons the same. Deﬁnition 1.1 A lattice polygon is the convex hull in R2 of ﬁnitely many points in Z2. Here, the word lattice refers to Z2 whose elements we call lattice points. In other words, in dimension 2, lattice points are simply the points on the grid given by the vectors with all coordinates integral. A lattice polygon is the smallest convex set that contains a given ﬁnite set of lattice points. Restricting the vertices to lattice points is quite restrictive. In particular, a lattice polygon cannot be arbitrarily small. We bound the volume with the next proposition. We will see that even more is true. Any polygon with this volume will be a triangle, and any such triangles are equivalent to each other in some sense we develop below. Proposition 1.2 (Pick’s Theorem) Any lattice triangle with only three lattice points (which must be its vertices) has area 1/2. This Proposition may seem unspectacular, at ﬁrst. But it is remarkable in several ways. First, the corresponding statement is plain wrong in — 6 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 1. An invitation to lattice polytopes (draft of October 7, 2020) higher dimensions (see (1.1) below). Second, it is the key ingredient in the proof of Pick’s Formula (Theorem 1.8) answering whether or not the triple (5, 2, 4) from Section 1.1 comes from a lattice polygon. Lastly, its proof uses methods we will come across in several places throughout this book, and it leads us to the notion of lattice equivalence that we develop in Deﬁnition 1.4. Proof (of Proposition 1.2). A translation by a lattice vector preserves the number of lattice points as well as the area. Thus, we can assume that our triangle ∆is the convex hull of the origin 0 together with two linearly independent lattice vectors v and w. Consider the set Π(v, w) := {λv + µw : λ, µ ∈[0, 1)} (cf. Figure 1.6). This is a half-open parallelogram, where the segments between the origin and v and w (but not v, w itself!) belong to the set, the other two bounding segments do not. Its area, \| det[v, w]\|, the absolute value of the determinant of the matrix with columns v and w, equals twice the area of ∆. Fig. 1.6: The half-open parallelo-gram in the proof of Pick’s Theorem (Proposition 1.2) We claim that Π(v, w) ∩Z2 = {0}. Suppose u = λv + µw ∈ Π(v, w) ∩Z2. Then either we have λ + µ ≤1 so that u is a lattice point in ∆which leaves only u = 0 as v, w ̸∈Π(v, w). Or we have λ + µ > 1 so that the reﬂection v + w −u of u in the parallelogram’s center is a lattice point in the interior of ∆; a contradiction. Now, every lattice point z ∈Z2 can be expressed in terms of v, w: z = λv + µw = ⌊λ⌋v + ⌊µ⌋w + u where u = (λ −⌊λ⌋)v + (µ −⌊µ⌋)w ∈Π(v, w) but also u = z −⌊λ⌋v −⌊µ⌋w ∈Z2 . By the above we must have u = 0, that is, z is an integer linear combi-nation of v and w. If we apply this to the standard basis vectors e1 and e2, we obtain integral coeﬃcients ﬁtting into the matrix equation [v, w] λ λ′ µ µ′ ! = 1 0 0 1 ! . Because the determinant is multiplicative, the reciprocal of the integer det[v, w] is an integer. Hence, det[v, w] = ±1 and ∆has area 1/2. ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 7 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) We will dwell on this proof a little longer. The method of decomposing an arbitrary lattice point z into an integral linear combination of v and w plus a lattice point from the parallelogram is an instance of a general decomposition, Lemma 3.26, which we will use over and over again. Moreover, this proof shows more than what we set out to prove. It shows that, if ∆= conv(0, v, w) is such a lattice triangle with only three lattice points, then every lattice point is an integral linear combination of v and w and thus the matrix [v, w] has an intergral inverse and its determinant is ±1. This deserves a deﬁnition. Deﬁnition 1.3 A vector space basis v, w of R2 is a lattice basis of Z2 if the set of integral linear combinations equals Z2: {λv + µw : λ, µ ∈Z} = Z2 . In other words, a lattice basis of Z2 consists of two integral vectors so that every other integral vector is an integral linear combination of these two. A change of basis matrix A must have integer entries (Why?), and so must its inverse A−1. We deﬁne Gl2(Z) := {A ∈Gl2(R) : A, A−1 ∈Z2×2}. A change of lattice basis corresponds to a linear map R2 →R2 represented by the change of basis matrix in Gl2(Z). We do not care which lattice basis we use to coordinatize a given lattice polygon. So we will consider to lattice polygons the same if they are related by such a Gl2(Z)-map. Also, it should not matter which lattice point we declare to be the origin. Therefore, we also allow translations by lattice vectors, leading us to consider aﬃne maps x 7→Ax + b, with A ∈Gl2(Z) and b ∈Z2 . We call such maps aﬃne lattice automorphisms of Z2 or unimodular transformations. Deﬁnition 1.4 Two lattice polygons P and P ′ are isomorphic (also called unimodularly equivalent or lattice equivalent), if there is an aﬃne lattice isomorphism mapping P onto P ′. Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 A more general deﬁnition is given in the next chapter (see Deﬁnition 2.74). With the new nomenclature, we can formulate a stronger version of Pick’s Theorem (Proposition 1.2) which follows from our proof. For this let us denote by ∆2 := conv(0, e1, e2) the standard or unimodular triangle. — 8 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 1. An invitation to lattice polytopes (draft of October 7, 2020) Fig. 1.7: Lattice Triangles Proposition 1.5 Any lattice triangle with only three lattice points is isomorphic to ∆2. It is important to build an intuition what these unimodular transforma-tions are, and what they do. Figure 1.7 shows three examples of lattice triangles. The three triangles look quite diﬀerent: their vertices have diﬀerent Euclidean distances and diﬀerent angles. Still, the top one is considerably distinguished from the lower two: it has four lattice points, while the others have only three. They cannot be lattice equivalent, as unimodular transformations preserve the number of lattice points by design. But the lower two triangles are equivalent by Proposition 1.5. If we pick the ﬁlled lattice point as the origin and the other two vertices of the third triangle as basis vectors (1, 0) and (0, 1). Then the second and third triangles are indeed isomorphic by the following aﬃne lattice isomorphism: Z2 →Z2 : x 7→ −1 1 −1 2 ! x + 1 4 ! . There are ﬁve more lattice isomorphisms carrying the third triangle to the second. The reader is invited to ﬁnd them. Exercise 1.6 So angles and Euclidean distances are not preserved by unimodular transformations. But, as we observed at the end of the proof of Pick’s Theorem (Proposition 1.2), a change of lattice basis has determinant ±1 (cf. Exercise 1.7). So unimodular transformations do preserve the area. And there is also a replacement for Euclidean length. Exercise 1.7 Deﬁnition 1.6 The lattice length of a lattice segment e = conv(v, w) ⊂ Rd is length(e) := \|e ∩Zd\| −1 . As this is the only reasonable notion of length in our context, we often refer to the lattice length of a segment simply as its length. The following is a summary of what we know thus far. Proposition 1.7 Equivalent lattice polygons contain the same number of lattice points, they have the same area and the same lattice perimeter. ⊓ ⊔ Pick’s Theorem (Proposition 1.2) about the area of triangles is at the heart of the proof of Pick’s Formula (Theorem 1.8) in the next section. Be Haase, Nill, Paffenholz: Lattice Polytopes — 9 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) warned, however, that the corresponding version fails in higher dimensions. This is famously shown by the so called Reeve simplices: Rd(m) := conv 0, e1, e2, . . . , ed−1, d−1 X i=1 ei + med ! (1.1) Here, d is the dimension of the simplex, and m is a positive integer (its normalized volume, see Deﬁnition 2.80 in the next chapter for this notion). Figure 1.8 depicts the 3-dimensional Reeve simplex R3(4). As the reader is encouraged to prove in Exercise 1.9, any such simplex has only its vertices as lattice points. This shows that in dimension d ≥3 there are inﬁnitely many lattice simplices containing only d + 1 lattice points each and which have pairwise diﬀerent volumes. In particular they are non-isomorphic. As we will discuss later in Section 4.6, this makes life considerably more interesting in higher dimensions, and it highlights once more how remarkable Pick’s Theorem (Proposition 1.2) really is. Exercise 1.9 Fig. 1.8: Reeve’s Tetrahedron 1.3 Triangulations and Pick’s formula The goal of this section is to prove an elegant formula for the computation of the area of a lattice polygon just by counting lattice points. We have seen this miracle already for triangles (Pick’s Theorem (Proposition 1.2)) and we will now generalize it to arbitrary lattice polygons. We thus ﬁnd a ﬁrst relation among the parameters b, i, a from Section 1.1. Keep in mind that the Reeve simplices (1.1) exclude a straightforward generalization to higher dimensions. To ﬁnd some generalization is the topic of Chapter 3 on Ehrhart Theory. Theorem 1.8 (Pick’s Formula) Let P be a lattice polygon with i in-terior lattice points, b lattice points on the boundary, and (Euclidean) area a. Then a = i + b 2 −1 . This shows, for instance, that the triple (5, 2, 4) from Section 1.1 can never come from a lattice polygon. Fig. 1.9: Splitting P into pieces. The ﬁrst image is for the case b ≥4, the second for i ≥1. Proof. We prove this by induction on the number l := b + i of lattice points in the polygon P. The smallest case b = 3 and i = 0 is covered by Pick’s Theorem (Proposition 1.2). There are two cases to consider for the induction: (1) either P has b ≥4 lattice points on the boundary, or (2) b = 3 and we have at least one interior lattice point, i.e. i ≥1. — 10 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 1. An invitation to lattice polytopes (draft of October 7, 2020) If P has at least four lattice points on the boundary then we can cut P into two lattice polygons Q1 and Q2 by cutting along a chord e through the interior of P given by two boundary lattice points. Let Qj, j = 1, 2 have area aj, bj boundary lattice points, and ij interior lattice points, respectively. Let e have ie interior lattice points (and two boundary lattice points). Both Q1 and Q2 have less than l lattice points, so by induction Pick’s Formula holds for Q and Q′, i.e. a1 = i1 + b1 2 −1 , a2 = i2 + b2 2 −1 . Further i = i1 + i2 + ie , b = b1 + b2 −2ie −2 , so a = a1 + a2 = i1 + i2 + 1 2(b1 + b2) −2 = i −ie + 1 2(b + 2ie + 2) −2 = i + b 2 −1 . If b = 3 and i ≥1 then we can split P into three pieces Q1, Q2, and Q3 by coning over some interior point of P. See Figure 1.9. Again, all three pieces have fewer lattice points than P, so we know Pick’s Formula for those by our induction hypothesis. A similar compu-tation as the one above shows that Pick’s Formula also holds for P (see Exercise 1.10). ⊓ ⊔ Exercise 1.10 The reader is invited to prove that the same relation is true for non-convex lattice polygons in Exercise 1.13. One can also prove that this theorem is equivalent to the Euler relation (see Exercise 1.11). Note that the same induction shows that any lattice polygon can be subdivided into triangles which are isomorphic to ∆2 (called unimodular triangles). As the Reeve simplex shows, this is not true in higher dimensions. Questions about the existence of such unimodular triangulations will be discussed in the last Chapter 8 of this book. See also Exercise 1.12. Exercise 1.11 Exercise 1.12 The idea of subdiving a convex object into triangles, or simplices in dimensions 3 and above, is quite inﬂuential. We will devote a whole chapter to this (Chapter 8). Deﬁnition 1.9 (Triangulation) Let P be a lattice polygon. A (lattice) triangulation of P is a collection S of lattice triangles such that (1) Any two triangles ∆d, simplex′ ∈S intersect in an edge of both, and (2) the union (as point sets) of all triangles is P. Fig. 1.10: Not allowed in a triangula-tion Figure 1.10 shows two conﬁgurations of triangles that are not allowed in a triangulation. In the ﬁrst, the intersection is not a full edhe in both, in Haase, Nill, Paffenholz: Lattice Polytopes — 11 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) the second, the triangles intersect in more than an edge. A proper lattice triangulation is shown in Figure 1.11. In fact, Pick’s Formula (Theorem 1.8) also holds in a non-convex setting. We can generalize it to any (non-convex) lattice polygon with the property, that any vertex is incident to exactly two edges and no edges intersect. You will prove this in Exercise 1.13. Exercise 1.13 1.4 A Classiﬁcation of Lattice Polygons Scott’s theorem gives a precise bound on how large a lattice polygon can be, given the number of interior lattice points. It thus provides the answer to all questions (triples) from the end of Section 1.1. Problems like these, which relate information about lattice points of a convex body to its geometry and its invariants are subject of the ﬁeld of Geometry of Numbers. We deal with such questions again in Chapter 4. Theorem 1.10 (Scott, 1976 ) Let P ⊂R2 be a lattice polygon with i ≥1 interior lattice points and Euclidean volume a. Then either (1) P ∼ = 3 ∆2 and hence, a = 9/2 and i = 1, or (2) a ≤2(i + 1). Proof. Let b be the number of boundary lattice points. Using Pick’s Formula (Theorem 1.8) we can reformulate the condition to b ≤a + 4 unless P = 3 ∆2, in which case b = 9 and a = 9/2. p′ = 5 p = 3 Fig. 1.12: P in a box Lattice isomorphisms preserve a, b and i, so we can place P tightly into a rectangle R := [0, p′] × [0, p] and p is the smallest possible among all lattice equivalent P. Then p ≥2 as i ≥1. Exchanging coordinates is a lattice isomorphism, so we have 2 ≤p ≤p′ . (1.2) The polygon P intersects the bottom and top edge of the rectangle in edges of length q and q′, see Figure 1.12. q′ = 3 q = 1 Fig. 1.13: The bot-tom and top edge of P Fig. 1.14: Bounding the volume At most 2(p −1) boundary lattice points of P are not on the two horizontal edges of R, so b ≤q + 1 + p −1 + q′ + 1 + p −1 = q + q′ + 2p. (1.3) Subdividing the convex hull of the top edge and the bottom edge into two triangles as in Figure 1.14 we get a ≥1 2pq + 1 2q′p = p 2(q + q′). (1.4) — 12 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 1. An invitation to lattice polytopes (draft of October 7, 2020) Using Pick’s Formula (Theorem 1.8) and i ≥1 we also know that a ≥b/2 . (1.5) We split the proof into four cases: (1) p = 2 or q + q′ ≥4, (2) p = q + q′ = 3, (3) p = 3 and q + q′ ≤2, (4) p ≥4 and q + q′ ≤3 For (1) we can combine (1.3) and (1.4) into 2b −2a ≤2(q + q′ + 2p) −p(q + q′) = (q + q′ −4)(2 −p) + 8 , (1.6) which is at most 8 as the ﬁrst summand is at most 0. This implies b ≤a + 4 as desired. For (2) we can use the same inequality to obtain 2b −2a ≤9, i.e. b ≤a + 9/2. Now, if P has at least one vertex not on the upper or lower edge of R, then (1.4) and hence also (1.6) become strict inequalities, so that in this case b < a + 9/2. As a ∈1 2Z≥0, this implies b ≤a + 4. If, on the other hand, all vertices of P are on the upper and lower edge of R, then a = p(q+q′)/2 = 9//2 and b ≤a + 9/2 = 9. If b < 9 then b ≤a + 4. Otherwise, all vertices are on the upper and lowe redge of R, b = 9, a = 9/2 and thus i = 1. A simple geometric consideration shows that then either q = 3 or q′ = 3 and P must be the triangle 3 ∆2 (Exercise 1.14). In case (3) the inequlity (1.3) implies b ≤8 and (1.5) L U U′ L′ X Y δ Fig. 1.15: The δ-gap Exercise 1.14 shows Fig. 1.16: Estimating the area of P from below: The triangle XUL b −a ≤b −b/2 ≤4 . Now assume we are in the forth case, so p ≥4 and q + q′ ≤3. We choose points L = (l, 0), U = (u, p), X = (0, x), and Y = (p′, y) in P such that δ := \|u −l\| is minimal. See Figure 1.15. We can assume that u ≤l (otherwise we ﬂip the polygon). Let L′ = (u, 0) and U′ = (l, p). The triangle SL spanned by A, L′, and U bounds the volume of the triangle spanned by X, L, and U (look at the height over the edge XU). Similarly, the triangle SU spanned by Y , U′ and L bounds the volume of the triangle Y , U, and L from below. Hence, a ≥1 2p(p′ −δ) . The shearing v 7→ 1 k 0 1 ! v Haase, Nill, Paffenholz: Lattice Polytopes — 13 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) (a) Realization for b = 3 (b) Realization for 4 ≤b ≤i + 5 (c) Realization for i + 5 ≤b ≤ 2i + 6 Fig. 1.18: Realizations of pairs (i, b) according to Scott’s Theorem is a lattice isomorphism that, if applied to P, leaves p, q, and q′ invariant. Hence, we can transform P such that δ ≤1 2(p −q −q′) . (1.7) As p was chosen to be minimal, we also still have p ≤p′ after this transformation. This implies a ≥1 4p(p + q + q′) , so that 4(b −a) ≤8p + 4q + 4q′ −p(p + q + q′) = p(8 −p) −(p −4)(q + q′) . We have p ≥4, so that the left hand side is bounded by 16. This shows b ≤a + 4. ⊓ ⊔ Note that for polygons that are not lattice polygons, there is no such upper bound on their areas. Figure 1.17 shows why. Scott’s theorem deﬁnes a polyhedral set L given by the inequalities i ≥1 a ≥ 3/2 a ≤2(i −1) such that any pair (a, i) coming from a lattice polygon is either (a, i) = (9/2, 1) or inside L. What about the converse? Is every point (a, i) ∈1 2Z × Z ∩L the volume an number of interior lattice points of some lattice polygon? This is easier to answer when we move from the pair (a, i) used in Scott’s Theorem to the pair (b, i) using Pick’s Theorem. With this transformation our inequalities read i ≥1 b ≥3 b ≤2(i + 3) . We can indeed realize all of these pairs of integers as the number of lattice points in the interior and the boundary of a lattice polygon. Figure 1.18 shows the construction. Exercise 1.15 Exercise 1.16 Exercise 1.17 — 14 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 1. An invitation to lattice polytopes (draft of October 7, 2020) 1.5 Dilations If we know the number of interior and boundary lattice points of a lattice polygon P, then we know its area by Pick’s Theorem. Can we also say something about dilates of P, i.e. about the number of interior or boundary lattice points of k · P for some k ∈Z≥0? The interior points cannot just scale: Even for lattice polygons without interior lattice points all suﬃciently high multiples will contain a lattice point in their interiors. However, the volume clealy scales with k2, and the number of boundary lattice points scales with k. We can plug this into Pick’s Theorem to obtain the number i(k) of interior lattice points in k · P: i(k) = ak2 −b 2k + 1 . which is a polynomial of degree 2 in k with coeﬃcients a, −b/2 and 1. We can reformulate this for the number l(k) = i(k) + b(k) of total lattice points in k · P to obtain l(k) = i(k) + b(k) = k2a −k b 2 + 1 + kb = ak2 + b 2k + 1 , which is again a polynomial of degree 2 in k. Furthermore, we observe that i(k) = l(−k). We will see that this observation is a special case of two much more general and funcamental theorems, the Theorems of Ehrhart and Ehrhart-Macdonald, which we will study in detail in Chapter 3. The catch is that in any dimension the number of lattice points in the k-th dilate of a polytope is given by a polynomial in k, and that the number of interior lattice polytopes is given (up to sign) by evaluating this same polynomial at −k. Exercise 1.18 1.6 Problems included on page 8 1.1. Let P be the lattice polygon with the vertices A, B, C, D (and P ′ respectively with the vertices A′, B′, C′, D′) as given in Figure 1.19. (1) Compute the areas of P and P ′. (2) Are P and P ′ isomorphic? If yes, ﬁnd an explicit unimodular transformation mapping P to P ′. 0 A D C B C′ B′ A′ D′ Fig. 1.19: Two lattice polygons i l d d 8 1.2. HNF in 2d geometrically with Anleitung 1.3. HNF in 2d matrix from geometry with Anleitung Haase, Nill, Paffenholz: Lattice Polytopes — 15 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 1.4. SNF in 2d geometrically with Anleitung included on pa 1.5. SNF in 2d matrix from geometry with Anleitung included on pa 1.6. A vector v ∈Z2 (or in any lattice) is called primitive if it is not a non-trivial integer multiple of some other lattice vector. (1) Show that any primitive v ∈Z2 is part of a lattice basis. (2) Show that every rational simplicial 2-dimensional cone is uni-modularly equivalent to a cone spanned by ( 1 0 ) and ( p q ) for integers 0 ≤p < q. included on page 9 1.7. Show that an integral matrix A ∈Z2×2 has an integral inverse if and only if det A = ±1, that is, Gl2(Z) = A ∈Z2×2 : det A = ±1 . included on page 9 1.8. Show that the converse of Proposition 1.7 is wrong. included on page 10 1.9. Show that Rd(m) are d-dimensional simplices of volume m/d! with d + 1 lattice points. Hint: projection map. included on page 11 1.10. Finish the induction in the proof of Pick’s Formula (Theorem 1.8) for the missing case b = 3 and i ≥1. included on page 11 1.11. Euler’s Formula states that a ﬁnite planar graph with v nodes, e edges and f bounded faces satisﬁes v −e + f = 1 Show that this is equivalent to Pick’s Formula. included on page 11 1.12. UT from maximal collection of non-crossing segments included on page 12 1.13. We have seen in Pick’s Formula (Theorem 1.8) that there is a simple relation between the area and the lattice points of a convex lattice polygon. Prove, that the same relation also holds for non-convex lattice polygons, where a non-convex polygon is a connected subset of R2 bounded by straight noncrossing segments starting and ending in lattice points (the vertices) such that to any vertex there are precisely two incident segments. Hint: The proof for the convex case essentially works. The part that needs consideration is the induction step, where we assume that we can subdivide our polygon with a diagonal. — 16 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 1. An invitation to lattice polytopes (draft of October 7, 2020) 1.14. Show that a polygon with volume 9/2, one interior lattice points, 9 boundary lattice points, and whose vertices are on parallel lines at distance 3 must be the simplex 3 ∆2. 1.15. Describe as precisely as you can which pairs (b, i) can be realized for lattice polygons. Hint: Consider long and ﬂat quadrilaterals in which the interior points are lined up on a straight line. We will study this in detail in Chapter 3 included on page 14 1.16. Let P be a lattice polygon, b(P) the number of boundary lattice points, and i(P) the number of interior lattice points. Prove that a given pair (b, i) of nonnegative integers equals the (b(P), i(P)) for some lattice triangle P if and only if there exist integers A, B, C ∈ Z with A > 0 and 0 ≤B < C such that b = A + gcd(B, C) + gcd(B −A, C) and i = (AC −b)/2 + 1. In this case, the triangle with vertices (0, 0), (A, 0), (B, C) can be chosen. Hint: Move a vertex of the triangle into the origin and use Exer-cise 1.6. included on page 14 1.17. Following up on Exercise 1.16 one can plot (b(P), i(P)) for all lattice triangles P in a given range, see Figure 1.20. Prove that the region at the bottom, denoted by σ◦ 1, is given by Scott’s theorem (the special point (9, 1) is not visible in the very dense plot). Can you also describe the other prominent regions σ◦ i (for i ≥1)? included on page 15 1.18. open problems e.g. from LDP polygon paper. Haase, Nill, Paffenholz: Lattice Polytopes — 17 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 0 500 1000 1500 2000 0 500 1000 1500 2000 σ1o σ2o σ3o σ4o σ5o σ6o i (interior points) b (boundary points) Fig. 1.20: Number of boundary and in-terior lattice points of lattice triangles — 18 — Haase, Nill, Paffenholz: Lattice Polytopes Polytopes and Lattices 2 Contents 2.1 Polyhedra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.1.1 Cones and Polytopes . . . . . . . . . . . . . . . . . . . 20 2.1.2 Convex hulls and half-spaces. . . . . . . . . . . . . 23 2.1.3 The face lattice . . . . . . . . . . . . . . . . . . . . . . . . 26 2.2 Decompositions of Polytopes . . . . . . . . . . . . . 29 2.2.1 Polyhedral Complexes . . . . . . . . . . . . . . . . . . 30 2.2.2 Regular Subdivisions and Triangulations . . 31 2.3 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.3.1 Discrete Subgroup and Lattice Bases . . . . . 33 2.3.2 Coordinates and Normal Forms . . . . . . . . . . 41 2.3.3 Metric Geometry . . . . . . . . . . . . . . . . . . . . . . . 45 2.3.4 Hilbert Bases . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.4 Lattice polytopes . . . . . . . . . . . . . . . . . . . . . . . . 48 2.4.1 Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 2.4.2 Examples of Lattice Polytopes and Constructions. . . . . . . . . . . . . . . . . . . . . . . . . . 49 2.4.3 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 2.5 Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Notation: inner normals running example: cube, hypersimplex Check proof missing/should be re-moved You have seen in the previous chapter how polygons and integer points interact nicely and produce some nice and useful classiﬁcation results. We will show that all these results can, with appropriate modiﬁcations, Lecture Notes Lattice Polytopes (draft of October 7, 2020) actually be carried over to general dimensions. However, before we can start with this in the chapter on Ehrhart Theory (Chapter 3) we should take a closer look at the objects we are considering. We will do this in the next sections, but only very brieﬂy. Most of the topics are covered in other courses, and we will give pointers to other books where appropriate. 2.1 Polyhedra Polyhedral cones are the intersection of a ﬁnite set of linear half spaces. Generalizing to intersections of aﬃne half spaces leads to polyhedra. We are mainly interested in the subset of bounded polyhedra, the polytopes. Specializing further, we will deal with integral polytopes. In the second part of this chapter we link integral polytopes to lattices, which are discrete subgroups of the additive group Rd. This gives a connection to commutative algebra by interpreting a point v ∈Zd as the exponent vector of a monomial in d variables. 2.1.1 Cones and Polytopes We use Z, Q, R and C to denote the integer, rational, real and complex numbers. For X ∈{Z, Q, R} we use X>0 := {x ∈X \| x > 0} X≥0 := {x ∈X \| x ≥0} and similarly X<0 and X≤0. We are mostly concerned with objects that can be deﬁned from a, usually ﬁnite, subset X ⊆Rd. We can study spaces generated by such a set. The most commonly studied notion here is the linear span of X. Deﬁnition 2.1 Let X ⊆Rd. A linear combination of X is a sum v := X x∈X λxx where λx = 0 for all but ﬁnitely x ∈X. The linear hull or linear span lin(X) of X is the set of all linear combinations of X, lin(X) := ( X x∈X λxx : λx ∈R and λx = 0 for all but ﬁnitely many x ) . The set X is a linear space if X equals its linear span. A linear combination is an aﬃne combination if addtionally the sum of the coeﬃcients λx is 1. The aﬃne hull aﬀ(X) of X is the set of all aﬃne combinations, aﬀ(X) := ( X x∈X λxx : λx ∈R with P x∈X λx = 1 and λx = 0 for all but ﬁnitely many x ) . — 20 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) The linear span of X is the smallest linear space containing X and the common intersection of all linear spaces containing X. Similarly, the aﬃne hull of X is the smallest aﬃne space containing X and the intersection of all aﬃne spaces containing X. For a matrix A ∈Rd×n with column vectors a1, . . . , an we also write lin(A) := lin ({a1, . . . , an}) and aﬀ(A) := aﬀ({a1, . . . , an}) A set of points X is linearly or aﬃnely independent if no point of X can be written as a linear or aﬃne combination of the other points. Linear spaces can always be spanned by a ﬁnite subset of X. All minimal such sets, the bases of lin X, have the same size, which is the dimension of lin X. The translation of a subset Y ⊆Rd by a vector t ∈Rd is Y −t := {y −t : y ∈Y } For any aﬃne space A = aﬀX we can consider its translation by a vector x ∈A. This is a linear space. The dimension of A is the dimension of A −x. Hence, any point in the aﬃne hull of X can be written as an aﬃne combination of at most d + 1 points in X. Deﬁnition 2.2 A linear combination is conic if all coeﬃcients are non-negative, and it is convex if it is conic and aﬃne. The set of all conic combinations of a set X is the cone over X, denoted by cone(X). The set of all convex combinations of X is the convex hull conv(X). X is a cone if X = cone(X) and X is a convex set if X = conv(X). A polyhedral cone is the cone of a ﬁnite subset of Rd. A polytope is the convex hull of ﬁnitely many points in Rd. The dimension of a cone is the dimension of its linear span. The dimension of a polytope is the dimension of the aﬃne space it spans. See Figure 2.1 for an example. Again, we sometimes write cone(A) and conv(A) for the conic or convex hull of the set of column vectors of a matrix A ∈Rd×n. We are mostly interested in cones and convex sets deﬁned by a ﬁnite set X. Clearly, if the dimension of a polytope is less Fig. 2.1: Cone (blue) and Polytope (red) for the point set of the red points. than the dimension of the ambient space, then we can restrict to that aﬃne space. Hence, we may assume that the dimension of our polytopes coincides with the dimension of the space (we will see later that it will be useful to also consider lower dimensional polytopes, though). For polytopes in dimension 2 we have already seen the polygons in the previous chapter. Those are all 2-dimensional polytopes. We need at least d + 1 aﬃnely independent points in Rd to aﬃnely span Rd, so any full-dimensional polytope has at least d + 1 points in its deﬁning set. Any polytope deﬁned by precisely d + 1 aﬃnely independent points is called simplex. Any two simplices can be identiﬁed via a bijective Haase, Nill, Paffenholz: Lattice Polytopes — 21 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Fig. 2.2: Simplex, Cube, Cross Poly-tope, Dodecahedron, Icosahedron aﬃne map (if you translate both simplices such that one point is in the origin this a just a change of basis). In dimension 3 there are the famous regular polytopes, which are the cube, the tetrahedron, the octahedron, the dodecahedron, and the icosahedron, see Figure 2.2. Three of them can be generalized to higher dimensions. We have seen the simplex above, which is a tetrahedron in dimension 3. The unit cube Cd is the convex hull of the set X := {0, 1}d. The octrahedron can be realized as the special case d = 3 of the polytope deﬁned as the convex hull of ±ei for 1 ≤i ≤d, where ej is the j-th unit vector in Rd. In general, those polytopes are called cross polytopes. Let us also look at a slightly more complicated, but highly intersting group of polytopes, the hypersimplices. The hypersimplex h(d, k) ⊂Rd for 1 ≤k ≤d −1 is most easily deﬁned as a polytope of one dimension less than its ambient space. It is the convex hull of all vertices of the unit cube whose coordinates sum up to k: h(d, k) := conv x ∈{0, 1}d : d X i=1 xi = k ! = Cd ∩ ( x ∈Rd : d X i=1 xi = k ) . For k = 1 and k = d −1 we obtain a (d −1)-dimensional simplex. You can of course extend the deﬁnition to k = 0 and k = d, but these are just single points in Rd. Deﬁnition 2.3 (boundary and interior points) Let K be a convex set. A point x ∈K is an interior point of K if there is some ε > 0 such that Bx(ε) ⊆K. Otherwise x is a boundary point. x ∈K is a relative interior point of K if it is an interior point of K if considered as a subset of aﬀK. See also Figure 2.3. Bε(b) Bε(i) Fig. 2.3: An interior point i and a boundary point b. Similar to the linear and aﬃne spaces above any point x in a cone can be written as the conic generation of at most d elements of X, and a point in the convex hull as the convex combination of at most d + 1 elements of X. Diﬀently from above, however, the choice of these points — 22 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) depends on x. The followinf theorem, whose proof ist left as Exercise 2.1, makes this precise. Theorem 2.4 (Carathéodory’s Theorem) Let X ⊆Rd, C = cone(X), and y ∈C. Then there are x1, x2, . . . , xd ∈X and λ1, λ2, . . . , λd ≥0 such that y = d X i=1 λixi . Similarly, for P = conv(X) and z ∈P there are x0, x1, . . . , xd ∈X and λ0, λ1, . . . , λd ≥0 such that z = d X i=0 λixi and d X i=0 λi = 1 . See also Figure 2.4 v1 v3 v5 x v2 v4 v6 Fig. 2.4: x can be written as a convex combination of v1, v3 and v5. Exercise 2.1 2.1.2 Convex hulls and half-spaces There is a second deﬁnition of a polytope that we want to introduce now. Deﬁnition 2.5 (hyperplanes and half-spaces) For any non-zero func-tional a ∈(Rd)⋆and β ∈R the set H := {x \| ⟨a, x ⟩≤β} is the aﬃne hyperplane deﬁned by a and β. An aﬃne hyperplane is a linear hyperplane if β = 0. The (negative) half-space corresponding to an aﬃne hyperplane is H−:= {x \| ⟨a, x ⟩≤β} . We say that a point y ∈Rd is beneath H if ⟨a, y ⟩< β and beyond H if ⟨a, y ⟩> β. Note that λa, λβ for any λ ̸= 0 deﬁnes the same hyperplane as a, β, and the same aﬃne half space if λ > 0. Hence, the deﬁning functional for a hyperplane or half space is unique only up to a non-zero and positive factor, respectively. Deﬁnition 2.6 A polyhedron P is the intersection of ﬁnitely many aﬃne half spaces, P = \ {x \| ⟨ai, x ⟩≤βi} = {x \| ⟨a1, x ⟩≤β1, . . . , ⟨am, x ⟩≤βm} for ai ∈Rd and βi ∈R and 1 ≤i ≤k. This is often written in the more consise form Haase, Nill, Paffenholz: Lattice Polytopes — 23 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) P = { x \| Ax ≤b } where A ∈Rk×d whose rows are the functionals a1, a2, . . . , ak and b is the vector with entries β1, β2, . . . , βk. Example 2.7 We look at some simple examples. (1) The unit cube Cd is deﬁned by the inequalities xi ≥0 xi ≤1 for 1 ≤i ≤d . (2) The inequalities x1 ≥0 x1 −x2 ≤2 x1 + 3x2 ≤10 3x1 −x2 ≥0 deﬁne a polygon with vertices " 0 0 # " 2 0 # " 4 2 # " 1 3 # . see Figure 2.5. x1 ≥0 x1 −x2 ≤2 x1 + 3x2 ≤10 3x1 −x2 ≥0 Fig. 2.5: The poly-gon of Example 2.7(2) Deﬁnition 2.8 A polyhedron P is a (polyhedral) cone if all deﬁning inequalities are linear, that is, P = \ H− ai,0 (2.1) for some a1, a2, . . . , ak ∈(Rd)⋆. We have already deﬁned a cone over a set X as the set of all conic combinations in the previous section. We will see below that this and the newly deﬁned notion of a polyhedral cone coincide if X is a ﬁnite set, i.e. any polyedral cone can equally be described as the cone over some suitably chosen ﬁnite set X, and any cone over a ﬁnite set is polyhedral. We will not encounter non-polyhedral cones, that is, cones deﬁned as the set of conic combinations over an inﬁnite set X, in this book. Therefore, we will often omit the word polyhedral and just speak of cones in the text, and only stress this restriction in deﬁnitions and theorems. The dimension of such a polyhedron deﬁned by half spaces is again deﬁned as the dimension of its aﬃne hull. We sometimes use the notion d-polytope for a d-dimensional polyhedron. A polyhedron is full dimensional if dim P = d. Deﬁnition 2.9 Let P = T H− ai,βi be a polyhedron. The recession cone and lineality space of P are rec P = \ H− ai,0 and lineal P = \ Hai,0 . A polytope is pointed if lineal P = ∅. — 24 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) Example 2.10 1 We can associate a cone to each polyhedron P ⊆Rd that esssentially has the same combinatorial and geometric properties. This is the homoge-nization of P or just the cone over P deﬁned by C(P) := cone({1} × P) ⊆Rd+1 , (2.2) so if P := {x \| ⟨a1, x ⟩≤β1, . . . , ⟨am, x ⟩≤βm} ⊆Rd with ai ∈(Rd)⋆, βi ∈R for i ∈[m] then C(P) = {(x0, x) \| −β1x0 + ⟨a1, x ⟩≤0, . . . , −βmx0 + ⟨am, x ⟩≤0} . It is often convenient to look at the homogenization of the polyhedron instead of the polyhedron itself as it is deﬁned by linear instead of aﬃne inequalities. We can recover the polyhedron by intersecting the cone with the hyperplane x0 ≡1 (and projecting). Deﬁnition 2.11 (Minkowski sum) The Minkowski sum of two sets X, Y ⊆Rd is the set X + Y := {x + y \| x ∈X, y ∈Y } . Fig. 2.6: The orange polygon is the Minkowski sum of the red and blue polygons A set X is ﬁnitely generated if it can be written as a Minkowski sum of a polytope, a cone, and a linear space, that is, there are vi ∈Rd, i = 1, . . . , r, rj ∈Rd, j = 1, . . . , s, vbk ∈Rd, k = 1, . . . , t such that X =    r X i=1 λivi + s X j=1 µiri + r X k=1 νibi : λi, µj, νk ∈R, λi, µj ≥0, Pr i=1 λi = 1   . (2.3) Theorem 2.12 (Weyl-Minkowski Theorem) Let P ⊆Rd. Then P is a polyhedron if and only if it is ﬁnitely generated. A proof of this theorem can be found in . Projections of polyhedra are again polyhedra, ﬁnitely generated by the projections of the generators. Example 2.13 In the notation of Weyl-Minkowski Theorem (Theo-rem 2.12) we can deﬁne a polyhedron with v1 := " 1 3 # v2 := " 1 2 # v3 := " 2 1 # r1 := " 1 3 # r2 := " 3 1 # , see Figure 2.7. It is deﬁned by the inequalities 3x1 −x2 ≥3 x1 ≥1 x1 + x2 ≥3 x1 −3x2 ≤−1 . Haase, Nill, Paffenholz: Lattice Polytopes — 25 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) r1 v1 v2 v3 The Minkowski sum of a polytope with a cone C or a linear space L is unbounded if C or L have positive dimension. Hence, we can deduce the following duality for polytopes from the Weyl-Minkowski Theorem. Corollary 2.14 (Weyl-Minkowski-Duality) A bounded set P ⊆Rd is a polytope if and only if it is the bounded intersection of a ﬁnite number of aﬃne half spaces. ⊓ ⊔ From this theorem we obtain two equivalent descriptions of a polytope: (1) as the convex hull of a ﬁnite set of points in Rd, (2) as the bounded intersection of a ﬁnite set of aﬃne half spaces. The ﬁrst is called the interior or V-description, The second is the exterior or H-description. Both are important in polytope theory, as some things are easy to describe in one and may be diﬃcult to deﬁne in the other. 2.1.3 The face lattice Throughout this section let P := {x \| ⟨a1, x ⟩≤β1, . . . , ⟨am, x ⟩≤βm} be a polyhedron deﬁned by ai ∈(Rd)⋆, βi ∈R for i ∈[m]. We have seen in the examples above that some intersections of the hyperplanes distin-guish lower dimensional subsets of a polyhedron. We want to formalize this observation in this section. A hyperplane H := {x \| ⟨a, x ⟩≤β} for some a ∈(Rd)⋆and β ∈R deﬁnes a valid hyperplane if P is contained in the negative half space of H, that is, if ⟨a, x ⟩≤β for all x ∈P. A valid hyperplane is supporting if P ∩H is non-empty. Deﬁnition 2.15 (faces) Let P be a polytope. A face F of P is either P itself or the intersection of P with a valid linear hyperplane. If F ̸= P then F is a proper face. Observe that the empty set is also a face of P. For any face F we have F ∩P = aﬀF ∩P , so faces of polyhedra are again polyhedra and a face of a face of the polyhedron is a face of the polyhedron. The dimension of a face of a polyhedron P is its dimension as a polyhedron, dim F := dim aﬀF . We sometimes use the notion k-face for a k-dimensional face of a polytope P. If for a polyhedron P and a functional a ∈(Rd)⋆the value β of rephrase max{⟨a, x ⟩\| x ∈P} — 26 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) is ﬁnite then H := {x \| ⟨a, x ⟩≤β} is a supporting hyperplane of P and P ∩H is a face of P, the face deﬁned by a. The functionals deﬁning a face are exactly those in the negative dual of the recession cone. Any functional ai in the deﬁnition of P for some i ∈[m] is an implied equality if ⟨a, x ⟩= βi for all x ∈P. The set of all implied equalities of P is eq(P) := {j ∈{1, . . . , m} \| ⟨aj, x ⟩= βj for all x ∈P} . Observe that this is a property of the speciﬁed hyperplane description, not of the polytope itself. The aﬃne hull of P is then given by the intersection of the implied equalities, aﬀ(P) = \ j∈eq(P ) {x \| ⟨aj, x ⟩= βj} . The hyperplane description is irredundant if no proper subset of the half spaces deﬁnes the same polytope, and redundant otherwise. Let P := be a polyhedron. A point x ∈P is a relative interior point of P if ⟨ai, x ⟩= βi for all i ∈eq(P) ⟨ai, x ⟩< βi for all i ̸∈eq(P) . Any polytope of dimension d ≥1 has a relative interior point. Observe that this notion of relative interior points coincides with the one given in Deﬁnition 2.3. So we have two diﬀerent ways to check whether a point is in the realitve interior of a polyhedron. If F is a proper face of P, then F = {x \| ⟨ai, x ⟩= βi for i ∈I} ∩P for a subsystem I ⊆[m] of the inequalities of P. In particular, P has only a ﬁnite number of faces. A proper face F of P is a facet if it has dimension dim P −1. Now assume that P is full dimensional and the deﬁning functionals α1, . . . , αm are irredundant. Then F is a facet of P if and only if F = {x \| Q aix = βi} ∩P for some i ∈[m]. Furthermore, if P is full dimensional, then a1, . . . , am are unique up to scaling with a positive factor. Also, any proper face of P is contained in a facet, and if F1, F2 are proper faces, then F1 ∩F2 is a proper face of P. A face F of P is minimal if there is no non-empty proper face G of P with G ⊊F. F is minimal if and only if F = aﬀF if and only if it is a translate of lineal P. The minimal faces of a pointed polyhedron are called vertices. They are points in Rd. The set of all vertices is denoted by V(P). If P is pointed, then F is an edge of P, if e is a segment, and a extremal ray otherwise. If P is a cone, then F is called a minimal proper face. Two vertices of P are adjacent if they are contained in the same edge. Faces of a polyhedron are ordered by inclusion. Hence, the set faces(P) of all faces of P (including the empty set and P itself) is a poset, which is actually an Eulerian lattice, the face Haase, Nill, Paffenholz: Lattice Polytopes — 27 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) lattice of P. A k-face F and a j-face G of P are incident if either F is a face of G or vice versa. The f-vector (or face vector) of P is the vector f(P) := (f0(P), . . . , fd−1(P)) , where fi(P) is the number of i-dimensional faces of P, for 0 ≤i ≤d −1. Exercise 2.5 Exercise 2.6 Let P ⊆Rd be a full-dimensional polyhedron with 0 ∈int P. The polar dual of a polytope P is Theorem 2.16 Let P ⊆Rd be a d-dimensional polytope with 0 ∈int P and vertices v1, . . . , vm. Then P ⋆is deﬁned by the inequalities P ⋆:= n a ∈(Rd)⋆: ⟨a, v1 ⟩≤1, . . . , ⟨a, vm ⟩≤1 o and its vertices are 1/bjaj for each facet deﬁning inequality ⟨aj, x ⟩≤bj of P. ⊓ ⊔ You will show in Exercise 2.7 that dualizing twice gives back the original polytope. Exercise 2.7 Corollary 2.17 Let P ⊆Rd be a d-dimensional polytope with 0 ∈ int P. Then we have a bijective correspondence between k-faces of P and (d −1 −k)-faces of P ⋆for 0 ≤k ≤d −1 and if a k-face F of P is contained in a (k + 1)-face G of P, then the face corresponding to G in P ⋆is contained in the face corresponding to F in P ⋆. ⊓ ⊔ Corollary 2.18 If f = (f1, . . . , fd−1) is the f-vector of P and f′ = (f′ d−1, . . . , f′ 0) that of P ⋆then fi = f′ d−1−i for 0 ≤i ≤d −1 . ⊓ ⊔ For the next observations we switch to the interior description of a polytope. Let ϕ : Rd →Re be an aﬃne map and P = conv(v1, . . . , vn) + cone(w1, . . . , wl) a polyhedron for some v1, . . . , vn, w1, . . . , wl ∈Rd. Then ϕ   d X i=1 λivi + l X j=1 µjwj  = d X i=1 λiϕvi + l X j=1 µjϕwj , so ϕP is again a polyhedron. Deﬁnition 2.19 (aﬃne equivalence) Let P ⊆Rd and Q ⊆Re be two polytopes. P and Q are aﬃnely equivalent if there are aﬃne maps ϕ : Rd →Re and ψ : Re →Rd such that ϕP = Q ψQ = P . — 28 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) A polytope is simplicial if all faces are simplices. It is simple if all k-dimensional faces are incident to exactly dim P −k faces. It suﬃces to check this condition for the vertices. Simplicial and simple are dual notions, that is, the dual of a full dimensional simplicial polytope is simple, see Exercise 2.8. f-vectors of simplicial polytopes have been completely charecterized in the g-Theorem of Billera, Lee and Stanley following a conjec-ture of McMullen . This theorem can best be described with a linear The following needs more background transformation of the f-vector, which we introduce now. We can write the f-vector as a polynomial in the form f(t) := (t −1)d + d X i=1 fi−1(t −1)d−i . Writing this polynomial in the basis 1, t, t2, . . . , td gives the h-polynomial f(t) = d X j=0 hjtj . We also need the following notion of an M-sequence. For any integers hi as in-/out-edges in a simple poly-tope n, k ≥1 there is a unique way to express n in the form n = ak k + ak−1 k −1 + · · · + ai i with ak ≥ak−1 ≥. . . ≥ai ≥i ≥1. We deﬁne n⟨k⟩:= ak−1 k −1 + ak−2 k −2 + · · · + ai−1 i −1 and 0⟨0⟩= 0. A nonnegative sequence (m0, m1, m2, . . .) is an M-sequence if m0 = 1 and m⟨k⟩ k ≤mk−1 for all k ≥2. With this notion we have the following theorem. Theorem 2.20 (g-Theorem) A sequence h = (h0, . . . , hd) is the h-vector of a simplicial polytope if and only if (1) hi = hd−i for i = 0, . . . , ⌊d/2⌋ (2) (g0, g1, . . . , g⌊d/2⌋) is an M-sequence, where g0 = h0 and gi = hi − hi−1 for i = 1, . . . , ⌊d/2 ⌋. Example 2.21 2 2 example missing 2.2 Decompositions of Polytopes We start our considerations with subdivisions of polytopes into smaller pieces and study polyhedral complexes and triangulations. Haase, Nill, Paffenholz: Lattice Polytopes — 29 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 2.2.1 Polyhedral Complexes F is n Th pu Deﬁnition 2.22 (polyhedral complex) A polyhedral complex C is a ﬁnite family of polyhedra (the cells of the complex) such that for all P, Q ∈C (1) if P ∈C and F is a face of P then F ∈C, and (2) F := P ∩Q is a face of both P and Q. A cell P is maximal if there is no Q ∈C strictly containing it. The dimension of C is the maximal dimension of a cell of the complex. A complex is pure if all maximal cells have the same dimension. In this case the maximal cells are the facets of the complex. We will denote by C[k] the set of k-dimensional faces of C. A polyhedral complex S is a subcomplex of C if its cells are a subset of the cells of C. Example 2.23 Here are some examples of a polyhedral complex. See also Figure 2.8. (1) Any polytope or cone can be viewed as a polyhedral complex. This complex has one maximal cell, the cone or polytope itself. This is also called the trivial subdivision of the cone or polytope. In general, subdivisions are deﬁned with the next deﬁnition below. (2) The boundary complex of a d-dimensional polytope naturally has the structure of a pure polyhedral complex. The maximal cells are the facets of the polytope, and its dimension is d −1, the dimension of the facets of the polytope. (3) See the middle ﬁgure in Figure 2.8 for a non-pure polyhedral complex. It has three 2-dimensional maximal cells and one 1-dimensional maximal cell. Deﬁnition 2.24 (face vector) The face vector of a pure d-dimensional polyhedral complex C is the vector f(C) = (f−1, f0, . . . , fd) where fk counts the number of k-dimensional faces of C. Observe that this is completely analogous to our earlier deﬁnition of the face vector of a polytope. Further, f−1 corresponds to the empty face, hence, f−1 = 1 for any polyhedral complex. We will see later that the entries of the face vector satisfy a linear relation, the Euler equation. The Euler characteristic of the complex C is χ(C) := −f−1 + f0 −f1 + . . . + (−1)dfd . — 30 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) This satisﬁes some addition formula. Let C and C′ be two polyhedral complexes such that C ∩C′ is a subcomplex of both. Then there union is also a polyhedral complex and χ(C) + χ(C′) = χ(C ∪C′) −χ(C ∩C′) . (2.4) Deﬁnition 2.25 (fan) A fan is a pure connected polyhedral complex such that all cells of the complex are cones. Deﬁnition 2.26 (normal cone) The normal cone NP (F) of a face F of a polytope P is the set of linear functionals a such that there is some β with ⟨a, F ⟩= β and ⟨a, P ⟩≤β. Proposition 2.27 The normal cone is a polyhedral cone spanned by the facet normals deﬁning the face F. ⊓ ⊔ Deﬁnition 2.28 (normal fan) The normal fan of a polytope P is the collection of all normal cones of proper faces of P. Fans naturally have the structure of a polyhedral complex. In this case all cells are cones. Deﬁnition 2.29 (tangent cone) Let P be a d-polytope and F a face of P. The tangent cone T F P of F is the cone T F P := {p + v ∈Rd \| p ∈F, p + εv ∈P for some ε > 0} . The tangent cone is the common intersection of all supporting half-spaces at F. Note that the tangent cones are not cones in the usual sense, as their apex is not in the origin. We call them aﬃne cone if we want to emphasize this. We can use a point w ∈F to shift the cone into the origin. The following proposition is proved in Exercise 2.9. Proposition 2.30 The shifted cone T F P −w is dual to the normal cone of F. Exercise 2.9 Exercise 2.10 2.2.2 Regular Subdivisions and Triangulations (regular) triangulations of cones and fans. half-open decomposition Lots of stuﬀmissing in here Often it is useful to subdivide a polytope into smaller pieces and look at the pieces separately. It will turn out that the most useful subdivisions are those where all pieces are simplices. Such subdivisions are called triangulations of the polytope. The next deﬁnition formalizes this notion. Deﬁnition 2.31 (Subdivision and Triangulation) A subdivision of a polytope P is a pure polyhedral complex S such that P = S C∈S C. A subdivision is a triangulation of P if all cells are simplices. Haase, Nill, Paffenholz: Lattice Polytopes — 31 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Example 2.32 3 3 example A subdivision or triangulation is without new vertices, if V(∆d) ⊆V(P) for any ∆d ∈T. We will use the basic fact that for every ﬁnite V ⊂Rd the polytope conv V has a triangulation with vertex set V . Similarly, the cone pos V has a triangulation with rays {R≥0v : v ∈V } . Deﬁnition 2.33 (regular subdivision) A subdivision S of a polytope with vertices {v1, . . . , vm} (of the subdivision) is regular if there is a weight vector w such that S is the projection of the lower hull of conv((wi, vi) \| 1 ≤i ≤m) , where the lower hull is the polyhedral complex of those facets whose normal has negative ﬁrst coordinate. Given a set of points V := {v1, . . . , vm} and a weight vector w ∈Rm we denote by Sw(V ) the regular subdivision obtained as the lower hull of lift(w) := conv((wi, vi) \| 1 ≤i ≤m). You will show in Exercise 2.11 that all subdivisions of a polygon using only the vertices of the polygon are regular. WISHLIST: convex piecewise Exercise 2.11 linear function Ψw(x) = min{h : (h, x) ∈lift(w)} Deﬁnition 2.34 (polyhedral sphere, polyhedral ball) Theorem 2.35 Every d-polytope P has a regular triangulation using only the vertices of the polytope. Proof. Let V := V(P) be the vertices of the polytope. We can assume that P is full dimensional. We claim that any suﬃciently generic vector w induces a regular triangulation. The subdivision induced by w is a triangulation if and only if for each facet of the lower hull of lift(w) is a d-simplex, i.e. if at most d + 1 of the points (w1, v1), . . . , (wd, vd) lie on a common hyperplane. For any (d + 2)-tuple (wi1, vi1), . . . , (wid+2, vid+2) being on a common hyperplane means that the determinant det    1 1 · · · 1 wi1 wi2 · · · wid+2 vi1 vi2 · · · vid+2    vanishes. We can view this determinant as a linear functional in the entries of w. There are ( m d+2) diﬀerent such functionals, hence, the complement Zc of the union of the zero sets of these functionals is not empty. Choosing any w ∈Zc satisﬁes our requirements. ⊓ ⊔ — 32 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) It is important to realize that not all triangulations of a polytope are regular. See e.g. Figure 2.9 for a simple example. You will prove that it is indeed not regular in Exercise 2.12 (1) intersections (2) reﬁnements Corollary 2.36 Every pointed cone C can be triangulated into simplicial cones without introducing new generators. Proof. If C is pointed, then there is a functional u such that utx > 0 for all x ∈C . Then P := C ∩{x \| utx = 1} is a polytope, and C is the cone over P. By the previous Theorem 2.35 P has a regular triangulation T without new vertices. The cones over the cells in this triangulation give a triangulation of the cone C without using new generators. ⊓ ⊔ Exercise 2.12 Fig. 2.9: A non-regular subdivision 2.3 Lattices We introduce the central tool for this book. It will link our geometric objects, the polytopes, to algebraic objects, namely toric ideals and toric varieties. Throughout this section, V will be a ﬁnite-dimensional real vector space equipped with the topology induced by a norm ∥. ∥and with a translation invariant volume form. motivate this shift from Rd to V 2.3.1 Discrete Subgroup and Lattice Bases Lattices can be deﬁned in two diﬀerent (but equivalent) ways. On the one hand as the integral generation of a linearly independent set of vectors, on the other hand as a discrete abelian subgroup of the vector space. We will start with the latter characterization of a lattice, This is often very useful to describe lattices without the explicit choice of a basis. We will deduce the other representation in a sequence of propositions that introduce some interesting structure ofr lattices. Recall that a subset Λ ⊆V is an additive subgroup of V if (1) 0 ∈Λ (2) x + y ∈Λ for any x, y ∈Λ (3) −x ∈Λ for any x ∈Λ . Haase, Nill, Paffenholz: Lattice Polytopes — 33 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Deﬁnition 2.37 (lattice, rank) A lattice Λ in V is a discrete additive subgroup Λ of V : for all x ∈Λ there is ε > 0 such that Bε(x) ∩Λ = {x}. The rank of Λ is the dimension of its linear span, that is, rank Λ := dim lin Λ. Exercise 2.13 Exercise 2.14 Note that this notion of a lattice is not connected to the face lattices that we looked at earlier. You will show in Exercise 2.15 that this deﬁnition is independent of the chosen norm, and in Exercise 2.16 that one can choose the same ε for all x ∈Λ. Exercise 2.15 Exercise 2.16 Example 2.38 (1) The standard integer lattice is the lattice spanned by the d standard unit vectors e1, . . . , ed. It is commonly denoted by Zd. We will later see that essentially any lattice looks like this integer lattice. See Figure 2.10. Fig. 2.10: The lattice Z2 (2) Root systems are a famous class of lattices. We introduce some of them here, and you can explore more in the exercises. a) We can identify Rd with the linear subspace L := ( x ∈Rd+1 : d X i=0 xi = 0 ) . The set Ad := L ∩Zd+1 is a lattice in L. Ad is clearly discrete, as it is a subset of a discrete set, and the addition of any two elements in A stays in L, as this is a linear subspace. The same is true for the multiplication by −1. This is the root lattice Ad. See Figure 2.11. Fig. 2.11: The lattice A2 b) Let Dd be the set Dd := ( x ∈Zd : d X i=1 xi is even ) . (2.5) Again, this is a discrete set and addition and multiplication by −1 stay inside the set. This is the root lattice Dd. (3) Subgroups of lattices are again lattices. To make a concrete example for this, the set Λ2;3 := x ∈Z2 : x1 + x2 ≡0 mod 3 is a lattice. (4) Let Λ be a lattice and L ⊂V be a proper linear subspace of V . Then Λ ∩L is a lattice in L. Exercise 2.17 For a set B = {b1, . . . , bd} ⊂V of linearly independent vectors we deﬁne the subgroup Λ(B) := ( d X i=1 λibi \| λi ∈Z, 1 ≤i ≤d ) — 34 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) Deﬁnition 2.39 (lattice basis) A linearly independent subset B ⊂V is called a lattice basis (or Λ-basis) if it generates the lattice: Λ = Λ(B). Example 2.40 B := (" 3 0 # , " −1 1 #) is a basis of the lattice in Ex-ample 2.38(3). Deﬁnition 2.41 (parallelepiped) For a ﬁnite subset A = {v1, . . . , vk} ⊂ Rd the half-open zonotope Π(A ) spanned by these vectors is the set Π(A ) := ( k X i=1 λivi \| 0 ≤λi < 1 for 1 ≤i ≤k ) . If A is linearly independent, the zonotope is a parallelepiped. Here is one of the most fundamental deﬁnitions for lattices. Deﬁnition 2.42 (fundamental parallelepiped) Let Λ be a lattice with basis B = {b1, . . . , bd}. The parallelepiped Π(B) is the fundamental parallelepiped of the lattice with basis B. See Figure 2.12 for an example. Clearly, the fundamental parallelepiped 0 u v Fig. 2.12: A parallelepiped spanned by some vectors depends on the chosen basis. However, its volume, the determinant of the lattice, does not, and we obtain a very nice representation of points in the underlying vector space with the following proposition. Proposition 2.43 Let Λ be a lattice in V and assume it has a basis B = {b1, . . . , bd}. Then any point x ∈lin Λ has a unique representation x = a + y for a ∈Λ and y ∈Π(B). Proof. There are unique λ1, . . . , λd ∈R such that x = Pd i=1 λibi. Set a := Pd i=1 ⌊λi⌋bi and y := Pd i=1 {λi} bi. Then y ∈Π(B), a ∈Λ, and x = a + y. Now assume that there is a second decomposition x = a′ + y′ with a ̸= a′ (and thus also y ̸= y′). We can write y and y′ as y = d X i=1 αibi y′ = d X i=1 α′ ibi for some 0 ≤αi, α′ i < 1, 1 ≤i ≤d. Hence, \|αi −α′ i\| < 1. From a′ −a = y −y′ = d X i=1 (αi −α′ i)bi and a′ −a ∈Λ we know that αi −α′ i ∈Z for 1 ≤i ≤d. Hence, αi −α′ i = 0, so y = y′. Hence, also a = a′. ⊓ ⊔ From this theorem it follows immediately that the parallelepipeds of a lattice with basis B tile the space, see also Exercise 2.18. Haase, Nill, Paffenholz: Lattice Polytopes — 35 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Corollary 2.44 Let Λ be a lattice in Rd and assume it has a basis B := {b1, . . . , bd} and let Π := Π(b1, . . . , bd) be the fundamental parallelepiped. Then Rd is the disjoint union of all translates of Π by vectors in Λ. ⊓ ⊔ Exercise 2.18 We now show that any set B of linearly independent vectors actually generates a lattice with basis B. Lemma 2.45 Let B = {b1, . . . , bd} ⊂V be linearly independent. Then the subgroup Λ(B) := ( d X i=1 λibi : λi ∈Z, 1 ≤i ≤d ) generated by B is a lattice. Proof. The linear map Rd →lin B given by λ 7→Pd i=1 λibi is bijective, and hence a homeomorphism. It maps the discrete set Zd ⊂Rd onto Λ(B). Let z ∈Rd ∩Π(b1, . . . , bd) be an interior point of Π(b1, . . . , bd). Then there is ε > 0 such that Bε(z) ⊆Π(b1, . . . , bd). We claim that Bε(x) ∩Λ = {x} for all x ∈Λ. Indeed, if y ∈Bε(x) ∩Λ = {x} and y ̸= x, Then x′ := x −y ∈Λ and x′ + z ∈Π(b1, . . . , bd), a contradiction to Proposition 2.43. ⊓ ⊔ Theorem 2.46 Every lattice has a basis. For the proof we need some prerequisites. The following lemma is imme-diate from the deﬁnition. Lemma 2.47 If K ⊂V is bounded, then K ∩Λ is ﬁnite. ⊓ ⊔ Deﬁnition 2.48 (Λ-rational subspace) A subspace U ⊆V is Λ-rational if it is generated by elements of Λ. Proposition 2.49 Let V be a ﬁnite-dimensional real vector space, let Λ ⊂V be a lattice, and let U ⊆V be a Λ-rational subspace. Denote the quotient map π : V →V /U . (1) Then π(Λ) ⊂V /U is a lattice. (2) Furthermore, if Λ ∩U has a basis b1, . . . , br, and π(Λ) has a basis c1, . . . , cs, then any choice of preimages ˆ ci ∈Λ of the ci for 1 ≤i ≤s yields a Λ-basis b1, . . . , br, ˆ c1, . . . , ˆ cs. In the situation of the proposition, we will often write Λ/U for π(Λ). Proof. (1) π(Λ) is the image of a group under a homomorphism. Hence, it is a subgroup of V /U. The hard part of the proposition is to prove that π(Λ) is discrete in V /U. The space U is Λ-rational. So we can choose a vector space basis {v1, . . . , vr} ⊂Λ ∩U of U. We can extend this basis to a vector space — 36 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) basis B = {v1, . . . , vd} ⊂Λ of lin Λ. These bases yield maximum norms d X i=1 λivi := max ({\|λi\| : i = 1, . . . , d}) on lin Λ and d X i=1 λivi ! + U ′ := max ({\|λi\| : i = r + 1, . . . , d}) on lin Λ/U. Denote the unit ball of lin Λ by W. By Lemma 2.47, the set W ∩Λ is ﬁnite. Set ε := min {1} ∪{∥v + U∥′ : v ∈W ∩Λ \ U} . This minimum over a ﬁnite set of positive numbers is positive. Now suppose v = d X i=1 λivi ∈Λ with ∥v + U∥′ < ε. Then v′ := r X i=1 (λi −⌊λi⌋)vi + d X i=r+1 ⌊λi⌋vi ∈Λ represents the same coset: v + U = v′ + U, and v′ ∈W ∩Λ. We conclude v′ ∈U and thus v′ + U = 0 ∈V /U. (2) Let b1, . . . , br, ˆ c1, . . . , ˆ cs be as in the proposition, and let v ∈Λ. Be-cause the cj form a lattice basis of π(Λ), there are integers λ1, . . . , λs so that π(v) = Ps j=1 λjcj. Thus, v −Ps j=1 λj ˆ cj ∈ker π = U. Be-cause the bi form a lattice basis of Λ ∩U, there are integers µ1, . . . , µr so that v −Ps j=1 λj ˆ cj = Pr i=1 µibi. So b1, . . . , br, ˆ c1, . . . , ˆ cs generate Λ. They must be linearly independent for dimension reasons. ⊓ ⊔ Deﬁnition 2.50 (primitive vector) A non-zero lattice vector v ∈Λ is primitive if it is not a positive multiple of another lattice vector, i.e. conv(0, v) ∩Λ = {0, v}. Proof (of Theorem 2.46). We proceed by induction on r := rank Λ. For r = 0, the empty set is a basis for Λ. For r = 1, a primitive vector yields a basis. Assume r ≥2. Let b ∈Λ be primitive, and set U := lin b. Then {b} is a basis for U ∩Λ, and Λ/U is a lattice by the ﬁrst statement of Proposition 2.49. Because rank Λ/U = r −1, it has a basis by induction. By the second statement of Proposition 2.49, we can lift to a basis of Λ. ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 37 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Exercise 2.19 Deﬁnition 2.51 (unimodular transformation) Let Λ and Λ′ be lat-tices. A linear map T: lin Λ →lin Λ′ which induces a bijection Λ →Λ′ is called unimodular or a lattice transformation. T is a lattice isomorphism if Λ = Λ′. Deﬁnition 2.52 (sublattice and index) Let Λ ⊂Rd be a lattice. Any lattice Γ ⊆Λ is a sublattice of Λ. Sets of the form a + Γ := {a + x \| x ∈Γ} for some a ∈Λ are cosets of Γ in Λ. The set of all cosets is Λ/Γ. The size \|Γ/Λ\| is the index of Γ in Λ. Theorem 2.53 Let Λ′ ⊆Λ be lattices with lin Λ = lin Λ′. Then there is a basis b1, . . . , br of Λ and integers k1, . . . , kr ∈Z>0 with λi\|λi+1 for 1 ≤i ≤d −1 such that k1b1, . . . , krbr is a basis of Λ′. Proof. We proceed by induction on r := rank Λ = rank Λ′. For r = 1, a Λ-primitive vector has a positive integral multiple which is Λ′-primitive. Assume r ≥2. Because lin Λ = lin Λ′, for every v ∈Λ there is a positive integer k so that kv ∈Λ′. Choose br ∈Λ and kr ∈Z>0 so that br is Λ-primitive, and so that kr is minimal. Set U := lin br. Then br is a basis for U ∩Λ, and krbr is a basis for U ∩Λ′. By Proposition 2.49, Λ′/U ⊆Λ/U are lattices of rank r −1. By induction, there is a basis b1, . . . , br−1 of Λ/U together with positive integers k1, . . . , kr−1 so that k1b1, . . . , kr−1br−1 is a basis for Λ′/U. Let ˆ bi ∈Λ be representatives of the bi for i = 1, . . . , r −1. Then there are representatives ci ∈Λ′ of the kibi. By Proposition 2.49, b1, . . . , br is a basis for Λ, and c1, . . . , cr−1, krbr is a basis for Λ′. By adding a suitable multiple of krbr ∈Λ′ to the ci, we may assume that ci = kiˆ bi + libr for 0 ≤li < kr and for all i = 1, . . . , r −1. But then, ci is a positive integral multiple of some Λ-primitive vector: ci = miai. The two expressions for ci together imply li = 0 or mi ≤li < kr in contradiction to the minimality of kr. Altogether, we obtain li = 0 for all i, and hence, ci = kiˆ bi as required. ⊓ ⊔ Exercise 2.20 Exercise 2.21 Corollary 2.54 Let Λ′ ⊆Λ be lattices with lin Λ = lin Λ′, and let B′ be a basis of Λ′. Then \|Λ/Λ′\| = \|Π(B′) ∩Λ\| = detΛ Λ′ . Proof. The quotient map π : Λ →Λ/Λ′ induces a bijection Π(B′) ∩Λ → Λ/Λ′ by Proposition 2.43. So the ﬁrst two quantities are equal, and in particular the second one is independent of the chosen Λ′-basis. — 38 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) That means, for the proof that the last two quantities agree, we can choose bases as in Theorem 2.53. Then the change of bases matrix is diagonal with determinant k1 · . . . · kr, while the set Π(B′) ∩Λ consists of the points P i libi for 0 ≤li ≤ki −1. ⊓ ⊔ Exercise 2.22 Exercise 2.23 Exercise 2.24 Exercise 2.25 Exercise 2.26 In dimensions d ≥2 there are inﬁnitely many unimodular matrices. Hence, there are also inﬁnitely many diﬀerent bases of a lattice. In Section 6.4 we deal with the problem of ﬁnding bases of a lattice with some nice properties. We will e.g. construct bases with “short” vectors. Exercise 2.27 Exercise 2.28 Exercise 2.29 Deﬁnition 2.55 (dual lattice) Let Λ ⊂V be a lattice with lin Λ = V . Then set Λ⋆:= {α ∈V ⋆\| α(a) ∈Z for all a ∈Λ} is the dual lattice to Λ. If b1, . . . , bd is a basis of Λ and α1, . . . , αd is the corresponding dual basis (i.e. αi(bj) = 1 if i = j, and αi(bj) = 0 otherwise), then Λ⋆is spanned by α1, . . . , αd as a lattice. Hence, the dual lattice is indeed a lattice. Further, dualizing twice gives us back the original lattice, Λ⋆⋆= Λ, as b1, . . . , bd is a dual basis to α1, . . . , αd. The following observation is left to the reader as as Exercise 2.32. Exercise 2.30 Exercise 2.31 Lemma 2.56 det(Λ) det(Λ⋆) = 1 . Exercise 2.32 Recall the distance function in Rd, d(x, y) := ∥x −y∥ and d(x, S) := inf z∈S(d(x, z)) for any x, y ∈Rd, S ⊆Rd. Lemma 2.57 Let Λ ⊆Rd be a lattice and v1, . . . , vk ∈Λ, k < d, linearly independent. Deﬁne V := lin(v1, . . . , vk). Then there is v ∈Λ −V and x ∈V such that d(v, x) ≤d(w, y) for any y ∈V , w ∈Λ −V . Proof. Let Π := Π(v1, . . . , vk). Then Π is a compact subset of Rd. Choose any a ∈Λ −V and set r := d(a, Π). Let Br(Π) := {x \| d(x, Π) ≤r} . Then a ∈(Br(Π) −V ) ∩Λ. Further, Br(Π) is bounded, so Br(Π) ∩Λ is ﬁnite by Lemma 2.47. Hence, we can choose some v ∈(Br(Π) −V ) ∩Λ that minimizes d(v, Π). Choose some x ∈Π such that d(v, x) attains this Haase, Nill, Paffenholz: Lattice Polytopes — 39 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) minimal distance. We will show that these choices satisfy the requirements of the proposition. Let w ∈Λ −V and y ∈V . By deﬁnition of V there are coeﬃcients λ1, . . . , λk ∈R such that y = k X i=1 λivi . Set z := k X i=1 ⌊λ⌋i vi , and z′ := k X i=1 {λ}i vi , Then z, w −z ∈Λ and z′ = y −z ∈Π. Further, w −z ̸∈V . Hence, d(y, w) = d(y −z, w −z) ≥d(w −z, Π) ≥d(v, Π) = d(v, x) . ⊓ ⊔ We obtain a second proof that any lattice has a basis. For this, let v1, . . . , vd ∈Λ be any linearly independent set in Λ. We consider the chain of subspaces L0 := {0} and Li := lin (v1, . . . , vi) L0 ⊊L1 ⊊. . . ⊊Ld . By Lemma 2.57 we can ﬁnd b1 ∈L1 closest to 0. Let w be any other lattice vector in L1. Then there is λ such that w = λb1, and 0 ≤∥w −⌊λ⌋b1∥< 1 . By our choice of b1 now λ ∈Z. Now assume by induction that we habe a lattice basis of Lk−1. Choose any bk ∈Łk \ Lk−1 closest to Lk−1 via Lemma 2.57. Let w ∈Λ ∩Lk. Then there are µi, ηi ∈R for 1 ≤i ≤k such that bk := X µivi and w := X ηivi . Potentially ﬂipping bk we can assume than µk > 0. We can ﬁnd some ℓ∈Z such that 0 ≤η′ k := ηk −ℓµk < µk. Then d(w −ℓbk, Lk−1) = d(η′ kvk, Lk−1) < d(µkvk, Lk−1) = d(bk, Lk−1) . Hence, by our choice of bk we conclude that w −ℓbk ∈Lk−1 −capΛ, so that ℓ∈Z. As we already know a lattice basis b1, . . . , bk−1 of Lk−1 we ontain an integral representation of w in b1, . . . , bk. Hence, wen we habe reached Ld, then we have constructed a lattice basis of Λ. This reproves Theorem 2.46. So far, we have considered lattices in linear spaces. We can shift all deﬁnitions to aﬃne spaces. Deﬁnition 2.58 (aﬃne lattice) Let Λ be a subset of an aﬃne space A. Λ is an aﬃne lattice if for some x ∈Λ the set Λ −x is a lattice. A subset B ⊆Λ is an aﬃne lattice basis of Λ if B −x is a lattice basis of Λ −b. An aﬃne lattice isomorphism is a map on Λ that comes from a lattice isomoprhism on Λ −x. — 40 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) 2.3.2 Coordinates and Normal Forms So far, all our considerations about lattices did not depend on a particular basis and a representation of transformations in coordinates w.r.t. to such a basis. However, sometimes, in particular for explicit computations in examples, it is more convenient to consider lattices and transformations in a given basis. We now reconsider some notions in the presence of a basis and introduce the Hermite and Smith normal form. Those allow us to compute bases and reprove Theorem 2.53. Lemma 2.59 Let B and B′ be bases of the lattices Λ and Λ′ respectively. Then a linear map T: lin Λ →lin Λ′ is unimodular if and only if the matrix representation A of T with respect to the bases B and B′ is integral and satisﬁes \| det A\| = 1. Proof. The matrix A has only integral entries if and only if T(Λ) ⊆Λ′. Similarly, if T is unimodular, then the inverse transformation exists, and its matrix A−1 also has integral entries. Thus, det A and det A−1 are integers with product 1. Conversely, if A is integral with \| det A\| = 1, then, by Cramer’s rule A−1 exists and is integral. ⊓ ⊔ Lemma 2.60 Let A ∈Zd×d be non-singular. Then Aλ = µ has an integral solution λ for any integral µ ∈Zd if and only if \| det A\| = 1. Proof. “⇒”: By Cramer’s rule, the entries of λ are λi = ± det(Ai), where Ai is the matrix obtained from A by replacing the i-th column with µ. “⇐”: If \| det A\| > 1, then 0 < \| det A−1\| < 1, so A−1 contains a non-integer entry aij. If ej ∈Zm is the j-th unit vector, then Aλ = ej has no integer solution. ⊓ ⊔ The set of such matrices is denoted by Gl(d, Z). Corollary 2.61 An integral matrix A ∈Zd×d is the matrix representa-tion of a unimodular transformation of a lattice if and only if \| det A\| = 1. ⊓ ⊔ Corollary 2.62 Let Λ be a lattice with basis b1, . . . , bd lin Λ. Then c1, . . . , cd ∈Λ is another basis of Λ if and only if there is a unimod-ular transformation T :∈Λ →lin Λ such that T(bi) = ci for 1 ≤i ≤d. ⊓ ⊔ We are now ready to deﬁne an important invariant of a lattice. Deﬁnition 2.63 (Determinant of a lattice) Let Λ′ ⊆Λ be lattices with lin Λ = lin Λ′, and let B and B′ be bases of Λ and Λ′ respectively. Let A be the matrix representation of the identity lin Λ′ →lin Λ with respect to the bases B′ and B. Then the determinant of Λ′ in Λ is the integer Haase, Nill, Paffenholz: Lattice Polytopes — 41 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) detΛ Λ′ := \| det A\| . If Λ = Zd, we will often write det Λ′ for det ZdΛ′. By Lemma 2.59 and Corollary 2.62 this deﬁnition is independent of the chosen bases. Exercise 2.33 Exercise 2.34 Next we study a way to obtain a nice basis for a lattice generated by a set of (not necessarily linearly independent) vectors in Qd. Deﬁnition 2.64 (Hermite normal form) Let A = (aij) ∈Qd×m with m ≥d be of full row-rank. The matrix A is in Hermite normal form if ▶aij = 0 for j > i and ▶ajj > aij ≥0 for i > j . So a matrix in Hermite normal form is an lower triangular matrix, and the largest entry in each row is on the diagonal. Depending on the context we sometimes use the transposed matrix, i.e. we claim that a matrix is in Hermite normal form if it has at least as many rows as columns, it is upper triangular, and the largest entry in each column is on the diagonal (and if the matrix is square we can also consider lower triangular matrices).    5 0 0 0 1 2 0 0 3 1 4 0    A matrix in Hermite normal form. Theorem 2.65 (Hermite normal form) Let A ∈Qd×m of full row-rank. Then there is a unimodular matrix U ∈Zm×m such that AU is in Hermite normal form. Proof. Let g be the common denominator of all entries of A. Then gA is an intergral matrix, and if H is in Hermite normal form with a unimodular transformation U such that gA = HU, then also 1/gH is in Hermite normal form and A = 1/gHU. Hence, in the following we can replace A by gA and assume that A is an integral matrix. Now observe that the following three transformations on the columns of a matrix A can be realized by a multiplication with suitably chosen unimodular matrix T from the right: (1) Exchanging two columns, and (2) multiplying a column by −1, and (3) adding an integral multiple of one column to another column. These operations are called elementary transformations for a matrix. Any succession of such operations is then realized by the product of the corresponding transformation matrices, which is again unimodular. In the following, we will show that we can transform A into its Hermite normal form using only such elementary transformations. The unimodular matrix U in the theorem is then given by the product of the corresponding tranformation matrices. — 42 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) We show that we can tranform A into its Hermite normal form by induction on the rows of A. So assume that A already has the form A = " B 0 M C # (2.6) for matrices B, C, M where B ∈Zk×k is in Hermite normal form and k ≥0. Consider the ﬁrst row (c11, . . . , c1,m−k) of the matrix C. Using elementary column operations we can transform C such that (1) c11 ≥c12 ≥. . . , c1,m−k ≥0 and (2) c := c11 + c12 + · · · + c1,m−k is as small as possible. Then c11 > 0 as A has full row rank. Further, if c12 ̸= 0, then we can subtract the second from the ﬁrst column and reorder the columns if necessary to obtain a smaller total sum c. Hence, c12 = c13 = . . . = c1,m−k = 0. The column operations on C clearly extend to A without aﬀecting B and M, so we can apply them to A to obtain a matrix A =    B 0 0 m c11 0 M′ c′ 1 C′   , where m′ is a row vector of length k, the ﬁrst row of the matrix M. By adding or subtracting multiples of the (k + 1)st column (the one containing c11) to the ﬁrst k columns of A we can assume that all entries of m are nonnegative and smaller than c11. In this way we have again reached a matrix of the form (2.6), but this time B has size (k + 1) × (k + 1). After d steps A is in Hermite normal form using only elementary operations. ⊓ ⊔ Remark 2.66 Using only elementary column operations in the proof was convenient as this directly provides a proof that the transformation matrix turning A into its Hermite normal form H is unimodular. However, this is ineﬃcient for cumputations. Here one usually does the following. To transform the ﬁrst row of C into one where all but the ﬁrst elements are zero one does the following steps: (1) swap a column with a non-zero entry in the ﬁrst position to the front, possibly multiply by −1 to make it positive (2) for any column cj with non-zero ﬁrst entry c1j one computes the greatest common divisor g of c11 and c1j and two integers x, y such that g = xc11 + y + c1j. This can be done with the extended Euclidean algorithm. Now we replace the ﬁrst column c1 by xc1 + ycj and the column cj by 1/g(c1jc1 −c11cj). Note that in the second linear combinations the coeﬀcients 1/gc1j and 1/gc11 are both integral. Haase, Nill, Paffenholz: Lattice Polytopes — 43 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) A simple consideration shows that the transformation matrix correspond-ing to the transformation used in the second step has determinant ±1 and thus is unimodular. Using this approach implies that a Herminte normal form of any rational matrix can be computed in polynomial time in the size of the input matrix A. Theorem 2.67 The Hermite normal form of a matrix A ∈Qd×m is unique. ⊓ ⊔ Remark 2.68 We can use the Hermite normal form to eﬃciently per-form various tasks on lattices. For this, Let B and B′ be matrices whose columns generate lattices Λ := Λ(B) and Λ′ := Λ(B′). (1) The ﬁrst d colunms of the Hermite normal form of B give a basis of the lattice Λ. (2) The lattices Λ and Λ′ are equal if and only if the Hermite normal forms of B and B′ coincide. (3) lattice′ is a sublattice of Λ if and only if the Hermite normal forms of B and the matrix obtained by adding the columns of B′ to B coincide. For the Hermite normal form we have used elementary column trans-formations, which we can realize by multiplication with a unimodular matrix from the right. Clearly, we can study the same transformations also for the rows of a matrix, and we can realize them by multiplications with a unimodular matrix from the left. This leads to another important normal form of a matrix, which we explain with the next theorem. Theorem 2.69 (Smith normal form) Let A ∈Zd×m be a matrix of full row rank. Then there are unimodular matrices L ∈Zd×d and R ∈Zm×m such that S = (sij)1≤i≤d,1≤j≤m := LAR satisﬁes (1) sij = 0 for i ̸= j, (2) sii > 0 for 1 ≤i ≤d, and (3) si−1,i−1 divides sii for 2 ≤i ≤d. The matrix S is unique, the companion matrices L and R are not. The last statement about the non-uniqueness of L and R follows from the obsevation that there are unimodular matrices that commute with S. When actually computing smith normal forms with their companions, this fact can be used for an attempt to keep entries in L and R small. Proof. As in the proof of the Hermite normal form it suﬃces to show that we can transform A into its Smith normal form using elementary row and column operations. The existence of the companions then follows. We again use induction. Suppose that after some elementary trans-formations A has the form — 44 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) A = " S 0 0 C # (2.7) where S is a diagonal matrix with positive entries s11, . . . , skk for k ≥0 on the diagonal such that sj−1,j−1 divides sjj for 2 ≤j ≤k, and skk devides all entries of C. Among all transformations of C that we can reach with elementary row and column operations we pick one such that min(\|cij\| \| 1 ≤i ≤ d, 1 ≤j ≤mandcij ̸= 0) is minimal. We can also assume that this minimum is attained by a11. Then clearly c11 is the only non-zero element in the ﬁrst row and column, as otherwise we can obtain a smaller entry by a suitable row of column operation. Further, a similar consideration shows that c11 must divide all other entries of C. We have extended our induction form (2.7) from k to k + 1. Uniqueness of S follows from the observation that in each step the element c11 that we construct is the greatest common divisor of the elements in C. ⊓ ⊔ You will use the Smith normal form to reprove Theorem 2.53 using bases of the lattices in a representation w.r.t. to a basis of the vector space in Exercise 2.35. Exercise 2.35 2.3.3 Metric Geometry In this section, we will give a short discussion about lattices and metric geometry (mainly following ). This is the ﬁrst point in the book where Λ really is meant to be a (non-standard) lattice in Rd. An interesting geometric application can be found in the next section. Usually, when dealing with lattice polytopes we start with an abstract lattice Λ ∼ = Zd and associate an abstract vector space Λ ⊗Z R ∼ = Rd with the volume form which evaluates as 1/d! on a fundamental domain of Λ. In particular, note that the ’length’ of a vector is not well-deﬁned. In general, we deﬁne the dual lattice as Λ⋆:= HomZ(Λ, Z) and the dual vector space as (Λ ⊗Z R)⋆:= HomR(Λ ⊗Z R, R) . Note that the dual lattice naturally sits inside of the dual vector space. While these deﬁnitions are abstract, they stress the point that in general it is not necessary and often misleading to identify dual spaces or lattices. 4 4 example missing In contrast, in lattice theory the viewpoint is opposite to ours. The starting point is an euclidean vector space, say, Rd with the usual scalar Haase, Nill, Paffenholz: Lattice Polytopes — 45 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) product ⟨·, · ⟩. Now, the choice of the embedded lattice matters! For instance, their determinants diﬀer. In this section, we will follow this convention. So, let Λ ⊂Rd be a lattice of full rank, and we assume that we have a scalar product ⟨·, · ⟩. Now, we can identify Rd and (Rd)⋆: Rd ∼ = (Rd) ⋆, x →⟨·, x ⟩. In particular, we get under this identiﬁcation Λ⋆= {x ∈Rd : ⟨x, y ⟩∈Z ∀y ∈Λ} ⊆Rd . Note that while Λ⋆⋆= Λ, it may happen that Λ⋆̸= Λ. For instance, if Λ = Zd/2, then Λ⋆= 2Zd. 2.3.4 Hilbert Bases Let v1, . . . , vn ∈Λ. Then C := cone(v1, . . . , vn) is a polyhedral cone. Let SC := C ∩Λ Then SC with addition is a semi-group, the semi-group of lattice points in C. Indeed, 0 ∈SC and if x, y ∈SC, then x + y ∈SC. A set H ⊆SC generates SC as a semigroup if for any x ∈SC there are λh ∈Z≥0 for h ∈H such that x = X h∈H λhh . Such a set is a Hilbert basis of SC. A Hilbert basis is minimal if any other Hilbert basis of SC contains this basis. Observe that in general an inclusion-minimal Hilbert basis is not unique. Consider e.g. the cone C = R2. Then both H1 := {e1, e2, −(e+e2)} and H2 := {±e1, ±e2} are minimal Hilbert bases, but they diﬀer even in size. A vector a ∈Zd is primitive if gcd(a1, . . . , ad) = 1. Theorem 2.70 Let v1, . . . , vn ∈Λ, C := cone(v1, . . . , vn), and S := C ∩Zd the semi-group of lattice points in C. Then SC has a Hilbert basis. If C is pointed, then SC has a unique minimal Hilbert basis. Proof. Deﬁne the parallelepiped Π := ( k X i=1 λiyi \| 0 ≤λi ≤1, 1 ≤i ≤k ) . Let H := Π ∩Λ. We will prove that H is a Hilbert basis. (1) H generates C, as y1, . . . , yk ∈H. — 46 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) (2) Let x ∈C ∩Λ be any lattice vector in C. Then there are η1, . . . , ηk ≥0 such that x = Pk i=1 ηiyi. We can rewrite this as x = k X i=1 (⌊ηi⌋+ {ηi}) yi , so that x − k X i=1 ⌊ηi⌋yi = k X i=1 {ηi}yi . The left side of this equation is a lattice point. Hence, also the right side is a lattice point. But h := k X i=1 {ηi}yi ∈Π , so h ∈Π ∩Zd = H. This implies that x is a integral conic combina-tion of points in H. So H is a Hilbert basis. Now assume that C is pointed. Then there is b ∈Rd such that btx > 0 for all x ∈C −{0} . Let K := n y ∈C ∩Zm \| y ̸= 0, y not a sum of two other integral vectors in C o . Then K ⊆H, so K is ﬁnite. Assume that K is not a Hilbert basis. Then there is x ∈C such that x ̸∈Z≥0K. Choose x such that btx is as small as possible. Since x ̸∈K, there must be are x1, x2 ∈C such that x = x1 + x2. But btx1 ≥0 , btx2 ≥0, btx ≥0 and btx = btx1 + btx2 , so btx1 ≤btx , btx2 < btx . By our choice of x we get x1, x2 ∈Z≥0K, so that x ∈Z≥0K, a contra-diction. ⊓ ⊔ From the above proof it follows that for a simplicial cone all Hilbert basis elements ecept for the generators of the cone are contained in the fundamental parallelepiped of the cone. Computing these points in the parallelepiped can be done by computing the Smith normal form. This, together with the generators of the cone is only a generating set G for the integer points in the cone. So we need to check for all of the (ﬁnitely many points) whether it is a sum of two other elements in G and in this case remove it from G. For non-simplicial cones we can obtain a Hilbert basis in three steps: Haase, Nill, Paffenholz: Lattice Polytopes — 47 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) (1) Triangulate the cone (2) Compute a Hilbert basis in each simplicial cone (3) Combine all Hilbert bases. This is a generating set for the integer points in the original cone. Reduce this set to a Hilbert basis by removing all points from it that are the sum of two other points in the set. Deﬁnition 2.71 (homogeneous) Let Λ ⊂Rd be a lattice and C ⊆Rd a ﬁnitely generated cone with generators v1, . . . , vd ∈Λ. C is homogeneous with respect to some linear functional c ∈Zd if there is λ ∈Z such that ctvj = λ for 1 ≤j ≤d. The height of an integer point x ∈C is ht(x) := ctx Deﬁnition 2.72 (normal cone) A ﬁnitely generated cone C which is homogeneous w.r.t. a functional c is normal if all Hilbert basis elements have height 1. 2.4 Lattice polytopes Deﬁnition 2.73 A lattice polytope is a polytope in Rd with vertices in a given lattice Λ ⊂Rd. Note that dim(P) ≤rank(Λ). Usually we will consider full-dimensional lattice polytopes, i.e., dim(P) = rank(Λ). However, we note that we can always consider P as a full-dimensional lattice polytope with respect to its ambient lattice aﬀ(P) ∩Λ of rank dim(P) in its ambient aﬃne space aﬀ(P). Note that we need to be careful when considering lattices in faces of a polytope w.r.t. to the lattice of the polytope, see Exercise 2.36. Exercise 2.36 Throughout (except when explicitly noted otherwise), the reader should assume Λ = Zd. In this case, a lattice polytope is also called integral polytope. We will use more general lattices only at very few places in the chapter on Geometry of Numbers (Chapter 4). 2.4.1 Equivalence Having introduced the objects of our interest, we should next state when two of them are considered isomorphic. Figure 2.13 shows three examples of lattice triangles in dimension two. As the reader should Fig. 2.13: Lattice triangles notice, all three triangles look quite diﬀerent: their vertices have diﬀerent Euclidean distances and diﬀerent angles. Still, the top one is considerably distinguished from the lower two: it has four lattice points, while the others have only three. Actually, more is true: the second and third are isomorphic. — 48 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) Deﬁnition 2.74 Two lattice polytopes P ⊂Rd and P ′ ⊂Rd′ (with respect to lattices Λ ⊂Rd and Λ′ ⊂Rd′) are isomorphic or unimodularly equivalent, if there is an aﬃne lattice isomorphism of the ambient lattices Λ ∩aﬀ(P) →Λ′ ∩aﬀ(P ′) mapping the vertices of P onto the vertices of P ′. Recall from Deﬁnition 2.51 that a lattice isomorphism is just an iso-morphism of abelian groups. Moreover, an aﬃne lattice isomorphism is an isomorphism of aﬃne lattices. Here, note that an aﬃne lattice does not need to have an origin (e.g., consider the set of lattice points in a hyperplane). However, if we ﬁx some lattice point to be the origin, an aﬃne lattice isomorphism can be deﬁned as a lattice isomorphism followed by a translation, i.e., x 7→Tx + b where T : Λ →Λ′ is a (linear) lattice isomorphism and b ∈Λ′. Luckily, by Corollary 2.61, in our usual situation Λ = Zd = Λ′ there is an easy criterion to check when a linear map Rd →Rd is a lattice automorphism of Zd. The matrix corresponding to the map must have integal entries and its determinant is 1 or −1. Again, as we have seen from the example above, it is very important to realize that in our setting isomorphisms do not preserve angles or distances! Let us note an immediate consequence of Corollary 2.61 Corollary 2.75 Unimodularly equivalent lattice polytopes have the same number of lattice points and the same volume. ⊓ ⊔ 2.4.2 Examples of Lattice Polytopes and Constructions. A theory only comes to life through its examples and counterexamples. Luckily, there are many interesting lattice polytopes, reﬂecting the many mathematical ﬁelds where lattice polytopes play a role. Repeatedly, the same polytope comes by several diﬀerent names, due to the fact that it has been (re-)discovered from diﬀerent points of departure. In the examples of multiply named poly-topes; cut/correlation, . . . –CH following we collect only the most important constructions and examples as provisions for the road through this text. More examples are treated in the exercises. Deﬁnition 2.76 ∆d := conv(0, e1, . . . , ed) is called the standard or unimodular d-simplex. We also call any polytope isomorphic to ∆d a unimodular d-simplex. In other words, a lattice polytope is a unimodular simplex if and only if its vertices form an aﬃne lattice basis. This is the simplest possible lattice polytope. Exercise 2.37 We discuss ways to construct new lattice polytopes from given ones. We have seen already the Minkowski sum construction introduced in Haase, Nill, Paffenholz: Lattice Polytopes — 49 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Deﬁnition 2.11. Clearly, if in this construction both summands are lattice polytopes, then so is their sum. One of the most useful constructions is the product of two (or more) polytopes. Given two lattice polytopes P in Rd and Q in Re we can construct a lattice polytope P × Q, the product of P and Q in Rd+e via P × Q := n (p, q) ∈Rd+e : p ∈P and q ∈Q o . The product has vertices (p, q) for vertices p of P and q of Q. Thus P × Q is a lattice polytope. The prism over P is a special case of a product where Q is just an interval. Mostly, one takes Q = [0, 1] if nothing else is speciﬁed. Let P = conv(v1, . . . , vn) be a polytope. P is a pyramid with apex v1 if there is an aﬃne hyperplane H such that v2, . . . , vn ∈H and v1 ̸∈H. Given a polytope P, we can construct a pyramid over P by embedding P into {0} × Rd ⊆R × Rd and taking the convex hull with any x ̸∈ {0} × Rd. We deﬁne Pyr(P) := conv({0} × P, e0) , where e0 is the ﬁrst standard unit vector in R × Rd. Let P ⊆{0} × Rd be a polytope and x, y ̸∈{0} × Rd, such that the segment between x and y intersects P in the interior of P. Then BiPyr(P) := conv(P, x, y) , is the bipyramid with apices x and y. A polytope Q is a bipyramid if it can be written as an (aﬃne image) of the bipyramid over a polytope P. The join of two lattice polytopes P and Q of dimensions d and e is P ⋆Q := conv (P × 0e × {0}, 0d × Q × {1}) , where 0d and0e are the zero vectors in dimension d and e. This is clearly again a lattice polytope. Note that a pyramid is a special case of this, where we take Q to be a single point. Further constructions, e.g. Cayley polytopes and Lawrence prisms will be discussed at the relevant places in the next chapters. Lattice polytopes also play an important role in various branches of mathematics. We give a few examples. (1) In enumerative combinatorics one can study the order polytope (2) cut polytopes or traveling salesperson polytopes in combinatorial opti-mization (3) hypersimplex (4) Birkhoﬀpolytope and the permutation polytopes — 50 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) 2.4.3 Volumes This section is devoted to a fundamental result on lattice polytopes. In Section 1.2 we have shown that for polygons the number of interior lattice points and the volume are connected. Here we will prove that in any dimension d there are only ﬁnitely many isomorphism classes of d-dimensional lattice polytopes of ﬁxed volume. We have seen that for polygons this implies that there are only ﬁnitely many isomorphism types with a ﬁxed number of lattice points. Unfortunately, no such result is true in dimensions three and above and you have constructed examples in Exercise 1.9. It is an extremely important point to realize that starting already in dimension three, having information about the volume of a lattice polytope is much stronger than just knowing the number of its (interior) lattice points. Remark 2.77 Note that here we always take the volume as induced by the lattice Λ, i.e., the volume of a fundamental parallelepiped equals one. For instance, [0, 1]d is a fundamental parallelepiped for Zd. Consider the standard simplex ∆d deﬁned in Deﬁnition 2.76. Note that vol(∆d) = 1/d! The following observation shows that the standard simplex deﬁned in is indeed the smallest possible lattice polytope: Proposition 2.78 Let P ⊂Rd be a d-dimensional lattice simplex. Then there is an aﬃne lattice homomorphism ϕ : Zd →Zd, x 7→Ax + b mapping the vertices of ∆d onto the vertices of P. In this case, d! vol(P) = \| det(ϕ)\| ∈Z≥1. Proof. We may assume that P = conv(0, v1, . . . , vd). In this case, ϕ is given by ei 7→vi for i = 1, . . . , d. Hence, vol(P) = \| det(ϕ)\| vol(∆d) = det v1 · · · vd 1 d!. ⊓ ⊔ Corollary 2.79 Let P ⊂Rd be a d-dimensional lattice polytope. Then d! vol(P) ∈Z≥1 . We have d! vol(P) = 1 if and only if P is a unimodular simplex. Proof. By Theorem 2.35 we can triangulate P into simplices without introducing additional vertices apart from those of P. In particular, any simplex is a d-dimensional lattice simplex. Now, the statement follows from the previous proposition. ⊓ ⊔ This motivates the following deﬁnition. Haase, Nill, Paffenholz: Lattice Polytopes — 51 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Deﬁnition 2.80 The normalized volume of a d-dimensional lattice poly-tope P ⊂Rd is deﬁned as the positive integer nvol(P) := d! vol(P) . Remark 2.81 Note that it makes sense to extend the previous deﬁni-tion also to low-dimensional lattice polytopes by considering them as full-dimensional polytopes with respect to their ambient lattice. Hence, nvol(P) ≥1 for any lattice polytope. Note that, if P = conv(0, v1, . . . , vd) is a d-dimensional lattice simplex in Rd, then by Proposition 2.78 the normalized volume of P equals the volume of the parallelepiped spanned by v1, . . . , vd. Noch Bemerkung dass anderer Beweis per Quotientengruppe As we have seen, lattice polytopes have normalized volume at least 1. Given a triangulation of a lattice polytope P of normalized volume V into lattice simplices, we see that this triangulation can have at most V simplices. This observation gives us an empirical reason why there should be only ﬁnitely many lattice polytopes of given volume and dimension (of course, up to unimodular transformations). Finally, let us give the formally correct proof. Theorem 2.82 Let P ⊂Rd be a d-dimensional lattice polytope, nvol(P) = V . Then there exists some lattice polytope Q ⊂Rd such that Q ⊆ [0, d · V ]d and P ∼ = Q. Moreover, if P is a simplex, then d · V may be substituted by V . Corollary 2.83 There exist only ﬁnitely many isomorphism classes of lattice polytopes of given dimension and volume. ⊓ ⊔ We will ﬁrst prove Theorem 2.82 for simplices. We need the following useful observation to extend this result to arbitrary polytopes. The proof is left as Exercise 2.38. The centroid of a simplex with vertices v0, . . . , vd is 1 d+1 Pd i=0 vi. Lemma 2.84 Let P ⊂Rd be a d-dimensional polytope. Then there exists a d-dimensional simplex S ⊆P whose vertices are vertices of P such that S ⊆P ⊆(−d)(S −x) + x , where x is the centroid of S. In other words, if v0, . . . , vd are the vertices of S, then S ⊆P ⊆(−d)S + d X i=0 vi. ⊓ ⊔ Exercise 2.38 — 52 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) Proof (of Theorem 2.82). We can assume that one vertex of P is the origin 0. First, let P = conv(0, v1 . . . , vd) be a simplex. Let V ∈Zd×d be the matrix whose columns are the coordinate vectors of v1, . . . , vd. By the Hermite normal form (Theorem 2.65) (where we transpose the left and right side of the equation) there exists U ∈Gld(Z) such that UV = H and H is an upper triangular matrix with non-negative integer entries such that in each column the maximal element is on the diagonal. We denote the columns of the right matrix by h1, . . . , hd. Therefore, U deﬁnes a unimodular transformation mapping P to Q := conv(0, h1, . . . , hd) ⊆[0, det H]d , where the last inclusion follows, as det H ≥hii for all 1 ≤i ≤d and 0 ≤hij ≤max hii. This proves the claim for simplices, as nvol(P) = nvol(Q) = det H. In general, there exists a lattice d-simplex S ⊆P as in Lemma 2.84. Then the previous part of the proof shows that there exists a unimodular transformation ϕ : Zd →Zd such that ϕ(S) ⊆[0, nvol(S)]d . Let S have vertices v0, . . . , vd. Then P ∼ = ϕ(P) ⊆(−d)ϕ(S) + d X i=0 ϕ(vi) ⊆[0, −d nvol(S)]d + d X i=0 ϕ(vi) . Since nvol(S) ≤nvol(P), the statement follows after an aﬃne unimodular transformation (translating by −Pd i=0 ϕ(vi) and multiplying by −1). ⊓ ⊔ 2.5 Software We can du actual computations with polytopes, cones and fans using the software framework polymake. polytope> $c=cube(3); polytope> print $c->VERTICES; 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 1 Haase, Nill, Paffenholz: Lattice Polytopes — 53 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 2.6 Problems included on pa 2.1. Prove Carathéodory’s Theorem (Theorem 2.4). included on page 27 2.2. Show that the preimage of the projection of a face F is again a face (but not necessarily the original one). included on page 27 2.3. If some x ∈Rd is in the relative interior of two faces of a convex set, then the two faces coincide. included on page 27 2.4. Let π : Rd →Rm be a projection that maps a d-dimensional polytope P onto a m-dimensional polytope Q. Then, if x is a point in the interior of Q, relint(π−1(x) ∩P) ⊆int (P). included on page 28 2.5. Show that a d-dimensional polytope has faces in any dimension 0 ≤k ≤d −1. included on page 28 2.6. Prove that any (d−2)-dimensional face of a d-dimensional polytope is contained in precisely two facets. included on page 28 2.7. Let P ⊆Rd be a d-dimensional polytope with 0 ∈int P. Prove that dualizing the dual polytope P ⋆gives you back the original polytope P. included on page 29 2.8. Show that simple and simplicial are dual notions. included on page 31 2.9. Prove Proposition 2.30. included on page 31 2.10. Prove that the tangent cone of a face of a polytope is precisely the intersection of the half spaces deﬁning F. included on page 32 2.11. Show that any subdivision S of a polygon P such that V(S) = V(P) is regular. included on page 33 2.12. Prove that the subdivision in Figure 2.9 is not regular. included on page 34 2.13. Show that a discrete additive subgroup of Rd is closed. included on page 34 2.14. Let Λ be a discrete closed subset of Rd (for instance, a discrete additive subgroup by Exercise 2.13) and B a bounded subset of Rd. Then Λ ∩B is a ﬁnite set. Give an example that shows that this is not correct for general discrete subsets. — 54 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) 2.15. Show that the deﬁnition of a lattice in Deﬁnition 2.37 does not depend on the norm chosen to deﬁne the balls. 2.16. Show that in Deﬁnition 2.37 we can choose the same ε for all x ∈Λ. 2.17. Prove that the following subsets of Rd are lattices. (1) Bd (2) Recall the deﬁnition of Dd from (2.5). Let us deﬁne D 1/2 d := Dd +( 1 21 + Dd) for even d. Show that this is a lattice. For d = 8 this is the root system E8. (3) E7 (4) E6 These are the so called root systems. included on page 36 2.18. Let Λ ⊆Rd be a lattice wit fundamental parallelepiped Π. Show that the lattice translates of Π cover Rd without overlap, i.e. [ x∈Λ (x + Π) = Rd and (x + Λ) ∩(y + Λ) = ∅for x, y ∈Λ, x ̸= y. included on page 38 2.19. Show that any full-dimensional cone contains a lattice basis. Hint: Use induction over the dimension. included on page 38 2.20. Let b1, . . . , bd be linearly independent lattice points in a lattice Λ of rank d. Show that the closed fundamental parallelepiped spanned by b1, . . . , bd contains a lattice basis of Λ. included on page 38 2.21. Let Λ ⊆Rd be a lattice and v1, . . . , vd ∈Λ be such that vol Π(v1, . . . , vd) = det Λ. Then v1, . . . , vd is a basis of Λ. included on page 39 2.22. Let Λ′ be a sublattice of Λ ⊆Rd with rank d. Let Π be the fundamental parallelepiped of a lattice basis of Λ. Show that v 7→v + Λ′ is a bijektion from Π ∩Λ to Λ/Λ′ ist. included on page 39 2.23. Show that Z2/ < k1e1, k2e2 >Z ∼ = Z/k1Z ⊕Z/k2Z for k1, k2 ∈ Z included on page 39 2.24. Let Λ be a lattice of rank d in Rd and let L be a linear subspace in Rd of dimension n. Show that, if L ∩Λ is a lattice of rank n, then any lattice basis of L ∩Λ′ can be extended to a lattice basis of Λ. Haase, Nill, Paffenholz: Lattice Polytopes — 55 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) included on pa 2.25. Let Λ be a lattice and v ∈Λ prtimitive. Show that v is part of some lattice basis. included on pa 2.26. Let Λ be a lattice of rank d in Rd and B := {b1, . . . , bd} ⊂Λ linearly independent. Show that B is a lattice basis of Λ if and only if Π(B) as volume det Λ. included on page 39 2.27. The following map is a (canonical) isomorphism ψ : Rd →((Rd)∗)∗, x 7− →(u 7→u(x)) . included on page 39 2.28. The map ψ ψ from Exercise 2.27 induces a natural isomorphism between Λ and (Λ∗)∗. included on page 39 2.29. Let Λ be a lattice of rank d in Rd, and T : Rd →Rd a linear map with T(Λ) ⊆Λ. Show that for T ∗: (Rd)∗→(Rd)∗, ϕ 7→(v 7→ϕ(T(v))) we have that T ∗(Λ∗) ⊂Λ∗. included on page 39 2.30. let b1, . . . , bd be a basis of Rd. For x = Pd i=1 λibi ∈Rd we deﬁne b∗ i (x) = λi. Show that b∗ 1, . . . , b∗ d is a basis of (Rd)∗. included on page 39 2.31. Let b1, . . . , bd be a lattice basis of the lattice Λ in Rd. Then b∗ 1, . . . , b∗ d is a lattice basis of the lattice Λ∗in (Rd)∗. included on page 39 2.32. Prove Lemma 2.56. included on page 42 2.33. Let Λ ⊆Rd be a lattice, v1, . . . , vk ∈Λ, L := lin(v1, . . . , vk) and L⊥its orthogonal complement with orthogonal projection π : Rd →L⊥. (1) Show that Γ := π(Λ) is a lattice in L⊥. (2) Show that Γ ⋆⊆Λ⋆. included on page 42 2.34. Let Λ ⊆Zd be a sub-lattice of rank d, and let v1, . . . , vd be a basis of Λ with fundamental parallelepiped Π(v1, . . . , vd) = nX λibi \| λi ∈[0, 1) o . Show that \|Zd/Λ\| = \|Π(v1, . . . , vd) ∩Zd\| = det Λ . — 56 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) 2.35. Reprove Theorem 2.53 using the Smith normal form (Theo-rem 2.69). 2.36. Construct a Λ-polytope P and a face F where aﬀF ∩Λ ̸= aﬀF ∩⟨P ∩Λ⟩̸= aﬀF ∩⟨F ∩Λ⟩. included on page 49 2.37. Show that ∆d has volume 1/d!. Hint: think of ∆d as iterated pyramids or subdivide [0, 1]d into d! simplices. included on page 52 2.38. Let P ⊆Rd be a d-dimensional lattice polytope (or any convex body) Show that there is a simplex S ⊆P with vertices v0, . . . , vd such that P ⊆(−d)(S −x) + x = (−d)S + (d + 1)x (2.8) and P ⊆(d + 2)(S −x) −x = (d + 2)S −(d + 1)x (2.9) where x := 1 d + 1 d X i=0 vi is the centroid of S. Hint: Choose S := conv(v0, . . . , vd) ⊆P with maximal volume in P. For any 0 ≤i ≤d let Hi be the facet hyperplane of the facet of S not containing vi, ri := d(vi, Hi) and Ri := {x : d(x, Hi) ≤ri}. Show that P ⊆Ri for 0 ≤i ≤d. Express Td i=0 Ri, (−d)(S −x) + x, and (d + 2)(S −x) −vx in barycentric coordinates with respect to v0, . . . , vd and compare. included on page 9 2.39. Sei K eine konvexe Menge in Rd, und a, b ∈Z≥0. Zeige (a + b)K = aK + bK := {ax + by : x, y ∈K}. (Gilt das auch für a, b < 0?) included on page 9 2.40. Recall the isolation theorem : Given an open convex set S in Rd. muss angepasst werden Then any point outside of S can be strictly separated from S. Finde explizite Beispiele in Dimension 2, dass (1) dies die Voraussetzung konvex braucht (2) dies die Voraussetzung oﬀen braucht Haase, Nill, Paffenholz: Lattice Polytopes — 57 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) (3) eine abgeschlossene konvexe Menge kann einen Extrempunkt haben, der keine Seite (Ecke) ist (4) eine kompakte Menge K eine Seite F haben kann, die eine Seite G hat, die aber keine Seite von K ist included on pa 2.41. Wie kann man die aﬃne Hülle mithilfe von ‘Aﬃn-Kombinationen’ beschreiben? kleinster ist! Aﬃne Kombinationen sollten schon vorher deﬁniert worden sein. Deﬁniere aﬃner Unterraum als Verschiebung von linearem UR, und aﬃne Huelle als kleinster ... included on page 9 2.42. (Härter) Sei S ⊂Rd, x ∈Rd. Zeige: x ist ein Extrempunkt von conv(S) g.d.w. x ∈S und x ̸∈conv(S{x}). nur falls Extrempunkte deﬁniert wur-den, auch oben included on page 9 2.43. Check that Weyl-Minkowski for cones implies that for polytopes muss man anpassen included on page 9 2.44. Man erinnere sich, wieso die Determinante einer linearen Abbildung unabhängig von der Basiswahl ist. Nun beweise für ein Gitter Λ ⊂Rd, dass wenn T eine lineare Abbildung von Rd nach Rd ist, so dass die Einschränkung TΛ : Λ →Λ wohldeﬁniert ist und surjektiv, dann ist TΛ bijektiv. (Tipp: wieso ist T bijektiv?) Zeigen Sie allgemeiner(?), dass ein Homomorphismus von Λ →Λ surjektiv ist g.d.w. bijektiv. included on page 9 2.45. Dualizing non-polyhedral cones. included on page 9 2.46. Existence of a Hilbert basis included on page 9 2.47. Man mache sich an einem Beispiel plausibel (oder beweise für i = 1), dass für gegebenes i ∈{1, . . . , n} und für ganzzahlige n × n-Matrizen der ggT aller Determinanten von i × i-Untermatrizen bei Multiplikation mit unimodularen Matrizen invariant bleibt. Wieso impliziert dies die Eindeutigkeit der Smith-Normalform? included on page 9 2.48. Man berechne die Smith-Normalform für 3 0 0 14 ! und für eine 3 × 3-Matrix Ihrer Wahl. included on page 9 2.49. Sei P ein d-dim. Gitterpolytop in Rd mit 0 ∈int (P) und F eine Facette von P. Zeige dass es eine lineare unimodulare Transforma-tion gibt, die F auf eine Teilmenge von Rd−1 × {m} für m ∈Z≥1 abbildet. qDie Zahl m ist der ganzzahlige Abstand des Ursprunges von F). Hinweis: Betrachten Sie ηF und bilden Sie es auf e∗ d ab. dieser Begriﬀsollte wohl in Text rein statt in Exercises?! — 58 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 2. Polytopes and Lattices (draft of October 7, 2020) 2.50. Ist ein Skalarprodukt ⟨·, ·⟩auf Rd gegeben, so ist ϕ : Rd →(Rd)∗, y 7→(x 7→⟨y, x⟩) ein Isomorphismus. included on page 9 2.51. Ein Gitter-Isomorphismus Zn →Zn wurde deﬁniert als eine Z-lineare Abbildung, die bijektiv ist. Man mache sich klar, dass hier eine Z-lineare Abbildung nichts anderes als ein Gruppenhomo-morphismus ist (wenn man von Gruppen und Homomorphismen schon gehört hat). Prüfe, dass die inverse Abbildung eines Gitteriso-morphismus auch wieder ein Gitterisomorphismus ist. Kennen Sie Beispiele z.B. aus der Analysis wo ‘bijektiv’ und ‘Isomorphismus’ nicht das gleiche sind? (Idee: stetige Funktionen). included on page 9 2.52. Man mache sich klar: Jeder Gitter-Isomorphismus Zn →Zn deﬁniert auch einen Vektorraum-Isomorphismus Rn →Rn (Hinweis: Betrachte die Standardba-sis). Jeder Gitter-Isomorphismus ist durch eine Gln(Z)-Matrix gegeben. Dies ergibt eine Bijektion zwischen Gitterisomorphismen und Gln(Z)-Matrizen. included on page 9 2.53. Zeige, dass für v1, . . . , vd ∈Rd das Volumen des aufgespannten Parallelepipeds gleich d! mal dem Volumen der konvexen Hülle von 0, v1, . . . , vd ist. included on page 9 2.54. Zeige, dass unter einer aﬃn-linearen Abbildung die konvexe Hülle des Bildes einer Menge gleich dem Bild der konvexen Hülle ist. included on page 9 2.55. Hmm, kann man eigentlich einen Isomorphiebegriﬀfür rationale Polytope (also solche, deren Ecken alle rationalen Koordinaten besitzen) formulieren? included on page 9 2.56. Finde das normalisierte Volumen der konvexen Hülle von (2, 0, 4), (1, 1, 0), (0, 2, −2). included on page 9 2.57. Ist der Endlichkeitssatz für Gittersimplizes scharf? (Oﬀen: was für Gitterpolytope?) Haase, Nill, Paffenholz: Lattice Polytopes — 59 — Ehrhart Theory 3 Contents 3.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 3.2 Old Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.2.1 Examples of Ehrhart polynomials . . . . . . . . 65 3.3 Generating Functions for Lattice Points . . . 67 3.4 Ehrhart’s theorem . . . . . . . . . . . . . . . . . . . . . . . 73 3.5 Stanley’s theorem . . . . . . . . . . . . . . . . . . . . . . . 76 3.5.1 Half-Open Decompositions of Cones . . . . . . 76 3.5.2 The integer point generating function of half-open cones . . . . . . . . . . . . . . . . . . . . . . . . 79 3.5.3 Stanley’s theorem and the h∗-polynomial of a lattice polytope . . . . . . . . . . . . . . . . . . . . 79 3.5.4 Where does the h∗-notation come from? . . 81 3.6 Reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 3.6.1 Stanley reciprocity for cones . . . . . . . . . . . . . 83 3.6.2 Ehrhart-Macdonald reciprocity for lattice polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 3.7 Properties of the h∗-polynomial . . . . . . . . . . . 87 3.7.1 Degree and codegree of lattice polytopes . . 87 3.7.2 Ehrhart polynomials of lattice polygons . . . 90 3.7.3 Polytopes with Small Degree . . . . . . . . . . . . 91 3.8 Brion’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . 91 3.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Lecture Notes Lattice Polytopes (draft of October 7, 2020) In this chapter we will be concerned with counting lattice points in polytopes. The central theorem of this chapter gives a very beautiful relation between geometry and algebra. It is due to Eugène Ehrhart and tells us that the function counting the number of lattice points in dilates of a polytope P ⊆Rd, the Ehrhart counting function, ehrP (k) := \|k · P ∩Zd\| , is the evaluation of a polynomial ehrP (t) of degree d in t = k. This polynomial ehrP (t) is called the Ehrhart polynomial of P, and it is at the heart of the theory of lattice polytopes. 3.1 Motivation Pick’s Theorem and Reciprocity Monomials of degree d in R[x1, . . . , xd] Still, these are not lattice polytopes Semi-Magic Squares A magic square is an n by n grid ﬁlled with n2 positive integers, such that the sum of each row, each column, and the two main diagonals is the same for all. This number is the magic constant. Sometimes it is additionally required that the entries in each row, column and diagonal are pairwise distinct and the total set of entries is {1, . . . , n2}. See Table 3.1 for a famous example that already appeared in the painting Melancholia I by Albrecht Dürer in 1514. 16 3 2 13 5 10 11 8 9 6 7 12 4 15 14 1 Table 3.1: Dürers magic square of 1514. The number of possible n × n magic squares with magic constant b is given by the set of positive integer solutions to n−1 X i=0 xij = n−1 X i=0 xji = n−1 X i=0 xii = n−1 X i=0 xi,n−i = b . for 0 ≤j ≤n −1. Volumes This section is a bit short The most important natural invariant of a convex body is its volume. Computing the volume of a convex body is in general a complicated problem. Counting lattice points in multiples of a polytope is directly related to it. Let P ⊂Rd be a convex body. As illustrated in Figure 3.1, we can approximate the volume by counting the volume of little cubes centered at the more and more reﬁned lattice Zd/k (for k →∞). We rigoros ist das eigentlich vom masstheoretischen Standpunkt aus? Fig. 3.1: Approximating a convex body by smaller and smaller cubes vol(P) = Z P dx = lim k→∞ 1 kd \|P ∩(Zd/k)\| = lim k→∞ 1 kd \|kP ∩Zd\| = lim k→∞ 1 kd ehrP (k) (3.1) — 62 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) We see that knowing inﬁnitely many values of the Ehrhart counting function allows to determine the volume. However, if we would know that ehrP (k) is actually a polynomial function, then by Equation (3.1) it must have degree d and leading coeﬃcient vol(P). In particular, the Ehrhart polynomial would be determined by knowing d + 1 many values of it. It follows that if the Ehrhart counting function of P is polynomial with constant term 1 (as we will show later for lattice polytopes P), then vol(P) can be explicitly computed from \|kP ∩Zd\| for k = 1, . . . , d. This looks already very much like a generalization of Pick’s formula (which we have seen in Pick’s Formula (Theorem 1.8))! In Exercise 3.1, the reader is invited to work this out explicitly in dimension three. Exercise 3.1 Exploiting the important reciprocity principle which we will also learn about later in this chapter, one can even show that (and this is just one possibility of such a generalized Pick’s formula), the volume of a d-dimensional lattice polytope can be determined from knowing the number of lattice points in kP for k = 1, . . . , ⌈d/2⌉together with the number of interior lattice points of kP for k = 1, . . . , ⌊d/2⌋. This gives a nice formula for the volume of a three-dimensional lattice polytope (see Exercise 3.2) that should satisfy the curiosity of the reader for a Pick’s formula in dimension three and convey the usefuleness of Ehrhart’s theorem. Exercise 3.2 Of course, there are many more arguments why Ehrhart polynomials are important: ▶they allow to read oﬀimportant properties of the polytope, ▶their coeﬃcients form a basis for the space of ..., ▶they have algebro-geometric analogues, ▶... TODO: more and better 3.2 Old Motivation Before we delve into the theory and compute Ehrhart polynomials of poly-topes we want to introduce some problems where counting, enumerating or sampling lattice points appear naturally. Knapsack type problems Assume that you are given a container C of size ℓ(the knapsack), and k goods of a certain sizes s1, . . . , sk and values v1, . . . , vk. Two important variants of a knapsack problem are the tasks to ﬁll the container either with goods of the highest possible total value, i.e. to solve the problem max x1v1 + · · · + xkvk subject to x1s1 + · · · + xksk ≤ℓ xi ∈{0, 1} for 1 ≤i ≤k , Haase, Nill, Paffenholz: Lattice Polytopes — 63 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) or to ﬁll the container completely with goods of a prescribed total value v, i.e. to solve x1v1 + · · · + xkvk = v x1s1 + · · · + xksk = ℓ xi ∈{0, 1} for 1 ≤i ≤k . Here is a simple example of such a problem. The U.S. currency has four diﬀerent coins that are in regular use, the penny (1¢), nickel (5¢), dime (10¢), and the quarter (25¢). You may wonder how many Is this still true? diﬀerent ways there are to pay 1$ using exactly ten coins. With a little consideration you probably come up with the solution 100 = 5 · 1 + 0 · 5 + 2 · 10 + 3 · 25 = 0 · 1 + 6 · 5 + 2 · 10 + 2 · 25 = 0 · 1 + 3 · 5 + 6 · 10 + 1 · 25 = 0 · 1 + 0 · 5 + 10 · 10 + 0 · 25 , so there are essentially four diﬀerent ways. This is exactly the number of lattice points in the polytope P :=      x ∈R4 : x1, x2, x3, x4 ≥0, x1 + x2 + x3 + x4 = 10, x1 + 5x2 + 10x3 + 25x4 = 100      . This may look like a much more complicated approach than just testing with some coins. But what if you want to ﬁnd the 182 ways to pay 10$ using 100 coins, or the 15876 ways to pay 100$ with 1000 coins? You could also have goods with diﬀerent quantities not containing a single unit like the penny in the coin problem. For example, your favorite sweets may come in boxes fo size ﬁve, nine, and 16. If you want to buy one for yourself and all your coworkers, you will necessarily have some left when you give everyone one sweet if you have three, six, or ten coworkers. In mathematical terms, the equation 5x1 + 9x2 + 16x3 = b has no non-negative integral solution in x1, x2 and x3 if b is 4, 7, or 11. On the other hand 34 = 5 · 5 + 9 35 = 7 · 5 36 = 4 · 5 + 16 37 = 2 · 5 + 3 · 9 38 = 4 · 5 + 2 · 9 — 64 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) so that any b ≥24 can be obtained as a non-negative integer combination of 5, 9 and 16. Hence, there is a largest b that cannot be obtained in this way. The general case of this question, i.e. the question to ﬁnd, for given postive integers a1, . . . , _d the largest b such that a1x2 + · · · + adxd = b has no non-negative integral solution in x1, . . . , xd is called the Frobenius problem, and the solution b is called the Frobenius number. You can check that in our example the Frobenius number is 33. Contingency Tables Consider Table 3.2 (which is a simpliﬁed version of a table produced by the Statistische Landesamt Berlin for academic degrees awarded at Berlin universities in 2005) Very nice! Haben wir aktuellere Daten :-) Diploma PhD Teacher FU 1989 1444 299 3732 TU 1868 421 115 2404 HU 920 441 373 1774 4817 2306 787 Table 3.2: Degrees awarded in Berlin in 2005 You may ask how likely it is to have exactly this distribution of the entries of the table, if its margins, i.e. the totals of degrees awarded at each university, and the totals of diﬀerent degrees awarded, are given. Assuming a uniform distribution, you would need to know the number of possible tables with these margins. This is the number of lattice points in the polytope P :=      X ∈R3×3 ≥0 : x11 + x12 + x13 = 3732, x21 + x22 + x23 = 2404, x31 + x32 + x33 = 1774, x11 + x21 + x31 = 4817, x12 + x22 + x32 = 2306, x13 + x23 + x33 = 787      . There are 714,574,663,432 of them. Further potential examples: (1) Integer Linear Programming: Chickens, Nuggets, and others, Network ﬂows (2) Frobenius numbers (3) GUT: (semi-)magic squares!!!!! Integer Linear Programming Magic squares 3.2.1 Examples of Ehrhart polynomials In this short section, we compute for some simple examples the counting function ehrP (k) of a polytope directly. We will observe that it is indeed given by a polynomial, and that evaluating at negative integers gives the number of interior points. Before we start, we want to give a formal deﬁnition of the counting function. For this, let S ⊆Rd, and let k ∈Z>0. The k-th-dilation of a set of S is the set kS := {kx : x ∈S} . We introduce the following counting function. Deﬁnition 3.1 The Ehrhart counting function of a bounded subset S ⊆Rd is the function Haase, Nill, Paffenholz: Lattice Polytopes — 65 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) ehrS(k) : Z≥1 − →Z≥1 k 7− → kS ∩Zd . With this deﬁnition we can look at our ﬁrst examples. compare deLoera Semesterberichte Let L := [a, b] ⊂R, a, b ∈R be an interval on the real line. Here, counting is easy, L contains ⌊b⌋−⌈a⌉+ 1 integers. The k-th dilate of P is [ka, kb]. By the same argument it contains ⌊kb⌋−⌈ka⌉+ 1 integral points, so L 2L 3L Fig. 3.2 ehrL(k) = ⌊kb⌋−⌈ka⌉+ 1 . Figure 3.2 shows the interval I = [0, 3 2] and its second and third dilation. If the boundary points a and b are integral and a ≤b, then we can simplify the formula. In this case also all multiples of a and b are integral, and we can omit the ﬂoor and ceiling operations to obtain ehrL(k) = k(b −a) + 1 . We observe that this is a polynomial of degree 1 in k. We will see that this observation is a very special case of the Theorem of Ehrhart that we will prove below. Now we turn to some examples of polytopes in general dimension d ≥0. Let us ﬁrst consider the standard simplex ∆d := conv(0, e1, . . . , ed) introduced in Deﬁnition 2.76. See Figure 3.3 for the lattice points in a multiple of this ismplex. Proposition 3.2 Let ∆d be the d-dimensional standard simplex. Then ehr∆d(k) = d + k d = (d + k) · (d + k −1) · . . . · (k + 1) d! . Fig. 3.3: Lattice points in a standard simplex. Observe that this is a polynomial in the variable k of degree d with leading coeﬃcient 1/d! . Exercise 3.3 Proof. There is a bijection between the lattice points in k ∆d and se-quences of k dots and d bars: to each such sequence, assign the vector x ∈Rd whose ith coordinate equals the number of dots between the ith bar and the (i + 1)st bar for 1 ≤i ≤d −1 (we don’t write down the number of dots after the last bar, it is determined by the rest): · · \| · · · \| \| · ← → x = (2, 3, 0) This yields a bijection between the sequences and lattice points with non-negative coordinates and with P xi ≤k. ⊓ ⊔ Another simple, but very important example is the unit cube deﬁned in Example 2.7(1). The k-th dilate of the cube is kCd = k · [0, 1]d = [0, k]d. Hence, the Ehrhart counting function is given by ehrCd(k) = (k + 1)d . Note again that this is a polynomial in k of degree d. — 66 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) 3.3 Generating Functions for Lattice Points The examples in the previous sections show that we really want to investigate lattice points S := P ∩Zd in polytopes or polyhedra P ⊆Rd, i.e. count them (if the nuber is ﬁnite), enumerate them, explore structure on this set S, or explain their interactions with polyhedral geometry, algebra and other ﬁeld. All of this requires us to ﬁrst ﬁnd a way to distinguish lattices points in a polyhedron from all others, i.e. a way to encode them, preferably in an eﬃcient and explicit way, that we can easily write down in a short and concise form. It should be simple from our notation to decide whether a point is in our list or not. We have already seen two more indirect ways already above, directly from the interior and exterior description of a polytope. A lattice point x is in our set S if it is either a convex combination of the vertices of P or satisﬁes all deﬁning inequlities of P. This description is ﬁne for a single particular lattice point. But it does not tell us much about the whole set of points, nor about the structure of the set. For this, we need to ﬁnd a way to make our description of the set S more explicit. In a ﬁrst, rather naïve approach, we could now be tempted to explicitely list all lattice points in our polytope (this clearly only works well for bounded objects). To make a simple example, look at the polytope P3 := [0, 3]. This is the simple segment shown in Figure 3.4. The naïve approach gives us the list: Fig. 3.4: The polytope P3. 0, 1, 2, 3 . This works well in this small example, but consider the structurally similar example P10002 := [0, 10002]. Here, our plain list 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, . . . easily exceeds a line, and also the length of this book. To get a compact encoding of the points we need a better idea. Here is one that might look really strange at ﬁrst, but will prove to be very powerful. We can replace each point k ∈P3 with its monomial tk. With this we deﬁne a polynomial that contains precisely the monomial corresponding to points in our polytope: 1 + t + t2 + t3 = 3 X i=0 ti , The option to write our polynomial as a sum already shows a quite compact way to encode the lattice points. Observe, that the representation is not really more complicated for P10002. However, it is pretty obvious Haase, Nill, Paffenholz: Lattice Polytopes — 67 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) that this particular compact notation as a sum is only possible in very special cases, so we need to look further. If you look at the polynomial you may realize that there is another option to write this more condensed than using a sum: We can also write this as the geometric series GP3 (t) := 1 −t4 1 −t . whose expansion is again our polynomial. Again, doing the same for P10002 does not really make this notation more complicated: G[0,10002] (t) := 1 −t10003 1 −t . We will see that this idea of using a geometric series to specify the lattice points in a polyhedron is both suﬃciently ﬂexible to work for all polyhedra, and eﬃcient enough that we can use it to really study the structure of the set of lattice points. Her comes another surprising and powerful property of our last observation. If we try to do write down the lattice points of the unbounded polyhedron P∞:= [0, ∞), then our ﬁrst two approaches obviously become infeasible. However, the third works and turns out to be even shorter and more appealing5! As a geometric series we can consisely describe all 5 If the reader feels slightly wary about what happens at t = 1, be assured that in our approach here we will not deal with any analytic convergence issues and will not evaluate at certain values. lattice points in P∞via the monomials in G[0,∞) (t) := 1 1 −t . As this extended example suggests, the generating function we used to encode the lattice points will indeed provide a powerful bookkeeping tool for counting and enumerating lattice points in polytopes. It will soon become apparent it is indeed quite useful and natural to encode lattice points not only in polytopes, but more generally in any bounded or unbounded subset of Rd, as in the last example of a ray in R1. You should keep this in mind for the following considerations. In the above example of the one-dimensional cone x ≥0 ⊆R we have seen that we can use rational functions in one variable t to describe the inﬁnite series of all monomials corresponding to the lattice points in the cone. We now want to formalize this idea, and directly generalize it to arbitrary dimensions. Let \| be some ground ﬁeld (you can just think of \| = C, if you like). We assign the monomial ta := ta1 1 ta2 2 · · · tad d in d variables to a lattice point a = (a1, . . . , ad) ∈Zd. In the above example all lattice points were non-negative and thus lead to the “usual kind” of monomials. In general, the coordinates of a are allowed to be — 68 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) negative, so this is a Laurent polynomial living in the Laurent polyonmial ring L = \|[t±1 1 , . . . , t±1 d ] . Moreover, note that the sum of monomials for the cone x ≥0 is inﬁnite. Since we do not care about convergence, we will actually consider our sums not as Laurant polynomials, but as series in a subset of the L-module b L := \|t±1 1 , . . . , t±1 d of formal Laurent series. We give an example before we write down the proper deﬁnition. Example 3.3 Let P be the polygon P := conv " 0 2 2 3 1 −1 2 0 # (see Figure 3.5). Recall that the convex hull of a matrix is deﬁned to be the convex hull of the column vectors of the matrix. We list the lattice points as monomials in the Laurent polynomial Fig. 3.5: The polygon of Example 3.3. t2 1t2 2 + t2 + t1t2 + t2 1t2 + t1 + t2 1 + t3 1 + t2 1/t2 . Deﬁnition 3.4 (integer point series) For S ⊂Rd the integer point series b GS is the formal Laurent series b GS (t) := X a∈S∩Λ ta ∈b L . Translating a set S ⊆Rd by some integral vector a ∈Zd amounts to multiplication of its generating series with ta, b Ga+S (t) = tab GS (t) . Remark 3.5 (Warning) When dealing with formal power series and rational series, one has to be very careful in order not to make a mistake. Therefore, we would like to give an example here (just for one variable) that justiﬁes this caution: Consider the following expression of formal Laurent series b GR (t) (1 −t) = (· · · + t−2 + t−1 + 1 + t + t2 + · · · )(1 −t) = (· · · + t−2 + t−1 + 1 + t + t2 + · · · ) −(· · · + t−1 + 1 + t + t2 + t3 · · · ) = 0 Haase, Nill, Paffenholz: Lattice Polytopes — 69 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Clearly, we deduce b GR (t) = · · · + t−2 + t−1 + 1 + t + t2 + · · · = 0 1 −t = 0 (3.2) However, this is wrong! The left side of the last equation is deﬁnitely not zero as a Laurent series. So, where is our mistake? In case the reader hasn’t already found it, we will explain this apparent riddle in Remark 3.10. Actually, not all Laurent series appear as a generating series for lattice points in polyhedra. The ones we will encounter have a nice additional structure that we will work out with the next deﬁnitions and theorems. Deﬁnition 3.6 (summable Laurent series) A Laurent series b G ∈b L is summable if there is a Laurent polynomial g ∈L such that the series gb G is a Laurent polynomial. Clearly all Laurent polynomials are summable. On the other hand, the series 1 + t2 + t3 + t5 + t7 + t11 + t13 + t17 + . . . = 1 + X k prime tk cannot be summable. We will denote the set of all summable Laurent series by Lsum. We leave the proof of the following proposition to the reader as Exercise 3.4. Exercise 3.4 Proposition 3.7 Lsum is a L-submodule of b L. ⊓ ⊔ Example 3.8 Before we continue we want to work out some simple, but quite important examples of summable series coming from polyhedra. (1) Let us ﬁrst consider the polyhedron P∞= [0, ∞) that we introduced above. The integer point series is b GP∞(t) = X a∈Z≥0 ta = 1 + t + t2 + t3 + · · · . Using the polynomial g(t) := (1 −t) we obtain g(t)b GP∞(t) = 1, so b GP∞(t) is a summable series. (2) Now let C := cone(e1, e2) for the standard unit vectors e1, e2 ∈R2. Then b GC (t, s) = X a,b∈Z≥0 tasb = X a ta ! X b sb ! = 1 + t + s + t2 + s2 + ts + t3 + · · · . Similar to the previous case we can use the polynomial — 70 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) g(t, s) := (1 −t)(1 −s) to obtain g(t, s) · b GC (t, s) = (1 −t) ·  X a∈Z≥0 ta  · (1 −s) ·  X b∈Z≥0 sb  = 1. Hence, b GC (t, s) is a summable series. (3) Finally, let V := {e1, . . . , ed} and C = cone(V ). In the same way, we see d Y i=1 (1−tei) · X z∈Zd ≥0 tz = d Y i=1 (1−tei) · X (n1,...,nd)∈Zd ≥0 (te1)n1 · · · (ted)nd = 1 (3.3) Hence, b GC (t) is summable. Proposition 3.9 There is a natural homomorphism from summable series to rational functions Φ : Lsum − →R := \|(t1, . . . , td) , mapping b G to f/g if gb G = f in b L. We will abbreviate this also by writing b G Φ 7→f g The proof of this proposition is left as Exercise 3.5. Exercise 3.5 Remark 3.10 (Resolving the warning of Remark 3.5) We can now explain where the mistake was. The equation (3.2) should be replaced by the following correct expression: Φ(b GR (t)) = Φ(· · · + t−2 + t−1 + 1 + t + t2 + · · · ) = 0 1 −t = 0. In other words, not the Laurent series is zero but only its associated rational function! While it is often very convenient to use the equality sign ‘=’ between a summable Laurent series and a rational function (instead of using a cumbersome and non-standard notation such as Φ), one cannot stress enough that one must be aware that such an equality only holds on the level of rational functions and not on the level of Laurent series. We hope to make this point clear by using the Φ 7− →symbol instead in these situations. In particular, we see from the previous example that Φ is not an injective map. However, it clearly is for Laurent polynomials (check!). In other words, L is a submodule of Lsum, and Φ\|L is the identity map. A more general criterion on injectivity is proven in Exercise 3.6. Exercise 3.6 Haase, Nill, Paffenholz: Lattice Polytopes — 71 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Deﬁnition 3.11 (integer point generating function) Suppose S ⊆ Rd is a set so that b GS (t) is summable. The integer point generating function of S is GS (t) := Φ(b GS (t)) . If S ⊆Rd is bounded, then we are allowed to identify (see also Exercise 3.6) GS (t) = X a∈S∩Zd ta . In the one-dimensional example P∞= [0, ∞) above we have already computed the image of the generating series in R, it is GP∞(t) = 1 1−t. We can now generalize this observation to rational simplicial cones (which will be generalized to half-open simplicial cones later). Let D be a simplicial rational cone in Rd with primitive ray generators V := {a1, . . . , ad}. We recall the fundamental parallelepiped of V from Deﬁnition 2.42 Π(V ) := (X v∈V µvv : µv ∈[0, 1) for v ∈V ) We know from Corollary 2.44 that the fundamental parallelepipeds tile the space without overlap (strictly, there we talked about lattice bases, however, the same argument works for the generating set V ). Proposition 3.12 In this notation, b GD (t) is summable with GD (t) = P y∈Π(D)∩Zd ty Qd i=1(1 −tai) Proof. Let Z≥0V stand for the set of Z≥0-linear combinations of V . By replacing the Laurent monomial tei by tai in (3.3), we get d Y i=1 (1 −tai) · X z∈Z≥0V tz = d Y i=1 (1 −tai) · X (n1,...,nd)∈Zd ≥0 (ta1)n1 · · · (tad)nd = 1. By Corollary 2.44 we get d Y i=1 (1 −tai) · X x∈D∩Zd tx = d Y i=1 (1 −tai) · X y∈Π(D)∩Zd X z∈Z≥0V ty+z = d Y i=1 (1 −tai) · X z∈Z≥0V tz · X y∈Π(D)∩Zd ty = X y∈Π(D)∩Zd ty . ⊓ ⊔ — 72 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) Our integer point generating series contain a monomial for every lattice point in a set. If we have the series for two sets S, S′, then we can obtain the series for the union S ∪S′ by adding the two series and subtracting all lattice points that we encoded in both series, i.e. the generating series for the lattice points in the intersection S ∩S′. This principle clearly extends to the union of any ﬁnite number of sets. We can compute the generating series from the generating series of the sets and all partial intersections if we keep track of the multiplicities a partial intersection appears in the total sum. This is called the principle of inclusion-exclusion. You will study this in more detail in Exercise 3.9. Triangulating a rational polyhedral cone into rational simplicial cones (see Section 2.2)) and using inclusion-exclusion (see also Figure 3.6) yields the following general result. Corollary 3.13 The integer point generating series of a rational poly-hedral cone is summable. P1 P2 Q Fig. 3.6: Let Q be the dotted chord of the polygon and P1, P2 the two polygons obtained by cutting P along Q. Then \|P ∩Z2\| = \|P1 ∩Z2\| + \|P2 ∩ Z2\| −\|Q ∩Z2\|. 3.4 Ehrhart’s theorem After these preparations, let us prove Ehrhart’s theorem. Theorem 3.14 (Ehrhart’s Theorem) The Ehrhart counting func-tion given by k 7→ehrP (k) for k ∈Z≥1 extends to a polynomial function t 7→ehrP (t) of degree d and leading coeﬃcient vol(P). Deﬁnition 3.15 (Ehrhart polynomial) For a polytope P the polyno-mial ehrP (t) as in the previous theorem is the Ehrhart polynomial of P. We have already seen in § 2.1.2 that it is convenient to homogenize a polytope and work with the cone over P instead of P. Recall that we have deﬁned C(P) in (2.2) via C(P) := cone ({1} × P) ⊆Rd+1 , We usually write a vector x ∈Rd+1 with indices starting from 0 and use x0 for the special coordinate. See Figure 3.7 for the cone over a triangle. In our setting the especially convenient property of this representation of our polytope is the fact that we can recover all dilates of P from C(P). More precisely, for any k ≥0 we get the k-th dilate of P by intersecting C(P) with the hyperplane x0 = k, and the lattice points in kP by intersecting with {k} × Zd. Hence, Fig. 3.7 b GC(P ) (t, 1, . . . , 1) = X k≥0 kP ∩Zd tk = 1 + X k≥1 ehrP (t) tk. Haase, Nill, Paffenholz: Lattice Polytopes — 73 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) As substituting variables clearly keeps summability, by Corollary 3.13 the following deﬁnitions make sense. Deﬁnition 3.16 Let P be a lattice d-polytope. The Ehrhart series of P is the summable formal Laurent series b EhrP (t) := 1 + X k≥1 ehrP (t) tk ∈\|t in one variable t. The corresponding rational function will be denoted EhrP (t) := Φ(b EhrP (t)) ∈\|(t) . To proceed we consider some well-known results on generating functions. Lemma 3.17 For j ∈Z≥0, X Z≥0 k + d −j d zk Φ 7− → zj (1 −z)d+1 . The proof will be given in Exercise 3.7. Exercise 3.7 Proposition 3.18 Let f, g : R →R be such that ∞ X t=0 f(t )zt Φ 7− → g(z) (1 −z)d+1 . Then f(t) is a polynomial of degree at most d if and only if g(z) = P k∈Z≥0 gkzk is a polynomial of degree at most d. In this case: f(t) = g0 t + d d + g1 t + d −1 d + . . . + gd t d . and the leading coeﬃcient of f is g(1) d! . In particular, f has degree d if and only if g(1) ̸= 0. Exercise 3.8 Proof. We deﬁne the polynomials fj(t) := (t+d−j d ) for 0 ≤j ≤d. The set {fo, . . . , fd} is a basis of R[t]≤d (Exercise 3.8). Let f be a polynomial of degree at most d. Then there are g0, . . . , gd such that f(t) = d X j=0 gjfj(t) = d X j=0 gj t + d −j d . The coeﬃcient of td is 1 d! P gj. We compute X t≥0 d X j=0 gj t + d −j d zk = d X j=0 gj X t≥0 t + d −j d zk Φ 7− → Pd j=0 gjzj (1 −z)d+1 = g(z) (1 −z)d+1 — 74 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) For the converse direction, injectivity of Φ on polynomials implies f(t) = Pd j=0 gj(t+d−j d ). Now, we use again the basis property. ⊓ ⊔ Now, let us compute the Ehrhart generating function for lattice simplices. Proposition 3.19 Let S be a d-simplex. Then EhrS(t) = h⋆(t) (1 −t)d+1 where h⋆is a polynomial of degree ≤d. Further, for h⋆(t) = Pd k=0 h⋆ ktk, we have h⋆ k = #(Π(C(S)) ∩Zd+1 ∩{x \| x0 = k}) ∈Z≥0. In particular, h⋆ 0 = 1 and h⋆(1) ̸= 0. Proof. Let {a0, a1, . . . , ad} be the vertex set of {1} × S, with ai = (1, vi) for i = 0, . . . , d. Applying the substitution (t0, t1, . . . , td) by (t0, 1, . . . , 1) to Proposition 3.12 we obtain that EhrS(t) = h⋆(t0) (1 −t0)d+1 for the polynomial h⋆(t0) = P (y0,y)∈Π(C(S))∩Zd+1 ty0 0 . Let (y0, y) = Pd i=0 λi(1, vi) ∈Π(C(S)) ∩Zd+1, so 0 ≤λi < 1 for i = 0, 1, . . . , d. In particular, y0 < d + 1, so y0 ≤d. Moreover, y0 ≥0 with equality if and only if also y = (0, 0, . . . , 0). ⊓ ⊔ We have now collected all necessary tools and deﬁnitions to prove Ehrhart’s Theorem (Theorem 3.14). Proof (of Ehrhart’s Theorem (Theorem 3.14)). Combining Proposi-tion 3.19 with Proposition 3.18 we get that the Ehrhart counting function of an n-dimensional lattice simplex in Rd uniquely extends to a polyno-mial function of degree at most n. For a general polytope P we triangulate it into maximal-dimensional simplices Fi and consider the triangulation of C(P) into the associated simplicial cones C(Fi). Then we apply inclusion-exclusion (e.g. Exer-cise 3.9). Finally, we use (3.1). ⊓ ⊔ Fig. 3.8: The boundary complex of a triangle + + − − − Fig. 3.9: Inclusion-Exculsion on the boundary complex Exercise 3.9 Remark 3.20 At this point it is intuitive, but wrong, to conclude ehrP (0) = 0P ∩Zd = 1 . As we will see later in Corollary 3.35, ehrP (0) = 1 does hold if P is a polytope. But the interpretation as 0P ∩Zd is wrong as the following example shows. Haase, Nill, Paffenholz: Lattice Polytopes — 75 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) We can count lattice points in dilations of complexes of lattice poly-topes. The entire chain of arguments given carries over to this setting. We obtain a counting function which is the evaluation of a polynomial. Consider, for example, C to be the boundary of a standard triangle, see Figure 3.8. Then our counting polynomial turns out to be ehrC(k) = 3k with constant coeﬃcient zero! See Figure 3.9. We will come back to this example in Remark 3.36. Exercise 3.10 Exercise 3.11 Exercise 3.12 Exercise 3.13 Exercise 3.14 Exercise 3.15 Exercise 3.16 3.5 Stanley’s theorem 3.5.1 Half-Open Decompositions of Cones The goal of this chapter is to deduce more information about the Ehrhart polyonomial. For instance, as we have seen in Remark 3.20, we haven’t even determined what its constant term is! And while we know that the numerator of the Ehrhart series of a d-dimensional lattice simplex is a polynomial of degree at most d whose coeﬃcients are all nonnegative integers, we weren’t able to conclude these strong statements for arbitrary polytopes from our naive inclusion-exclusion proof of Ehrhart’s theorem. In fact, this is the content of Stanley’s celebrated theorem, and in order to prove it we will need a more reﬁned way of decomposing our polytopes into simplices. The goal is to have no overlap in order to avoid any overcounting (and thus subtraction). This is called half-open decomposition [9, 32]. There are various ways how to do this. We will use a generic reference point as an arbiter to decide which points belong to which cells. As above, the right setting to do this is to consider cones instead of polytopes. Deﬁnition 3.21 (half-open decomposition) Given a vector ξ ∈Rd, we deﬁne the half-open cone Cξ with respect to ξ ∈Rd Cξ := {y ∈C : y + εξ ∈C for all ε > 0 small enough} . We say ξ ∈Rd is generic with respect to C (respectively, a triangulation T of C) if ξ is not in the linear hull of a (d −1)-dimensional face of C (respectively, any simplicial (d −1)-cone in T). Cξ can also be described as precisely the set of elements in C that are not visible from ξ (Exercise 3.17). See Figure 3.10 for an example. Let Exercise 3.17 ξ Fig. 3.10: Making a cone half open. The right face and the origin are not part of the half open cone. us note some properties: ▶If ξ is generic with respect to C, then Cξ = {y ∈C : y + εξ ∈int C for all ε > 0 small enough} , — 76 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) ▶If ξ ∈C, then Cξ = C, and if ξ is additionally generic, then C−ξ = int C . ▶If ξ ∈C is generic with respect to a triangulation T of C, then it is also generic with respect to C. In this case, ξ ∈int D ⊂int C for a unique D ∈T[d]. Our ﬁrst goal is to show that making cones half-open is compatible with decompositions. Proposition 3.22 Let T be a triangulation of the d-cone C, and let ξ ∈Rd be generic with respect to T. Then we have the following disjoint union: Cξ = G D∈T[d] Dξ. where T[d] is the set of d-dimensional faces of the triangulation (see Deﬁnition 2.22). In particular, if ξ ∈C is generic with respect to T, then C = G D∈T[d] Dξ and int C = G D∈T[d] D−ξ. A half-open decomposition in this way is illustrated in Figure 3.11, where this shows a slice through C containing ξ. x x Fig. 3.11: A triangulation and its half open decomposition. Proof. Let y ∈Dξ. Then for any ε > 0 small enough y + εξ ∈int (D) ⊂ int (C), so y ∈Cξ. Conversely, let y ∈Cξ, so y + εξ ∈int (C) for any ε > 0 small enough. This implies that there exists a unique D ∈T[d] so that y + εξ ∈int D for small enough ε > 0. The uniqueness argument implies disjointness of the union on the right hand side. ⊓ ⊔ We remark that there is also a beautiful generalization of the previous result using indicator functions described in the book of Hemmecke et al. . Let us now focus on simplicial d-cones D ⊂Rd. Deﬁnition 3.23 Let D be a simplicial d-cone in Rd and V = {v1, . . . , vd} ⊂ Rd the primitive ray generators. Let us note that ξ ∈Rd is generic with respect to D if and only if all coeﬃcients λv in the unique representation ξ = P λvv are non-zero. We deﬁne I+(ξ) := {v ∈V : λv > 0} and I−(ξ) := {v ∈V : λv < 0} . Using this notation, let us note the following alternative description of a half-open simplicial cone (Exercise 3.18). Exercise 3.18 Haase, Nill, Paffenholz: Lattice Polytopes — 77 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Lemma 3.24 Let ξ ∈Rd be generic with respect to a simplicial d-cone D with primitive ray generators V . Then Dξ = (X v∈V µvv : µv ≥0 for v ∈I+(ξ) and µv > 0 for v ∈I−(ξ) ) . We have seen in Corollary 2.44 how to translate Rd with translates of parallelepipeds. In the following, we will use this idea for the half-open cones Dξ. 0 v1 v2 ξ’ ξ Πξ(D) = Π−ξ′(D) Fig. 3.12: Half open cone and funda-mental parallelepiped for ξ and −ξ′. The dashed lines and the vertices with withe points are not part of the cone or fundamental parallelepiped. Deﬁnition 3.25 Let D be a simplicial d-cone with primitive ray genera-tors V . In case ξ ∈Rd is generic, we deﬁne the half-open parallelepiped Πξ(D) with respect to ξ as Πξ(D) := (X v∈V µvv : µv ∈[0, 1) for v ∈I+(ξ) and µv ∈(0, 1] for v ∈I−(ξ) ) Note that Πξ(D) ⊂Dξ. See Figure 3.12 for an illustration. For x strictly in the interior of D we recover the usual half-open fundamental paral-lelepiped of D with generating set V . The following result generalizes Corollary 2.44 and is left as Exercise 3.19. Exercise 3.19 Lemma 3.26 Let V = {v1, . . . , vd} ⊂Rd be linearly independent, and suppose ξ ∈Rd is generic with respect to the simplicial cone D := cone V . Denote by Λ the lattice generated by V . Then any point w ∈Rd has a unique representation w = y + z with y ∈Λ and z ∈Πξ(D). We can further decompose each of the half-open simplicial cones into half-open boxes. Recall that Z≥0V stands for the set of Z≥0-linear combinations of V . Proposition 3.27 Let V be the set of primitive ray generators of a simplicial d-cone D ⊂Rd, and let ξ ∈Rd be generic with respect to D. Then we have the following disjoint union: Dξ = G w∈Z≥0V w + Πξ(D) Proof. The fact that the translates by Λ-vectors are pairwise disjoint follows from the uniqueness in Lemma 3.26. From the existence part we see that Rd is covered by all Λ-translates of Πξ(D). It remains to observe that for w ∈Λ Dξ ∩ w + Πξ(D) =    w + Πξ(D) for w ∈Z≥0V ∅ else, We leave the veriﬁcation of this identity to the reader (Exercise 3.20). ⊓ ⊔ Exercise 3.20 Exercise 3.21 Exercise 3.22 — 78 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) 3.5.2 The integer point generating function of half-open cones Let us compute the integer point generating series and generating function for half-open cones. Corollary 3.28 Let V = {v1, . . . , vd} ⊂Zd be a linearly independent set of primitive vectors, let D = cone V , and let ξ ∈Rd be generic with respect to V . Then the integer point generating function of the half-open cone Dξ is summable, and GDξ (t) = GΠξ(D) (t) (1 −tv1)(1 −tv2) · · · (1 −tvd) . (3.4) Using Proposition 3.27 the proof follows precisely along the lines of the proof of Proposition 3.12 (just replace Corollary 2.44 by Lemma 3.26). Together with Proposition 3.22 we get the following nice formula. Corollary 3.29 Let C be a rational cone in Rd, let T be a triangulation of C into rational simplicial cones, and let ξ ∈C be generic. Then b GC (t) = X S∈T[d] b GSξ (t) , and b Gint C (t) = X S∈T[d] b GS−ξ (t) . (3.5) In particular, both series are summable, and (3.5) also holds on the level of rational functions. Proof. Equation (3.5) is a translation of Proposition 3.22 into generating functions. By Corollary 3.28, all the summands are summable Laurent series. ⊓ ⊔ 3.5.3 Stanley’s theorem and the h∗-polynomial of a lattice polytope Let us apply the previous results to cones over lattice polytopes. Proposition 3.30 Later: Beweis Let P ⊂Rd be a lattice polytope, let T be a triangu-lation of the cone C(P) which is induced by a lattice triangulation of P, and let ξ ∈C(P) be generic. Then b EhrP (t) is summable with sum EhrP (t) = GC(P ) (t, 1) = P S∈T[d+1] GΠξ(C(S)) (t, 1) (1 −t)d+1 . (3.6) Exercise 3.23 Now, we are nearly done. It remains to show the following lemma, which we leave to the reader as Exercise 3.24. Lemma 3.31 Let S ⊂Rd be a d-dimensional lattice simplex, and let ξ ∈Rd+1 be generic. We deﬁne h⋆ i,C(S),ξ := {y ∈Πξ(C(S)) ∩Zd+1 : y0 = i} Then Haase, Nill, Paffenholz: Lattice Polytopes — 79 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) (1) ξ ∈int C(S) if and only if h⋆ 0,C(S),ξ = 1. Otherwise, h⋆ 0,C(S),ξ = 0. (2) −ξ ∈int C(S) if and only if h⋆ d+1,C(S),ξ = 1. Otherwise, h⋆ d+1,C(S),ξ = 0. In particular, Pd+1 i=0 h⋆ i,C(S),ξ ≥1. Exercise 3.24 Here is the main result of this chapter. Theorem 3.32 (Stanley’s Non-Negativity Theorem) Let P be a d-dimensional lattice polytope. Then EhrP (t) = h⋆ 0 + h⋆ 1t + h⋆ 2t2 + · · · + h⋆ dtd (1 −t)d+1 , h⋆ 1, . . . , h⋆ d ≥0 and h⋆ 0 = 1. In particular, ehrP (t) = t + d d + h⋆ 1 t + d −1 d + · · · + h⋆ d−1 t + 1 d + h⋆ d t d , (3.7) Proof. We simply apply Proposition 3.30. In this notation h⋆ 0 + h⋆ 1t + h⋆ 2t2 + · · · + h⋆ dtd = X S∈T[d+1] GΠξ(C(S)) (t, 1) From Lemma 3.31 and ξ ∈int (C(P)) (as ξ is generic), we conclude that h⋆ 0 = 1 and h⋆ d+1 = 0. The last statement follows now from Proposi-tion 3.18. ⊓ ⊔ Deﬁnition 3.33 (h⋆-polynomial) The polynomial h⋆that appears in the numerator of the rational generating function of the Ehrhart series of P is the h⋆-polynomial of P. Example 3.34 Why do we consider the h∗-polynomial and don’t stick to the original description? For this, let us consider the Reeve simplex from (1.1) for m = 13, i.e. R := conv(0, e1, e2, e1 + e2 + 13e3) . Then the Ehrhart polynomial is 1 −1/6t + t2 + 13/6t3, however, it’s h∗-polynomial is 1 + 12t. This shows why working with the h∗-polynomial is so much more convenient: the values are integers and they are nonneg-ative. As the proof shows, the reason is that they have a nice counting interpretation as the number of lattice points in half-open parallelepipeds. Corollary 3.35 Let P be a d-dimensional lattice polytope for d ≥0. Then the following holds: (1) The constant term of the Ehrhart polynomial is 1 if P is non-empty. (2) h⋆(1) = Pd i=0 h⋆ i = d! vol(P) = nvolZd(P). — 80 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) (3) h⋆ 1 = ehrP (1) −d −1 = \|P ∩Zd\| −d −1. Proof. (1) This follows from Equation (3.7) by plugging in t = 0. (2) This follows from Proposition 3.18 (or directly from Equation (3.7)). (3) This follows from plugging in t = 1 into Equation (3.7). ⊓ ⊔ Remark 3.36 We already raised the issue of the constant coeﬃcient in Remark 3.20. Coming back to the boundary C of the standard triangle, there are two diﬀerent half-open decompositions. Fig. 3.13: Two half open decomposi-tions of the boundary of a triangle. The one on the left yields an h⋆-polynomial 1 + t + t2 while the one on the right yields 3t. Which one is “correct”? Looking at the cone over C, we see that the decomposition on the left contains the origin (once), while the one on the right does not. This explains the diﬀerence: 1 + t + t2 (1 −t)2 = 3t (1 −t)2 + 1 . We also see that there is exactly one choice for the multiplicity of the origin (in this case zero) so that the h⋆-polynomial has degree ≤1 which we need if we want the counting function to agree with the evaluation of a polynomial. We will identify this choice with the Euler characteristic in Remark 3.46 below. Exercise 3.25 Finally, let us note the following theorem proved by Stanley in . A completely diﬀerent proof appears in (Beck, Sottile ). The reader can move Beck/Sottile reference to begin-ning of half-open decomposition try to give a proof using the methods developed above in Exercise 3.26. Should maybe not be an exercise. Choosing the right decomposition if dim P < dim Q has some subtleties. More than if we work with irrational decomposition –AP Theorem 3.37 (Stanley’s Monotonicity Theorem) Let P and Q be two lattice polytopes such that P ⊆Q, d = dim Q and let h⋆ P and h⋆ Q be their h⋆-polynomials. Then h⋆ P,i ≤h⋆ Q,i 6for all 0 ≤i ≤d. 6 bad notation Exercise 3.26 Proof. easy if dim P = dim Q, else, choose ξ′ ∈relint Q and wiggle to generic ξ ∈relint P. then half-open decomposition from Cξ induces on Q the same half-open decomposition as ξ′. 7proof missing 7 proof missing Corollary 3.38 For two lattice polytopes P and Q with Q ⊆P we have deg Q ≤deg P. In particular, any face of P has degree at most deg P. ⊓ ⊔ 3.5.4 Where does the h∗-notation come from? Example 3.39 h⋆of ∆d. do half-open simplex as well ∆d with k facets removed has (d+1−k j+1−k) many j-faces. The origin for the funky notation h⋆is its close connection to the h-vector from enumerative combinatorics. Suppose C is a simplicial complex cite Stanley and others Haase, Nill, Paffenholz: Lattice Polytopes — 81 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) with f0 vertices, f1 edges, and so forth. Then the combinatorial h-vector make ref to C deﬁned there of C is deﬁned by the implicit equation fj = d X k=0 hk d + 1 −k j + 1 −k (3.8) This is, in fact, an invertible linear transformation between f-vector and h-vector. Its relation to the h∗-vector is given by the following result exercise: compute h from f Diese subsection noch ausformulieren. : Theorem 3.40 (Betke and McMullen 1985) Let P be a lattice poly-tope with a triangulation T. Let h⋆be the h⋆-vector of P and h the h-vector of the triangulation. Then hk ≤h⋆ k for 0 ≤k ≤d with equality if and only if the triangulation is unimodular. Proof. The number of simplices with k facets removed in a half-open decomposition satisﬁes the h-vector equation. Such a simplex has a box point at height k, so it contributes one to h⋆ k. It contributes more if the simplex was not unimodular. ⊓ ⊔ expand 3.6 Reciprocity explain the goal of this section before making a concrete example The interior of L = [a, b] is int L = (a, b). For integers a, b we can count the lattice points inside int L: ehrint L(k) = k(b −a) −1 . Evaluating ehrL(k) at −k for some positive integer k gives ehrL(−k) = (−k)(b −a) + 1 = −((−k)(b −a) −1) = −ehrint L(−k) . So for intervals the Ehrhart polynomial evaluated at negative integers counts (up to a sign) the lattice points in the interior of the interval. This would be a nice property, but maybe the example of an interval is too special to conjecture such a relation in general. So let us compute the interior lattice points in a more complicated example. We consider the d-dimensional standard simplex ∆d that we have already seen in the beginning of this chapter. We use the following observation to count lattice points in the interior ∆d. As we only want to count the lattice points in the interior of the k-th dilate of the simplex, we can ﬁrst consider all lattice points and then leave out lattice points (1) that have a 0 among their coordinates, or (2) whose coordinates sum up to k. — 82 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) This just means that we only want to count lattice points that satisfy the inequalities xi ≥1 for 1 ≤i ≤d, and whose coordinates sum up to at most k −1. Hence, we want to count lattice points in the set deﬁned by the inequalities xi ≥1 and d X i=1 xi ≤k −1 . Translating this by 1 ∈Rd gives the simplex deﬁned by the inequalities xi ≥0 and d X i=1 xi ≤k −d −1 , and this simplex clearly contains the same number of lattice points. We have computed this number in Proposition 3.2, so ehrint ∆d(k) = k −1 d . We see that also the number of interior lattice points is a polynomial in k of degree d. From d −k d = (−1)d k −d + d −1 d = (−1)d k −1 d we can conclude that ehrint ∆d(k) = (−1)dehr∆d(−k) . We can make the same observation as for the interval: The lattice points in the interior of the k-th dilation of the simplex are (up to a sign) the evaluation at −k of the Ehrhart polynomial! Let us check one more example, before we attempt to prove our observation. Consider the standard unit cube Cd. Counting the interior points in this case is rather simple. We obtain ehrint C(k) = (k −1)d = (−1)d((−k) + 1)d = (−1)dehrC(−k) , and again, the number of lattice points in the interior is given by the Ehrhart polynomial evaluated at negative values. 3.6.1 Stanley reciprocity for cones Let x = (x1, . . . , xd) ∈(Rd −{0})d. Then 1 x denotes the vector 1 x1 , . . . , 1 xd . Haase, Nill, Paffenholz: Lattice Polytopes — 83 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Lemma 3.41 Let D ⊂Rd be a simplicial cone with primitive generators V = {v1, . . . , vd}, and let ξ ∈Rd be generic. Then the map α : Πξ(V ) ∩Zd+1 − →Π−ξ(V ) ∩Zd+1 ξ 7− → d X i=0 vi −ξ is a bijection. Proof. proof uses wrong decomposition! Let y ∈Πξ(V ), so y has a representation of the form y = X v∈I λvv + X v∈J µvv for 0 < λv ≤1, 0 ≤µv < 1 Hence d X i=0 vi −y = X v∈I (1 −λv)v + X v∈J (1 −µv)v ∈Π−ξ(V ) ∩Zd+1 , which proves the claim. ⊓ ⊔ It follows from Corollary 3.29 (pick −ξ ∈int C) that b Gint C (t) is also a summable Laurent series. Theorem 3.42 (Stanley’s Reciprocity Theorem) Let C be a d-di-mensional polyhedral cone with rational generators. Then GC (t) = (−1)d Gint C 1 t . Proof. Let T by a triangulation of C and ξ ∈C generic as above. For S ∈T[d] let V (S) be the set of primitive generators of S, and let s(S) = P v∈V (S) v denote their sum. Then, Lemma 3.41 implies GΠξ(S) (t) = X a∈Πξ(S)∩Zd ta = X a∈Π−ξ(S)∩Zd ts(S)−a = ts(S) GΠ−ξ(S) 1 t . By Corollary 3.29 and Corollary 3.28 we can just sum up this equation over all maximal cones to obtain the desired result: GC (t) = X S∈T[d] GSξ (t) = X S∈T[d] GΠξ(S) (t) Q v∈V (S)(1 −tv) = X S∈T[d] ts(S) GΠ−ξ(S) 1 t Q v∈V (S)(1 −tv) = (−1)d X S∈T[d] GΠ−ξ(S) 1 t Q v∈V (S)(1 −1 tv ) = (−1)d X S∈T[d] GS−ξ 1 t = (−1)d Gint C 1 t . ⊓ ⊔ It is important to note that Stanley’s theorem is clearly wrong on the level of Laurent series! — 84 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) 3.6.2 Ehrhart-Macdonald reciprocity for lattice polytopes Finally, we can formalize our observation from the beginning of this section. Theorem 3.43 (Ehrhart-Macdonald Reciprocity) Let P ⊆Rd be a d-dimensional lattice polytope with Ehrhart polynomial ehrP (t), and let k ∈Z>0. Then ehrP (−k) = (−1)d \| int kP ∩Zd\| . The proof needs a little fact about the map Φ that maps summable Laurent series to rational functions. Lemma 3.44 Let f be a polynomial. Then X k∈Z f(k)tk Φ 7− →0. Proof. It suﬃces to prove this for the basis fm := (t+m m ), m ∈Z≥0, of References for binomial stuﬀmissing R[t]. So pick some m. Then X k≥0 fm(k)tk = X k≥0 k + m m tk Φ 7− → 1 (1 + t)m+1 . We compute the other sum: Diskussionsbedarf: = problematik ob Reihe oder Funktion problematisieren; und noch besseres Lemma (nicht nur exercise) wann wirklich Gleichheit auch als Reihe ist X k≤−1 fm(k)tk = X k≤−1 k + m m tk = X k≥1 −k + m m t−k = X k≥1 (−1)m k −1 m t−k = X k≥m+1 (−1)m k −1 m t−k = (−1)mt−(m+1) X k≥0 k + m m t−k = (−1)mt−(m+1) 1 (1 −1 t )m+1 = (−1)m tm+1(1 −1 t )m+1 = (−1)m (t −1)m+1 = (−1)m+1(−1)m (1 −t)m+1 = − 1 (1 −t)m+1 . ⊓ ⊔ Using this we can ﬁnally prove our reciprocity theorem. Haase, Nill, Paffenholz: Lattice Polytopes — 85 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Proof (Ehrhart-Macdonald Reciprocity (Theorem 3.43)). We make two computations: Φ  X k≥1 \| int kP ∩Zd\| tk  = Gint C(P ) (t, 1, . . . , 1) , and, using Lemma 3.44 for the second equation, wieder als eine Gleichung schreiben Φ  (−1)d X k≥1 ehrP (−k) tk  = Φ  (−1)d X k≤−1 ehrP (k) 1 tk   = Φ  (−1)d+1 X k≥0 ehrP (k) 1 tk   = (−1)d+1EhrP 1 t = (−1)d+1GC(P ) 1 t , 1, . . . , 1 Theorem 3.42 and Exercise 3.6 imply X k≥1 \| int kP ∩Zd\| tk = (−1)d X k≥1 ehrP (−k) tk . Comparing coeﬃcients of these two Laurent series gives the desired result. ⊓ ⊔ As an immediate application we can compute the Euler characteristic. Proposition 3.45 (Euler-Characteristic) Let S be a subdivision of the rational polytope P ⊂Rd into rational polytopes. Then X ∅̸=F∈S (−1)dim F = 1 . The restriction to rational objects is an artefact of our method and is not necessary for the validity of the assertion. Proof. Scaling P and S by a positive integer, we can assume that S contains only integral polytopes. Using the disjoint decomposition of P into the relative interiors of faces of S we see that for all k ∈Z≥1 \|kP ∩Zd\| = X F∈S \| relint kF ∩Zd\| . Thus, by Ehrhart-Macdonald Reciprocity (Theorem 3.43), we have an equality ehrP (t) = X F∈S (−1)dim F ehrF (−t) of polynomials. Evaluating at t = 0 and applying Corollary 3.35(1) yields the desired identity. ⊓ ⊔ — 86 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) Remark 3.46 The same argument shows that the constant coeﬃcient of the counting polynomial of a complex of lattice polytopes equals the Euler-characteristic of the complex: ehrC(0) = χ(C). This resolves the riddle raised in Remarks 3.20 and 3.36. 3.7 Properties of the h∗-polynomial As an immediate application of the results about counting lattice points that we have obtained so far we prove some facts about the relation between the geometry or combinatorics of lattice polytopes and their h⋆-vector. We will obtain more such results in the following chapters. 3.7.1 Degree and codegree of lattice polytopes Let us give some more applications regarding the h⋆-polynomial. Deﬁnition 3.47 (Degree and Codegree) The degree of P is deﬁned as deg(P) := max(k ∈Z≥0 : h⋆ k ̸= 0). The codegree of P is deﬁned as codeg(P) := d + 1 −deg(P). Ehrhart’s theorem implies 0 ≤deg(P) ≤d, so 1 ≤codeg(P) ≤d + 1. The degree of a lattice polytope can be seen as an algebraic measure of the complexity of a lattice polytope. Its concrete geometric interpretation is given by the codegree. Corollary 3.48 The codegree of a d-dimensional lattice polytope equals the smallest positive integer k such that kP contains an interior lattice point. Proof. This follows from Lemma 3.49 and the Ehrhart-Macdonald Reci-procity (Theorem 3.43). ⊓ ⊔ Exercise 3.27 Lemma 3.49 Let p be a polynomial of degree d with rational generating function X t≥0 p(t)zt = h⋆ 0 + h⋆ 1t + h⋆ 2t2 + · · · + h⋆ dtd (1 −t)d+1 Then h⋆ d = h⋆ d−1 = . . . = h⋆ k+1 = 0 and hk ̸= 0 if and only if p(−1) = p(−2) = . . . = p(−(d −k)) = 0 and p(−(d −k + 1)) ̸= 0. In this case, h⋆ k = p(−(d + 1 −k)). Haase, Nill, Paffenholz: Lattice Polytopes — 87 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) The proof is left as Exercise 3.28. Applied to our situation this has the following immediate consequence. Exercise 3.28 Corollary 3.50 Let P be a lattice polytope. The highest non-zero coeﬃ-cient h⋆ deg(P ) of h⋆equals the number of lattice points in int((codeg P)P). ⊓ ⊔ Finally, using reciprocity it is possible to compute the second highest coeﬃcient of the Ehrhart polynomial. Exercise 3.29 Proposition 3.51 Let P be a lattice polytope with Ehrhart polynomial ehrP (t) = c0 + c1t + c2t2 + · · · + cdtd. Then cd−1 equals half of the normalized surface area of the boundary of P. You will prove this result in Exercise 3.29. The following proposition is immediate from the fact that the sum of the coeﬃcients of the h⋆-polynomial is the lattice volume and that the linear coeﬃcient counts the number of lattice points minus (d + 1). Proposition 3.52 A d-dimensional lattice polytope P has lattice volume nvolZd(P) = \|P ∩Zd\| −d if and only if deg(P) ≤1. In particular, P has lattice nvolZd(P) = 1 if and only if its degree deg(P) is 0. In this case P is the standard simplex. ⊓ ⊔ Proposition 3.53 Let P be a lattice polytope of degree deg P ≤k. Then int F ∩Zd = ∅for all faces F of P of dimension at least k + 1. Proof. Assume there is a face F of P of dimension k + 1 that contains a relative interior lattice point v. Then there are k + 2 vertices u1, . . . , uk+2 of F such that v = k+2 X i=1 λiui k+1 X i=1 λi = 1 and λi > 0 for 1 ≤i ≤k + 2 . Further, we can ﬁnd d −1 −k vertices uk+3, . . . , ud+1 of P such that u1, . . . , ud+1 is aﬃnely independent. Consider v′ := k+2 X i=1 1 d −k λiui = d+1 X i=k+3 1 d −k ui . This is a point in int P. Now (d −k)v′ is integral, so (d −k) int P ∩Zd ̸= ∅. Thus, codeg P ≤(d −k), or deg P ≥k + 1. In particular, this implies that for deg P = 0 only the vertices are lattice points, and for deg P = 1 the only lattice points that are not vertices are on the edges of P. Let X = {x1, . . . xk} ⊆P ∩Λ be a set of k lattice points that is not entirely contained in a proper face of P. Then x := x1 + · · · + xk is an interior lattice point of kP. This proves the next propisition. — 88 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) Proposition 3.54 Let P be a d-dimensional lattice polytope of degree at most s. Then any subset W ⊆P ∩Λ of at most d −s lattice points is contained in a proper face of P. ⊓ ⊔ We have viewed these propositions so far by ﬁrst ﬁxing the dimension. We could swap this view and ﬁx the degree. Then the proposition for example tells us that in a polytope of dimension d ≥s + 2 any two lattice points must be in a common face. Let P be a d-dimensional lattice polytope in Rd. We deﬁne the lattice pyramid over P as Pyr(P) := conv(P × {0}, ed+1) ⊆Rd+1 , where e1, . . . , ed+1 is the standard basis of Rd+1. See Figure 3.14 for an example. In Exercise 3.30 you will show the following proposition. Fig. 3.14: A lattice pyramid over a unit square Proposition 3.55 For a lattice polytope P the lattice pyramid Pyr(P) of P has the same h⋆-polynomial as P. ⊓ ⊔ Exercise 3.30 Exercise 3.31 Deﬁnition 3.56 (Lawrence Prism) A Lawrence prism with heights h1, . . . , hd ≥0 and h1 + · · · + hd ≥2 is the polytope Law(h1, . . . , hd) := conv 0, e1, . . . , ed−1, e1 + h1ed, . . . , ed−1 + hd−1ed, hded . Let ∆2 2 := 2 ∆2. This is sometimes called the exceptional triangle. We have the following proposition. Proposition 3.57 Let P be a k-fold lattice pyramid over a lawrence prism or the exceptional triangle for some k ≥0. Then P has degree 1. Proof. The h⋆-polyonmial of the exceptional triangle is 1 + 3t, which proves this part of the proposition. Further, lattice pyramids do not change the h⋆-polynomial, so we only have to check the proposition for lawrence prisms. For a lawrence prism L := Law(h1, . . . , hd) we have a natural lattice projection π : L →∆d−1. Hence, if kL has an interior lattice point, then so has k ∆d−1. Hence k ≥d, and deg L ≤1. As by assumption h1 + · · · + hd−1 ≥2 for L we can ﬁnd d lattice points not contained in a common face of L, so dL contains an interior lattice point. Hence, deg L = 1. ⊓ ⊔ Deﬁnition 3.58 A lattice polytope P is empty if the vertices are the only lattice points in P. Haase, Nill, Paffenholz: Lattice Polytopes — 89 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Proposition 3.59 Let P be a lattice polytope of degree at most 1 such that no lattice point is strictly between two other lattice points. Then P is empty. Proof. By Corollary 3.38 any face of P has degree at most 1. Hence, no lattice point can be in the interior of a k-dimensional face for k ≥2. ⊓ ⊔ Exercise 3.32 Proposition 3.60 Let P be a lattice polytope of degree deg P = 1 such that V(P) = P ∩Zd. Then P is a simplex or there are u1, u2, u3, u4 ∈ V(P) such that u1 + u2 = u3 + u4 . Proof. 3.7.2 Ehrhart polynomials of lattice polygons Fig. 3.15 (0, 0) (2, 0) (2, b −4) (1, i + 1) Fig. 3.16: Lattice polygons re-alizing 4 ≤ b ≤ 2i + 6 As an example, we will completely classify Ehrhart polynomials of lattice polygons in this section. Essentially, the main work was already done in the Chapter 1 by proving Scott’s inequlity (Theorem 1.10). Now, we just have to exploit the properties of the h⋆-polynomial. Proposition 3.61 A polynomial h⋆ 2t2 + h⋆ 1t + 1 for h⋆ 1, h⋆ 2 ∈Z≥0 is the h⋆-polynomial of a lattice polygon if and only if (1) h⋆ 2 = 0 and h⋆ 1 is arbitrary. Then P has no interior lattice points. (2) h⋆ 2 = 1 and h⋆ 1 = 7. Then P ∼ = 3∆2. (3) 1 ≤h⋆ 2 ≤h⋆ 1 ≤3h⋆ 2 + 3. Then P has interior lattice points. See Exercise 3.34 for concrete examples. Proof. Let us ﬁrst show that these conditions are necessary. Note that h⋆ 2 is the number of interior lattice points i, while h⋆ 1 = b+ i−3, where b is the number of boundary lattice points. Moreover, vol(P) = nvolZd(P)/2 = (1 + h⋆ 1 + h⋆ 2)/2. Hence, Scott’s theorem tells us that, if i ≥1 and P ̸∼ = 3∆2, then h⋆ 1 ≤3h⋆ 2 + 3. Finally, if i ≥1, then h⋆ 2 = i ≤h⋆ 1 = b + i −3, since b ≥3. It suﬃces to realize lattice polygons satisfying each of these conditions. For i = 0, any b ≥3 can be realized by lattice polygons of the form as depicted in Figure 3.15. In fact, it is not diﬃcult to show that these In fact, as Exercise 3.33 shows these are precisely the lattice polygons without interior lattice points. Let i ≥1. The condition h⋆ 2 ≤h⋆ 1 ≤3h⋆ 2 + 3 is equivalent to 3 ≤b ≤ 2i + 6. The case b = 3 is easy to realize, so let b ≥4. Then any of these cases is realized by Figure 3.16. ⊓ ⊔ — 90 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) ice polygons All possible pairs (h⋆ 1, h⋆ 2) are depicted in Figure 3.17. Let us now deduce all Ehrhart polynomials c2t2 + c1t + 1 of lattice polygons. By Pick’s Theorem 1.8 c2 equals the area of P, and by Proposition 3.51, c1 is half the number of boundary lattice points of P. The following theorem characterizes all pairs (c1, c2) that correspond to an Ehrhart polynomial of a polygon. Corollary 3.62 A polynomial c2t2 + c1t + 1 with c1, c2 ∈1/2Z and c1 ≥3/2 deﬁnes the Ehrhart polynomial of a lattice polygon P if and only if one of the following three conditions is satisﬁed: (1) c1 −c2 = 1. Then P has no interior lattice points. (2) c1 = c2 = 9/2. Then P is 3∆2. (3) c1 ≤c2/2 + 2. Then P has interior lattice points. Exercise 3.34 commented: classiﬁcation of empty polygons 3.7.3 Polytopes with Small Degree 3.8 Brion’s theorem The goal of this ﬁnal section is the celebrated Theorem of Brion. It relates for any lattice d-polytope the integer point generating functions of all vertex cones of P to the integer point generating function of the polytope. Let P be a rational d-dimensional polytope and F a face of P. Recall the tangent cone of F in P from Deﬁnition 2.29: T F P := {v ∈Rd : ∃w ∈F, ε > 0 : w + ε(v −w) ∈P} . The tangent cone is the common intersection of all supporting half-spaces at F. We have seen in Proposition 2.30 that the shifted cone T F P −x for some x ∈F is dual to the normal cone of F. We can use the generating series of tangent cones to compute the generating series of the polytope. Theorem 3.63 (Brianchon-Gram Theorem) Let P be a rational d-polytope. Then b GP (t) = X F⪯P (−1)dim F b GT F P (t) , where the sum is over all non-empty faces of P. The Brianchon-Gram identity is valid more generally on the level of indicator functions. In order to prove it, we need to study the complex of visible faces. We say that a face F of a polytope P is visible from a point v ̸∈P if for some (equivalently every) w ∈relint F the segment conv(v, w) intersects P only in w. Haase, Nill, Paffenholz: Lattice Polytopes — 91 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Lemma 3.64 F is visible from v if and only if v ̸∈T F P . Proof. If v ̸∈T F P, there is a separating linear functional a: ⟨a, v ⟩> max⟨a, T F P ⟩≥max⟨a, P ⟩. But then, this strict inequality is true for every point of the form (1 − λ)v + λw for λ ∈[0, 1). So none of them can belong to P. That is, F is visible from v. If, conversely, v ∈T F P, we know that there are w′ ∈F and an ε > 0 so that v′ := w′ + ε(v −w′) ∈P. Further, taking a smaller ε if necessary, we can assume that w′′ := w + ε(w −w′) ∈F, because w ∈relint F. But then the point ε2v + (1 −ε2)w = (1 −ε)w′′ + εv′ belongs to both conv(v, w) and to P, and F is not visible from v. ⊓ ⊔ Corollary 3.65 If P is a rational polytope and v ̸∈P a rational point, then the set visibleP (v) of faces of P visible from v is a polyhedral complex, and as such is isomorphic to a rational subdivision of a rational polytope. add to notes: can drop “rational” ev-erywhere. Proof. From the deﬁnition of visibility we see that G ⪯F ∈visibleP (v) implies G ∈visibleP (v). So visibleP (v) is a subcomplex of the boundary of P. Let H be a rational hyperplane separating v from P, and consider the rational polytope Q := H ∩conv(P ∪{v}). Then H ∩conv(F ∪{v}) : F ∈visibleP (v) is a subdivision of Q which is combinatorially isomorphic to visibleP (v). ⊓ ⊔ m Fig. 3.18: m ̸∈ P. The complex S is drawn in red. Proof (of the Brianchon-Gram Theorem (Theorem 3.63)). Think of the Laurent polynomial on the left hand side as an inﬁnite Laurent series that contains all possible monomials, but most coeﬃcients are 0. To prove this relation we compare coeﬃcients of an arbitrary monomial tm on both sides. We have to distinguish the two cases m ∈P and m ̸∈P. (1) m ∈P: Then m ∈T F P for every non-empty face F of P. Hence, the coeﬃcient of tm on the right hand side is X ∅̸=F⪯P (−1)dim F = 1 , using Euler’s relation. cross-reference — 92 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) (2) m ̸∈P: See Figure 3.18. By Lemma 3.64, the coeﬃcient of tm on the right hand side is X ∅̸=F⪯P (−1)dim F − X ∅̸=F∈visibleP (m) (−1)dim F = 1 −1 = 0 , using the fact that the Euler-characteristic of visibleP (m) is 1 by Corollary 3.65 and Remark 3.46. ⊓ ⊔ Now recall the map Φ : b L − →R that we introduced in Section 3.3. There we have only applied it to pointed polyhedral cones. We now want to study this map also in the case of cones that have a nontrivial lineality space. Recall that for a cone C the lineality space is deﬁned as lin(C) := C ∩(−C). It is the maximal linear subspace contained in C. We start with a simple example that explains the basic idea of our next theorem. Consider the sets C+ := [0, ∞) ⊂R C−:= 3 −C+ = (−∞, 3] P := [0, 3] . C0 is a one-dimensional cone, and P is the intersection of C0 and C−, P = C+ ∩C−. We compute the integer point generating function and the image under Φ for C+ and C−. The series are b GC+ (t) = X k≥0 tk b GC−(t) = X k≤3 tk = t3 X k≤0 tk = t3 X k≥0 t−k , so we obtain the functions GC+ (t) = Φ(b GC+ (t)) = 1 1 −t GC−(t) = Φ(b GC−(t)) = t3 1 1 −1 t = −t4 1 −t The integer point generating function of P is the ﬁnite geometric series b GP (t) = GP (t) = 1 −t4 1 −t = 1 + t + t2 + t3 . We observe that GP (t) = GC+ (t) + GC−(t) . Using the construction of the map Φ we can make the following symbolic calculation GP (t) = Φ(b GC+ (t)) + Φ(b GC−(t)) = Φ(b GC+ (t) + b GC−(t)) = Φ(b GR+P (t)) = Φ(b GR (t)) + Φ(b GP (t)) This can only hold if Φ(b GR (t)) = 0, i.e. if Φ maps the inﬁnite series P k∈Z tk to 0. The following proposition shows that this indeed holds in This is the content of Lemma 3.44 general for cones with nontrivial lineality space. Haase, Nill, Paffenholz: Lattice Polytopes — 93 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Proposition 3.66 Let C ⊆Rd be a polyhedral cone with integer point series b GC (t). If lineal C ̸= {0} then Φ(b GC) = 0. Proof. Let v ∈lineal(C) −{0}. Then Rv ⊆C, so that tvb GC (t) = b GC (t) . Applying the map Φ gives tvΦ(b GC (t)) = Φ(b GC (t)) ⇐ ⇒ (1 −tv)Φ(b GC (t)) = 0 . v ̸= 0 implies Φ(b GC) = 0. ⊓ ⊔ We can apply the observation of this proposition to obtain a very simple formula for the integer point generating function of a polytope. Theorem 3.67 (Brion’s Theorem) Let P be a rational d-polytope. Then GP (t) = X v vertex of P GT vP (t) . Proof. Apply the map Φ to both sides of the Brianchon-Gram Identity of Brianchon-Gram Theorem (Theorem 3.63). The only non-pointed tangent cones are those originating from a vertex of P, so by Proposition 3.66 only the contributions of the vertices are non-zero on the right hand side. ⊓ ⊔ Exercise 3.35 Exercise 3.36 (0, 0) (1, 0) (1, 1) (0, 1) Fig. 3.19 Example 3.68 Let P be the = 0/1-square in R2. See Figure 3.19. Then GP (x, y) = 1 (1 −x)(1 −y) + x (1 −1 x)(1 −y) + y (1 −x)(1 −1 y ) + xy (1 −1 x)(1 −1 y ) = 1 (1 −x)(1 −y) + −x2 (1 −x)(1 −y) + −y2 (1 −x)(1 −y) + x2y2 (1 −x)(1 −y) = (1 −x2)(1 −y2) (1 −x)(1 −y) = 1 + x + y + xy So GP (1, 1) = 1 + 1 + 1 + 1 = 4. The theorem provides us with a general method to explicitly compute the function GP (t). We have seen in Corollary 3.28 how we can compute the I don’t understand this paragraph Explain use of Brion’s Theorem in Barvinok’s algorithm integer point generating series of a simplicial cone. To use this formula in the Theorem of Brion we triangulate the polytope P, and compute the generating function of each simplex in the triangulation (including the lower dimensional ones). We then sum up the generating functions using the principle of inclusion-exclusion. — 94 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) 3.9 Problems included on page 63 3.1. Determine a formula for the volume of a 3-dimensional lattice polytope using the number of lattice points in the k-multiple for k = 1, 2 and 3. included on page 63 3.2. Determine a formular for the volume of a 3-dimensional lattice polytope using \|P ∩Z3\|, \| int (P) ∩Z3\|, \|2P ∩Z3\|. The reader is also invited to ﬁnd many more such formulas by using diﬀerent values of the Ehrhart polynomial. included on page 66 3.3. Show ∞ X k=0 k + d d xk = 1 (1 −x)d+1 . (Why is the equality sign justiﬁed here?) included on page 70 3.4. The goal of this exercise is to give a proof of Proposition 3.7. (1) Show that the set Lsum of summable Laurent series is an L-submodule of ˆ L, i.e. show that for f ∈L and g, h ∈Lsum also f · g and g + h are summable. (2) Prove that this turns Φ into a homomorphism of L-modules, i.e. show that Φ(f · g) = fΦ(g) and Φ(f + g) = Φ(f) + Φ(g). included on page 71 3.5. Prove that there is a natural homomorphism from summable series to rational functions Φ : b L − →R := \|(x1, . . . , xd) , mapping b G to f/g if gb G = f in b L. included on page 71 3.6. Let S, S′ be subsets of the a (possibly translated) pointed cone in Rd. Then GS (t) = GS′ (t) implies b GS (t) = b GS′ (t). included on page 74 3.7. Prove Lemma 3.17. Hint: do j = 0 ﬁrst included on page 74 3.8. Zeige dass (t+d−j d ) mit j = 0, . . . , d eine Basis des Vektorraums der Polynome vom Grad ≤d ist. included on page 75 3.9. Let subsets S1, . . . , Sm of Rd be given. Then b GS i∈[m] Si (t) = X ∅̸=I⊆[m] (−1)\|I\|+1b GT i∈I Si (t) . Remark: This is just the usual inclusion-exclusion formula for sets. Haase, Nill, Paffenholz: Lattice Polytopes — 95 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) included on pa 3.10. Determine the Ehrhart polynomial of the reeve tetrahedron deﬁned in (1.1) for m ∈Z≥1. What do you observe for m = 20? included on page 76 3.11. Let P be a lattice polytope with Ehrhart polynomial ehrP (t). Compute the Ehrhart polynomial of the bipyramid over P. included on page 76 3.12. Compute the Ehrhart polynomial of the cross polytope. included on page 76 3.13. Let P be a d-dimensional lattice polytope with Ehrhart polynomial Pd k=0 cktk. Show that cd−1 = 1 2 vol(∂P). Here, vol(∂P) denotes the surface area of P, namely, vol(∂P) := X F∈F(P ) vol(F), where F(P) is the set of facets of P and vol(F) denotes the (non-normalized) volume with respect to the lattice aﬀ(F) ∩Zd. For instance, note that vol(conv((1, 0), (0, 1))) equals 1 and not √ 2. Hence, vol(∂conv((1, 0), (0, 1), (−1, 0), (0, −1))) = 4 . included on page 76 3.14. A simplex which is unimodularly equivalent to the standard sim-plex is called unimodular. A triangulation is unimodular if all its simplices are. (1) For a k-dimensional unimodular simplex ∆and t ∈Z≥1 show that \|Zk ∩relint(t∆)\| = t −1 k . (2) Suppose P admits a unimodular triangulation T with f0(T) vertices, f1(T) edges, . . . , fd(T) d-simplices. Show that ehrP (t) = d X k=0 fk(T) t −1 k . (3) Conclude that any two unimodular triangulations have the same f-vector (f0, . . . , fd) . included on page 76 — 96 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) 3.15. Prove that the coeﬃcients of the Ehrhart polynomial of a d-dimensional lattice polytope are in Z/d!. 3.16. For integers p,q with gcd(p, q) = 1 deﬁne the tetrahedron ∆pq = conv 0 1 0 1 0 0 1 p 0 0 0 q . (1) Argue that its vertices are its only lattice points. (White proved a converse: every lattice tetrahedron with only four lattice points is unimodularly equivalent to a ∆pq.) (2) Compute the Ehrhart polynomial and the h⋆-polynomial of ∆pq. (3) For which parameters are ∆pq and ∆p′q′ unimodularly equiva-lent? included on page 76 3.17. Let ξ ∈Rd and C a d-cone. Then Cξ = {y ∈C : yε ∈C for all ε > 0 small enough} , where yε := (1 −ε)y + εξ. included on page 77 3.18. Prove Lemma 3.24. included on page 78 3.19. Prove Lemma 3.26. included on page 78 3.20. Check carefully and rigorously the last identity in the proof of Proposition 3.27. included on page 78 3.21. Show directly that C\C[x) is a union of faces of C. included on page 78 3.22. Let T be a triangulation of a full-dimensional cone C. Show that there is always a generic element ξ ∈int (C). included on page 79 3.23. Consider the polygon with vertices (0, 0), (1, 0), (0, 2), (2, 4). Com-pute, using a half-open decomposition GC(P ) (t1, t2, t3). included on page 80 3.24. Prove Lemma 3.31. included on page 81 3.25. Let Q, P be lattice polytopes with Q ⊆P. Show that there exists a triangulation of P that restricts to a triangulation of Q. Hint: Let V denote the set of vertices. Choose ﬁrst a generic regular triangulation w : V(Q) →R, leading to linear functions lσ on simplices σ of the triangulation. Now, choose generic values of w on V(P)\V(Q) such that w(v) > lσ(v) for all σ in the triangulation of Q and vertices v ∈V(P)\V(Q). Haase, Nill, Paffenholz: Lattice Polytopes — 97 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) included on pa 3.26. Let Q, P be lattice polytopes with Q ⊆P. Show Stanley’s Mono-tonicity Theorem (Theorem 3.37) h∗ Q ≤h∗ P coeﬃcientwise. Hint: Choose a triangulation as in Exercise 3.25. . Needs more hints. See exercise sheets?! included on page 87 3.27. Give a direct geometric proof using Carathéodory’s Theorem (The-orem 2.4) that codeg(P) ≤d + 1 for a d-dimensional lattice poly-tope. included on page 88 3.28. Prove Lemma 3.49. included on page 88 3.29. Prove Proposition 3.51. included on page 89 3.30. Show that h⋆ Pyr(P ) = h⋆ P for a lattice polytope P. included on page 89 3.31. Let m ∈Z≥1. Use Exercise 3.30 to show that fm(k) := k X j=1 jm is a polynomial in k. What is its degree and leading coeﬃcient? included on page 90 3.32. Calculate the h∗-polynomial of an empty 3-dimensional lattice polytope P with a vertices and of normalized volume b. Here empty means that any lattice point in P is a vertex of P. Deduce the h∗-polynomials of the tetrahedra ∆pq of Exercise 2.2. Check that you get the same solution for the Ehrhart polynomial as before :-) included on page 91 3.33. Let P be a lattice polygon. Show that P has no interior lattice points if and only if P is unimodular equivalent to 2∆2 or it is unimodularly equivalent to conv((0, 0), (a, 0), (0, 1), (0, b)) for some a, b ≥0. included on page 91 3.34. Are (1) 1 + 8t + t2 (2) 1 + 9t + t2 (3) 1 + 2t + 2t2 (4) 1 + t + 2t2 — 98 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 3. Ehrhart Theory (draft of October 7, 2020) h∗-polynomials of a lattice polygon? 3.35. Compute the Ehrhart generating function of P = [0, 1]2 using the Brion’s Theorem (Theorem 3.67). 3.36. Apply Brion’s identity to P := conv " 0 2 2 3 1 −1 2 0 # and verify that both rational functions coincide (you may want to use a computer for this). Haase, Nill, Paffenholz: Lattice Polytopes — 99 — Geometry of Numbers 4 Contents 4.1 Minkowski’s Theorems . . . . . . . . . . . . . . . . . . . 102 4.2 Coverings and Packings . . . . . . . . . . . . . . . . . . 106 4.3 Flatness Theorem . . . . . . . . . . . . . . . . . . . . . . . 110 4.4 Finiteness of lattice polytopes with few interior lattice points . . . . . . . . . . . . . . . . . . . . 111 4.4.1 Finiteness of barycentric coordinates of lattice simplices . . . . . . . . . . . . . . . . . . . . . . . . 112 4.4.2 Coeﬃcient of asymmetry . . . . . . . . . . . . . . . . 115 4.4.3 Bounding the volume . . . . . . . . . . . . . . . . . . . 116 4.5 Lower Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 4.6 Empty lattice simplices . . . . . . . . . . . . . . . . . . 119 4.7 Lattice polytopes without interior lattice points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 4.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 Geometry of numbers deals with the relation between two objects: convex bodies on the one hand, and lattices one the other hand. A typical question in this area is whether and how the volume and the number of lattice points of convex body are related. The term “geometry of numbers” was coined by Minkowski who used convex geometric methods, in particular his fundamental theorem Corol-lary 4.3, in order to bound class numbers in algebraic number theory. In the 20th century geometry of numbers has grown into an established ﬁeld of research with connections into many branches of mathematics. Lecture Notes Lattice Polytopes (draft of October 7, 2020) While most of the theory treats general convex bodies, in these notes we will focus on those tools which we need to prove results that apply only to lattice polytopes. 4.1 Minkowski’s Theorems Minkowski’s two theorems are the basis of this whole branch of discrete mathematics. Both essentially tell us something about generators of the lattice and prove that we can ﬁnd such generators with a bounded Euclidean length. The theorems, of which we will only prove the ﬁrst, are however not constructive. We will remedy this in Chapter 6. Throughout, Λ ⊂Rd is a lattice of rank d (the reader may think of Zd). A set K ⊆Rd is centrally symmetric if x ∈K implies −x ∈K. Deﬁnition 4.1 A subset K ⊆Rn is a convex body if K is bounded and convex. The set of convex bodies in Rd is deneted by C. The subset of centrally symmetric convex bodies is C0. Note that the deﬁnition of the term convex body varies in the literature. The following theorem establishes a fundamental correspondence between lattice points in a centrally symmetric convex body and its volume. Theorem 4.2 (van Der Corput, 1935) Let K ⊂Rd be a centrally symmetric convex set. Then vol(K) ≤2d \|K ∩Λ\| det Λ . If K is compact, then the inequality is strict. Minkowski’s First Theorem, that he proved almost forty years earlier, is now a direct corollary of this. This result is the fundamental theorem in this area and it is considered to be the starting point of the theory. Corollary 4.3 (Minkowski’s First Theorem, 1898) Let K ⊆Rd be convex and centrally-symmetric with vol K > 2d det Λ. Then there exists a ̸= 0 in K ∩Λ. If K is also compact, then it suﬃces to assume vol K ≥2d det Λ. ⊓ ⊔ Exercise 4.1 Exercise 4.2 For the proof of these results we need the following lemma, which uses a beautiful pidgeonhole-style argument to prove that the intersection of a suﬃciently large set with some aﬃne translate of the lattice is large. Lemma 4.4 (Generalized Blichfeldt’s Theorem, 1914) Let S ⊆ Rd be a (Lebesgue measurable) set with vol(S) > m det(Λ) for a positive integer m. Then there exist m + 1 pairwise distinct points p1, . . . , pm+1 ∈S such that pi −pj ∈Λ for all i, j. — 102 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Proof. By considering a suﬃciently large subset, we may assume that S is bounded. Choose a closed fundamental parallelepiped (see Deﬁnition 2.42) Π := Π(Λ) of Λ. Note that det Λ = vol Π. For any x ∈Λ let Sx := {y ∈Π \| x + y ∈S} = Π ∩(S −x) Note that Sx ̸= ∅if and only if x ∈(S −Π) ∩Λ. As S −Π is bounded, Exercise 2.14 implies that there are only ﬁnitely many x ∈Λ with Sx ̸= ∅. This implies that the function x S x + S 0 Fig. 4.1 f := X x∈Λ idx , where idx is the indicator function on Sx (i.e., it evaluates to 1 on Sx and 0 elsewhere), is well-deﬁned. Using that fundamental parallelepiped tile the space (Corollary 2.44) we compute Z Π f dx = X x∈Λ Z Π idx dx = X x∈Λ vol(Sx) = X x∈Λ vol(Π ∩(S −x)) = X x∈Λ vol(S ∩(x + Π)) = vol(S) > m det Λ = Z Π m dx Hence there is y ∈Π with f(y) > m. Since f only evaluates to integers, we get f(y) ≥m + 1. In particular, there exist x1, . . . , xm+1 ∈Λ such that y ∈Sx1 ∩· · · ∩Sxm+1. Therefore, deﬁning pi := y + xi ∈S for i = 1, . . . , m + 1 yields m + 1 points which have the desired properties. ⊓ ⊔ (a) The set S and the fun-damental parallelepiped Π y (b) And the shifted intersections with the fun-damental parallelepiped Fig. 4.2: Illustrating the proof of Generalized Blichfeldt’s theorem (Lemma 4.4) Now, we can easily prove van der Corput’s Theorem (Theorem 4.2). Proof (of van der Corput’s Theorem (Theorem 4.2)). We will give an indirect proof. Let us assume that vol(K) > m2d det(Λ) for a positive integer m. Our goal is to show that there exist m distinct pairs of non-zero lattice points ±x1, . . . , ±xm in K. Together with the origin this will give 2m + 1 lattice points in K. Let T := 1 2K. Then vol T = vol K 2d > m det Λ. Hence, by Gen-eralized Blichfeldt’s theorem (Lemma 4.4), there are m + 1 distinct points p1, . . . , pm+1 ∈T such that pi −pj ∈Λ for all i, j. Choose xi := pi −pm+1 for i = 1, . . . , m as the desired lattice points. Note that xi = pi + (−pm+1) ∈T + T = K. Let K be compact and vol K = 2d det Λ. Since K is compact, for each x ∈2K\K there exists 0 < ϵx < 1 such that x ̸∈(1 + ϵx)K. Haase, Nill, Paffenholz: Lattice Polytopes — 103 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Boundedness of 2K implies that 2K has only ﬁnitely many lattice points. Let ϵ be the minimum over ϵx for all x ∈(2K\K) ∩Λ. This choice ensures that (1 + ϵ)K and K have the same set of lattice points (note that αK ⊆α′K for 0 < α < α′ as K is centrally-symmetric and convex). Since vol((1 + ϵ)K) > 2d det Λ, the result follows. ⊓ ⊔ Centrally-symmetric convex bodies with the origin as their only interior lattice point which have maximal volume 2d det(Λ) are also called extremal bodies. Minkowski’s theorem does not tell us how to ﬁnd the integral point, it just tells us it exists. There are polynomial time algorithms to explicitly ﬁnd such a point, but only for a much larger volume bound. See Section 6.4 on the LLL-Algorithm for a method. Finding a short lattice vector is a very important problem in integer optimization and in cryptography, see e.g. [25, 50, 51]. Although we cannot easily compute a shortest vector of a lattice, Minkowski’s Theorem at least allows us to estimate the length of such a vector. 0 Fig. 4.3: A triangle in the plane together with two scaled copies with scaling factors λ1 and λ2. Proposition 4.5 Let Λ ⊂Rd be a lattice. Then there is a vector v ∈ Λ \ {0} such that ∥v∥≤ √ d(det Λ) 1/d . Proof. Let Vd be the volume of the d-dimensional unit ball Bd and choose α := 2 det Λ Vd 1/d . (4.1) Then vol(α Bd) = αd Vd ≥2d det Λ . By Minkowski’s First Theorem (Corollary 4.3) there is a non-zero lattice point v in α Bd, hence, of length at most α. We need to estimate the size of α. The volume of the unit ball is Vd := π⌊d/2⌋2⌈d/2⌉ Q 0≤i0{dim lin(λK ∩Λ) ≥k}. For K = Bd we call λk := λk(Bd) the k-th successive minimum of the lattice Λ. See also Figure 4.3. — 104 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Then λ1 ≤λ2 ≤· · · ≤λd . Note that λ1 > 0 as Λ is discrete. The following corollary is equivalent to Minkowski’s First Theorem (Corollary 4.3). Corollary 4.7 Let K ∈C0. Then λd 1 vol K ≤2d det Λ . Note that a compact centrally symmetric convex body deﬁnes a norm ∥.∥K on Rd via ∥x∥K := max µ ( µx ∈K ) , and any norm is of this form. Proposition 4.8 Let K ∈C0 be compact and Λ ⊆Rd be a lattice with successive minima λ1, . . . , λd with respect to K. Then there is a (vector space) basis v1, . . . , vd ∈Λ such that ∥vi∥K = λi for 1 ≤i ≤d. extend proof to show it is a basis Proof. Pick some index 1 ≤j ≤d. By deﬁnition of λj there is a se-quence (wi)i≥1 ⊆Λ of lattice vectors such that limi→∞∥wi∥= λj. For suﬃciently large i we have wi ∈2K. K is compact, so we can ﬁnd a convergent sub-sequence wik, converging to some vector w. We need to prove that w ∈Λ. By deﬁnition, limk→∞∥w −wik∥K = 0, so for suﬃciently large k ∥w −wik∥K < λ1/2 . The triangle inequality then implies for suﬃciently large k, l ∥wik −wil∥K ≤∥w −wik∥K + ∥w −wil∥K < λ1 . But wil −wil is a lattice vector, so wik = wil for suﬃciently large k, l. Hence, wik = w for suﬃciently large k, and w is a lattice vector. ⊓ ⊔ Remark 4.9 The vectors found in the previous proposition need not be a basis of the lattice Λ. For an example, the lattice polytope P := conv (±e1, ±e2, ±(e1 + e2 + 2e3)) in the lattice Z3 is centrally symmetric and its lattice points are the vertices and the origin. Hence, the successive minima are λ1 = λ2 = λ3 = 1, but no subset of the vertices is a lattice basis of Z3. The following result is a cornerstone of the theory of successive minima. We will not prove this much stronger theorem here. A proof of the upper bound can be found in . Haase, Nill, Paffenholz: Lattice Polytopes — 105 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Theorem 4.10 (Minkowski’s Second Theorem, 1896) Let K ∈C0. Then 1 d! · 2d det Λ ≤λ1 · · · λd vol K ≤2d det Λ. Theorem 4.11 (Minkowski, 1910) Let K ⊂Rd be a centrally sym-metric convex set with int (K) ∩Λ = {0}. Then \|K ∩Λ\| ≤3d. Proof. We may choose Λ = Zd. Assume the statement fails. We consider the map ϕ : Zd →(Z/3Z)d given by assigning each coordinate its congruence class modulo 3. This is a homomorphism (so ϕ(x ± y) = ϕ(x) ± ϕ(y)). Note that (Z/3Z)d has 3d elements. Hence, by the pigeon hole principle there exist two distinct lattice points x, y ∈Zd with ϕ(x) = ϕ(y). Therefore, ϕ(x −y) = 0, thus p := x −y 3 ∈Λ . Since K is centrally symmetric, 0 ̸= p = x 3 + −y 3 ∈2 3K . This contradicts the assumption in the theorem. ⊓ ⊔ Recently, it was shown that up to unimodular transformations the stan-dard cube [−1, 1]d is the only centrally-symmetric lattice polytope with int (K) ∩Λ = {0} and \|K ∩Λ\| = 3d . Theorem 4.12 (Betke, Henk, Wills, 1993 ) Let K ∈C0. Then \|K ∩Zn\| ≤ 2 λ1 + 1 d Proof. 8proof missing 8 proof missing Conjecture 4.13 (Betke, Henk, Wills, 1993 ) Let K ∈C0. Then \|K ∩Zn\| ≤ d Y i=1 2 λi + 1 . 4.2 Coverings and Packings (a) Packing of the square lattice (b) Packing of the hexagonal lattice Fig. 4.4: The packing ra-dius for diﬀerent lattices For r > 0 and z ∈Rd let Br(z) := {x ∈Rd \| ∥x −z∥< r} be the open ball of radius r around z. In this section we consider the conﬁguration of all translates of such a ball to all lattice points. We want — 106 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) to determine for which radii these translates are pairwise disjoint or cover the whole space, and relations between these two. We start with the ﬁrst and introduce the packing radius of a lattice, which is the largers redius of a ball such that any two translates to a lattice point either coincide or are disjoint. Deﬁnition 4.14 (packing radius) Let Λ be a lattice in Rd. The pack-ing radius is ϱ(Λ) := sup r>0 (Br(x) ∩Br(y) = ∅for all x, y ∈Λ) , i.e. the largest r > 0 such that the open balls of radius r around any two distinct lattice points do not intersect. Exercise 4.4 See Figure 4.4 for some examples. You will prove the folloing lemma in Exercise 4.4. Proposition 4.15 Let Λ ⊆Rd be a lattice and v a shortest non-zero lattice vector in Λ. Then ϱ(Λ) = 1 2∥v∥. ⊓ ⊔ Exercise 4.5 Recall that the dual of a lattice Λ is deﬁned to be the set of all linear functionals that map lattice points to integers. This is itself a lattice Λ⋆ in Rd⋆. Proposition 4.16 Let Λ be a lattice in Rd with dual lattice Λ⋆. Then ϱ(Λ) · ϱ(Λ⋆) ≤ d/4 . Proof. By Proposition 4.15 the packing radius is half the length of a shortest non-zero lattice vector, and by Proposition 4.5 we can bound this length with ϱ(Λ) ≤1 2 √ d(det Λ) 1/d ϱ(Λ⋆) ≤1 2 √ d(det Λ⋆) 1/d both for Λ and its dual. The proposition now follows as det Λ · det Λ⋆= 1. ⊓ ⊔ Exercise 4.6 Now we switch the view and want to ﬁnd out how large we need to make the radius of our balls so that the translates cover the wohle space. This is captured with the next deﬁnition. Deﬁnition 4.17 (Covering Radius) Let Λ be a lattice in Rd. The covering radius is µ(Λ) := max x∈Rd d(x, Λ) , i.e. the largest possible distance between any point in Rd and its nearest lattice point. Haase, Nill, Paffenholz: Lattice Polytopes — 107 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) See Figure 4.5 for an example. The reader should convince her- or himself that the covering radius is indeed well-deﬁned and ﬁnite. In particular, the maximum is attained for some point x ∈Rd (by a standard compactness argument). Fig. of the Exercise 4.7 Lemma 4.18 Let Λ be a lattice with successive minima λ1, . . . , λd and linearly independent vectors v1, . . . , vd such that λi = ∥vi∥for 1 ≤i ≤d. Then µ(Λ) ≥1 2∥vi∥ for 1 ≤i ≤d . Proof. Let u = 1/2vd. Assume there is w ∈Λ such that d(u, w) < 1/2∥vd∥. Then ∥w∥≤∥u∥+ d(u, w) < ∥vd∥, so u cannot be linearly independent of v1, . . . , vd−1 by the choice of vd. Hence, w is in the span of v1, . . . , vd−1. But then 2w −vd is linearly independent, and ∥2w −vd∥= ∥2(w −u)∥< ∥vd∥ again contradicting the choice of vd. Hence, d(u, Λ) = d(u, 0) = 1 2∥vd∥. This implies that µ(Λ) ≥1 2∥vd∥≥1 2∥vi∥ for 1 ≤i ≤d, where the latter follows from ∥vd∥≥∥vi∥for all i. ⊓ ⊔ Proposition 4.19 Let Λ be a lattice in Rd with dual lattice Λ⋆. Then 4 µ(Λ) · ϱ(Λ⋆) ≥1 Proof. Let Λ be a lattice with successive minima λ1, . . . , λd and linearly independent vectors v1, . . . , vd ∈Λ such that λi = ∥vd∥for 1 ≤i ≤d. Let u be a shortest non-zero lattice vector in Λ⋆. Proposition 4.15 and Lemma 4.18 imply for any 1 ≤i ≤d 4 µ(Λ) · ϱ(Λ⋆) = 2 µ(Λ) · ∥u∥≥∥vi∥· ∥u∥. (4.2) The vectors v1, . . . , vd are a basis, so for at least one i we have \|vi(u)\| ≥1. Hence, for that i ∥vi∥· ∥u∥≥1 , which, together with (4.2) implies the claim. The following theorem is the key ingredient for the ﬂatness theorem that we will prove in Section 4.3. — 108 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Theorem 4.20 Let Λ be a lattice in Rd with dual lattice Λ⋆. Then 4 µ(Λ) · ϱ(Λ⋆) ≤d 3/2 . Our proof of this theorem is based on an argument by Schnorr, Lagarias, and Lenstra . Proof. We use induction over d. For d = 1 we have, for some λ > 0, Λ = λZd and Λ⋆= λ−1Zd . Thus, µ(Λ) = λ/2 and ϱ(Λ⋆) = λ−1/2, so that 4 µ(Λ) ϱ(Λ⋆) = 1. Now let d > 1. We choose a shortest non-zero lattice vector v ∈Λ. Then ∥v∥= 2 ϱ(Λ). Let L be the orthogonal complement of v with projection π : Rd →L and Γ := π(Λ). Then Γ is a lattice in L and Γ ⋆⊆Λ⋆by Exercise 2.33. Hence ϱ(Γ ⋆) ≥ϱ(Λ⋆) . (4.3) We now want to bound µ(Λ). For this, let x ∈Rd, y = π(x) and u a closest point to y in Γ. Then ∥u −y∥≤µ(Γ) . Consider the line π−1(u). Any two neighboring lattice points of Λ on this line have distance ∥v∥. Hence, we can pick a point w ∈Λ ∩π−1(u) such that d(x, w + (y −u)) ≤1 2∥v∥. Using the right angled triangle x, w, w + (y −u) we compute ∥x −w∥2 ≤∥x −(w + (y −u))∥2 + ∥y −u∥2 . Now x was chosen arbitrary, so we can assume it is a point with maximum distance to the lattice and we can estimate (note that w need not be a lattice point closest to x) µ(Λ)2 ≤∥x −w∥2 ≤µ(Γ)2 + 1 4∥v∥2 = µ(Γ)2 + ϱ(Λ)2 . Hence, we obtain µ(Λ)2 · ϱ(Λ⋆)2 ≤µ(Γ)2 · ϱ(Λ⋆)2 + ϱ(Λ)2 · ϱ(Λ⋆)2 ≤µ(Γ)2 · ϱ(Γ ⋆)2 + ϱ(Λ)2 · ϱ(Λ⋆)2 ≤(d −1)3 + 1 16d2 ≤d3 , where the second inequality follows from (4.3), the third from Proposi-tion 4.16 and the fourth by induction. This proves the theorem. ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 109 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 4.3 Flatness Theorem Playing around with two-dimensional convex sets the reader may get the impression that a convex body without interior lattice points cannot be arbitrarily wide. Indeed this is a fundamental fact in the geometry of numbers. The following considerations are based on an argument given in . Deﬁnition 4.21 (width) Let Λ ⊆Rd be a lattice with dual lattice Λ⋆. Let K ⊂Rd be a full-dimensional convex body. The width of K with respect to a non-zero lattice vector a ∈Λ⋆is deﬁned as width(K; a) := max x∈K a(x) −min x∈K a(x). We deﬁne the width of K with respect to Λ as widthΛ(K) := inf(width(K; a) : a ∈Λ⋆\ {0}). You will show in Exercise 4.8 that for full-dimensional convex bodies the inﬁmum is actually a minimum, and in Exercise 4.9 that the width of convex bodies with dimension less than the ambient dimension is actually 0. Recall that an ellipsoid is the image of a ball (in some norm) Exercise 4.8 Exercise 4.9 under an aﬃne linear map. See Deﬁnition A.2 for a full deﬁnition and the whole Appendix A for properties. Our approach to bound the lattice width of empty convex bodies will proceed in three steps. We ﬁrst prove it for balls, then extend to ellipsoids and ﬁnally use Theorem A.5 to approximate an arbitrary convex body with ellipsoids from the interior and the exterior. The following lemma does the ﬁrst two steps. Lemma 4.22 Let Λ be a lattice, v ∈Λ⋆a shortest non-zero lattice vector and E an ellipsoid such that E ∩Λ = ∅. Then widthv(E) ≤d3/2. Proof. We prove this ﬁrst for the case that E is a ball. In this case we know by Proposition 4.15 that ∥v∥= 2 ϱ(Λ⋆). Let r be the radius of the ball. Then r ≤µ(Λ). Now widthv(E) = r∥v∥= 2 ϱ(Λ⋆) µ(Λ) , and the latter is at most d3/2 by Theorem 4.20. For the extension to ellipsoids we use that the bound d3/2 obtained is independent of the lattice. Further, any ellipsoid is a linear image of a ball and the image of a lattice Λ for a non-singular linear map T is a lattice. More precisely, let x 7→Tx + t be the aﬃne map such that T(E) = B is a ball, and let Λ′ := T(Λ). Then Λ′ is a lattice in Rd and B ∩Λ′ = ∅. Hence, for a shortest non-zero vector v′ ∈Λ′, its preimage c := T −1v and a shortest non-zero vector vw ∈Λ we have widthw(E) ≤widthv′(E) = widthv(B) ≤d 3/2 . ⊓ ⊔ — 110 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) We can extend our bound for the width of a convex body from balls and ellipsoids to general convex bodies with empty interior, albeit only with a weaker right hand side. The key observation for this is Theorem A.5, which tells us the we can estimate any convex body from the interior and exterior with a suitably chosen ellipsoid. Theorem 4.23 Let K ⊂Rd be a convex body with K ∩Λ = ∅. Then widthΛ(K) ≤d 5 2 . Proof. Let E be a maximum volume ellipsoid in K with center z. Then deﬁne maximum volume ellipsoid also E ∩Λ = ∅. Let v be a shortest non-zero lattice vector in Λ such that width(E; v) ≤d3/2 by the previous Lemma 4.22. Clearly, the width of K is translation invariant, so we can assume that z is the origin. By Theorem A.5 we deduce K ⊆dE, and thus widthv(K) ≤d widthv(E) ≤d · d 3/2 = d 5/2 . ⊓ ⊔ Remark 4.24 In fact, the bound of the previous theorem can be strength-need reference ened to be of order d3/2, so that widthΛ(K) ≤O d 3 2 . Note that the upper bound only depends on the dimension and not on the given lattice. It is unknown and an active subject of current research, whether the sharp bound is actually of the form O (d). Rudelson-Paper noch? Examples for best possible bound? 2d + 1 ??? O(d log(d)) for simplices 4.4 Finiteness of lattice polytopes with few interior lattice points If a lattice polytope does not have interior lattice points, its volume Auch hier ueberall Λ statt Zd (ist klarer im Gegenzug zu Gitter der barycentr. co-ord.) can be arbitrarily large. However, if the polytope is centrally symmetric, Minkowski’s First Theorem (Corollary 4.3) shows that its volume is bounded, if it contains only one interior lattice point. The same statement is wrong without central symmetry. The examples in Figure 1.17 have one non-lattice vertex. The reader will not be able to construct such examples of arbitrary large volume and only one interior lattice point using lattice polytopes. The reason for this is one of the arguably most important ﬁniteness result about lattice polytopes. We prove here a qualitative version, following and extending an idea of Borisov & Borisov (see also [13, Theorem 4.1]). Fig. 4.6: An arbitrarily big rational triangle with one interior lattice point 0 1 2 3 n −1 n −2 n −3 n −4 n Fig. 4.7: An arbitrarily big triangle without interior lattice points Theorem 4.25 Given positive integers d, i, there is a bound V (d, i) so that every lattice d-polytope with exactly i interior lattice points has volume less than V (d, i). Haase, Nill, Paffenholz: Lattice Polytopes — 111 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) For dimension d = 2, Scott’s theorem Theorem 1.10 makes this result precise. In general, this is still subject of current research. From Corol-lary 2.83 we derive a powerful ﬁniteness theorem. Corollary 4.26 Given positive integers d, i, there are, up to lattice equivalence, only ﬁnitely many lattice d-polytopes with exactly i interior lattice points. 4.4.1 Finiteness of barycentric coordinates of lattice simplices For the proof of Theorem 4.25 we need to investigate the barycentric coordinates of interior lattice points in simplices, if the number of interior lattice points is bounded. The trick is to consider a quite general setup. Let T d := Rd/Zd. Any element in T d has a unique representative in the half-open standard square [0, 1)d, see Figure 4.8. We denote the quotient map Rd →T d by x 7→[x]. We consider the open simplex Fig. 4.8: The torus T d ◦ ∆d := {x ∈Rd : xi > 0 for all i = 1, . . . , d and d X i=1 xi < 1} ⊂Rd and its (bijective) image in T d ∆d := {[x] ∈T d : x ∈◦ ∆d} ⊂T d. For y ∈T d, let us deﬁne the generated subgroup ⟨y⟩= {ky : k ∈Z} ⊂ T d. For the considerations in the next paragraph the following is the crucial deﬁnition: M d i := {y ∈∆d : \|⟨y⟩∩∆d \| ≤i} ⊂T d Example 4.27 Let d = 2, y := [(1/2, 1/3)] ∈∆2. Then ⟨y⟩= {[(0, 0)], [(1/2, 1/3)], [(0, 2/3)], [(1/2, 0)], [(0, 1/3)], [(1/2, 2/3)]}. See Figure 4.9 for an illustation. Hence, ⟨y⟩∩∆d = {[(1/2, 1/3)]}, thus, y ∈M 2 1 . Fig. 4.9: An element y ∈M 2 1 and ⟨y⟩ The main result of this section will be the following proposition. Proposition 4.28 M d i is ﬁnite. Note that so far there is no lattice involved! This is all just about subgroups of the torus of ﬁnite order. In order to relate this to our lattice polytope problem, let us recall the notion of barycentric coordinates. If a d-dimensional simplex S has vertices v0, . . . , vd, then any point x in S can be uniquely written as x = β0v0 + · · · + βdvd with β0 + · · · + βd = 1 and β0, . . . , βd > 0. Here, x is in the interior of S if and only if β0, . . . , βd > 0. Now, the relation to our problem about lattice polytopes is given by the following observation. — 112 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Corollary 4.29 Let S ⊂Rd be a d-dimensional lattice simplex with i := \| int (S) ∩Λ\| ≥1. For vertices v0, . . . , vd of S, we deﬁne the bijective map β : int (S) →∆d, d X j=0 βivj 7→[(β1, . . . , βd)] . Then β(int (S) ∩Λ) ⊆M d i . So the set of barycentric coordinates an interior lattice point of such a simplex can have is ﬁnite. Proof. The map β is called barycentric coordinates. It is linear and thus yields a homomorphism Rd →T d. As k times a lattice point w ∈int(S) ∩Λ is a lattice point we have ⟨β(w)⟩∩∆d is contained in β(int (S) ∩Λ). which proof? more high-brow proof? We may assume that after a lattice translation v0 = {0}. Let x = Pd i=1 βivi ∈int (S) ∩Λ, and y := β(x) = [(β1, . . . , βd)] ∈∆d. If for k ∈Z≥0, ky ∈∆d, then there exists x′ = Pd i=1 β′ ivi ∈int (S) with β(x′) = [(β′ 1, . . . , β′ d)] = ky = [(kβ1, . . . , kβd)]. This implies that β′ i − kβi ∈Z for i = 1, . . . , d, thus x′ = d X i=1 (β′ i −kβi)vi ! + k d X i=1 βivi ! ∈Λ. Hence, the number of elements in ⟨y⟩∩∆d is at most the number of elements in int (S) ∩Λ which is i. This proves y ∈M d i . ⊓ ⊔ In other words, there are only ﬁnitely many barycentric coordinates possible for interior lattice points in a lattice simplex which contains a certain, non-zero number of interior lattice points overall. Let us give some preparations for the proof of Proposition 4.28. We need a natural translation-invariant distance function on T: d(y, y′) := min{∥x −x′∥: y = [x], y′ = [x′] for x, x′ ∈Rd} . Look at Figure 4.10 to get a better intuition for this deﬁnition. Note Fig. 4.10: Illustrating the deﬁnition of our metric that there are two kinds of elements of the group T d: the rational points have ﬁnite order, and the irrational points have inﬁnite order. Lemma 4.30 For x ∈{0}r × Rd−r (with 0 ≤r ≤d) and ε > 0 there is a positive integer k and z ∈{0}r × Rd−r with [z] = [kx] and \|\|z\|\| < ε. Proof. If x is rational, then there exists a positive integer k such that [kx] = 0, so deﬁne z = 0 ∈Rd. If x is irrational, then ⟨[x]⟩is inﬁnite. Haase, Nill, Paffenholz: Lattice Polytopes — 113 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) As T d is compact, there must be an accumulation point. Hence, there are natural numbers k′ > k′′ so that d((k′ −k′′)[x], ) = d(k′[x], k′′[x]) < ε . Choose k := k′ −k′′. As kx ∈{0}r × Rd−r, the deﬁnition of the metric d([kx], ) implies the existence of z ∈{0}r × Rd−r, as desired. (Note that we do not claim that z ∈[0, 1]d, see Figure 4.11.) ⊓ ⊔ Fig. 4.11: z doesn’t have to be in [0, 1]d Proof (Proposition 4.28). As ∆d is in bijection with ◦ ∆dd, every element in M d i ⊂∆d corresponds to an element in ˆ M d i ⊂◦ ∆dd. Deﬁne ˆ M d i Let us deﬁne the (compact) closure of ◦ ∆dd in Rd: ∆d := {x ∈Rd : xi ≥0 for all i = 1, . . . , d and d X i=1 xi ≤1} ⊂Rd We assume that M d i (thus, ˆ M d i ) is not ﬁnite. Then there exists an accumulation point x∗∈∆d of ˆ M d i . Changing the coordinate system on T d if necessary, we may assume that x∗ 1 = . . . = x∗ r = 0, and x∗ r+1, . . . , x∗ d, 1 −Pd j=1 x∗ j > 0. In particular, x∗∈{0}r × Rd−r >0 . We can assume this by permuting just the coordinates, since M d i can be characterized in a symmetric way, see Exercise 4.10. Exercise 4.10 Let us choose ε > 0 so that Bε(x∗) ∩(Rr >0 × Rd−r) ⊂◦ ∆dd, see Figure 4.12. Fig. 4.12: Bε(x∗) ∩ (Rr >0 × Rd−r) ⊂ ◦ ∆dd Now, Lemma 4.30 implies the existence of a positive integer k and and z∗∈{0}r × Rd−r so that [kx∗] = [z∗] and ∥z∗∥< ε/2i. With x∗ being an accumulation point of ˆ M d i , there is an x ∈ ˆ M d i such that \|\|x −x∗\|\| < ε/(2(ik + 1)) and x ̸= x∗−z∗/k. Note that we have x −x∗∈Rd >0 × Rd−r. We deﬁne for j = 0, 1, . . . , i wj := x∗+ jz∗+ (jk + 1)(x −x∗) ∈Rd. Let us show that [wj] ∈⟨[x]⟩∩∆d for j = 0 . . . , i, where all these (i + 1) elements are pairwise diﬀerent. This would show [x] ̸∈M d i , which is a contradiction. First, we observe that ▶wj ∈Rr >0 × Rd−r. ▶\|\|wj −x∗\|\| < j ε 2i + (jk + 1) ε 2(ik+1) ≤ε. — 114 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Hence, the deﬁnition of ε implies that wj ∈◦ ∆dd. Next, let us use [kx∗] = [z∗] to deduce [wj] = j[z∗] + (jk + 1)[x] −jk[x∗] = [(jk + 1)x] ∈⟨[x]⟩. Finally, let us note wj = x + j(z∗+ k(x −x∗)). As z∗+ k(x −x∗) ̸= 0 by the choice of x, we see that w0, . . . , wi are pairwise diﬀerent. This ﬁnishes the proof. ⊓ ⊔ 4.4.2 Coeﬃcient of asymmetry Now that we have braved the technical core in the proof of Theorem 4.25, the machinery will give us a volume bound V (d, i) in terms of a parameter ε(d, i) that we introduce below. It will depend only on the dimension and the number of interior lattice points. The smallest barycentric coordinate of a point inside a simplex is a measure for how far in the interior of the simplex the point sits. To measure the same thing for points in more general polytopes (or convex bodies), we use the convex-geometric notion of coeﬃcient of asymmetry. Deﬁnition 4.31 Let K ⊂Rd be a d-dimensional convex body, w ∈ int K, then ca(K; w) := sup η∈Rd{0} max{λ > 0 : w + λη ∈K} max{λ > 0 : w −λη ∈K} is the coeﬃcient of asymmetry. We have ca(K; w) ≥1. Note that ca(K; w) = 1 if and only if K is centrally symmetric with respect to w. So, the closer ca(K; w) is to 1 the more w lies in the ’center’ of K (the converse may not be true). See Figure 4.13 for two examples. We now extend Proposition 4.28 from w + λη λ λ′ w K (a) An example with ca(K; w) = 1 w K λ λ′ (b) An example with ca(K; w) ≫1 Fig. 4.13: Coeﬃcient of Asymmetry simplices to polytopes. Deﬁnition 4.32 (Minimal barycentric coordinates) For positive in-tegers d and i, the minimal barycentric coordinate of any interior lattice point in any lattice d-simplex with precisely i interior lattice points will be denoted by sbc(d, i). This deﬁnition yields a well-deﬁned positive number sbc(d, i) because of Corollary 4.29, a consequence of the main result of the previous section. You will prove some simple properties in Exercise 4.11. Here is a simple Exercise 4.11 fact that you will prove in Exercise 4.12. Lemma 4.33 For d ≥1 and i ≥1 we have sbc(d + 1, i) ≤sbc(d, i)/2. ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 115 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Exercise 4.12 Proposition 4.34 Let P ⊆Rd be a d-dimensional lattice polytope with i > 0 interior lattice points, and let w ∈int P ∩Zd. Then ca(P; w) ≤max 1 sbc(d, i′) −1 : 1 ≤i′ ≤i . is sbc(d, i) monotone in i? The proof uses two auxiliary results. You will prove the following lemma in Exercise 4.13. Lemma 4.35 Let P ∈Rd be a polytope, and let w ∈int P. Then the coeﬃcient of asymmetry is attained at a vertex. That is, there is a vertex v of P so that ca(P; w) = 1 max{λ > 0 : w −λ(v −w) ∈P} . (4.4) ⊓ ⊔ Fig. 4.14: The coeﬃcient of asym-metry is attained at a vertex Exercise 4.13 The following corollary justiﬁes the claim that the coeﬃcient of asymmetry of a point in a polytope is a qualitative generalization of the smallest barycentric coordinate of a point in a simplex. Corollary 4.36 Let S = conv(v0, . . . , vd) ⊂Rd be a d-simplex, and let 0 < β0 ≤. . . ≤βd with Pd j=0 βj = 1. Set w := Pd j=0 βjvj. Then ca(S; w) = 1 β0 −1. You will give a proof of this in Exercise 4.14. Exercise 4.14 Exercise 4.15 Proof (of Proposition 4.34). Let v be a vertex of P as in (4.4). Let c := ca(P; w), and denote the opposite point by v′ := w −1 c (v −w). There is a face F of P which contains v′ in its relative interior. In a Fig. 4.15: ca(P; w) ≤ ca(S; w) for some simplex S lattice triangulation of F there must be a lattice simplex S′ which contains v′ in its relative interior. Hence, the lattice simplex S := conv(v, S′) of dimension 1 ≤d′ ≤d contains w in its relative interior, and the segment conv(v, v′) certiﬁes ca(S; w) ≥c. Furthermore, every relative interior point of S is an interior point of P, so S contains j interior lattice points for 1 ≤j ≤i. The statement follows now from the previous corollary. ⊓ ⊔ Exercise 4.16 4.4.3 Bounding the volume We are ﬁnally in the position to ﬁnish the proof of Theorem 4.25. For this we need the following observation. You will prove this in Exercise 4.17. — 116 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Lemma 4.37 Let K be a d-dimensional convex body with 0 ∈int K. Set c := ca(K; 0). Then −1 cK ⊆K. Proposition 4.38 Let K ⊂Rd be a d-dimensional convex body. If w ∈int K ∩Zd, then vol(K) ≤2d−1 ca(K; w)d(\| int K ∩Zd\| + 1) . Proof. We set i := \| int K ∩Zd\| and c := ca(K; w). We may assume that w = 0. Consider Q := conv(−1 cK ∪1 cK). We have Q = −Q, and 1 cK ⊆Q ⊆K by the previous lemma. Hence, \| int Q ∩Zd\| ≤i so that by van der Corput’s Theorem (Theorem 4.2) vol Q ≤(i + 1)2d−1. This yields vol(K) ≤cd vol(Q) from which the statement follows. ⊓ ⊔ The bound is tight, as you will show in Exercise 4.18. This implies Exercise 4.18 Theorem 4.25 with V (d, i) = (i + 1)2d−1 1 sbc(d, i) −1 d . Gilt nur wenn Monotonicity bewiesen wurde (exercise?) Let us ﬁnish by presenting without proof the currently best and most general result. The following theorem is a combination of results obtained by Hensley , Lagarias and Ziegler and Pikhurko . Theorem 4.39 (Hensley; Lagarias & Ziegler; Pikhurko) Let P ⊆ Rd be a d-dimensional lattice polytope such that Il(P) ̸= ∅. Then vol(P) ≤(8dl)d(8l + 7)d·22d+1\| int P ∩lZd\| 4.5 Lower Bounds We can make the bound V (d, i) of the previous theorem more precise. Similar to the volume we can also look at the total number of lattice points of a lattice polytope when the number of interior lattice points is ﬁxed. For this, let L(d, i) be the maximal number of lattice points of a d-dimensional lattice polytope with exeactly i interior lattice points. For integers m1, . . . , md we consider the following simplex Sd (m1,...,md) := x ∈Rd \| xi ≥0 , X xi mi ≤1 . This is a lattice simplex with vertices 1 and miei for the unit basis vectors ei. Hence, it is a lattice simplex. See Figure 4.16 for an example. Fig. 4.16: Sd (2,3,12) We compute its volume and number of lattice points for a particular choice of the parameters mi. We deﬁne a sequence (aj)j ≥1 via Haase, Nill, Paffenholz: Lattice Polytopes — 117 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) a1 := 2 and aj := j−1 Y l=1 al + 1 for j ≥2 and for ﬁxed d ≥3 and i ≥1 we set kj := aj for 1 ≤j ≤d −1 and kd := (i + 1)(ad −1) . The sequence of the ai is the Sylvester sequence. It satisﬁes the following properties, which you will prove in Exercise 4.19. Lemma 4.40 (1) For d ≥1 we have 1 − d X i=j 1 aj =   d Y j=1 aj   −1 (2) for d ≥4 we have Qd i=1 ai ≥22d−a for a constant a ≈0.5856 . . .. We deﬁne the Sylvester simplices by Sd i := Sd (k1,...,kd) . It follows from Lemma 4.40(1) that for i = 1 Exercise 4.19 d X j=1 1 kj ≤1 1 kd + d X j=1 1 kj = 1 so that (1, 1, . . . , 1)t is in the interior of Sd 1, while (1, 1, . . . , 2)t is not. As the ki are strictly increasing we conclude that Sd 1 has exactly one interior lattice point. Similarly we can show that (1, 1, . . . , s)t for 1 ≤s ≤i are the only interior lattice points of Sd i . The volume of Sd i is vol(Sd i ) = 1 d! d Y j=1 kj = i + 1 d! (ad −1)2 . (4.5) We estimate the number of lattice points. For this, let R be the ridge of Sd i deﬁned by x1 = x2 = 0. Exercise 4.20 Exercise 4.21 Exercise 4.22 It follows from Exercise 4.20 that L(S + t) ≤L(S) with equality if t ∈Zd. By Exercise 4.21 know that vol(R) = Z Cd−2 \|(R + t) ∩Zd−2\| dt = Z Cd−2 L(R + t) dt ≤L(R) ≤L(Sd i ) . Thus L(Sd i ≥vol(R) ≥d(d −1) 2 · 3 vol(Sd i ) (4.6) — 118 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Theorem 4.41 (Perles, Zaks, Wills ) Let i ≥1. We have V (3, i) ≥6(i + 1) L(3, i) ≥16i + 23 V (4, 1) ≥147 L(4, 1) ≥680 (4.7) and for d ≥4 V (d, i) ≥i + 1 d! 22d−a L(d, i) ≥ i + 1 6(d −2)!22d−a (4.8) where a ≈0.5856 . . .. Proof. For the special cases in dimensions 3 and 4 in (4.7) see Exer-cise 4.23. We know that the volume of Sd i is vol(Sd i ) = i + 1 d! (ad −1)2 by (4.5). Using Lemma 4.40(2) we estimate ad −1 ≥22d−1−a from which the lower bound on the voume follows. The bound for the lattice points now follows from this and (4.6). ⊓ ⊔ Exercise 4.23 4.6 Empty lattice simplices Every lattice polytope has a triangulation which uses all the lattice points. Its simplices have the property that the only lattice points they contain are their vertices. Deﬁnition 4.42 (empty) A lattice polytope P is empty if the vertices of P are the only lattice points in P. Empty simplices in this sense can be regarded as the fundamental building blocks of lattice polytope theory. Because of its number theoretic nature, the study of these objects is in general very hard. In this section, we will present what we know (in low dimension) and what we do not yet know (in higher dimensions). Let us start with the simplest cases d = 1 and d = 2. There is (up to equivalence) only one empty segment, [0, 1]. By Pick’s Formula (Theorem 1.8), an empty triangle must be unimodular as well. The only other empty polygon is the unit square [0, 1]2 up to unimodular equivalence (Exercise 4.24). The situation is signiﬁcantly more subtle in dimension ≥3. The Reeve simplices Rd(m) that we have seen in (1.1) are examples of empty lattice simplices in any dimension ≥3 of arbitrary large normalized volume m. Exercise 4.24 Surprisingly, in dimension 3 an astonishing and completely not obvi-ous result still holds. Haase, Nill, Paffenholz: Lattice Polytopes — 119 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Theorem 4.43 (Howe (in Scarf 1985)) If P ⊂R3 is an empty lat-tice polytope in the lattice Z3, then widthZ3(P) = 1. Here widthZ3(P) is the lattice width deﬁned in Deﬁnition 4.21. We will only prove the theorem for the crucial case of an empty tetrahedron. Here the previous result takes the following form. Theorem 4.44 (White 1964) If T ⊂R3 is an empty lattice tetrahe-dron, then T is equivalent to Tpq := conv    1 0 0 1 0 1 0 p 0 0 0 q   . Note that the ﬁrst coordinate is always 0 or 1 on the vertices of Tpq, hence, the lattice width is one. By now, there are several proofs known (White 1964 , Morrison & Stevens 1984 , Howe (in Scarf 1985) , Sebö 1999 )9Before we prove these two theorems we need the following 9 The history of these proofs is interesting useful observation by Scarf which is true in arbitrary dimensions, not just in dimension three. For this, let us deﬁne the following function. For a ∈Zd with D := Pd j=1 aj −1 > 0 deﬁne f : {1, . . . , D −1} →Z, h 7→ d X j=1 ajh D . Lemma 4.45 (Scarf’s criterion) For a ∈Zd the d-dimensional sim-plex T = conv(e1, . . . , ed, a) is empty if and only if f(h) > h + 1 for h = 1, . . . , D −1. If d = 3, this is equivalent to f(h) = h + 2 for h = 1, . . . , D −1 and gcd(aj, D) = 1 for j = 1, . . . , d. Proof. First, observe that f(h) ≥Pd j=1 ajh D = D+1 D h > h, and thus f(h) ≥h + 1 for all h. Next, we have the following inequality description of T {e1, . . . , ed, a} (Exercise 4.25). ( x ∈Rd : 0 ≤Pd j=1 xj −1 < D and ai D −1 < −xi + ai D Pd j=1 xj ≤ ai D for i = 1, . . . , d ) Let us show that T is not empty if and only if there is an h ∈ Im Bild sollte noch Gerade mit Wert 0 und D eingezeichnet sein Fig. 4.17: Inequality de-scription of T \ {e1, e2, a} {1, . . . , D −1} with f(h) = h + 1. If there is an x ∈Zd ∩T \ {e1, . . . , ed, a}, set h := Pd j=1 xj −1. The above inequality description yields 0 ≤h < D. This implies h ∈ {1, . . . , D −1}, as for h = 0, x would be a non-vertex lattice point in conv e1, . . . , ed which is not possible. Moreover, the conditions for — 120 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) i = 1, . . . , d are equivalent to aih D ≤xi < aih D + 1 so that necessarily xi = ⌈aih D ⌉, and we obtain f(h) = Pd j=1 xj = h + 1. If, on the other hand, there is an h ∈{1, . . . , D −1} with f(h) = h + 1, set xi = ⌈aih D ⌉for i = 1, . . . , d. This deﬁnes x ∈Zd that satisﬁes all above inequalities, hence, T is not empty. Finally, if d = 3, we consider f(h) + f(D −h) = Pd j=1⌈aj h D⌉+ ⌈aj D−h D ⌉. Using ai h D + ai D −h D =    aj if D divides ajh aj + 1 if D does not divide ajh we obtain f(h) + f(D −h) ≤Pd j=1(aj + 1) = D + 1 + d = D + 4. If T is empty, our previous considerations show that f(h) + f(D −h) ≥ h + 2 + D −h + 2 = D + 4 so that f(h) = h + 2 for all h = 1, . . . , D −1, and D never divides ajh for j = 1, . . . , d. This also immediately implies gcd(aj, D) = 1, since otherwise D/ gcd(aj, D) ∈{1, . . . , D −1}, thus, D wouldn’t divide ajD/ gcd(aj, D) = lcm(aj, D), a contradiction. ⊓ ⊔ Exercise 4.25 Let us note that if D = 1, then the simplex T in Scarf’s criterion is automatically empty and of width one (the sum of all coordinates equals 1 for e1, . . . , ed, and 2 for a). Proof (of the Theorem of White (Theorem 4.44)). Our proof is based on an argument by Scarf, and proceeds in two steps. CHECKEN OB HOWE? Oder wer? (1) We show ﬁrst that if T is empty, then T is, for some a1, a2, a3 ≥0, equivalent to conv    1 0 0 a1 0 1 0 a2 0 0 0 a3    (2) Then we use Scarf’s criterion (Lemma 4.45) together with the lower bound a1, a2, a3 ≥2 to obtain a contradiction. Step 1 (Standard form): We can assume that T = conv    0 v11 v12 v13 0 0 v22 v23 0 0 0 v33    is in Hermite normal form, i.e. v11 > 0, v22 > v12 ≥0, v33 > v13, v23 ≥0 (cf. Deﬁnition 2.64, note that here we look at the transposed form). The triangle spanned by the ﬁrst three vertices is empty and hence unimodular Haase, Nill, Paffenholz: Lattice Polytopes — 121 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) by Pick’s Formula (Theorem 1.8). This implies v11 = v22 = 1, v12 = 0. Now, T ∼ = T +    0 0 1   = conv    0 1 0 v13 0 0 1 v23 1 1 1 v33 + 1   ∼ = conv    0 1 0 a1 0 0 1 a2 1 0 0 a3    with a1 := v13 ≥0, a2 := v23 ≥0, and a3 := v33 −v13 −v23 + 1. (For the last equivalence we have subtracted the ﬁrst two coordinates from the third.) We abbreviate D := a1 + a2 + a3 −1 = v33 > 0. If a3 were negative, then let us consider the following aﬃne combination    1 1 0   = D −a1 D    1 0 0   + D −a2 D    0 1 0   + −a3 D    0 0 1   + 1 D    a1 a2 a3   . Here, D −a1 = v33 −v13 > 0, D −a2 = v33 −v23 > 0, so this would be a proper convex combination. Therefore, (1, 1, 0)t ∈int T, a contradiction to T being empty. h 2h 3h 7h 1 2 3 7 2 4 6 3 3 6 9 10 4 8 1 6 5 10 4 2 6 1 7 9 7 3 10 5 8 5 2 1 9 7 5 8 10 9 8 4 Table 4.1: Table for a = (2, 3, 7)t mod D = 11 Step 2: Recall the function f used in Scarf’s criterion: For a ∈Zd with D := Pd j=1 aj −1 > 0 we have deﬁned f : {1, . . . , D −1} →Z via h 7→Pd j=1 l ajh D m . Let d = 3 and D = a1 + a2 + a3 −1 ≥1 . Assume a1, a2, a3 ≥2 and f(h) = h + 2 for all h. Make a D −1 by 3 table of the numbers hai reduced mod D for h = 1, . . . , D −1 as exempliﬁed in Table 4.1. Later we will sometimes use the notation [k]D for the remainder in {0, . . . , D −1} of an integer k modulo D. Note that the Dth-row consists simply of zeroes. We will refer to a row for h ∈{2, . . . , D −1} (mod D) as a proper row. The ai are coprime to D, so the three columns will be permutations of the set {1, . . . , D −1}, (compare Exercise 4.26). Also, we can add the ith and the jth row of the table and reduce it mod D to obtain the (i + j)th (mod D) row. Roughly speaking, in each ith column we add ai on the entry above until this number exceeds D in which case we reduce this entry, see Table 4.1. Exercise 4.26 In each column we underline the entries where the reduction takes place (a ‘jump’). These can be characterized by any of the following equivalent conditions for an entry in a proper row: (Exercise 4.27): ▶the entry is smaller than the entry directly above ▶the entry (in the ith column) is (necessarily strictly) smaller than ai ▶the entry (in the ith column and hth row) satisﬁes aih D = ai(h −1) D + 1 . — 122 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) The last condition allows us to make the following crucial observation: As for h = 2, . . . , D −1 f(h) = 3 X i=1 aih D = h + 2 , we have f(h) = f(h −1) + 1. Hence, in each proper row precisely one entry gives a contribution ⌈aih D ⌉to this sum which is one larger than in the row above. In other words, each proper row has precisely one underlined entry. We see that a = (2, 3, 7)t yields a simplex which is not empty as there is no underlined entry in row 3 of Table 4.1. We will use these restrictions in order to get a contradiction. For this, we consider the special entries D −1. Let us recall that 1 < a1, a2, a3 < D −1. (1) Assume there is a (necessarily proper) row h with two entries D − 1. In this case, the row D −h has two entries 1. Therefore, we have a (necessarily proper) row with two (underlined) entries 1, a contradiction. (2) Assume there is a (necessarily proper) row with entries D −1 and 1 (say, in the ﬁrst and second column). Multiplying by a2 yields another row with entries D −a2 and a2. Hence, we’re in the very ﬁrst row with entries a1, a2, a3, so D −a2 = a1, a contradiction (recall a3 > 1). (3) Next, assume there is a two by two conﬁguration D−1 α β D−1 By symmetry, let these be in the ﬁrst and second column. Summing these two rows yields again a row with entries β −1 < β < a1 and α −1 < α < a2. Hence, as this cannot be the 1-st row, and there is no proper row with two underlined entries, we see that this must be the D-th row, so β −1 = 0, thus β = 1. Therefore, the ﬁrst of above two rows has an entry D −1 and 1. This gives a contradiction to the previous point. (4) This leads to the last case, where we are left with dealing with a sub-table of proper rows of the form h1 D −1 α β h2 γ D −1 δ h3 ε ζ D −1 (4.9) We note that α, β, γ, δ, ε, ζ > 1. Consider h1 + h2 \| γ −1 α −1 [β + δ]D , Haase, Nill, Paffenholz: Lattice Polytopes — 123 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) As this is not the 1-st row (α −1 < a2) or D-th row (α −1 ̸= 0), [β + δ]D cannot be underlined, so [β + δ]D > a3. In particular, [β + δ −1]D ≥a3. In fact, as a3 < β and δ < a3 we see [β + δ −1]D > a3. By symmetry, this implies that h1 + h2 + h3 \| [ε + γ −1]D [α + ζ −1]D [β + δ −1]D , is neither the ﬁrst row, nor the D-th row, and it also does not have an underlined entry which is a contradiction. ⊓ ⊔ Exercise 4.28 Exercise 4.29 The reader may wonder whether the previous result also holds in higher dimension. Unfortunately, it fails already in dimension 4, as we will show now. For this, let us deﬁne a lattice simplex S(a, b, c, d) ⊂R4 as the convex hull of the columns of the following matrix:      1 0 0 0 a 0 1 0 0 b 0 0 1 0 c 0 0 0 1 d      It was shown in that for integers D ≥9 and coprime to 6 the 4-simplex S(2, 2, 3, D −6) is empty, of normalized volume D and has width 2. Hence, there exist inﬁnitely many non-isomorphic empty lattice 4-simplices with width > 1. On the other hand, it is known that there are only ﬁnitely many empty lattice 4-simplices with width > 2, as shown by Barile, Bernardi, Borisov, and Kantor . Up to now, the maximal possible width known is 4, realized by S(6, 14, 17, 65). CHRISTIAN: Hier sollte man noch etwas zu den neueren Entwicklungen sagen! 4.7 Lattice polytopes without interior lattice points Focusing again on lattice polytopes, let us make the following deﬁnition. Deﬁnition 4.46 A lattice polytope P is called hollow if it does not contain any interior lattice points. Are points hollow? Sollten nicht. Alle Beweise checken, dass alles korrekt ist. From the viewpoint of geometry of numbers, hollow lattice polytopes are a somewhat more natural class to consider than empty simplices. In this section, we will prove a structural result on hollow lattice polytopes that will imply the ﬂatness theorem for this class of convex sets. Let us ﬁrst look at low dimensions. In dimension one, there is only one hollow lattice segment, [0, 1]. In dimension two, there are two kinds of hollow lattice polygons. They are depicted in Figure 4.18, see also Exer-cise 3.33. Observe that one way to construct hollow lattice d-polytopes Fig. 4.18: “All” hol-low lattice polygons — 124 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) is to take a hollow polytope Q ⊂Rd−1 and choose any subpolytope of the inﬁnite prism Q × R. Observe as well that having width one is equivalent to allowing an integral projection onto the hollow polytope [0, 1]. Maybe we should have deﬁned integral/lattice projection much earlier? More generally, the reader should convince oneself that the following observation holds: If a lattice polytope has a lattice projection onto a hollow lattice polytope (of smaller dimension, not a point), then it was hollow to begin with. The following theorem states that this is the only way to get hollow polytopes – up to ﬁnitely many exceptions. Theorem 4.47 There are only ﬁnitely many (equivalence classes of) hollow lattice d-polytopes which do not allow an integral projection onto a hollow lattice (d −1)-polytope. Notation muss besser werden; hollow projection wird benutzt, muss deﬁniert werden; checken ob smaller dimension und kein Punkt geschrieben werden muss, oder im-pliziert drin ist The proof is based on an argument by Averkov et al. It proceeds with the following four steps, which we will explain in detail below: (1) If P is hollow and contains a “long” segment, then P projects onto a hollow polytope. (2) If P has “many” lattice points, then P contains a long segment. (3) If P is hollow and does not project, then there are lattice polytopes P ⊆P ′ ⊂P ′′ so that P ′ is hollow, P ′′ is not hollow, but int P ′′ ∩Zd ⊂ P ′. (4) nvolZd P ≤nvolZd P ′′ ≤V (d, \|P ′ ∩Zd\|) which is bounded, since by (2) and (3) the number of lattice points of P ′ as a non-projectable polytope is bounded. Step 1: For the ﬁrst step we need the following lemma. Lemma 4.48 Let Q ⊂Rd be a lattice polytope with interior lattice points, and let ¯ v ∈∂Q ∩Zd. Then there is a lattice simplex S ⊆Q with ¯ v ∈V(S) and \| relint S ∩Zd\| = 1. Fig. 4.19: The simplex in Lemma 4.48 may be low-dimensional The cross polytope shows that we cannot assume that S is full-dimensional (see Figure 4.19). Proof. Consider F := {R ⊆Q lattice polytope : ¯ v ∈R, R not hollow} . Then F is a ﬁnite non-empty family of lattice polytopes. Hence, there is an inclusion-minimal element R ∈F. We claim that R satisﬁes the assertion of the lemma. As R ∈F, there is a relative interior lattice point w. First, let us prove that R must be a simplex. The line through ¯ v and w intersects P in a segment conv(¯ v, v). Let F be the face of Q which Fig. 4.20: R is a pyramid with apex ¯ v Haase, Nill, Paffenholz: Lattice Polytopes — 125 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) contains v in its relative interior. Because w ∈relint(conv(¯ v, F)) and R is minimal, we must have R = conv(¯ v, F). Also, if F was not a simplex, the carrier of v in a triangulation of F would be a proper subpolytope F ′ ⊂F containing v in its relative interior – again, contradicting minimality of R. Next, let us show that relint(R) ∩Zd = {w}. We can assume that among all lattice points in relint(R) w has minimal barycen-tric ¯ v-coordinate. Denote the vertices of F by v1, . . . , vr, and set Fi := V(F) \ {i} for i = 1, . . . , r. The simplices S0 := conv(w, F), Si := conv(¯ v, w, Fi) (i = 1, . . . , r) together with all their faces triangu-late R. Fig. 4.21: Triangulation of R and the region of all points having larger barycentric ¯ v-coordinate than w A face of this triangulation meets relint(R) if and only if it contains w. If such a face also contains ¯ v, it cannot contain relative interior lattice points because R was minimal. If it does not contain ¯ v, it is a face of S0, and all points in S0 have smaller barycentric ¯ v-coordinate than w. ⊓ ⊔ With this tool at hand, we can establish the ﬁrst step. ε im Beweis nochmal deﬁnieren! Lemma 4.49 Let P ⊂Rd be hollow, u ∈Zd primitive and v ∈P ∩Zd so that v + Nu ∈P for some N > sbc(d −1, 1)−1. Consider the projection π : Rd →Rd/Ru. Then Q := π(P) is hollow with respect to π(Zd) = Zd/Zu. Proof. If Q had interior lattice points, according to Lemma 4.48, there would be a simplex S ⊆Q so that ¯ v := π(v) ∈V(S), and relint(S) ∩ Zd/Zu = {w}. Fig. 4.22: A long edge and a one-point-simplex If α is the barycentric ¯ v-coordinate of w in S, then the length of the segment π−1(w) ∩P is at least αN ≥sbc(dim S, 1)N ≥sbc(d−1, 1)N > 1. Therefore, this segment must contain an interior lattice point of P in contradiction to P being hollow. ⊓ ⊔ Step 2: Lemma 4.50 Let P ⊂Rd be a lattice polytope. If \|P ∩Zd\| > Nd, then P contains a segment of length N. Proof. compare “Rabinowitzs Lemma” [arXiv:1103.0103 Zhong §2] There must be two diﬀerent points v, v′ ∈P ∩Zd whose coordinates agree mod N. Then u := 1 N (v −v′) ∈Zd and v, v′ = v + Nu ∈P. ⊓ ⊔ Step 3: Lemma 4.51 Suppose the hollow polytope P does not allow a hollow projection. Then there are lattice polytopes P ⊆P ′ ⊂P ′′ so that P ′ is hollow, P ′′ is not hollow, but int P ′′ ∩Zd ⊂P ′. Proof. If there is no v1 ∈Zd \ P so that P1 := conv(P, v1) is hol-low, set P ′ := P. Otherwise, construct recursively hollow polytopes — 126 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) Pi+1 := conv(Pi, vi+1) until no such extension is possible. The process has to terminate because superpolytopes of non-projectable polytopes are non-projectable and the total number of lattice points is bounded by Lemmas 4.49 and 4.50. Let P ′ be that last, non-extendable hollow polytope in the sequence. Now choose v1 ∈Zd \ P ′ and set P 1 := conv(v1, P ′). Because P ′ is non-extendable, P 1 is not hollow. If there is no v2 ∈int P 1 ∩ Zd \ P ′, choose P ′′ := P 1. Otherwise, construct recursively P i+1 := conv(vi+1, P ′) with vi+1 ∈int P i ∩Zd \ P ′. The process has to terminate because there are only ﬁnitely many lattice points in P 1 \ P ′. The last polytope P ′′ in this series will be a strict superpolytope of P ′ and as such not hollow. But, being the last in the series, we have int P ′′ ∩Zd ⊂P ′. ⊓ ⊔ Check: Ist es nicht (sbc(d −1, 1) + 1)−d ? Step 4: Let P ⊆P ′ ⊂P ′′ be as in Lemma 4.51. Then P ′′ is not hollow and \| int P ′′ ∩Zd\| ≤\|P ′ ∩Zd\| ≤sbc(d −1, 1)−d. This holds again, since otherwise by Lemma 4.50 and Lemma 4.49, P ′ and thus P were projectable. Now the Theorem 4.25 from the previous section kicks in: nvolZd P ≤ nvolZd P ′′ ≤V (d, sbc(d −1, 1)−d) is bounded. Then the number of pos-sible equivalence classes for P is ﬁnite by Corollary 2.83. This ﬁnially proves Theorem 4.47. Erwaehnen: 3d - Jaron/Weismantel -result We say a hollow lattice polytope P is inclusion-maximal if it is not strictly contained in a hollow lattice polytope. Corollary 4.52 There exist only ﬁnitely many inclusion-maximal hollow lattice d-polytopes (up to unimodular equivalence). Moreover, any hollow lattice d-polytope that does not admit a hollow projection is contained in an inclusion-maximal hollow lattice d-polytope. Proof. If a lattice polytope admits a lattice projection onto a hollow lattice polytope of smaller dimension (not a point), then it is clearly contained in a larger lattice polytope that also projects onto the same lattice polytope, hence, is also hollow. Therefore, inclusion-maximal bild hollow lattice polytopes do not admit hollow projections, hence, they are only ﬁnitely many of these by Theorem 4.47. Iif a hollow lattice polytope P is not contained in any inclusion-maximal hollow lattice polytope, then there exists an inﬁnite chain of hollow lattice polytopes containing P. In particular, Theorem 4.47 there is a hollow lattice polytope P ′ ⊃P that admits a hollow projection, hence, P does. ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 127 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) We can now easily deduce the ﬂatness theorem for hollow lattice polytopes. For this we simply note that under lattice projection the width cannot decrease (Exercise 4.31). Exercise 4.31 Corollary 4.53 The maximal width of a hollow lattice d-polytope equals the maximal width of the ﬁnitely many (up to unimodular equivalence) inclusion-maximal lattice d′-polytopes with d′ = 1, . . . , d. Moreover, there are only ﬁnitely many hollow lattice d-polytopes whose width is strictly larger than the maximal width of hollow lattice (d −1)-polytopes. noch unterschied zu inclusion-maximal convex bodies erwaehnen; noch kruempelmann etc. 3d-resultat erwaehnen 4.8 Problems Exercise 4.32 included on page 102 4.1. Let K ⊆Rd be a centrally symmetric convex body with int (K) ∩ Zd = {0} and nvolZd(K) = 2d. Show that K is a polytope and each facet of K contains at least one lattice point in its relative interior. Hint: For any lattice point x choose a half space Hx containing both x and K. Let Sx := Hx ∩−Hx. Now consider the intersection of all Sx and prove that this satisﬁes the assumptions of Minkowski’s First Theorem (Corollary 4.3). included on page 102 4.2. Let K ⊆Rd be a centrally symmetric polytope with int (K) ∩Zd = {0}. Show that K has at most 2(2d −1) containing a lattice point in their relative interior. Hint: Choose lattice points in the interior of facets. Consider their coordinates module 3 (i.e., their image under Zd → (Z/3Z)d). Look at the diﬀerence of two points having the same image and use the pidgen-hole principle. included on page 104 4.3. Prove that the unit ball in dimension d has volume Vd := π⌊d/2⌋2⌈d/2⌉ Q 0≤2i≤d(d −2i) . included on page 107 4.4. Show that the packing radius is ﬁnite and equals half of the length of a shortest non-zero lattice vector. included on page 107 4.5. Let Λ be a lattice in Rd. Show that there is a non-zero x ∈Λ such that ∥x∥∞≤(det Λ) 1/d], . included on page 107 4.6. Let Λ0 ⊆Λ ⊆Rd be lattices. Show that ϱ(Λ) ≤ϱ(Λ0) ≤ Λ/Λ0 ϱ(Λ) . — 128 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) 4.7. Prove (1) µ(Zd) = √ d/2 (2) µ(D3) = 1 (3) µ(Dn) = √n/2 for n ≥4 (4) µ(E8) = 1 included on page 110 4.8. Show that in the deﬁnition of lattice width we can replace the inﬁmum with a minimum for a full-dimensional convex body K. Thus, the width is strictly positive. included on page 110 4.9. Show that the lattice width of a low dimensional convex body is 0. included on page 114 4.10. Let x ∈◦ ∆dd as in § 4.4.1. Deﬁne x0 := 1 −Pd i=1 xi. Prove that [x] ∈M d i if and only if ( k ∈Z≥1 : d X i=0 {kxi} = 1 and {kxi} > 0 ∀i = 0, . . . , d ) has at most i elements. Here {·} denotes the fractional part in [0, 1). included on page 115 4.11. (1) Compute sbc(2, i) (2) sbc(d + 1, i) ≤1/2 sbc(d, i), (3) sbc(d, i) ≤1/i2−2d−2 using Kahn-Wills-Zaks simplices For a solution see page 225 included on page 116 4.12. Prove that sbc(d + 1, i) ≤sbc(d, i)/2 for d, i ≥1. Hint: Let S be a d-simplex with vertices v0, . . . , vd and interior lattice point x = P λivi realizing sbc(d, i) such that λd = sbc(d, i). Consider the (d + 1)-simplex with vertices (vi, 0) for 0 ≤i ≤d −1 and (vd, ±1). included on page 116 4.13. Prove Lemma 4.35. For a solution see page 225 included on page 116 4.14. Prove Corollary 4.36. included on page 116 4.15. Show that the coeﬃcient of asymmetry of an interior point of a d-simplex equals at least d, with equality only for the centroid. included on page 116 Haase, Nill, Paffenholz: Lattice Polytopes — 129 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 4.16. Let K, K′ d-dimensional convex bodies satisfying K′ ⊆K ⊆µK′ + v for some µ > 0, v ∈Rd. Show that we have for all w ∈int (K′) ca(K; w) ≤µ ca(K′; w) + µ −1 . included on page 117 4.17. Prove Lemma 4.37. included on page 117 4.18. Prove the the bound in Proposition 4.38 is tight. included on page 118 4.19. Let (ai)i ≥1 be the Sylvester sequence given by a1 := 2 and ai := i−1 Y j=1 aj + 1 for i ≥2 . Prove that (1) for d ≥1 we have 1 − d X i=j 1 aj =   d Y j=1 aj   −1 (2) for d ≥4 we have Qd i=1 ai ≥ 22d−a for a constant a ≈ 0.5856 . . .. included on page 118 4.20. Let Sd i be the Sylvester simplex, l its number of lattice points and x ∈R. Show that, if the translate Sd i has l lattices points then x ∈Zd. Hint: Show that if one of the facets parallel to a coordinate hyper-plane does not contain a lattice point then you can translate the simplex so that it does, while the number of lattice points does not decrease. included on page 118 4.21. Let P be a lattice polytope and Cd the unit cube. Show that vol(P) = Z Cd \|(P + t) ∩Zn\| dt . included on page 118 4.22. recursively we deﬁne t1 := 2, tn := 1 + n−1 Y j=1 tj for n ≥2 . Show that (1) tn = t2 n−1 −tn−1 + 1 for n ≥2. — 130 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 4. Geometry of Numbers (draft of October 7, 2020) (2) Pn−1 j=1 1 tj = 1 − 1 tn−1 for n ≥2. (3) 22n−1 ≥tn ≥22n−2. 4.22. let d, i ∈Z≥1. Deﬁne the d-dimensional lattice simplex Sd,i mit vertices v0, . . . , vd: v0 := 0, vj := tjej for j = 1, . . . , d −1, vd := (i + 1)(td −1)ed . Show that int (Sd,i) ∩Zd = {(1, 1, . . . , 1, j) : j = 1, . . . , i} . and compute the volue (which is conjectured to be the largest possible for dimensions d ≥4). Hint: Consider barycentric coordinates (α0, . . . , αd) and show by induction αj = 1 tj for j = 1, . . . , d −1. included on page 119 4.23. Prove the low dimensional cases in Theorem 4.41. included on page 119 4.24. Show that ∆2 and [0, 1]2 are up to unimodular equivalence the only empty lattice polygons. included on page 121 4.25. Verify the inequality description in the proof of Scarf’s criterion (Lemma 4.45). included on page 122 4.26. Show that multiplying with a element coprime to D in Z induces a group automorphism of Z/DZ. included on page 123 4.27. Check the equivalent descriptions of a jump in the second part of the proof of Theorem of White (Theorem 4.44). included on page 124 4.28. Show, using the table in the proof of the Theorem of White (Theo-rem 4.44) that T is not empty for (a1, a2, a3) = (3, 4, 5). included on page 124 4.29. Use the proof of the Theorem of White (Theorem 4.44) to show the following: Let P be an empty lattice tetrahedron with one vertex in the origin. Then the subgroup (or sublattice) of Z3 that is spanned by the vertices of P is a cyclic quotiont group. included on page 125 4.30. Find a d-dimensional lattice simplex in Rd without interior points and lattice width larger than d. included on page 128 4.31. Prove that projection cannot decrease the width of a convex body. included on page 128 4.32. Show that Pd i=0 k2 = (d+2 3 ) + (d+1 3 ) Hint: You should try to do this via Ehrhart Theory. Consider a (d −1)-fold pyramid over a square. Haase, Nill, Paffenholz: Lattice Polytopes — 131 — Minkowski meets Ehrhart 5 Contents 5.1 Lattice Polytopes of given h∗-Polynomial . . 133 5.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 5.1.2 The pyramid theorem for lattice simplices . 135 5.1.3 Proof of the pyramid theorem. . . . . . . . . . . . 137 5.2 Lattice polytopes of small degree . . . . . . . . . 139 5.2.1 Cayley-Polytopes. . . . . . . . . . . . . . . . . . . . . . . 139 5.2.2 Small Degree . . . . . . . . . . . . . . . . . . . . . . . . . . 142 5.2.3 Normal Form . . . . . . . . . . . . . . . . . . . . . . . . . . 142 5.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 5.1 Lattice Polytopes of given h∗-Polynomial 5.1.1 Introduction With regard to the fundamental results in Ehrhart theory we have seen in Chapter 3 that the Ehrhart polynomial can be more compactly encoded in terms of the h⋆-polynomial. In this section, we will give a qualitative answer to the following natural question: how many lattice polytopes have the same h⋆-polynomial? In other words, how much information is contained in this polynomial, and to which extent does it determine the polytope? We start with some general observations. Let P be a Fig. 5.1: Two lattice polytopes with the same h⋆-polynomial 1 + 3t d-dimensional lattice polytope. Lecture Notes Lattice Polytopes (draft of October 7, 2020) (1) The sum over the coeﬃcients of its h⋆-polynomial equals the (nor-malized) volume of P. Since, by Theorem 3.32, all the coeﬃcients are non-negative integers, the question about the number of lattice polytopes with the same h⋆-polynomial is essentially equivalent to the problem of ﬁnding all lattice polytopes with given normalized volume and degree (of the h⋆-polynomial). (2) The simplest h⋆-polynomial consists only of the constant coeﬃcient 1. In this case, P has normalized volume 1, so it is a unimodular simplex (of arbitrary dimension). Fig. 5.2: Unimodular simplices in dimensions 0, 1, 2 and 3 (3) More generally, lattice pyramids with base P have the same h⋆-polynomial as P, see Exercise 3.30. Therefore, it makes sense to consider lattice polytopes ‘modulo’ lattice pyramid constructions. For instance, one may think of unimodular simplices as (successive) lattice pyramids over a point, see Figure 5.2. In this respect, the degree deg(P) seems to be a more natural invariant to consider than the dimension of a lattice polytope. (4) It is also insightful to recall that not only the volume but any coeﬃcient of the h⋆-polynomial behaves monotonically with respect reference to inclusions: Q ⊆P = ⇒h⋆ i (Q) ≤h⋆ i (P) for any i. In particular, deg(Q) ≤deg(P). Here is the main result of this section. Let us deﬁne the function f(c, s) := c(2s + 1) + 4s −2 . Theorem 5.1 (Pyramid theorem) Let P be a d-dimensional lattice polytope of degree s := deg(P). If d > f(\|V(P)\| −d −1, s) then P is a (possibly successive) lattice pyramid over a lattice polytope of dimension ≤f(\|V(P)\| −d −1, s). The proof will be given in the next two subsections. From this result we get the desired ﬁniteness result. We note an important consequence of this theorem. Corollary 5.2 There are (up to lattice pyramid constructions) only ﬁnitely many lattice polytopes of given h⋆-polynomial. Proof. Note that h⋆ 1(P) = \|P ∩Zd\| −(d + 1) ≥\|V(P)\| −d −1 for any d-dimensional lattice polytope. Let s be the degree of the given polynomial and c its linear coeﬃcient. Therefore, by the pyramid theorem it suﬃces to prove the theorem for lattice polytopes up to dimension f(c, s). However, in ﬁxed dimension there is only a ﬁnite number of lattice polytopes of given normalized volume by Corollary 2.83. ⊓ ⊔ — 134 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 5. Minkowski meets Ehrhart (draft of October 7, 2020) 5.1.2 The pyramid theorem for lattice simplices We will ﬁrst consider the situation where P is a d-dimensional lattice simplex of degree s := deg(P) embedded in Rd+1 as P × 1. In this case \|V(P)\| −d −1 = 0, so it remains to show that P is a lattice pyramid, if d ≥4s −1. Let v1, . . . , vd be the vertices of P. Let m be a point in the half-open fundamental parallelepiped Π(P) given as m = Pd i=0 λivi. Then we deﬁne the support of m as supp m := {i : λi ̸= 0} and the height of m as ht(m) := d X i=0 λi. Note that ht(m) is a non-negative integer. We will need a criterion how to check whether a lattice simplex is a lattice pyramid. For this, let us deﬁne the support of the simplex as supp P := [ m∈Π∩Zd+1 supp m . The proof of the following observation is left as Exercise 5.1. Exercise 5.1 Lemma 5.3 P is a lattice pyramid with apex vi for some i ∈{0, . . . , d + 1} if and only if i ̸∈supp P. ⊓ ⊔ In other words, if the support of P is a proper subset of {0, . . . , d + 1}, then P is a (successive) lattice pyramid over the convex hull conv(vi : i ∈supp P) of those vertices that are contained in the support. Hence, to prove the theorem for simplices we have to show that for any lattice simplex of degree s its support is bounded by 4s −1, i.e. \| supp P\| ≤4s −1 . Let us ﬁrst bound the support of one lattice point in Π(P). Lemma 5.4 Let m ∈Π(P) ∩Zd+1. Then supp m ≤2s . {1} × P m m⋆ Fig. 5.3: The central symmetry m ↔m⋆about the ‘green’ center of Π(P ) Proof. We deﬁne m⋆:= X i∈supp m (1 −λi)vi =   X i∈supp m vi  −m . Haase, Nill, Paffenholz: Lattice Polytopes — 135 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Then m⋆∈Π(P) ∩Zd+1. We observe \| supp m\| = ht   X i∈supp m vi  = ht (m + m⋆) = ht (m) + ht (m) . By Proposition 3.19, h⋆ i (P) = \|{ht (x) : x ∈Π ∩Zd+1}\|, in particular, ht (m) , ht (m⋆) ≤s. ⊓ ⊔ Here is another proof, which uses the notion of the codegree (see Deﬁni-tion 3.47 and Corollary 3.48). Proof (Alternative proof). Let P ′ := conv(vi : i ∈supp m). By the monotonicity theorem, deg(P ′) ≤s. Now, m ∈int ( ht(m)P ′) ∩Zd+1, hence, codeg(P ′) ≤ht(m) ≤s, again by Proposition 3.19. Therefore, \| supp P ′\| = dim(P ′) + 1 = codeg(P ′) + deg(P ′) ≤2s. ⊓ ⊔ Let us choose m0 ∈Π(P) ∩Zd+1 whose support I0 := supp m0 has the maximal possible number of elements (at most 2s). Now, we consecutively deﬁne a ‘greedy’ sequence {m}k=0,...,r of lattice points in Π(P) such that the size of each set Ik := supp mk \ [ j<k supp mj is maximal. Note that the support of P is completely covered by these disjoint sets Ik’s. Lemma 5.5 For k = 1, . . . , r, we have \|Ik\| ≤\|Ik−1\|/2 . In particular, \|Ik\| ≤\|I0\|/2k . Proof. The following arguments are illustrated in Figure 5.4. Let us deﬁne a := \|Ik−1\ supp mk\| , b := \|Ik−1 ∩supp mk\| , and c := \|Ik\| . Note that \|Ik−1\| = a + b. By the greedy choice of mk−1, we necessarily have b + c ≤\|Ik−1\|, so c ≤a. (5.1) Let us write — 136 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 5. Minkowski meets Ehrhart (draft of October 7, 2020) mk = d X i=0 λk i vi . Recall that any real number x ∈R can be written as x = k + {x} for Ik pp(mk) supp(u) ? c w on Ik−1 ∪ k ∈Z and {x} ∈[0, 1). In this notation, we deﬁne u := n mk + mk−1o := d X i=0 n λk i + λk−1 i o vi . Note that u ∈Π(P) ∩Zd+1. Since, supp mk−1 ∪supp mk \ supp mk−1 ∩supp mk ⊆supp u , the ‘greediness’ of mk−1 again implies that a + c ≤a + b, so c ≤b. Together with (5.1) this implies \|Ik\| = c ≤a + b 2 = \|Ik−1\|/2 . ⊓ ⊔ For our lattice simplex P we can now compute \| supp P\| = r X k=0 \|Ik\| ≤ r X k=0 1 2r ! \|I0\| < ∞ X k=0 1 2r ! \|I0\| = 2 · \|I0\| . So Lemma 5.4 shows the desired statement \| supp P\| < 4s . This ﬁnished the proof of Theorem 5.1 in the case of lattice simplices. 5.1.3 Proof of the pyramid theorem In order to deal with general lattice polytopes, we need the important notion of circuits. Deﬁnition 5.6 A circuit A ⊂Rd is a minimal aﬃnely dependent subset, i.e., A itself is aﬃnely dependent, while any proper subset is aﬃnely independent. − + − + (a) Circuit in a square − − − + + (b) Circuit in a triangle bipyramid Fig. 5.5: A two-dimensional and a three-dimensional circuit Here are some facts about circuits A (Exercise 5.2): Exercise 5.2 ▶The dimension of the convex hull of a circuit A equals \|A\| −2. ▶There is an (up to scalar multiplication) unique aﬃne relation X v∈A λv v = 0 X v∈A λv = 1 , where λv ̸= 0 for any v ∈A. Haase, Nill, Paffenholz: Lattice Polytopes — 137 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) ▶The previous relation yields a partition of A into two non-empty subsets A+, A−depending on the signs of the corresponding λv. Note that the convex hull of the elements in A+ (respectively, in A−) is not contained in the boundary of conv(A). Here is the crucial observation: Lemma 5.7 Let C be the convex hull of a circuit A. Then \|A\| ≤2 deg(C) + 2 . Proof. As observed above, the sum of the lattice points in A+ is a lattice point in the interior of \|A+\|C. Therefore, \|A+\| ≥codeg(C) = dim(C) + 1 −deg(C) = (\|A+\| + \|A−\| −2) + 1 −deg(C) . This implies \|A−\| ≤deg(C) + 1. The same bound also holds for \|A+\|. ⊓ ⊔ We are now in the position to ﬁnish the proof. Fig. 5.6: The induction step Proof (Proof of Theorem 5.1 in the general case). Let c := \|V(P) −d−1\|. We prove the statement by induction on c. The case c = 0 was dealt with in the previous section. So, we may assume c > 0, and d > f(c, s) = c(2s + 1) + 4s −1. Since P is not a simplex, there exists a vertex v such that P ′ := conv (V(P){v}) is d-dimensional (see Exercise 5.3). Note that \|V(P ′)\| −dim(P ′) −1 < c by construction, and deg(P ′) ≤s by the monotonicity theorem. Hence, the induction hypothesis yields that P ′ is a (possibly successive) lattice pyramid over a lattice polytope B′ of dimension dim B′ ≤f(\|V(P ′)\| −d −1, deg(P ′)) ≤(c −1)(2s + 1) + 4s −1 . Since v ∈aﬀ(P ′), there exists a circuit A among v ∪V(P ′) which contains v, see also Figure 5.6. By the monotonicity theorem and Lemma 5.7, \|A\| ≤2s + 2. We deﬁne B as the convex hull of the vertices of B′ and the elements in A. Since v ∈aﬀ(A\v), we compute dim(B) ≤dim(B′) + \|A\| −1 ≤(c −1)(2s + 1) + 4s −1 + 2s + 1 = f(c, s) . Since v ∈B, P is a lattice pyramid over B, as desired. ⊓ ⊔ Exercise 5.3 — 138 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 5. Minkowski meets Ehrhart (draft of October 7, 2020) 5.2 Lattice polytopes of small degree In the last section of this chapter we explain a construction of high-dimensional lattice polytopes of small degree. We will state a general result that shows that any lattice polytope of small degree is given this way. As an application in Ehrhart theory, we prove a ﬁniteness result generalizing the ﬁniteness of lattice polytopes with a given non-zero number of interior lattice points. 5.2.1 Cayley-Polytopes For this, we will need the notion of a Cayley polytope. Deﬁnition 5.8 Given lattice polytopes P0, . . . , Pt in Rk, the Cayley sum is deﬁned as P0 ∗· · · ∗Pt := conv((P0 × 0) ∪(P1 × e1) · · · ∪(Pt × et)) ⊂Rk × Rt , where e1, . . . , et denotes the standard lattice basis of Rt. We say P ⊆Rd is a Cayley polytope of length t + 1 if there exists an aﬃne lattice basis of Zd ∼ = Zk × Zt identifying P with the Cayley sum P0 ∗· · · ∗Pt for some lattice polytopes P0, . . . , Pt in Rk. Let us note that dim(P0 ∗· · · ∗Pt) = dim aﬀ(P0, . . . , Pt) + t −1 . Aequivalente deﬁnition per e0, . . . , et in die Uebungsaufgaben: insb. auch von Cayleykegel (ohne zusaetzliche 1, dadurch wird alles sehr viel schoene; hier nicht gemacht, da Goren-steinkegel noch nicht deﬁniert) There is a convenient and useful characterization of Cayley polytopes. Lemma 5.9 ([7, Proposition 2.3]) Let us deﬁne u ∈(Zd+1)∗such that ⟨u, {1} × P ⟩= 1. Let C be the cone over {1} × P. Then the following statements are equivalent: Notation ist hier noch nichtstandard verglichen mit Rest (1) P is a Cayley polytope P0 ∗· · · ∗Pt of length t + 1 (2) There is a lattice projection P onto a unimodular t-simplex ist lattice projection schon deﬁniert? (3) There are nonzero x0, . . . , xt ∈C⋆∩(Zd+1)∗such that x0 + · · · + xt = u Proof. (1) ⇒(2): The deﬁnition of a Cayley polytope implies the existence of a surjective lattice homomorphism Zd →Zt, mapping the vertices of P onto {0, e1, . . . , et}. (2) ⇒(1): This follows also directly from the deﬁnition. (1) ⇒(3): In Deﬁnition 5.8, let b1, . . . , bd, e1, . . . , et be the stan-dard basis of Zd, and b∗ 1, . . . , b∗ d, e∗ 1, . . . , e∗ t its dual lattice basis. As Haase, Nill, Paffenholz: Lattice Polytopes — 139 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) u = (1, 0, . . . , 0), setting xi := (0, e∗ i ) for i = 1, . . . , t and x0 := (1, −e∗ 1 −· · · −e∗ t ) satisﬁes the desired conditions. (3) ⇒(2): Let us deﬁne ϕ : Zd →Zt+1 m 7→(⟨x0, m ⟩, . . . , ⟨xt, m ⟩) Let v ∈V(P). Then t X i=0 ⟨xi, v ⟩= ⟨w∨, v ⟩= 1 . Since, ⟨xi, v ⟩≥0 for i = 0, . . . , t we see that ϕ(v) ∈{e0, . . . , et}, where e0, . . . , et denotes the standard basis of Zt+1. Assume there exists i ∈{0, . . . , t} which is not an image of a vertex of P. In this case, we would get ⟨xi, v ⟩= 0 for all v ∈V(P), thus, xi = 0, a contradiction. Hence, P projects onto S := conv(e1, . . . , et+1), a unimodular t-simplex (up to unimodular equivalence). ⊓ ⊔ Since the t-th multiple of a unimodular t-simplex contains no interior lattice points, we conclude from (2) that codeg(P0 ∗· · · ∗Pt) ≥t + 1 . The following result (which we state without proof) shows that any lattice polytope in high dimensions of small degree decomposes into small-dimensional lattice polytopes. Theorem 5.10 Any lattice polytope of degree s is a Cayley polytope of length ≥d + 1 −(s2 + 19s −4)/2 (equivalently, P is a Cayley polytope of lattice polytopes in dimension ≤(s2 + 19s −4)/2). It is conjectured that (s2 + 19s−4)/2 can be replaced by a linear function in s. Here is an application in Ehrhart theory. Theorem 5.11 There exist only ﬁnitely many h∗-polynomials of lattice polytopes of degree s and with given leading coeﬃcient h∗ s. This result follows immediately from the following statement combined with Theorem 5.10. Here, let V (q, k) be given as in Theorem 4.25. Proposition 5.12 Check ob uebereinstimmt mit Gorenstein kapitel If P is a d-dimensional lattice polytope of degree s that is a Cayley polytope of length c + 1 ≥d + 1 −N, then nvolZd(P) ≤N! NN V (N, h∗ s)N . — 140 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 5. Minkowski meets Ehrhart (draft of October 7, 2020) Proof. We denote by S the unimodular simplex ∆c = conv(e0, . . . , ec), where e0 = 0. Let d = c + q and π : Zd = Zc+q →Zc be the projection onto the ﬁrst c coordinates. For i = 0, . . . , c we set Pi ⊂Rq such that ei × Pi = π−1(ei) ∩P. Set r = d + 1 −s, so rP contains exactly k := h∗ s interior lattice points. The interior lattice points in rP are exactly the interior lat-tice points in π−1(λ) ∩rP, for interior lattice points λ ∈rS. Say λ = (λ1, . . . , λc) is an interior lattice point of rS such that π−1(λ) contains an interior lattice point of P. Then λ1, . . . , λc and λ0 = r −λ1 −· · · −λc are positive integers. Now, any ’ﬁber’ of π can be identiﬁed with a Minkowski sum of P0, . . . , Pc: π−1(λ) ∩P = λ × (λ0P0 + · · · + λcPc) . Let ωij be the width of Pj with respect to the i-th coordinate on Rq, the diﬀerence between the maximum and the minimum of the i-th coordinates of points in Pj. By Theorem 2.82 there is a choice of coordinates on Rq such that λ0P0 + · · · + λcPc is contained in the standard cube [0, C]q with the side length C being equal to q times the normalized volume of λ0P0 + · · · + λcPc. Now, we may choose by Theorem 4.25 C = q V (q, k). Since widths are additive and each λj is a positive integer, it follows that ωi0 + · · · + ωic ≤C, for 1 ≤i ≤q. Now P projects onto S, so we can express the normalized volume of P as an integral nvolZd(P) = d! · Z S vol π−1(λ) ∩P dλ, where vol is the ordinary euclidean volume. The volume of π−1(λ) ∩P is bounded by the product of its coordinate widths, which is Qc i=1(λ0ωi0 + · · · + λcωic). Expanding the product and substituting into the integral above gives nvolZd(P) ≤d! · X j1,...,jq ω1j1 · · · ωqjq Z S λj1 · · · λjq dλ , (5.2) where the sum is over (j1, . . . , jq) ∈{0, . . . , c}q. Now it follows from Hölder’s inequality that the integral over S of the monomial λj1 · · · λjq is bounded above by the integral of λq 1, and a straightforward induction shows that Z S λq 1 dλ = q!/d!. Haase, Nill, Paffenholz: Lattice Polytopes — 141 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Substituting into (5.2) then gives nvolZd(P) ≤q! · X j1,...,jq (ω1j1 · · · ωqjq). The sum on the right hand side may be written as Qq i=1(ωi0 + · · · + ωic), which is bounded above by Cq since, for each i, ωi0 + · · · + ωic is less than or equal to C. We conclude that nvolZd(P) is bounded above by q! · Cq. Now the theorem follows, since q = d −c ≤d −(d −N) = N. ⊓ ⊔ TODO: degree one als Bild rein (ohne Beweis) noch Bild! 5.2.2 Small Degree 5.2.3 Normal Form In many applications, the question arises whether two lattice polytopes P and Q are isomorphic. Of course, isomorphism here means the existence of a unimodular transformation x 7→Ax + b for A ∈Gld(Z) and b ∈Zd. A brute force way would be to go through all of the ﬁnitely many combinatorial isomorphisms between P and Q and check whether the corresponding aﬃne transformations are unimodular. However, given a large list of lattice polytopes this procedure is not feasible. This was the problem faced by Maximilian Kreuzer and Harald Skarke in their large-scale classiﬁcation of all reﬂexive polytopes up to dimension four [33, 34] (see § 7.5.2). In this section we present their solution as implemented in the software package PALP 1 Kreuzer and Skarke showed that it is possible to associate to any lattice polytope P ⊂Rd a normal form N(P) = (w1 · · · wl) ∈Mat Zd × l(Z) where l is the number of vertices of P such that P ∼ = conv(w1, . . . , wl) and the following property holds: P ∼ = Q ⇐ ⇒N(P) = N(Q). Hence, any total ordering on the set of integral matrices induces a total ordering on all unimodular equivalence classes of d-dimensional lattice polytopes. We will describe the various ingredients in this construction step by step. They will also play an important role in the classiﬁcation algorithm of reﬂexive polytopes which is described in § 7.5.2. — 142 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 5. Minkowski meets Ehrhart (draft of October 7, 2020) 5.2.3.1 Lattice distance Let us recall the notion of lattice distance of a vertex v of a full-dimensional polytope P ⊂Rd from a facet F of P. The hyperplane H spanned by F is given as H = {x ∈Rd : ⟨ηF , x ⟩= b} for ηF ∈(Z⋆)d a primitive inner normal and b ∈Q. Then the lattice ist primitive schon deﬁniert? distance d(v, F) of v from F is deﬁned as ⟨ηF , v ⟩−b. Note that d(v, F) ≥ 0 with equality if and only if v ∈F. The lattice distance is illustrated in the following ﬁgure: Here is an equivalent way of viewing these numbers. Deﬁne the cone C(P) := cone(P × 1) ⊂Rd+1 over P. Then the rays of the dual cone C(P)⋆are in one-to-one correspondence to facets F of P. Moreover, the ray corresponding to a facet F is generated by a primitive lattice point uF ∈(Z⋆)d+1) such that uF = (ηF , −bF ). Hence, the lattice distance d(v, F) equals ⟨uF , (v, 1) ⟩. Lemma 5.13 If P ∼ = Q via an aﬃne unimodular transformation ϕ, then the lattice distance of a vertex v from a facet F of P equals the lattice distance of the vertex ϕ(v) from the facet ϕ(F) of Q. Proof. The statement seems obvious, however, it doesn’t hurt to give a detailed proof. Exercise missing Note that CP ∼ = CQ via a linear unimodular transfor-mation ψ (see Exercise 5.4) such that ψ((v, 1)) = (ϕ(v), 1). Moreover, C⋆Q ∼ = C⋆P via ψ⋆, thus, uϕ(F ) = (ψ⋆)−1(uF ). Hence, d(ϕ(v), ϕ(F)) = ⟨uϕ(F ), (ϕ(v), 1) ⟩= ⟨(ψ⋆)−1(uF ), ψ((v, 1)) ⟩= ⟨uF , (v, 1) ⟩= d(v, F). ⊓ ⊔ ⊓ ⊔ Exercise 5.4 5.2.3.2 The vertex-pairing matrix Throughout, we ﬁx a total ordering on the set of integral matrices. We will use the lexicographic ordering, i.e., for a m × n-matrix {ai,j} we proceed in this order a1,1, a1,2, . . . , a1,n, a2,1, a2,2, . . . , am,n Let P have the vertex set V(P) = {v1, . . . , vl} (with \|V(P)\| = l) and the set of facets F(P) = {F1, . . . , Fk} with (\|F(P)\| = k). Then the matrix of vertex-facet lattice distances {d(vi, Fj)}i=1,...,l; j=1,...,k is not unique. It depends on the choosen ordering of the vertices and facets of P (corresponding to permuting the rows and columns of the 1 We remark that there are several choices of such an implementation, and we do not claim that ours is precisely the one used in PALP. Haase, Nill, Paffenholz: Lattice Polytopes — 143 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) previous matrix). In order to make it unique, we use our total ordering and deﬁne the vertex-pairing matrix VPM(P) of P as the (unique) largest matrix among the set of l × k-matrices obtained in this way. Note that VPM(P) encodes the vertex-facet-incidences of P and hence its complete combinatorial structure. Lemma 5.13 implies P ∼ = Q = ⇒ VPM(P) = VPM(Q) . It is important to observe that the converse does not hold. The two non-isomorphic lattice polygons in Figure 5.7 have the same vertex-pairing-matrix    3 0 0 0 3 0 0 0 3    Fig. 5.7: Two lattice polygons with the same vertex-pairing-matrix 5.2.3.3 The group Ord(P) We can ﬁnd an ordering of the set of vertices and facets of P such that VPM(P) = {d(vi, Fj}i=1,...,l; j=1,...,k Of course, there may be several choices how to do so. Let us make this precise. We deﬁne Ord(P) as the set of bijections σ : {1, . . . , l} →V(P) such that there exists some bijection π : {1, . . . , k} →F(P) with VPM(P) = {d(σ(i), π(j))}i=1,...,l; j=1,...,k By construction, Ord(P) only depends on VPM(P). It is an exercise for the reader to check that Exercise ▶For each σ ∈Ord(P) there is only one such choice for π. ▶Identifying V(P) with {1, . . . , l} realizes Ord(P) as a subgroup of the full symmetric group on {1, . . . , l}. ▶Ord is invariant under umimodular equivalence. In Figure 5.7, Ord(P) is the full symmetric group with 6 elements. However, Ord(P) may often be much smaller, see in Figure 5.8. — 144 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 5. Minkowski meets Ehrhart (draft of October 7, 2020) Fig. 5.8: Lattice triangles with Ord(P) of size 1, respectively, 2 5.2.3.4 Linear and aﬃne normal forms For σ ∈Ord(P) we deﬁne the d × l-matrix Aσ := vσ(1) · · · vσ(l) . We will need the Hermite normal form theorem for integral matrices in its full generality. For the proof we refer to Exercise ??? Exercise missing . Theorem 5.14 (General Hermite normal form) Given a matrix A ∈ Matd×l(Z) whose column space has full dimension d, there exists a unique unimodular matrix U ∈Gld(Z) such that H := UA is in Hermite normal form, i.e., there exists an r ∈{1, . . . , l} and a strictly increasing function f: {1, . . . , d} →{1, . . . , r} such that for i = 1, . . . , d (1) Hi,j = 0 for j < f(i) and j > r (2) Hi,f(i) > Hi′,f(i) ≥0 for 1 ≤i′ < i EXISTIEREN GUTE ALGORITH-MEN DAZU.? For σ ∈Ord(P) let us denote by Hσ the unique matrix UAσ. Then we deﬁne the linear normal form LNF(P) of P as the smallest matrix (with respect to the total ordering) among all Hσ for σ ∈Ord(P). By construction, the linear normal form has the property that P and P ′ are mapped onto each other by a lattice-preserving linear automorphism if and only if LNF(P) = LNF(P ′). Example 5.15 Let us consider the triangle P on the left of Figure 5.8. Here, VPM(P) =    6 0 0 0 3 0 0 0 2    \| Ord(P)\| = 1, and for σ ∈Ord(P) we have Aσ = −1 2 −1 −1 −1 1 ! Therefore, Hσ = 1 1 −1 0 3 −2 ! is the linear normal form of P. Haase, Nill, Paffenholz: Lattice Polytopes — 145 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Example 5.16 Let us consider the triangle P on the right of Figure 5.8. Here, VPM(P) =    3 0 0 0 3 0 0 0 1    Ord(P) = {σ1, σ2}, and we have Aσ1 = 0 0 1 −1 2 2 ! and Aσ1 = 0 0 1 2 −1 2 ! Therefore, Hσ1 = 1 −2 0 0 0 1 ! and Hσ2 = 2 −1 0 0 0 1 ! Hence, Hσ1 equals the linear normal form of P. CHECK PALP In order to get the desired aﬃne normal form we apply (5.14) to the d × l-matrix A′ := 0 v1 −v0 · · · vl −v0 . This yields a unique matrix U ∈Gld(Z) such that UA′ is in Hermite normal form. Let us denote for σ ∈Ord(P) by H′ σ the unique matrix U(Aσ)′. Then we deﬁne the (aﬃne) normal form N(P) of P as the smallest matrix among all d × l-matrices H′ σ for σ ∈Ord(P). By construction, the aﬃne normal form has the desired property that P and P ′ are mapped onto each other by a lattice-preserving aﬃne automorphism if and only if N(P) = N(P ′). Example 5.17 Let us continue with Example 5.15. For the unique σ ∈ Ord(P) we have (Aσ)′ = 0 3 0 0 0 2 ! This is already in Hermite normal form, hence H′ σ = 0 3 0 0 0 2 ! is the aﬃne normal form of P. — 146 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 5. Minkowski meets Ehrhart (draft of October 7, 2020) Example 5.18 Let us continue with Example 5.16. Here, Ord(P) = {σ1, σ2}, and we have Komplexeres Bsp.? (Aσ1)′ = 0 0 1 0 3 3 ! and (Aσ1)′ = 0 0 1 0 −3 0 ! Therefore, H′ σ1 = 0 3 0 0 0 1 ! and H′ σ2 = 0 3 0 0 0 1 ! Hence, H′ σ1 = H′ σ2 equals the aﬃne normal form of P. CHECK PALP The reader may wonder whether we really needed Ord(P) in order to deﬁne these normal forms. Indeed, it would have been possible to take the group of all permutations of V(P). However, in general Ord(P) is much smaller and therefore computationally more feasible. Moreover, these concepts will be useful when dealing with reﬂexive polytopes, see Section ???. 5.2.3.5 The algorithm of Kasprzyk and Grinis See 5.3 Problems Exercise 5.5 included on page 135 5.1. Prove Lemma 5.3. included on page 137 5.2. Prove all the basic facts on circuits needed in § 5.1.3. some exercises should be in diﬀerent chapters? included on page 138 5.3. Prove that for any d-dimensional polytope P which is not a simplex there exists a vertex such that the convex hull of the other vertices is full-dimensional. included on page 143 5.4. pb:aﬃne-to-cone included on page 147 5.5. A dummy exercise. Haase, Nill, Paffenholz: Lattice Polytopes — 147 — Short Rational Generating Functions 6 Contents 6.1 Hermite and Smith Normal Forms . . . . . . . . 150 6.2 Computing Short Rational Generating Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 6.2.1 Polynommial Time Evaluation . . . . . . . . . . . 155 6.2.2 Integer Linear Programming via Evaluation157 6.3 The Shortest Vector Problem . . . . . . . . . . . . 157 6.4 Short Lattice Bases . . . . . . . . . . . . . . . . . . . . . . 160 6.4.1 Applications of LLL . . . . . . . . . . . . . . . . . . . . 170 6.5 Variations of Barvinok’s Algorithm . . . . . . . 170 6.6 The Closest Vector Problem . . . . . . . . . . . . . . 171 6.7 Integer Programming in Fixed Dimension . 172 6.8 Software . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 6.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 In Section 4.1 we have shown that any lattice has a non-zero vector of length at most √ d det Λ. However, the proof was not constructive and we have left open the question how one can actually compute such a vector in polynomial time. We will solve this problem in Section 6.3. This algorithm hinges on the fact that we need to compute a reduced basis eﬃciently. We will give an algorithm for one special reduced basis, the δ-reduced basis using the LLL-method in Section 6.4. This will ﬁnally give the complete algorithm for counting lattice points. Using this basis algorithm we can also solve the problem of computing a rational generating function for Lecture Notes Lattice Polytopes (draft of October 7, 2020) a cone in polynomial time in ﬁxed dimension, left open in Section 3.4. In Section 6.5 we discuss some improvements of the algorithm. Finally, we will show that we can solve an integer program in polynomial time in ﬁxed dimension. 6.1 Hermite and Smith Normal Forms Smith normal forms using Kannan and Bachem . See also Gerald Jäger, A new algorithm for computing the Smith normal form and its implementation on parallel machines 6.2 Computing Short Rational Generating Functions In Chapter 3 we have seen that we can express the lattice points in a cone as a rational function. The proof somehow also gives a method to compute such a function by enumerating the lattice points in the fundamental parellelipiped and deriving the appropriate numerator from this. Yet, as we will see below, this is in general not a task that can be done in polynomial time in the size of the input and the dimension. In this section we will develop a diﬀerent approach using signed decompositions of a cone to compute the multivariate rational generating function GP (t) of a simplicial cone in polynomial time, at least for ﬁxed dimension. This extends to all cones by using a triangulation without new vertices. We looked at this for polytopes in Theorem 2.35. The same construction works for cones. Note that this construction of a triangulation is polynomial in the size of the input. The h⋆-polynomial, and thus the Ehrhart polynomial can be obtained from the generating function of the cone over the polytope by specializing the rational function at x = (t, 1). Similarly, we may count lattice points in a polytope via the generating functions of the vertex cones specialized at x = 1, using Brion’s Theorem (Theorem 3.67). However, these values are (removable) poles of the generating functions. We will discuss a possible approach to evaluate the functions eﬃciently in § 6.2.1. The initial idea for the algorithm presented in this section is due to Barvinok and Barvinok and Pommersheim . For ﬁxed dimension d the algorithm is polynomial in the input size. Practical implemen-tations can be found in the software package LattE and its successors LattE macchiato and LattE integrale [S1, S2] or in barvinok [S3, 60]. See Section 6.8 for some examples of computations of short generating functions. Yet, the algorithm needs Minkowski’s First Theorem (Corollary 4.3) to ﬁnd a short vector in the lattice as a (frequent) intermediate step. — 150 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) We have already dicussed that its proof as presented in Chapter 4 is not constructive. We solve this in Section 6.3. This in turn relies on a method to obtain a lattice basis that approximates an orthogonal vector space basis, which is the topic of Section 6.4, where we discuss the LLL-algorithm of Arjen Lensta, Hendrik Lensta, and László Lovász. Eﬃcient implentations also use the half open decompositions introduced in § 3.5.1 to get rid of lower dimensional cones in the decomposition or a decomposition of the dual cone to turn lower dimensional cones into cones with lineality. We discuss these improvements in Section 6.5. In the case of a polytope P, the generating function of the lattice points is actually a polynomial (the sum of all monomials for all lattice points in P). Observe however, that we usually cannot write this down in polynomial time in the size of the input, just because there may be more than polynomially many lattice points compared to the number of vertices. Thus, we should not try to expand a rational function to get rid of the poles. We aim for a polynomial method (in ﬁxed dimension) to compute integer point generating functions of simplicial cones. We theoretically know how to write down the function, the formula is just given by Corollary 3.28. But from an algorithmic point of view this may not be eﬃcient, as its numerator may need too many monomials, compared to the size of the input. An example, which already works in dimension 2, is the following. Consider the cone C := conv(e1, e1 + ke2) for some k ∈Z>. We need k monomials in the numerator or k cones in a unimodular triangulation, which is not polynomial in the input, which are just the two generators of the cone. See also Figure 6.1 Fig. 6.1: The cone spanned by e1 and e1 + kek needs k unimodular cones in its decomposition Hence, if we want a polynomial time algorithm we need a better way to subdivide. The key idea for this is to use signed decompositions, which This needs a ﬁgure and an example are decompositions where we may take new rays outside of the original cone and use addition and subtraction of rational generating functions to obtain the desired rational generating function of the original cone. It was the achievement of Barvinok to show that with this method you can get away with a polynomial number of (even unimodular) cones. To make this precise, let C be a d-dimensional cone spanned by primitive rays v1, . . . , vd. We deﬁne the index of the cone C to be Index(C) := #(Π(v1, . . . , vd) ∩Zd) = \| det(v1, . . . , vd)\| = vol Π(v1, . . . , vd) This needs a ﬁgure The cone C is unimodular if and only if Index(C) = 1. In a triangula-tion T of our cone C we will record the index of each cone in T (note that, Haase, Nill, Paffenholz: Lattice Polytopes — 151 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) for the generating series we also need to take lower dimensional cones into account to account for overcounting in intersections via inclusion-exclusion). This collection of indices is both a measure of how far we are still from a unimodular triangulation and it gives an indicator whether we are already done or which cones we need to subdivide further. Oberserve that the index of a face F of a cone C is bounded by the index of C, i.e. Index(F) ≤Index(C), so we actually only need to track indices of maximal cones and subdivide those if the index is still larger than one. This is, of course, only useful if we can provide a method that subdivides a cone into cones of smaller index. Here the idea of signed decompositions is needed to make this eﬃciently. The basic tool in the construction is Minkowski’s First Theorem (Corollary 4.3), which tells us that for any compact, convex, and centrally symmetric K ⊆Rd with vol K ≥2d there exists a ̸= 0 in K ∩Zd. We use this theorem in the following way. If, for a cone C, the index Index(C) is still larger than 1, then K := 1 d √ Index C X λ1vi : −1 ≤λi ≤1 is a compact, convex, and centrally symmetric body with volume ﬁgure vol(K) = 2d . Hence, we can conclude that there is w ∈K ∩Zd diﬀerent from 0. We can write w as a linear combination of the cone generators and, as w is contained in K, we obtain a bound on the size of the coeﬃcients, that is, we know that w = λ1v1 + λ2v2 + · · · + λdvd for 0 ≤\|λi\| ≤(Index(C))−1/d . (6.1) As already observed before, the proof of Minkowski’s First Theorem (Corollary 4.3) is not constructive, and it is generally diﬃcult to compute such a point w. See Section 6.3 for an approach. Consider the vector w obtained in (6.1). By replacing w with −w if necessary we can assume that w, v1, . . . , vd lie in a common half-space. Additionally we may clearly assume that w is primitive. By construction, \|λi\| ≤(Index C)−1 d . We deﬁne new cones Cj := cone(v1, . . . , vj−1, w, vj+1, . . . , vd) for 1 ≤j ≤d by replacing vj by w. We compute the index of these new cones to be — 152 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) Index Cj = \| det(v1, . . . , vj−1, w, vj+1, . . . , vd)\| = d X k=1 \|λk\| · \| det(v1, . . . , vj−1, vk, vj+1, . . . , vd)\| = \|λj\| · \| det(v1, . . . , vd)\| = \|λj\|(Index C) ≤(Index C)−1 d (Index C) = (Index C) d−1 d and the right hand side is strictly less than Index C if Index C ≥2. As the index is an integral number we see that the index actually drops by at least one. We deﬁne a corresponding sign function to make a signed subdivision of C with the cones Cj. For 1 ≤j ≤d let εj :=        0 if dim Cj < d 1 if det(v1, . . . , vd) · det(v1, . . . , vj−1, w, vj+1, . . . , vd) > 0 −1 otherwise. Using our decomposition and the corresponding sign function we may write the integer point generating series as the signed sum b GC (t) = d X j=1 εjb GCj (t) + lower dimensional contributions. In our decomposition we have ▶at most d d-dimensional cones, ▶at most 2dd cones of any dimension. We repeat this decomposition for each cone of index ≥2 in our trian-gulation successively until there is no cone of index greater than one left. Let us compute how many iterations we need in this procedure. After n decomposition steps, a cone D in the decomposition has index at most Index D ≤(Index C) d−1 d n and we need to check for which n this number drops below 2 (recall that the index is integral, so it must be 1). We take the binary logarithm twice to solve this expression for n: lg2 lg2 (Index C) d−1 d n = lg2 d −1 d n lg2(Index C) = n lg2 d −1 d + lg2 lg2(Index C) = −n lg2 d d −1 + lg2 lg2(Index C) . (6.2) Haase, Nill, Paffenholz: Lattice Polytopes — 153 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Algorithm 6.1: Barvinok’s Algorithm: Original Version Input: A polyhedron P = {x \| Ax ≤b} with vertices v1, . . . , vk. Output: The integer point generating function for P as GP (t) = X i∈I εi tai (1 −tvi1) · · · (1 −tviki ) for εi ∈{−1, 1}, ai, vij ∈Zd. for i ←1 to k do Compute vertex cone Ci at vertex vi; Triangulate Ci into ki simplicial cones Cij; for j ←1 to ki do do a signed decomposition of Cij into unimodular cones Ck ij.; compute the unique interior point ak ij in Ck ij; end for sum up the contributions using the inclusion-exclusion principle; end for sum up the contributions using the inclusion-exclusion principle; Hence, if n > lg2 lg2(Index C) lg2 d d−1 = O (d lg2 lg2 Index C) , then the right hand side of (6.2) is negative, so that Index D ≤ (Index C) d−1 d n ≤(Index C) d−1 d n < 2 , i.e. Index D = 1 . So the number of iterations until we reach unimodular cones is indeed polynomial. However, we also need to check that the number of cones we produce is polynomial. In n steps we produce at most (d2d)n = 2nd lg2 d ≤2Md2 lg2 d lg2 lg2 Index C = (lg2 Index C)Md2 lg2 d = (lg2 Index C)O (d2 lg2 d). cones. Hence, we conclude that with this approach indeed, in ﬁxed dimension, the number of cones is bounded by a polynomial in lg2 Index C. This is in the order of the input size of our cone in binary encoding. We summarize the algorithm in Algorithm 6.1 and the following theorem. Theorem 6.1 Let d ∈Z>0 be ﬁxed. Then there is a polynomial time algorithm that computes the integer point generating function GP (t) in the form — 154 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) GP (t) := X i∈I εi tai (1 −tvi1) · · · (1 −tvisi ) , where εi ∈{−1, 1}, a ∈Zd, vij ∈Zd \ {0} for all i, j and si ≤d, for any d-dimensional polyhedron P given in its exterior description. ⊓ ⊔ Note that, in this theorem, the index set I runs over all cones, including lower dimensional ones, in the subdivision. 6.2.1 Polynommial Time Evaluation We can use Theorem 6.1 to compute h⋆-polynomials or Ehrhart polyno-mials by computing the generating function of the cone over the polytope. Similarly, using Brion’s Theorem (Theorem 3.67), we can count lattice points in P and multiples of P (and thus also comput the Ehrhart poly-nomial via interpolation) by computing the rational generating functions of all vertex cones. However, this usually requires us to evaluate the generating function at x = (t, 1) or x = 1. Although we now from the theory that these are regular points of the rational functions, they are poles in the representation we obtain. We use tools from analysis to evaluate them (Note again, that expansion into a Taylor series, though theoretically a method to remove the pole, is not an option if we want evaluate this in polynomial time.). One possible approach is to deﬁne a curve γ(s) for a real parameter s ≥0 such that γ(0) is (t, 1) or 1 (depending on whether we look at the cone over P or at a vertex cone of P) and the only pole of our generating function on that curve occours for s = 0. Then we take the limit s →0. We need the following lemma. Lemma 6.2 Let v1, . . . , vk ∈Rd. Then there is m ∈Rd such that ⟨m, vj ⟩̸= 0 for 1 ≤j ≤k. Proof. We use the moment curve m(λ) := 1, λ, λ2, . . . , λd−1 . The map λ 7− → d Y j=1 ⟨m(λ), vj ⟩ is a nonzero polynomial of degree (d −1)k. Hence, it has at most (d −1)k zeros and we can try a polynomial number of values to ﬁnd a λ such that m(λ) gives the claim. ⊓ ⊔ Using this lemma we can ﬁnd m = (m1, . . . , md) ∈Rd such that, in the notation of Theorem 6.1, Haase, Nill, Paffenholz: Lattice Polytopes — 155 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) ⟨m, ai ⟩̸= 0 ⟨m, vij ⟩̸= 0 for i ∈I and 1 ≤j ≤si . Now consider (we look at evaluation at 1, the case for (t, 1) is similar) x(r) := (erm1, . . . , ermd) . We get the desired evaluation as lim r→0 GP (x(r)) . Let αi := ⟨m, ai ⟩ νij := ⟨m, vij ⟩. Then GP (x(r)) = X i∈I εi eαir Qsi j=1(1 −euijr) , and the summands are all rational functions in one variable r which are deﬁned for all r > 0. We want to compute the constant term of the Laurent expansion of all summands at r = 0. Now consider a single such fraction. We can transform it to obtain eαir Qsi j=1(1 −euijr) = 1 rsi eαir si Y j=1 r 1 −euijr . (6.3) Now each factor r 1 −euijr is deﬁned for all r and we can compute the Laurent expansion up to degree si + 1: r (1 −euijr) = Tij(r) + Rij(rsi+1) , and similarly we get eαir = Si(r) + R′ ij(rk+1) . We compute the product up to degree si + 1: Pi(r) := Si(r) si Y j=1 Tij(r) + R′′ i (rsi+1) . Let ci be the coeﬃcient of rsi (note that (6.3) has an additional factor of 1 rsi , so for this product ci is the constant coeﬃcient). We sum them up with the given signes to obtain c := X i∈I εici . This is the desired limit and thus the evaluation at 1. — 156 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) Remark 6.3 Using the Todd-polynomials tdm(ξ1, . . . , ξd) deﬁned by k Y i=1 xξi 1 −e−xξi = ∞ X m=0 tdm(ξ1, . . . , ξd)xm one may obtain a closed formula for the evaluation of GC (1), see e.g. [18, Thm. 7.2.1]. However, this requires us to evaluate the Todd polynomials. 6.2.2 Integer Linear Programming via Evaluation You can use GP (t) also to solve linear programs. If you want to maximize over a functional c ∈Zd, then you can just substitute t = (zc1, zc2, . . . , zcd). The highest degree of a monomial in the result is the optimal solution. 6.3 The Shortest Vector Problem In this section we consider Minkowski’s problem of ﬁnding a non-zero vector of shortest length in a lattice. (SVP). Let Λ be a lattice with lattice basis v1, v2, . . . , vd. The Shortest Vector Problem (SVP) asks to ﬁnd, in polynomial time in the dimension and the input size of the lattice basis, a non-zero vector u ∈Λ \ {0} of shortest possible length. This is a quite famous and important optimization problem. Observe that give some reasons, add some history a shortest non-zero lattice vector has length λ1(Λ), the ﬁrst successive minimum introduced in Deﬁnition 4.6. An important relaxation of the shortest vector problem is the ap-proximate shortest Vector problem (SVP)γ that asks, for a given bound γ = γ(d) ≥1 to ﬁnd a non-zero lattice vector u of length ∥u∥≤γλ1 . In fact, our algorithm for (SVP) is based on a solution for the approximate problem by constructing a reduced basis. We show that the coeﬃcients of a shortest vector in such a basis are bounded by a constant depending on the dimension only. Hence, we can solve the shortest vector problem by enumerating over all possible coeﬃcients (in ﬁxed dimension). In this section we explore the diﬃculties in the computation of a short basis vector, introduce reduced bases and show how such bases bound the size of coeﬃcients for a shortest vector. This proves that, in ﬁxed dimension and given a reduced basis, we can solve (SVP) in polynomial time. Haase, Nill, Paffenholz: Lattice Polytopes — 157 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) In the following Section 6.4 we introduce a special type of reduced bases, the δ-reduced bases of Arjen Lenstra, Hendrik Lenstra and László Lovász from 1982 together with a polynomial time algorithm for their computation, again in ﬁxed dimension. Let us ﬁrst discuss why the computation of a short vector, or a basis with short vectors is more involved for lattice bases than for vector space bases. We know from Proposition 4.5 that there is a vector of length at most √ d det Λ, but the proof did not give a method to actually construct such a vector, not even by enumeration. However, the problem would be easy to solve if vectors in the lattice basis actually were pairwise othogonal. In this case, the shortest of the lattice basis vectors is in fact a non-zero lattice vector of shortest length. If we disregard the lattice structure for a moment, any vector space basis can be transformed into an orthogonal one using Gram-Schmidt orthogonalization. This algorithm runs in polynomial time depending on the dimension. However, it clearly does not respect the lattice structure. The example of the hexagonal lattice shown in Figure 6.2 shows in fact that, in general, there even does not exist a basis with pairwise orthogonal vectors, and we have also already seen bases with vectors of necessarily diﬀerent lengths. Fig. 6.2: The hexagonal lattice For an approximate solution of the shortest vector problem we want to discuss how close we can get to an orthogonal basis for lattice vectors (while being able to compute the basis in polynomial time). Let Λ ⊆Rd be a lattice with a basis v1, v2, . . . , vd ∈Rd. The order of the basis vectors is important for the following considerations. We consider the increasing chain of subspaces V0 := {0} Vk := lin(v1, . . . , vk) for 1 ≤k ≤d . together with the induced lattices Λk := Λ ∩Vk on these spaces. For 0 ≤k ≤d let πk : Rd →Vk be the orthogonal projection onto the subspace Vk and deﬁne a new set of vectors via wk := vk −πk−1(vk) = vk − k−1 X j=1 λjkwj with λjk := ⟨vk, wj ⟩ ∥wj∥2 . See also Figure 6.3. This is the Gram-Schmidt-orthogonalization of the Fig. 6.3: Orthogonal projec-tion of a lattice generator lattice basis v1, . . . , vd. So in particular we have ⟨wi, wj ⟩= 0 for 1 ≤i < j ≤d and d(vk, Vk−1) = ∥wk∥. See also Figure 6.4. Fig. 6.4: Distance of vk from the sub-space Vk is given by the norm of wk. — 158 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) Example 6.4 Let v1 := 2 1 ! , v2 := 1 2 ! . This is a basis of R2 with Gram-Schmidt-orthogonalization w1 = v1 and w2 = −3/5 6/5 ! = −4/5v1 + v2 , see Figure 6.5. v2 v1 = w1 w2 Fig. 6.5: The lattice basis of Exam-ple 6.4 We can write the original lattice basis in terms of the Gram-Schmidt basis as vk = wk + k−1 X j=1 λjkwj for 1 ≤k ≤d . (6.4) The vectors w1, . . . , wd are pairwise orthogonal, so det Λ = d Y j=1 ∥wj∥. and det Λk = k Y j=1 ∥wj∥. The Gram-Schmidt vectors give a lower bound for the length of a shortest vector of the lattice. Proposition 6.5 Let Λ be a d-dimensional lattice with basis v1, v2, . . . , vd and Gram-Schmidt-orthogonalization w1, w2, . . . , wd. Then ∥u∥≥min(∥w1∥, . . . , ∥wd∥) for all u ∈Λ \ {0}. Proof. We can write u as a linear combination u = d X j=1 ηjwj . Let k be the highest index such that ηk ̸= 0. We can rewrite u = ηkwk + k−1 X j=1 µjwj for some coeﬃcients µj, 1 ≤j ≤k −1. By othogonality, \|ηk\| ≥1, so that ∥u∥≥\|ηk\|∥wk∥≥∥wk∥. ⊓ ⊔ Deﬁnition 6.6 (orthogonality defect) The orthogonality defect of the lattice basis v1, v2, . . . , vd is MΛ := 1 det Λ d Y j=1 ∥vj∥. Haase, Nill, Paffenholz: Lattice Polytopes — 159 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Observe that the basis vectors v1, v2, . . . , vd are pairwise orthogonal if and only if MΛ = 1. In all other cases MΛ is strictly larger than 1. With this notion we have a measure how close a basis is to orthogo-nality. Deﬁnition 6.7 A lattice basis is reduced if its orthogonality defect is bounded by a constant Md depending on the dimension only. This deﬁnition leaves open whether there actually exists such a constant Md and how large we should choose it so that there is a reduced basis. We anwser this in the next Section 6.4 with an algorithm to compute a reduced basis and an explicit value for Md. The usefulness of notion is demonstrated by the following proposition, which shows that we can compute a shortest non-zero lattice vector from a reduced lattice basis in polynomial time. Theorem 6.8 Let Λ be a lattice with reduced basis v1, . . . , vd and let u ∈Λ{0} be a shortest non-zero lattice vector. Then u = d X j=1 λjvj with \|λj\| ≤ √ dM for 1 ≤j ≤d. We can reduce the bound in this the-orem to M instead of √ dM, see [21, Thm. 14.13]. Worth it? Proof. Let v1 be the shortest vector among v1, . . . , vd, and let V = (v1, . . . , vd). Then u = V λ for λ = (λ1, . . . , λd). Hence, λ = V −1u. By Cramer’s rule all entries of V −1 are determinants of (d −1) × (d −1)-minors of V , divided by det V . So each entry of V −1 is bounded by ∥v2∥· . . . · ∥vd∥· 1 det V ≤ M ∥v1∥. So \|λj\| ≤ X \|ui\| M ∥v1∥≤ √ d∥u∥M ∥v1∥≤ √ dM , where the last inequality uses ∥u∥≤∥v1∥. ⊓ ⊔ Hence, we can solve the shortest vector problem in polynomial time by enumerating all N := (2 √ dM)d choices for the coeﬃcients of the shortest vector, once we can ﬁnd a reduced basis in polynomial time. Theorem 6.9 In ﬁxed dimension d we can ﬁnd w ∈Π0 ∩Zd, w ̸= 0 in time polynomial in log3 max(∥vi∥). ⊓ ⊔ 6.4 Short Lattice Bases We have seen in the previous section that reduced bases allow the compu-tation of a shortest vector, and thus count lattice points in a polytope, in — 160 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) polynomial time. Yet, so far we don’t even know that such bases exist for our lattice Λ. We address this question in the following by introducing a very special type of reduced basis together with a polynomial time algorithm ﬁrst described by Arjen Lenstra, Hendrik Lenstra and László Lovász in 1982 . The algorithm is known as as the LLL-algorithm in honour of the three authors. Deﬁnition 6.10 (δ-LLL-reduced basis) Let Λ be a lattice with lattice basis v1, . . . , vd and Gram-Schmidt orthogonalization w1, . . . , wd. Let λjk be the coeﬃcients of the representation of the vj’s in the Gram-Schmidt basis as in (6.4). The basis v1, . . . , vd is δ-reduced for some 1/4 < δ < 1 if (1) \|λjk\| ≤1/2 for all 1 ≤j < k ≤d and (2) for all 1 ≤k ≤d −1 δ d(vk, Vk−1)2 ≤d(vk+1, Vk−1)2 (6.5) We mostly consider this deﬁnition for the particular choice δ = 3/4. We say that a coeﬃcient λjk is weakly reduced if it satisﬁes condition (1) of the deﬁnition. A lattice basis is weakly reduced if all coeﬃcients are others call this normalized. Change? weakly reduced. Vk−1 vk+1 vk Fig. 6.6: The second condition for LLL Geometrically a basis is δ-reduced if the vector vk+1 is not much closer to the subspace spanned by the ﬁrst k −1 basis vectors than the vector vk, see Figure 6.6. Observe that we could equally write this condition with the Gram-Schmidt-vectors w1, w2, . . . , wd as δ∥wk∥2 ≤∥wk+1 + λk,k+1wk∥2 for all 1 ≤k ≤d −1. v1 = w1 v2 v3 w2 w3 v′ 1 = w′ 1 v′ 2 v′ 3 = w′ 3 w′ 2 Fig. 6.7: A lattice basis that is not 3/4-reduced and one for the same lattice that is reduced. Example 6.11 Here is an example of a 3/4-reduced basis and one that is not. Consider the three vectors v1 :=    2 1 0    v2 :=    0 1 1    v3 :=    1 −1 3    Their Gram-Schmidt-orthogonalization is w1 :=    2 1 0    w2 :=    −2/5 4/5 1   = v2 −2 5w1 w3 :=    1 −2 2   = v3 −1 5w1 −w2 Haase, Nill, Paffenholz: Lattice Polytopes — 161 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Then λ13 violates condition (1) of the deﬁnition, and v1 and v2 violate condition (2), as d(v1, V0)2 = 5 and d(v1, V0)2 = 9/5 but 3/4 · 5 > 9/5. However, the basis v′ 1 :=    0 1 1    v′ 2 :=    2 1 0    v′ 3 :=    1 −2 2   , that spans the same lattice, is LLL-reduced. It has Gram-Schmidt-orthogonalization w′ 1 :=    0 1 1    w′ 2 :=    2 1/2 −1/2   = v′ 2 −1 2w′ 1 w′ 3 :=    1 −2 2   = v′ 3 . For an orthogonal basis w1, . . . , wd of Rd we can ﬁnd a permutation such that the distances d(wk, Vk−1) strictly increase for 2 ≤k ≤d. The condition for a δ-reduced basis is a relaxation of this. We only require that, if a basis vector is shorter than a previous one, it is shorter by at most a factor of δ. The following considerations will show that this relaxation is suﬃcient to compute such a basis in polynomial time (in the dimension and the input size), and it will lead to a universal bound as required in the approximate shortest vector problem. Hence, we can use this together with Theorem 6.8 to actually compute the shortest vector. It is important to note that the notion of reduced bases and the algorithm we are going to present has many more important applications apart from the computation of a shortest vector. We will see some of them in the following sections. For more applications see for example . Here is the main theorem of this section. add some more examples and refer-ences Theorem 6.12 (Lenstra, Lenstra, Lovász, 1982) Λ ⊂Rd a lattice. Then we can ﬁnd a basis v1, . . . , vd of Λ in polynomial time such that ∥v1∥, . . . , ∥vd∥≤M det Λ the statement of the theorem is in-complete: the basis is also reduced etc We need some preparations for the proof of this theorem. Consider the representation of our given basis in (6.4). Assume that for some pair of indices i < k the absolute value of the coeﬃcient λik is larger than 1/2. Then there is a unique µik ∈R and aik ∈Z such that \|µik\| ≤1/2 λik = aik + µik . We set v′ k := vk −aikvi and v′ j := vj for j ̸= k .. This leaves the subspaces Vj invariant and v′ 1, . . . , v′ j is still a basis of the lattice Λj for 1 ≤j ≤d. The new basis has the same Gram-Schmidt — 162 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) orthogonalization w1, . . . , wn. We want to compute the coeﬃcients λ′ jk for the new basis in the representation (6.4). We consider a ﬁxed k between 1 and d. Clearly, the adjustment of vk at most aﬀects the coeﬃcients λjk for this k and 1 ≤j ≤k −1, so λ′ jl = λjl for l ̸= k and j < l. We compute the new coeﬃcients for v′ k. For this, we can write v′ k = vk −aikvi as v′ k = wk + k−1 X j=1 λjkwj −aikvi . The vector vi is in the subspace Vi, so it has a representation vi = i X j=1 ηjwj , So we can write v′ k = wk + i X j=1 (λjk −aikηj)wj + k−1 X j=i+1 λjkwj , Now vi −wi ∈Vi−1, so ηi = 1 and thus \|λik −ηiaik\| = \|λik −aik\| = \|µik\| ≤ 1/2 . So the coeﬃcient λ′ ik := λik −ηi is weakly reduced, and the new coeﬃ-cients for v′ k are λ′ jk := λjk −aikηj forj ≤i λ′ jk := λjk forj > i We want to use this to make all coeﬃcients weakly reduced, that is, to make the lattice basis weakly reduced. However, we have to be careful, as making λ′ ik weakly reduced also aﬀects λjk for j < i. So, for ﬁxed k we apply this procedure to all λik with i in decreasing order. Doing this for all 1 ≤k ≤d gives us a weakly reduced basis. This procedure is formalized in Algorithm 6.2. Do we really want to write out algo-rithms in pseudo-code? There are (d 2) coeﬃcients, and reducing λik with the above procedure we have to touch at most i ≤d other coeﬃcients, so this process terminates after at most O (d3) steps. We summarize this in the following proposition. Proposition 6.13 Any lattice Λ has a weakly reduced basis. More pre-cisely, we can transform any basis into a weakly reduced one in O (d3) steps using Algorithm 6.2. ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 163 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Algorithm 6.2: Weakly Reduced Basis Input: Λ lattice in Rd of rank d with lattice basis v1, . . . , vd. Output: weakly reduced lattice basis v′ 1, . . . , v′ d Compute the Gram-Schmidt orthogonalization w1, . . . , wd; Compute coeﬃcients λjk, 1 ≤j < k ≤d such that vk = wk + Pk−1 j=1 λjkwj; for k from 2 to d do for j from k −1 to 1 do Determine µjk ∈R, ajk ∈Z such that λjk = µjk + ajk and \|µjk\| ≤1/2,; vk ←vk −ajkvj; end for end for return v1, . . . , vd; Do we need t include a run time also dep How can we transform a weakly reduced basis into a reduced one? We will show that repeatedly applying the following two steps lead to a successful algorithm. (1) If a weakly reduced basis has a pair of vectors vj, vj+1 violating condition (6.5), then we exchange vj and vj+1. (2) The new basis will usually not be weakly reduced anymore. We ﬁx this by applying Algorithm 6.2 again. We repeat these two steps until we reach a reduced basis. This is formalized in Algorithm 6.3. Algorithm 6.3: LLL Input: A lattice basis v1, . . . , vd. Output: A reduced lattice basis v′ 1, . . . , v′ d Make v1, . . . , vd weakly reduced (Algorithm 6.2); while V not δ-reduced do Find a pair vj, vj+1 that violates (6.5) and exchange the two vectors; Make v1, . . . , vd weakly reduced (Algorithm 6.2); end while return v1, . . . , vd; Formulation of the algorithms is quite sloppy. How formal do we want to be? We clearly obtain a δ-reduced basis if this procedure terminates. We introduce a new numerical invariant of a lattice basis. The potential of a lattice basis v1, . . . , vd is deﬁned as D(v1, . . . , vd) := d Y j=1 det Λj . (6.6) — 164 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) We can express D(v1, . . . , vd) also in the Gram-Schmidt orthogonaliza-tion. D(v1, . . . , vd) = d Y k=1 k Y j=1 ∥wj∥. Note that D(v1, . . . , vd) depends on the order of the basis vectors. It puts more weight on those vectors coming ﬁrst in the basis. Hence, it will change if we exchange two of the basis vectors. Let us determine the change in D after one iteration of Algorithm 6.3. We have observed above that making a basis weakly reduced does not change the subspaces Vi and the lattices Λi, hence, it also does not change D. Assume that vj and vj+1 for 1 ≤j ≤d −1 violate the condition of Deﬁnition 6.10(1). Let v′ 1, . . . , v′ d be the basis obtained by exchanging vj and vj+1, i.e. v′ j+1 := vj v′ j := vj+1 v′ i := vi for i ̸= j, j + 1 . Let V ′ i , Λ′ i be the new subspaces and lattices, 1 ≤i ≤d. Then V ′ i = Vi Λ′ i = Λi for i ̸= j , while V ′ j := lin(v′ 1, . . . , v′ j−1, v′ j) = lin(v1, . . . , vj−1, vj+1) . The new Gram-Schmidt vectors are w′ i := wi for i ̸= j, j + 1 w′ j := vj+1 − j−1 X i=1 λi,j+1wi = vj+1 −πj−1(vj+1) w′ j+1 := vj − j−1 X i=1 λijwi − ⟨vj, w′ j ⟩ ∥w′ j∥2 w′ j = vj −πj−1(vj) − ⟨vj, w′ j ⟩ ∥w′ j∥2 w′ j By our assumption on vj and vj+1 we have δ · d(vk, Vk−1)2 > d(vk+1, Vk−1)2 Consequently, the vectors wj and w′ j satisfy δ · ∥w′ j∥2 > ∥wj∥2 = β · ∥w′ j∥2 Haase, Nill, Paffenholz: Lattice Polytopes — 165 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) for some β < δ < 1. Hence, 1/β∥w′ j+1∥2 = ∥wj+1∥2 as for all other i we have w′ i = wi and det Λ = d Y i=1 ∥wi∥= d Y i=1 ∥w′ i∥ is invariant. This implies that det Λ′ j < √ δ det Λj and det Λ′ i = det Λi for i ̸= j , so also D(v′ 1, . . . , v′ d) = √ δD(v1, . . . , vd) < D(v1, . . . , vd) . (6.7) Hence, in each iteration, D(v1, . . . , vd) drops be at least a factor of √ δ < 1, so Algorithm 6.3 runs in polynomial time if D(v1, . . . , vd) has a polynomial lower bound in terms of the dimension. We provide such a bound with the following lemma. this is only true if we give a lower bound to λ1 depending polynomially on the dimension Lemma 6.14 Let Λ ⊂Rd be a lattice with lattice basis v1, . . . , vd and ﬁrst successive minimum λ1 := minu∈Λ{0}(∥u∥) introduced in Deﬁni-tion 4.6. Then D(v1, . . . , vd) ≥λ d(d+1)/2 1 d Y j=1 j−j/2 . Proof. Clearly λ1 ≤minu∈Λj{0}(∥u∥) for all 1 ≤j ≤d. Let v be the shortest non-zero vector in Λ. Then a shortest vector in Λj has length at least ∥v∥. Thus, from Proposition 4.5 we conclude det Λj ≥∥v∥j √ji ≥ λj 1 √jj for all 1 ≤j ≤d . Multiplying this inequality for all 1 ≤j ≤d gives the lemma. ⊓ ⊔ Let us return to the orthogonality defect M deﬁned in Deﬁnition 6.6 before we determine the precise running time of the LLL algorithm. We have seen in Theorem 6.8 that a polynomial upper bound for the orthogonality defect of a lattice basis allows us to solve the shortest vector problem (SVP) in polynomial time. We show now that M of a δ-reduced basis v1, . . . , vd is bounded by Md := 2 1 2 d 2 for δ = 3/4. Hence, we can use this M in deﬁnition Deﬁnition 6.7 and are guaranteed that bases satisfying the deﬁnition actually do exist. Proposition 6.15 Let Λ ⊂Rd be a lattice with reduced basis v1, . . . , vd. Let w1, . . . , wd be its Gram-Schmidt orthogonalization. Then ∥wj∥2 ≤∥vj∥2 ≤∥wj∥2 + 1 4 j−1 X i=1 ∥wi∥2 for 1 ≤j ≤d ∥wj∥2 ≥(δ −1 4)∥wj−1∥2 for 2 ≤j ≤d . — 166 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) Proof. By the deﬁnition of a weakly reduced basis we have to the above vj = wj + j−1 X k=1 λjkwk for coeﬃcients −1/2 ≤λkj ≤1/2. Taking the norm and using that the scalar product of any two of the wi’s is 0 implies ∥vj∥2 = ∥wj∥2 + j−1 X k=1 λ2 jk∥wk∥2 ≤∥wj∥2 + 1 4 j−1 X k=1 ∥wk∥2 . This proves the ﬁrst inequality. Further, we have d(vj, Vj−1)2 = ∥wj∥ d(vj+1, Vj−1)2 = ∥wj+1∥2 + λ2 j+1,j∥wj∥2 ≤∥wj+1∥2 + 1 4∥wj∥2 . By assumption d(vj+1, Vj−1)2 ≥δ · d(vj, Vj−1)2 so ∥wj+1∥2 + 1 4∥wj∥≥δ · ∥wj∥2 The second inequality follows. ⊓ ⊔ We deﬁne α := 1 δ−1 4 . Proposition 6.16 Let Λ ⊂Rd be a lattice with δ-reduced basis v1, . . . , vd. Then (1) ∥v1∥≤α d−1 2 λ1 (2) ∥v1∥≤α d−1 4 (det Λ) 1 d Proof. Let e1, . . . , wd be the Gram-Schmidt orthogonalization of v1, . . . , vd. By Proposition 6.15 ∥wj+1∥2 ≥(δ −1 4)∥wj∥2, so ∥wj∥2 ≤α∥wj+1∥2. By induction we obtain ∥wj∥2 ≤α∥wk∥2 for 1 ≤j < k ≤d . (6.8) Hence, we obtain for all 1 ≤j ≤d ∥v1∥2 = ∥w1∥2 ≤αj−1∥wj∥2 ≤αd−1∥wj∥2 , (6.9) so ∥v1∥2 ≤αd−1 min j ∥wj∥2 and by Proposition 6.5 Haase, Nill, Paffenholz: Lattice Polytopes — 167 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) ∥v1∥≤α d−1 2 λ −1 . This proves the ﬁrst item. Taking the product of (6.9) for all j gives ∥v1∥2d ≤ d Y i=1 αi−1∥wi∥2 = α d(d−1) 2 ∥w1∥2 · · · ∥wd∥2 = α d(d−1) 2 (det Λ)2 . ⊓ ⊔ For the following estimate we spacialize to the case δ = 3 4, so α = 2. Corollary 6.17 Let Λ ⊂Rd be a lattice with 3/4-reduced basis v1, . . . , vd. Then d Y i=1 ∥vi∥≤2 1 2 d 2 det Λ . Proof. We compute ∥vj∥2 ≤∥wj∥2 + 1 4 j−1 X k=1 ∥wk∥2 ≤∥wj∥2 1 + 1 4 j−1 X k=1 2j−k ! ≤2j−1∥wj∥2 and d Y j=1 ∥vj∥≤ d Y j=1 2j−1∥wj∥2 = 2 1 2 d 2 Y j = 1d∥wj∥≤2 1 2 d 2 det Λ . ⊓ ⊔ Corollary 6.18 The orthogonality defect M of a 3/4-reduced basis is at most 2 1 2 d 2 . ⊓ ⊔ Proposition 6.19 Let Λ ⊆Rd be a lattice with a reduced basis v1, . . . , vd. Let λ1 := mina∈Λ−{0} ∥a∥be the ﬁrst successive minimum of the lattice and assume we have a vector x ∈Λ with ∥x∥≤αλ1 for some α ≥1 and x = Pd i=1 ηivi. Then \|ηi\| ≤2 d−1 2 3 2 d−i α ≤3dα for 1 ≤i ≤d . what is this proposition good for? Proof. 10proof missing ⊓ ⊔ 10 proof missing Now let us return to the problem of determining the runtime of the LLL algorithm. For simplicity we assume in the following that the initial lattice basis satisﬁes vj ∈Zd. In this case we have D := D(v1, . . . , vd) ≥1 . (6.10) — 168 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) By (6.7), the value of D drops by at least a factor of √ δ each time we swap a pair of basis vectors that violate condition Deﬁnition 6.10(1), so we need at most k = ⌈ 1 log 1/ √ δ log D⌉swaps. Now ∥wj∥≤∥vj∥for 1 ≤j ≤d, so we can bound the initial value of D by D ≤ max j ∥vj∥ d(d+1)/2 , and we can bound the number of iterations by k = 1 log 1/ √ δ d(d + 1) 2 log max j ∥vj∥ . (6.11) The input size of the LLL algorithm is bounded from below by I := ⟨v1, . . . , vd⟩≤max d, max j ∥vj∥ , so the right hand side of (6.11) is polynomial in I for δ < 1. More precisely, we habe that the number of swaps we have to perform in the worst case satisﬁes k = O d2(log d + s) where s is the largest binary encoding length of an entry of a vector in the lattice basis. After each swap we have to restore the property that our lattice basis is weakly reduced. By Proposition 6.13 this takes at most O (d3) steps, so overall, the lattice beasis reduction needs at most O d5(log d + s) steps. To show that LLL actually runs in polynomial time we also have to show that all numbers computed in intermediate steps of the algorithm have a binary encoding size bounded by a polynomial in I. prove this! Remark 6.20 The additional assumption that all lattice basis vectors all have integral entries can be removed. We just scale the initial basis with the common denominator of all entries of the basis vectors. how much do we loose by this? Make this precise? Remark 6.21 In the form we have discussed the LLL algorithm needs at most O d5(log d + s) steps on numbers of size O (d + s). This cor-responds to the original form of the algorithm given by Arjen Lenstra, Hendrik Lenstra and László Lovász . There habe been found several improvements that run signiﬁcantly faster, for example these improvements need more expla-nation (1) Claus-P. Schorr has improved the bound to O (d4(log d + s)) (2) Arne Storjohann has shown that we can compute a reduced basis in time O (n3(log d + s)) on integers of the same size . should we discuss this in more detail? Haase, Nill, Paffenholz: Lattice Polytopes — 169 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 6.4.1 Applications of LLL include stand (1) factoring polynomials (2) knapsack (3) quadratic equations (4) . . . 6.5 Variations of Barvinok’s Algorithm This subsection is still pretty sketchy... In the previous section we gave a description of the basic version of Barvinok’s algorithm. Since its publication in 1994 there have been found various improvements. We will shortly discuss some of them. A fundamental problem that inﬂuences the running time of the algo-rithm are the lower dimensional cones that we have to consider for the inclusion-exclusion step. In § 3.5.1 we have seen a way to avoid lower dimensional faces by looking at a half-open decomposition of our cone into simplicial cones. Using such a decomposition signiﬁcantly reduces the number of cones we have to consider. Namely, only d instead of d2d new cones in each step of the decomposition. This half open decomposi-tion works nicely also with the signed decompositions we have used in Barvinok’s algorithm. You can work out the details in Exercise 6.1. Exercise 6.1 A slightly diﬀerent idea is the following. We have seen that the rational generating function of cones with nontrivial lineality space are 0. We can exploit this to remove the contribution of lower dimensional cones by ﬁrst dualizing and subdividing on the dual side. Then intersections on the dual side give unions on the primal side, and those vanish in the rational generating function, as thez contain a line. Let us make this precise. If C ⊆Rd is a polyhedral cone, then let S := {Di : i ∈I} be the maximal cones of a subdivision of the dual cone C∨i.e. C∨= [ i∈I Di Then C = \ i∈I D∨ i and on the level of generating series b GC (x) = b GT i∈I D∨ i (x) We can rewrite the right hand side to — 170 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) Algorithm 6.4: Barvinok’s Algorithm: Polarized Version Input: A polyhedron P = {x \| Ax ≤b} with vertices v1, . . . , vk. Output: The integer point generating function for P as GP (t) = X i∈I εi tai (1 −tvi1) · · · (1 −tviki ) for εi ∈{−1, 1}, ai, vij ∈Zd. for i ←1 to k do Compute vertex cone Ci at vertex vi; Dualize the cones to C⋆ i ; Triangulate C⋆ i into ki simplicial cones C⋆ ij; for j ←1 to ki do do a signed decomposition of C⋆ ij into unimodular cones (Ck ij) ⋆.; Dualize back to Ck ij; compute the unique interior point ak ij in Ck ij; end for sum up the contributions using the inclusion-exclusion principle; end for sum up the contributions using the inclusion-exclusion principle; b GT i∈I D∨ i (x) = X b GD∨ i (x) + X J⊆I δJ b GS j∈J D∨ j (x) for δI ∈{0, ±1}. Observe that any nontrivial union of two dual cones D∨ i and D∨ j for i ̸= j contains a line, as there is a linear hyperplane H spanned by some vector u in primal space (the dual of the dual) that intersects the cones at most in their boundary and has the two cones on diﬀerent sides. Then Ru ∈D∨ j ∪D∨ j . Hence, these serices vanish if we pass to the generating functions via our map Φ, so that GC (x) = X i∈I GD∨ i (x) , where we have lost all contributions except for the dual maximal cones. The following observation shows that this is a decomposition into cones of index 1. You will prove this in Exercise 6.2. Lemma 6.22 Index C = 1 ⇔Index C∗= 1 Exercise 6.2 Hence, with this dualization trick we obtain a decomposition into uni-modular cones where we only have to consider the maximal cones. This is a signiﬁcant improvement over our original version where we had to keep track of all lower dimensional cones introduced by the inclusion-exclusion principle. We summarize the algorithm in Algorithm 6.4. 6.6 The Closest Vector Problem We could skip this and the section on integer programming for the next version Haase, Nill, Paffenholz: Lattice Polytopes — 171 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) 6.7 Integer Programming in Fixed Dimension The integer programming problem is the task to ﬁnd an integral point in a polyhedron given by inequalities that maximizes a linear functional, i.e. ﬁnd, for A ∈Zm×d, b ∈Zm and c ∈Zd, max(⟨c, x ⟩) subject to Ax ≤b x ∈Zd . Diﬀerent from the linear programming problem, where we drop the requirement that x ∈Zd, this problem is known to be in the class NP, but not known to be in P. However, for ﬁxed dimension, we can give a polynomial time algorithm for the integer programming problem. This algorithm is due to Hendrik Lenstra. It is based on ﬂatness and its proof via ellipsoids, together with the computation of reduced bases with the LLL-algorithm. As for ﬂatness, the method has three steps. We ﬁrst consider balls and extend to ellipsoids. Then we use the John ellipsoids to squeeze a polytope P between two ellipsoids and use the method for ellipsoids to ﬁnd a feasible integer point in P or assert that there is none. Flatness is used to slice a polytope with a polynomial number of integral translates of a lattice hyperplane and descend in dimension if we cannot yet decide whether P has an integral point. Let us make this precise and start with the construction for balls in the next proposition. Proposition 6.23 Let Λ be a lattice with basis v1, . . . , vd and Bd := Bd(z) be the unit ball with center z. Then we can ﬁnd, in polynomial time, (1) either a lattice point u ∈Bd ∩Λ, or (2) a lattice functional c ∈Λ⋆with ∥c∥≤2O (2d) such thatBd ∩Λ is covered by at most O (2d) hyperplanes of the form {x : ⟨c, x ⟩= β} . for some β ∈Z. Proof. Assume that v1, . . . , vd is a reduced basis with orthogonality defect bounded by M ≤ d Y j=1 ∥vi∥ ∥wi∥≤2 d(d−1) 2 , where w1, . . . , wd is the associated Gram-Schmidt basis. Reorder the basis such that — 172 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) ∥v1∥≤∥v2∥≤. . . ≤∥vd∥. This may destroy reducedness of the basis, but does not aﬀect the orthogonality defect. We distinguish two cases (1) if ∥vd∥≤1 d, then we consider the representation z = d X i=1 λivi of the center in our basis. The point u = d X i=1 ⌊λi⌋vi is a lattice point and ∥u −z∥≤ d X i=1 ∥vi∥≤d∥vd∥≤1 . Hence u ∈Bd ∩Λ. (2) Otherwise we have ∥vd∥> 1 d. Let H := lin(v1, . . . , vd−1). Then H + vd = H + wd, so Λ ⊆ [ β∈Z H + βvd = [ β∈Z H + βwd . Now normvd ∥wd∥ ≤M ≤2 d(d−1) 2 , so ∥wd∥≥2−d(d−1) 2 ∥vd∥≥1 d2−d(d−1) 2 . Further max β∈Z (H + βvd ∩Bd ̸= ∅) −max β∈Z (H + βvd ∩Bd ̸= ∅) ≤2 . So H + βvd ∩Λ ̸= ∅for at most 2d2 d(d−1) 2 diﬀerent β’s. Let c := wd ∥wd∥2 . Any u ∈Λ can be written as u = µwd + h for h ∈H and µ ∈Z. This implies ⟨c, u ⟩= µ ∈Z . so that c ∈Λ⋆and ∥c∥≤ 1 ∥wd∥≤2O (2d). ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 173 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Algorithm 6.5: Computing an enclosing ellipsoid Input: A polyhedron P = {x \| Ax ≤b} for A ∈Qm×n and b ∈Qm. Output: An ellipsoid E and a vector a such that a + E ⊆P ⊆a + 2dE Compute a ball E0 containing P and translate so that the cetner is in the origin; if 1 2dE0 ⊆P then return E0; else ﬁnd x ∈ 1 2dE0 \ P; ﬁnd halfspace H+ containing P but not x via separation algorithm; let E′ 0 := E0 ∩H+; compute smallest ellipsoid E1 containing E′ 0; ﬁnd aﬃne transformation T0 such that T0E1 is a ball centered at the origin; recursively call this algorithm; end if Geometrically, this proposition tells us that either we can ﬁnd a reduced basis with only short vectors, or there must be a short dual vector. extend statem corollary Corollary 6.24 The proposition is also true if we replace the ball by an ellipsoid E = T Bd +a for an invertible linear transformation T. Proof. Pull back to a ball with T −1 and consider the lattice T −1Λ. ⊓ ⊔ By Theorem A.5 we know that for a convex body K there is an ellipsoid E centered at the origin such that a + E ⊆K ⊆a + dE . However, the proof of this theorem given in the appendix was not con-structive. Note ﬁrst, that for us it is suﬃcient to ﬁnd an approximation of E where we relax the scaling factor on the right hand side. It suﬃces to ﬁnd E such that a + E ⊆K ⊆a + 2dE . For general convex bodies the computation of E is an SDP-Problem. However, in our case the convex body is a polyhedron and the algorithm becomes simpler. It is the same algorithm that is used for the ellipsoid method of linear programming. We sketch an approach below. For a full algorithm with proof the reader sould consult any book on linear programming, e.g. [51, Ch. 13]. The algorithm for the ellipsoid is given in Algorithm 6.5. It can be shown that in each recursive call the volume of the ellipsoid drops by a factor of at least (1 −1 d). Hence, the algorithm terminates after a polynomial number of steps. — 174 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) Proposition 6.25 For a polytope P ⊆Rd we can compute in polynomial time an ellipsoid E and a vector a such that a + E ⊆K ⊆a + 2dE . ⊓ ⊔ This proposition allows us to extend Corollary 6.24 to an algorithm that computes an integer point in a polytope or asserts that the polytope contains no integer point. Theorem 6.26 (Lenstra) For a polytope P we can decide, for ﬁxed dimension in polynomial time in the input size, whether P contains an integer point. Proof (sketch). The algorithm is given in Algorithm 6.6. Its correctness follows from the considerations above. It remains to check the running time. For this, let T(d) denote the number of recursive calls in dimension d. Then the total running time is T(d) times a polynomial in d. By the previous corollary on ellipsoids we know that T(d) ≤T(d −1) · d · 2 d(d−1) 2 + 1 , where the ﬁrst d comes from the ﬂatness theorem and the rest from the bound given for ellipsoids. This implies T(d) ≤ d Y k=1 k · 2 k(k−1) 2 + 1 ≤2O (d3) . ⊓ ⊔ 6.8 Software We give examples for the computation of short generating functions in LattE and polymake. 6.9 Problems included on page 170 6.1. Show that we can use a half open decomposition in Barvinok’s algorithm. included on page 171 6.2. Prove Lemma 6.22. For a solution see page 225 Haase, Nill, Paffenholz: Lattice Polytopes — 175 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Algorithm 6.6: Integer Feasibility in ﬁxed dimension Input: A polyhedron P = {x \| Ax ≤b} for A ∈Qm×n and b ∈Qm. Output: An integer point in P ∩Zd, or the assertion that P ∩Zd is empty. Compute an ellipsoid E such that z + E ⊆P ⊆z + 2dE; Use the LLL-algorithm to compute a lattice basis v1, . . . , vd with Gram-Schmidt orthogonalization w1, . . . , wd such that the orthogonality defect with respect to the norm ∥.∥E is bounded.; reorder the basis by increasing norm; Determine λ1, . . . , λd such that z = Pd i=1 λivi and let u = Pd i=1⌊λi⌋vi; if u ∈P then return u; end if Set c := wd; foreach β ∈{⌊min(⟨c, x ⟩: x ∈P)⌋, . . . , ⌈max(⟨c, x ⟩: x ∈P)⌉} do use the Hermite Normal Form algorithm to compute a lattice basis V ′ := {v′ 1, . . . , v′ d−1} ⊆Qd and d ∈Qd such that d + Λ(V ′) = {x ∈Zd : ⟨c, x ⟩= β}; recursively solve the integer feasibility problem on x′ : A(d + P x′ jv′ j) ≤b ; if recursive call produces a lattice point u then return u; end if end foreach return no lattice point found; — 176 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 6. Short Rational Generating Functions (draft of October 7, 2020) Would be nice to insert section on polar sets here! Haase, Nill, Paffenholz: Lattice Polytopes — 177 — Reﬂexive and Gorenstein polytopes 7 Contents 7.1 Reﬂexive polytopes . . . . . . . . . . . . . . . . . . . . . . 180 7.1.1 Dimension 2 and the number 12. . . . . . . . . . 182 7.1.2 Dimension 3 and the number 24. . . . . . . . . . 185 7.2 The combinatorics of simplicial reﬂexive polytopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 7.2.1 The maximal number of vertices . . . . . . . . . 187 7.2.2 The free sum construction . . . . . . . . . . . . . . . 188 7.2.3 The addition property . . . . . . . . . . . . . . . . . . 189 7.2.4 Vertices between parallel facets . . . . . . . . . . 189 7.2.5 Special facets . . . . . . . . . . . . . . . . . . . . . . . . . . 191 7.3 Gorenstein polytopes . . . . . . . . . . . . . . . . . . . . 193 7.4 Finiteness of Gorenstein polytopes of given degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 7.5 Classiﬁcation of reﬂexive polytopes . . . . . . . 198 7.5.1 Smooth reﬂexive polytopes . . . . . . . . . . . . . . 198 7.5.2 All reﬂexive polytopes . . . . . . . . . . . . . . . . . . 198 7.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 Reﬂexive polytopes were introduced by Victor Batyrev in the context of mirror symmetry, a fascinating phenomenon in string theory. Their striking feature is that they always appear in dual pairs. Since then these special lattice polytopes have been intensively studied and classiﬁed by mathematicians and physicists alike. By now, all isomorphism classes Lecture Notes Lattice Polytopes (draft of October 7, 2020) of reﬂexive polytopes in dimension 4, nearly half a billion, are known! Despite all these eﬀorts, still many questions remain open. Amazingly, from the viewpoint of Ehrhart theory reﬂexive polytopes (and their slightly more general relatives, Gorenstein polytopes) can be recognized from having a symmetric h⋆-vector. What else is there to discover? Here is a plan for this chapter. We start bydeﬁning reﬂexive polytopes and explore some of their basic features. Next, we present some of their surprising properties in dimensions 2 and 3. In Section 7.2 we explore the combinatorics of reﬂexive polytopes in the more tractable situation, where all facets are simplices. In Section 7.3 we also consider ‘divisors’ of reﬂexive polytopes, called Gorenstein polytopes, and show that this is a natural class of lattice polytopes to work with. Finally, we consider the question how many Gorenstein polytopes exist using results in Ehrhart theory and the geometry of numbers developed in the previous chapters. Verbessern 7.1 Reﬂexive polytopes Hier Verweis auf Kapitel 2 fuer duale Vektorraeume und warum besser als Skalarprodukt. Nimm dazu Beispielk-lasse von Gitterpolytope bei denen alle innere Facettennormalen im Poly-top sind. Dann sieht man dass diese Eigenschaft nicht invariant unter un-modularer Aequivalenz ist, also keine intrinsische Eigenschaft in unserem Sinne. Let us recall the deﬁnition of a dual polytope of a d-dimensional polytope P ⊂Rd from Theorem 2.16.Let P ⊆Rd be a full-dimensional polytope with 0 ∈P. The dual polytope or polar of P is P ⋆:= {α ∈(Rd)⋆: α(x) ≥−1 for all x ∈P} Check Exercises 7.1 to 7.4 for some examples and properties. Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 It is well-known that the vertices of P ⋆correspond one-to-one to facets of P. More precisely, the vertices are the unique inner facet normals evaluating as −1 on facets. Recall from Chapter 2 that for full-dimensional Referenz zu Kapitel 2 fuer die allge-meine Bijektion!!! lattice polytopes there is a canonical choice for the facet normals as a primitive lattice vector in the dual lattice. We will in the following alsways makre ref precise assume that normal vectors are primitive dual lattice vectors. The most Add exercise to properties of polar bodies in general, bipolaritaetssatz important result is the duality theorem (which holds more generally for convex bodies containing the 0 in their interior): P ⋆⋆= P Let P ⊂Rd be a d-dimensional lattice polytope (with respect to Λ = Zd, and 0 ∈int P. Deﬁnition 7.1 Let F(P) be the set of facets of P, F ∈F(P), then there exists a unique primitive inner normal ηF ∈Λ∗and a unique integer cF ∈Z such that ⟨ηF , x ⟩= −cF ∀x ∈F ⟨ηF , x ⟩≥−cF ∀x ∈P — 180 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) In this case, cF is called (signed) integral distance of the origin from the facet F. The origin is in the polytope if cF ≥0. we need a convention for normals and signs of the right hand side! Recall that the lattice distance of a lattice point v from a lattice hyperplane H is given by \|⟨u, x ⟩−⟨u, v ⟩\| (7.1) if the hyperplane H has primitive normal u and x is any point in H. Give reference to previous deﬁnition Deﬁnition 7.2 A polytope P is called reﬂexive, if there exists w ∈ int P ∩Λ such that all facets have lattice distance 1 from w. The second part of the deﬁnition sounds complicate. Why not directly start with the facet normals we have already deﬁned? Equivalently, for any facet F ∈F(P) there exists a lattice point u ∈Λ⋆such that ⟨u, x ⟩= ⟨u, w ⟩+ 1 for any x ∈V(F). Note that in this case u is necessarily primitive, so u = ηF . As the following observation shows, there is no ambiguity about the interior point w. Proposition 7.3 Let P be a reﬂexive polytope with respect to w, then int P ∩Λ = {w} cF + 1 cF no lattice points F ηF ∈Λ∗ Fig. 7.1: Proof of Proposition 7.3. Proof. Let F ∈F(P). By deﬁnition, no lattice point lies strictly be-tween the hyperplanes aﬀ(F) and its parallel hyperplane through w. See Figure 7.1. Therefore, conv(w, F) ∩Λ = {w} ∪(F ∩Λ). Since P = S F∈F(P ) conv(w, F), the statement follows. ⊓ ⊔ Usually, the unique interior lattice points of a reﬂexive polytope is assumed to be the origin. We give the deﬁnition of a reﬂexive polytope in this generality in order to allow reﬂexive polytopes to be invariant under (aﬃne) unimodular transformations. It is also more natural in the study of Gorenstein polytopes, as we will see later. Reﬂexive polytopes were introduced because of their beautiful duality property. As a consequence, here is the promised characterization of reﬂexive polytopes (Exercise 7.5). Exercise 7.5 Proposition 7.4 Let P be a d-dimensional lattice polytope in Rd with 0 ∈int P. Then the following are equivalent: ▶P reﬂexive, ▶P ∗lattice polytope ▶P ∗reﬂexive. ⊓ ⊔ ﬁgure missing in git Fig. 7.2: (Non-)reﬂexive polygons This result is illustrated in Section 7.1. Coming back to Proposi-tion 7.3, the reader will prove in Exercise 7.6 that any lattice polygon with one interior lattice points is also a reﬂexive polygon (see Exercise 7.14 for the complete list). However, this is not true in dimension 3 and higher Haase, Nill, Paffenholz: Lattice Polytopes — 181 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) (Exercise 7.7). Yet, this property suﬃces to compute Ehrhart polynomials of 3-dimensional lattice polytopes (Exercise 7.8). Exercise 7.6 Exercise 7.7 Exercise 7.8 From Theorem 4.25 and Proposition 7.3 we can immediately deduce the ﬁniteness of these polytopes. Corollary 7.5 There exist only ﬁnitely many reﬂexive d-polytopes up to isomorphism. ⊓ ⊔ Surprisingly, these ﬁnite sets of polytopes together with their faces ‘cover’ the space of all lattice polytopes. Proposition 7.6 Any lattice polytope is isomorphic to the face of a reﬂexive polytope (of higher dimension). Proof. Any lattice polytope with no interior lattice points is isomorphic to the face of a lattice polytope with interior lattice points. Therefore, we may assume that P ⊆Rd is a d-dimensional lattice polytope with 0 ∈int (P). Let F1, . . . , Fk be the facets of P, so in our notation P = {x ∈Rd : ⟨ηFi, x ⟩≥cFi for i = 1, . . . , k} , for cFi ∈Z≤−1. If ck ≤−2, we deﬁne P ′ as the set of (x, xd+1) ∈Rd+1 satisfying ▶⟨ηFi, x ⟩≥cFi for i = 1, . . . , k −1 ▶⟨ηFk, x ⟩−xd+1 ≥cFk + 1 ▶xd+1 ≥−1 Then P ′ is a lattice polytope with (0, 0) ∈int (P ′) that is combinatorially a wedge over P, see Figure 7.3. All the previous inequalities describe facets of P’. In particular, P ′ has P × {−1} as a facet. Let us deﬁne Fig. 7.3: P and P ′ the invariant f(P) := X F∈F(P ) : cFi<−1 \|cFi\| . Then f(P ′) < f(P). In particular, iterating this construction yields after ﬁnitely many steps a lattice polytope P ′′ such that f(P ′′) = 0, i.e., P ′′ is reﬂexive. ⊓ ⊔ 7.1.1 Dimension 2 and the number 12 We turn to a remarkable result about reﬂexive polygons. Theorem 7.7 The numbers of boundary lattice points of a reﬂexive polygon and its dual add up to 12. — 182 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) At least ﬁve diﬀerent proofs appear in [29, 47]: by exhaustion, by a walk in the space of polygons, using toric varieties, using modular forms, or via relations in SL2(Z). by a reﬂex-We will pursue the walk-in-the-space-of-polygons strategy. It yields a more general version of the 12 for unimodular fans. But this needs some preparation. Two adjacent lattice points on the boundary of a reﬂexive polygon form a lattice basis by Pick’s theorem. The cones these lattice points generate a complete unimodular fan. Lemma 7.8 Let Σ be a complete unimodular fan in R2. Every ray τ ∈Σ with primitive generator v is contained in precisely two 2-cones σ = cone(v, w) and σ′ = cone(v, w′) in Σ. In this situation, there is a unique integer a(τ) such that w + w′ = a(τ)v . Proof. Since w, v form a lattice basis of Z2, we have w′ = k1w + k2v. Since v, w′ form a lattice basis, we deduce k1 = ±1. Hence, k1 = −1 by our assumption on the cones. Therefore, w + w′ = k2v. ⊓ ⊔ Lemma 7.9 Let P be a reﬂexive polygon, and let v1, v2, v3 be consecutive lattice points on the boundary of P with v1 + v3 = av2, for a ∈Z. If v2 is a vertex of P, then the edge of P ⋆dual to v2 has length 2 −a. (If v2 is not a vertex, then a = 2.) Fig. 7.5: The dual edge has length 2 −a Proof. Let {v⋆ 1, v⋆ 2} be the basis dual to the basis {v1, v2}. Then the vertices of P ⋆dual to the edges at v2 containing v1 and v3 respectively are v⋆ 1 + v⋆ 2 and (a −1)v⋆ 1 + v⋆ 2, respectively. ⊓ ⊔ In light of this lemma, Theorem 7.7 follows from the following theorem. Theorem 7.10 Let Σ be a complete unimodular fan in R2. Then X τ∈Σ (3 −a(τ)) = 12 , (7.2) where a(τ) is the paramter deﬁned in Lemma 7.8. This is the theorem we will prove by walking in the space of fans. Fig. 7.6: Smooth blow-up of a 2-dimensional fan Fig. 7.7: Not a smooth blow-up Proof (of Theorem 7.7). Let P be a reﬂexive polygon with vertex set V and boundary points B. Then B is the set of generators of a uniomdular complet fan, and we get 12 = X u∈B 3 −a(u) = X u∈B 1 + X u∈V 2 −a(u) = \|∂P ∩Λ\| + \|∂P ⋆∩Λ⋆\| ⊓ ⊔ Here are the steps in our walk. Haase, Nill, Paffenholz: Lattice Polytopes — 183 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Deﬁnition 7.11 Let Σ be a unimodular fan in R2, and let σ ∈Σ with primitive generators v1, v2. Set ϱ := cone(v1 + v2), and pull(Σ; ϱ) := Σ {σ} ∪{ϱ, cone(ϱ, v1), cone(ϱ, v2)} . We say that the fan pull(Σ; ϱ) is obtained as a smooth blow-up of σ in Σ. The deﬁning property of such a smooth blow-up is the fact that the new ray ϱ has ray parameter a(ϱ) = 1. As the reader can verify in Exercise 7.9, these steps preserve the validity of equation (7.2). You will classify 2-dimensional fans which are minimal with respect to blow ups in Exercise 7.10. You will consider the 3-dimensional setting in Exercise 7.11. Exercise 7.9 Exercise 7.10 Exercise 7.11 Lemma 7.12 If the complete unimodular fan Σ in R2 satisﬁes (7.2), and Σ′ is a smooth blow-up of Σ, then Σ′ also satisﬁes (7.2). It remains to show that these steps connect the space of fans. Theorem 7.13 Let Σ and Σ′ be two complete unimodular fans in R2. Then there is another complete unimodular fan Σ′′ which can be obtained by a sequence of smooth blow-ups from both Σ and Σ′. The corresponding statement in general dimension has been conjectured by Oda. This Strong Oda Conjecture is still wide open. This question is stronger than the question of connectivity of the space of complete unimodular fans, where we can have intermediate blow-donws. This weaker statement has been shown in all dimensions. It was the foundation of the celebrated Weak Factorization Theorem. REFERENCES - LATER EINFUEGEN For the proof of Theorem 7.13 we need three lemmas. Fig. 7.8: Diﬀerent λ’s Lemma 7.14 In the notation of Lemma 7.8 let P := conv(0, v, w, w′). (1) 0 is a vertex of conv(0, w, v, w′) if and only if a(τ) ≤1, but (2) v is a vertex of conv(0, w, v, w′) if and only if a(τ) ≤1. Proof. 0 is a vertex if and only if 1 2(w + w′) = a 2v for some a 2 > 0, so a ≥1. Similarly, v is a vertex if and only if a 2 < 1, so a ≤1. ⊓ ⊔ Using this fact we can prove a crucial step towards connectivity: Lemma 7.15 Let Σ be a unimodular fan in R2 which reﬁnes a unimod-ular fan Σ′. Then Σ can be obtained from Σ′ by a sequence of smooth blow-ups. Proof. Use induction on r := \| Σ \ Σ′ \|. If r = 0, we have Σ = Σ′. If r ≥1, all ray parameters of rays of Σ that do not belong to Σ′ must be ≥1 by Lemma 7.14(1). On the other hand, if σ ∈Σ′ contains — 184 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) rays of Σ in the interior, then the convex hull of the primitive generators has a vertex in the interior of σ. The corresponding ray falls into case (2) of Lemma 7.14 and hence must have parameter = 1. ⊓ ⊔ Fig. 7.9: Unimodularly subdividing a 2-cone Lemma 7.16 Let σ ⊂R2 be a pointed 2-cone. Then there is a unimod-ular fan Σ with support σ. We will prove the corresponding statement in general dimension in Chap-ter 8. Proof. Consider the polyhedron P := conv(σ ∩Z2 \ {0}). The bounded segments of P generate cones which form a fan with support σ. For any such segment, the triangle it forms with 0 does not contain any other lattice points. Hence, it is unimodular by Pick’s theorem. ⊓ ⊔ Proof (Theorem 7.13). The collection Σ := {σ ∩σ′ : σ ∈Σ, σ′ ∈Σ′} is a complete fan. By Lemma 7.16, there is a complete unimodular fan Σ′′ reﬁning Σ. Now, we use the previous lemma. ⊓ ⊔ Putting it all together, Lemma 7.12 and Theorem 7.13 imply Theo-rem 7.10. 7.1.2 Dimension 3 and the number 24 In dimension three, there is a possibly even more striking result. Theorem 7.17 If P is a 3-dimensional reﬂexive polytope, then X e edge of P length e · length e⋆= 24 . This result was ﬁrst proved by Dimitrios Dais as follows. By , a general anticanonical hypersurface Z in the toric variety associated with P must be a 2-dimensional Calabi–Yau, i.e., a K3-surface which has Euler characteristic χ(Z) = 24. By , the above sum computes χ(Z). For about a decade, this remained the only proof (apart from exhaustion). We will provide an elementary proof in the present section. Sadly, the story does not continue in dimensions ≥4. But in dimension three, we can carry out a similar program as we did in dimension two. First, we describe parameters for unimodular fans which will replace dual edge lengths (Exercise 7.12). Exercise 7.12 Lemma 7.18 Let Σ be a complete unimodular fan in Rd. Every (d −1)-cone τ ∈Σ[d −1] with primitive generators v1, . . . , vd−1 is contained in precisely two d-cones σ = cone(τ, vd) and σ′ = cone(τ, v′ d) in Σ. In this situation, there are unique integers a(τ, vi) so that vd + v′ d = d−1 X i=1 a(τ, vi)vi . ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 185 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Next, we construct a complete unimodular fan from a reﬂexive 3-polytope. Proposition 7.19 Let P be a 3-dimensional reﬂexive polytope, and let T be a full lattice triangulation of its boundary. Then Σ := {cone σ : σ ∈T} is a complete unimodular fan. Further, if conv(v1, v2) ∈T is contained in an edge e of P, then the dual edge e⋆of P ⋆has length 2 −a(τ, v1) −a(τ, v2) where τ = cone(v1, v2) ∈Σ. (Otherwise a(τ, v1) + a(τ, v2) = 2.) Proof. For Σ, we only need to prove unimodularity. Every triangle σ in a full triangulation is unimodular in its aﬃne span by Pick’s theorem. Because P is reﬂexive, this aﬃne span has distance one to the origin, so that conv(0, σ) is unimodular. In order to compute the length of e⋆, consider the two 3-cones σ = cone(τ, v3) and σ′ = cone(τ, v′ 3) of Σ containing τ. By deﬁnition of the parameters we have v′ 3 = −v3 + a(τ, v1)v1 + a(τ, v2)v2 . As in dimension two, let {v⋆ 1, v⋆ 2, v⋆ 3} be the basis dual to the basis {v1, v2, v3}. Then the vertices of P ⋆dual to the facets of P containing {v1, v2, v3} and {v1, v2, v′ 3} are v⋆ 1 + v⋆ 2 + v⋆ 3 and v⋆ 1 + v⋆ 2 + (a(τ, v1) + a(τ, v2) −1)v⋆ 3, respectively. ⊓ ⊔ Thus, Theorem 7.17 follows from the following fan version. Theorem 7.20 Let Σ be a complete unimodular fan in R3. Then X τ∈Σ with primitive generators v1,v2 (2 −a(τ, v1) −a(τ, v2)) = 24 . We could, again, prove this theorem using a walk in the space of fans. The invariance under smooth blow-ups is elementary. But connectivity of the space of fans is a deep theorem, way beyond the scope of these notes. Luckily, one can deduce the 24 from the 12 by double counting. Where is the star deﬁned? Lemma 7.21 Let Σ be a complete unimodular fan in R3, and let ϱ ∈ Σ with primitive generator v. Then the projection π : R3 →R3/Rv maps star(ϱ; Σ) to a complete unimodular fan Σ /ϱ. If τ ∈star(ϱ; Σ) is a 2-cone with primitive generators v, v′, then the corresponding ray π(τ) of Σ /ϱ has parameter a(π(τ)) = a(τ, v′). Fig. 7.10: Quotient fan Σ /ϱ Proof. Let v1 and v2 be the additional primitive generators of the 3 cones containing τ. Then v1 + v2 = a(τ, v)v + a(τ, v′)v′. Applying π yields π(v1) + π(v2) = a(τ, v′)π(v′). ⊓ ⊔ — 186 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) Proof (Theorem 7.20). Let us ﬁrst collect what we need. Our fan Σ gives rise to a triangulation of the 2-sphere with vertex set Σ, edge set Σ, and triangle set Σ. As such, we have 3\| Σ\| −\| Σ\| = 6 from Euler’s formula and double counting of edge-triangle-incidences. Also, we have P v∈Σ deg v = 2\| Σ\|, where deg v denotes the number of edges containing the vertex v. Armed with these formulas we compute X τ∈Σ with primitive generators v1,v2 (2 −a(τ, v1) −a(τ, v2)) = X cone(v)∈Σ X τ∈Σ τ=cone(v,w) (1 −a(τ, w)) = X cone(v)∈Σ X τ∈Σ τ=cone(v,w) ((3 −a(τ, w)) −2) = X cone(v)∈Σ 12 −2 deg v = 12\| Σ\| −4\| Σ\| = 24 . Here we have used Theorem 7.10 for the quotient fans Σ /ϱ in the third equality. ⊓ ⊔ 7.2 The combinatorics of simplicial reﬂexive polytopes Fig. 7.11: a reﬂexive 3-polytope Throughout, let P ⊂Rd be a d-dimensional reﬂexive polytope with the origin of the lattice Λ = Zd in its interior. 7.2.1 The maximal number of vertices It is a natural question to ask for the maximal number of vertices of a d-dimensional reﬂexive (or Gorenstein) polytope. Let us look at small dimension d ≤4, where the answer is known by the classiﬁcation of Kreuzer and Skarke. H Fig. 7.12: The hexagon Example 7.22 d = 2: \|V(P)\| ≤6, only attained by the reﬂexive hexagon H , see Figure 7.12. d = 3: \|V(P)\| ≤14, only attained by the polytope in Figure 7.11 d = 4: \|V(P)\| ≤36, only attained by H × H . Based upon these observations we state the following daring conjec-ture. Conjecture 7.23 \|V(P)\| ≤6 d 2 , equality holds for d even and P ∼ = H d 2 . Haase, Nill, Paffenholz: Lattice Polytopes — 187 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) This question is still wide open. It has been shown to hold for simple centrally symmetric reﬂexive polytopes, since this class of reﬂexive poly-topes can be completely classiﬁed. In the following we will present some of the techniques used to prove this. 7.2.2 The free sum construction Our main motivation is to determine the maximal number of vertices of a simplicial reﬂexive polytope and to ﬁnd out how the extremal polytopes look like. We will see that free sums of reﬂexive polytopes are reﬂexive, but let us ﬁrst recall the deﬁnition. For two polytopes Pi ⊆Rdi, di ∈Z≥0, with 0 ∈int (Pi) (i = 1, 2) the free sum is the polytope P1 ◦P2 := conv (P1 × {0}, {0} × P2) ⊆Rd1+d2 Example 7.24 42 (1) Let P1 = P2 = [−1, 1] be a segment of length 2 around the origin. Then P1 ◦P2 is the convex hull of ±e1, ±e2. See Figure 7.13. (2) For P1 = H and P2 = [−1, 1] the free sum P1 ◦P2 is BiPyr(H ). See Figure 7.14. Fig. 7.13: The free sum of P1 = P2 = [−1, 1] Fig. 7.14: The free sum of a hexagon and a segment. You will examine the relation between products and sums in Exercise 7.13. It follows directly from the deﬁnition that the free-sum construction is Exercise 7.13 the dual operation to products: (P1 ◦P2)∗= P ∗ 1 × P ∗ 2 . In particular, if P1, P2 are reﬂexive, then P1 ◦P2, P1 × P2 are reﬂexive. There are nice formulas how the Ehrhart- and h∗-polynomials behave under the free sum and product construction. Proposition 7.25 Let P1, P2 be reﬂexive. Then ehrP1×P2(k) = ehrP1(k) ehrP2(k) , h⋆ P1◦P2(t) = h⋆ P1(t)h⋆ P2(t) . While the ﬁrst result follows directly from the construction, the second one is not obvious. We will leave this as an exercise. We will prove it in the last chapter (Corollary 8.27). Here is our main theorem. Its proof will occupy the remainder of this section. Theorem 7.26 Let P be reﬂexive and simplicial. Then \|V(P)\| ≤3d, and equality holds only if d is even and P ∼ = H ◦. . . ◦H \| {z } d 2 . Simplicial reﬂexive d-polytopes with 3d −1 vertices are also completely known. — 188 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) 7.2.3 The addition property Lattice points in reﬂexive polytopes satisfy a partial addition property. For this let us deﬁne a relation. Deﬁnition 7.27 Let x, y ∈∂P ∩Λ, x ̸= y. Then x ∼y, if there exists a facet of P containing x and y. Here is the main observation: Proposition 7.28 Let x, y ∈∂P ∩Λ, x ̸= y. Then (1) either x ∼y (2) or x + y = 0 (3) or x + y ∈∂P ∩Λ. If (3) holds, then x ∼x + y or y ∼x + y. Moreover, there exist a, b ∈Z≥1 such that z := ax + by ∈∂(P) ∩Λ such that x ∼z ∼y. In this case, a = 1 or b = 1. Proof. Assume (1) and (2) do not hold and (3) is wrong. Then duality yields that there exists a facet F ∈F(P) such that −1 > ⟨ηF , x + y⟩∈Z. Hence, −2 ≥⟨ηF , x + y⟩= ⟨ηF , x⟩+ ⟨ηF , y⟩where ⟨ηF , x⟩, ⟨ηF , y⟩≥1. This would imply x, y ∈F, a contradiction. Now, let F ∈F(P) such that −1 = ⟨ηF , x + y⟩= ⟨ηF , x⟩+ ⟨ηF , y⟩. Since ⟨ηF , x⟩, ⟨ηF , y⟩∈Z≥−1, we get either x ∈F and ⟨ηF , y⟩= 0 or y ∈F, ⟨ηF , x⟩= 0. Let us assume the ﬁrst case. We consider the pair x + y, y. If x + y ∼y, we are done. So assume not. Then we get (x + y) + y ∈∂P ∩Λ. Hence, since ⟨ηF , y ⟩= 0, we still have x + 2y ∈F. Now, we consider the pair x + 2y, y. Since \|P ∩Λ\| < ∞, we cannot repeat this argument ad inﬁmum, so there has to exist some b ∈Z≥1 such that x + by ∼y. ⊓ ⊔ −1 0 y x F x + y x + 2y . . . x + by You can use this in Exercise 7.14 to construct the complete list of 16 reﬂexive polygons. Exercise 7.14 Note that even if x, y are vertices, z does not have to be a vertex again, if the dimension of P is larger than two. This result has many applications. As an immediate result we deduce the following constraints on the combinatorics of a simplicial reﬂexive polytope (Exercise 7.15). Exercise 7.15 Corollary 7.29 The diameter of the vertex-edge graph of a simplicial reﬂexive polytope is at most three. 7.2.4 Vertices between parallel facets In this section, we use the partial addition of lattice points to deduce the precise form of vertices that lie between two parallel facets of a simplicial reﬂexive polytope. Haase, Nill, Paffenholz: Lattice Polytopes — 189 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Let us ﬁx a facet F of a simplicial reﬂexive d-polytope P. We denote the vertices of F by b1, . . . , bd. Let Fi ∈F(P) such that Fi ∩F = conv(b1, . . . , bi−1, bi+1, , . . . , bd) . Then there exists a unique mi ∈V(P) such that V(Fi) = {b1, . . . , mi, . . . , bd} for i = 1, . . . , d. Note that b1, . . . , bd is in general not a lattice basis. We can still deﬁne the dual (vector space) basis b∗ 1, . . . , b∗ d ∈(Rd)∗. These are in general no lattice points. The next lemma shows in particular, that there are at most d vertices which lie on the adjacent parallel hyperplane to a facet. Lemma 7.30 Let vv be a vertex of a simplicial reﬂexive polytope with ⟨ηF , v⟩= 0. (1) For 1 ≤d v ∈Fi ⇐ ⇒ v = mi ⇐ ⇒ ⟨b∗ i , v⟩< 0. In particular, there is always such a facet Fi containing v. (2) If ⟨b∗ i , mi⟩= −1 and ⟨ηF , mi⟩= 0 for all i = 1, . . . , d, then b1, . . . , bd is a lattice basis. Proof. i ∈{1, . . . , d}. Let αi := −1 −⟨ηF , mi⟩ ⟨b∗ i , mi⟩ , where ⟨b∗ i , mi⟩< 0 since 0 ∈int(P). Since ⟨ηF , mi⟩≥0, we have αi > 0. We claim that ηFi = ηF + αib∗ i . It suﬃces to check this equality for the vertices of Fi, where the left side always evaluates to −1: j ̸= i: ⟨ηF , bj⟩+ αi⟨b∗ i , bj⟩= −1, ⟨ηF , mi⟩+ αi⟨b∗ i , mi⟩= −1. Now, we can prove (1) and (2). (1) ηF = −b∗ 1 −. . . −b∗ d, ⟨ηF , v⟩= 0 ⇒∃i : ⟨b∗ i , v⟩< 0. Moreover, ⟨b∗ i , v⟩< 0 ⇐ ⇒⟨ηF + αib∗ i , v⟩< 0 ⇐ ⇒⟨ηFi, v⟩< 0 ⇐ ⇒v ∈Fi ⇐ ⇒v = mi. (2) Here: αi = 1 ∀i = 1, . . . , d, hence b∗ i = ηFi −ηF ∈Λ∗∀i = 1, . . . , d. Therefore x = Pd i=1 λibi ∈Λ ⇒λi = ⟨b∗ i , x⟩∈Z ⇒b1, . . . , bd is a lattice basis. ⊓ ⊔ Combining this lemma with the addition property for the lattice points in the dual reﬂexive polytope yields our desired result. — 190 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) Proposition 7.31 v ∈V(P), ⟨ηF , v⟩= 0. If −F ∈F(P), then there are I, J ⊆{1, . . . , d}, I ∩J = ∅, \|I\| = \|J\| such that v = X j∈J bj − X i∈I bi. Proof. We use the notation in the proof of the previous lemma. Let I := {i ∈{1, . . . , d} such that⟨b∗ i , v⟩< 0} ̸= ∅. Then i ∈I (1) = ⇒v = mi. −F ∈F(P) = ⇒Fi ∩F = ∅ = ⇒ηFi ̸∼η−F = −ηF Addition = ⇒ ηFi −ηF ∈∂P ∗∩Λ∗ = ⇒−1 ≤⟨α∗ i b∗ i , ±b∗ i ⟩= ±αi ∈Z αi>0 = ⇒αi = 1 = ⇒⟨b∗ i , v⟩= ⟨ηFi −ηF , v⟩= ⟨ηFi, v⟩ \| {z } =−1 −⟨ηF , v⟩ \| {z } =0 = ⇒⟨b∗ i , v⟩= −1 Using the same argument for −F shows that ⟨b∗ j, v⟩> 0 = ⇒⟨b∗ j, v⟩= 1. ⊓ ⊔ 7.2.5 Special facets We can now prove Theorem 7.26. The key idea is the following notion (due to Øbro). Deﬁnition 7.32 (Special Facet) Let F ∈F(P) such that P v∈V(P ) v ∈ cone(F). Such a facet is called special. From now on, let F be a special facet. Obviously, special facets exist. Let us slice the polytope (for i ∈{−1, 0, 1, . . .}): HP (F, i) := {v ∈V(P) : ⟨ηF , v ⟩= i} ∀i ∈Z≥−1 Clearly, \|HP (F, 0)\| = d . Moreover, Lemma 7.30 yields \|HP (F, 1)\| ≤d. Haase, Nill, Paffenholz: Lattice Polytopes — 191 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) By deﬁnition of a special facet we have the following inequality: 0 ≥⟨ηF , X v∈V(P ) v⟩= X i≥−1 i\|HP (F, i)\| = −d + X i≥1 i\|HP (F, i)\|. (7.3) Hence, we can simply count the vertices: \|V(P)\| = X i≥−1 \|HP (F, i)\| ≤\|HP (F, −1)\| \| {z } =d + \|HP (F, 0)\| \| {z } ≤d + X i≥1 \|HP (F, i)\| \| {z } ≤d ≤3d. (7.4) It remains to consider the equality case (\|V(P)\| = 3d). In this case, equality in (7.3) yields that ⟨ηF , X v∈V(P ) v⟩= 0. Hence, P v∈V(P ) v = 0, so any facet of P is special. Moreover, equalities in (7.3) and show that any vertex of P lies in HP (F, i) for i = −1, 0, 1. Therefore, −ηF ∈P ⋆. So, −P ⋆⊆P ⋆, and thus −P ⋆= P ⋆. In other words, P is centrally symmetric. Since \|V(P) = 3d and \|HP (F, −1)\| = \|HP (F, 1)\| = d, we have \|HP (F, 0)\| = d. Lemma 7.30(1) yields HP (F, 0) = {m1, . . . , md}, as deﬁned in the previous subsection. Let k ∈{1, . . . , k}. By Propo-sition 7.31 there are I, J ⊆{1, . . . , d}, I ∩J = ∅, \|I\| = \|J\| such that mk = P j∈J bj −P i∈I bi. Since all m1, . . . , md are pairwise diﬀerent, Lemma 7.30(1) yields that mk = bjk −bk for some jk ∈{1, . . . , d}, jk ̸= k. In other words, σ : {1, . . . , d} →{1, . . . , d} k 7→jk is a ﬁxed-point free (σ(i) ̸= i) involution (σ2 = 1, jjk = k). We may b2 b1 b1 −b2 Fig. 7.15: The situation for d = 2. assume that this permutation is of the form σ = (1 2)(3 4) · · · (d −1 d). In particular, d is even. Moreover, P = conv(±b1, ±(b1 −b2), ±b2, . . . , ±bd, ±(bd−1 −bd), ±bd). It remains to show that b1, . . . , bd is a lattice basis, since it that case — 192 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) P ∼ = H ◦· · · ◦H \| {z } d 2 . See Figure 7.15. We observe that for any i = 1, . . . , d ⟨b∗ i , mi⟩= ⟨b∗ i , bji −bi⟩= −1 ⟨ηF , mi⟩= 0 Hence, Lemma 7.30(2) ﬁnishes the argument. This proves Theorem 7.26. 7.3 Gorenstein polytopes In this section, we will generalize the deﬁnition and duality of reﬂexive polytopes in a setting which is more natural from the viewpoint of cones as well as from Ehrhart theory. For this, let P be a full-dimensional lattice polytope in Rd. Let Λ = Zd and ¯ Λ = Zd+1. Recall from Chapter 2 that the cone over P (or the homogenization of P) is CP := cone({1} × P) ⊆Rd+1 with dual cone C⋆ P := {u ∈(Rd+1)∗⟨u, x⟩≥0 ∀x ∈CP } The unique primitive inner normals of facets F ∈F(P) are uF = (−cF , ηF ), since ⟨(−cF , ηF ), (x, 1) ⟩= ⟨ηF , x ⟩−cF    = 0 x ∈F ≥0 x ∈P By duality of cones we have the correspondence: facets of P ↔rays of C∨ P F ↔cone(uF ) We are going to denote cones associated to lattice polytopes as Gorenstein cones. Deﬁnition 7.33 Let C ⊆Rd+1 be a (d + 1)-dimensional pointed ratio-nal cone. Then C is called Gorenstein cone, if there exists a d-dimensional lattice polytope P ⊆Rd such that C ∼ = CP . Simpler: A cone is Gorenstein if it is equivalent to a cone over a polytope? TODO-LATER: Erklaeren woher das Wort Gorenstein kommt! Equivalently, C is a Gorenstein cone if and only if there exists a lattice point uC ∈¯ Λ⋆such that ⟨uC, x ⟩= 1 for all primitive generators of the rays of C. In this case, uC is necessarily primitive. Haase, Nill, Paffenholz: Lattice Polytopes — 193 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) See Exercise 7.16 for some properties. Exercise 7.16 Our main result gives a complete characterization of lattice polytopes whose cones have dual Gorenstein cones in terms of Ehrhart theory. The reader may have to recall the deﬁnition of the degree and codegree of a lattice polytopes from Chapter 3. Theorem 7.34 The following are equivalent for a d-dimensional lattice polytope P ⊆Rd of degree s and codegree r: (1) rP reﬂexive (2) ∀k ≥r: int(kP) ∩Λ = w + (k −r)P ∩Λ for some w ∈int(rP) ∩Λ (3) C∨ P Gorenstein cone In this case: uC⋆ P = {r} × w and r is the unique k ∈Z≥1 such that kP is reﬂexive. Proof. Let x′ = (k, x) for x ∈kP, u′ F = (−βF , uF ) for a facet F := {x : ⟨uF , x ⟩≥βF of P, and ¯ Λ := Ze0 × Λ. The cone CP over P has facets F ′ = cone(F) deﬁned by ⟨u′ F , x′ ⟩≥0 for facets F of P. (1) ⇒(2): We prove both inclusions in (2). The direction "⊇" is clear. We prove the other direction. Let w be the unique interior point of rP, so that ⟨u′ F , w′ ⟩= 1 for all facets F of CP . Let x ∈int(kP) ∩Λ for some k ≥r. Then ⟨u′ F , x′ −w′ ⟩= ⟨u′ F , x ⟩ \| {z } ≥1 −⟨u′ F , w′ ⟩ \| {z } =1 ≥0 ∀F ∈F(P) In particular, x′ −w′ ∈(C⋆ P )⋆= CP and ⟨u′ P , x′ −w′ ⟩= k −r Hence, x −w ∈(k −r)P ∩Λ, as desired. (2) ⇒(3): We want to show that ⟨uF , w ⟩= 1 for all facets F of P. For this let us deﬁne C′ := cone(w, F ′). Choose a lattice basis b1, . . . , bd+1 such that b1, . . . , bd ∈cone(F) and bd+1 / ∈cone(F). By translating bd+1 with a multiple of Pd i=1 bd we can assume that bd+1 ∈C′. Therefore, bd+1 ∈int CP ∩¯ Λ. In particular, there exists some k ≥r such that bd+1 ∈int({k} × kP) ∩¯ Λ. By assumption, there exists some m ∈(k −r)P ∩Λ such that bd+1 = w′ + m′ and ⟨b∗ d+1, F ′ ⟩= 0 = ⟨uF , F ′ ⟩ for the dual lattice vector to bd+1. As b∗ d+1 is primitive, we see b∗ d+1 = uF . Therefore, — 194 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) 1 = ⟨uF , bd+1 ⟩= ⟨uF , w ⟩+ ⟨uF , m′ ⟩, hence, ⟨u′ F , w′ ⟩= 1, as desired. rP kP + w (k −r)P (3) ⇒(1): Let w′ ∈int CP ∩¯ Λ such that ⟨u′ F , w′ ⟩= 1 for all facets F of P . Then there exists k ≥r such that w′ ∈int kP ∩¯ Λ. Let F ∈F(P), so, 1 = ⟨w′, u′ F ⟩= −kβF + ⟨uF , w ⟩. Hence, kP is reﬂexive (w.r.t. w). Let us show the additional last statement in the theorem. Assume k > r, then \| int kP ∩¯ Λ\| ≥\|w + (k −r)P ∩¯ Λ\| = \|(k −r)P ∩¯ Λ\| ≥\|P ∩¯ Λ\| > 1 This is a contradiction. ⊓ ⊔ CP (0, 1) P C∨ P u = (1, 2), ⟨u, ·⟩= 3 (−1, 2) (0, 1) P ∨ Fig. 7.17: The cone over P = [−1, 2] and its Gorenstein dual. P = [0, 1]2 Fig. 7.18: The unit square Exercise 7.17 Exercise 7.18 See Exercise 7.17 for an example of a non-Gorenstein cone. This motivates our main deﬁnition. Deﬁnition 7.35 (Gorenstein polytope) P is a Gorenstein polytope of index r if it satisﬁes the conditions of Theorem 7.34. In other words, a lattice polytope P is a Gorenstein polytope if some multiple kP is a reﬂexive polytope. This multiple k is necessarily equal to the codegree by Proposition 7.3. For instance, reﬂexive polytopes are precisely Gorenstein polytopes of codegree 1. You will prove a criterion similar to Proposition 7.28 in Exercise 7.19 Exercise 7.19 Example 7.36 42 (1) See Figure 7.17. C∨ P is not a Gorenstein cone, ⇒P is not a Goren-stein polytope. r = codeg P = 1, int P ∩Λ = 2 > 1; h∗ 0 = 1, h∗ 1 = 2. (2) See Figure 7.18. P = [0, 1]2 is a Gorenstein polytope of codegree r = codeg(P) = 2. (2P is reﬂexive, see Figure 7.19). (3) The Birkhoﬀpolytope Bn is a famous polytope which is deﬁned as the convex hull of all n × n-permutation matrices. It is a Gorenstein polytope of codegree n, see Exercise 7.20. Exercise 7.20 Corollary 7.37 For a Gorenstein polytope of index r we have ehrP (−k) = (−1)dehrP (k −r) ∀k ∈Z Proof. This follows from Theorem 7.34(2). ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 195 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Theorem 7.38 Let P ⊆Rd be a d-dimensional lattice polytope of degree s and codegree r. Then the following three conditions qgree. (1) P is a Gorenstein polytope of index r. (2) b EhrP t−1 = (−1)d+1trb EhrP (t) (3) h⋆ i = h⋆ s−i ∀i = 0, . . . , s 2P (2P −( 1 1 ))∗ Fig. 7.19: Twice the unit square and its dual. Proof. For a Gorenstein polytope of index r we have X k≥1 \| int kP ∩¯ Λ\|tk = X k≥1 (−1)dehrP (−k) tk = X k≥1 ehrP (k −r) tk = X k≥r ehrP (k −r) tk = tr X k≥0 ehrP (k) tk So by Stanley’s Reciprocity Theorem (Theorem 3.42) b EhrP 1 t = (−1)d+1b Ehrint P (t) = (−1)d+1trb EhrP (t) . (7.5) This proves the ﬁrst implication. On the level of rational functions we get Ps i=0 h⋆ i ti (1 −t)d+1 = (−1)d+1 1 tr Ps i=0 h⋆ i 1 t i (1 −1 t )d+1 = (−1)d+1 1 tr · 1 ts Ps i=0 h⋆ i ts−i (1 −1 t )d+1 = Ps i=0 h⋆ i ts−i (1 −t)d+1 = Ps i=0 h⋆ s−iti (1 −t)d+1 , where the ﬁrst equality follows from (7.5). Comparing the coeﬃcients yields (3). Conversely, if (3) holds, then rP w kP (k −r)P + w (k −r)P w Fig. 7.20 (−1)d+1b EhrP 1 t = trb EhrP (t) , and by reciprocity (−1)d+1b EhrP 1 t = b Ehrint P (t) = (−1)d X k≥1 ehrP (−k) tk , while trb EhrP (t) = X k≥0 ehrP (k) tk+r = X k≥r ehrP (k −r) tk . Again comparing coeﬃcients gives ehrP (−k) = (−1)dehrP (k −r) for k ≥r and ehrP (−k) = 0 for 1 ≤k ≤r. Hence ehrP (−k) = (−1)dehrP (k −r) for k ≥1 . ⊓ ⊔ improve ﬁgure! There is a natural duality of Gorenstein polytopes extending the one of reﬂexive polytopes. Since Gorenstein polytopes do not have interior lattice points, if r > 1, we have to use the duality of cones. — 196 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) Proposition 7.39 Let P be a Gorenstein polytope (as in Theorem 7.34). Then P ∨:= {x ∈C⋆ P : ⟨uC⋆ P , x⟩= 1} = conv((−cF , ηF ) : F ∈F(P)) is also a Gorenstein polytope of the same dimension, degree and codegree as P, called the dual Gorenstein polytope. Proof. uC⋆ P = w, hence, ⟨uC⋆ P , uP ⟩= r. Let G ∈F(P ⋆). Then ⟨uG, G ⟩= 0 and ⟨uG, uCP ⟩= 1. Therefore, rP ⋆is reﬂexive. ⊓ ⊔ Note that this duality is quite subtle. For instance, for r > 1, P ∨does not lie in the hyperplane Rd × 1. Thus, it is not intrinsically embedded in Rd. It is merely given as a d-dimensional polytope in Rd+1. Moreover, except for codegree 1, P ∨is not isomorphic to (rP −w)∗, as one might guess at ﬁrst. For instance, in Example 7.36(2) with r = 2, P ∨is just isomorphic to P. See also Exercise 7.21. Exercise 7.21 Let us consider the case of a reﬂexive polytope P, say, 0 ∈int(P). Then we recover the duality of reﬂexive polytopes: P ∨= {x ∈(Rd+1)∗: ⟨x, (y, 1)⟩≥∀y ∈P, ⟨x, (0, 1)⟩= 1} = {(x, 1) : ⟨x, y⟩≥−1 ∀y ∈P} = P ⋆× 1. This gives another proof of Proposition 7.4. 7.4 Finiteness of Gorenstein polytopes of given degree In this section we are going to prove the following result in Ehrhart theory: Theorem 7.40 There exist only ﬁnitely many symmetric h∗-polynomials of lattice polytopes of degree s. Equivalently, the normalized volume of any Gorenstein polytope of degree s is bounded by a number depending only on s. We remark that Theorem 7.40 follows from Theorem 5.11, since for Gorenstein polytopes the leading coeﬃcient h∗ s = h∗ 0 = 1 is ﬁxed! Then, why do we prove Theorem 7.40? The answer is that we can give a complete, self-contained and insightful proof for Gorenstein polytopes without having to rely on Theorem 5.10 whose proof is out of reach of this book. Corollary 5.2 lets us deduce an interesting consequence: Haase, Nill, Paffenholz: Lattice Polytopes — 197 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Corollary 7.41 There exist only ﬁnitely many Gorenstein polytopes of degree s up to isomorphism and lattice pyramid constructions. For s = 1 and s = 2 these ﬁnite lists are completely known. The proof of Theorem 7.40 proceeds in the same way as that of Theorem 5.11. We use Proposition 5.12 to reduce the problem to the existence of Cayley decompositions of Gorenstein polytopes with large dimension and small degree. Next, instead of invoking Theorem 5.10, we can prove this claim directly. P × 1 w w∨ S Fig. 7.21: P and P ∨. Proposition 7.42 If P is a Gorenstein d-polytope of degree s ≥1 and d ≥2s, then P is a Cayley polytope of lattice polytopes in dimension ≤2s −1. Proof. See Figure 7.21. Let us identify P and P × 1. We denote by w∨= uCP and by w = uC⋆ P . Then we can choose a lattice d-simplex S ⊆P ∨such that w∨∈cone(S). We can ﬁnd x1, . . . , xk ∈V(S) such that w∨= k X i=1 xi ! + xk+1 , where xk+1 ∈Π(S). Let us recall d + 1 −s = r = ⟨w, w∨⟩. We consider two cases: If xk+1 = 0, then d + 1 −s = k, so k = d + 1 −s ≥d + 2 −2s. If xk+1 ̸= 0, then we get d + 1 −s = k + ⟨w, xk+1 ⟩≤k + deg(S) ≤ k + deg(P ∨) = k + s by Proposition 3.19, Theorem 3.37, Proposition 7.39. Referenzen checken Therefore, in this case, k + 1 ≥d + 2 −2s. In any case, Lemma 5.9 implies that P ∨is a Cayley polytope of length ≥d + 2 −2s. das mit der Dimension vielleicht ex-plizit vorher erklaeren The proof of Theorem 7.40 follows now from applying Proposition 5.12 with N = 2s −1. 7.5 Classiﬁcation of reﬂexive polytopes 7.5.1 Smooth reﬂexive polytopes Øbro 7.5.2 All reﬂexive polytopes Kreuzer and Skarke 7.6 Problems included on page 180 7.1. Show that for ∅̸= A ⊂Rd (equipped with some scalar product) we always have ((A◦)◦)◦= A. — 198 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) 7.2. Let P be a d-dimensional lattice polytope containing 0 in its interior. Show that V(P ∗) ⊂Qd. included on page 180 7.3. Let T : Rd →Rd be a linear map that sends a d-dimensional polytop P containing 0 in its interoir to Q (i.e. Q = T(P)). What is the relation between Q∗and P ∗? included on page 180 7.4. Show that Sd,1 := conv(e1, . . . , ed, −1) is reﬂexive. included on page 181 7.5. Prove Proposition 7.4. included on page 182 7.6. Show that a lattice polygon is reﬂexive if and only if it contains precisely one interior lattice point. included on page 182 7.7. Take [−1, 1]d, remove one vertex and take the convex hull of the remaining lattice points. Show that it still contains precisely one interior lattice point. Is this a reﬂexive polytope? included on page 182 7.8. Compute h∗-polynomial and Ehrhart polynomial of a 3-dimensional reﬂexive polytope having b many lattice points. included on page 184 7.9. Prove Lemma 7.12. included on page 184 7.10. Classify unimodular fans in R2 which are minimal with respect to blow-ups. included on page 184 7.11. Deﬁne smooth blow-ups for three-dimensional unimodular fans (or higher-dimensional); and show the invariance of the equality in Theorem 7.20 under smooth blow-ups. included on page 185 7.12. Prove Lemma 7.18. included on page 188 7.13. Let P, Q be any polytopes (not necessarily with vertices in the lattice) that contain 0 in the interior. Show that the dual of the product is the free sum of the duals. included on page 189 7.14. The complete list of unimodular equivalence classes of reﬂexive polygons is in Figure 7.22. Reference! (1) Pick your favorite and compute its dual polytope. To which representative in the list is it unimodularly equivalent? Haase, Nill, Paffenholz: Lattice Polytopes — 199 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Fig. 7.22: The 16 reﬂexive polygons. (2) Use Proposition 7.28 to prove that this list is complete. included on pa 7.15. Prove Corollary 7.29, i.e. show that any pair of vertices of a simplicial reﬂexive polytope can be connected by at most three edges. included on pa 7.16. Let C be a (d + 1)-dimensional polyhedral cone. Show that C ∼ = CP if and only if there is a lattice point w in the dual lattice with ⟨w, x ⟩= −1 for all primitive generators of C. Further, in this case w is primitive and in the interior of C∗. included on page 195 7.17. Show explicitely that for P = conv(0, 3) the dual cone C∗ P is not Gorenstein. included on page 195 7.18. Let Λ ⊂Rd be a lattice of rank d. We may choose (why ?) v1, . . . , vd ∈Λ recursively such that vi+1 has the minimal non-zero distance from the subspace ⟨v1, . . . , vi⟩. Show that v1, . . . , vd — 200 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 7. Reﬂexive and Gorenstein polytopes (draft of October 7, 2020) is a lattice basis of Λ. In particular, any primitive vector of Λ can be completed to a lattice basis. included on page 195 7.19. Show that any pair of vertices of a Gorenstein polytope of index r > 1 either is in a common face or is an antipodal pair (with respect to the point w, where w is the unique interior point of rP). included on page 195 7.20. Prove that the Birkhoﬀpolytope Bn (the convex hull of n × n-permutation matrices) is a Gorenstein polytope of codegree n. What is its dimension and degree? What is the unique interior lattice point of nBn ? included on page 197 7.21. Find the dual Gorenstein polytope for the Gorenstein polytope P = [0, 1]2 and show that it is isomorphic to P. Haase, Nill, Paffenholz: Lattice Polytopes — 201 — Unimodular Triangulations 8 Contents 8.1 Regular Triangulations . . . . . . . . . . . . . . . . . . . 204 8.2 Pulling Triangulations . . . . . . . . . . . . . . . . . . . 206 8.3 Compressed Polytopes . . . . . . . . . . . . . . . . . . . 208 8.4 Special Simplices in Gorenstein Polytopes . 210 8.5 Dilations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 8.5.1 Composite Volume . . . . . . . . . . . . . . . . . . . . . 215 8.5.2 Prime Volume . . . . . . . . . . . . . . . . . . . . . . . . . 216 8.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 ▶unimodular cover ▶integer Carathéodory + BGHMW ▶IDP ▶vectors u, v, w; covectors a, b, c; scalars α, β, γ ▶\psubdiv S ▶\regsubdiv{\vw}{V} Sw(V ) ▶\reglift{\vw}{V} lift(w) ▶\regfunction{\vw}{V} Ψw ▶\pull{\psubdiv}{\vv} pull(S; v) It this chapter, we study under which assumptions a lattice polytope admits a triangulation into unimodular simplices. Such unimodular trian-gulations of lattice polytopes arise in algebraic geometry, commutative Lecture Notes Lattice Polytopes (draft of October 7, 2020) algebra, integer programming and, of course, combinatorics. Because of the nice implications, having a unimodular triangulation is a desirable property. But presumably, “most” lattice polytopes do not admit a uni-modular triangulation. (Yet, there is no theorem to date that would pin down what the previous statement actually means.) We will deﬁne regular triangulations, and prove our ﬁrst theorem for cones in Section 8.2. Then, we deﬁne the particularly nice class of compressed polytopes in Section 8.3, and give a few examples of important triangulations. Section 8.4 can be regarded as a capstone section of these notes. We use unimodular triangulations to prove unimodality of the h⋆ coeﬃcients of Gorenstein polytopes with regular unimodular triangulation, tying together ideas from Chapters 3 and 7 with this one. Finally, in Section 8.5, we prove a mysterious theorem from the early days of toric geometry which, to this day, raises more questions than it answers. Some further examples and motivation for (regular) unimodular triangulations are hidden in the exercises. 8.1 Regular Triangulations Regular subdivisions form a well-behaved class of subdivisions of a given point conﬁguration. Regular subdivisions are partially ordered by re-ﬁnement (see Deﬁnition 8.2). Hence, they form a poset. This poset is isomorphic to the face lattice of the so called secondary polytope. This was proved by Gel′fand, Kapranov, and Zelevinsky , see also . The proof is beyond the scope of these notes. The maximal elements in the poset are the regular triangulations, which in particular proves that every point conﬁguration has a triangulation (see Proposition 8.7 in the next section). In this and the following section we will work with point conﬁgu-rations. We may think of A = P ∩Zd, but the results with workmore generally for any point set and produce a subdivision of its convex hull. Though we will use this only at one place, the whole theory easlily carries over to vector conﬁgurations and subdivisions of fans. alles ausformulieren There is a close connection to linear and integer programming as well as to Gröbner bases of toric ideals. We will not cover this here, but the interested reader can ﬁnd this in or . Deﬁnition 8.1 (regular subdivision) A subdivision S of the point conﬁguration A is regular if there is a weight vector ω ∈RA such that S is the projection of the lower hull of the polyhedron lift(ω) := conv(v × [ωv, ∞) : v ∈A ) — 204 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 8. Unimodular Triangulations (draft of October 7, 2020) in Rd+1. We write Sω(A ) or Sω(P) for the regular subdivision induced by ω. A subdivision is a triangulation if all cells are simplices. Here, the lower hull is the polyhedral complex of those facets whose normal has negative ﬁrst coordinate. The faces of Sω(A ) are the domains of linearity of the function Ψω : P →R given by v 7→min h : (v, h) ∈lift(ω) . Less formally, a regular subdivision can be thought of as a subdivision that can be realized as a “convex folding” of the polytope (Figure 8.1 on the left). All three triangulations in Figure 8.2 are regular while the triangulation on the right in Figure 8.1 is not. Fig. 8.1: A regular triangulation via folding Fig. 8.2: Regular triangulations Deﬁnition 8.2 Reﬁnement of a subdivision. Lemma 8.3 Given a point conﬁguration A ⊂Rd and a weight vector ω ∈RA a set F is a face of Sω(A ) if and only if there is a functional aF ∈(Rd)⋆and an αF ∈R such that for all v ∈A ⟨aF , v ⟩+ ωv ≥αF and F is the convex hull of those v that attain equality. Proof. This is a direct translation of the deﬁnition. ⊓ ⊔ Lemma 8.4 Given a point conﬁguration A ⊂Rd and a weight vector ω ∈RA deﬁne ω′ ∈RA by ω′ v := Ψω(v). Then lift(ω) = lift(ω′) (and hence, Sω(A ) = Sω′(A ) as well as Ψω = Ψω′), and for all F ∈Sω′(A ) and all v ∈A we have v ∈F ⇐ ⇒⟨aF , v ⟩+ ω′ v = αF , (8.1) where we use the certiﬁcates (aF , αF ) from Lemma 8.3. We will call a pair (Sω′(A ), ω′) satisfying (8.1) for all F ∈Sω′(A ) tight. If all v ∈A are vertices of Sω(A ), then (Sω(A ), ω) is automatically tight. Fig. 8.3: A lift of a regular triangula-tion. Fig. 8.4: A non-regular triangulation. Lemma 8.5 Let Sω(A ) and Sω′(A ) be two regular subdivisions of the same point set A . For ε > 0 small enough Sω−εω′(A ) is a regular reﬁnement of Sω(A ). In particular, if Sω′(A ) is a triangulation, then Sω−εω′(A ) is one too. Proof. If ⟨aF , a ⟩+ ω(a) −αF > 0 Haase, Nill, Paffenholz: Lattice Polytopes — 205 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) for some face F of Sω(A ) and a ∈A , then also ⟨aF , a ⟩+ (ω −εω′)(a) −αF > 0 for ε small enough. Thus, in Sω(A ), the weight function at most subdi-vides within faces. Hence, ω −εω′ produces is a reﬁnement of Sω(A ). Clearly, if ω′ produces a triangulation of A , then this is true also for the restriction to all faces of Sω(A ). ⊓ ⊔ 8.2 Pulling Triangulations Pulling reﬁnements are a useful tool for constructing regular triangula-tions. Deﬁnition 8.6 (Pulling Reﬁnements) Consider a point conﬁgura-tion A ⊂Rd and s subdivision S of P = conv(A ). For u ∈A we obtain the pulling reﬁnement pull(S; u) when we replace every face F ∈S which contains u by the pyramids conv(u, F ′) where F ′ runs over all faces F ′ ⪯F which do not contain u. The deﬁnition works mutatis mutandis for vector conﬁgurations and subdivisions of fans. Fig. 8.5: A pulling reﬁnement. EXAMPLES HERE Here are some facts about the structure of pulling subdivisions. We say that a vertex v of a polytope P is an apex of P if all other vertices are contained in a facet of P. Proposition 8.7 Let A be a point conﬁguration in Rd and S a subdi-vision of A (not necessarily regular). (1) Pulling preserves regularity. (2) Pulling all points in A in some order results in a triangulation whose vertex set is all of A . (3) If only vertices of P are pulled, then every maximal cell is the join of the ﬁrst pulled vertex v1 with a maximal cell in the pulling subdivisions of the facets not containing v1. In particular, we see that every (regular) lattice subdivision of a lattice polytope has a (regular) reﬁnement which is a full triangulation. Part (3) identiﬁes the triangulation constructed back in Theorem 2.35 with the triangulation obtained by pulling the vertices in the given order. Proof. (1): Let S = Sω(A ) be a regular subdivision of P = conv(A ) induced by tight weights ω ∈RA . Let u ∈A . Set ω′ u := ωu −ε and ω′ v := ωv for all v ∈A \ {u}. We claim that then, for small enough ε > 0, the pulling reﬁnement pull(S; u) is induced by the weights ω′. exercise: make example where ω′ is no longer tight. To prove the claim, we will show that every face of pull(S; u) lifts to a lower face of lift(ω′). Then we show that these faces cover P, and hence, all lower faces of lift(ω′) project to faces of pull(S; u). — 206 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 8. Unimodular Triangulations (draft of October 7, 2020) A face F of pull(S; u) is either ▶a face of S not containing u or ▶it is of the form conv(u, F ′) where F ′ ⪯G ∈S with u ∈G \ F ′. In the former case we have ω\|F = ω′\|F so that lift(ω\|F ) = lift(ω′\|F ) which, for small enough ϵ, still is a lower face of lift(ω′). (This imposes ﬁnitely many upper bounds on ϵ.) In the latter case, there are certiﬁcates (aF ′, αF ′) and (aG, αG) ∈(Rd)⋆× R . For the new weight ω′ (and ε small enough) we have ⟨aF ′, u ⟩+ ω′ u −αF ′ > 0 and (8.2) ⟨aG, u ⟩+ ω′ u −αG < 0 . (8.3) Thus, there are coeﬃcients λ, µ > 0 with λ + µ = 1 such that λ ⟨aF ′, u ⟩+ ω′ u −αF ′ + µ ⟨aG, u ⟩+ ω′ u −αG = 0 . Then λ(aF ′, αF ′) + µ(aG, αG) is a certiﬁcate for the face F := conv(F ′, u) of Sω′(A ): for all v ∈A we have λ ⟨aF ′, v ⟩+ ω′ v −αF ′ + µ ⟨aG, v ⟩+ ω′ v −αG ≥0 with equality for v ∈F ′ and for v = u, while for v ̸∈F ′ and v ̸= u we have a stricly positive summand. To ﬁnish the proof of our claim, we take a point v ∈P and show that it belongs to a face of pull(S; u). Let G ∈S be the face which contains v in its relative interior. If u ̸∈G, we have G ∈pull(S; u) and we are home. If u ∈G, consider the ray from u through v. It hits the boundary of G in the relative interior of some face F ′ ⪯G. But then v ∈F := conv(F ′, u) by construction, and F ∈pull(S; u). (2): By deﬁnition, every face of pull(S; u) which contains u is a pyramid with apex u. (In particular, u is a vertex of pull(S; u).) If Q ∈S has v as an apex, then every face of Q containing v has v as an apex. Consequently, every face of pull(S; u) inside Q and containing v still has v as an apex. After pulling all lattice points, all lattice points are vertices of the subdivision, and the cells have each of their vertices as apices. Hence, they are simplices. (3): If we apply the previous argument to the trivial subdivision of P, we see that v1 is an apex of every maximal cell. ⊓ ⊔ Using pulling reﬁnements (the fan version), we get unimodular trian-gulations of cones — this corresponds to resolution of singularities for toric varieties. It works like a charm, in arbitrary characteristic. Haase, Nill, Paffenholz: Lattice Polytopes — 207 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Theorem 8.8 Every rational cone has a regular unimodular triangula-tion. Note that the remarkable claim in this theorem is the fact that we can make the triangulation unimodular. We already know that there are regular triangulations. Proof. Given a rational cone C, we can, using pulling subdivisions on the vector conﬁguration of primitive generators of C, ﬁnd a triangulation T of C into simplicial cones. We will proceed by induction on lexicographically ordered pairs (D, N) where D is the maximal determinant D occuring in the triangulation and N is the number of cones of determinant D. If D = 1 then the triangulation is unimodular. If D > 1, we choose an inclusion-minimal cone F ∈T of determinant D. Let v1, . . . , vk be the primitive generators of F. Pick one of the D −1 non-zero lattice vectors in the fundamental parallelepiped Π(F): u = P λjvj. (We have chosen F minimal so that all λj are positive.) If v1, . . . , vd are the primitive generators of a face G of T containing F, then G is subdivided into cones Gj (j = 1, . . . , k) where the generator vj is replaced by u. As det(Gj) = λj det(G) = λjD < D all the cones in pull(T; u) have determinant < D. ⊓ ⊔ 8.3 Compressed Polytopes The notion of compressed polytopes was coined by Richard Stanley . Surprisingly many well-known polytopes fall into this category. Deﬁnition 8.9 A lattice polytope P ⊂Rd is compressed if all lattice points in P are vertices, and all pulling triangulations are unimodular. Compressed polytopes admit several characterizations. A lattice polytope P has width 1 with respect to a facet F, if it lies between the hyperplane spanned by this facet and the next parallel lattice hyperplane, that is every point of P has lattice distance at most one from F. The main implication of the following Theorem is due to Francisco Santos. The proof we present here is the original one (MSRI 1997, unpub-lished). The result was subsequently also proven by Ohsugi and Hibi and by Sullivant . Theorem 8.10 Let P be a lattice polytope. Then the following is equiv-alent: (1) P is compressed. (2) P has width one with respect to all its facets. (3) P is lattice equivalent to the intersection of a unit cube with an aﬃne space. — 208 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 8. Unimodular Triangulations (draft of October 7, 2020) As a pulling triangulation of P induces a pulling triangulation on all faces of P, faces of compressed polytopes are compressed. In the same way, the facet-width-one property is inherited by faces. Lemma 8.11 If P has width one with respect to all its facets then the same is true for all faces of P. Proof. By induction on the codimension it is enough to consider a facet F of P. If G is a facet of F, there is a facet F ′ of P so that G = F ∩F ′. Denote the primitive inner normal of F ′ to P by η. The restriction of η to aﬀ(F) is an integral inner normal of G to F. (It might, a priori, not be primitive.) But at every point of F ⊂P, η takes values in [⟨η, G ⟩, ⟨η, G ⟩+ 1]. That is, F has width one with respect to G. ⊓ ⊔ Proof (of Theorem 8.10). (2) = ⇒(1): Choose an ordering of the lattice points in P and let T be the corresponding pulling triangulation. The restriction of T to a face of P is the pulling triangulation induced by the restricted ordering of the lattice points. Thus, by induction on the dimension, the triangulations of the facets are unimodular. Using the recursive description of Proposition 8.7(3) we see that every maximal simplex in T is the join of a unimodular simplex S in some facet with the ﬁrst lattice point v1 that was pulled. The facet width assumtion guarantees that v1 is at distance one from S so that conv(S, v1) is also unimodular. The other implications are easy. ⊓ ⊔ exercises Example 8.12 Examples of compressed polytopes include (1) the Birkhoﬀpolytope, (2) order polytopes and hypersimplices, (3) stable set polytopes of perfect graphs. In Buch beschreiben und hier darauf verweisen. We can apply the above characterization of compressed polytopes to triangulate bigger polytopes using hyperplane arrangements. Deﬁnition 8.13 A collection A := {a1, . . . , ar} ⊂(Zd)⋆of functionals that span (Rd)⋆is said to form a unimodular matrix A if all (d × d)–minors of the r × d-matrix with rows ai are either 0, 1 or −1. Such a collection induces an inﬁnite arrangement of hyperplanes {v ∈Rd : ⟨ai, v ⟩= k} for i = 1, . . . , r and k ∈Z , The induced subdivision of Rd is the lattice dicing of Rd by A. The notion lattice dicing was coined by . Deﬁnition 8.14 We call a lattice polytope P whose collection of primi-tive facet normals forms a unimodular matrix facet unimodular. The above hyperplane arrangement slices P into dicing cells. We call this subdivision the canonical subdivision of a facet unimodular polytope. Haase, Nill, Paffenholz: Lattice Polytopes — 209 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Lemma 8.15 Let P ⊆Rd be a facet unimodular polytope. (1) The cells of a lattice dicing are lattice polytopes. exercise: totally unimodular implies integral dicing (2) Canonical subdivisions are regular. proof (3) Every face of a facet unimodular polytope is again facet unimodular in its own aﬃne lattice. Canonical subdivisions induce canonical exercise subdivisions on faces. The proof of this lemma ist left as Exercise 8.1. Exercise 8.1 Theorem 8.16 Suppose that P ⊂Rd is a facet unimodular lattice poly-tope. Then P has a regular unimodular triangulation. Proof. The dicing cells have width one with respect to all their facets by construction. Thus, any pulling reﬁnement of the canonical subdivision will be unimodular. ⊓ ⊔ As a direct application of Theorem 8.16, ﬂow polytopes as well as poly-where to mention these examples ﬁrst? topes with facets in the root system of type A have regular unimodular triangulations. This method also shows that if P has a (regular) unimod-exercise: do the details ular triangulation then so do all its integral dilates cP. You will work out the details in Exercise 8.2. Theorem 8.17 If P has a (regular) unimodular triangulation T then its dilation cP has one too, for every positive integer c. ⊓ ⊔ Exercise 8.2 exercise: prove using dicing — a fact which we will prove with a diﬀerent method in Theorem 8.17 below. 8.4 Special Simplices in Gorenstein Polytopes The goal of this section is to prove the following theorem of Bruns and Römer. Theorem 8.18 The h⋆-vector of a Gorenstein polytope with a regular unimodular triangulation is unimodal. That is, 1 = h⋆ 0 ≤. . . ≤h⋆ ⌊s/2⌋≥ . . . ≥h⋆ s. This theorem and its proof are due to Bruns and Roemer . For general Gorenstein polytopes the theorem fails, as shown by Payne and Mustata . However, it is still open whether the following property exercise: non-unimodal example might suﬃce. Deﬁnition 8.19 A lattice d-polytope P ⊆Rd is said to possess the integer decomposition property (IDP) if for every k ∈Z≥2 and for every lattice point u ∈kP ∩Zd there exist v1, . . . , vk ∈P ∩Zd such that u = v1 + · · · + vk. — 210 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 8. Unimodular Triangulations (draft of October 7, 2020) Equivalently, let C ⊂Rd+1 be the cone spanned by P × {1}. Then P is IDP if and only if the semigroup of lattice points in C is generated by lattice points in P × {1}. As Exercise 8.3 shows, being IDP is weaker than having a unimodular triangulation. Polytopes having the IDP-property are also called integrally closed or sometimes normal. exercises: more notions (very ample, etc.), more examples; much of the hierarchy The central tool in the proof of Theorem 8.18 is the notion of a special simplex. The use of special simplices in this context had been move to Gorenstein Cayley pioneered by Athanasiadis . Note that this notion of a special simplex is diﬀerent from the special facets introduced in the classiﬁcation of smooth reﬂexive polytopes. (a) A 1-dimensional special simplex (b) A 2-dimensional special simplex Fig. 8.6: Two special simplices in the unit cube Deﬁnition 8.20 A simplex S = conv(v1, . . . , vk) for v1, . . . , vk ∈⊆P ∩ Λ inside a polytope P is special if S ∩F is a facet of S for all facets F of P. Example 8.21 Here are some simple examples of special simplices. (1) The only special simplex in the unit simplex ∆is ∆itself (2) See Figure 8.6 for two diﬀerent special simplices in the unit cube. The reﬂexive cube [−1.1]d has also the origin as a special simplex. (3) Figure 8.7 shows a special simplex in the tetrahedron with vertices 0, e1, e2 and 2e3. Note that also the tetrahedron itself is special. (4) Figure 8.8 shows two special simplices in the bipyramid over a trian-gle. (5) A special simplex in the Birkoﬀpolytope for (n × n) −matrices is, for instance, the simplex spanned by the vertices corresponding to the permutations matrices for the permutations i 7− →i + k mod n 1for 1 ≤l ≤n . Example 8.22 If S is am (r −1)dimensional special simplex in a polytope P of codegree r, then S is necessarily unimodular. Otherwise, kS ⊆kP for some k < r contains a lattice point. Lemma 8.23 Let P be a Gorenstein polytope. If P has the IDP property, then P has a special simplex with codeg(P) many vertices. If P has a special simplex with r = codeg(P) many vertices, then its vertices have lattice distance zero or one from all facets. Fig. 8.7: A special simplex in a Lawrence prism Proof. Let P ⊂Rd be a polytope with the IDP property and Gorenstein adjust proof to new statement with degree s. Let uP ∈CP be the Gorenstein point in the cone over P. Because P has the IDP property, we can write uP = v1 + . . . + vr for v1, . . . , vr ∈({1} × P) ∩Zd+1. We claim that S := conv(v1, . . . , vr) is a special simplex. Haase, Nill, Paffenholz: Lattice Polytopes — 211 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Every facet F of P is dual to a vertex w of the Gorenstein dual {1} × P ∨. Then ⟨w, vi ⟩≥0 for all i = 1, . . . , r, and ⟨w, uP ⟩= P⟨w, vi ⟩= 1. Thus, ⟨w, vj ⟩= 1 for exactly one index j. Hence vj ̸∈F) for this j while ⟨w, vi ⟩= 0, that is, vi ∈F for all i ̸= j. Hence, S ∩F contains all but one of the vi. ⊓ ⊔ Fig. 8.8: Special simplices in a triangle bipyramid. Note that there may be more than one such. Fig. 8.9 The punchline in the proof of Theorem 8.18 will be that we project the polytope along a special simplex, and obtain a reﬂexive polytope with the same h∗-vector which inherits a regular unimodular triangulation from P. The following deﬁnition describes a subcomplex of P which will project bijectively onto the boundary of that reﬂexive polytope. Deﬁnition 8.24 Let S = conv(v1, . . . , vr) ⊆{1} × P be a special sim-plex. Denote by Γ(P, S) the subcomplex of ∂P generated by faces of the form F1 ∩. . . ∩Fr where Fi is a facet of CP with vi ̸∈Fi for i = 1, . . . , r. Lemma 8.25 Let S ⊆P be a special simplex in a Gorenstein polytope, and let T be a triangulation of Γ(P, S). Then the complex T ⋆S generated by {conv(S ∪F) : F ∈T} is a triangulation of P. This triangulation is unimodular if T was, and it is regular if T is the restriction to Γ(P, S) of a regular triangulation of P. Proof. We work in the cone Cp ⊂Rd+1 over {1} × P. Add a 1-coordinate to the simplex vertices {1} × S = conv(v1, . . . , vr). The set of primitive facet normals of CP is partitioned into sets A1, . . . , Ar so that ⟨a, vj ⟩= δij for a ∈Ai. Consider the concave piecewise linear function ω(x) := min ( r X i=1 ⟨ai, x ⟩: ai ∈Ai for i = 1, . . . , r ) . We will prove the following. (1) The domains of linearity of ω are cone(F ∪S) for F ∈Γ(P, S). This shows that these polytopes form a regular subdivision Γ(P, S) ⋆S of P. (2) If F is a (unimodular) simplex in Γ, then conv(S ∪F) is a (unimod-ular) simplex. (3) The conv(S ∪F) cover P. (4) The conv(S ∪F) and their faces form a polyhedral complex. (5) If T is the restriction to Γ(P, S) of a regular triangulation of P then T ⋆S is regular. (1): To compute ω(x), we choose, for i = 1, . . . , r, an element ai ∈Ai minimizing ⟨·, x ⟩. So the domains of linearity of ω are indexed by tuples (a1, . . . , ar) ∈Qr i=1 Ai and characterized by inequalities ⟨ai, x ⟩≤⟨a′ i, x ⟩for all a′ i ∈Ai . (8.4) — 212 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 8. Unimodular Triangulations (draft of October 7, 2020) A choice (a1, . . . , ar) determines a face F of Γ(P, S). We claim that the set of points x ∈{1} × P for which ω(x) = Pr i=1⟨ai, x ⟩agrees with conv(F ∪S). This then shows that these polytopes form a regular subdivision of P. Observe that the elements of Ai agree along {1} × S, and ⟨a′ i, x ⟩≥ ⟨ai, x ⟩= 0 along x ∈F. This implies that ω(x) = Pr i=1⟨ai, x ⟩along cone(F ∪S). Conversely, suppose x ∈{1} × P so that ω(x) = Pr i=1⟨ai, x ⟩. Set xΓ := x −Pr j=1⟨aj, x ⟩vj. For any a′ i ∈Ai we have ⟨a′ i, xΓ ⟩= ⟨a′ i, x ⟩−⟨ai, x ⟩≥0 by (8.4). Thus, xΓ ∈C. On the other hand, ⟨ai, xΓ ⟩= 0 for all i so that xΓ ∈cone F and x = xΓ + Pr i=1⟨ai, x ⟩vi ∈cone F + cone S. (2): If w1, . . . , ws form a simplex T in Γ(P, S), we want to show that the set v1, . . . , vr, w1, . . . , ws of vectors in Zd+1 generates the lattice. Let x ∈Zd+1. Then x = y + X i λivi for y ∈T and λi ≥0 for 1 ≤i ≤r. As conv(w1, . . . , ws) belongs to a face of Γ(P, S), there are ai ∈Ai for i = 1, . . . , r so that ⟨ai, wj ⟩= 0 and ⟨ai, vj ⟩= δij. Hence, λi = ⟨ai, x ⟩∈Z . Thus, y = x − X i λivi ∈T ∩Zd+1 and, as T is unimodular, y = P j µjwj for integral µj, 1 ≤j ≤s. (3): Every vertex u is either in S or in Γ as otherwise all facets containing u would also contain vi for some ﬁxed i. However, all facets containing vi describe the tangent cone T viP which has only one vertex. (4): This follows immediately from the previous considerations. (5): Suppose ω′ ∈RA induces a triangulation of P which restricts to T along Γ(P, S). Then the weights ω + εω′ for ε > 0 small enough will induce the triangulation T ⋆S: by the pertubation Lemma 8.5, the resulting subdivision is a reﬁnement of Γ(P, S) ⋆S (induced by ω) so that every conv(F ∪S) = F ⋆S is subdivided according to ω′. Every subdivision of this join is the join of S with its restriction to F (cf. Exercise 8.4). ⊓ ⊔ Haase, Nill, Paffenholz: Lattice Polytopes — 213 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Exercise 8.4 Lemma 8.26 Let T be a regular triangulation of the polytope P so that all maximal simplices have a common vertex v ∈int P. Then T has the same h-vector as a simplicial polytope. ⊓ ⊔ The proof of Theorem 8.18 uses h-vectors of triangulations (see § 3.5.4) and Theorem 3.40 by Betke-McMullen. Proof (of Theorem 8.18). If the Gorenstein polytope P has a regular unimodular triangulation, then it is IDP (Exercise 8.3). By Lemma 8.23 P contains a special simplex S, and we can deﬁne the complex Γ(P, S). By Lemma 8.25 we can modify the given triangulation of P, if necessary, to obtain a regular unimodular triangulation of the form T ⋆S for a unimodular triangulation T of Γ(P, S). Thus h⋆(P) = h⋆(Γ(P, S)) = h(T) by Theorem 3.40. It remains to show that T is combinatorially isomorphic to the boundary complex of a simplicial polytope. Then, the g-theorem implies that h(T) is unimodal. For this, let Φ be a strictly convex piecewise linear function on T ⋆S. As S is a face of the triangulation, there is a linear functional u such that ⟨u, v ⟩= Φ(v) for all v ∈S and ⟨u, v ⟩< Φ(v) for all v ̸∈S. Now let L be an aﬃne space meeting S transversally in its relative interior. Then, for small ε > 0, Q := {x ∈L : Φ(x) −⟨u, x ⟩≤ε} is a polytope whose boundary complex has the same combinatorics as Γ(P, S). ⊓ ⊔ ﬁll in details! As a corollary we can prove now the missing part of Proposition 7.25, see also Exercise 8.5. Exercise 8.5 Corollary 8.27 Let P1, P2 be reﬂexive. Then h⋆ P1◦P2 = h⋆ P1h⋆ P2. Proof. In this situation, P := P1 ∗P2 is also called the free join of P1 and P2. It’s h⋆-polynomial is given by the product of those of P1 and P2 (Exercise 8.5). The origins in P1 and P2 form a special simplex S of P. As remarked in the proof (Exercise 8.6), projecting along the aﬃne span of S does not change the h⋆-polynomial. Its image is the reﬂexive polytope P1 ◦P2. ⊓ ⊔ Exercise 8.6 8.5 Dilations One of the ﬁrst theorems about unimodular triangulations was proved in the early days of toric geometry by Knudsen, Mumford, and Water-— 214 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 8. Unimodular Triangulations (draft of October 7, 2020) man . They were interested in semi-stable reduction of families of algebraic varieties. Theorem 8.28 () There is a factor c = c(P) ∈Z>0 such that the dilation c · P admits a regular unimodular triangulation. We say that c(P) is a KMW-number of P. The KMW-theorem raises more questions than it answers, such as: ▶What is the minimum c(P) for a given polytope P? Is there a c(d) that is a KMW-number for every polytope of dimension d? ▶What is the structure of the set of KMW-numbers of a given P? Is it a monoid? Theorem 8.17 implies it is closed under taking multiples of an element, but it is not clear whether it is closed under taking sums. On the other end, no polytope P and integer c are known so that c is a KMW-number for P but c + 1 is not. For the proof of Theorem 8.28 we follow the strategy of the original ingenious proof (we omit the regularity bit). Compare also [15, §§3.A&3.B]. Proof (Proof of Theorem 8.28). The theorem is true for lattice polyhedral complexes: every cell F is a lattice polytope in its own lattice ΛF , and these lattices are compatible along intersections. In fact, the additional ﬂexibility oﬀered by this structure is used in the proof. Every triangulation of P carries two distinguished lattice structures: the one given by the embedding P ⊂Rd on the one hand, and the one which declares every simplex to be unimodular on the other. I really don’t understand what you are trying to say and why this would give a proof Starting from a full triangulation of P, the proof proceeds by in-duction on the maximal normalized volume V of a cell. If V is a prime number, the diﬀerent cells of volume V do not interfere. They can be subdivided independently. But if V is composite, then this very fact is used to interpolate between the unimodular lattice structure and a multiple of the given one. The two cases of the induction step are treated in Lemmas 8.29 and 8.33 below. The proofs occupy the remainder of this section. ⊓ ⊔ 8.5.1 Composite Volume For the induction step, we need some preparation. It is convenient to embed our lattice simplicial complex S on vertices v1, . . . , vN into RN via vi 7→ei. For every face F ∈S this yields a linear map ϕF : ΛF →RN, and we denote ˆ Λ the sum of the images of these lattices. Observe that ϕF (ΛF ) is generated by convex combinations of unit vectors, and therefore every element has integral coordinate sum. If vi ∈F, call xi an F-coordinate Haase, Nill, Paffenholz: Lattice Polytopes — 215 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) of x. In this setting, we can actually dilate S by a positive integer (and keep the lattice ˆ Λ). For each F ∈S, the fundamental parallelepiped of F is the half open cube Π(F) := {x ∈RN : xi ∈[0, 1) if vi ∈F, and xi = 0 if vi ̸∈F}. A box point of F is an element of Π(F) ∩ˆ Λ. It is in the relative interior if all its F-coordinates are strictly positive. The box points of F represent the elements of the ﬁnite abelian group (ZN + ϕF (ΛF ))/ZN; their number, the index [ZN + ϕF (ΛF ) : ZN], equals the normalized volume of F. Lemma 8.29 Let V be a composite integer, and suppose that for every lattice simplicial complex S all whose cells have volume less than V there is a factor c ∈Z>0 such that cS has a unimodular triangulation. Then the same is true for all lattice simplicial complexes all whose cells have volume no more than V . Proof. Let F1, . . . , FM be the volume V faces of S. For each of them choose non-zero box points mi ∈Π(Fi) ∩ˆ Λ of order strictly less than V in ˆ Λ/ZN. Deﬁne lattices Λ0 := ZN, Λi := Λi−1 + Zmi for i = 1, . . . , M, and ΛM+1 := ˆ Λ. To begin with, S is unimodular with respect to Λ0. The maximal volume of a simplex of S with respect to Λ1 is bounded by the index [Λ1 : Λ0] which by choice of m1 is less than V . By induction, there is a c1 ∈Z>0 so that c1S has a unimodular triangulation with respect to Λ1. In Λ2, this triangulation can only have simplices of volume [Λ2 : Λ1] which by choice of m2 is less than V . Continuing this way, we obtain a ΛM-unimodular triangulation of cM · . . . · c1S. But now, the index [ΛM+1 : ΛM] is also less than V . So some cM+1 · . . . · c1S has a ˆ Λ-unimodular triangulation. ⊓ ⊔ 8.5.2 Prime Volume Throughout the remainder of this section V is a prime number, and S is a lattice simplicial complex with maximal simplex volume V . The (open) star, star(S; F), of a face F of a simplicial complex S is the set of all faces that contain F. The closed star, star(S; F), contains additionally all faces of elements of star(S; F). The boundary, ∂star(S; F), of star(S; F) is the diﬀerence star(S; F) \ star(S; F). Lemma 8.30 The set of volume V simplices is a pairwise disjoint union of open stars of inclusion minimal volume V simplices. Each inclusion minimal volume V simplex has V −1 relative interior box points. — 216 — Haase, Nill, Paffenholz: Lattice Polytopes Chapter 8. Unimodular Triangulations (draft of October 7, 2020) Proof. Suppose F ∈S has volume V , and G is a face of F with a relative interior box point m. Since V is prime, m generates the group (ZN + ϕF (ΛF ))/ZN. As all non-G-coordinates of m vanish, the same is true for all multiples of m, and therefore for all box points of F. ⊓ ⊔ Lemma 8.31 If F ∈S is an inclusion minimal simplex of volume V , then there is a c ≤d so that c · star(S; F) has a subdivision which induces the standard hypersimplicial subdivision on c · ∂star(S; F) with the property that all simplices in any pulling triangulation have volume < V . Proof. Let m be a box point of F. Set c := P i mi so that m ∈relint cF. As all non-F-coordinates of m vanish and all F-coordinates are less than one, we have c < dim F + 1. Integrality implies c ≤d. (We could use the symmetry of Π(F) to obtain c ≤⌈d/2⌉.) Subdivide the facets of c · ∂star(S; F) canonically into hypersimplices. Subdivide c · star(S; F) into pyramids over these hypersimplices with apex m. Now, let G be a cell of a pulling triangulation reﬁning this subdivision. Then G = conv(m, G′) where G′ lives inside cF ′ for some facet F ′ of ∂star(S; F). There is a unique vertex vj of F not in F ′, and the normalized volume of G equals mj · V < V . ⊓ ⊔ Lemma 8.32 d!S has a triangulation into simplices of volume < V . Proof. Subdivide every simplex of volume less than V canonically into hypersimplices. For every inclusion minimal simplex F of volume V , choose c and subdivide c · star(S; F) as in Lemma 8.31. Now, d! · star(S; F) = d! c · (c · star(S; F)) has a canonical subdivision into pyramids over hyper-simplices. (Need to say something about this.) It restricts to the canonical subdivision on the boundary. Now pull all the lattice points. ⊓ ⊔ This end is surprising... Corollary: Lemma 8.33 Let V be a prime number, and suppose that for every lattice simplicial complex S all whose cells have volume less than V there is a factor c ∈Z>0 such that cS has a unimodular triangulation. Then the same is true for all lattice simplicial complexes all whose cells have volume no more than V . 8.6 Problems included on page 210 8.1. Prove Lemma 8.15 Haase, Nill, Paffenholz: Lattice Polytopes — 217 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) included on pa 8.2. Prove Theorem 8.17 by using Theorem 8.16. included on pa 8.3. Show that a lattice polytope is integrally closed if it admits a unimodular triangulation. included on page 214 8.4. Prove the last claim in the proof of Lemma 8.25. included on page 214 8.5. Let P ⊂Rn and Q ⊂Rm be lattice polytopes. Show that the product of their h⋆-polynomials equals the h⋆-polynomial of the convex hull of P × {0} × {0} and {0} × Q × {1}. included on page 214 8.6. Use the methods of proof of Theorem 8.18 to show that the pro-jection of a Gorenstein polytope of codegree r along a special simplex of dimension r −1 yields a reﬂexive polytope with the same h⋆-polynomial. — 218 — Haase, Nill, Paffenholz: Lattice Polytopes Some Convex Geometry A Contents A.1 Convex Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . 219 A.2 Ellipsoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219 A.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223 We provide some results from general convex geometry. A.1 Convex Bodies Deﬁnition A.1 (Convex Body) A convex body is a compact convex set K ⊆Rd such that int K ̸= ∅. It is centrally symmetric if for any x ∈K also −x ∈K. A.2 Ellipsoids Let Bd := {x ∈Rd \| ∥x∥= 1} be the unit ball. Deﬁnition A.2 (Ellipsoid) Let T : Rd →Rd be an invertible linear transformation and t ∈Rd. The set E := E(T, t) := T(B) + t is the ellipsoid with center t. Lecture Notes Lattice Polytopes (draft of October 7, 2020) See Figure A.1 for an example. We can write the ellipsoid explicitely as E = n x ∈Rd \| ⟨T −1(x −t), T −1(x −t) ⟩≤1 o = n x ∈Rd \| ⟨Q(x −t), x −t ⟩≤1 o for the positive semideﬁnite matrix Q = (TT t)−1. We can also assume Fig. A.1: The ellipsoid for T = 2 0 0 1 and t = 0 that T is positive deﬁnite, as any linear transformation T decomposes into a product T = US for an orthogonal matrix U and a positive deﬁnite matrix S, and U(Bd) = Bd. In a basis of eigenvectors u1, . . . , ud with eigenvalues λ1, . . . , λd for Q this takes the form E = n x ∈Rd \| λ1(x1 −t1)2 + · · · + λd(xd −td) ≤1 o . The volume of the ellipsoid is vol E = \| det T\| vol Bd = vol Bd √det Q . Let K be a convex body and η := sup(vol E : E ⊆K ellipsoid) . (A.1) Theorem A.3 Let K ⊆Rd be a convex body. Then the supremum of (A.1) is attained, i.e. there is an ellipsoid E ⊆K such that vol E = η. E is called a maximum volume ellipsoid. Proof. Let Bd be the unit ball. We deﬁne the set S := n (T, a) ∈Gl(d) × Rd : T(Bd) + a ⊆K o . Any ellipsoid E ⊆K is of the form R = T(Bd) + a and vol(E) = \| det T\| · vol(Bd) . (A.2) K is compact, so there is r > 0 such that ∥x∥≤r for all x ∈K. Hence, ∥a∥≤r and ∥T∥≤2r for all (T, a) ∈S . (A.3) Hence, S is a closed and bounded subset of Gl(d) × Rd , and the map (T, a) 7− →\| det T\| attains its maximum at some (T0, a0) ∈S. As K is non-empty, we have \| det T\| > 0 and E0 := T0(Bd) + a0 is an ellipsoid of maximum volume. — 220 — Haase, Nill, Paffenholz: Lattice Polytopes Appendix A. Some Convex Geometry (draft of October 7, 2020) Remark A.4 The ellipsoid obtained in the previous theorem is in fact unique and is also called the Löwner-John ellipsoid or John ellipsoid. Essentially, for the uniqueness one shows ﬁrst that, if there were two points (T1, a1) and (T2, a2) attaining the maximum, then, if T1 ̸= T2, 1 2(T1 + T2) generates an ellipsoid of strictly larger volume. So T1 = T2, and now, if a1 ̸= a2, then conv(T1(Bd) + a1, T2(Bd) + a2) contains an ellipsoid of strictly larger volume. We can use the maximum volume ellipsoid of a convex body K to aaproximate K ap to a factor depending on the dimension alone. The following theroem is the key result of this section. Theorem A.5 Let K ⊆Rd be a convex body and E a maximum volume ellipsoid in K. If the center of E is the origin, then K ⊆d · E. Using a translation we can of course always assume the the center of E is the origin. Proof. By deﬁnition there is an invertible linear transformation T such that E = T(Bd). We can apply T −1 to both K and E, so that in the following we can assume that E = Bd. We then need to show that there is no point z ∈K with ∥z∥≥d. Assume on the contrary that there is such a point z ∈K with ∥z∥> d and let L := conv(Bd ∪{z}) ⊆K . We construct an ellipsoid inside L of volume larger than vol Bd. Using a linear transformation we can assume that z = me1. For parameters a, b and ε we consider the ellipsoid Fd := ( x ∈Rd \| 1 a2 (x1 −ε)2 + 1 b2 d X i=2 x2 i ≤1 ) . This is symmetric in the last d −1 coordinated. Hence, it suﬃces to consider the case d = 2, i.e. F := x ∈Rd \| 1 a2 (x1 −ε)2 + x2 2 b2 ≤1 . See Figure A.2 for a sketch of the setting. Now clearly an ellipsoid of Fig. A.2: A sketch of the setting for the proof of Theorem A.5. maximum volume in F will touch at the point (−1, 0). Plugging this into the deﬁning equation we obtain a = ε + 1 . (A.4) The tangent to F at a point (u, v) is given by the equation u −ε a2 (x1 −ε) + v b2 x2 = 1 . (A.5) Haase, Nill, Paffenholz: Lattice Polytopes — 221 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) and, as (u, v) is on the boundary of F, 1 = (u −ε)2 a2 + v2 b2 (A.6) Now we want to determine the particular tangent to the ellipsoid that also passes through me1 and touches the unit ball, i.e. the boundary segment of L added in the convex hull of Bd with me1. This line touches the unit ball in a point (p, q) and can thus also be written as px1 + qx2 = 1 . (A.7) It passes through me1 and p2 + q2 = 1, so u −ε a2 = 1 m −ε p = 1 m q = √ m2 −1 m . (A.8) From (A.7) we deduce that the slope of the tangent is −1/ p m2−1. Com-puting the slope from (A.5) we obtain − 1 √ m2 −1 = −u −ε a2 b2 v . Squaring and ﬁrst using (A.6) and then the ﬁrst equation of (A.8) we obtain 1 m2 −1 = (u −ε)2 a4 b2 1 −(u −ε)2 a2 −1 = 1 (m −ε)2 b2 1 − a2 (m −ε)2 −1 Using (A.4) and solving for b2 gives b2 = (m −ε)2 −(1 + ε)2 m2 −1 . Now let us return to the ellipsoid Fd in dimension d. Its volume is vol Fd = abd−1 vol Bd . Now abd−1 = (1 + ε) (m −ε)2 −(1 + ε)2 m2 −1 (d−1)/2 As a function in ε its derivative at 0 is 1 −d −1 2 2 m −1 = m −d m −1 . Hence, for m > d and small ε the volume of Fd is larger than that of Bd. ⊓ ⊔ — 222 — Haase, Nill, Paffenholz: Lattice Polytopes Appendix A. Some Convex Geometry (draft of October 7, 2020) From the proof of Theorem A.6 it becomes pretty obvious that we can get a larger ellipsoid inside L if K is centrally symmetric and we use the knowledge that also −z is in K and we can thus replace L by L := conv(Bd ∪{±z}). With essentially the same proof this leads to the following approximation of a centrally symmetric convex body K by an ellipsoid. Theorem A.6 Let K ⊆Rd be a centrally symmetric convex body and E a maximum volume ellipsoid in K. If the center of E is the origin, then K ⊆ √ d · E. Exercise A.4 The proof is left to the reader as Exercise A.4. A.3 Problems included on page 223 A.1. Let Cd be the cube deﬁned by \|xi\| ≤1. Prove that the maximum volume ellipsoid is the unit ball. included on page 223 A.2. Let ∆d ⊆Rd+1 be the d-dimensional simplex deﬁned as the convex hull of the unit vectors. Prove that a maximum volume ellipsoid is the ball in the aﬃne hull of ∆d with center 1/(d+1)1. Deduce that the bound of Theorem A.6 is best possible. included on page 223 A.3. Let K be a centrally symmetric convex body. Prove that then also a maximum volume ellipsoid is centrally symmetric (and in particular centered at the origin). included on page 223 A.4. Prove Theorem A.6. Haase, Nill, Paffenholz: Lattice Polytopes — 223 — Solutions to some Exercises B B.1 Solutions for Chapter 4 4.13. ca(P; w) attained for η, can assume v := w + η ∈∂P. If v is not a vertex of P, then there is a v′ ∈Rd \ {0} so that v ± v′ ∈P. compute the deﬁning quotient . . . can decrease the dimension of the carrier face of v. Alternative proof: max{λ > 0 : w −λ(v −w) ∈P} is a linear optimization problem. It has a basic optimal solution. 4.11. (1) Let A, B and C be the vertices of the triangle realising sbc(2, i) for an interior point L and assume the smallest coeﬃcient is at vertex C. We can transform the triangle so that A is the origin and B is on the x-axis and C = (c1, c2) is in the positive orthant. Then smallestbary(2, i) = 1 c2 , so that sbc(2, i) is realized by a triangle with maximal height and i interior lattice points. This is obtained if B = (2, 0) and c1 = 0. In this case c2 = 2i + 2, so sbc(2, i) = 1 2i+2. (2) (3) B.2 Solutions for Chapter 6 6.2. This is equivalent to det(Λ) · det (Λ∗) = 1. References (draft of October 7, 2020) References to software packages (citations starting with an “S”) are listed in the software section of the references. Christos A. Athanasiadis. Ehrhart polynomials, simplicial polytopes, magic squares and a conjecture of Stanley. J. Reine Angew. Math., 583 (2005), pp. 163–174. doi: 10.1515/crll.2005.2005.583.163. url: org/10.1515/crll.2005.2005.583.163 (cit. on p. 211) Margherita Barile, Dominique Bernardi, Alexander Borisov, and Jean-Michel Kantor. On empty lattice simplices in dimension 4. Proc. Amer. Math. Soc., 139:12 (2011), pp. 4247–4253. doi: 10.1090/S0002-9939-2011-10859-1. arXiv: 0912.5310 [math.AG]. url: 10.1090/S0002-9939-2011-10859-1 (cit. on p. 124) A. I. Barvinok. Computing the Ehrhart polynomial of a convex lattice polytope. Discrete Comput. Geom., 12:1 (1994), pp. 35–48. doi: 10.1007/ BF02574364. url: (cit. on pp. 150, 151) Alexander Barvinok. A course in convexity. Graduate Studies in Mathe-matics (vol. 54). American Mathematical Society (Providence, RI), 2002, x+366 pages (cit. on p. 45) Alexander Barvinok. Integer points in polyhedra. Zurich Lectures in Ad-vanced Mathematics. European Mathematical Society (EMS), Zürich, 2008, viii+191 pages. doi: 10.4171/052 (cit. on p. 110) Alexander I. Barvinok and James E. Pommersheim. An algorithmic theory of lattice points in polyhedra. In: New perspectives in algebraic combinatorics (Berkeley, CA, 1996–97). Cambridge Univ. Press (Cambridge), 1999, pp. 91–147 (cit. on p. 150) Victor Batyrev and Benjamin Nill. Combinatorial aspects of mirror sym-metry. In: Integer points in polyhedra — geometry, number theory, repre-sentation theory, algebra, optimization, statistics. Ed. by Matthias Beck, Christian Haase, Bruce Reznick, Michèle Vergne, Volkmar Welker, and Ruriko Yoshida. Contemp. Math. (Vol. 452). Papers from the AMS-IMS-SIAM Joint Summer Research Conference held in Snowbird, UT, June 11–15, 2006. Amer. Math. Soc. (Providence, RI), 2008, pp. 35–66. doi: 10 . 1090 / conm / 452 / 08770. arXiv: math / 0703456 [math.CO] (cit. on p. 139) Victor V. Batyrev. Dual polyhedra and mirror symmetry for Calabi–Yau hypersurfaces in toric varieties. J. Alg. Geom., 3 (1994), pp. 493–535 (cit. on p. 185) Matthias Beck and Frank Sottile. Irrational proofs for three theorems of Stanley. European J. Combin., 28:1 (2007), pp. 403–409. doi: 10.1016/j. ejc.2005.06.003. arXiv: math/0501359 [math.CO] (cit. on pp. 76, 81) Ulrich Betke, Martin Henk, and Jörg M. Wills. Successive-minima-type inequalities. Discrete Comput. Geom., 9:2 (1993), pp. 165–175. doi: 10. 1007/BF02189316. url: (cit. on p. 106) Haase, Nill, Paffenholz: Lattice Polytopes — 227 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Ulrich Betke and Peter McMullen. Lattice points in lattice polytopes. Monatsh. Math., 99:4 (1985), pp. 253–265. doi: 10.1007/BF01312545 (cit. on p. 82) Louis J. Billera and Carl W. Lee. Suﬃciency of McMullen’s conditions for f-vectors of simplicial polytopes. Bull. Amer. Math. Soc. (N.S.), 2:1 (1980), pp. 181–185 (cit. on p. 29) Alexandr A. Borisov. Convex lattice polytopes and cones with few lat-tice points inside, from a birational geometry viewpoint. arXiv:math/ 0001109[math.AG]. 2000 (cit. on p. 111) Alexandr A. Borisov and Lev A. Borisov. Singular toric Fano varieties. Mat. Sb., 183:2 (1992), pp. 134–141. doi: 10.1070/SM1993v075n01ABEH003385. url: (cit. on p. 111) Winfried Bruns and Joseph Gubeladze. Polytopes, Rings, and K-Theory. Monographs in Mathematics. XIV, 461 p. 52 illus. Springer-Verlag, 2009 (cit. on p. 215) Winfried Bruns and Tim Römer. h-vectors of Gorenstein polytopes. J. Combin. Theory Ser. A, 114:1 (2007), pp. 65–76. doi: 10.1016/j.jcta. 2006.03.003. url: (cit. on p. 210) Vladimir I. Danilov and Askol’d G. Khovanskii. Newton polyhedra and an algorithm for computing Hodge–Deligne numbers. Math. USSR Izvestiya, 29:2 (1987), pp. 279–298 (cit. on p. 185) Jesús A. De Loera, Raymond Hemmecke, and Matthias Köppe. Algebraic and geometric ideas in the theory of discrete optimization. MOS-SIAM Series on Optimization (vol. 14). Society for Industrial and Applied Math-ematics (SIAM) (Philadelphia, PA), 2013, xx+322 pages (cit. on pp. 77, 157) Jesus deLoera, Francisco Santos, and Jörg Rambau. Triangulations. Algo-rithms and Computation in Mathematics (vol. 25). Springer, 2010 (cit. on pp. 32, 204) Jan Draisma, Tyrrell B. McAllister, and Benjamin Nill. Lattice width directions and Minkowski’s 3d-theorem (Jan. 2009). eprint: 0901.1375 (cit. on p. 106) Friedrich Eisenbrand. Integer programming and algorithmic geometry of numbers. English. In: ed. by Michael Jünger, Thomas Liebling, Denis Naddef, George Nemhauser, William Pulleyblank, Gerhard Reinelt, Gio-vanni Rinaldi, and Laurence Wolsey. From the early years to the state-of-the-art, Papers from the 12th Combinatorial Optimization Workshop (AUSSOIS 2008) held in Aussois, January 7–11, 2008. Berlin: Springer (Berlin), 2010, pp. xx+804. doi: 10.1007/978-3-540-68279-0_14. url: (cit. on p. 160) Robert M. Erdahl and Sergej S. Ryshkov. On lattice dicing. English. Eur. J. Comb., 15:5 (1994), pp. 459–481. doi: 10.1006/eujc.1994.1049 (cit. on p. 209) Israel M. Gel′fand, Michael M. Kapranov, and Andrei V. Zelevinsky. Dis-criminants, resultants, and multidimensional determinants. Mathematics: Theory & Applications. Birkhäuser Boston Inc. (Boston, MA), 1994, x+523 pages. doi: 10.1007/978-0-8176-4771-1 (cit. on p. 204) — 228 — Haase, Nill, Paffenholz: Lattice Polytopes References (draft of October 7, 2020) Roland Grinis and Alexander Kasprzyk. Normal forms of convex lattice polytopes. Jan. 2013. arXiv: 1301.6641 [math.CO] (cit. on p. 147) Martin Grötschel, László Lovász, and Alexander Schrijver. Geometric algorithms and combinatorial optimization. Second. Algorithms and Com-binatorics (vol. 2). Springer-Verlag (Berlin), 1993, xii+362 pages (cit. on p. 104) Christian Haase and Günter M. Ziegler. On the maximal width of empty lattice simplices. Eur. J. Comb., 21:1 (2000), pp. 111–119 (cit. on p. 124) Martin Henk. Successive minima and lattice points. Rend. Circ. Mat. Palermo (2) Suppl.,: 70, part I (2002). IV International Conference in “Stochastic Geometry, Convex Bodies, Empirical Measures & Applications to Engineering Science”, Vol. I (Tropea, 2001), pp. 377–384. eprint: math. MG/0204158 (cit. on p. 105) Douglas Hensley. Lattice vertex polytopes with interior lattice points. Paciﬁc J. Math., 105:1 (1983), pp. 183–191. doi: 10.2140/pjm.1983.105.183 (cit. on p. 117) Lutz Hille and Harald Skarke. Reﬂexive polytopes in dimension 2 and certain relations in SL2(Z). English. J. Algebra Appl., 1:2 (2002), pp. 159– 173. doi: 10.1142/S0219498802000124 (cit. on p. 183) Ravindran Kannan and Achim Bachem. Polynomial algorithms for com-puting the Smith and Hermite normal forms of an integer matrix. SIAM J. Comput., 8:4 (1979), pp. 499–507. doi: 10.1137/0208040. url: https: //doi.org/10.1137/0208040 (cit. on p. 150) George R. Kempf, Finn F. Knudsen, David Mumford, and Bernard Saint– Donat. Toroidal Embeddings I. Lecture Notes in Mathematics (vol. 339). Springer–Verlag, 1973 (cit. on p. 215) Matthias Köppe and Sven Verdoolaege. Computing Parametric Rational Generating Functions with a Primal Barvinok Algorithm. Electronic journal of Combinatorics, 15 (2008) (cit. on p. 76) Maximilian Kreuzer and Harald Skarke. Classiﬁcation of reﬂexive polyhedra in three dimensions. Adv. Theor. Math. Phys., 2:4 (1998), pp. 853–871 (cit. on p. 142) Maximilian Kreuzer and Harald Skarke. Complete classiﬁcation of reﬂex-ive polyhedra in four dimensions. Adv. Theor. Math. Phys., 4:6 (2000), pp. 1209–1230 (cit. on pp. 142, 198) Maximilian Kreuzer and Harald Skarke. PALP: a package for analysing lattice polytopes with applications to toric geometry. Comput. Phys. Comm., 157:1 (2004), pp. 87–106. doi: 10.1016/S0010-4655(03)00491-0. url: (cit. on p. 142) J. C. Lagarias. Knapsack public key cryptosystems and Diophantine approx-imation (extended abstract). In: Advances in cryptology (Santa Barbara, Calif., 1983). Plenum, New York, 1984, pp. 3–23 (cit. on p. 170) J. C. Lagarias, H. W. Lenstra Jr., and C.-P. Schnorr. Korkin-Zolotarev bases and successive minima of a lattice and its reciprocal lattice. Combinatorica, 10:4 (1990), pp. 333–348. doi: 10.1007/BF02128669. url: org/10.1007/BF02128669 (cit. on p. 109) Jeﬀrey C. Lagarias and Günter M. Ziegler. Bounds for lattice polytopes containing a ﬁxed number of interior points in a sublattice. English. Can. J. Math., 43:5 (1991), pp. 1022–1035 (cit. on p. 117) Haase, Nill, Paffenholz: Lattice Polytopes — 229 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) A. K. Lenstra, H. W. Lenstra Jr., and L. Lovász. Factoring polynomials with rational coeﬃcients. Math. Ann., 261:4 (1982), pp. 515–534. doi: 10.1007/BF01457454. url: (cit. on pp. 161, 169, 170) P. McMullen. The numbers of faces of simplicial polytopes. Israel J. Math., 9 (1971), pp. 559–570. doi: 10.1007/BF02771471. url: org/10.1007/BF02771471 (cit. on p. 29) David R. Morrison and Glenn Stevens. Terminal quotient singularities in dimensions three and four. Proc. Amer. Math. Soc., 90:1 (1984), pp. 15–20. doi: 10.2307/2044659 (cit. on p. 120) Mircea Musta and Sam Payne. Ehrhart polynomials and stringy Betti numbers. Math. Ann., 333:4 (2005), pp. 787–795. doi: 10.1007/s00208-005-0691-x. eprint: math/0504486 (math.AG) (cit. on p. 210) Phong Q. Nguyen and Brigitte Vallée, eds. The LLL algorithm. Information Security and Cryptography. Survey and applications. Springer-Verlag, Berlin, 2010, xiv+496 pages. doi: 10.1007/978-3-642-02295-1. url: (cit. on p. 162) Mikkel Øbro. Classiﬁcation of smooth Fano polytopes. PhD thesis. Uni-versity of Aarhus, 2007. url: pure.au.dk/portal/files/41742384/imf_ phd_2008_moe.pdf (cit. on p. 198) Hidefumi Ohsugi and Takayuki Hibi. Convex polytopes all of whose reverse lexicographic initial ideals are squarefree. English. Proc. Am. Math. Soc., 129:9 (2001), pp. 2541–2546 (cit. on p. 208) Oleg Pikhurko. Lattice points in lattice polytopes. English. Mathematika, 48:1-2 (2001), pp. 15–24 (cit. on p. 117) Bjorn Poonen and Fernando Rodriguez-Villegas. Lattice polygons and the number 12. Amer. Math. Monthly, 107:3 (2000), pp. 238–250 (cit. on p. 183) Herbert E. Scarf. Integral polyhedra in three space. Math. Oper. Res., 10 (1985), pp. 403–438 (cit. on p. 120) Claus-P. Schnorr. A hierarchy of polynomial time lattice basis reduction algorithms. Theoret. Comput. Sci., 53:2-3 (1987), pp. 201–224. doi: 10. 1016/0304-3975(87)90064-8. url: (cit. on p. 169) Alexander Schrijver. Combinatorial optimization. Polyhedra and eﬃciency (3 volumes). English. Algorithms and Combinatorics 24. Berlin: Springer., 2003 (cit. on p. 104) Alexander Schrijver. Theory of linear and integer programming. English. Wiley-Interscience Series in Discrete Mathematics. A Wiley-Interscience Publication. Chichester: John Wiley & Sons Ltd., 1986 (cit. on pp. 25, 104, 174) P. R. Scott. On convex lattice polygons. Bull. Austral. Math. Soc., 15:3 (1976), pp. 395–399 (cit. on p. 12) András Seb. An introduction to empty lattice simplices. In: Integer pro-gramming and combinatorial optimization (Graz, 1999). Ed. by Gérard Cornuéjols, Rainer E. Burkard, and Gerhard J. Woeginger. Lecture Notes in Comput. Sci. (Vol. 1610). Springer (Berlin), 1999, pp. 400–414. doi: 10.1007/3-540-48777-8_30. url: (cit. on p. 120) — 230 — Haase, Nill, Paffenholz: Lattice Polytopes References (draft of October 7, 2020) Richard P. Stanley. A monotonicity property of h-vectors and h∗-vectors. European J. Combin., 14:3 (1993), pp. 251–258. doi: 10.1006/eujc.1993. 1028. url: (cit. on p. 81) Richard P. Stanley. Decompositions of rational convex polytopes. Ann. Discrete Math., 6 (1980). Combinatorial mathematics, optimal designs and their applications (Proc. Sympos. Combin. Math. and Optimal Design, Colorado State Univ., Fort Collins, Colo., 1978), pp. 333–342 (cit. on p. 208) Richard P. Stanley. The number of faces of a simplicial convex polytope. Adv. in Math., 35:3 (1980), pp. 236–238. doi: 10.1016/0001-8708(80) 90050-X (cit. on p. 29) Arne Storjohann. Faster Algorithms for Integer Lattice Basis Reduction. 1996 (cit. on p. 169) Bernd Sturmfels. Gröbner bases and convex polytopes. Univ. Lecture Series (vol. 8). American Mathematical Society (Providence, RI), 1996, xii+162 pages (cit. on p. 204) Seth Sullivant. Compressed polytopes and statistical disclosure limitation. Tohoku Math. J. (2), 58:3 (2006). Preprint arXiv:math.CO/0412535, pp. 433–445. url: tmj/1163775139 (cit. on p. 208) Sven Verdoolaege, Rachid Seghir, Kristof Beyls, Vincent Loechner, and Maurice Bruynooghe. Counting integer points in parametric polytopes using Barvinok’s rational functions. Algorithmica, 48:1 (2007), pp. 37–66. doi: 10.1007/s00453-006-1231-0. url: (cit. on p. 150) G. K. White. Lattice tetrahedra. Canadian J. Math., 16 (1964), pp. 389–396. doi: 10.4153/CJM-1964-040-2. url: (cit. on p. 120) J. Zaks, M. A. Perles, and J. M. Wilks. On lattice polytopes having interior lattice points. Elem. Math., 37:2 (1982), pp. 44–46 (cit. on p. 119) [S1] Jesús A. De Loera, Raymond Hemmecke, Ruriko Yoshida, and Jeremy Tauzer. lattE. 2005 (cit. on p. 150) [S2] Matthias Köppe. Latte macchiato – An improved version of Latte. http: //www.math.uni-magdeburg.de/~mkoeppe/latte/. 2007 (cit. on p. 150) [S3] Sven Verdoolaege. barvinok. 2007 (cit. on p. 150) Haase, Nill, Paffenholz: Lattice Polytopes — 231 — H-description, 26 M-sequence, 29 V -description, 26 δ-reduced basis, 161 h⋆-polynomial, 73, 80, 81, 87, 90, 197 Λ-rational subspace, 36 d-polyhedron, 24 g-Theorem, 29 k-face, 26 h-vector, 81 Pick’s formula, 10 Pick’s theorem, 10 f-vector, 28 addition property, 189, 190 additive subgroup, 33 adjacent, 27 aﬃne combination, 20 aﬃne equivalence of polyhedra, 28 aﬃne hull, 20 aﬃne lattice, 40 aﬃne lattice automorphism, 8 aﬃne lattice basis, 40 aﬃne lattice isomorphism, 40 aﬃne space, 20 aﬃnely independent, 21 algorithm Barvinok’s ∼, 150, 154 Intgeger Feasibility in ﬁxed dimension, 176 Intgeger Programming in ﬁxed dimension, 176 LLL, 151, 162 Polarized Barvinok’s ∼, 171 apex, 206 of a bipyramid, 50 of a pyramid, 50 automorphism lattice, 8 Barvinok’s algorithm, 150, 154 barycentric coordinates minimal, 115 basic properties h⋆-polynomial, 80 basis δ-reduced, 161 LLL-reduced, 161 of a lattice, 8, 35 reduced, 149, 157, 160, 161 weakly reduced, 161 beneath, 23 beyond, 23 bipyramid, 50 apex, 50 Birkhoﬀpolytope, 50, 195 boundary complex, 30 of a polytope, 30 boundary point, 22 Brianchon-Gram identity, 91 — 233 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) Brion Theorem of ∼, 150 Brion’s Theorem, 94, 150 canonical subdivision, 209 Caratheodory’s Theorem, 23 Cayley polytope, 198 cell maximal, 30 of a polyhedral complex, 30 centally symmetric, 102 circuit, 137 codegree of a lattice polytope, 87 coeﬃcient of asymmetry, 115 combination conic, 21 convex, 21 complex cell of a polyhedral ∼, 30 dimension of a polyhedral ∼, 30 polyhedral dimension, 30 facets, 30 maximal cell, 30 pure, 30 subcomplex, 30 polyhedral ∼, 30 compressed polytope, 208 cone, 21 face of, 26 fundamental parallelepiped, 78 Gorenstein, 193 half-open, 76, 84, 170 height, 48 homogeneous, 48 index, 151 minimal proper face, 27 normal, 48 over a polytope, 25, 73 polyedral, 24 polyhedral, 21 proper face of, 26 conic combination, 21 conic hull, 21 contingency table, 65 convex body, 102, 219 centally symmetric, 102 convex combination, 21 convex hull, 21 convex set, 21 boundary point, 22 interior point, 22 relative interior point, 22 coset, 38 counting function, 65 covering radius, 107 cross polytope, 22 cube, 22 cut polytope, 50 degree of a lattice polytope, 87 determinant of a lattice, 41 dicing, 209 dilation of a set, 65 dimension of a face, 26 of a polyhedral complex, 30 of a polyhedron, 24 distance function, 39 dual lattice, 39 edge, 27 Ehrhart counting function, 65 Ehrhart polynomial, 62, 73, 73, 82, 85, 90 Ehrhart series, 74, 80 Ehrhart’s theorem, 73 Ehrhart-Macdonald reciprocity, 85 ellipsoid, 110, 219 empty lattice polytope, 119 empty polytope, 89 equivalence lattice, 8 unimodular, 8 Euler characteristic, 30 Euler-Characteristic, 86 exterior description, 26 extremal ray, 27 face dimension, 26 minimal, 27 minimal proper, 27 of a polytope, 26 proper, of a polytope, 26 tangent cone, 31 face lattice, 27 face vector, 28, 30 facet special, 191 facet unimodular lattice polytope, 209 facets of a polyhedral complex, 30 fan, 31 smooth blow-up, 184 far half-open cone, 76 far half-open parallelepiped, 76 ﬁnitely generated, 25 ﬁniteness Gorenstein polytopes, 197 — 234 — Haase, Nill, Paffenholz: Lattice Polytopes Index (draft of October 7, 2020) formal Laurent series, 69 free sum, 188 Frobenius number, 65 Frobenius problem, 65 full dimensional, 24 fundamental parallelepiped, 35, 78 Generalized Blichfeldt’s Theorem, 102 generic reference point, 76 Gorenstein cone, 193 Gorenstein polytope, 180, 187, 193, 195, 196, 197 Gram-Schmidt orthogonalization, 158 half-open cone, 76, 84, 170 half-open decomposition, 76, 84, 170 half-open parallelepiped, 76 half-open simplex, 76 half-space aﬃne, 23 linear, 23 Hermite normal form, 42 Hilbert basis, 46 minimal, 46 homogeneous, 48 homogenization, 25 hull aﬃne, 20 conic, 21 convex, 21 linear, 20 hyperplane aﬃne, 23 linear, 23 supporting, 26 valid, 26 hypersimplex, 22, 50 IDP, 210 implied equality, 27 incident, 28 face, 28 inclusion-exclusion principle of ∼, 73 independent aﬃnely, 21 linearly, 21 index of a cone, 151 of a lattice, 38 inner normal primitive, 180 integer decomposition property, 210 Integer Feasibility in ﬁxed dimension, 176 integer point generating function, 72, 79, 91, 94 integer point series, 69, 72 summable, 70 Integer Programming in ﬁxed dimension, 176 integral polytope, 48 integrally closed, 210 interior description, 26 interior point, 22, 27 irredundant, 27 isomorphic polygon, 8 John ellipsoid, 221 join, 50 KMW number, 215 KMW Theorem, 215 Knapsack problem, 63 Löwner-John ellipsoid, 221 lattice, 34, 45 δ-reduced basis, 161 aﬃne, 40 basis, 40 isomorphism, 40 basis, 35 covering radius, 107 determinant, 41 fundamental parallelepiped, 35 index, 38 isomorphism, 38 LLL-reduced basis, 161 orthogonality defect, 159, 166 packing radius, 107 rank, 34 reduced basis, 161 standard integer ∼, 34 sublattice, 38 transformation, 38 unimodular, 41 weakly reduced basis, 161 lattice automorphism, 8 lattice basis potential, 164 lattice basis, 8 lattice dicing, 209 lattice equivalence, 8 lattice isomoprhism, 38 lattice isomorphic, 49 lattice isomorphism, 49 lattice length, 9 lattice polygon, 6 lattice polytope, 48 Cayley, 198 codegree, 87 compressed, 208 degree, 87 empty, 89, 119 facet unimodular, 209 Haase, Nill, Paffenholz: Lattice Polytopes — 235 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) hollow, 124 IDP, 210 lattice isomorphic, 49 normal, 210 normalized volume, 52 unimodularly equivalent, 49 lattice pyramid, 198 lattice transformation, 38 lattice triangulation, 11 lattice width, 110 Laurent polynomial, 69 Laurent polynomial ring, 69 Laurent series, 69, 69, 72 summable, 70 Lawrence prism, 89 lineality space, 24 linear combination, 20 linear hull, 20 linear space, 20 linear span, 20 linearly independent, 21 LLL algorithm, 151 LLL-reduced basis, 161 matrix unimodular, 209 maximum volume ellipsoid, 220 minimal barycentric coordinates, 115 minimal face of a polytope, 27 minimal proper face, 27 Minkowski sum, 25 Minkowski’s First Theorem, 102 Minkowski’s Second Theorem, 106 mirror symmetry, 179 near half-open cone, 76 near half-open parallelepiped, 76 normal cone, 48 normal cone of a polytope, 31 normal fan of a polytope, 31 normal form Hermite ∼, 42 normal lattice polytope, 210 normalized volume, 52, 197 octahedron, 22 order polytope, 50 orthogonality defect, 159, 166 orthogonalization Gram-Schmidt∼, 158 packing radius, 107 parallelepiped, 35 fundamental, 78 half open, 35 half-open, 76 parallepiped fundamental, 35 permutation polytope, 50 Pick’s Theorem, 119, 122, 183 point boundary, 22 interior, 22 relative interior, 22 polar polytope, 180 Polarized Barvinok’s algorithm, 171 polarized, 171 polygon lattice ∼, 6 polyhedral ball, 32 polyhedral complex, 30 cell, 30 dimension, 30 face vector, 30 facets, 30 maximal cell, 30 pure, 30 subcomplex, 30 polyhedral cone, 21, 24 polyhedral sphere, 32 polyhedron, 23 dimension, 24 face of, 26 homogenization, 25 interior point, 27 lineality space, 24 pointed, 24 proper face of, 26 recession cone, 24 vertex of, 27 polynomial h⋆, 73, 80, 81, 87, 90 Ehrhart, 62, 73, 73, 82, 85, 90 Laurent, 69 Todd∼, 157 polynomial ring Laurent, 69 polytope, 21 k-face, 26 ∼Birkhoﬀ, 50 ∼cut, 50 ∼order, 50 ∼permutation, 50 ∼traveling salesperson, 50 aﬃnely equivalent, 28 bipyramid, 50 boundary complex, 30 Cayley, 198 compressed, 208 edge, 27 — 236 — Haase, Nill, Paffenholz: Lattice Polytopes Index (draft of October 7, 2020) empty, 89 extremal ray, 27 face tangent cone, 31 face of, 26 facet unimodular, 209 free sum, 188 Gorenstein, 180, 187, 193, 195, 196, 197 homogenization, 25 hypersimplex, 50 integral, 48 integrally closed, 210 interior point, 27 join, 50 lattice, 48 minimal face, 27 normal cone, 31 normal fan, 31 normalized volume, 52 pointed, 24 polar, 180 prism, 50 product, 50 proper face of, 26 pyramid, 50 reﬂexive, 179, 181, 181, 182, 185, 187–189, 193 simple, 29 simplicial, 29 special simplex, 211 vertex of, 27 potential of a lattice basis, 164 primitive, 37 primitive inner normal, 180 principle of inclusion-exclusion, 73 prism, 50 product, 50 prosm Lawrence, 89 pulling reﬁnement, 206 pyramid, 50 apex, 50 lattice, 198 over a polytope, 50 Pyramid Theorem, 134 rank, 34 rational subspace, 36 recession cone, 24 reduced basis, 149, 157, 160, 161 redundant, 27 Reeve simplex, 10 reﬂexive polytope, 179, 181, 181, 182, 185, 187–189, 193 addition property, 189, 190 regular subdivision, 32, 204 relative interior point, 22 root system, 34 Ad, 34 Dd, 34 Scott’s theorem, 12 series Ehrhart, 74, 80 formal Laurent, 69 Laurent, 69, 69, 72 summable, 70 simple, 29 simplex, 21 half-open, 76 Reeve, 10 standard, 49, 66 unimodular, 49 unit, 66 simplicial, 29 smooth blow-up of a fan, 184 space aﬃne, 20 linear, 20 span linear, 20 special facet, 191 special simplex, 211 standard simplex, 49, 66 Stanley Reciprocity, 84 Stanley’s Monotonicity theorem, 81 subcomplex, 30 subdivision, 31 canonical, 209 regular, 32, 204 trivial, 30 without new vertices, 31 subgroup additive, 33 sublattice, 38 subspace Λ-rational, 36 rational, 36 successive minimum, 104 summable, 70 tangent cone of a face, 31 tetrahedron, 22 Theorem Caratheodory, 23 Generalized Blichfeldt’s ∼, 102 KMW, 215 Lenstra, Lenstra, Lovász, 162 Minkowski’s First ∼, 102 Minkowski’s Second ∼, 106 of Brianchon-Gram, 91 Haase, Nill, Paffenholz: Lattice Polytopes — 237 — Lecture Notes Lattice Polytopes (draft of October 7, 2020) of Brion, 94 Pick’s ∼, 119, 122 Pick’s ∼, 183 Stanley’s Monotonicity, 81 van der Corput’s ∼, 102 Weyl-Minkowski, 25 theorem Pick’s, 10 Ehrhart-Macdonald, 85 ﬂatness, 111 Howe, 120 of Ehrhart, 73 Scott’s, 12 Stanley Reciprocity, 84 Stanley’s nonnegativity∼, 80 White, 120 Theorem of Brion, 150 Theorem of Lenstra, Lenstra, Lovász, 162 tight regular subdivision, 205 Todd-polynomial, 157 transformation lattice, 38 unimodular, 38 unimodular ∼, 8 traveling salesperson polytope, 50 triangle standard, 8 unimodular, 8 triangulation, 11, 31, 79 lattice, 11 pulling reﬁnement, 206 unimodular, 203 without new vertices, 31, 32 trivial subdivision, 30 unimodular of a lattice, 41 triangle, 8 triangulation, 203 unimodular equivalence, 8 unimodular matrix, 209 unimodular simplex, 49 unimodular tranformation, 8 unimodular transformation, 38 unimodularly equivalent, 49 unit simplex, 66 university degrees, 65 vertex of a polyhedron, 27 of a polytope, 27 vertex-edge graph, 189 visible complex, 92 visible face, 91 volume normalized, 52, 197 weakly reduced basis, 161 Weyl-Minkowski-Theorem, 25 width, 110 zonotope, 35 half open, 35 — 238 — Haase, Nill, Paffenholz: Lattice Polytopes Athanasiadis, 211 Barvinok, Alexander, 150 Betke, 82 Blichfeldt, 102 Brianchon, 91 Brion, 94, 150 Bruns, 210 Ehrhart, 65, 73, 74, 85 Euler Leonhard, 30 Gram, 91 Gram, Jørgen Pedersen, 158 Hensley, 117 Hermite, 42 Hibi, 208 Hilbert, 46 Howe, 120 Knudson, 215 Lagarias, 117 Lenstra, Arjen, 151, 162 Lenstra, Hendrik, 151, 162 Lovász, László, 151, 162 Macdonald, 85 McMullen, 82 Minkowski, 102, 106 Mumford, 215 Oshugi, 208 Pick, 10, 183 Pikhurko, 117 Pommersheim, 150 Römer, 210 Santos, 208 Scarf, 120 Schmidt, Erhard, 158 Schorr, Claus-P., 169 Scott, 12 Stanley, 80, 81, 84, 208 Storjohann, Arne, 169 Sullivant, 208 Todd, 157 van der Corput, 102 Waterman, 215 White, 120 Ziegler, 117 — 239 —
17031	https://www.justintimemedicine.com/curriculum/dysk-aortic-stenosis	Aortic Stenosis \| DYSK \| JustInTimeMedicine Search the site College of Human Medicine Need Search Help? JustInTimeMedicine Curriculum Resources Contacts LOG IN SDC Components Educational Program Objectives M1 Content Intersessions Middle Clinical Experience Late Clinical Experience Chief Complaints and Concerns Class Specific Information Resources Diseases You Should Know Aortic Stenosis Aortic Stenosis \| DYSK Updated 01/06/2025 2:06 PM - Kate Baird Aortic Valve Anatomy and Pathology Short Axis View Of Aortic Valve Of Heart Cardiac Valves Summary Aortic stenosis is abnormal narrowing of the aortic valve Aortic sclerosis is calcification and thickening of the aortic valve without narrowing Aortic stenosis is the most common valvular disease in the US and Europe, affecting nearly 4% of patients ≥75 years, and 2% to 7% of patients >65 years. Aortic stenosis is the second most frequent cause for cardiac surgery. Aortic stenosis is classified as mild, moderate or severe depending on the assessment of three parameters of aortic valve including aortic valve area, aortic jet velocity, and transvalvular pressure gradient Risk factors for AS include: congenitally bicuspid valve (age 30-40), degeneration (older age), hypercholesterolemia (high LDL-cholesterol), rheumatic heart disease, radiation therapy among others Patients may be asymptomatic (common) or present with chest pain, exertional dyspnea, syncope, signs of heart failure and gastrointestinal bleeding due to angiodysplasia and the physical exam findings depend on the severity of the stenosis, however all patients with have a systolic ejection murmur is heard at the second intercostal space in the right upper sternal border radiates to carotid arteries. The primary means of diagnosing AS is with Echocardiography. The primary treatment for AS is valve replacement or valvulotomy, medical treatment is less effective. Asymptomatic AS has an excellent prognosis, however prognosis falls after the development of symptoms without valve replacement Definition Abnormal narrowing of the aortic valve The normal aortic valve area is 3-4 cm 2 General/Background Aortic stenosis is classified as mild, moderate or severe depending on the assessment of three parameters of aortic valve Aortic valve area Aortic jet velocity, and Transvalvular pressure gradient The prevalence of calcific AS is 1.3% in patients aged 65–74; 4% in those aged 85 or older (J Am Coll Cardiol 1997 Mar 1;29(3):630-4) In individuals with high risk factors (see below), the aortic valve area narrows gradually, which in turns increases transvalvular gradient and the velocity of the blood through the valve. Aortic stenosis is the most common valvular disease in the US and Europe, affecting nearly 4% of patients ≥75 years, and 2% to 7% of patients >65 years. Aortic stenosis is the second most frequent cause for cardiac surgery. Significant coronary artery disease (CAD) is present in 50%pts with AS Risk Factors/Causes Congenitally bicuspid valve (age 30-40); estimated prevalence is 13.7 cases/1000 persons (Circulation 2015 Jan 27;131(4):e29-322) Drug induced disease Rheumatic heart disease is most common in developing countries Congenitally bicuspid valve (age 30-40); estimated prevalence is 13.7 cases/1000 persons (Circulation 2015 Jan 27;131(4):e29-322) Drug induced disease Rheumatic heart disease is most common in developing countries Radiation therapy Calcific aortic stenosis shares the same risk factors as atherosclerosis, including: Age > 60 years. Smoking Hypertension Hypercholesterolemia (high LDL-cholesterol) Pathogenesis Normal aortic valves have three thin leaflets. The pathology of aortic stenosis includes processes similar to those in atherosclerosis, including lipid accumulation, inflammation, and calcification LV mass increases to compensate for increased obstruction, producing LV failure With aging, the collagen of those leaflets is destroyed. When calcium is built up on the valve, it makes it thicker, reduces the leaflet motility, and causes valve stenosis. Bicuspid valves do not open as widely as normal valves with three cusps. As a result, a turbulence blood flow creates across the bicuspid valves, which leads to an increased scarring, thickness and stenosis of the valve. Rheumatic fever damages the aortic valve producing fibrosis and fusion of the commissures of the leaflets, which causes aortic stenosis StageDefinitionValve AnatomyValve HemodynamicsConsequencesSymptoms A At risk of AS Bicuspid anatomy or AV sclerosis Aortic velocity < 2.meters/second None None B Progressive AS Leaflet calcification (mild-to-moderate) with some ↓ in systolic motion Mild:Mean gradient < 20 mm HG OR aortic velocity 2.5–2.9 meters/second Moderate:Mean gradient 20-39 mm HG OR aortic velocity 3.0–3.9 meters/second Early diastolic dysfunction \|Nml EF \|None C Asymptomatic severe AS Leaflet calcification (severe) with severe ↓ leaflet opening Severe: Mean gradient> 40 mm Hg OR aortic velocity >4 meters/second OR Very severe: Mean gradient> 60 mm Hg OR aortic velocity >5 meters/second OR Aortic valve area: <1 cm 2/m 2 Diastolic dysfunction\| Mild LVH \| Nml to low EF None D Symptomatic severe AS Leaflet calcification (severe) with severe ↓ leaflet opening Aortic valve area: <1 cm 2/m 2 WITH mean gradient< 40 mm Hg OR aortic velocity <4 meters/second Diastolic dysfunction\| LVH \| EF may be < 50% \| Pulmonary HTN may be present Exertional angina, (pre)syncope, dyspnea or ↓ exercise tolerance Adapted from \|Journal of the American College of Cardiology 2014 63:e57-e185 Clinical Findings History/Symptoms Specifically ask about: Many patients with aortic stenosis are asymptomatic even with severe LV outflow obstruction, however symptoms may include: Chest pain (angina —even in absence of CAD) Exertional dyspnea Syncope Signs of heart failure: fatigue, weakness, and dyspnea Gastrointestinal bleeding due to angiodysplasia PE Specifically look for or establish: JVD (jugular venous pulse: may show prominent "a" waves) Carotid pulse: While the carotid pulse is still normal in mild aortic stenosis, it becomes weak, with delayed and plateaued peak in advanced disease which is called pulsus parvus et tardus Crackles Lower extremity edema Heart sounds: S 1 is usually normal or soft. S 2 is physiologically split in mild disease. In advance disease paradoxical splitting of the S 2 occurs, resulting from late closure of A 2. In severe stenosis: S 2 becomes single (because the aortic component of the second heart sound, A 2, is absent when the valve becomes immobile) S 4 also can be heard An ejection click can be heard immediately after S 1 in children and young adults with congenital aortic stenosis. Systolic ejection murmur is heard at the second intercostal space in the right upper sternal border radiates to carotid arteries. This murmur has a diamond shape with mid to late systolic peak depending on the severity of valve stenosis. In advance disease, murmur intensity lessens, as the stroke volume becomes reduced. Suprasternal notch thrill also may be felt in patients with aortic stenosis. Differential Diagnosis Include aortic stenosis, the differential diagnosis of: Angina CHF or pulmonary edema Syncope Presence of a systolic murmur (e.g. mitral or tricuspid insufficiency, hypertrophic cardiomyopathy) Dx Management Laboratory Consider ordering: Brain natriuretic peptide (BNP): is useful in asymptomatic patients with severe aortic stenosis to predict symptom onset and their need for surgery. It also can be used to predict mortality in patient with severe aortic stenosis Definition BNP is a biomarker of CHF The natriuretic peptides ANP (atrial natriuretic peptide) and BNP (brain or B-type natriuretic peptide) are synthesized in the heart. ANP and BNP PHYSIOLOGICALLY decrease blood pressure, by: Vasodilation A reduction in cardiac preload by: Increasing venous capacitance Suppression of renin-angiotensin system diuresis natriuresis Suppression of sympathetic tone Pathophysiology BNP is activated, synthesized and released in heart failure triggered by wall stretch (dilation and ↑ pressure within the atrial & ventricle myocardial cells) Blood concentrations of BNP increase with age Women have higher BNP levels than men Patients with GFR < 60 ml/min have higher concentrations, additionally patients with the following may have increased levels: DM HTN Severe lung disease Drugs used in treating heart failure affect BNP levels: Diuretics and vasodilators ↓ BNP levels (through decreasing filling pressures in the atria and ventricles) ACE inhibitors, Angiotensin receptors blockers and spironolactone all ↓ BNP levels Use in Screening Among 1374 patients at risk for heart failure, a strategy of yearly BNP-based screening (compared to usual care) was associated with a decrease in the incidence of LV dysfunction with or without heart failure (5.3% vs 8.7%) and a decrease in the incidence of heart failure (1.0% vs 2.1%), and a 40% decrease in the rates of emergency hospitalizations for major cardiovascular events. (JAMA. 2013 Jul 3;310(1):66-74) Use in Diagnosis The accuracy of BNP in the diagnosis of CHF in primary care is high 70% positive predictive value higher with higher values (recall the PPV is the probability that a patient has disease when restricted to those who test positive) Levels greater than 400 positive predictive value of 98% 98% negative predictive value (cut off of 76 pg/ml) (recall the NPV is the probability that a patient does not have disease when restricted to only those patients who test negative) Low or normal levels of BNP can confidently rule out CHF as a diagnosis Acute setting BNP levels < 100 pg/ml are very unlikely to have CHF NPV of values < 50 pg/ml is 96% (New Engl J Med 2002;347:161) BNP is most useful for the exclusion of HF in ER patients presenting with acute dyspnea. BNP is not a stand-alone test, but it complements clinical judgment and other testing. Use in Therapeutic Management Treatment decisions guided by symptoms AND BNP levels may be associated with improved clinical outcomes. (ACC 2010;55:645) BNP guided therapy reduces all-cause mortality ~ 24% in patients < 75 with chronic HF in patients compared to patients managed without the use of BNP (i.e. targeting a specific BNP with CHF medications). (Arch Int Med 2010;170:507) BNP use in the emergency setting in patients presenting with acute dyspnea is associated with a decreased length of hospital stay by 1 day, and reduced admission, but not hospital mortality rates. (Ann Int Med 2010:153:728) References The N-terminal Pro-BNP investigation of dyspnea in the emergency department (PRIDE) study. Am J Cardio 2005;95:948 Natriuretic peptide-based screening and collaborative care for heart failure: the STOP-HF randomized trial. JAMA. 2013 Jul 3;310(1):66-74. Radiology/Cardiology Consider ordering: Echocardiogram: Is key in the diagnosis of aortic stenosis, and for assessment of severity and left ventricular function. Echocardiogram may show left ventricular hypertrophy, thickened and immobile aortic valve and dilated aortic root Cardiac catheterization: is used when noninvasive tests are inadequate for diagnosis, and for the preoperative assessment of risk for coronary artery disease ECG: may show left ventricular hypertrophy, left atrial enlargement, ST-T changes, AV block, hemiblock, or bundle branch block. LVH as measured on chest leads \| Scroll to see arrows Left ventricular hypertrophy and tall peaked T waves(video 06:52) \| G.Ferenchick, MD Resources for electrocardiogram interpretation Chest x-ray: usually nonspecific, but it may show: Calcification of aortic cusps in lateral view Left atrial enlargement, pulmonary congestion Poststenotic dilatation of aorta. ECG exercise stress testing: it may be considered in asymptomatic patients with severe aortic stenosis for risk assessment (i.e. to "unmask" symptoms); EST is contraindicated in symptomatic patients with severe aortic stenosis. The presence of limiting symptoms (e.g. dyspnea) is associated with a 1-year symptom free survival of 49% vs 89% of those with AS who do not experience limiting symptoms with EST. (Eur Heart J 2005;26:1309) Other prognostic parameters that can be used in EST include: the occurrence of angina, or near syncope, insufficient increase in BP during exercise, complex arrhythmias and repolarization abnormalities. The present any of these factors is associated with event free survival of 20% at 2 years vs > 80% in those without AS who do not have these features on EST. (Heart 2001;86:381) Normal EST is associated with a very low risk of death. (Am J Cardiol 2009;104:972) Radionuclide studies: can be used to evaluate myocardial perfusion and LV function at rest and exercise in AS patients. Multidetector computed tomography (MDCT)can be used in the work up of AS patients to assess aortic valve area accurately. Aortic valve calcification (AVC) load predicts excess mortality in the presence of AS (J Am Coll Cardiol. 2014 Sep 23;64(12):1202-13) Rx Management Chronic Management Avoidance of strenuous activity is recommended in moderate or severe AS Although treatment of aortic stenosis is not always recommended in asymptomatic patients, echocardiography is necessary in those patients to monitor the progression of valve stenosis: Every 3-5 years in mild aortic stenosis Every 1-2 years in moderate stenosis Annually in severe cases The onset of symptoms in patients with AS is the most important variable in deciding on aortic valve replacement (AVR) The mortality rate in patients with AS after the development of symptoms if ~ 2% per month (Lancet 2015 Jun 20;385(9986):2485-91) Aortic valve replacement may include: Surgical aortic valve replacement (SAVR), or transcatheter aortic valve replacement (TAVR) According to the American College of Cardiology, "The choice of proceeding with surgical AVR versus TAVR is based on multiple factors, including the surgical risk, patient frailty, comorbid conditions, and patient preferences and values" (J Am Coll Cardiol 2017 Jul 11;70(2):252-289.) AVR is the only effective treatment for older adults with severe symptomatic aortic stenosis, and severe symptomatic AS is a "Class I" recommendation in many guidelines. AVR indicated for patients in severe aortic stenosis with: Symptoms Left ventricular ejection fraction < 50% Patients having coronary artery bypass graft (CABG) surgery or any valvular surgery AVR may be considered for patients with moderate aortic stenosis having CABG, aorta or other valvular surgery Asymptomatic patients with severe aortic stenosis and abnormal stress test (symptoms, fall in BP and complex ventricular arrhythmias) or with high risk of rapid progression (moderate to severe calcification, Extremely severe aortic stenosis (aortic valve area < 0.6 cm2, mean gradient > 60 mm Hg, jet velocity > 5 meters/second) Severe ventricular hypertrophy (> 15 mm) without HTN Medical treatment in aortic stenosis generally does not affect progression of aortic valve disease nor improves survival. However, medications are used to carefully control concurrent cardiac conditions: If↑ blood pressure– medications should be initiated at low doses including angiotensin-converting enzyme (ACE) inhibitors, and second-generation dihydropyridine calcium channel blockers. Diuretics should be used with caution because of the potential frisk of reducing cardiac output. If pulmonary congestion — cautious treatment with digitalis, diuretics, angiotensin-converting enzyme (ACE) inhibitors If acute pulmonary edema — nitroprusside infusion can be used but should be done in intensive care unit (ICU) with invasive homodynamic monitoring If angina — nitrates or beta blockers Statins have not clearly been shown to slow progression of aortic stenosis Prognosis/Clinical Course Aortic stenosis has the worst prognosis of all valvular lesions, and it is associated with increased morbidity and mortality in elderly. Asymptomatic patients have an excellent prognosis for survival, with mortality rate of less than 1% per year. However, survival notably worsens after symptoms appear, and symptomatic severe aortic stenosis is associated with an expected annual death rate of 25% without aortic valve replacement, (Heart 2011;97:253) In patients with symptomatic aortic stenosis who are treated with only medical therapy; median survival depends on clinical presentation being 1.5-2 years for heart failure, 3 years for syncope, and 5 years for angina. Seventy percent of symptomatic patients without valve replacement therapy die within 3 years (Ann Intern Med 2017 Jan 3;166(1):ITC1-ITC16) AVR may improve survival in asymptomatic and symptomatic patients with severe aortic stenosis. In patients with Heyde's syndrome (aortic stenosis associated with colonic angiodysplasia), aortic valve replacement decreased the rate of re-bleeding by 81% among patients with gastrointestinal hemorrhage. (Am J Gastroenterol 2014 Apr;109(4):474-83) Resources Calcified and thickened aortic valve Calcified aortic stenosis Clinical Practice Guideline 2017 AHA/ACC Focused Update of the 2014 AHA/ACC Guideline for the Management of Patients With Valvular Heart Disease: A Report of the American College of Cardiology/American Heart Association Task Force on Clinical Practice Guidelines. J Am Coll Cardiol 2017 Jul 11;70(2):252-289\|Full text 2014 AHA/ACC guideline for the management of patients with valvular heart disease: a report of the American College of Cardiology/American Heart Association Task Force on Practice Guidelines. J Am Coll Cardiol 2014 Jun 10;63(22):e57-185. \|Full text ACC/AHA 2006 guidelines for the management of patients with valvular heart disease. Circulation 2006;114:e84\| Full text Guidelines for the management of valvular heart disease. Eur Heart J 2007;28:230\|Full text References Aortic Stenosis.Ann Intern Med 2017 Jan 3;166(1):ITC1-ITC16 Calcific aortic stenosis: from bench to bedside — emerging clinical and cellular concepts. Heart 2003;89:801\|Full text The medical management of valvular heart disease. Heart 2002;87:395\|Full text Essential Evidence Plus: Aortic Stenosis (requires MSU net ID log-in) 2017 AHA/ACC Focused Update of the 2014 AHA/ACC Guideline for the Management of Patients With Valvular Heart Disease: A Report of the American College of Cardiology/American Heart Association Task Force on Clinical Practice Guidelines. Circulation 2017 Jun 20;135(25):e1159-e1195. The video image of the aortic valve is used with permission from Wikimedia commons\|By Valveguru — Own work, CC BY-SA 3.0, Contributors 1 Contributor Gary Ferenchick, M.D., M.S., F.A.C.P. Department of Medicine Author Table of Contents Summary Definition General/Background Risk Factors/Causes Pathogenesis History/Symptoms PE Differential Diagnosis Dx Management Laboratory Radiology/Cardiology Rx Management Chronic Management Prognosis/Clinical Course Resources Clinical Practice Guideline References Call us: (517) 432-6200Contact InformationPrivacy StatementSite Accessibility Call MSU: (517) 355-1855Visit: msu.eduNotice of Nondiscrimination SPARTANS WILL.© Michigan State University
17032	https://easingthehurrysyndrome.wordpress.com/tag/illustrative-mathematics/	Illustrative Mathematics \| Easing the Hurry Syndrome Easing the Hurry Syndrome A blog about taking time to learn from my students Home About Jennifer Wilson Math Practices Learning Progressions#LL2LU Mathematics Teaching Practices#LL2LU RSS Tag Archives: Illustrative Mathematics What I Learned Today: Scale Drawings and Maps 07 Sep I asked my 15-year-old what she learned today at school. She paused for a moment and then answered my question by asking me what I learned at school today. It took me a while to think about what I had learned [which will make me more patient when I ask her the question again tomorrow], and then I remembered and shared with her: We are working with some teachers who are using the Illustrative Mathematics 6–8 Math curriculum. The 7 th grade teachers are in Unit 1, Scale Drawings. They are working with Scale Drawings and Maps.Today I learned to look more closely at the scale given for a map. Look at the following for a moment. What’s the same? What’s different? The last two are from Illustrative Mathematics, which you can download for free at openupresources.org. What’s different about the scales on the last two? Attend to precision, MP6, says, “Mathematically proficient students try to communicate precisely to others. … They state the meaning of the symbols they choose, including using the equal sign consistently and appropriately.” I’m not sure that we would have noticed a difference, except that we were trying to find some assessment items from another source and saw that many aligned to 7.G.A.1 included a scale in the form of “1 cm = 100 miles”. I’ve looked at lots of maps and I never noticed the incongruity of saying that 1 cm equals 100 miles. We don’t really mean that 1 cm equals 100 miles, right? Not in the same sense that we say 4 quarters equals $1 or 3+4=7. Is there any wonder that our students misuse the equal sign? And so the journey continues, grateful for the authors of this curriculum who make me pay closer attention to attending to precision and grateful for my daughter who makes me think and share about what I’m learning, too … Leave a comment Posted by jwilson828 on September 7, 2017 in Geometry, Grade 7, Illustrative Mathematics 6–8 Math curriculum Tags: attend to precision, CCSS-7.G.A.1, equal sign, Illustrative Mathematics, MP6, Open Up Resources, scale drawings and maps Using Rigid Motions for Parallel Lines Angle Proofs 20 Nov CCSS-M.G-CO.C.9. Prove theorems about lines and angles. Theorems include: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. After proving that vertical angles are congruent, we turned our attention towards angles formed by parallel lines cut by a transversal. My students come to high school geometry having experience with angle measure relationships when parallel lines are cut by a transversal. But they haven’t thought about why. We make sense of Euclid’s 5 th Postulate (wording below from Cut the Knot): If a straight line crossing two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if extended indefinitely, meet on that side on which are the angles less than the two right angles. We use dynamic geometry software to explore Parallel Lines and Transversals: And then traditionally, we have allowed corresponding angles congruent when parallel lines are cut by a transversal as the postulate in our deductive system. It makes sense to students that the corresponding angles are congruent. Then once we’ve allowed those, it’s not too bad to prove that alternate interior angles are congruent when parallel lines are cut by a transversal. But we wonder whether we have to let corresponding angles in as a postulate. Can we use rigid motions to show that the corresponding angles are congruent? One student suggested constructing the midpoint, X, of segment BE. Then we created a parallel to lines m and n through X. That didn’t get us very far in showing that the corresponding angles are congruent. (image on the top left) Another student suggested translating line m using vector BE. So we really translated more than just line m. We really translated the upper half-plan formed by line m. We used took a picture of the top part of the diagram (line m and above) and translated it using vector BE. We can see in the picture on the right, that m maps to n and the transversal maps to itself, and so we conclude (bottom left image) that ∠CBA is congruent to ∠DEB: if two parallel lines are cut by a transversal, the corresponding angles are congruent. Once corresponding angles are congruent, then proving alternate interior (or exterior) angles congruent or consecutive interior (or exterior) angles supplementary when two parallel lines are cut by a transversal follows using a mix of congruent vertical angles, transitive and/or substitution, Congruent Supplements. But can we prove that alternate interior angles are congruent when parallel lines are cut by a transversal using rigid motions? Several students suggested we could do the same translation (translating the “top” parallel line onto the “bottom” parallel line). ∠2≅∠2’ because of the translation (and because they are corresponding), and we can say that ∠2’≅∠3 since we have already proved that vertical angles are congruent. ∠2≅∠3 using the Transitive Property of Congruence. We conclude that when two parallel lines are cut by a transversal, alternate interior angles are congruent. Another team suggested constructing the midpoint M of segment XY (top image). They rotated the given lines and transversal 180˚ about M (bottom image). ∠2 has been carried onto ∠3 and ∠3 has been carried onto ∠2. We conclude that when two parallel lines are cut by a transversal, alternate interior angles are congruent. Another team constructed the same midpoint as above with a line parallel to the given lines through that midpoint. They reflected the entire diagram about that line, which created the line in red. They used the base angles of an isosceles triangle to show that alternate interior angles are congruent. Note 1: We are still postulating that through a point not on a line there is exactly one line parallel to the given line. This is what textbooks I’ve used in the past have called the parallel postulate. And we are postulating that the distance between parallel lines is constant. Note 2: We haven’t actually proven that the base angles of an isosceles triangle are congruent. But students definitely know it to be true from their work in middle school. The proof is coming soon. Note 3: Many of these same ideas will show that consecutive (or same-side) interior angles are supplementary. We can use rigid motions to make the images of two consecutive interior angles form a linear pair. After the lesson, a colleague suggested an Illustrative Mathematics task on Congruent angles made by parallel lines and a transverse, which helped me think through the validity of the arguments that my students made.As the journey continues, I find the tasks, commentary, and solutions on IM to be my own textbook – a dynamic resource for learners young and old. Advertisement Leave a comment Posted by jwilson828 on November 20, 2014 in Angles & Triangles, Geometry, Rigid Motions Tags: CCSS-M.G-CO.C.9, Geometry Nspired, Illustrative Mathematics, Parallel Lines Cut by a Transversal, proving alternate interior angles congruent using rigid motions, proving corresponding angles congruent using rigid motions, rigid motions, TI-Nspire Technology Two Wheels and a Belt 09 Aug What do you do on the last day of class? I needed to spend some time talking with students about their grades. So they worked on a task while I met with individual students. Which means I wasn’t able to orchestrate as productive of a mathematical discussion as I would have liked. We started with the pictures, absent of any explanation or measurements. What question could we explore? What is the circumference of the circles? 1 What is the volume of the cylinder? 1 are circles equal? 1 are the 2 circles the same size what are their radii 1 are the areas of the two circles equal 1 are the bases the same size 1 are the bases the same size? 1 could the one of the bases of the cylinder have the same diameter as the other? dilation 1 does the highlighted line have any significance to the cylinder 1 does this shape qualify for a cylider 1 is it a cylinder or 2d lines and circles resembling a bike chain? 1 is it a cylinder or gears 1 is it a cylinder 1 is it a cylinder? is it 2 similar circles with a sring around them? 1 is it a cylinder? 1 what are the wheels connected by the belt being used for? 1 what is the length of the darker line? 1 what is the length of the line around the wheels? 1 what is the radius of both circes? 1 what is the relationship between the lines and the circles 1 what is the volume of the cylinder 1 what is the volume of the figure? 1 whether its a cylinder or a dialation 1 why are the circles not touching 1 Does is help to know that these are two wheels and a belt? Does looking at a picture of some gears help to make sense of the picture? I asked the question again. What information do you need to determine the length of the belt? Teams worked together to make a list of measurements. Then they started working. I checked in every few minutes in the midst of meeting with students about grades. I knew some of the misconceptions that students would have because we did this task last year. Students had the opportunity to look for and make use of structure. Is the quadrilateral a rectangle? Many students were calculating as if it were. Someone convinced us that the quadrilateral is only a trapezoid. Students corrected calculations and made more progress. I checked in again. Is the part of the belt highlighted in green a semicircle? Many students were calculating as if it were. Someone convinced us that the arc is greater than a semicircle. It wasn’t the ideal way to have class and discuss student work. But for the last day of class, it wasn’t bad. And now the journey continues to a new school year … 2 Comments Posted by jwilson828 on August 9, 2014 in Circles, Geometric Measure & Dimension, Geometry, Modeling with Geometry Tags: Illustrative Mathematics, look for and make use of structure, Quick Poll, TI-Nspire Navigator, what question can we explore Making Time for Tasks 11 Mar I co-presented a power session at the T3 International Conference Sunday morning, and I’ve posted the stories that I shared of Running around a Track I and Running around a Track II. I have also posted Running around a Track III during which I processed the feedback we received during and after the sessions. Several participants asked about the time that rich tasks take, and so I thought it might be helpful to think about making time for tasks in this separate post. I enjoy teaching on a block schedule. I know I don’t feel as rushed as I would otherwise. But a friend asked me how often we do these types of tasks in our classes. How often do you do this type of rich task? There are some tasks through which the math content can be initially and naturally be learned. I think of Placing a Fire Hydrant(and last year’s notes here) and Locating a Warehouse(with last year’s notes here). There are some that are more culminating tasks for a unit, especially the modeling tasks. A typical unit for us lasts about 8 days. (See some of the Unit Student Reflections posts to know more about how we organize content:Special Right Triangles, Dilations.) We do the culminating performance type tasks 1-2 of those days towards the end of the unit. On the other days, we are learning the content using practices such as look for and make use of structure and look for and express regularity in repeated reasoning, but not always through tasks. For example, when our learning goal is G-C.A.2 Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle. I find that exploring the relationships using technology first, verifying conjectures using formative assessment questions (more skill-based practice), and then moving towardswhy works best for my students. If you know of a good task for learning G-C.A.2, I would love to know. But for now, at least, we end this unit with a few culminating tasks instead of beginning the unit with them. The culminating tasks are Circles in Triangles,Inscribing and Circumscribing Right Triangles,Temple Geometry.I will note that while we didn’t actually do the tasks until the end of the unit, I did show students one of the diagrams from the beginning so that they could keep in mind throughout the unit where we were heading. So now that you know that I don’t give a rich task every day, let’s think about what steps we might take to reduce the class time for the tasks. [These questions and comments will make more sense if you have looked at how the lessons for Running around a Track I and Running around a Track II played out with my students.] One of the participants in our Sunday session wondered what would have happened if I had given the students a teaser of what was to come the day before this lesson. What if I had shown them the picture and asked what they wondered then? Or sent a link to a Google form for them to submit their question outside of class? Or what if I actually showed them the tasks and questions (blacking out information in I that would give away II or vice versa) just so students could begin the process of thinking about the structure of the track before they did any calculations in class? I think these are great ideas – having students spend the “alone” time for processing the questions being asked could definitely make the time spent on the task take less class time. Another question that tied into the idea of planning lessons with rich tasks was whether we use a textbook. I think of the textbook as a source of information and practice problems for students. We don’t have our units arranged like our textbook, but I do reference page numbers for each lesson on the student syllabus so that they will know where to look for extra help and practice. (Our textbook isn’t CCSS at this point, and I’m not convinced that the new ones have geometry written in the spirit of CCSS. When I’m reviewing a new textbook, I first look at the Table of Contents. When transformations is still the topic of chapter 8, I find myself skeptical.) Someone else asked about how rich tasks complement some of the skills practice that students need. We don’t get to every practice problem that we include on our student handouts, but I have finally had the time to work through them, and so I post the worked handouts online for students to check problems they do outside of class for additional practice. We also give online practice assignments with two chances to students through Canvas so that they can get immediate feedback on what they know and don’t know. We are trying to teach our students how to use formative assessment. [I’ve been reading Transformative Assessment and Transformative Assessment in Action, both by James Popham. According to Popham, the first level of formative assessment is when it is used by the teacher to make instructional adjustments as needed to further student learning. The second level of formative assessment is when it is used by the student to make learning adjustments as needed to further learning. The third level of formative assessment is when it is used by the class as a whole to help all students meet the learning goals of the lesson, and the fourth level of formative assessment is a transformation of the school – all teachers are learning about formative assessment in school wide PD and practicing it in their classrooms.] My students and I have explicitly discussed in class that if you take the online practice assignment and miss every problem, you should make a learning adjustment before trying again. So if I don’t use a textbook, how do I plan my lessons? The top three sites that I use are Illustrative Mathematics, the Mathematics Assessment Project, and the Math Nspired lessons at TI’s site. I also follow a lot of bloggers. It helps that I’ve taught geometry for 20 years, and it helps that I’ve been making an effort for students to learn by doing for at least 17 of those years. (I owe a huge apology to all of the students I had the first 3 years.) But it does take time. I am lucky to work with a great team of geometry teachers who are willing to help and try tasks and use formative assessment with their students. We taught our CCSS Geometry course last year for the first time, and our administrator worked it out so we could have 4 teachers and 30 students in our first block class together. We worked through the lessons together with the students and each other, and then the other 3 teachers had a planning block after that class so that they could correct everything we had done wrong the first time before they taught it in their own classrooms the rest of the day. This year, we have a team of Algebra 2 teachers doing the same thing, and next year, our Algebra 1 teachers will teach one class together. I didn’t know what my administrator would say when I proposed this idea to him, but I’ve learned it doesn’t hurt to ask. It was a total scheduling pain, and some of our other classes were more crowded, but that was worth the sacrifice for what we learned teaching the first class together. I’ve just told you what works for us in our efforts to include more rich tasks, but it’s only one perspective. What works for you? What additional sites do you use to find rich tasks? Do you start with them more often than end with them? We’d love to learn more from you as the journey continues … easing the hurry syndrome, one task at a time. 1 Comment Posted by jwilson828 on March 11, 2014 in Geometry, Standards for Mathematical Practice Tags: #NspiredatT3, dynamic geometry software, formative assessment, geometry, Illustrative Mathematics, incorporating rich tasks, TI-Nspire Technology Running around a Track III 11 Mar I co-presented a power session at the T3 International Conference Sunday morning, and I’ve posted the stories that I shared of Running around a Track I and Running around a Track IIfrom the Illustrative Mathematics tasks Running around a Track Iand Running around a Track II. In this post, I want to process the feedback that we received from the participants during and after the session. (There is not really an IM task called Running around a Track III, although there could be one from some of the suggestions participants gave for extensions!) Several participants asked about the time that it takes to do rich tasks in class. I’m going to address that conversation in my next blog post, Making Time for Tasks. The tasks have students make sense of the lanes on an Olympic Track. One participant didn’t understand why I changed tasks with the two classes. I neglected to explain that during the session. I was really just doing an experiment to see how the tasks were different and how students approached them depending on the given information. Was one easier for students than the other? Did I need to scaffold the tasks for my students differently than they were written? Were students more successful with one task than with the other? On Saturday, I went to a session by @bamentj from Darwin, Australia and learned about TodaysMeet. I wondered about using it as a backchannel during the session. A lot of participants were using Twitter throughout the conference, but we wanted a place where participants could interact with each other more than usual in a large session – and not get lost in the conference hashtag (or use two hashtags to make a subset of tweets for our session) being used by the other power sessions as well. Even though this meant that others at the conference wouldn’t find out as much as they might have otherwise about our session, we still wanted to try it.We made a room called CCSSPower. The link will only be live through March 15, 2014. Several times we asked participants to use the protocol “I like, I wish, I wonder …” to provide feedback. So it turned out that we didn’t use TodaysMeet as effectively as we could have. In fact, one of the first posts I read was from Joe: I wonder what the purpose of today’s meet.com is. I did not see the use other than to post comments/questions that never were answered. My reply, after the session (in 3 posts): Hi, Joe. Thank you for your comment. If we were to use TodaysMeet again, we would have a second projector to observe and use the comments. I like that the back channel can give participants a chance to communicate, whether or not the instructors are able to address the comments. But I will definitely use it differently if I use it again in a session. I’ve also thought since my post, that since we had co-presenters, whether or not we had the second projector, one of us could have been monitoring a second computer with the backchannel while the other spoke. We learn from our mistakes, right? As I have read through the other comments so that I could address some of the questions in this blog post that we didn’t get to address in the session, I will say that while we could have used the backchannel more effectively, it wasn’t a disaster. Those who were on the backchannel were having their own conversation. Several of the questions did come out in our whole group discussion, and some questions were answered by others in the backchannel without the presenters having to get involved. One participant liked the pre-assessment ACT question being the same as the post-assessment. I am glad that one of the teachers on our geometry team had the idea to include the question in the lesson. I would not have thought of that myself. I looked back at Running around the Track I this week and noticed that one commenter suggested that the content in parts a and b of the task was on a grade 7 level. That might be true, but especially in this first year of CCSS implementation, the data I received from sending the ACT question without choices at the beginning of class (around 50% correct for both classes) indicated that our students had not thought through the mathematical content in the task before. Several ideas for extensions came out during the session. What about having students also calculate area of the track for the ACT question? What about having student calculate the amount of paint needed for the lanes? Could you have students measure a local track with a trundle or wheel? It occurred to me that it would be nice to take students to the track for the lesson. And then it also occurred to me that it would add at least 30 minutes to the time of the lesson for a visit to the track. I will note that one student in particular was engaged by this lesson more than any other this year. I asked him recently what his career pathway was, and he answered in all serious that is was to be a professional football player. He was an expert in class during this lesson. Another suggestion was to use a video. I agree. If you find the right clip, please share it with me! I started by looking for a clip, but those that I found were longer than I wanted to use, and longer than I had time to search for the perfect segment to watch. A few more comments from the back channel: With regards to everyone missing the question about whether the lane lines were similar: If everyone got the quick poll wrong, I wonder how they would respond if you told them the answer is “no”, could they rethink their reasoning. I wonder how you selected the student to present his“wrong” answer? Perseverance is a best practice that we have to facilitate in our classrooms. FYI while we were doing the 400 meter question…US women won the 4×400 meter gold at worlds in Poland. What do you in your classroom when everyone gets a wrong answer? If you decide to have someone explain their work anyway to correct incorrect thinking, how do you select students to present their work? Do you use a random student generator (we have one where we check roll in our PowerTeacher grade book)? Or “equity sticks” (usually tongue depressors with student names – some teachers replace and some teachers don’t replace when you call on students)? Or keep a clipboard with notes about whom you’ve chosen for whole class discussions? I’ve been trying the latter this year. It’s not perfect, but it is a start to at least paying attention to how often I call on students. And so the journey continues … collaborating with educators from all over the world to improve our classroom practices. Thank you for the opportunity to learn from you, and thank you to Ellen from Illustrative Mathematics for sharing such great resources with us during the session and giving us a preview of what is coming soon to their website. 1 Comment Posted by jwilson828 on March 11, 2014 in Circles, Geometric Measure & Dimension, Geometry, Professional Learning & Pedagogy, Standards for Mathematical Practice Tags: "I like ... I wish ... I wonder" protocol, #NspiredatT3, backchannel during professional development session, Illustrative Mathematics, providing productive feedback Running around a Track II 10 Mar I gave a different class of students Running around a Track II recently. I started by showing them the same picture of the start of the 2012 Olympic Women’s 400 M race. I changed the prompt, though. What do you wonder? I wonder why they start at different lines instead of at the same place. 1 I wonder who 1st realized that the smaller of the concentric circles would be shorter 1 I wonder whats special about this picture? 1 I wonder why if you are running on the outter circle its the same distance as running on the inner circle 1 I wonder what the distance is that they are running 1 I wonder what event they are running? 1 I wonder why tracks are ovals 1 I wonder if each one of those arcs are similar by using dialations 1 I wonder what the exact distance of all of the lanes are 1 I wonder what is happening. 1 I wonder if the circles are cocentric 1 I wonder how big the race track is as a whole. 1 I wonder why all of the starting points are in different places 1 I wonder what the ratio of the innermost ring is to the rest of the rings… or are they all congruent? 1 I wonder why the runners start at different starting positions. 1 I wonder why there are flags that are the same 1 I wonder if its better to be closer up on the outside lane than farther back on the inside lane 1 I wonder if that’s the olympics 1 I wonder which lane is the shortest distance 1 I wonder how they decided how far back to put the first runner 1 I wonder how I can figure out how to find the different measures of each of those arcs of the circle 1 [And from my cynic:] I wonder what question you’re going to ask about this image. 1 I wonder why are there so many americans 1 I wonder what shape the track is 1 I wonder why there are 3 people from us and why they are not starting from same side? 1 I wonder who the runners are 1 I wonder if the distance in the inner circle is different than the distance in the outer circle even if they start in differert places 1 I wonder if all of the starting points are marked off by distance or angle. 1 I wonder why the track runners are spaced apart rather than running all together. 1 I wonder what the arc measure of the track is? 1 Next, I sent out a Quick Poll of a question from a released ACT that students should be able answer as a result of the lesson. Who already knows how to solve this? Who knows how to solve it quickly, since the ACT is timed? I removed the choices, just to see how students answered without them. We didn’t talk about the results. I told them we would revisit the question at the end of the lesson. Just less than half of the students had it correct initially: 14/31. Before we started analyzing what was happening with each lane, I asked the students whether the lane lines were similar. Just for the record, I wouldn’t have thought to ask this question on my own. It was suggested in the commentary for the task on Illustrative Math, so I thought I would try it. I’m not sure I really had a backup plan for this formative assessment check. Not one student got this correct? How should we proceed? Understand that at this point, the students don’t know if everyone is correct or no one is correct. There is a great feature of Navigator that lets me “show correct answer” or not. While I monitor student responses with my projector screen frozen, I decide whether I will show the correct answer. What does it mean for two figures to be similar? Several students wanted to answer. I called on F.K., because F.K. loves dilations, and I knew that she would explain similarity in terms of transformations. Two figures are similar if there is a sequence of transformations including a dilation that will map one figure onto the other. Okay – so can we describe a dilation that will map one lane onto another lane? What would be the center of the dilation? At this point, we moved to the technology. I had created the track on a Graphs page of TI-Nspire. Mainly because I wanted to prove that I could without asking my technical author friends Bryson and Jeff. It’s not beautiful, but it is functional. I didn’t even use this document in the class with Running around a Track I, but maybe it would be helpful now. I have a slider set up to change the radius of the circular part of the track. Are the lanes of the track similar? Does the technology help you see it? What happens if we dilate the straight part of the track about the center of the track by a scale factor of 1.1? And then let’s dilate the circular part of the track using the same center. (Bryson and Jeff could probably figure out how to make this happen all at once – my track is in 3 separate parts.) Is there a dilation to show that the lanes of the track are similar? The straight part of our track is a problem. There is no line that contains the center of dilation and the endpoint of each straight part of the track. One of the students wondered if each one of those arcs are similar by using dilations. Now if we are just talking the arcs, then yes, they are similar using a dilation about the center of the arc – but not using the center of the track. The technology helped us make sense of whether the lane lines are similar. Now back to the real task. This task is a bit different – it has less scaffolding than the first task. Students jump right in to calculating the perimeter of the track 20 cm inside lane 1 and 30 cm inside lane 2 so that they can determine how far ahead the runner in lane 2 needs to be ahead of the runner in lane 1. I sent a poll to collect the results. It was disastrous, and the bell was about to ring. How in the world could I recover the lesson? We went back to the picture, and ended class by talking in more detail about the diagrams. We started the next class calculating the perimeter of the paint on the most inside lane. Even then, we didn’t have everyone with us. Don’t you love that 5 of my students just assumed the inside lane was 400 m without doing any calculations? We went back to the questions on Running around a Track II, though and really did make progress. To close the lesson, I sent the Quick Poll from the start of class: We are up to 80% correct, from just under 50% at the beginning of the lesson. And with choices, we have 90% correct. I finally showed the students the results and we talked about the misconception for choice C. I think it is interesting to ask students which practices they used when working on a task. And which would you choose if you could only choose one Math Practice? I have shared before that my goal isn’t just to provide opportunities for my students to use the Math Practices in class – but also for them to recognize when they are using them. I ask my students to write a journal reflection each quarter on using a math practice. N.R. writes about this task: In class today and yesterday, we worked on a problem about the track of the 400 meter dash in the Olympics. While working on this problem, we used the math practice of modelling with mathematics. We applied what we learned about circles in this unit to figuring out how far apart runners in different lanes have to start in order to run the same distance and to end at the same area in the straightway section. When solving this problem, I had to use the equations for circumference and perimeter and combine them. Once I finished working the problem, I decided that the runner in lane 1 has to start 7.037 meters behind the runner in lane 2. I also found that the runner in lane 2 has to start 7.666 meters in behind the runner in lane 3. This problem has helped me to be very attentive to detail. In this problem I had to be very careful that I worked everything correctly and completely. 3 Comments Posted by jwilson828 on March 10, 2014 in Circles, Geometric Measure & Dimension, Geometry, Professional Learning & Pedagogy, Standards for Mathematical Practice Tags: formative assessment, Illustrative Mathematics, look for and make use of structure, Quick Poll, similarity using dilation, TI-Nspire Technology Seven Circles …Again 24 Feb We tried Seven Circles Ifrom Illustrative Mathematics a few weeks ago. At the end of class one day, I showed students the diagram and what question they might explore with it. I collected their responses using an Open Response Quick Poll and have shown the results below. What does this figure have to do with geometry? 1 if we connected each top vertex of the triangles, will it make a hexagon? 1 whats the area of all the circles 1 Why are the circles in this shape? 1 what are the circles forming? 2 what is the area of all of the circles 1 why are we looking at circles? 1 are the spaces between the circles triangles? 1 what are the seven circles forming? 1 do all the circles have the same diameter? 1 do all the circles have the same diameter 1 Are the circles’ diameters the same? 1 Can the measures of the triangles that can be drawn through circles be calculated quickly 1 why are all the circles touching? 1 Why are the circles in that certain arrangement? 1 Can you find the area for that? 1 Is there a way to solve non 90° triangles? (With sin, cos, tan, or the other trig functions) 1 Can the circles be mapped onto each other with a rigid motion? 1 when you look at the image what do you see? 1 are the 6 figures that look like triangles in the gaps of the circles considered triangles since their sides arent straight 1 what are the circles for 1 are all of the circles congruent to each other? 1 What is the significance of the circular pattern? 1 How can you find the measurement of each circle 1 What are the triangular looking spaces in between the triangles called? 1 Some students were interested in the space between the circles. Other students wondered whether the circles were congruent. The task is given below. My students felt like it was pretty obvious that this could work with 7 congruent circles. I gave them different sized coins so that they could play. What if the circle in the middle is not congruent to the others? Will this work for 6 congruent circles? Or 8 congruent circles? After students played for a few minutes, I sent them a TNS document that a friend made to explore this task. I used Class Capture to watch while students used the technology to make sense of the necessary and sufficient conditions for 6 circles and 7 circles in the given arrangement. Who had something interesting to discuss with the whole class? Many students saw the regular pentagon or regular hexagon with vertices at the centers of the outside circles and used that to make sense of the mathematics. While I was watching them, I was trying to figure out how we should proceed as a class. We started with Claire’s work. What do we know? We saw a dilation. We saw central angles of a regular pentagon. We saw isosceles triangles, which we bisected to make right triangles. We saw an opportunity to use right triangle trigonometry. We looked for and made use of structure. We reasoned abstractly and quantitatively. And before the bell rang, we looked back at the picture with 7 circles and recognized that the 30-60-90 triangles require that the radius of the center circle equal the radius of the outer circles. We only touched the surface of what we can learn from this task. Last year, we didn’t even do that.Last year, I shared the task with students during their performance assessment lesson, but we spent all of our time on Hopewell Triangles. This year, we got to it, but I know that our exploration could have been better. We began to answer what are the necessary and sufficient conditions for 6 circles. And in the process, we came across an argument for why the 7 circles must be congruent. But we didn’t really solve the conditions for 6 circles. I wanted to write about this as a reminder that we are all learning. In this journey, I am finding good tasks out there to try with my students. And I am more confident about some than about others. Even though I don’t know exactly how the tasks should play out in the classroom, I am going to keep trying them. And I’m not going to throw out the task just because we didn’t get as deep into the mathematics as I wish we had. I will try again next year as the journey continues … Leave a comment Posted by jwilson828 on February 24, 2014 in Circles, Geometry, Right Triangles Tags: Class Capture, Illustrative Mathematics, look for and make use of structure, Quick Poll, reason abstractly and quantitatively, TI-Nspire, TI-Nspire Navigator, What's your question Placing a Fire Hydrant 14 Oct This task is from Illustrative Mathematics.Students are asked to place a fire hydrant equidistant from three locations. My students worked on paper first. I was impressed with their work this year. More so than last year, which could have to do with us swapping Units 1 and 2 from last year to this year. They used paper, rulers, folding, and compasses. Several of them realized that if they could find the circle that contained all three locations, the center would be equidistant (and thus the location of the fire hydrant). However, their methods for finding a circle to contain all three points were not very precise (which meant they didn’t already know everything they needed to know about triangle centers). After students worked on paper & a few students presented their ideas (I deliberately sequenced them, after reading Smith & Stein’s 5 Practices for Orchestrating Productive Mathematics Discussions), we tried to solve a simpler problem. I asked two students in the room to stand up front. I asked another student to stand equidistant from the two students. The student stood at the midpoint of the segment containing student 1 and student 2 as endpoints. Is student 3 the only person who can be equidistant from students 1 and 2? A student in the back of the room said that she could be equidistant as well. She stood in the back. Another student in the middle of the room said that he could be equidistant as well. He stood in the middle. What is true about students 3, 4, and 5? They are all the same distance from student 1 and student 2. What else is true about students 3, 4, and 5? They are in a line. Yes, they are collinear. What else is true about students 3, 4, and 5? What would happen if we drew the line containing those students? What relationship would that line have to the segment containing students 1 and 2 as endpoints? Somehow, we got perpendicular bisector out of the conversation. Note: I’m pretty sure I read about this idea in a Mathematics Teache r years ago, but I do not have a reference for the article. What happens if we add a 3 rd person to our original problem? Are students 3, 4, and 5 equidistant from students 1, 2 and our 3 rd person? No. Can anyone be equidistant to all three of our people? The students recognized the significance of taking the points two at a time – and then eventually determined that the perpendicular bisectors would be concurrent at our point of interest. We moved to our dynamic geometry technology. Students constructed the perpendicular bisectors of the given triangle and moved the triangle. We created the circumscribed circle. We paid attention to what happened to the circumcenter. Is the circumcenter always a good location for the fire hydrant? I used Class Capture to monitor student progress. Can the location of the circumenter give us information about the type of triangle? I made a student the Live Presenter to show us what she found about the type of triangle and the location of the circumcenter. Finally, we ended with the straightedge and compass construction for the perpendicular bisector, focusing on the question “What segments, angles, arcs, and triangles are always congruent in the construction”. For whatever reason, I made the list on the non-interactive whiteboard, but I marked the student findings on the interactive whiteboard. We got into isosceles triangles, congruent base angles, rhombi, and more. And so the journey continues… 3 Comments Posted by jwilson828 on October 14, 2013 in Angles & Triangles, Geometry, Tools of Geometry Tags: circumcenter, Class Capture, Illustrative Mathematics, Live Presenter, perpendicular bisector, TI-Nspire Navigator Reflected Triangles 02 Sep We used a task from Illustrative Mathematics as part of a performance assessment on Rigid Motions. On the next page, △ABC has been reflected across a line into the blue triangle. Construct the line across which the triangle was reflected. Justify your conclusion. Students used TI-Nspire to construct the line of reflection, and we asked them to explain their construction on paper. I monitored their work using Class Capture to see what approaches the students were using. This year, almost everyone created segments BB’ and CC’. Some also created segment AA’. Some then constructed the midpoint of the segments then created the line through those midpoints. Some used the perpendicular bisector tool. Our class discussion focused on what were the fewest number of objects we could construct to get the line of reflection. And then I asked H.K. if she would share her work. She had the same idea as most of the others, but instead of drawing a segment with a vertex and its image as the endpoints (AA’, BB’, or CC’), she constructed the midpoint of segment BC and the corresponding midpoint of segment B’C’. Then she joined those image and pre-image points with the segment tool (actually vector tool) and constructed the perpendicular bisector of the segment. That’s a big deal. H.K. reminded us of an important property of rigid motions. Most of the segments and angles that we found congruent as we were exploring focused on the vertices of the triangle and their images. But the same is true from every point on the pre-image to its corresponding point on the image – not just from the vertices. Thanks for the reminder, H.K. And so the journey continues …. 2 Comments Posted by jwilson828 on September 2, 2013 in Geometry, Rigid Motions Tags: Class Capture, Illustrative Mathematics, line of reflection, rigid motions, TI-Nspire, TI-Nspire Navigator Two Wheels and a Belt 18 Jun Two Wheels and a Belt(and Why I am Convinced That the Standards for Mathematical Practice Must Be How We Do Math) One of the last tasks that we gave our geometry students this year wasTwo Wheels and a Belt from Illustrative Mathematics. A certain machine is to contain two wheels, one of radius 3 centimeters and one of radius 5 centimeters, whose centers are attached to points 14 centimeters apart. The manufacturer of this machine needs to produce a belt that will fit snugly around the two wheels, as shown in the diagram below. How long should the belt be? The correct answer is 53.42 cm, which several students got, in more than one way. Some used correct mathematical reasoning to get the correct answer. Others used incorrect mathematical reasoning to get the correct answer. It is unfortunate that the incorrect reasoning produced the correct answer. What is more unfortunate is that this incorrect reasoning goes unobserved when our focus is only on answers. When our focus is on how we do the math and not just on what we get as an answer, students and teachers can learn more about mathematics. As soon as the students began to construct viable arguments and critique the reasoning of others, the misconceptions in the incorrect solution became evident. And so the journey to help students know and understand mathematics continues … 2 Comments Posted by jwilson828 on June 18, 2013 in Circles, Geometry, Right Triangles Tags: CCSS Mathematical Practices, construct viable arguments, Illustrative Mathematics ← Older posts Search Top Posts & Pages Partitioning a Segment Which One Doesn't Belong? Conditional Statements & Instructional Adjustments Proving Segments Congruent First Sides of a Rectangle Recent Posts #NCTMLive and #T3Learns webinar: Implement tasks that promote reasoning and problem solving, and use and connect mathematical representations. I can elicit and use evidence of student thinking Webinar: Establish Mathematics Goals to Focus Learning, and Elicit and Use Evidence of Student Thinking. I can establish mathematics goals to focus learning What I Learned Today: Scale Drawings and Maps Categories Categories #### Follow Blog via Email Enter your email address to follow this blog and receive notifications of new posts by email. Email Address: Follow RSS - Posts RSS - Comments Create a free website or blog at WordPress.com. Entries (RSS) and Comments (RSS) Easing the Hurry Syndrome Blog at WordPress.com. SubscribeSubscribed Easing the Hurry Syndrome Join 187 other subscribers Sign me up Already have a WordPress.com account? Log in now. Easing the Hurry Syndrome SubscribeSubscribed Sign up Log in Report this content View site in Reader Manage subscriptions Collapse this bar Loading Comments... Write a Comment... Email (Required) Name (Required) Website Privacy & Cookies: This site uses cookies. By continuing to use this website, you agree to their use. To find out more, including how to control cookies, see here: Cookie Policy Design a site like this with WordPress.com Get started Advertisement
17033	https://public.csusm.edu/aitken_html/Essays/Algebra/Binary_operations.pdf	Binary Operations, Monoids, and Groups W. E. Aitken September 2022 Edition∗ This is one in a planned series of documents that survey the foundations of algebra. The documents of this series aim to cover the same sort of topics one would see in an introductory abstract algebra class, but to explore these topics perhaps a bit more abstractly and in more depth than one would necessarily want in a ﬁrst course. So this series should be thought of as foundational but not introductory. As such the target audience is a reader who has already had some exposure to abstract algebra, but wishes to explore or review the foundations of the subject. This document covers ideas related to the concept of binary operations. This includes examples, various properties (commutative, associative) that binary op-erations can have, the ideas of identity and inverse, and so on. This leads to the deﬁnitions of monoids and groups. It also covers basic laws of exponents, the general associativity laws, and the general commutative law. The ideas related to binary operations are ﬁrst step to algebra, and so this document can be thought of as the “ground ﬂoor” for modern algebra built on a solid (but elementary) set-theoretical base. As such it is not intended to cover group theory in any degree beyond an introduction to a few basic notions, and it does not cover ring theory at all. 1 Background and Notation I assume familiarity with some basic set theory including the idea of a Cartesian product A×B of two sets A and B, basic ideas related to functions, the notion of a family, and the notion of a total ordering of a set. I will assume as known the basic number systems (see, for example, in my Number Systems text). These constitute the logical prerequisites, but, as mentioned above, some previous exposure to groups or rings, say, is also useful. Otherwise, some of this document might seem a bit abstract and unmotivated. I use matrices in some examples as well, and the notion of continuous and diﬀerentiable functions come up in a few exercises, but these concepts are not central to the development of the theory. In particular, as part of the set-theoretic background and notion of function, I use the concept of domain and codomain of a function, and use the notation A →B to indicate that the function f has domain A and codomain B (we sometimes say ∗Version of September 16, 2022. Copyright c ⃝2017–2022 by Wayne Edward Aitken. This work is made available under a Creative Commons Attribution 4.0 License. Readers may copy and redistribute this work under the terms of this license. This document is an modiﬁed and expanded version of a set of handouts written for my Math 470 class, introduction to abstract algebra, during the period 2017–2020. 1 that A →B is the type of f). When it comes time to deﬁne exponentiation, I will take as established the concept and basic properties of iteration of functions of the form f : S →S. Such properties are established in my Number Systems text. I will use N, Z, Q, R, C for the set of natural numbers, the set of integers, the set of rational numbers, the set of real numbers, and the set of complex numbers respectively. We have the inclusion N ⊊Z ⊊Q ⊊R ⊊C. There are two deﬁnitions of N that are fairly common. I use the version where N includes 0 as an element: N = {0, 1, 2, . . .}. The other popular version of the natural numbers excludes 0 and starts at 1; I will write this set as Z+. The set Z diﬀers from N by including negative integers. 2 Binary Operations Deﬁnition 1. Let S be a set. A binary operation on S is just a function S×S →S. Example 1. Let S = R. Multiplication ×: R×R →R is a binary operation since it takes as input two real numbers (thought of as an ordered pair) and outputs a real number. Addition and subtraction also give binary operations on R, but division does not. Let S = N. Multiplication and addition are binary operations on N. (Subtrac-tion and division are not). There are many, many other examples. For example, the function f : N × N →N deﬁned by the rule f(x, y) = 17 is a binary operation. This operation ignores the inputs x and y and always outputs the constant value of 17. Such constant operations are not as interesting as the usual operations in arithmetic, but they do satisfy the oﬃcial deﬁnition. Example 2. Let Xn = {1, . . . , n}. Let S = F(Xn) be the set of functions Xn →Xn. Thus S has nn elements. Then composition is a binary operation. To see this, suppose g, f ∈S. In other words, f, g are both functions Xn →Xn. Then the composition g ◦f is also a function Xn →Xn. In other words, if g, f ∈S then also g ◦f ∈S. So composition ◦is a binary operation on S = F(Xn). Let Sn be the set of bijective functions Xn →Xn. We also call such functions permutations. If g, f ∈Sn then g◦f is also a bijection, so is in Sn. Thus composition is a binary operation Sn × Sn →Sn. Note that Sn has n! elements. Example 3. Let Mm,n(R) be the set of real matrices with m rows and n columns. For example, two elements of M2,3(R) are 1 0 √ 2 −1 1 2 11 and 1 e π 2 2e 2π . Then matrix addition + is a binary operation on Mm,n(R). When m = n we write Mn(R) instead of Mm,n(R). Matrix multiplication is a binary operation on Mn(R). 2 Remark. There are two possible notations we can use with binary operations: (i) or-dinary functional notation or (ii) inﬁx notation. Inﬁx notation is the standard notation for binary operations, but it is sometimes illuminating to use functional notation from time to time. First we describe ordinary functional notation. If f : S × S →S is a binary operation, and if a, b ∈S, then in ordinary functional notation we write f(a, b) for the result or output of the function. The element f(a, b) ∈S is often called the value. (Ordinary functional notion is sometimes called “preﬁx” notation, and sometimes we write f a b as a variant for f(a, b) when using this notation.) For inﬁx notation we put the function name in between the two inputs. Typi-cally the function name is not a letter, but a symbol such as ∗, +, or ×. Suppose, for example, that ∗: S × S →S is a binary operation, and that a, b ∈S. Then instead of writing the value as ∗(a, b) or ∗a b, in inﬁx notation we write a ∗b where we put ∗between the two inputs. In inﬁx notation we require parenthesis in what is otherwise an ambiguous expression. For example, a ∗b ∗c could be interpreted as (a∗b)∗c or a∗(b∗c). If we have established grouping conventions we can leave the parentheses oﬀand let the reader mentally supply them. For instance the inﬁx expression x + y · z in various numbers systems (or rings in general) is interpreted as x + (y · z), not as (x + y) · z. We use inﬁx notation in the next deﬁnition: Deﬁnition 2. Let ∗: S ×S →S be a binary operation. We say that this operation is commutative if a ∗b = b ∗a for all a, b ∈S. Remark. To show that ∗: S × S →S is not commutative, you just need to ﬁnd a speciﬁc counterexample. For example, you probably suspect that subtraction −: Z × Z →Z is not commutative. To prove it you need to give a speciﬁc counterexample. We choose 1 and 0 as our inputs in Z. This is a counterexample since 1 −0 ̸= 0 −1. (Note that choosing 1 for both does not give a counterexample, since 1 −1 = 1 −1. So not every choice of a, b ∈S will give you a counterexample). Remark. If we use ordinary functional notation instead of inﬁx notation for a binary operation f : S × S →S, the deﬁnition of commutative can be written as requiring f(a, b) = f(b, a) for all a, b ∈S. An example of such a commutative binary operation (with ordinary functional notation) would be the function deﬁned by the equation f(x, y) = sin(x2y2) 3 where S = R. On the other hand the function g(x, y) = sin(xy2) does not look commutative because of the asymmetry between x and y in the formula. But to prove it is not commutative you need a speciﬁc counterexample. (In this case it is not diﬃcult to ﬁnd such a counterexample). Remark. Suppose Sn is the set of permutations Xn →Xn. Then the binary opera-tion of composition ψ : Sn × Sn →Sn applied to (g, f) can be written in two ways: as ψ(g, f) or as g ◦f. The ﬁrst is ordinary functional notation The second is inﬁx notation using the symbol ◦to denote the (inﬁx version) of the operation. 2.1 Closure Sometimes people describe a binary operation as “closed” to indicate that if a, b ∈S then a ∗b ∈S. For us this term is redundant for a binary operation since ∗is required to be a function S × S →S, so if a, b ∈S then a ∗b must always be in the codomain S. But the concept is useful if we are in a slightly diﬀerent situation. Suppose you have a function mapping ordered pairs in S to values in some set T. In other words, suppose that we have a function ∗: S × S →T. Suppose also that S is a subset of T. We say this function is closed on S if a∗b ∈S for all a, b ∈S. In other words, for any two inputs in S the value “lands in” S and never at a value which is in T but outside S. When we are in this situation, we can restricted the codomain to S and then we would have a true binary operation.1 Informally it is common for people to says that “binary operations must be closed”, but as mentioned above closure build into the deﬁnition of binary operation, and not an extra condition that needs to be imposed on binary operations. However, when you deﬁne a binary operation S × S →S, you do need to check that the values are always in S, and you may call this “checking for closure” if you would like. This is part of making sure the deﬁnition is well-deﬁned, and often this check is left (tacitely) to the reader. The concept of closed becomes more important when we work with submonoids and subgroups. Suppose ∗: S × S →S is a binary operation on S and that A is a subset of S. We say that ∗is closed on A if a ∗b ∈A for all a, b ∈A. If that holds we get a binary operation A × A →A by restricting the domain and codomain of the original operation. Example 4. Subtraction N × N →Z is not closed since the image of the function contains negative integers which are outside N. So subtraction cannot be thought of as a binary operation on N. Subtraction is a binary operation on Z, though. 1In mathematics, restriction of codomain is often done tacitly, or without much fanfare, since the restricted operation is in some sense “the same” and is represented by the same collection of ordered pairs. But the “type” of the operation is changed since the codomain is changed. More often authors will be more explicit about restriction of domain, but even this more standard type of restriction is sometimes done tacitly, even though the “type” of the function has changed. 4 3 Associativity Suppose that ∗: S × S →S is a binary operation. An expression such as a ∗b ∗c that involves more than two elements is ambiguous since a binary operation com-bines only two elements at a time. To avoid ambiguity, you need parentheses to show what two elements are being combined. For example a∗b∗c can be interpreted as (a ∗b) ∗c, which really means d ∗c where d = a ∗b. On the other hand, a ∗b ∗c can be interpreted as a ∗(b ∗c), which really means a ∗d where d = b ∗c. It turns out that many important operations are associative, and the two interpretations of a ∗b ∗c give equal values. Remark. For general operations, that may or may not be associative, a common convention is to group from left to right. So a ∗b ∗c ∗d would be interpreted as ((a ∗b) ∗c) ∗d. For example, subtraction is not associative (for Z), so what does 3 −4 −(−10) −5 mean? According to the left-to-right convention, it is equal to 4. The grouping convention is very important for non-associative operations such as subtraction. For associative operations it doesn’t really make a diﬀerence since the value is the same. Deﬁnition 3. Suppose ∗: S × S →S is a binary operation. Then ∗is said to be associative if (a ∗b) ∗c = a ∗(b ∗c) for all a, b, c ∈S. Remark. If we use ordinary functional notation, and if f : S × S →S is the binary operation, then associativity means f(f(a, b), c) = f(a, f(b, c)) for all a, b, c ∈S. 4 Operation Tables Suppose ∗: S×S →T is given where S is a ﬁnite set. We can describe the operation fully by giving a square table giving all the values of ∗. This involves (1) ordering the set S, (2) making a row and column for each element of S in the order from step 1, (3) for each a, b ∈S writing the value a ∗b in the a-row and b-column. If all elements of the table are in S then we conclude that ∗is closed and so (by choosing T = S), the table can be regarded as describing a binary operation S × S →S. Exercise 1. Given an operation table for addition modulo 4 for the set S = {0, 1, 2, 3}. 5 Exercise 2. Given an operation table for multiplication modulo 7 for the set S = {1, 2, 3}. is this a binary operation (is it closed)? Exercise 3. Given an operation table for multiplication modulo 7 for the set S = {1, 2, 4}. is this a binary operation (is it closed)? Exercise 4. What has to be true about the operation tables for a binary operation to be commutative? Remark. There is no quick criteria that I know of to decide if an operation is associative by just looking at the operation table. Exercise 5. Suppose S = {T, F} be a set with two distinct elements called “true” and “false”. This set is important in logic. Deﬁne the “and” the “or” and the “exclusive or” binary operations. Find the tables for these operators. Exercise 6. Give a table for the composition operation on S3. (Use cycle notation if you are familiar with this). 5 Identity Elements Deﬁnition 4. Let ∗: S × S →S be a binary operation. An identity element for ∗ is an element e ∈S such that e ∗a = a and a ∗e = a for all a ∈S. Exercise 7. Give examples of identity elements. Not all binary operations have identity elements. Can you give examples of this as well? Exercise 8. What are the identity elements (if any) for the logical operations of Exercise 5. Theorem 1. Let ∗: S × S →S be a binary operation. If there is an identity for ∗, then it is unique. Proof. Suppose that e1, e2 ∈S are both identities and consider e1 ∗e2. Since e1 is an identity, we have e1 ∗e2 = e2. Since e2 is an identity, we have e1 ∗e2 = e1. Thus e1 = e2. 6 Remark. From now on if we know that an element is an identity for a given binary operation, then we can call this element the identity because of the above uniqueness theorem. Identities are unique, but left or right identities are not necessarily unique. We begin with the deﬁnition of left and right identities. Deﬁnition 5. Let ∗: S × S →S be a binary operation. A left identity element for ∗is an element e ∈S such that e ∗a = a for all a ∈S. Deﬁnition 6. Let ∗: S × S →S be a binary operation. A right identity element for ∗is an element e ∈S such that a ∗e = a for all a ∈S. Remark. Observe that, by deﬁnition, every identity element is a left identity and a right identity. In fact our deﬁnitions make it clear that an element is an identity if and only if it both a left-identity and a right identity. Exercise 9. Suppose that ∗: S × S →S is given by a operation table. Suppose that e1 ∈S. What has to be true about the e1 row in order for e1 to be a left identity? Suppose e2 ∈S. What has to be true about the e2 column for e2 to be a right identity? What has to be true about rows and columns for e to be a (two-sided) identity? Exercise 10. Deﬁne a binary operation with an operation table so that the op-eration has exactly three distinct left-identities. This shows that left identities are not unique. Does your operation have any right identities? Is it commutative? Exercise 11. Suppose that ∗: S × S →S is commutative. Show that every left identity is automatically an identity (so left identities are unique in this case since identities are unique). As seen above, it is possible in a noncommutative binary operation to have more than one left identity. In this case there will be no right identities. In fact, the following theorem gives uniqueness whenever there exists both left and right identities (even if at ﬁrst we do not know if the right and left identities are the same element). Theorem 2. Let ∗: S × S →S be a binary operation. Suppose e1 ∈S is a left identity and e2 ∈S is a right identity. Then e1 = e2. In particular, e1 and e2 are the unique identity for ∗. Proof. Use the same type of argument as used for Theorem 1. (In fact, we could have started with the above theorem, and then made Theorem 1 a corollary). 7 6 Monoids Deﬁnition 7. A monoid M is a set with a given choice of operation ∗: M×M →M such that (1) ∗is associative, and (2) ∗has an identity element in M. Deﬁnition 8. A commutative monoid M is a monoid whose binary operation is commutative. Example 5. We have many examples of monoids: • The natural numbers N is a commutative monoid under addition. In this case the unique identity is 0. (In this document, we accept 0 as a natural number.) • The natural numbers N is a commutative monoid under multiplication. In this case the unique identity is 1. • The set of functions F(Xn) is a monoid under composition. What is the identity? • The set of permutations Sn is a monoid under composition. What is the identity? • The integers Z form a commutative monoid under addition. What is the identity? • The integers Z form a commutative monoid under multiplication. What is the identity? • The set Q is a commutative monoid under addition. What is the identity? • The set Q is a commutative monoid under multiplication. What is the iden-tity? • The set R is a commutative monoid under addition. What is the identity? • The set R is a commutative monoid under multiplication. What is the iden-tity? • The set Mn,m(R) of commutative matrices is a commutative monoid under addition. What is the identity? • The set Mn(R) of matrices is a monoid under multiplication. What is the identity? • The set C is a commutative monoid under addition. What is the identity? • The set C is a commutative monoid under multiplication. What is the iden-tity? • The set {T, F} is a commutative monoid under the “and” operation. What is the identity? • The set {T, F} is a commutative monoid under the “or” operation. What is the identity? 8 • The set {T, F} is a commutative monoid under the “exclusive or” operation. What is the identity? • The set Zn of integers modulo n is a commutative monoid under addition. What is the identity? • The set Zn of integers modulo n is a commutative monoid under multiplica-tion. What is the identity? Remark. Two common (inﬁx) notations for monoids are multiplicative notation (using juxtaposition or some sort of product sign for the binary operation) and additive notation (using + for the binary operation). In multiplicative notation 1 is a common notation for the unique identity. In additive notation 0 is the common notation for the identity element. Sometimes, for example in logic, we might want to avoid either of these standard conventions. Remark. You may also see the term “semigroup”. A semigroup S is a set with a given choice of operation ∗: S × S →S such that ∗is associative. Note that every monoid is a semigroup, but it is easy to ﬁnd examples of semigroups that are not monoids. Remark. A monoid has two pieces of data: (i) the underlying set M and (ii) the binary operation ∗on M. We sometimes write this package (or “structure”) as an ordered pair ⟨M, ∗⟩. If you change either component, you have a diﬀerent monoid. For example ⟨Z, +⟩is distinct from ⟨N, +⟩. Likewise, ⟨Z, +⟩is distinct from ⟨Z, ·⟩. If the operation on M is clear from context, we sometimes denote the monoid ⟨M, ∗⟩simply as M. In other words, it is common to use the name of the underlying set as also the name for the monoid as well. 7 Inverses Warning: we cannot talk about inverses until we ﬁrst have an identity element. So in this section we will always assume we have an identity element; often we will work in a monoid where we also have associativity. Deﬁnition 9. Let ∗: M × M →M be a binary operation with an identity ele-ment e ∈M. Let a ∈M. We say that a is invertible if there is an element b ∈M such that a ∗b = b ∗a = e. Theorem 3. Let M be a monoid with identity element e ∈M. If a ∈M is invertible then there is a unique element b ∈M such that a ∗b = b ∗a = e. Proof. Suppose there are two such elements b1, b2 ∈M. Consider the resulting elements b1 ∗(a ∗b2) and (b1 ∗a) ∗b2 in M. We have b1 ∗(a ∗b2) = b1 ∗e = b1. and (b1 ∗a) ∗b2 = e ∗b2 = b2. The associativity law now implies that b1 = b2. 9 Deﬁnition 10. Let M be a monoid with identity element e ∈M. Suppose a ∈M is invertible. Then the inverse of a is deﬁned to be the unique element b ∈M such that a ∗b = b ∗a = e. In additive notation (where the operation is written +), we write −a for the inverse of a. If we are not using additive notation, we usually write a−1 for the inverse of a. Exercise 12. Suppose that ∗: M × M →M is given by a operation table. What do you have to check about the table to verify that b ∈M is the inverse of a ∈M? Exercise 13. When talking about inverses, we usually want S to be a monoid (so that inverses are unique). But we can deﬁne the notion of an inverse, even if S is not a monoid: all we need is that S has an identity. However, inverses are not necessarily unique when S is not a monoid. Give an example of non-uniqueness of inverses. Hint: write an operation table for a set S = {e, a, b, c} where e is an identity, but where b and c are are both inverses of a. Exercise 14. There is a notion of left inverse and right inverse. In this case you do not always have uniqueness even in a monoid. Give the deﬁnition of left inverse. Give the deﬁnition of right inverse. Exercise 15. Assume that a ∈M where M is a monoid. Show that if a has a left inverse, and that a has a right inverse, then the left inverse is equal to the right inverse and a is invertible. Here are some basic laws about inverses. (The proofs are straightforward): Theorem 4. Suppose e ∈M is the identity element of a monoid. Then e is invertible, and its inverse is itself: e−1 = e. Theorem 5. Suppose c ∈M is an invertible element of a monoid. Then c−1 is invertible, and (c−1)−1 = c. Corollary 6. Suppose c ∈M is an invertible element of a monoid with c−1 = d. Then d is invertible with d−1 = c. Theorem 7. Suppose c, d ∈M be invertible elements of a monoid. Then c ∗d is invertible, and (c ∗d)−1 = d−1 ∗c−1. Deﬁnition 11. We sometimes call invertible element in a monoid units. We can rephrase Theorems 4, 5, and 7 in terms of units: Theorem 8. Suppose M is a monoid. Then the identity element is a unit, the inverse of a unit is a unit, and the product of units is a unit (where here we are using multiplicative notion for convenience). 10 8 Cancellation laws Theorem 9 (Left-cancellation Law). Suppose M is a monoid and c ∈M is in-vertible. Suppose that c ∗x = c ∗y with x, y ∈M. Then x = y. Proof. Since c ∗x = c ∗y we get the equation c−1 ∗(c ∗x) = c−1 ∗(c ∗y). By the associative law, we can rewrite this equation as (c−1 ∗c) ∗x = (c−1 ∗c) ∗y. By the deﬁnition of inverses, we can write this equation as e ∗x = e ∗y. where e ∈M is the identity. By the deﬁnition of identity, we get x = y. Similarly we have the following: Theorem 10 (Right-cancellation Law). Suppose M is a monoid, and c ∈M is invertible. Suppose that x ∗c = y ∗c with x, y ∈M. Then x = y. Remark. For the left-cancellation law, it is enough that c is left-invertible. For the right-cancellation law, it is enough that c is right-invertible. Example 6. In the monoid ⟨R, +⟩, every element is invertible (it turns out to be a “group”). So the cancellation law for addition holds for all c ∈R: c + x = c + y = ⇒x = y. However, in the monoid ⟨R, ×⟩, only the nonzero elements are invertible. So the cancellation law cx = cy = ⇒x = y only applies for c ̸= 0. 9 Groups Deﬁnition 12. A group is a monoid such that every element is invertible. 11 Deﬁnition 13. An Abelian group is a group whose operation is commutative. In other words, it is a commutative monoid such that every element has an inverse. We use the term “Abelian” in honor of the mathematician Abel (1802–1829).2 Example 7. Some examples: • The set of natural numbers N is not a group under addition nor is it a group under multiplication. • The monoid of integers Z under addition is an Abelian group, but the corre-sponding monoid Z under multiplication is not a group (since 2, for example, has no inverse). • The set of permutations Sn is a group under composition. The inverse of a permutation α ∈Sn is the inverse function α−1. This group is non-Abelian if n > 2. (In contrast, the monoid of functions F(Xn) is not a group if n > 1). • The set Mn,m(R) of matrices is a group under addition. Describe inverses. Is it Abelian? • The set Mn(R) of matrices is not a group under multiplication. However, the subset GLn(R) of matrices with determinate not equal to zero is a group. The inverse of A is the matrix A−1. If n > 1 then GLn(R) is a non-Abelian group. • The set R is a group under addition, but not under multiplication. However, if we remove 0 the set that remains R× = R −{0} is a group. • The set Q is a group under addition, but not under multiplication. What element (or elements) of Q under multiplication do not have inverses? What if you remove just 0 from Q? • The set C is a group under addition, but not under multiplication. If you remove 0 from C you do get a group under multiplication. • The set {T, F} is a group under the “exclusive or” operation. Observe that the inverse of any element is itself. (Are there any other logical operators on {T, F} for which we get a group?) • The set Zn of integers modulo n is a group under addition, but not under multiplication. Is it Abelian? Is there a connection between Z2 under addition and {T, F} under exclusive or? • Let p be a prime. Then the set Z× p of integers modulo p not congruent to zero turns out to be an Abelian group under multiplication with p −1 elements. For groups we have the cancellation law: Theorem 11. Let c ∈G where G is a group. If c ∗x = c ∗y with x, y ∈G, then x = y. Similarly, if x ∗c = y ∗c with x, y ∈G, then x = y. 2I capitalize “Abelian” to honor Abel, but many authors write “abelian” with a lower-case ‘a’. 12 Proof. This follows from cancellation laws for monoids. Theorem 12. If G is a group with identity element e then e−1 = e For all c ∈G (c−1)−1 = c. For all c, d ∈G (c ∗d)−1 = d−1 ∗c−1. If G is an Abelian group then for all c, d ∈G (c ∗d)−1 = c−1 ∗d−1. Proof. This follows from the inverse laws for monoids. Recall that if the operation of a monoid M is written additively using +, then it is customary to write the identity element as 0 and the inverse of a ∈M as −a, if it exists. We can rephrase various results using additive notation. For example, the above theorem can be stated as follows: Corollary 13. If G is a group with additive notation then −0 = 0 For all c ∈G −(−c) = c. For all c, d ∈G −(c + d) = (−d) + (−c) If G is an Abelian group then for all c, d ∈G −(c + d) = (−c) + (−d). Remark. When we use additive notation we are usually working in an Abelian group or at least a commutative monoid. Deﬁnition 14. Let G be a group with additive notation. Given a, b ∈G we deﬁne the subtraction operation as follows: a −b def = a + (−b). Corollary 14. If G is an Abelian group with additive notation then for all c, d ∈G −(c + d) = (−c) −d. Exercise 16. Suppose M is monoid such that every element has a left-inverse. Show that M is a group. 13 9.1 The One-Side Inverse Shortcut According to the deﬁnition of inverse we must show both a ∗b = e and b ∗a = e in order to conclude that b is the inverse of a. If we are working in a group then this is overkill, and there is a short-cut: we only need to show one of the two equations. (Of course if ∗is commutative one equation is enough; but the point here is that in a group G one equation is enough even if G is non-Abelian.) Theorem 15. Let G be a group with identity element e. If a ∗b = e where a, b ∈G then b ∗a = e and so b = a−1 and a = b−1. Proof. Note that a ∗b = e = a ∗a−1. By the cancellation law, we get b = a−1. In particular, b ∗a = a−1 ∗a = e and so a = b−1. 10 Translation Functions Think of R2 as the collection of vectors in the plane under the operation of vector addition. If you ﬁx a vector (c1, c2) then the function (x, y) 7→(c1, c2) + (x, y) = (c1 + x, c2 + y) corresponds to the geometric idea of translation: each point in R2 maps to its translation by the vector (c1, c2). This operation is used to move points, and so subsets of R2, in a way that preserves lengths and angles and so on. In particular, this map is an example of what we call an “isometry”. We can generalize this translation map to any monoid ⟨M, ∗⟩. To do so ﬁx an element c ∈M and deﬁne a function Tc : M →M by the rule Tc(x) def = c ∗x. We call this left-translation by c. We deﬁne right-translation in a similar manner. Of course if the monoid is commutative, as in the case of R2 under addition, left and right translations by c ∈M agree. If unspeciﬁed, a translation will be understood as left-translation. The following two results are straightforward: 14 Lemma 16. Let e be the identity in the monoid M. Then the translation func-tion Te is the identity function M →M. Lemma 17. Let Ta and Tb be translation functions M →M associated to ele-ments a, b ∈M where ⟨M, ∗⟩is a monoid. Then the composition Ta ◦Tb : M →M is also a translation function. In fact Ta ◦Tb = Ta∗b. Exercise 17. Show that had we deﬁned Ta and Tb as right-translations, we would get Ta ◦Tb = Tb∗a instead. We can use the above two Lemmas to prove the following: Lemma 18. Suppose that c ∈M is an invertible element of a monoid M. Then the translation functions Tc and Tc−1 are inverse functions to each other. In other words the compositions Tc ◦Tc−1 and Tc−1 ◦Tc are both the identity function. We can express this fact nicely with the identity T −1 c = Tc−1. Corollary 19. Suppose that c ∈M is an invertible element of a monoid M. Then the translation functions Tc deﬁned by Tc(x) = c ∗x is a bijection. We can apply this result to operation tables of ﬁnite monoids M: Theorem 20. Suppose ⟨M, ∗⟩is a ﬁnite monoid with invertible element c. Sup-pose M is ordered once and for all, and consider the associated operation table for ∗. Then the row associated to c has every element of M appearing exactly once. Similarly the column associated to c has every element of M appearing exactly once. Proof. Suppose a1, a2, . . . , an are the elements of M listed in order. Then the row associated to c consists of the following ﬁnite sequence of elements: c ∗a1, c ∗a2 . . . , c ∗an. Think of this sequence as Tc(a1), Tc(a2), . . . , Tc(an). Every element appears at most once because Tc is injective. Every element appears at least once because Tc is surjective. The argument for the c-column is similar. 11 Exponentiation in a Monoid Throughout this section, let M be a monoid with operation ∗and identity element e. For most of what follows we will employ multiplicative notation, allowing us to write a ∗b simply as ab. Our goal is to introduce powers au where n ∈N and, in fact, when a is invertible for all n ∈Z. (We translate all the main results to additive notation later on where we write na). 15 Let a ∈M be an element of the monoid. Informally, if n ∈N then we deﬁne an to be a ∗a ∗· · · ∗a where a is repeated n times. If a ∈M is invertible, we deﬁne a−n for negative integers −n < 0 by deﬁning a−n either as (a−1)n or (an)−1. If one follows this path, then one can derive the usual laws of exponentiation, even for negative powers (if a is invertible). However, proving them often requires a bunch of diﬀerent cases which depends on the sign (positive or negative) of each of the exponents involved in the given law. We take another approach to exponentiation using iteration of translation func-tions. This approach treats positive and negative powers in a more uniﬁed manner. Our plan is to iterate the translation function x 7→ax starting with x = e yielding the sequence e a ∗e a ∗a ∗x a ∗a ∗a ∗e . . . . The nice thing about this approach is that we can use negative iteration when the translation function is invertible. In what follows I will take the concept of iteration, and some of the basic results for iterated functions, as part of the set-theoretical background for abstract algebra.3 Here are three propositions summarizing what we need: Proposition 21. Suppose f : S →S is a function from a set S to itself. Then for any nonnegative integer n ∈N we have the nth iteration f n of f. Properties of iteration for such a function f : S →S include the following: • For all n ∈N, the iteration f n is a function S →S. • The zeroth iteration f 0 is the identity map S →S. • The ﬁrst iteration f 1 is just f : S →S itself. • For m, n ∈N, we have f m ◦f n = f m+n. In particular, for n ∈N we have f n+1 = f n ◦f = f ◦f n. • For m, n ∈N, we have (f m)n = f mn. Proposition 22. Suppose f : S →S is a bijection from a set S to itself. Then we can extend the deﬁnition of interation f u to all integers u ∈Z. Properties of iteration for such a bijection f : S →S include the following: • For all u ∈Z, the iteration f u is also a bijection S →S. • The iteration f −1 is, as the notation suggests, the inverse of f. • For u, v ∈Z, we have f u ◦f v = f u+v. • For u, v ∈Z, we have (f u)v = f uv. • If f is the identity function, then f u is the identity function for all u ∈Z. 3See my number systems textbook for a formal account of iteration and a rigorous development of the properties of iteration. 16 Proposition 23. Suppose f, g: S →S are commuting functions. In other words, suppose f ◦g = g ◦f. Then (f ◦g)n = f n ◦gn for all n ∈N. If in addition f and g are both bijections S →S, then (f ◦g)u = f u ◦gu for all u ∈Z. With this we are ready for the formal deﬁnition of exponentiation. Recall that from Corollary 19 that the translation function Ta is a bijection if a is invertible. Deﬁnition 15 (Exponentiation in a multiplicative monoid). Let M be a monoid that employs multiplicative notation. If n ∈N then an def = T n a (e). where e ∈M is the identity element, and where Ta is the translation map x 7→a∗x. If a is invertible in M, then we extend the above deﬁnition for all u ∈Z: au def = T u a (e) (This is well-deﬁned since Ta is a bijection in this case). Theorem 24. Suppose that a ∈M where M is a multiplicative monoid. Then a0 = e where e ∈M is the identity element, and a1 = a. If a is invertible in M then a−1 is the inverse of a in M. Proof. Observe that a0 = T 0 a (e) = e since T 0 a is the identity map (Proposition 21). Next, by Proposition 21 and the above deﬁnition, a1 = T 1 a (e) = Ta(e) = ae = a. Finally, suppose a is invertible, and let b be the inverse of a in M. By Lemma 18, the translation Tb is the inverse of Ta, so T −1 a = Tb by Proposition 22 (where T −1 a here mean the −1 ∈Z iteration of Ta). Thus a−1 = T −1 a (e) = Tb(e) = be = b where a−1 means the −1 power of a (not, a priori, the inverse of a which we denote simply as b). Remark. The last part of the above shows that both meanings of a−1 agree: as an inverse and as a negative power. 17 To prove additional properties of exponentiation, we will need a few lemmas about translations functions. The ﬁrst states that the iteration of translation is a translation. Lemma 25. Consider the iteration T u a of the translation Ta where a ∈M and where u ∈N or more generally, if a is invertible, where u ∈Z. Then there is an element c ∈M such that T n a = Tc. Proof. Fix a ∈M. First we show that the claim holds when u ∈N by induction. The base case holds since T 0 a = Te where e is the idenitity. Suppose that T n a = Tc for a particular n ∈N and c ∈M. Then T n+1 a = T n a ◦Ta = Tc ◦Ta = Tca. (Prop 21 and Lemma 17) Thus the claim holds for n+1 with ca as the associated element of M. By induction we accept the claim for u ≥0. We have established the claim for for all a ∈M and all u ≥0, and wish to establish the claim for negative u. So let a ∈M be invertible with inverse b ∈M, and suppose u = −n for some n ∈N. We have Tb = T −1 a by Lemma 18. So, using the identity (f u)v = f uv from Proposition 22, T u a = T −n a = T −1 a n = T n b . We have established that T n b = Tc for some c ∈M. Thus T u a = Tc as desired. We can actually be more speciﬁc about what c is in the previous lemma. Lemma 26. Suppose that a ∈M. Then T u a = Tau for all u ∈N. If a is invertible, then this holds more generally for all u ∈Z. Proof. By Lemma 25, T u a = Tc for some c ∈M. Apply T u a and Tc to the identity element e ∈M: au def = T u a (e) = Tc(e) = ce = c. Now we are ready to prove some important identities. Theorem 27. Let a ∈M where M is a multiplicative monoid. If m, n ∈N then am+n = aman. More generally, if a is invertible in M and if u, v ∈Z then au+v = auav. 18 Proof. Observe that Tam+n = T m+n a (Lemma 26) = T m a ◦T n a (Proposition 21) = Tam ◦Tan (Lemma 26) = Taman (Lemma 17) From Tam+n = Taman we get am+n = Tam+n(e) = Taman(e) = aman. The above proof is for m, n ∈N. To generalize to u, v ∈Z, replace m with u, replace n with v, and replace the reference to Proposition 21 with Proposition 22 in the above argument. We are not assuming that M is a commutative monoid. However, we have the following: Corollary 28. Let a ∈M. Then a power of a commutes with any power of a. Proof. This follows since addition is commutative in Z. Theorem 29. Let a ∈M where M is a multiplicative monoid. If m, n ∈N then (am)n = amn. If a ∈M is invertible and if u, v ∈Z then (au)v = auv. Proof. We have Tam = T m a (Lemma 26), so (Tam)n = (T m a )n = T mn a where the last equality is based on a property of iterations (Proposition 21). This, together with Deﬁnition 15, yields (am)n = (Tam)n(e) = T mn a (e) = amn. To generalize to u, v ∈Z, when a is invertible, replace m with u, n with v, and the reference to Proposition 21 with Proposition 22 in the above argument. Theorem 30. Let a ∈M where M is a monoid. If a is invertible, then so is au for all u ∈Z. Proof. Observe that a−u is the inverse of au by previous exponentiation laws. In a commutative monoid we have that (ab)n = anbn. We will show this, but ﬁrst we need a simple lemma. Lemma 31. Let a, b ∈M. Suppose ab = ba (which holds, for example, when M is a commutative monoid). Then Ta and Tb commute as functions: Ta ◦Tb = Tb ◦Ta. 19 Proof. This follows from Lemma 17 since Tab = Tba in this case. Theorem 32. Let a, b ∈M in a multiplicative monoid. If ab = ba (which is true, for example, if M is a commutative monoid or an Abelian group) then for all n ∈N (ab)n = anbn. If in addition a and b are invertible, then so is ab, and for all u ∈Z (ab)u = aubu. Proof. By Lemma 31, Ta and Tb commute. So Tab n = Ta ◦Tb)n (Lemma 17) = T n a ◦T n b (Prop. 23) = Tan ◦Tbn (Lemma 26) = Tanbn (Lemma 17) This, together with Deﬁnition 15, yields (ab)n def = Tab n(e) = Tanbn(e) = anbne = anbn. If a and b are invertible, then ab is invertible since it has inverse b−1a−1. The above argument for n ∈N generalizes to u ∈Z in this case. Theorem 33. Let e ∈M be the identity element in a multiplicative monoid. Then eu = e for all u ∈Z. Proof. We know that Te is the identity map by Lemma 16. So, by Proposi-tion 22, T u e is also the identity map. Thus eu def = T u e (e) = e. 11.1 Exponentiation in Additive Notation When M is a monoid or group written in additive notation, we use a diﬀerent notation for exponentiation. The results are the same, but are just expressed in a diﬀerent notation. When M is a monoid with its binary operation written as + then the identity element is generally written as 0. For n ∈N and a ∈M we write na for the additive version of exponentiation. Similarly if a is invertible and u ∈Z, we write ua for exponentiation. All the results of the previous section hold for M, but one just has to translate to additive notation. For the convenience of the reader, we summarize the main results when the monoid is a group. Theorem 34 (Exponentiation in an additive group). Suppose A is a group written in additive notation. Suppose a ∈A. Then we have the following 20 • 0 a = 0. (Here the ﬁrst 0 is in Z and the second is in G). • 1 a = a • (−1)a = −a • (u + v)a = ua + vb for all u, v ∈Z • (uv)a = u(va) for all u, v ∈Z • ua commutes with va for all u, v ∈Z. If a, b ∈G commute, a + b = b + a, then • u(a + b) = ua + ub for all u ∈Z Finally, • u 0 = 0 for all u ∈Z (here 0 is in G) Remark. We use the normal elementary algebra notational conventions for supress-ing parenthesis. For example, the expression ua + vb really means (ua) + (vb) (and not u(a + (vb)) say). Remark. Most groups that use additive notation are Abelian, so the law u(a + b) = ua + ub holds for all u ∈Z and a, b ∈G. 12 The Induced Operation on Closed Subsets Deﬁnition 16. Let ∗: S ×S →S be a binary operation. If A is a subset of S then we say that A is closed under ∗if the following holds: ∀a, b ∈A, a ∗b ∈A. Example 8. Consider N under addition, and let A be the subset of even numbers greater than 11. Then A is closed under +. Let B be the set of odd natural numbers. Then B is not closed under + since 7 + 3 is not in B. Deﬁnition 17. Let ∗: S × S →S be a binary operation. Let A ⊆S be a subset that is close under ∗. Then we deﬁne a binary operation on ∗A : A × A →A using the following rule: if a, b ∈A then a ∗A b is deﬁned to be a ∗b. This operation is called the induced operation on A. We warn the reader that the term induced operation can refer to diﬀerent things depending on context. The above is what is meant when A is a subset of S. There are also induced operation corresponding to sets of functions, or sets of cosets. The common idea is that an induced binary operation closely matches, in some sense, a previously given binary operation, but often has a diﬀerent, albeit related, domain or codomain. 21 The induced operation of Deﬁnition 17 is often thought of as the operation obtained by restricting the operation ∗from S to the subset A. The assumption that A is closed plays a key role here. It is always legal to restrict the domain of a function. So we can always restrict ∗: S × S →S to A × A to obtain a function A × A →S. It is not always legal, however, to restrict a codomain. You can only restrict to a smaller codomain if the this proposed codomain contains the image. So we can only restrict from A × A →S to A × A →A if the image of the restriction A × A →S is a subset of A. This is exactly the requirement that A is closed under ∗: when we say “A is closed” we just mean that the image of ∗ restricted to A × A is contained in A. By deﬁnition we have the law a ∗A b = a ∗b for all a, b ∈A. So ∗and ∗A look like the same operation; the only real diﬀerence is the domain and codomain. The bottom line is that, as long as a, b ∈A, the expressions a ∗b and a ∗A b are completely interchangeable. The operation ∗A : A × A →A is sometime called the restriction of ∗to A. We also say that the operation S × S →S an extension of the operation A × A →A to S. It is common practice to use the same symbol for both operations S ×S →S and A × A →A. When we do so, we rely on context to tell us which is meant by the symbol in any given usage. (When applied to elements in A, both notions give the same result, so one can be be safely ambiguous. For example, if A = N and S = Z then 7 + 3 is 10 whether or not you are thinking of +: Z × Z →Z or the induced operation +: N × N →N.) Lemma 35. Let ∗: S × S →S be a binary operation. Let A ⊆S be a subset that is close under ∗. If ∗: S × S →S is associative, then the induced operation A × A →A is also associative. If ∗: S × S →S is commutative, then the induced operation A × A →A is also commutative. Proof. We will prove the claim related to the associative law. The proof of the commutative claim is similar. So let a, b, c ∈A. We have (a ∗A b) ∗A c def = (a ∗b) ∗c by deﬁnition of ∗A. Similarly, we have a ∗A (b ∗A c) def = a ∗(b ∗c). Since a, b, c ∈A, and since A is a subset of S, we have a, b, c ∈S. Since ∗is associative on S by assumption, we have (a ∗b) ∗c = a ∗(b ∗c). Putting these equations together we get the desired equation: (a ∗A b) ∗A c = a ∗A (b ∗A c). 22 From this we get the following: Theorem 36. Suppose M is a monoid with operation ∗and identity element e. If A is a subset closed under ∗and if A contains e then A is itself a monoid under the induced operation. The identity of A with the induced operation is the identity of M. 13 Submonoids Informally a submonoid is a subset A of a monoid M that is itself a monoid under the induced operation. This will only make sense if the subset A is closed under the operation of M, so this will always be required. One question is whether we want the identity of A to be the same element as the identity of M. Based on the answer to this question we have two notions of submonoid. Here we will pursue the notion where M and A have the same identity element: this will be a second requirement.4 As we saw in Theorem 36, these two requirements suﬃce. So we deﬁne submonoid as follows: Deﬁnition 18. Suppose ⟨M, ∗⟩is a monoid with identity element e. A submonoid is a subset A ⊆M such that (1) A is closed under ∗and (2) e ∈A. As we saw above, every submonoid is itself a monoid under the induced operation with identity e. Example 9. The set of even integers forms a submonoid of ⟨Z, +⟩. The set of odd integers fails to be a submonoid. If A is the set of even numbers greater than 11 then A is a closed subset of Z under addition, but is not a submonoid since it does not contain 0. (And in fact, this A has no identity element). We restate Theorem 36: Theorem 37. Suppose A is a submonoid of a monoid M. Then A is itself a monoid under the induced operation. The following is clear: Theorem 38. Every submonoid of a commutative monoid is itself a commutative monoid (under the induced operation). Exercise 18. Consider the monoid ⟨Z10, +⟩. Let A be the submonoid {0, 2, 4, 6, 8}. Make an operation table for A using the induced operation +A. Theorem 39. Suppose M is a monoid and that B is a submonoid of M. Let A be a subset of B. Then A is a submonoid of M if and only if A is a submonoid of B. Proof. Let e be the identity element of M. Since B is a submonoid, e is the identity of B. Let ∗be the binary operation on M. Then, as we have seen, the induced operation ∗B is the binary operation on the submonoid B. 4The set M = N × N under multiplication is a monoid under componentwise products with identity (1, 1). The subset A = N × {0} is a monoid under the induced operation with identity element (1, 0). However, by our oﬃcial deﬁnition, A is not a submonoid of M since the identity of A is diﬀerent from that of M. 23 Suppose that A is a submonoid of M. Then e ∈A. But e is the the identity of B. Suppose a, b ∈A. Then a ∗B b = a ∗b since ∗B is the induced operation. Since A is a submonoid of M, the element a ∗b must be in A. Thus a∗Bb is in A. We conclude that A is closed in B. We have established both conditions needed for A to be a submonoid of B. Suppose that A is a submonoid of B. Then the identity e of B is in A. So the identity of M is in A since e is also the identity of M. Suppose a, b ∈A. Then a ∗B b = a ∗b since ∗B is the induced operation. Since A is a submonoid of B, the element a ∗B b must be in A. Thus a ∗b is in A. We conclude that A is closed in M. We have established both conditions needed for A to be a submonoid of M. The following lemmas give compatibility between notions in a monoid and in a submonoid: Lemma 40. Suppose A is a submonoid of M. If a ∈A has an inverse b in the monoid A, then b is the inverse of a in M as well. Similarly, if a has inverse b in M, and if a, b ∈A, then a has inverse b in A. Proof. Let e be the identity of M, which is also the identity of A since A is a submonoid. Suppose a ∈A has inverse b in A. By deﬁnition of inverse a ∗A b = b ∗A a = e. But a ∗A b = a ∗b and b ∗A a = b ∗a since a, b ∈A. Thus a ∗b = b ∗a = e in M. Now suppose a has inverse b in M, so a ∗b = b ∗a = e. Suppose a, b ∈A. Then a ∗A b = a ∗b and b ∗A a = b ∗a. Thus a ∗A b = b ∗A a = e. Lemma 41 (Multiplicative Version). Suppose A is a submonoid of M. Let a ∈A. Then an as deﬁned in M is the same as an as deﬁned in A where here n ∈N. In particular A is closed under powers: an ∈A for all a ∈A and n ∈N where an is as deﬁned in M. If in addition a is invertible in A then the above extends to all n ∈Z. Proof. This follows by induction for n ∈N. The more general statement for a invertible follows from the properties of exponents. Exercise 19. Suppose that M is a multiplicative monoid. Let a ∈M. Show that A = {an \| n ∈N} is a submonoid of M. Show that this A is the smallest submonoid containing a in the sense that any submonoid of M containing a must contain A as a submonoid. Show that if a ∈M is invertible, then A′ = {au \| u ∈Z} is also a submonoid, and that in A′ is an Abelian group. Show that it is the smallest submonoid containing a that is a group. 14 The Unit Group In this section ⟨M, ∗⟩will be a monoid with identity e. There is a submonoid of M that is of particular interest: 24 Deﬁnition 19. Let ⟨M, ∗⟩be a monoid. Then M inv is deﬁned to be the set of elements of M that are invertible. Elements of M inv are called units. As we will see, M inv forms a group under the induced operation, and we call M inv the unit group of M. The following is just a restatement of the deﬁnition of group: Theorem 42. A monoid M is a group if and only if M inv = M. Example 10. (Availing ourselves of some linear algebra). Let M = Mn(R) be the set of n-by-n matrices under matrix multiplication. This is a monoid. The unit group M inv consists of invertible matrices is a group. It is an important group in mathematics called the general linear group, and is often denoted as GLn(R). As we learn in linear algebra, GLn(R) is the set of n-by-n matrices with determinant not equal to zero. Example 11. (Assuming familarity with rings) Let M = R be a ring. If we ignore addition, and use multiplication as the operation, we get a monoid. In this case Rinv consists of all elements in R with multiplicative inverses. We sometimes write this subset as R×, and call this the group of units of R. Note that if F is a ﬁeld, then F × is F −{0}. Exercise 20. Consider M = N, M = Z, and M = Q under multiplication. What is M inv in each of these cases? Note that we get a group under multiplication in each case. Now do the same exercise for addition: identify M inv in each of these cases. As mentioned above we will determine that M inv is a group. First we convince ourselves that M inv is a submonoid, and hence M inv is itself a monoid. Lemma 43. Let ⟨M, ∗⟩be a monoid. Then M inv is a submonoid. Proof. The follows from Theorem 8. Theorem 44. Let ⟨M, ∗⟩be a monoid. Then M inv is a submonoid of M that is in fact a group under the induced operation. Furthermore, M inv is the largest such group in the following sense: if G is a submonoid of M that is a group under the induced operation, then G ⊆M inv. Proof. Since M inv is a submonoid, it is in fact a monoid under the induced oper-ation. Does every element of M inv have an inverse in M inv? By deﬁnition, every element of M inv has an inverse in M, but Theorem 8 tells us more: the inverse is guaranteed to be in M inv itself. Suppose G is a submonoid of M that is a group under the induced operation. Then every element of G is a unit in M (see Lemma 40). Thus G ⊆M inv. Exercise 21. If ⟨M, ∗⟩be a commutative monoid, then can you conclude that M inv is an Abelian group under the induced operation? Exercise 22. Find the group of invertible elements in ⟨Z8, ×⟩. Make an operation table for this group. 25 15 Subgroups Suppose ⟨G, ∗⟩is a group with identity element e. Some subsets A ⊆G will be groups under the induced operation, but others will not. If we want A to be a group under the induced operation, we will certainly need A to be closed under ∗. If we want A and G to share the same identity element (as we did with monoids), we would want A to contain e, and then we would need A to be “closed under inverses” as well in the sense that a−1 ∈A for all a ∈A where here a−1 is the inverse of a in G. The following shows that these properties are indeed necessary and suﬃcient: Lemma 45. Suppose ⟨G, ∗⟩is a group with identity element e, and that A ⊆G is a subset closed under ∗. Then A is a group under the induced operation if and only if (1) e ∈A, in other words A is a submonoid, and (2) if a ∈A then a−1 ∈A where a−1 is the inverse of a in G. Proof. Suppose A is a group under the induced operation. Then A must contain an identity element e′. Note that e′ ∗e′ = e′ = e′ ∗e. So e′ = e by the cancellation law in G. So A is a submonoid of G. By assumption each a ∈A has an inverse in A. This inverse is a−1 (the inverse of a in G) by Lemma 40. Thus (1) and (2) both hold. Conversely suppose (1) and (2) hold. By (1) A is a submonoid of G, so is a monoid under the induced operation. By (2) and Lemma 40 every element of A is invertible. Thus A is a group. The above lemma motivates the following deﬁnitions: Deﬁnition 20. Let ⟨G, ∗⟩be a group. We say that a subset A ⊆G is closed under inverses if the inverse of a is in A for all a ∈A. Deﬁnition 21. Suppose ⟨G, ∗⟩is a group. A subgroup is a subset A ⊆G such that (1) A is closed under ∗, (2) the identity element of G is in A and (3) A is closed under inverses. By Lemma 45 together with Lemma 40 and Lemma 41 we have the following: Theorem 46. If A is a subgroup of G, then A is itself a group under the induced (restricted) operation. The identity of A is the identity in G. The inverse of an element a ∈A in A is equal to its inverse in G. For each a ∈A and n ∈Z the nth power of a in A is equal to the nth power of a in G. Exercise 23. Show that if A is a subgroup of an Abelian group, then A is itself an Abelian group under the induced operation. (Hint: Lemma 35). Remark. The idea of a subgroup makes it easy to construct examples of groups without proving all the equations that must hold in a group: the associative law equation, the identity law equations, and the inverse law equations. Now that we have the above theorem, we get these equations for free when dealing with a subgroup of a known group. 26 If A is not a subgroup of a known group, and you want to show it is a group, you have to do it the hard way and prove all the required equations (including associativity, identity equations, and inverse equations). There are some short-cuts to checking that A is a subgroup. The ﬁrst allows you to replace checking that the identity is in A with checking that A nonempty. Proposition 47. Suppose ⟨G, ∗⟩is a group. Then a subset A ⊆G is a subgroup if and only if (1) A is nonempty, (2) A is closed under ∗, (3) A is closed under inverses. Proof. First suppose that A is a subgroup. Then since the identity e is in A we know that it is (1) nonempty. Conditions (2) and (3) also hold by the deﬁnition of subgroup. Conversely suppose (1), (2), and (3) hold. Since A is nonempty, it contains an element a ∈A. By (3) we have a−1 ∈A. So by (2) we have a ∗a−1 ∈A. In other words e ∈A. This, together with (2) and (3), means that A is a subgroup by the deﬁnition of subgroup. The following gives you a way to check two conditions instead of three (we use multiplicative notation for convenience, but it can be translated into additive notation): Proposition 48. Suppose ⟨G, ∗⟩is a group. Then a subset A ⊆G is a subgroup if and only if (1) A is nonempty and (2) a ∗b−1 ∈A for all a, b ∈A. Exercise 24. Prove the above proposition. Hint: assuming (1) and (2), ﬁrst show that e ∈A, then show A is closed under inverses. Finally show A is closed under ∗. For ﬁnite groups the situation is easier and there is a nice shortcut. Actually the enclosing group G does not even need to be ﬁnite, just the subset A. Proposition 49. Suppose ⟨G, ∗⟩is a group, and that A is a ﬁnite subset of G. Then A is a subgroup of G if and only if (1) A contains the identity, and (2) A is closed under ∗. Proof. One direction is clear, so we focus on the other direction. Suppose a ∈A. Consider the translation function Ta : A →A given by the rule Ta(x) = a ∗x. This function is well-deﬁned since A is closed under ∗. This function is injective by the cancellation law. Since A is ﬁnite, this function must be surjective as well by properties of ﬁnite sets. Since e ∈A, there must be an element b mapping to e. Since Ta(b) = e we conclude that a∗b = e. Thus b is the inverse of a (Theorem 15). Thus A is closed under inverses. This together with (1) and (2) gives the result. Corollary 50. Every ﬁnite submonoid of a group is a subgroup. 27 Exercise 25. Suppose ⟨G, ∗⟩is a group, and that A is a ﬁnite subset of G. Suppose that (1) A is nonempty and (2) A is closed under ∗. Show that the identity must be in A. Conclude that A must be a subgroup. In other words, (1) in the above Proposition can be replaced by “(1) A is nonempty”. Hint: if a ∈A, then all its powers are in A. Two powers with distinct exponents are equal so the identity e is a power of a. Exercise 26. Suppose M is a monoid with identity e, then show that M and {e} are submonoids of M. Suppose G is a group with identity e. Then show that G and {e} are subgroups of G. 16 Products and Sums of Finite Sequences We will use the term ﬁnite ordered sequence in a set S for any family (ai)i∈I of elements of S such that I is a ﬁnite, totally ordered set. We deﬁne an equivalence relation for such ﬁnite ordered sequences: suppose (ai)i∈I and (bJ)i∈J are two ﬁnite ordered sequences in S then we say that these ﬁnite ordered sequences are order equivalent if (i) the cardinality of I and J are equal, and (ii) the unique order-preserving bijection γ : I →J has the property that bγ(i) = ai for all i ∈I. The reader can verify the following: Lemma 51. Order equivalence is an equivalence relation among ﬁnite ordered se-quences in a set S. The deﬁnition of order equivalence makes use of the following fact from set theory (that can be proved by induction): Lemma 52. If I and J are ﬁnite totally ordered sets of the same size, then there is a unique order-preserving bijection I →J. We say that a ﬁnite ordered sequence is nonempty if I is nonempty. We now deﬁne the product of ﬁnite ordered sequences: Deﬁnition 22. Fix a set S and a binary operation ∗: S × S →S. Then we deﬁne the product Q(ai)i∈I of any nonempty ﬁnite ordered sequence (ai)i∈I in S recursively as follows. If I has one element, call it α, then Q(ai)i∈I is deﬁned to be just aα. If I has more than one element then let ω be the last element of I, and let I′ be I −{ω}. Then we deﬁne the product by the recursive equation Y (ai)i∈I def = Y (ai)i∈I′ ∗aω where (ai)i∈I′ denotes the restriction of the given family to I′. We will sometimes follow the common convention and write Q(ai)i∈I as Y i∈I ai. If I = [m, n]Z is the interval of integers from m to n (inclusive) then we can employ the notation n Y i=m ai 28 for the product Q(ai)i∈I. If there is no danger of ambiguity, we can use abbrevia-tions for products such as Y ai or Y i ai. If the operation is written + then we usually use additive notation where we re-place Q with P and use the term sum instead of product. Deﬁnition 23. If M is a monoid or a group, then we extend the above deﬁnition to include the empty sequence: we deﬁne the product of the empty ﬁnite ordered sequence to be the identity element of M. Theorem 53. Let S be a set with binary operation ∗: S × S →S. If (ai)i∈I and (bJ)i∈J are order equivalent nonempty ﬁnite ordered sequences, then Y (ai)i∈I = Y (bj)i∈J. (If S = M is a monoid, then this extends to the empty sequence.) Proof. A straightforward induction argument suﬃcies. 17 The General Associativity Laws Now we consider various generalizations of the associativity law. These are all based on the following result: Theorem 54 (General Associative Law: First Form). Let S be a set with an asso-ciative binary operation ∗: S ×S →S. Suppose (ai)i∈I is a nonempty ﬁnite ordered sequence in S. Suppose also that I is the disjoint union of two nonempty subsets I1 and I2 and that every element of I2 is an upper bound of I1. Then Y i∈I ai = Y i∈I1 ai ∗ Y i∈I2 ai. Proof. If I2 = {ω} is a singleton then ω must be the maximum in I, and the result now follows from the recursive deﬁnition of the product. We proceed by induction and assume I2 has more than one element. Let ω be the maximum element of I2 and let I′ 2 = I2 −{ω}. Note also that ω is the maximum of all of I. Observe that, by the associative law and the induction hypothesis (on the size of I2)  Y i∈I1∪I′ 2 ai  ∗aω =  Y i∈I1 ai ∗ Y i∈I′ 2 ai  ∗aω = Y i∈I1 ai ∗    Y i∈I′ 2 ai  ∗aω   By deﬁnition of product, the left-hand side is just Q(ai)i∈I and the right-hand side simpliﬁes as desired: Y i∈I1 ai ∗    Y i∈I′ 2 ai  ∗aω  = Y i∈I1 ai ∗ Y i∈I2 ai. 29 For monoids we can extend this law to possibly empty sequences: Corollary 55 (General Associative Law: First Form for Monoids). Let ⟨M, ∗⟩be a monoid. Suppose (ai)i∈I is a ﬁnite ordered sequence in M. Suppose also that I is the disjoint union of two subsets I1 and I2 where every element of I2 is an upper bound of I1. Then Y i∈I ai = Y i∈I1 ai ∗ Y i∈I2 ai. Proof. If I1 and I2 are nonempty then use the theorem. If I1 or I2 are empty, replace the corresponding product with the identity element, and the result is clear. We can freely remove or add terms that are the identity element: Theorem 56. Let ⟨M, ∗⟩be a monoid with identity element. Suppose (ai)i∈I is a ﬁnite ordered sequence in S. Suppose I′ is a subset of I such that ai = e if i ∈I−I′, in other words, the ordered subsequence (ai)i∈I′ contains all the nontrivial terms of (ai)i∈I. Then Y i∈I ai = Y i∈I′ ai. In particular, if ai = e for all i ∈I then Q i∈I ai is equal to e, the product of the empty subsequence. Proof. This can be proved by induction on the size of I. We distinguish two cases: where I′ contains the maximum element of I, and where I′ does not contain this maximum. We can extend this law to more than two terms: Theorem 57 (General Associative Law: Second Form). Let S be a set with an associative binary operation ∗: S × S →S. Suppose (ai)i∈I is a nonempty ﬁnite ordered family in S. Suppose also that I is the disjoint union of k ≥1 nonempty subsets I1, . . . , Ik such that every element of a given It is an upper-bound for all Is with s < t. Then Y i∈I ai = k Y j=1 Y i∈Ij ai. Proof. If k = 1 then the result is clear by the deﬁnition of singleton products. We proceed by induction to k ≥2. Suppose I′ = I1 ∪· · · ∪Ik−1. Then by Theorem 54, and the induction hypothesis, Y i∈I ai = Y i∈I′ ai ∗ Y i∈Ik ai =   k−1 Y j=1 Y i∈Ij ai  ∗ Y i∈Ik ai. The right-hand side simpliﬁes to the desired express (using the deﬁnition of prod-uct). For monoids we can drop the requirement that each Ij is nonempty. We can even allow k = 0: 30 Corollary 58 (General Associative Law: Second Form for Monoids). Let ⟨M, ∗⟩ be a monoid with identity element e. Suppose (ai)i∈I is a nonempty ﬁnite ordered family in M. Suppose also that I as the disjoint union of k ≥0 subsets I1, . . . , Ik such that every element of a given It is an upper-bound for all Is with s < t. Let J be the set of nonnegative integers up to k. Then Y i∈I ai = Y j∈J Y i∈Ij ai. Proof. If I is empty, then we get a product Q j∈J e of identity elements which is the identity, and the result follows. If I is nonempty then we can remove from J all j such that Ij is empty (Theorem 56), so we reduce to the case where each Ij is nonempty. We then can use the above theorem to establish the result. Next we consider the idea that we can rearrange the parenthesis for associative operators without aﬀecting the result. Suppose ∗is associative, then we regard expressions such as ((a1 ∗a2) ∗a3) ∗a4 and (a1 ∗a2) ∗(a3 ∗a4) as representing diﬀerent calculations of the same result (by associativity). We formalize the notion of “calculation” as follows:5 Deﬁnition 24. Let S be a set with a binary operation ∗: S×S →S. We deﬁne the concept of a “calculation” of a nonempty ﬁnite sequence with terms in S recursively as follows: If I = {i0} is a singleton set, the only calculation of the sequence (ai)i∈I is ai0. If I has size n ≥2 then the calculations of (ai)i∈I are the elements of S of the form b1 ∗b2 where I1 and I2 are nonempty subsets partitioning I such that every element of I2 is an upper bound of I1, where b1 is a calculation of (ai)i∈I1, and where b2 is a calculation of (ai)i∈I2. Example 12. Consider the sequence 4, −3, 2, 1, 3 with terms in Z. Consider sub-traction as our binary operation. Then 4 − (−3) −(2 −1) −3 = 5 and (4 −(−3)) −((2 −1) −3)) = 9 are two calculations. The results are not equal, which illustrates that subtraction is not associative. Theorem 59 (General Associative Law: Third Form). Let S be a set with an as-sociative binary operation ∗: S × S →S. Suppose (ai)i∈I is a nonempty ﬁnite ordered family in S. Then all calculations of (ai)i∈I using ∗are equal to the prod-uct Q(ai)i∈I. In particular, all calculations of (ai)i∈I are equal. Proof. If I = {i0} then the only calculation of (ai)i∈I is ai0, which is equal to Q(ai)i∈I by deﬁnition. So we can assume the size of I is greater than one, and proceed by induction. Suppose b is a calculation of (ai)i∈I, so b = b1 ∗b2 5With a bit more work we could formalize “calculation” in terms of the binary treelike structure used to describe how to calculate the result. Then the theorem would say that any two such trees associated to a given sequence a1, a2, . . . , an. 31 where b1 = Q(ai)i∈I1, where b2 = Q(ai)i∈I2, where I1 and I2 partition I, and where every element of I2 is an upper bound of I1. By the induction hypothesis and by Theorem 54 b = b1 ∗b2 = Y i∈I1 ai ∗ Y i∈I2 ai = Y i∈I ai. Finally we mention that our deﬁnition of an is equivalent to the other popular deﬁnition:6 Theorem 60. Let a ∈M where M is a multiplicative monoid. If n is a positive integer then an = n Y i=1 a. Proof. We can prove this by induction on n. The base case reduces to the equa-tion a1 = a established above. The induction step can be proved as follows: an+1 = ana1 = n Y i=1 a ! a = n+1 Y i=1 a. Exercise 27. Translate the results of this section to additive notation. 18 General Commutativity Laws In this section we will develop general commutativity laws for multiplicative monoids. We use monoids since the statements and proofs are a bit more ele-gant than in a more general setting. But we can certainly generalize these results to “commutative semigroups” where we do not require an identity element (you just need to be careful about empty sequences). Also it is straightforward to convert the results to additive notation. First we show we can move any term to the end: Lemma 61. Let ⟨M, ∗⟩be a commutative monoid. If (ai)i∈I is a ﬁnite ordered sequence of elements in M and if i0 ∈I then Y i∈I ai = Y i∈I′ ai ! ∗ai0 where I′ = I −{i0} with the restricted ordering. 6We used iteration to deﬁne powers, but other authors such as Bourbaki in Algebra use the equation in Theorem 60 instead, at least for positive n. 32 Proof. Let I1 be the set of elements of I strictly smaller than i0 and let I2 be the set of elements of I strictly larger than i0. From Corollary 55 and the recursive deﬁnition of products we have Y i∈I ai = Y i∈I1∪{i0} ai ∗ Y i∈I2 ai = Y i∈I1 ai ! ∗ai0 ! ∗ Y i∈I2 ai. Using associativity and commutivity we have Y i∈I1 ai ! ∗ai0 ! ∗ Y i∈I2 ai = Y i∈I1 ai ∗ ai0 ∗ Y i∈I2 ai ! = Y i∈I1 ai ∗ Y i∈I2 ai ! ∗ai0 ! . Finally, by associativity and Corollary 55 we have Y i∈I1 ai ∗ Y i∈I2 ai ! ∗ai0 ! = Y i∈I1 ai ∗ Y i∈I2 ai ! ∗ai0 = Y i∈I′ ai ! ∗ai0. From this lemma we can prove the ﬁrst commutative law which states that the order of the index set does not aﬀect the product: Theorem 62 (General Commutative Law: First Form). Let ⟨M, ∗⟩be a commu-tative monoid. Let I and J be two ordered ﬁnite sets with the same underlying set which we call K. Let (ak)k∈K be a family of elements of M indexed by K, which we we can think of as a ﬁnite ordered sequence either as (ai)i∈I or as (aj)j∈J. Note these sequences have the same terms, but perhaps given in a diﬀerent order. Then Y i∈I ai = Y j∈J aj. Proof. The proof is by induction on the size of K. If K is empty, then both sides of the equation in question are the identity element by deﬁnition of the empty product. So assume K is not empty, and ﬁx an element k ∈K. Let I′ = I −{k} and J′ = J −{k} where these sets have the restricted order from I and J respectively. Note that I′ and J′ have the same underlying set. By the above lemma and the induction hypothesis we have Y i∈I ai = Y i∈I′ ai ! ∗ak =  Y j∈J′ aj  ∗ak = Y j∈J aj. Deﬁnition 25. Let ⟨M, ∗⟩be a commutative monoid and let (ak)k∈K be a family of elements of M indexed by a ﬁnite set K. Then we deﬁne Q(ak)k∈K to be the value resulting from any chosen total ordering of K. By the above theorem, all orderings give the same value, so this new type of product is well-deﬁned, and agrees with the old deﬁnition (Deﬁnition 22) for any ordering of K. 33 Recall that in Section 16 we considered an equivalence relation for ﬁnite ordered sequences called order equivalence, and showed that order equivalent sequences have the same product. Recall that two ﬁnite ordered sequence (ai)i∈I and (bJ)i∈J with values in a given set are order-equivalent if (i) the cardinality of I and J are equal, and (ii) the unique order-preserving bijection γ : I →J has the property that bγ(i) = ai for all i ∈I. We now deﬁne another equivalence relation where we drop the requirement that γ be order-preserving: we declare two ﬁnite fami-lies (ai)i∈I and (bJ)i∈J with values in a given set to be multiset-equivalent if there is a bijection γ : I →J with the property that bγ(i) = ai for all i ∈I. (Note: we can deﬁne a multiset as an equivalence class under this relation. However, we will not need multisets, per se, in this document.) Theorem 63 (General Commutative Law: Second Form). Let (ai)i∈I and (bj)j∈J be two ﬁnite families of elements of M where ⟨M, ∗⟩is a commutative monoid. If (ai)i∈I and (bj)j∈J are multiset-equivalent then Y i∈I ai = Y j∈J aj. Proof. Let γ be a bijection γ : I →J with the property that bγ(i) = ai for all i ∈I. Fix any ordering on I. If J comes with an order, then ignore it. Instead, impose the unique order on J such that γ is an order preserving map. Under this choice of order we have that (ai)i∈I and (bj)j∈J are order-equivalent sequences and so, using these orderings, Y i∈I ai = Y j∈J bj. Note that this equality continuous to hold when we drop the orderings on I and J (Theorem 62 and Deﬁnition 25). 19 Induced Operation on Functions Above in Section 12 we used the term “induced operation” for the restriction of a binary operation to a closed subset. We will switch gears a bit here and deﬁne another kind of “induced operations”, namely induced operations on functions. The motivation is observation that if we have a binary operation on a set M then we get a natural binary operation on the set of functions S →M where S is any given domain: Deﬁnition 26. Suppose S and M are sets. Then let F(S, M) be the set of func-tions S →M. Deﬁnition 27. Let M be a set with a binary operation ∗: M × M →M and let S be be a ﬁxed set to serve as a domain. Then we deﬁne a binary operation ∗F on the set of functions F(S, M) as follows: given a pair of functions f, g ∈F(S, M) deﬁne f ∗F g by the rule (f ∗F g)(x) def = f(x) ∗g(x). 34 We call ∗F the induced operation on functions. We will often write ∗F as just ∗ and use context to distinguish this operation from that on M. Remark. Observe that ∗F (or just ∗) gives a binary operation F(S, M) × F(S, M) →F(S, M). The following lemmas are proved in a straightforward manner. Lemma 64. Suppose that ∗: M × M →M is an associative operation. Then the induced operation on F(S, M) is also associative. Lemma 65. Suppose that ∗: M × M →M is a commutative operation. Then the induced operation on F(S, M) is also commutative. Lemma 66. Suppose that ∗: M × M →M is a binary operation with identity element e ∈M. Let e: S →M be the constant function deﬁned by the equa-tion e(x) = e for all x ∈S. Then e is an identify element for the induced operation on F(S, M). Theorem 67. Suppose M is a monoid and that S is a set. Then F(S, M) is a monoid under the induced operation. The identity element is the constant func-tion S →M whose values are given by the identity element of M. If M is a commutative monoid then so is F(S, M). Theorem 68. Suppose ⟨G, ∗⟩is a group and that S is a set. Then F(S, G) is a group under the induced operation. The identity element is the constant func-tion S →G whose values are given by the identity element of G. Proof. By Theorem 67, F(S, G) is a monoid. To show it is a group we need to show every element f ∈F(S, G) has an inverse. Let g: S →G be deﬁned by the rule x 7→f(x)−1 (for convenience we adopt multiplicative notation for G, and use e for the identity in G). Then for any x ∈S we have (f ∗g)(x) = f(x) ∗g(x) = f(x) ∗f(x)−1 = e. In other words, f ∗g is the constant function e whose values are e. Similarly, we can show g ∗f = e. Thus g is the inverse of f in F(S, G). Exercise 28. Suppose M monoid and that S is a set. Describe the elements in the group F(S, M)inv of invertible elements. Exercise 29. (Taking as given the notions of continuity and diﬀerentiability) Con-sider ⟨R, +⟩. Let G = F(R, R). As we have seen, G is a group under the induced addition +. Let H1 be the subset of continuous functions. Show that H1 is a subgroup of G. Let H2 be the subset of diﬀerentiable functions. Show that H2 is a subgroup of H1 and G. Exercise 30. (Taking as given the notions of continuity and diﬀerentiability) Con-sider ⟨R, ·⟩. Let S be the open interval (0, 1), and let M = F(S, R). As we have seen, M is a monoid under the induced multiplication. A typical element of M is the function deﬁned by f(x) = 1/x. Let H1 be the subset of continuous functions. Show that H1 is a submonoid of M. Let H2 be the subset of diﬀerentiable functions. Show that H2 is a submonoid of H1 and G. 35 19.1 Cartesian Powers of a Monoid We now consider the special case of sets of functions F(S, M) where S = {1, . . . , n} where n is a positive integer. Then an element f of F(S, M) can be completely described simply by specifying n values: f(1), f(2), . . . , f(n). In other words, elements of F(S, M) can be thought of as ﬁnite se-quences (a1, . . . , an) with ai ∈M. Here we identify (a1, . . . , an) with the func-tion i 7→ai which is an element of F(S, M) where S = {1, . . . , n}. Deﬁnition 28. Let M be a set, and let n be a positive integer. Then M n is deﬁned to be F({1, . . . , n}, M). As discussed above, we think of M n as the set of ﬁnite sequences of the form (a1, . . . , an) where each ai ∈M. We call M n the nth Cartesian power of M. If M has a binary operation ∗: M ×M →M, then we get an induced operation ∗ on M n using Deﬁnition 27. The induced operation can be written as (a1, . . . , an) ∗(b1, . . . , bn) = (a1 ∗b1, . . . , an ∗bn) where the operation ∗in the left-hand side is the induced operation, and the oper-ation in the right hand side is the operation in M. The following is just a special case of previous results for F(G, S): Theorem 69. Let n be a positive integer. If M is a monoid, then so is M n (under the induced operation). If M is a commutative monoid, then so is M n. If G is a group then so is Gn. If G is an Abelian group, then so is Gn. Exercise 31. (Taken as given the vector spaces R2 and R3). Consider R as a group under addition. Show that the induced operation on R2 and R3 are the usual vector addition. Conclude that R2 and R3 are Abelian groups under vector addition. Exercise 32. Suppose H is a subgroup of G. By extending codomains, you can identify every element of F(S, H) with an element of F(S, G), and so can iden-tify F(S, H) with a subset of F(S, G). Show that when you do this, the operation on F(S, H) is the restriction of the operation on F(S, G), and that F(S, H) can be regarded as a subgroup of F(S, G). Note: as a special cases we see Hn can be regarded as a subgroup of Gn for all positive n. (Note: the analogous property holds for monoids and submonoids, and this can be shown along the way.) Remark. For any set G it is common to deﬁne G0 to be a set with one element. When G is a group or monoid, it is common to regard G0 as a group using the unique binary operation on this one-point set G0. (This is consistent with thinking of G0 as the set F(∅, G) ). 20 Composition of Functions Above we considered one sort of operation on functions which gives a monoid, assuming the codomain is a monoid. Now we describe another way to form monoids 36 whose elements are functions. Let X be a given set and consider F(X, X), the set of functions from X to itself. In this case we write F(X, X) simply as F(X). Note that composition ◦gives a binary operation ◦: F(X) × F(X) →F(X) and that if X is ﬁnite of size n then F(X) has nn elements. The following theorems are straightforward consequences of basic set-theoretical facts: Theorem 70. Let X be a set. Then F(X) is a monoid under composition. The identity element is the identity function X →X. Theorem 71. Let X be a set. Then a function f ∈F(X) is invertible in the monoid F(X) if and only if it is invertible as a function X →X. The inverse of an invertible f in the monoid F(X) is the inverse function f −1 : X →X. Deﬁnition 29. Let X be a set. The group of units of F(X) is called the sym-metric group on X, which we can write as S(X) or SX, and its elements are called permutations of X. Remark. Note that if X is ﬁnite with n elements then the symmetric group on X has n! elements. If X = {1, . . . , n} then the symmetric group is sometimes just written as Sn (we sometimes write it as Sn) Remark. Observe that if A is a group or monoid then F(A, A) has two standard operations: composition and the induced operation. We have to be clear of what is meant in any given situation, especially since we sometimes use multiplicative notation fg for both operations. Similarly, f −1 can have two possible meanings depending on which monoid, and hence which binary operation, is under consider-ation. Exercise 33. Consider ⟨R, ·⟩, and let M = F(R, R). Let f, g ∈M be deﬁned by the equations f(x) = x + 1 and g(x) = x2. What is fg, gf, f ◦g, and g ◦f? 37
17034	https://www.scribd.com/document/575975866/Electrical-Machinery-PS-Bimbhra	Electrical Machinery PS Bimbhra \| PDF Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 100%(4)100% found this document useful (4 votes) 14K views 1,075 pages Electrical Machinery PS Bimbhra Nepal Full description Uploaded by Tarak Mondal Go to previous items Go to next items Download Save Save Electrical Machinery PS Bimbhra For Later Share 100%100% found this document useful, undefined 0%, undefined Print Embed Report Download Save Electrical Machinery PS Bimbhra For Later You are on page 1/ 1075 Search Fullscreen adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Electrical Machinery PS Bimbhra 83% (24) Electrical Machinery PS Bimbhra 1,000 pages Electrical Machines by Mr. S. K. Sahdev1 100% (9) Electrical Machines by Mr. S. K. Sahdev1 980 pages A Course in Power Systems-Gupta 100% (5) A Course in Power Systems-Gupta 570 pages EMAC Kothari Solutions 100% (4) EMAC Kothari Solutions 308 pages Electrical Machines I by U A Bakshi V U Bakshi 75% (8) Electrical Machines I by U A Bakshi V U Bakshi 355 pages Engineering Student Resources 100% (3) Engineering Student Resources 549 pages Electric Machines - Ashfaq Husain 100% (9) Electric Machines - Ashfaq Husain 642 pages Utilization of Electrical Power 57% (7) Utilization of Electrical Power 574 pages Power Sytem Neelakantan 83% (18) Power Sytem Neelakantan 366 pages S.N.Bhadra, D.Kastha, S.Banerjee - Wind Electrical Systems-Oxford University Press (2013) PDF 78% (9) S.N.Bhadra, D.Kastha, S.Banerjee - Wind Electrical Systems-Oxford University Press (2013) PDF 338 pages Electric Power System Components - Transformers and Rotating Machines No ratings yet Electric Power System Components - Transformers and Rotating Machines 485 pages Electric Machines - Principles, Applications, and Control Schematics 2nd Edition 100% (4) Electric Machines - Principles, Applications, and Control Schematics 2nd Edition 642 pages Books in The Ieee Press Series On Power Engineering 100% (1) Books in The Ieee Press Series On Power Engineering 490 pages R. Krishnan - Electric Motor Drives Modeling, Analysis, and Control 94% (32) R. Krishnan - Electric Motor Drives Modeling, Analysis, and Control 653 pages Electrical Machines GOOD TEXTBOOK No ratings yet Electrical Machines GOOD TEXTBOOK 72 pages Power Electronics - K.B Khanchandani PDF 83% (6) Power Electronics - K.B Khanchandani PDF 1,078 pages Electric Machinery, 6th Edition 25% (4) Electric Machinery, 6th Edition 5 pages Electrical Machines 100% (2) Electrical Machines 624 pages PS Bhimbra Generalised Machine Theory 100% (7) PS Bhimbra Generalised Machine Theory 582 pages Design and Testing of Electrical Machines 89% (9) Design and Testing of Electrical Machines 210 pages Electrical Machines - 2 (Book) 100% (2) Electrical Machines - 2 (Book) 330 pages Electrical Machinery Overview No ratings yet Electrical Machinery Overview 11 pages 0070459940 100% (2) 0070459940 226 pages Power Systems by C L Wadhwa 100% (4) Power Systems by C L Wadhwa 981 pages Electrical Measurements and Measuring Instruments 100% (1) Electrical Measurements and Measuring Instruments 317 pages 9780750320849 No ratings yet 9780750320849 382 pages Electric Machines - D. P. Kothari and I. J. Nagrath PDF 50% (4) Electric Machines - D. P. Kothari and I. J. Nagrath PDF 170 pages Electrical Machine TH No ratings yet Electrical Machine TH 22 pages Electrical Machine PDF No ratings yet Electrical Machine PDF 142 pages U E E by R K Rajput 100% (1) U E E by R K Rajput 574 pages Ee6401 em - I PDF No ratings yet Ee6401 em - I PDF 143 pages Electrical Machine Design - A. K. Sawhney 45% (11) Electrical Machine Design - A. K. Sawhney 97 pages Electric Machinery Fundamentals Power Energy 4th Edition Stephen J. Chapman Full Chapters Included 100% (2) Electric Machinery Fundamentals Power Energy 4th Edition Stephen J. Chapman Full Chapters Included 114 pages DC Machines and Synchronous Machines Bakshi 100% (2) DC Machines and Synchronous Machines Bakshi 497 pages M D Singh K B Khanchandani Power Electronics 86% (7) M D Singh K B Khanchandani Power Electronics 849 pages Induction & Synchronous Machines 100% (5) Induction & Synchronous Machines 373 pages Machines 1 PDF No ratings yet Machines 1 PDF 112 pages Fundamentals of Electrical Drives GK. Dubey 100% (2) Fundamentals of Electrical Drives GK. Dubey 166 pages Design of Electrical Machines 100% (6) Design of Electrical Machines 183 pages Kretyrx ELECTRICAL Machine Notes 100% (5) Kretyrx ELECTRICAL Machine Notes 873 pages Electrical Machines Course Guide No ratings yet Electrical Machines Course Guide 117 pages Electrical Machines Lecture Notes No ratings yet Electrical Machines Lecture Notes 90 pages Electrical Machines: S.K. Kataria & Sons 100% (1) Electrical Machines: S.K. Kataria & Sons 22 pages Dr. J.S.chitode - Power Electronics - III-Technical Publications 33% (3) Dr. J.S.chitode - Power Electronics - III-Technical Publications 323 pages Bhimbra-GENERALIZED THEORY ELECTRICAL MACHINES PDF No ratings yet Bhimbra-GENERALIZED THEORY ELECTRICAL MACHINES PDF 582 pages Electrical Machines Vol 2 4thed Ua Bakshi MV Bakshi PDF 100% (4) Electrical Machines Vol 2 4thed Ua Bakshi MV Bakshi PDF 547 pages Electrical Machines 100% (10) Electrical Machines 403 pages Electrical Machine Design - A. K. Sawhney No ratings yet Electrical Machine Design - A. K. Sawhney 134 pages Theory & Performance Of: (DC Machines and AC Machines) (DC Machines and AC Machines) No ratings yet Theory & Performance Of: (DC Machines and AC Machines) (DC Machines and AC Machines) 24 pages Special Electrical Machines PDF 75% (4) Special Electrical Machines PDF 1 page D. P. Sen Gupta, J. W. Lynn (Auth.) - Electrical Machine Dynamics-Macmillan Education UK (1980) 100% (8) D. P. Sen Gupta, J. W. Lynn (Auth.) - Electrical Machine Dynamics-Macmillan Education UK (1980) 310 pages PDF Electrical Machines Vol 2 4thed Ua Bakshi MV Bakshi DL 100% (1) PDF Electrical Machines Vol 2 4thed Ua Bakshi MV Bakshi DL 547 pages Machine Book by Bakshi 80% (5) Machine Book by Bakshi 476 pages Elements of Power System - J. B. Gupta PDF 33% (12) Elements of Power System - J. B. Gupta PDF 121 pages Electrical Machinery CH 1&2 P.S. Bimbhra 91% (11) Electrical Machinery CH 1&2 P.S. Bimbhra 238 pages Theory of Alternating Current Machinery - A. S. Langsdorf 0% (3) Theory of Alternating Current Machinery - A. S. Langsdorf 42 pages PDF 0% (1) PDF 55 pages J B Gupta Theory and Performance of Electrical Machines Book No ratings yet J B Gupta Theory and Performance of Electrical Machines Book 2 pages Electrical Machines 3rd Edition - S. K. Bhattacharya No ratings yet Electrical Machines 3rd Edition - S. K. Bhattacharya 113 pages Motor Duty Classes & Rating Guide No ratings yet Motor Duty Classes & Rating Guide 20 pages adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free adDownload to read ad-free ad Footer menu Back to top About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Support Help / FAQ Accessibility Purchase help AdChoices Legal Terms Privacy Copyright Cookie Preferences Do not sell or share my personal information Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Legal Terms Privacy Copyright Cookie Preferences Do not sell or share my personal information Support Help / FAQ Accessibility Purchase help AdChoices Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps Documents Language: English Copyright © 2025 Scribd Inc. We take content rights seriously. Learn more in our FAQs or report infringement here. We take content rights seriously. Learn more in our FAQs or report infringement here. Language: English Copyright © 2025 Scribd Inc. 576648e32a3d8b82ca71961b7a986505
17035	https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./03.E%3A_Stoichiometry_(Exercises)	Skip to main content 3.E: Stoichiometry (Exercises) Last updated : Apr 28, 2024 Save as PDF 2.E: Atoms, Molecules, and Ions (Exercises) 4.E: Aqueous Reactions (Exercises) Page ID : 43470 ( \newcommand{\kernel}{\mathrm{null}\,}) 3.1: Chemical Equations Conceptual Problems How does a balanced chemical equation agree with the law of definite proportions? What is the difference between S8 and 8S? Use this example to explain why subscripts in a formula must not be changed. What factors determine whether a chemical equation is balanced? What information can be obtained from a balanced chemical equation? Does a balanced chemical equation give information about the rate of a reaction? Numerical Problems Balance each chemical equation. a. KI(aq) + Br2(l) → KBr(aq) + I2(s) b. MnO2(s) + HCl(aq) → MnCl2(aq) + Cl2(g) + H2O(l) c. Na2O(s) + H2O(l) → NaOH(aq) d. Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s) e. SO2(g) + H2O(l) → H2SO3(aq) f. S2Cl2(l) + NH3(l) → S4N4(s) + S8(s) + NH4Cl(s) Balance each chemical equation. a. Be(s) + O2(g) → BeO(s) b. N2O3(g) + H2O(l) → HNO2(aq) c. Na(s) + H2O(l) → NaOH(aq) + H2(g) d. CaO(s) + HCl(aq) → CaCl2(aq) + H2O(l) e. CH3NH2(g) + O2(g) → H2O(g) + CO2(g) + N2(g) f. Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g) Balance each chemical equation. a. N2O5(g) → NO2(g) + O2(g) b. NaNO3(s) → NaNO2(s) + O2(g) c. Al(s) + NH4NO3(s) → N2(g) + H2O(l) + Al2O3(s) d. C3H5N3O9(l) → CO2(g) + N2(g) + H2O(g) + O2(g) e. reaction of butane with excess oxygen f. IO2F(s) + BrF3(l) → IF5(l) + Br2(l) + O2(g) Balance each chemical equation. a. H2S(g) + O2(g) → H2O(l) + S8(s) b. KCl(aq) + HNO3(aq) + O2(g) → KNO3(aq) + Cl2(g) + H2O(l) c. NH3(g) + O2(g) → NO(g) + H2O(g) d. CH4(g) + O2(g) → CO(g) + H2(g) e. NaF(aq) + Th(NO3)4(aq) → NaNO3(aq) + ThF4(s) f. Ca5(PO4)3F(s) + H2SO4(aq) + H2O(l) → H3PO4(aq) + CaSO4•2H2O(s) + HF(aq) Balance each chemical equation. a. NaCl(aq) + H2SO4(aq) → Na2SO4(aq) + HCl(g) b. K(s) + H2O(l) → KOH(aq) + H2(g) c. reaction of octane with excess oxygen d. S8(s) + Cl2(g) → S2Cl2(l) e. CH3OH(l) + I2(s) + P4(s) → CH3I(l) + H3PO4(aq) + H2O(l) f. (CH3)3Al(s) + H2O(l) → CH4(g) + Al(OH)3(s) Write a balanced chemical equation for each reaction. a. Aluminum reacts with bromine. b. Sodium reacts with chlorine. c. Aluminum hydroxide and acetic acid react to produce aluminum acetate and water. d. Ammonia and oxygen react to produce nitrogen monoxide and water. e. Nitrogen and hydrogen react at elevated temperature and pressure to produce ammonia. f. An aqueous solution of barium chloride reacts with a solution of sodium sulfate. Write a balanced chemical equation for each reaction. a. Magnesium burns in oxygen. b. Carbon dioxide and sodium oxide react to produce sodium carbonate. c. Aluminum reacts with hydrochloric acid. d. An aqueous solution of silver nitrate reacts with a solution of potassium chloride. e. Methane burns in oxygen. f. Sodium nitrate and sulfuric acid react to produce sodium sulfate and nitric acid. 3.2: Some Simple Patterns of Chemical Reactivity 3.3: Formula Masses Conceptual Problems What is the relationship between an empirical formula and a molecular formula Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition. Numerical Problems Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Numerical Problems. a. What is the formula mass of each species? a. ammonium chloride b. sodium cyanide c. magnesium hydroxide d. calcium phosphate e. lithium carbonate f. hydrogen sulfite ion b. What is the molecular or formula mass of each compound? a. potassium permanganate b. sodium sulfate c. hydrogen cyanide d. potassium thiocyanate e. ammonium oxalate f. lithium acetate What is the mass percentage of water in each hydrate? a. H3AsO4·5H2O b. NH4NiCl3·6H2O c. Al(NO3)3·9H2O What is the mass percentage of water in each hydrate? a. CaSO4·2H2O b. Fe(NO3)3·9H2O c. (NH4)3ZrOH(CO3)3·2H2O Which of the following has the greatest mass percentage of oxygen—KMnO4, K2Cr2O7, or Fe2O3? Which of the following has the greatest mass percentage of oxygen—ThOCl2, MgCO3, or NO2Cl? Calculate the percent composition of the element shown in bold in each compound. a. SbBr3 b. As2I4 c. AlPO4 d. C6H10O Calculate the percent composition of the element shown in bold in each compound. a. HBrO3 b CsReO4 c. C3H8O d. FeSO4 A sample of a chromium compound has a molar mass of 151.99 g/mol. Elemental analysis of the compound shows that it contains 68.43% chromium and 31.57% oxygen. What is the identity of the compound? The percentages of iron and oxygen in the three most common binary compounds of iron and oxygen are given in the following table. Write the empirical formulas of these three compounds. \| Compound \| % Iron \| % Oxygen \| Empirical Formula \| --- --- \| \| 1 \| 69.9 \| 30.1 \| \| \| 2 \| 77.7 \| 22.3 \| \| \| 3 \| 72.4 \| 27.6 \| \| What is the mass percentage of water in each hydrate? a. LiCl·H2O b. MgSO4·7H2O c. Sr(NO3)2·4H2O What is the mass percentage of water in each hydrate? a. CaHPO4·2H2O b. FeCl2·4H2O c. Mg(NO3)2·4H2O Two hydrates were weighed, heated to drive off the waters of hydration, and then cooled. The residues were then reweighed. Based on the following results, what are the formulas of the hydrates? \| Compound \| Initial Mass (g) \| Mass after Cooling (g) \| --- \| NiSO4·xH2O \| 2.08 \| 1.22 \| \| CoCl2·xH2O \| 1.62 \| 0.88 \| Which contains the greatest mass percentage of sulfur—FeS2, Na2S2O4, or Na2S? Given equal masses of each, which contains the greatest mass percentage of sulfur—NaHSO4 or K2SO4? Calculate the mass percentage of oxygen in each polyatomic ion. a. bicarbonate b. chromate c. acetate d. sulfite Calculate the mass percentage of oxygen in each polyatomic ion. a. oxalate b. nitrite c. dihydrogen phosphate d. thiocyanate The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula? A compound has the empirical formula C2H4O, and its formula mass is 88 g. What is its molecular formula? Mirex is an insecticide that contains 22.01% carbon and 77.99% chlorine. It has a molecular mass of 545.59 g. What is its empirical formula? What is its molecular formula? How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene? How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline? How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame? How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G? Combustion of a 34.8 mg sample of benzaldehyde, which contains only carbon, hydrogen, and oxygen, produced 101 mg of CO2 and 17.7 mg of H2O. a. What was the mass of carbon and hydrogen in the sample? b. Assuming that the original sample contained only carbon, hydrogen, and oxygen, what was the mass of oxygen in the sample? c. What was the mass percentage of oxygen in the sample? d. What is the empirical formula of benzaldehyde? e. The molar mass of benzaldehyde is 106.12 g/mol. What is its molecular formula? Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O. a. What is the mass of oxygen in the sample? b. What is the mass percentage of oxygen in the sample? c. What is the empirical formula of salicylic acid? d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula? Given equal masses of the following acids, which contains the greatest amount of hydrogen that can dissociate to form H+—nitric acid, hydroiodic acid, hydrocyanic acid, or chloric acid? Calculate the formula mass or the molecular mass of each compound. a. heptanoic acid (a seven-carbon carboxylic acid) b. 2-propanol (a three-carbon alcohol) c. KMnO4 d. tetraethyllead e. sulfurous acid f. ethylbenzene (an eight-carbon aromatic hydrocarbon) Calculate the formula mass or the molecular mass of each compound. a. MoCl5 b. B2O3 c. bromobenzene d. cyclohexene e. phosphoric acid f. ethylamine Given equal masses of butane, cyclobutane, and propene, which contains the greatest mass of carbon? Given equal masses of urea [(NH2)2CO] and ammonium sulfate, which contains the most nitrogen for use as a fertilizer? Conceptual Answers 1) What is the relationship between an empirical formula and a molecular formula An empirical formula refers to the simplest ratio of elements that is obtained from a chemical formula while a molecular formula is calculated to show the actual formula of a molecular compound. 2) Construct a flowchart showing how you would determine the empirical formula of a compound from its percent composition. Numerical Answers a. What is the formula mass of each species? a. 53.49146 amu b. 49.0072 amu c. 58.3197 amu d. 310.177 amu e. 73.891 amu f. 81.07 amu b. What is the molecular or formula mass of each compound? a.158.034 amu b. 142.04 amu c. 27.0253 amu d. 97.181 amu e. 124.1 amu f. 65.99 amu To two decimal places, the percentages are: a. 5.97% b. 37.12% c. 43.22% To two decimal places, the percentages of water are: a. 20.93% b. 40.13% c. 9.52% % oxygen: KMnO4, 40.50%; K2Cr2O7, 38.07%; Fe2O3, 30.06% % oxygen: ThOCl2, 5.02%; MgCO3, 56.93%; NO2Cl, 39.28% To two decimal places, the percentages are: a. 66.32% Br b. 22.79% As c. 25.40% P d. 73.43% C 6. a. 61.98% Br b. 34.69% Cs c. 59.96% C d. 21.11% S Cr2O3. Empirical Formulas Fe2O3 Fe4O4 Fe6O8 To two decimal places, the percentages are: a. 29.82% b. 51.16% c. 25.40% What is the mass percentage of water in each hydrate? a. 20.94% b. 36.25% c. 32.70% NiSO4 · 6H2O and CoCl2 · 6H2O FeS2 NaHSO4 Calculate the mass percentage of oxygen in each polyatomic ion. a. 78.66% b. 55.17% c. 54.19% d. 59.95% 15. a. 72.71% b. 69.55% c. 65.99% d. 0% The empirical formula of garnet, a gemstone, is Fe3Al2Si3O12. An analysis of a sample of garnet gave a value of 13.8% for the mass percentage of silicon. Is this consistent with the empirical formula? No, the calculated mass percentage of silicon in garnet is 16.93% C4H8O2 18. Empirical Formula: C10Cl12 Molecular Formula: C10Cl12 How many moles of CO2 and H2O will be produced by combustion analysis of 0.010 mol of styrene? Moles of CO2: 0.08 mol CO2 Moles of H2O: 0.04 mol H2O How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0080 mol of aniline? Mole of CO2: 0.048 mol CO2 Mole of H2O: 0.028 mol H2O Mole of N2: 0.004 mol N2 How many moles of CO2, H2O, and N2 will be produced by combustion analysis of 0.0074 mol of aspartame? Mole of CO2: 0.104 mol CO2 Mole of H2O: 0.666 mol H2O Mole of N2: 0.0074 mol N2 How many moles of CO2, H2O, N2, and SO2 will be produced by combustion analysis of 0.0060 mol of penicillin G? Mole of CO2: 0.096 mol CO2 Mole of H2O: 0.054 mol H2O Mole of N2: 0.060 mol N2 Mole of SO2: 0.060 mol SO2 23. a. 27.6 mg C and 1.98 mg H b. 5.2 mg O c. 15% d. C7H6O e. C7H6O Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen. Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg of H2O. a. What is the mass of oxygen in the sample? 70.4mg b. What is the mass percentage of oxygen in the sample? 61.70% c. What is the empirical formula of salicylic acid? C7H6O3 d. The molar mass of salicylic acid is 138.12 g/mol. What is its molecular formula? C7H6O3 hydrocyanic acid, HCN Calculate the formula mass or the molecular mass of each compound. a. 130.1849 amu b. 60.1 amu c. 158.034 amu d. 323.4 amu e. 82.07 amu f. 106.17 amu To two decimal places, the values are: a. 273.23 amu b. 69.62 amu c. 157.01 amu d. 82.14 amu e. 98.00 amu f. 45.08 amu Cyclobutene Urea 3.4: Avogadro's Number and the Mole Conceptual Problems Please be sure you are familiar with the topics discussed in Essential Skills 2 before proceeding to the Conceptual Problems. Describe the relationship between an atomic mass unit and a gram. Is it correct to say that ethanol has a formula mass of 46? Why or why not? If 2 mol of sodium reacts completely with 1 mol of chlorine to produce sodium chloride, does this mean that 2 g of sodium reacts completely with 1 g of chlorine to give the same product? Explain your answer. Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi3O8), a mineral used in the manufacture of porcelain. Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass. Numerical Problems Please be sure you are familiar with the topics discussed in Essential Skills 2 before proceeding to the Numerical Problems. Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass. Calculate the molecular mass or formula mass of each compound. KCl (potassium chloride) NaCN (sodium cyanide) H2S (hydrogen sulfide) NaN3 (sodium azide) H2CO3 (carbonic acid) K2O (potassium oxide) Al(NO3)3 (aluminum nitrate) Cu(ClO4)2 [copper(II) perchlorate] Calculate the molecular mass or formula mass of each compound. V2O4 (vanadium(IV) oxide) CaSiO3 (calcium silicate) BiOCl (bismuth oxychloride) CH3COOH (acetic acid) Ag2SO4 (silver sulfate) Na2CO3 (sodium carbonate) (CH3)2CHOH (isopropyl alcohol) Calculate the molar mass of each compound. a. b. c. d. e. Calculate the molar mass of each compound. a. b. c. d. For each compound, write the condensed formula, name the compound, and give its molar mass. a. b. For each compound, write the condensed formula, name the compound, and give its molar mass. a. b. Calculate the number of moles in 5.00 × 102 g of each substance. How many molecules or formula units are present in each sample? a. CaO (lime) b. CaCO3(chalk) c. C12H22O11 [sucrose (cane sugar)] d. NaOCl (bleach) e. CO2 (dry ice) Calculate the mass in grams of each sample. a. 0.520 mol of N2O4 b. 1.63 mol of C6H4Br2 c. 4.62 mol of (NH4)2SO3 Give the number of molecules or formula units in each sample. a. 1.30 × 10−2 mol of SCl2 b. 1.03 mol of N2O5 c. 0.265 mol of Ag2Cr2O7 Give the number of moles in each sample. a. 9.58 × 1026 molecules of Cl2 b. 3.62 × 1027 formula units of KCl c. 6.94 × 1028 formula units of Fe(OH)2 Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I2)? What is the total number of atoms in each sample? a. 0.431 mol of Li b. 2.783 mol of methanol (CH3OH) c. 0.0361 mol of CoCO3 d. 1.002 mol of SeBr2O What is the total number of atoms in each sample? a. 0.980 mol of Na b. 2.35 mol of O2 c. 1.83 mol of Ag2S d. 1.23 mol of propane (C3H8) What is the total number of atoms in each sample? a. 2.48 g of HBr b. 4.77 g of CS2 c. 1.89 g of NaOH d. 1.46 g of SrC2O4 Decide whether each statement is true or false and explain your reasoning. There are more molecules in 0.5 mol of Cl2than in 0.5 mol of H2. One mole of H2 has 6.022 × 1023 hydrogen atoms. The molecular mass of H2O is 18.0 amu. The formula mass of benzene is 78 amu. Complete the following table. \| Substance \| Mass (g) \| Number of Moles \| Number of Molecules or Formula Units \| Number of Atoms or Ions \| --- --- \| MgCl2 \| 37.62 \| a. \| b. \| c. \| \| AgNO3 \| d. \| 2.84 \| e. \| f. \| \| BH4Cl \| g. \| h. \| 8.93 × 1025 \| i. \| \| K2S \| j \| k. \| l. \| 7.69 × 1026 \| \| H2SO4 \| m. \| 1.29 \| n. \| o. \| \| C6H14 \| 11.84 \| p. \| q. \| r. \| \| HClO3 \| s. \| t. \| 2.45 × 1026 \| u. \| Give the formula mass or the molecular mass of each substance. Give the formula mass or the molecular mass of each substance. Conceptual Answers While both are units of mass, a gram is Avogadro’s number of atomic mass units so you would multiply the number of amu by 6.022x10^23 to find total number of grams The correct way to state formula mass of ethanol is to show the units of mass which is amu. No because moles and weight operate on different set of standards meaning that they’re not equal to each other. This means that moles of different compounds contain different weights. For example, 2 moles of Na = 2 x 22.989 g = 45.98g while 1 mole of Cl = 1 x 35.453 g = 35.453 g Cl. This makes the sodium react completely with chlorine. 2g of sodium would react with = (35.453/45.978) x 2 = 1.542 g Cl Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi3O8), a mineral used in the manufacture of porcelain . Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass. The information required to determine the mass of the solute would be the molarity of the solution because once that is achieved, volume of the solution and molar mass of the solute can be used to calculate the total mass. A derivatization that achieves this goes as: Molarity = moles of solute / volume of solution in liter -> Moles = molarity x volume in liter -> Mass= moles x molar mass. Numerical Answers Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass. Calculate the molecular mass or formula mass of each compound. 74.55 amu 49.01 amu 34.08 amu 65.01 amu 62.02 amu 94.20 amu 213.00 amu 262.45 amu Calculate the molecular mass or formula mass of each compound. 165.88 amu 116.16 amu 260.43 amu 60.05 amu 311.80 amu 105.99 amu 60.10 amu Calculate the molar mass of each compound. a. 153.82 g/mol b. 80.06 g/mol c. 92.01 g/mol d. 70.13 g/mol e. 74.12 g/mol Calculate the molar mass of each compound. a. 92.45 g/mol b. 135.04 g/mol c. 44.01 g/mol d. 40.06 g/mol For each compound, write the condensed formula, name the compound, and give its molar mass. a. C5H10O2, Valeric Acid, 102.13 g/mol b. H3PO3, Phosphorous acid, 82 g/mol For each compound, write the condensed formula, name the compound, and give its molar mass. a. C2H5NH2, Ethylamine, 45.08 g/mol b. HIO3, Iodic acid, 175.91 g/mol Calculate the number of moles in 5.00 × 102 g of each substance. How many molecules or formula units are present in each sample? a. 5.37 × 1024 mol b. 3.01 × 1024 mol c. 8.80 × 1023 mol d. 4.04 × 1024 mol e. 6.84 × 1024 mol Calculate the mass in grams of each sample. a. 47.85 grams b. 384.52 grams c. 536.57 grams Give the number of molecules or formula units in each sample. a. 7.83x1021 molecules b. 6.20x1023 molecules c. 1.60x1023 molecules Give the number of moles in each sample. a. 1590.8 moles b. 6011.3 moles c. 115244.1 moles Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I2)? 2.61 x1022 molecules What is the total number of atoms in each sample? a. 2.60x1023 atoms b. 1.01x1025 atoms c. 1.09x1023 atoms d. 2.41x1024 atoms What is the total number of atoms in each sample? a. 5.9x1023 atoms b. 2.8x1024 atoms c. 3.31x1024 atoms d. 8.15x1024 atoms What is the total number of atoms in each sample? a. 3.69x1022 atoms b. 1.13x1023 atoms c. 8.54x1022 atoms d. 3.50x1023 atoms 16. a. False, the number of molecules in 0.5 mol Cl2 are the same amount of molecules in H2 b. False, the number of molecules in H2 is 2 x (6.022 x10^23) H atoms c. True, 2 H (1.01 amu) + 1 O (16.01) = 18.0 amu d. True, C6H6 -> 12(6) + 1(6) = 78 amu Complete the following table a. 0.39 b. 2.36x10^23 c. 7.08x10^23 d. 482.8 e. 1.71x10^24 f. 8.55x10^24 g. 7932.7 h. 148.3 i. 5.36x10^26 j. 46938.5 k. 425.7 l. 1276.98 m. 126.5 n. 7.77x10^23 o. 5.44x10^24 p. 0.14 q. 8.27x10^22 r. 1.65x10^24 s. 34358 t. 406.8 u. 1.23x10^2 Give the formula mass or the molecular mass of each substance. 261.67 amu 301.04 amu 286.98 amu 551.73 amu Give the formula mass or the molecular mass of each substance. 273.21 amu 69.62 amu 330.04 amu 426.99 amu 3.6: Empirical Formulas from Analysis 3.6: Quantitative Information from Balanced Equations Conceptual Problems What information is required to determine the mass of solute in a solution if you know the molar concentration of the solution? Numerical Problems Refer to the Breathalyzer test described in Example 8. How much ethanol must be present in 89.5 mL of a person’s breath to consume all the potassium dichromate in a Breathalyzer ampul containing 3.0 mL of a 0.40 mg/mL solution of potassium dichromate? Phosphoric acid and magnesium hydroxide react to produce magnesium phosphate and water. If 45.00 mL of 1.50 M phosphoric acid are used in the reaction, how many grams of magnesium hydroxide are needed for the reaction to go to completion? Barium chloride and sodium sulfate react to produce sodium chloride and barium sulfate. If 50.00 mL of 2.55 M barium chloride is used in the reaction, how many grams of sodium sulfate are needed for the reaction to go to completion? How many grams of sodium phosphate are obtained in solution from the reaction of 75.00 mL of 2.80 M sodium carbonate with a stoichiometric amount of phosphoric acid? The second product is water; what is the third product? How many grams of the third product are obtained? How many grams of ammonium bromide are produced from the reaction of 50.00 mL of 2.08 M iron(II) bromide with a stoichiometric amount of ammonium sulfide? What is the second product? How many grams of the second product are produced? Conceptual Answers The information required to determine the mass of the solution with a known molar concentration are the molar mass of the solute and the volume of the solute. Numerical Answers Refer to the Breathalyzer test described in Example 8. How much ethanol must be present in 89.5 mL of a person’s breath to consume all the potassium dichromate in a Breathalyzer ampul containing 3.0 mL of a 0.40 mg/mL solution of potassium dichromate? 5.905 g Mg(OH)2 18.105 g Na2SO4 Third Product: CO2, Grams of Third Product: 9.24 g, Grams of Sodium Phosphate: 22.26g Grams Produced: 20.37 grams NH4Br2, Second Product: Iron (II) Sulfide, 9.15 grams FeS 3.7: LIMITING REACTANTS Conceptual Problems Engineers use conservation of mass, called a “mass balance,” to determine the amount of product that can be obtained from a chemical reaction. Mass balance assumes that the total mass of reactants is equal to the total mass of products. Is this a chemically valid practice? Explain your answer. Given the equation is it correct to say that 10 g of hydrogen will react with 10 g of oxygen to produce 20 g of water vapor? What does it mean to say that a reaction is stoichiometric? When sulfur is burned in air to produce sulfur dioxide, what is the limiting reactant? Explain your answer. Is it possible for the percent yield to be greater than the theoretical yield? Justify your answer. Numerical Problems Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems. Write a balanced chemical equation for each reaction and then determine which reactant is in excess. a. 2.46 g barium(s) plus 3.89 g bromine(l) in water to give barium bromide b. 1.44 g bromine(l) plus 2.42 g potassium iodide(s) in water to give potassium bromide and iodine c. 1.852 g of Zn metal plus 3.62 g of sulfuric acid in water to give zinc sulfate and hydrogen gas d. 0.147 g of iron metal reacts with 0.924 g of silver acetate in water to give iron(II) acetate and silver metal e. 3.142 g of ammonium phosphate reacts with 1.648 g of barium hydroxide in water to give ammonium hydroxide and barium phosphate Under the proper conditions, ammonia and oxygen will react to form dinitrogen monoxide (nitrous oxide, also called laughing gas) and water. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants. a. 24.6 g of ammonia and 21.4 g of oxygen b. 3.8 mol of ammonia and 84.2 g of oxygen c. 3.6 × 1024 molecules of ammonia and 318 g of oxygen d. 2.1 mol of ammonia and 36.4 g of oxygen When a piece of zinc metal is placed in aqueous hydrochloric acid, zinc chloride is produced, and hydrogen gas is evolved. Write a balanced chemical equation for this reaction. Determine which reactant is in excess for each combination of reactants. a. 12.5 g of HCl and 7.3 g of Zn b. 6.2 mol of HCl and 100 g of Zn c. 2.1 × 1023 molecules of Zn and 26.0 g of HCl d. 3.1 mol of Zn and 97.4 g of HCl Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced. a. NaI(aq) + Cl2(g) → NaCl(aq) + I2(s); 1.0 mol of NaCl b. NaCl(aq) + H2SO4(aq) → HCl(g) + Na2SO4(aq); 0.50 mol of HCl c. NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq); 1.5 mol of HNO3 Determine the mass of each reactant needed to give the indicated amount of product. Be sure that the chemical equations are balanced. a. AgNO3(aq) + CaCl2(s) → AgCl(s) + Ca(NO3)2(aq); 1.25 mol of AgCl b. Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s) + H2O(l); 3.8 g of PbSO4 c. H3PO4(aq) + MgCO3(s) → Mg3(PO4)2(s) + CO2(g) + H2O(l); 6.41 g of Mg3(PO4)2 Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.) a. KClO3(s) KCl(s)+O2(g);2.14 g of KClO3 produces 0.67 g of O2 b. Cu(s) + H2SO4(aq) → CuSO4(aq) + SO2(g) + H2O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide c. AgC2H3O2(aq) + Na3PO4(aq) → Ag3PO4(s) + NaC2H3O2(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction? A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction? Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting. A + 3B → 2C Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol. a. Write a balanced chemical equation for this reaction. b. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction: NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq) a. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO2? b. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide. The reaction of propane gas (CH3CH2CH3) with chlorine gas (Cl2) produces two monochloride products: CH3CH2CH2Cl and CH3CHClCH3. The first is obtained in a 43% yield and the second in a 57% yield. a. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion? b. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane? c. Use the actual percent yield to calculate how many grams of each product would actually be obtained. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions: a. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized? b. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction? c. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation? Aniline (C6H5NH2) can be produced from chlorobenzene (C6H5Cl) via the following reaction: C6H5Cl(l) + 2NH3(g) → C6H5NH2(l) + NH4Cl(s) Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia. a. Which is the limiting reactant? b. Which reactant is present in excess? c. What is the theoretical yield of ammonium chloride in grams? d. If 4.78 g of NH4Cl was recovered, what was the percent yield? e. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following: a. 9.36 × 1024 formula units of NaCl b. 8.5 × 104 mol of Br2 c. 3.7 × 108 g of NaCl Conceptual Answers It is not a chemically valid practice because the law of conservation of mass applies to closed/isolated systems only, so in this case of an open system converting mass into energy and allowing it to escape the system wouldn’t be valid. No it is not correct because in order to examine the molecular weight you need to take log of both molecules to see what there true weight is. In this case, log of O2 = 10g/(32 g/mol) = 0.3125 g/mol while log of H2 = 10g/(2 g/mol) = 5 g/mol. With that when hydrogen and oxygen react against each other ((2 x (0.3125 g/L)) x (18 g/mol H2O)) = 11.25 g of water vapor When a reaction is stoichiometric it means that the none of the reactants remain after being introduced leaving no excess or total leaving quantity. From a reaction of S + O2 -> SO2 we’re able to determine that the limiting reactant would be sulfur because the amount of O2 in the air is higher than the amount of sulfur that is burned in the reaction. Yes it is possible, given the equation of: percent yield = (actual yield)/(Theoretical Yield) its entirely possible to have the percent yield become greater than the theoretical yield when the product of the reaction has impurities that increase the mass compared if the product was pure. Numerical Answers Write a balanced chemical equation for each reaction and then determine which reactant is in excess. a. Ba(s) + Br2(l) = BaBr2(aq) Reactant in excess: Br2 b. Br2(l) + 2KI(s) = 2KBr + I2 Reactant in Excess: Br2 c. Zn + H2SO4 = ZnSO4 + H2 Reactant in excess: H2SO4 d. Fe + 2 AgC2H3O2 = Fe(C2H3O2)2 + 2 Ag Reactant in excess: AgC2H3O2 e. 2 (NH4)3PO4 + 3 Ba(OH)2 = 6 NH4OH + Ba3(PO4)2 Reactant in excess: Ba(OH)2 The balanced chemical equation for this reaction is 2NH3 + 2O2 → N2O + 3H2O a. NH3 b. NH3 c. O2 d. NH3 a. Excess: HCl b. Excess: HCl c. Excess: HCl d. Excess: Zn 15. a. 150 g NaI and 35 g Cl2 b. 29 g NaCl and 25 g H2SO4 c. 140 g NO2 and 27 g H2O 16. a. 2 AgNO3(aq) + CaCl2(s) = 2 AgCl(s) + Ca(NO3)2(aq); Mass of Reactant: 212.34g AgNO3 and 69.4 g CaCl2 b. Pb(s) + PbO2(s) + 2 H2SO4(aq) = 2 PbSO4(s) + 2 H2O(l); Mass of Reactant: 1.3g Pb, 1.5g PbO2, 1.2g H2SO4 c. 2 H3PO4(aq) + 3 MgCO3(s) = Mg3(PO4)2(s) + 3 CO2(g) + 3 H2O(l); Mass of Reactant: 6.17 g of MgCO3, 4.78g of H3PO4 a. 80% b. 30% c. 35.7% 81% 45%. (x/molecular mass of B) x (2/3) x (molecular mass of C) a. CO + 2H2 → CH3OH b. 58.28% 2 NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq) a. 11.9 g Cl2 b. (0.7458g= Grams of ClO2 = (x/(90.44 g/mol)) x (2/2) x (67.45 g ClO2/ 1 mol ClO2 23. a. 2.24 g Cl2 b. 4.95 g c. 2.13 g CH3CH2CH2Cl plus 2.82 g CH3CHClCH3 a. 4.14mg Pa b. 82.1% c. 0.0162 g PaI5 a. chlorobenzene b. ammonia c. 8.74 g ammonium chloride. d. 55% e. a. 6619.19g Cl2 b. 6.77 x 106 g Cl2 c. 2.52 x 108 g Cl2 Conceptual Problems II Engineers use conservation of mass, called a “mass balance,” to determine the amount of product that can be obtained from a chemical reaction. Mass balance assumes that the total mass of reactants is equal to the total mass of products. Is this a chemically valid practice? Explain your answer. Given the equation is it correct to say that 10 g of hydrogen will react with 10 g of oxygen to produce 20 g of water vapor? What does it mean to say that a reaction is stoichiometric? When sulfur is burned in air to produce sulfur dioxide, what is the limiting reactant? Explain your answer. Is it possible for the percent yield to be greater than the theoretical yield? Justify your answer. Numerical Problems II Please be sure you are familiar with the topics discussed in Essential Skills 2 () before proceeding to the Numerical Problems. Determine the percent yield of each reaction. Be sure that the chemical equations are balanced. Assume that any reactants for which amounts are not given are in excess. (The symbol Δ indicates that the reactants are heated.) a. KClO3(s) KCl(s)+O2(g);2.14 g of KClO3 produces 0.67 g of O2 b. Cu(s) + H2SO4(aq) → CuSO4(aq) + SO2(g) + H2O(l); 4.00 g of copper gives 1.2 g of sulfur dioxide c. AgC2H3O2(aq) + Na3PO4(aq) → Ag3PO4(s) + NaC2H3O2(aq); 5.298 g of silver acetate produces 1.583 g of silver phosphate Each step of a four-step reaction has a yield of 95%. What is the percent yield for the overall reaction? A three-step reaction yields of 87% for the first step, 94% for the second, and 55% for the third. What is the percent yield of the overall reaction? Give a general expression relating the theoretical yield (in grams) of product that can be obtained from x grams of B, assuming neither A nor B is limiting. A + 3B → 2C Under certain conditions, the reaction of hydrogen with carbon monoxide can produce methanol. a. Write a balanced chemical equation for this reaction. b. Calculate the percent yield if exactly 200 g of methanol is produced from exactly 300 g of carbon monoxide. Chlorine dioxide is a bleaching agent used in the paper industry. It can be prepared by the following reaction: NaClO2(s) + Cl2(g) → ClO2(aq) + NaCl(aq) a. What mass of chlorine is needed for the complete reaction of 30.5 g of NaClO2? b. Give a general equation for the conversion of x grams of sodium chlorite to chlorine dioxide. The reaction of propane gas (CH3CH2CH3) with chlorine gas (Cl2) produces two monochloride products: CH3CH2CH2Cl and CH3CHClCH3. The first is obtained in a 43% yield and the second in a 57% yield. a. If you use 2.78 g of propane gas, how much chlorine gas would you need for the reaction to go to completion? b. How many grams of each product could theoretically be obtained from the reaction starting with 2.78 g of propane? c. Use the actual percent yield to calculate how many grams of each product would actually be obtained. Protactinium (Pa), a highly toxic metal, is one of the rarest and most expensive elements. The following reaction is one method for preparing protactinium metal under relatively extreme conditions: a. Given 15.8 mg of reactant, how many milligrams of protactinium could be synthesized? b. If 3.4 mg of Pa was obtained, what was the percent yield of this reaction? c. If you obtained 3.4 mg of Pa and the percent yield was 78.6%, how many grams of PaI5 were used in the preparation? Aniline (C6H5NH2) can be produced from chlorobenzene (C6H5Cl) via the following reaction: C6H5Cl(l) + 2NH3(g) → C6H5NH2(l) + NH4Cl(s) Assume that 20.0 g of chlorobenzene at 92% purity is mixed with 8.30 g of ammonia. a. Which is the limiting reactant? b. Which reactant is present in excess? c. What is the theoretical yield of ammonium chloride in grams? d. If 4.78 g of NH4Cl was recovered, what was the percent yield? e. Derive a general expression for the theoretical yield of ammonium chloride in terms of grams of chlorobenzene reactant, if ammonia is present in excess. A stoichiometric quantity of chlorine gas is added to an aqueous solution of NaBr to produce an aqueous solution of sodium chloride and liquid bromine. Write the chemical equation for this reaction. Then assume an 89% yield and calculate the mass of chlorine given the following: a. 9.36 × 1024 formula units of NaCl b. 8.5 × 104 mol of Br2 c. 3.7 × 108 g of NaCl Numerical Answers II 17. a. 80% b. 30% c. 35.7% 45%. 21. a. CO + 2H2 → CH3OH b. 58.28% 23. a. 2.24 g Cl2 b. 4.95 g c. 2.13 g CH3CH2CH2Cl plus 2.82 g CH3CHClCH3 25. a. chlorobenzene b. ammonia c. 8.74 g ammonium chloride. d. 55% e. 2.E: Atoms, Molecules, and Ions (Exercises) 4.E: Aqueous Reactions (Exercises)
17036	https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational	Proof that π is irrational - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Lambert's proof 2 Hermite's proof 3 Cartwright's proof 4 Niven's proof 5 Bourbaki's proof 6 Laczkovich's proof 7 See also 8 References Proof that π is irrational [x] 17 languages العربية Deutsch Español فارسی Français 한국어 Bahasa Indonesia Italiano עברית Magyar 日本語 ଓଡ଼ିଆ Português Русский Українська Tiếng Việt 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Part of a series of articles on the mathematical constant π 3.14159 26535 89793 23846 26433... Uses Area of a circle Circumference Use in other formulae Properties Irrationality Transcendence Value Less than 22/7 Approximations Milü Madhava's correction term Memorization People Archimedes Liu Hui Zu Chongzhi Aryabhata Madhava Jamshīd al-Kāshī Ludolph van Ceulen François Viète Seki Takakazu Takebe Kenko William Jones John Machin William Shanks Srinivasa Ramanujan John Wrench Chudnovsky brothers Yasumasa Kanada History Chronology A History of Pi In culture Indiana pi bill Pi Day Related topics Squaring the circle Basel problem Six nines in π Other topics related to π v t e In the 1760s, Johann Heinrich Lambert was the first to prove that the number π is irrational, meaning it cannot be expressed as a fractiona/b,{\displaystyle a/b,} where a{\displaystyle a} and b{\displaystyle b} are both integers. In the 19th century, Charles Hermite found a proof that requires no prerequisite knowledge beyond basic calculus. Three simplifications of Hermite's proof are due to Mary Cartwright, Ivan Niven, and Nicolas Bourbaki. Another proof, which is a simplification of Lambert's proof, is due to Miklós Laczkovich. Many of these are proofs by contradiction. In 1882, Ferdinand von Lindemann proved that π{\displaystyle \pi } is not just irrational, but transcendental as well. Lambert's proof [edit] Scan of formula on page 288 of Lambert's "Mémoires sur quelques propriétés remarquables des quantités transcendantes, circulaires et logarithmiques", Mémoires de l'Académie royale des sciences de Berlin (1768), 265–322 In 1761, Johann Heinrich Lambert proved that π{\displaystyle \pi } is irrational by first showing that this continued fraction expansion holds: tan⁡(x)=x 1−x 2 3−x 2 5−x 2 7−⋱.{\displaystyle \tan(x)={\cfrac {x}{1-{\cfrac {x^{2}}{3-{\cfrac {x^{2}}{5-{\cfrac {x^{2}}{7-{}\ddots }}}}}}}}.} Then Lambert proved that if x{\displaystyle x} is non-zero and rational, then this expression must be irrational. Since tan⁡π 4=1{\displaystyle \tan {\tfrac {\pi }{4}}=1}, it follows that π 4{\displaystyle {\tfrac {\pi }{4}}} is irrational, and thus π{\displaystyle \pi } is also irrational. A simplification of Lambert's proof is given below. Hermite's proof [edit] Written in 1873, this proof uses the characterization of π{\displaystyle \pi } as the smallest positive number whose half is a zero of the cosine function and it actually proves that π 2{\displaystyle \pi ^{2}} is irrational. As in many proofs of irrationality, it is a proof by contradiction. Consider the sequences of real functionsA n{\displaystyle A_{n}} and U n{\displaystyle U_{n}} for n∈N 0{\displaystyle n\in \mathbb {N} _{0}} defined by: A 0(x)=sin⁡(x),A n+1(x)=∫0 x y A n(y)d y U 0(x)=sin⁡(x)x,U n+1(x)=−U n′(x)x{\displaystyle {\begin{aligned}A_{0}(x)&=\sin(x),&&A_{n+1}(x)=\int {0}^{x}yA{n}(y)\,dy\[4pt]U_{0}(x)&={\frac {\sin(x)}{x}},&&U_{n+1}(x)=-{\frac {U_{n}'(x)}{x}}\end{aligned}}} Using induction we can prove that A n(x)=x 2 n+1(2 n+1)!!−x 2 n+3 2×(2 n+3)!!+x 2 n+5 2×4×(2 n+5)!!∓⋯U n(x)=1(2 n+1)!!−x 2 2×(2 n+3)!!+x 4 2×4×(2 n+5)!!∓⋯{\displaystyle {\begin{aligned}A_{n}(x)&={\frac {x^{2n+1}}{(2n+1)!!}}-{\frac {x^{2n+3}}{2\times (2n+3)!!}}+{\frac {x^{2n+5}}{2\times 4\times (2n+5)!!}}\mp \cdots \[4pt]U_{n}(x)&={\frac {1}{(2n+1)!!}}-{\frac {x^{2}}{2\times (2n+3)!!}}+{\frac {x^{4}}{2\times 4\times (2n+5)!!}}\mp \cdots \end{aligned}}} and therefore we have: U n(x)=A n(x)x 2 n+1.{\displaystyle U_{n}(x)={\frac {A_{n}(x)}{x^{2n+1}}}.\,} So A n+1(x)x 2 n+3=U n+1(x)=−U n′(x)x=−1 x d d x(A n(x)x 2 n+1)=−1 x(A n′(x)⋅x 2 n+1−(2 n+1)x 2 n A n(x)x 2(2 n+1))=(2 n+1)A n(x)−x A n′(x)x 2 n+3{\displaystyle {\begin{aligned}{\frac {A_{n+1}(x)}{x^{2n+3}}}&=U_{n+1}(x)=-{\frac {U_{n}'(x)}{x}}=-{\frac {1}{x}}{\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {A_{n}(x)}{x^{2n+1}}}\right)\[6pt]&=-{\frac {1}{x}}\left({\frac {A_{n}'(x)\cdot x^{2n+1}-(2n+1)x^{2n}A_{n}(x)}{x^{2(2n+1)}}}\right)\[6pt]&={\frac {(2n+1)A_{n}(x)-xA_{n}'(x)}{x^{2n+3}}}\end{aligned}}} which is equivalent to A n+1(x)=(2 n+1)A n(x)−x 2 A n−1(x).{\displaystyle A_{n+1}(x)=(2n+1)A_{n}(x)-x^{2}A_{n-1}(x).\,} Using the definition of the sequence and employing induction we can show that A n(x)=P n(x 2)sin⁡(x)+x Q n(x 2)cos⁡(x),{\displaystyle A_{n}(x)=P_{n}(x^{2})\sin(x)+xQ_{n}(x^{2})\cos(x),\,} where P n{\displaystyle P_{n}} and Q n{\displaystyle Q_{n}} are polynomial functions with integer coefficients and the degree of P n{\displaystyle P_{n}} is smaller than or equal to ⌊1 2 n⌋.{\displaystyle {\bigl \lfloor }{\tfrac {1}{2}}n{\bigr \rfloor }.} In particular, A n(1 2 π)=P n(1 4 π 2).{\displaystyle A_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=P_{n}{\bigl (}{\tfrac {1}{4}}\pi ^{2}{\bigr )}.} Hermite also gave a closed expression for the function A n,{\displaystyle A_{n},} namely A n(x)=x 2 n+1 2 n n!∫0 1(1−z 2)n cos⁡(x z)d z.{\displaystyle A_{n}(x)={\frac {x^{2n+1}}{2^{n}n!}}\int _{0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z.\,} He did not justify this assertion, but it can be proved easily. First of all, this assertion is equivalent to 1 2 n n!∫0 1(1−z 2)n cos⁡(x z)d z=A n(x)x 2 n+1=U n(x).{\displaystyle {\frac {1}{2^{n}n!}}\int {0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z={\frac {A{n}(x)}{x^{2n+1}}}=U_{n}(x).} Proceeding by induction, take n=0.{\displaystyle n=0.} ∫0 1 cos⁡(x z)d z=sin⁡(x)x=U 0(x){\displaystyle \int {0}^{1}\cos(xz)\,\mathrm {d} z={\frac {\sin(x)}{x}}=U{0}(x)} and, for the inductive step, consider any natural numbern.{\displaystyle n.} If 1 2 n n!∫0 1(1−z 2)n cos⁡(x z)d z=U n(x),{\displaystyle {\frac {1}{2^{n}n!}}\int {0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z=U{n}(x),} then, using integration by parts and Leibniz's rule, one gets 1 2 n+1(n+1)!∫0 1(1−z 2)n+1 cos⁡(x z)d z=1 2 n+1(n+1)!((1−z 2)n+1 sin⁡(x z)x\|z=0 z=1⏞=0+∫0 1 2(n+1)(1−z 2)n z sin⁡(x z)x d z)=1 x⋅1 2 n n!∫0 1(1−z 2)n z sin⁡(x z)d z=−1 x⋅d d x(1 2 n n!∫0 1(1−z 2)n cos⁡(x z)d z)=−U n′(x)x=U n+1(x).{\displaystyle {\begin{aligned}&{\frac {1}{2^{n+1}(n+1)!}}\int {0}^{1}\left(1-z^{2}\right)^{n+1}\cos(xz)\,\mathrm {d} z\&\qquad ={\frac {1}{2^{n+1}(n+1)!}}{\Biggl (}\,\overbrace {\left.(1-z^{2})^{n+1}{\frac {\sin(xz)}{x}}\right\|{z=0}^{z=1}} ^{=\,0}\ +\,\int {0}^{1}2(n+1)\left(1-z^{2}\right)^{n}z{\frac {\sin(xz)}{x}}\,\mathrm {d} z{\Biggr )}\[8pt]&\qquad ={\frac {1}{x}}\cdot {\frac {1}{2^{n}n!}}\int {0}^{1}\left(1-z^{2}\right)^{n}z\sin(xz)\,\mathrm {d} z\[8pt]&\qquad =-{\frac {1}{x}}\cdot {\frac {\mathrm {d} }{\mathrm {d} x}}\left({\frac {1}{2^{n}n!}}\int {0}^{1}(1-z^{2})^{n}\cos(xz)\,\mathrm {d} z\right)\[8pt]&\qquad =-{\frac {U{n}'(x)}{x}}\[4pt]&\qquad =U_{n+1}(x).\end{aligned}}} If 1 4 π 2=p/q,{\displaystyle {\tfrac {1}{4}}\pi ^{2}=p/q,} with p{\displaystyle p} and q{\displaystyle q} in N{\displaystyle \mathbb {N} }, then, since the coefficients of P n{\displaystyle P_{n}} are integers and its degree is smaller than or equal to ⌊1 2 n⌋,{\displaystyle {\bigl \lfloor }{\tfrac {1}{2}}n{\bigr \rfloor },}q⌊n/2⌋P n(1 4 π 2){\displaystyle q^{\lfloor n/2\rfloor }P_{n}{\bigl (}{\tfrac {1}{4}}\pi ^{2}{\bigr )}} is some integer N.{\displaystyle N.} In other words, N=q⌊n/2⌋A n(1 2 π)=q⌊n/2⌋1 2 n n!(p q)n+1 2∫0 1(1−z 2)n cos⁡(1 2 π z)d z.{\displaystyle N=q^{\lfloor n/2\rfloor }{A_{n}}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=q^{\lfloor n/2\rfloor }{\frac {1}{2^{n}n!}}\left({\dfrac {p}{q}}\right)^{n+{\frac {1}{2}}}\int _{0}^{1}(1-z^{2})^{n}\cos \left({\tfrac {1}{2}}\pi z\right)\,\mathrm {d} z.} But this number is clearly greater than 0.{\displaystyle 0.} On the other hand, the limit of this quantity as n{\displaystyle n} goes to infinity is zero, and so, if n{\displaystyle n} is large enough, N<1.{\displaystyle N<1.} Thereby, a contradiction is reached. Hermite did not present his proof as an end in itself but as an afterthought within his search for a proof of the transcendence of π.{\displaystyle \pi .} He discussed the recurrence relations to motivate and to obtain a convenient integral representation. Once this integral representation is obtained, there are various ways to present a succinct and self-contained proof starting from the integral (as in Cartwright's, Bourbaki's or Niven's presentations), which Hermite could easily see (as he did in his proof of the transcendence of e{\displaystyle e}). Moreover, Hermite's proof is closer to Lambert's proof than it seems. In fact, A n(x){\displaystyle A_{n}(x)} is the "residue" (or "remainder") of Lambert's continued fraction for tan⁡x.{\displaystyle \tan x.} Cartwright's proof [edit] Harold Jeffreys wrote that this proof was set as an example in an exam at Cambridge University in 1945 by Mary Cartwright, but that she had not traced its origin. It still remains on the 4th problem sheet today for the Analysis IA course at Cambridge University. Consider the integrals I n(x)=∫−1 1(1−z 2)n cos⁡(x z)d z,{\displaystyle I_{n}(x)=\int _{-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz,} where n{\displaystyle n} is a non-negative integer. Two integrations by parts give the recurrence relation x 2 I n(x)=2 n(2 n−1)I n−1(x)−4 n(n−1)I n−2(x).(n≥2){\displaystyle x^{2}I_{n}(x)=2n(2n-1)I_{n-1}(x)-4n(n-1)I_{n-2}(x).\qquad (n\geq 2)} If J n(x)=x 2 n+1 I n(x),{\displaystyle J_{n}(x)=x^{2n+1}I_{n}(x),} then this becomes J n(x)=2 n(2 n−1)J n−1(x)−4 n(n−1)x 2 J n−2(x).{\displaystyle J_{n}(x)=2n(2n-1)J_{n-1}(x)-4n(n-1)x^{2}J_{n-2}(x).} Furthermore, J 0(x)=2 sin⁡x{\displaystyle J_{0}(x)=2\sin x} and J 1(x)=−4 x cos⁡x+4 sin⁡x.{\displaystyle J_{1}(x)=-4x\cos x+4\sin x.} Hence for all n∈Z+,{\displaystyle n\in \mathbb {Z} _{+},} J n(x)=x 2 n+1 I n(x)=n!(P n(x)sin⁡(x)+Q n(x)cos⁡(x)),{\displaystyle J_{n}(x)=x^{2n+1}I_{n}(x)=n!{\bigl (}P_{n}(x)\sin(x)+Q_{n}(x)\cos(x){\bigr )},} where P n(x){\displaystyle P_{n}(x)} and Q n(x){\displaystyle Q_{n}(x)} are polynomials of degree ≤n,{\displaystyle \leq n,} and with integer coefficients (depending on n{\displaystyle n}). Take x=1 2 π,{\displaystyle x={\tfrac {1}{2}}\pi ,} and suppose if possible that 1 2 π=a/b{\displaystyle {\tfrac {1}{2}}\pi =a/b} where a{\displaystyle a} and b{\displaystyle b} are natural numbers (i.e., assume that π{\displaystyle \pi } is rational). Then a 2 n+1 n!I n(1 2 π)=P n(1 2 π)b 2 n+1.{\displaystyle {\frac {a^{2n+1}}{n!}}I_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}=P_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}b^{2n+1}.} The right side is an integer. But 0<I n(1 2 π)<2{\displaystyle 0<I_{n}{\bigl (}{\tfrac {1}{2}}\pi {\bigr )}<2} since the interval [−1,1]{\displaystyle [-1,1]} has length 2{\displaystyle 2} and the function being integrated takes only values between 0{\displaystyle 0} and 1.{\displaystyle 1.} On the other hand, a 2 n+1 n!→0 as n→∞.{\displaystyle {\frac {a^{2n+1}}{n!}}\to 0\quad {\text{ as }}n\to \infty .} Hence, for sufficiently large n{\displaystyle n} 0<a 2 n+1 I n(π 2)n!<1,{\displaystyle 0<{\frac {a^{2n+1}I_{n}\left({\frac {\pi }{2}}\right)}{n!}}<1,} that is, we could find an integer between 0{\displaystyle 0} and 1.{\displaystyle 1.} That is the contradiction that follows from the assumption that π{\displaystyle \pi } is rational. This proof is similar to Hermite's proof. Indeed, J n(x)=x 2 n+1∫−1 1(1−z 2)n cos⁡(x z)d z=2 x 2 n+1∫0 1(1−z 2)n cos⁡(x z)d z=2 n+1 n!A n(x).{\displaystyle {\begin{aligned}J_{n}(x)&=x^{2n+1}\int {-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz\[5pt]&=2x^{2n+1}\int {0}^{1}(1-z^{2})^{n}\cos(xz)\,dz\[5pt]&=2^{n+1}n!A_{n}(x).\end{aligned}}} However, it is clearly simpler. This is achieved by omitting the inductive definition of the functions A n{\displaystyle A_{n}} and taking as a starting point their expression as an integral. Niven's proof [edit] This proof uses the characterization of π{\displaystyle \pi } as the smallest positive zero of the sine function. Suppose that π{\displaystyle \pi } is rational, i.e. π=a/b{\displaystyle \pi =a/b} for some integers a{\displaystyle a} and b{\displaystyle b} which may be taken without loss of generality to both be positive. Given any positive integer n,{\displaystyle n,} we define the polynomial function: f(x)=x n(a−b x)n n!{\displaystyle f(x)={\frac {x^{n}(a-bx)^{n}}{n!}}} and, for each x∈R{\displaystyle x\in \mathbb {R} } let F(x)=f(x)−f″(x)+f(4)(x)+⋯+(−1)n f(2 n)(x).{\displaystyle F(x)=f(x)-f''(x)+f^{(4)}(x)+\cdots +(-1)^{n}f^{(2n)}(x).} Claim 1:F(0)+F(π){\displaystyle F(0)+F(\pi )} is an integer. Proof: Expanding f{\displaystyle f} as a sum of monomials, the coefficient of x k{\displaystyle x^{k}} is a number of the form c k/n!{\displaystyle c_{k}/n!} where c k{\displaystyle c_{k}} is an integer, which is 0{\displaystyle 0} if k<n.{\displaystyle k<n.} Therefore, f(k)(0){\displaystyle f^{(k)}(0)} is 0{\displaystyle 0} when k<n{\displaystyle k<n} and it is equal to (k!/n!)c k{\displaystyle (k!/n!)c_{k}} if n≤k≤2 n{\displaystyle n\leq k\leq 2n}; in each case, f(k)(0){\displaystyle f^{(k)}(0)} is an integer and therefore F(0){\displaystyle F(0)} is an integer. On the other hand, f(π−x)=f(x){\displaystyle f(\pi -x)=f(x)} and so (−1)k f(k)(π−x)=f(k)(x){\displaystyle (-1)^{k}f^{(k)}(\pi -x)=f^{(k)}(x)} for each non-negative integer k.{\displaystyle k.} In particular, (−1)k f(k)(π)=f(k)(0).{\displaystyle (-1)^{k}f^{(k)}(\pi )=f^{(k)}(0).} Therefore, f(k)(π){\displaystyle f^{(k)}(\pi )} is also an integer and so F(π){\displaystyle F(\pi )} is an integer (in fact, it is easy to see that F(π)=F(0){\displaystyle F(\pi )=F(0)}). Since F(0){\displaystyle F(0)} and F(π){\displaystyle F(\pi )} are integers, so is their sum. Claim 2: ∫0 π f(x)sin⁡(x)d x=F(0)+F(π){\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx=F(0)+F(\pi )} Proof: Since f(2 n+2){\displaystyle f^{(2n+2)}} is the zero polynomial, we have F″+F=f.{\displaystyle F''+F=f.} The derivatives of the sine and cosine function are given by sin'=cos and cos'=−sin. Hence the product rule implies (F′⋅sin⁡−F⋅cos⁡)′=f⋅sin{\displaystyle (F'\cdot \sin {}-F\cdot \cos {})'=f\cdot \sin } By the fundamental theorem of calculus ∫0 π f(x)sin⁡(x)d x=(F′(x)sin⁡x−F(x)cos⁡x)\|0 π.{\displaystyle \left.\int {0}^{\pi }f(x)\sin(x)\,dx={\bigl (}F'(x)\sin x-F(x)\cos x{\bigr )}\right\|{0}^{\pi }.} Since sin⁡0=sin⁡π=0{\displaystyle \sin 0=\sin \pi =0} and cos⁡0=−cos⁡π=1{\displaystyle \cos 0=-\cos \pi =1} (here we use the above-mentioned characterization of π{\displaystyle \pi } as a zero of the sine function), Claim 2 follows. Conclusion: Since f(x)>0{\displaystyle f(x)>0} and sin⁡x>0{\displaystyle \sin x>0} for 0<x<π{\displaystyle 0<x<\pi } (because π{\displaystyle \pi } is the smallest positive zero of the sine function), Claims 1 and 2 show that F(0)+F(π){\displaystyle F(0)+F(\pi )} is a positive integer. Since 0≤x(a−b x)≤π a{\displaystyle 0\leq x(a-bx)\leq \pi a} and 0≤sin⁡x≤1{\displaystyle 0\leq \sin x\leq 1} for 0≤x≤π,{\displaystyle 0\leq x\leq \pi ,} we have, by the original definition of f,{\displaystyle f,} ∫0 π f(x)sin⁡(x)d x≤π(π a)n n!{\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx\leq \pi {\frac {(\pi a)^{n}}{n!}}} which is smaller than 1{\displaystyle 1} for large n,{\displaystyle n,} hence F(0)+F(π)<1{\displaystyle F(0)+F(\pi )<1} for these n,{\displaystyle n,} by Claim 2. This is impossible for the positive integer F(0)+F(π).{\displaystyle F(0)+F(\pi ).} This shows that the original assumption that π{\displaystyle \pi } is rational leads to a contradiction, which concludes the proof. The above proof is a polished version, which is kept as simple as possible concerning the prerequisites, of an analysis of the formula ∫0 π f(x)sin⁡(x)d x=∑j=0 n(−1)j(f(2 j)(π)+f(2 j)(0))+(−1)n+1∫0 π f(2 n+2)(x)sin⁡(x)d x,{\displaystyle \int {0}^{\pi }f(x)\sin(x)\,dx=\sum {j=0}^{n}(-1)^{j}\left(f^{(2j)}(\pi )+f^{(2j)}(0)\right)+(-1)^{n+1}\int _{0}^{\pi }f^{(2n+2)}(x)\sin(x)\,dx,} which is obtained by 2 n+2{\displaystyle 2n+2}integrations by parts. Claim 2 essentially establishes this formula, where the use of F{\displaystyle F} hides the iterated integration by parts. The last integral vanishes because f(2 n+2){\displaystyle f^{(2n+2)}} is the zero polynomial. Claim 1 shows that the remaining sum is an integer. Niven's proof is closer to Cartwright's (and therefore Hermite's) proof than it appears at first sight. In fact, J n(x)=x 2 n+1∫−1 1(1−z 2)n cos⁡(x z)d z=∫−1 1(x 2−(x z)2)n x cos⁡(x z)d z.{\displaystyle {\begin{aligned}J_{n}(x)&=x^{2n+1}\int {-1}^{1}(1-z^{2})^{n}\cos(xz)\,dz\&=\int {-1}^{1}\left(x^{2}-(xz)^{2}\right)^{n}x\cos(xz)\,dz.\end{aligned}}} Therefore, the substitutionx z=y{\displaystyle xz=y} turns this integral into ∫−x x(x 2−y 2)n cos⁡(y)d y.{\displaystyle \int _{-x}^{x}(x^{2}-y^{2})^{n}\cos(y)\,dy.} In particular, J n(π 2)=∫−π/2 π/2(π 2 4−y 2)n cos⁡(y)d y=∫0 π(π 2 4−(y−π 2)2)n cos⁡(y−π 2)d y=∫0 π y n(π−y)n sin⁡(y)d y=n!b n∫0 π f(x)sin⁡(x)d x.{\displaystyle {\begin{aligned}J_{n}\left({\frac {\pi }{2}}\right)&=\int {-\pi /2}^{\pi /2}\left({\frac {\pi ^{2}}{4}}-y^{2}\right)^{n}\cos(y)\,dy\[5pt]&=\int {0}^{\pi }\left({\frac {\pi ^{2}}{4}}-\left(y-{\frac {\pi }{2}}\right)^{2}\right)^{n}\cos \left(y-{\frac {\pi }{2}}\right)\,dy\[5pt]&=\int {0}^{\pi }y^{n}(\pi -y)^{n}\sin(y)\,dy\[5pt]&={\frac {n!}{b^{n}}}\int {0}^{\pi }f(x)\sin(x)\,dx.\end{aligned}}} Another connection between the proofs lies in the fact that Hermite already mentions that if f{\displaystyle f} is a polynomial function and F=f−f(2)+f(4)∓⋯,{\displaystyle F=f-f^{(2)}+f^{(4)}\mp \cdots ,} then ∫f(x)sin⁡(x)d x=F′(x)sin⁡(x)−F(x)cos⁡(x)+C,{\displaystyle \int f(x)\sin(x)\,dx=F'(x)\sin(x)-F(x)\cos(x)+C,} from which it follows that ∫0 π f(x)sin⁡(x)d x=F(π)+F(0).{\displaystyle \int _{0}^{\pi }f(x)\sin(x)\,dx=F(\pi )+F(0).} Bourbaki's proof [edit] Bourbaki's proof is outlined as an exercise in his calculus treatise. For each natural number b and each non-negative integer n,{\displaystyle n,} define A n(b)=b n∫0 π x n(π−x)n n!sin⁡(x)d x.{\displaystyle A_{n}(b)=b^{n}\int {0}^{\pi }{\frac {x^{n}(\pi -x)^{n}}{n!}}\sin(x)\,dx.} Since A n(b){\displaystyle A{n}(b)} is the integral of a function defined on [0,π]{\displaystyle [0,\pi ]} that takes the value 0{\displaystyle 0} at 0{\displaystyle 0} and π{\displaystyle \pi } and which is greater than 0{\displaystyle 0} otherwise, A n(b)>0.{\displaystyle A_{n}(b)>0.} Besides, for each natural number b,{\displaystyle b,}A n(b)<1{\displaystyle A_{n}(b)<1} if n{\displaystyle n} is large enough, because x(π−x)≤(π 2)2{\displaystyle x(\pi -x)\leq \left({\frac {\pi }{2}}\right)^{2}} and therefore A n(b)≤π b n 1 n!(π 2)2 n=π(b π 2/4)n n!.{\displaystyle A_{n}(b)\leq \pi b^{n}{\frac {1}{n!}}\left({\frac {\pi }{2}}\right)^{2n}=\pi {\frac {(b\pi ^{2}/4)^{n}}{n!}}.} On the other hand, repeated integration by parts allows us to deduce that, if a{\displaystyle a} and b{\displaystyle b} are natural numbers such that π=a/b{\displaystyle \pi =a/b} and f{\displaystyle f} is the polynomial function from [0,π]{\displaystyle [0,\pi ]} into R{\displaystyle \mathbb {R} } defined by f(x)=x n(a−b x)n n!,{\displaystyle f(x)={\frac {x^{n}(a-bx)^{n}}{n!}},} then: A n(b)=∫0 π f(x)sin⁡(x)d x=[−f(x)cos⁡(x)]x=0 x=π−[−f′(x)sin⁡(x)]x=0 x=π+⋯±[f(2 n)(x)cos⁡(x)]x=0 x=π±∫0 π f(2 n+1)(x)cos⁡(x)d x.{\displaystyle {\begin{aligned}A_{n}(b)&=\int {0}^{\pi }f(x)\sin(x)\,dx\[5pt]&={\Big [}{-f(x)\cos(x)}{\Big ]}{x=0}^{x=\pi }\,-{\Big [}{-f'(x)\sin(x)}{\Big ]}{x=0}^{x=\pi }+\cdots \[5pt]&\ \qquad \pm {\Big [}f^{(2n)}(x)\cos(x){\Big ]}{x=0}^{x=\pi }\,\pm \int {0}^{\pi }f^{(2n+1)}(x)\cos(x)\,dx.\end{aligned}}} This last integral is 0,{\displaystyle 0,} since f(2 n+1){\displaystyle f^{(2n+1)}} is the null function (because f{\displaystyle f} is a polynomial function of degree 2 n{\displaystyle 2n}). Since each function f(k){\displaystyle f^{(k)}} (with 0≤k≤2 n{\displaystyle 0\leq k\leq 2n}) takes integer values at 0{\displaystyle 0} and π{\displaystyle \pi } and since the same thing happens with the sine and the cosine functions, this proves that A n(b){\displaystyle A{n}(b)} is an integer. Since it is also greater than 0,{\displaystyle 0,} it must be a natural number. But it was also proved that A n(b)<1{\displaystyle A_{n}(b)<1} if n{\displaystyle n} is large enough, thereby reaching a contradiction. This proof is quite close to Niven's proof, the main difference between them being the way of proving that the numbers A n(b){\displaystyle A_{n}(b)} are integers. Laczkovich's proof [edit] Miklós Laczkovich's proof is a simplification of Lambert's original proof. He considers the functions f k(x)=1−x 2 k+x 4 2!k(k+1)−x 6 3!k(k+1)(k+2)+⋯(k∉{0,−1,−2,…}).{\displaystyle f_{k}(x)=1-{\frac {x^{2}}{k}}+{\frac {x^{4}}{2!k(k+1)}}-{\frac {x^{6}}{3!k(k+1)(k+2)}}+\cdots \quad (k\notin {0,-1,-2,\ldots }).} These functions are clearly defined for any real number x.{\displaystyle x.} Additionally, f 1/2(x)=cos⁡(2 x),{\displaystyle f_{1/2}(x)=\cos(2x),}f 3/2(x)=sin⁡(2 x)2 x.{\displaystyle f_{3/2}(x)={\frac {\sin(2x)}{2x}}.} Claim 1: The following recurrence relation holds for any real number x{\displaystyle x}: x 2 k(k+1)f k+2(x)=f k+1(x)−f k(x).{\displaystyle {\frac {x^{2}}{k(k+1)}}f_{k+2}(x)=f_{k+1}(x)-f_{k}(x).} Proof: This can be proved by comparing the coefficients of the powers of x.{\displaystyle x.} Claim 2: For each real number x,{\displaystyle x,} lim k→+∞f k(x)=1.{\displaystyle \lim {k\to +\infty }f{k}(x)=1.} Proof: The sequence x 2 n/n!{\displaystyle x^{2n}/n!} is bounded (since it converges to 0{\displaystyle 0}) and if C{\displaystyle C} is an upper bound and if k>1,{\displaystyle k>1,} then \|f k(x)−1\|⩽∑n=1∞C k n=C 1/k 1−1/k=C k−1.{\displaystyle \left\|f_{k}(x)-1\right\|\leqslant \sum _{n=1}^{\infty }{\frac {C}{k^{n}}}=C{\frac {1/k}{1-1/k}}={\frac {C}{k-1}}.} Claim 3: If x≠0,{\displaystyle x\neq 0,}x 2{\displaystyle x^{2}} is rational, and k∈Q∖{0,−1,−2,…}{\displaystyle k\in \mathbb {Q} \smallsetminus {0,-1,-2,\ldots }} then f k(x)≠0 and f k+1(x)f k(x)∉Q.{\displaystyle f_{k}(x)\neq 0\quad {\text{ and }}\quad {\frac {f_{k+1}(x)}{f_{k}(x)}}\notin \mathbb {Q} .} Proof: Otherwise, there would be a number y≠0{\displaystyle y\neq 0} and integers a{\displaystyle a} and b{\displaystyle b} such that f k(x)=a y{\displaystyle f_{k}(x)=ay} and f k+1(x)=b y.{\displaystyle f_{k+1}(x)=by.} To see why, take y=f k+1(x),{\displaystyle y=f_{k+1}(x),}a=0,{\displaystyle a=0,} and b=1{\displaystyle b=1} if f k(x)=0{\displaystyle f_{k}(x)=0}; otherwise, choose integers a{\displaystyle a} and b{\displaystyle b} such that f k+1(x)/f k(x)=b/a{\displaystyle f_{k+1}(x)/f_{k}(x)=b/a} and define y=f k(x)/a=f k+1(x)/b.{\displaystyle y=f_{k}(x)/a=f_{k+1}(x)/b.} In each case, y{\displaystyle y} cannot be 0,{\displaystyle 0,} because otherwise it would follow from claim 1 that each f k+n(x){\displaystyle f_{k+n}(x)} (n∈N{\displaystyle n\in \mathbb {N} }) would be 0,{\displaystyle 0,} which would contradict claim 2. Now, take a natural number c{\displaystyle c} such that all three numbers b c/k,{\displaystyle bc/k,}c k/x 2,{\displaystyle ck/x^{2},} and c/x 2{\displaystyle c/x^{2}} are integers and consider the sequence g n={f k(x)n=0 c n k(k+1)⋯(k+n−1)f k+n(x)n≠0{\displaystyle g_{n}={\begin{cases}f_{k}(x)&n=0\{\dfrac {c^{n}}{k(k+1)\cdots (k+n-1)}}f_{k+n}(x)&n\neq 0\end{cases}}} Then g 0=f k(x)=a y∈Z y and g 1=c k f k+1(x)=b c k y∈Z y.{\displaystyle g_{0}=f_{k}(x)=ay\in \mathbb {Z} y\quad {\text{ and }}\quad g_{1}={\frac {c}{k}}f_{k+1}(x)={\frac {bc}{k}}y\in \mathbb {Z} y.} On the other hand, it follows from claim 1 that g n+2=c n+2 x 2 k(k+1)⋯(k+n−1)⋅x 2(k+n)(k+n+1)f k+n+2(x)=c n+2 x 2 k(k+1)⋯(k+n−1)f k+n+1(x)−c n+2 x 2 k(k+1)⋯(k+n−1)f k+n(x)=c(k+n)x 2 g n+1−c 2 x 2 g n=(c k x 2+c x 2 n)g n+1−c 2 x 2 g n,{\displaystyle {\begin{aligned}g_{n+2}&={\frac {c^{n+2}}{x^{2}k(k+1)\cdots (k+n-1)}}\cdot {\frac {x^{2}}{(k+n)(k+n+1)}}f_{k+n+2}(x)\[5pt]&={\frac {c^{n+2}}{x^{2}k(k+1)\cdots (k+n-1)}}f_{k+n+1}(x)-{\frac {c^{n+2}}{x^{2}k(k+1)\cdots (k+n-1)}}f_{k+n}(x)\[5pt]&={\frac {c(k+n)}{x^{2}}}g_{n+1}-{\frac {c^{2}}{x^{2}}}g_{n}\[5pt]&=\left({\frac {ck}{x^{2}}}+{\frac {c}{x^{2}}}n\right)g_{n+1}-{\frac {c^{2}}{x^{2}}}g_{n},\end{aligned}}} which is a linear combination of g n+1{\displaystyle g_{n+1}} and g n{\displaystyle g_{n}} with integer coefficients. Therefore, each g n{\displaystyle g_{n}} is an integer multiple of y.{\displaystyle y.} Besides, it follows from claim 2 that each g n{\displaystyle g_{n}} is greater than 0{\displaystyle 0} (and therefore that g n≥\|y\|{\displaystyle g_{n}\geq \|y\|}) if n{\displaystyle n} is large enough and that the sequence of all g n{\displaystyle g_{n}} converges to 0.{\displaystyle 0.} But a sequence of numbers greater than or equal to \|y\|{\displaystyle \|y\|} cannot converge to 0.{\displaystyle 0.} Since f 1/2(1 4 π)=cos⁡1 2 π=0,{\displaystyle f_{1/2}({\tfrac {1}{4}}\pi )=\cos {\tfrac {1}{2}}\pi =0,} it follows from claim 3 that 1 16 π 2{\displaystyle {\tfrac {1}{16}}\pi ^{2}} is irrational and therefore that π{\displaystyle \pi } is irrational. On the other hand, since tan⁡x=sin⁡x cos⁡x=x f 3/2(x/2)f 1/2(x/2),{\displaystyle \tan x={\frac {\sin x}{\cos x}}=x{\frac {f_{3/2}(x/2)}{f_{1/2}(x/2)}},} another consequence of Claim 3 is that, if x∈Q∖{0},{\displaystyle x\in \mathbb {Q} \smallsetminus {0},} then tan⁡x{\displaystyle \tan x} is irrational. Laczkovich's proof is about the hypergeometric function. In fact, f k(x)=0 F 1(k−x 2){\displaystyle f_{k}(x)={}{0}F{1}(k-x^{2})}, and Gauss found a continued fraction expansion of the hypergeometric function using its functional equation. This allowed Laczkovich to find a new and simpler proof of the fact that the tangent function has the continued fraction expansion that Lambert had discovered. Laczkovich's result can also be expressed in Bessel functions of the first kind J ν(x){\displaystyle J_{\nu }(x)}. In fact, Γ(k)J k−1(2 x)=x k−1 f k(x){\displaystyle \Gamma (k)J_{k-1}(2x)=x^{k-1}f_{k}(x)} (where Γ{\displaystyle \Gamma } is the gamma function). So Laczkovich's result is equivalent to: If x≠0,{\displaystyle x\neq 0,}x 2{\displaystyle x^{2}} is rational, and k∈Q∖{0,−1,−2,…}{\displaystyle k\in \mathbb {Q} \smallsetminus {0,-1,-2,\ldots }} then x J k(x)J k−1(x)∉Q.{\displaystyle {\frac {xJ_{k}(x)}{J_{k-1}(x)}}\notin \mathbb {Q} .} See also [edit] Mathematics portal Proof that e is irrational Proof that π is transcendental References [edit] ^Lindemann, Ferdinand von (2004) , "Ueber die Zahl π", in Berggren, Lennart; Borwein, Jonathan M.; Borwein, Peter B. (eds.), Pi, a source book (3rd ed.), New York: Springer-Verlag, pp.194–225, ISBN0-387-20571-3. ^Lambert, Johann Heinrich (2004) , "Mémoire sur quelques propriétés remarquables des quantités transcendantes circulaires et logarithmiques", in Berggren, Lennart; Borwein, Jonathan M.; Borwein, Peter B. (eds.), Pi, a source book (3rd ed.), New York: Springer-Verlag, pp.129–140, ISBN0-387-20571-3. ^ Jump up to: abHermite, Charles (1873). "Extrait d'une lettre de Monsieur Ch. Hermite à Monsieur Paul Gordan". Journal für die reine und angewandte Mathematik (in French). 76: 303–311. ^Hermite, Charles (1873). "Extrait d'une lettre de Mr. Ch. Hermite à Mr. Carl Borchardt". Journal für die reine und angewandte Mathematik (in French). 76: 342–344. ^Hermite, Charles (1912) . "Sur la fonction exponentielle". In Picard, Émile (ed.). Œuvres de Charles Hermite (in French). Vol.III. Gauthier-Villars. pp.150–181. ^ Jump up to: abZhou, Li (2011). "Irrationality proofs à la Hermite". The Mathematical Gazette. 95 (534): 407–413. arXiv:0911.1929. doi:10.1017/S0025557200003491. S2CID115175505. ^Jeffreys, Harold (1973), Scientific Inference (3rd ed.), Cambridge University Press, p.268, ISBN0-521-08446-6 ^"Department of Pure Mathematics and Mathematical Statistics". www.dpmms.cam.ac.uk. Retrieved 2022-04-19. ^Niven, Ivan (1947), "A simple proof that π is irrational"(PDF), Bulletin of the American Mathematical Society, vol.53, no.6, p.509, doi:10.1090/s0002-9904-1947-08821-2 ^Bourbaki, Nicolas (1949), Fonctions d'une variable réelle, chap. I–II–III, Actualités Scientifiques et Industrielles (in French), vol.1074, Hermann, pp.137–138 ^Laczkovich, Miklós (1997), "On Lambert's proof of the irrationality of π", American Mathematical Monthly, vol.104, no.5, pp.439–443, doi:10.2307/2974737, JSTOR2974737 ^Gauss, Carl Friedrich (1811–1813), "Disquisitiones generales circa seriem infinitam", Commentationes Societatis Regiae Scientiarum Gottingensis Recentiores (in Latin), 2 Retrieved from " Categories: Pi Article proofs Irrational numbers Hidden categories: CS1 French-language sources (fr) CS1 Latin-language sources (la) Articles with short description Short description is different from Wikidata This page was last edited on 3 August 2025, at 19:07(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Proof that π is irrational 17 languagesAdd topic
17037	https://www.scaler.com/topics/nrow-in-r/	nrow() in R - Scaler Topics Experience Academy Data Science Neovarsity Topics Explore New Skill Test Courses Free Masterclass Search for Articles, Topics Experience Experience Scaler nrow() in R nrow() in R nrow() in R By Swarnabha Sinha 3 mins read Last updated: 17 Oct 2023 143 views Topics Covered Overview When it comes to data manipulation and analysis in R, having a clear understanding of various functions is essential. One such function that frequently comes into play is nrow() in R. This function allows you to efficiently count the number of rows in a DataFrame, providing a quick way to assess the size of your data and make informed decisions based on its dimensions. In this article, we’ll dive into the details of the nrow() function in R, exploring its syntax, parameters, return value, and providing several examples to illustrate its usage. nrow() Function in R When you're working with data in R, knowing the size of your dataset is a crucial piece of information. This is where the nrow() function comes into play. The nrow() in R is a handy tool that allows you to count the number of rows in a DataFrame. In simpler terms, a DataFrame is like a table containing your data, where each row corresponds to a unique observation or data point. Sometimes, you might have thousands or even millions of rows in your DataFrame, making it essential to quickly determine the count of rows. This is where the nrow() function proves its utility. To use the nrow() in R function, you simply provide it with the DataFrame you want to examine. The function then performs its magic and tells you exactly how many rows are present in that DataFrame. This information can be immensely valuable, helping you understand the scale of your data and make informed decisions about how to analyze or process it further. Syntax The syntax of the nrow() function is straightforward: Here, df represents the DataFrame for which you want to calculate the number of rows. Return Value The primary return value of the nrow() function is an integer representing the count of rows in the provided DataFrame. This count reflects the total number of observations present in the DataFrame. Examples Let's explore a variety of examples to understand how the nrow() in R can be utilized effectively. Counting Number of Rows in a DataFrame Output: In this example, we create a simple DataFrame named data with two columns: "ID" and "Name". Each row represents a person's ID and name. We then use the nrow() function to count the number of rows in the DataFrame, which is equivalent to the number of individuals in our dataset. The result is printed, and in this case, the output is 10, indicating there are 10 rows in the DataFrame. Using nrow in R with Condition Output: In this example, we first create a sample DataFrame named data with columns "ID" and "Name". We then add a new "Gender" column to the DataFrame, indicating the gender of each individual. Afterward, we use the nrow() function along with a condition to count the number of rows where the gender is "Female". Using nrow when Value is NA or NULL Output: In this example, we create a sample DataFrame named data with columns ID and Name. We then introduce NA values to the Age column using the NA keyword. After that, we use the nrow() function along with the !is.na() condition to count the number of rows with non-NA age values (i.e., excluding rows with NA values in the Age column). For Loop Using nrow in R Output: In this example, we create a list of DataFrames named data_frames. The list contains the original data DataFrame, a filtered DataFrame with only males, and another filtered DataFrame with ages greater than 25. We then use a for loop to iterate through each DataFrame in the list. For each DataFrame, we use the nrow() function to count the number of rows and print a message along with the row count. Conclusion The nrow() function in R is a powerful tool for accurately counting the number of rows in a DataFrame, providing a quick overview of data scale. With its straightforward syntax—simply providing the DataFrame as input—it is user-friendly for data analysts and programmers alike. One of its key strengths lies in facilitating conditional analysis, allowing users to count rows meeting specific criteria, thus enhancing data segmentation. Moreover, the function's ability to handle NA values adds a layer of flexibility when working with datasets containing missing information. Finally, the nrow() function's versatility extends to efficiently looping through multiple DataFrames, making it an excellent choice for batch row counting tasks. Good job on finishing the article! Its time to level up with a Challenge CHALLENGE Master the concepts you discovered from this article in 3 Questions! Boost your memory and retention Take on challenges and level up your skills Showcase your achievements and earn exciting rewards Start Challenge! Over 10k+ learners have completed this challenge and improved their skills Got suggestions? We would love to hear your feedback. Your feedback is important to help us improve CloseSubmit A Free learning platform made with by Explore Scaler Academy Data Science & ML Neovarsity Explore Topics Free Online Courses Challenges Contest Topics Articles Events Resources About Us Blog Careers Review Download the app! Get all scaler resources under one roof! 4.4 1.71 K Reviews 100K+ Downloads Popular Free Certification Courses Java Course for Beginners C++ Course with Certificate Python Course for Beginners Javascript Free Course for Beginners Data Science Course for Beginners DBMS Course Python and SQL for Data Science Course DSA Problem Solving for Interviews Instagram System Design Course Dynamic Programming Course All Free Online Courses Popular Tutorials Python Tutorial Java Tutorial DBMS Tutorial Javascript Tutorial C++ Tutorial SQL Tutorial Software Engineering Tutorial Data Science Tutorial Pandas Tutorial Deep Learning Tutorial All Tutorials Compilers Python Compiler Java Compiler Javascript Compiler C Compiler C++ Compiler Tools Json Validator SQL Formatter XML Formatter CSS Formatter JavaScript Formatter Copyright 2025 InterviewBit Technologies Pvt. Ltd. All Rights Reserved. Privacy PolicyTerms of UseContact Us
17038	https://www.schoenresearch.com/wp-content/uploads/2021/05/Equals-symbol_Dimensions-Spring-2021.pdf	Farfan, Guillermo, and Robert C. Schoen Elementary Students’ Understanding of the Equals Symbol: Do Florida Students Outperform Their Peers? Dimensions in Mathematics SPRING 2021, 41 (1), p. 27 - 38 27 SPRING 2021 Elementary Students’ Understanding of the Equals Symbol: Do Florida Students Outperform Their Peers? Guillermo Farfan and Robert C. Schoen If you were to ask a person whether “1 = 1” is a true statement, many of us would expect the answer to be “yes.” What may seem like a straightforward answer to us as mathematically literate adults, however, might not be so for children. Mathematics education and educational psychology journals, in fact, often signal a recurring concern that children do not understand the intended meaning of the equals symbol (=) in mathematics. Scholars report that children understand it to be a “do something” symbol rather than a symbol that indicates a bidirectional equivalence relation (Boggs et al., 2018; Hornburg et al., 2018). For example, elementary students often say that the unknown value in Equation 1—an item that is frequently used by researchers—is 12. (1) 8 + 4 =  + 5 That is, children commonly add the numbers on the left side of the equals symbol and write their sum (i.e., 12) in the box. Students also frequently determine the sum of all the given numbers and write 17 in the box (Falkner et al., 1999). Many studies conducted in the years since Falkner et al.’s (1999) findings continue to paint a grim picture of elementary students’ understanding of =. Scholars report low levels of understanding overall and limited increases in understanding as students enroll in higher grade levels—sometimes even reporting decreases in understanding. In our own work, we have encountered many elementary school children whose understanding does not conform to this picture, an experience that led us to wonder whether some of the published interpretations and conclusions have been limited by a reliance on small or biased samples. Here, to overcome some of these limitations, we use two large-scale sets of data with items intended to assess elementary students’ understanding of =. We seek to answer three questions: (1) What percentage of Florida elementary students (grades K–5) respond correctly to test items designed to assess their mathematical understanding of =? (2) Do Florida elementary students in higher grades perform better than those in lower grades students on questions designed to assess their mathematical understanding of =? (3) How does the performance of Florida elementary students compare with that of their peers elsewhere, as represented in published studies? In answering these questions, we refrain from formal hypothesis testing and limit our analysis to comparing the descriptive statistics provided by our two sets of data with some of the literature on this topic. For ease of comparison, we use students’ responses to two equals-symbol items that resemble Equation 1, which are part of a category we call Operation on Both Sides (OBS) items (Schoen et al., 2016a, 2016b). Dimensions in Mathematics 28 The Equals Symbol in Elementary Mathematics In elementary arithmetic, the equals symbol plays a fundamental role in students’ development of number sense, as seen in addition and subtraction problems, decomposition of two-digit numbers, and comparison between numbers (Bennett et al., 2016; Musser et al., 2014). In all these instances, the equals symbol is meant to denote a relationship where two mathematical expressions—the expressions on the left- and right-hand sides of the equals symbol—have the same value or are “equal to” each other. This definition is known in the literature as “relational understanding” of = (McNeil & Alibali, 2005; Molina & Ambrose, 2008), and it is seen as facilitating the transition from arithmetic to middle-school and high-school algebra (Bush & Karp, 2013; Knuth et al., 2006). Recent school mathematics standards also underscore the importance of a relational understanding of =. For example, the Common Core State Standards for Mathematics (CCSSM) explicitly mention the equals symbol in two first-grade standards (CCSSM 1.OA.7; 1.OA.8) and allude to it once in the Standards for Mathematical Practice (National Governors Association, 2010). The Mathematics Florida Standards (MAFS), which were adopted by the state of Florida in early 2014, were very similar to the CCSSM, but differed in some key ways. The MAFS included the same standards as the CCSSM in first grade for the equals symbol, but they added standards to continue to address student understanding of the equals symbol in grades 2 (MAFS.2.OA.1.a) and 4 (MAFS.4.OA.1.a, MAFS.4.OA.1.b). Signaling the importance of student understanding of the mathematical meaning of this symbol, the recently adopted Benchmarks for Excellent Student Thinking (BEST) mathematics standards mention the equals symbol 16 times, starting in grade K and continuing through grade 7 (Florida Department of Education, 2020). The CCSSM, MAFS, and BEST standards all define = as a relational symbol. As mentioned earlier, however, reports abound that many students in the United States read = not as a relational symbol but rather as marking an operation that must be performed. The frequently cited study by Falkner et al. (1999), for example, showed that fewer than 10% of first and second graders gave a mathematically correct response to the item “8 + 4 =  + 5.” More concerningly, their data indicated that the percentage of children who responded correctly to this item in fourth (n = 57) and fifth grades (n = 42) was lower than that in third grade (n = 208). Subsequent studies continued to reinforce the idea that children entering middle school interpret = operationally. In Table 1, we list eight such studies from outside Florida that made use OBS items. As shown in Table 1, the mean percentage of correct responses initially increased from 15% in first grade to 24% in second grade and then decreased to 19% in third grade. Only in fifth grade (62%) did more than half of the students answer correctly. Table 1 reveals gaps in the literature, however; second grade is represented in seven out of the eight studies listed, but no other grade level appears more than three times. Although this list is not exhaustive, the studies on it were chosen because they reported the percentage of correct responses by grade level and used the OBS item “8 + 4 =  + 5” or a similar item to assess children’s understanding of =. 29 SPRING 2021 Table 1 Percentages (and Sample Sizes) of Students Who Responded Correctly to Equals-Symbol Items by Grade Level in Studies Conducted Outside Florida Grade level Study 1 2 3 4 5 N Falkner et al. (1999)a† 0 (42) 6 (174) 10 (208) 7 (57) 7 (42) 668 Stephens et al. (2013)‡ 2 (104) 24 (108) 56 (78) 290 Bennett (2015)‡ 15 (213) 35 (233) 46 (267) 68 (469) 1,182 Powell et al. (2016)b† 16 (805) 27 (489) 1,294 Johannes et al. (2017)c‡ 50 (49) 49 Matthews & Fuchs (2018)a‡ 33 (153) 191 Johannes & Davenport (2019)b‡ 28 (406) 406 Chow & Wehby (2019)b‡ 24 (74) 74 Mean 15 (847) 24 (1,558) 19 (545) 35 (432) 62 (589) Note. Results for Operations on Both Sides (OBS) items (e.g., 8 + 4 =  + 5), rounded to nearest percent. Sample sizes shown in parentheses. aResults for other grade levels are not included in the table. bPercentage available for Spring 2013 and Fall 2012 only. cRandomized controlled trial. Percentage at pretest for control group only. †CCSSM not implemented at the time of data collection. ‡CCSSM partially or fully implemented at the time of data collection. Background Here, we report findings from the administration of the Mathematics Performance and Cognition Interview (MPAC; Schoen et al., 2016a, 2016b), and the Elementary Mathematics Student Assessment (EMSA; Schoen et al., 2017, 2018a, 2021). These instruments were administered as part of the evaluation of a program called Cognitively Guided Instruction (CGI) Cognitively Guided Instruction Cognitively Guided Instruction is a professional-development program for mathematics teachers that focuses teachers’ attention on their students’ mathematical thinking and encourages Dimensions in Mathematics 30 them to use this knowledge to drive their instructional decisions (Carpenter et al., 2015). Research shows that CGI can improve students’ mathematics achievement as well as teachers’ mathematics knowledge and instructional practice (Jacobs et al., 2007; Schoen et al., 2018b). Teachers involved in our study were randomly assigned either to receive CGI training (typically eight days of professional development per school year) or to serve as the waitlist control group. We report data from the use of selected OBS items for Florida students in the classrooms of teachers who had participated in the CGI program, but we separate them from those whose teachers had not yet participated in the program. Here, we refer to the former as the Florida CGI group and the latter as the Florida BAU group, because the latter represents “business-as-usual” in Florida. We also note that CGI-trained teachers in our data varied in amount of participation in the program. Data Sources Mathematics Performance and Cognition Interview (MPAC) The MPAC consists of a series of items designed to measure the mathematical thinking and achievement of first- and second-grade students with a focus on number, operations, and equality. MPAC is administered in a one-on-one interview format, where the interviewer poses mathematics problems, observes students solving the problems, and asks students to describe their thought processes when the key details of those processes are not observable. The MPAC was administered to a diverse sample of more than 1,400 students in 22 schools located in two Florida school districts during spring 2014 and spring 2015 (Schoen et al., 2016a, 2016b). Elementary Mathematics Student Assessment (EMSA) The EMSA is designed to serve as a mathematics achievement test administered in paper-and-pencil format to elementary students. The Florida data used in the current study were collected from tests administered to grades K–5 students near the end of the school year between spring 2016 and spring 2019. EMSA tests are designed to align with the core content domains described in the CCSSM and the MAFS (Schoen et al., 2021). EMSA test forms include items designed to measure student understanding of =. Equals-Symbol Items We chose two OBS items (i.e., 6 + 3 =  + 4, 5 + 3 =  + 4) from the MPAC and EMSA tests because they are the most similar to those that have been reported in the prior literature. Table 2 shows the total number of students (i.e., Florida BAU and Florida CGI together) and number of schools represented in the data. We note that we have a substantially larger sample size than what was reported for the studies outside Florida (Table 1) at most grade levels. We also provide results for students in Kindergarten, which was not reported in those studies. 31 SPRING 2021 Table 2 Total Number of Students Represented in the MPAC (Spring 2014–Spring 2015) and EMSA (Spring 2016–Spring 2019) Data by Grade Grade level Data source K 1 2 3 4 5 Total Schools Spring 2014 MPAC 332 278 610 22 Spring 2015 MPAC 442 420 862 22 Spring 2016 EMSA 950 1,821 1,764 1,057 744 953 7,289 66 Spring 2017 EMSA 610 1,077 1,021 2,708 77 Spring 2018 EMSA 109 55 37 46 247 3 Spring 2019 EMSA 638 655 690 19 2,002 29 Total 1,669 4,365 4,175 1,793 763 953 13,718 Data Analysis To determine the percentage of Florida students answering OBS items correctly, we first disaggregated our data by group (Florida BAU and Florida CGI), then calculated the percentage of correct responses to the two OBS items by grade level for each MPAC and EMSA test available. We then computed the global mean percentage (Mean) for each grade by adding the number of students who responded correctly to the two items across tests divided by the total number of students. This procedure was the same one used to calculate the global mean percentage of correct responses by grade level in Table 1. We also computed standard errors (SE) and 95% confidence intervals (CI) for our Florida BAU and Florida CGI group data and the data from outside Florida (Table 4). Results In our Florida BAU group sample, the average percentage of correct responses increased monotonically with grade level. The global mean percentage of correct responses in the early grades (K–2) was 10 percentage points higher than that in studies outside Florida (Table 1), starting with Kindergarten at 14%, followed by first grade at 25% and second grade at 34% (Table 3). Unlike the data in Table 1, however, average percentage of correct responses on these items did not decline in third grade but continued to increase, reaching 73% by fifth grade. We note an analogous trend across grade levels in the average of correct responses to OBS items for the Florida CGI group. The global mean percentage of correct responses in the early grades (K-2) was approximately 20 percentage points higher than what was reported in Table 1. The mean percentage in Kindergarten was 17%, followed by 35% in first grade and 47% in second grade. Once again, unlike the data in Table 1, average performance on OBS items did not decline in third grade but continued upward, reaching 75% by fifth grade (Table 3). Finally, we plotted a line of best fit using least-squares regression to aid visualization of the normative trends across grade levels and comparison of the mean performance of the three groups of students in Figure 1. In both cases, Florida elementary students performed better Dimensions in Mathematics 32 overall on equals-symbol items than did the groups for whom data are reported in Table 1, and average percentage of correct responses on these items appears strongly correlated with grade level for both the Florida BAU group and the Florida CGI group. Table 3 Percentage of Students Who Responded Correctly to Equals-Symbol Items by Grade Level for Florida BAU, Florida CGI, and Groups Outside Florida Grade level Data Source K 1 2 3 4 5 N Florida BAU Group Spring 2014 MPAC 9 15 334 Spring 2015 MPAC 13 15 503 Spring 2016 EMSA 10 34 46 62 71 73 3,036 Spring 2017 EMSA 19 66 100 Spring 2018 EMSA 23 40 48 83 247 Spring 2019 EMSA 22 34 47 85 2,002 Mean 14 25 34 55 72 73 Florida CGI Group Spring 2014 MPAC 21 15 276 Spring 2015 MPAC 21 15 359 Spring 2016 EMSA 14 39 46 63 67 75 4,253 Spring 2017 EMSA 21 36 66 2,608 Mean 17 35 47 63 67 75 Groups Outside Florida Falkner et al. (1999)a ¯ 0 6 10 7 7 668 Stephens et al. (2013) ¯ 2 24 56 290 Bennett (2015) ¯ 15 35 46 68 1,182 Powell et al. (2016)b ¯ 16 27 1,294 Johannes et al. (2017)c ¯ 50 49 Matthews & Fuchs (2018)a ¯ 33 191 Johannes & Davenport (2019)b ¯ 28 406 Chow & Wehby (2019)b ¯ 24 74 Mean 15 24 19 35 62 Note. Results for OBS items, rounded to nearest percent. No CGI-group data available for EMSA Spring 2018 and EMSA Spring 2019. aResults for other grade levels are not included in the table. bPercentage available for Spring 2013 and Fall 2012 only. cRandomized controlled trial. Percentage at pretest for control group only. Table 4 Standard Errors and 95% Confidence Intervals for Outside-Florida, Florida, and CGI Data by Grade Variable Outside Florida Florida group CGI group Grade K 1 2 3 4 5 K 1 2 3 4 5 K 1 2 3 4 5 n ¯ 847 1558 545 432 589 355 2012 1663 1277 418 497 1314 2353 2512 516 345 456 Mean ¯ .152 .244 .192 .354 .621 .141 .249 .344 .546 .715 .731 .168 .354 .469 .628 .672 .749 SE ¯ .012 .011 .017 .023 .020 .018 .010 .012 .014 .022 .020 .010 .010 .010 .021 .025 .020 Lower 95% CI ¯ .128 .223 .159 .308 .581 .105 .230 .321 .519 .671 .692 .148 .335 .450 .586 .622 .709 Upper 95% CI ¯ .176 .266 .225 .399 .660 .177 .268 .367 .574 .758 .770 .189 .374 .489 .670 .722 .789 Note. Results for OBS items. We converted the global mean percentage of correct responses (Mean) to a sample proportion (p̂) to calculate standard errors (e.g., 15.2% = .152). n = total number of students. SE = standard error: √( 𝑝̂(1−𝑝̂) 𝑁 ). CI = confidence interval: Mean ± (1.96 SE). Dimensions in Mathematics 34 Figure 1 Scatterplots for Average Percentage of Correct Responses to Equals-Symbol Items by Grade Level (K–5) for Outside-Florida, Florida BAU, and Florida CGI Data Note. Results for Operations on Both Sides (OBS) items (e.g., 8 + 4 =  + 5). Discussion Together, these results show that elementary students in Florida tend to perform better, on average, on items intended to assess their understanding of = than do their counterparts for whom results have been reported elsewhere in the literature. In our sets of data, the global mean percentage of correct responses in the early grades for Florida students is 10 to 20 percentage points higher than those in studies outside Florida. By third grade, more than half of Florida students in our sample responded correctly to the OBS items—and the CGI group performed even higher. We think it noteworthy that this higher level of performance appears in both the BAU and CGI groups, suggesting that these results are independent of teachers' CGI training. In addition, our findings clearly suggest that elementary students in higher grades outperform those in lower grades on the equals-symbol items—a phenomenon observed in all three groups of data. This results differs from the flatter (or up-and-down) trend shown in Table 1, though we cannot rule out this trend's being the result of the paucity of data available on third grade. 0 0.2 0.4 0.6 0.8 K 1 2 3 4 5 Grade Level Average Percent of Correct Responses Florida CGI Florida BAU Outside Florida 35 SPRING 2021 We suggest two plausible explanations of the differences between students’ performance in our data and those from outside Florida listed in Table 1. The first is the possibility of publication bias—a well-known phenomenon in in all fields of research (Hopewell et al., 2009; Kühberger et al., 2014). In this case, publication bias could make it more likely for scholars to report results that are shocking or that confirm an ongoing narrative in the corpus of literature regarding elementary students’ failure to understand the mathematical meaning of the equals symbol. A second explanation could be the influence of the CCSSM and MAFS, which specifically address understanding of =. Some states, including Florida, have a state-level process of review of instructional materials and their alignment with the curriculum standards. School districts in Florida are required to spend at least 50% of their budget for instructional materials on resources that are listed as approved. This process provides some amount of assurance that textbooks align with the adopted curriculum standards. The specific references to student understanding of the equals symbol in grade 1 in the CCSSM and then in grades 1, 2, and 4 in the MAFS—and the subsequent updates to curriculum materials that followed—led to positive effects on student performance. Limitations We believe one strength of our set of data is its size (Table 2), but as is often the case in scientific literature, our data do not represent a random sample of the general population of elementary students, so we cannot claim that our results generalize to the Florida student population. In addition, we use student responses to OBS items on the EMSA to infer comprehension. We acknowledge that this inference has limitations, and these are also present in previously published findings. Because the MPAC data were obtained in an interview setting (which allows for student feedback), we have more confidence in inferences drawn from them than in those from the EMSA data or the data available from previously published studies that we used in the secondary analysis. We note that student performance on the MPAC does appear to be lower than performance on the EMSA. Conclusion We entirely agree with previous research that = is a fundamentally important topic in elementary mathematics and that room remains for improvement in children’s understanding of it (Bush & Karp, 2013; Knuth et al., 2006). On the other hand, our results suggest that students’ performance on items similar to “8 + 4 =  + 5” is not as poor as the published literature suggests. These results may imply that policies and practices in Florida better support student understanding of = than do those in other states. Acknowledgements The research reported here was supported by the Institute of Education Sciences, U.S. Department of Education, through Grant No. R305A180429 to Florida State University. The opinions expressed are those of the authors and do not represent views of the Institute or the U.S. Department of Education. Dimensions in Mathematics 36 References Bennett, V. M. (2015). Understanding the meaning of the equal sign: An investigation of elementary students and teachers. [Doctoral dissertation, University of Louisville]. Electronic Theses and Dissertations, Paper 2303. Bennett, A. B., Burton, L. J., Nelson, L. T., & Ediger, J. R. (2016). Mathematics for elementary teachers: A conceptual approach (10th ed.). McGraw-Hill Education. Boggs, G., Whitacre, I., Schellinger, J. Champagne, Z., & Schoen, R. (2018). Contextual meanings of the equals sign as conceptual blends. For the Learning of Mathematics, 38(2), 34–39. Bush, S. B., & Karp, K. S. (2013). Prerequisite algebra skills and associated misconceptions of middle grade students: A review. The Journal of Mathematical Behavior, 32, 613–632. Carpenter, T. P., Fennema, E., Franke, M. L., Levi, L., & Empson, S. B. (2015). Children's mathematics: Cognitively Guided Instruction (2nd ed.). Heinemann. Chow, K. C., & Wehby, J. H. (2019). Effects of symbolic and nonsymbolic equal-sign intervention in second-grade classrooms. The Elementary School Journal, 119, 677–702. Falkner, K. P., Levi, L., & Carpenter, T. P. (1999). Children’s understanding of equality: A foundation for algebra. Teaching Children Mathematics, 6, 232–236. Florida Department of Education (2020). Florida’s B.E.S.T. Standards: Mathematics. Hopewell, S., Loudon, K., Clarke, M. J., Oxman, A. D., Dickersin, K. (2009). Publication bias in clinical trials due to statistical significance or direction of trial results. Cochrane Database Systematic Reviews, 21(1) (No.: MR000006). Hornburg, C. B., Wang, L., & McNeil, N. M. (2018). Comparing meta-analysis and individual person data analysis using raw data on children’s understanding of equivalence. Child Development, 89(6), 1983-1995. Jacobs, V. R., Franke, M. L., Carpenter, T. P., Levi, L., & Battey, D. (2007). Professional development focused on children’s algebraic reasoning in elementary school. Journal for Research in Mathematics Education, 38(3), 258–288. Johannes, K., Davenport, J., Kao, Y., Hornburg, C. B., & McNeil, N. M. (2017). Promoting children’s relational understanding of equivalence. Proceedings of the 39th Annual Meeting of the Cognitive Science Society, 600–605. Johannes, K., & Davenport, J. L. (2019). Targeted mathematical equivalence training lessens the effects of early misconceptions on equation encoding and solving. Proceedings of the 41st Annual Meeting of the Cognitive Science Society, 492–498. Kühberger, A., Fritz, A., & Scherndl, T. (2014). Publication bias in psychology: A diagnosis based on the correlation between effect size and sample size. PLOS One, 9(9), e105825. Knuth, E. J., Stephens, A. C., McNeil, N. M., & Alibali, M. W. (2006). Does understanding the equal sign matter? Evidence from solving equations. Journal for Research in Mathematics Education, 37, 297–312. 37 SPRING 2021 Matthews, P. G., & Fuchs, L. S. (2018). Keys to the gate? Equal sign knowledge at second grade predicts fourth-grade algebra competence. Child Development, 91, 14–28. McNeil, N., M., & Alibali, M. W. (2005). Why won’t you change your mind? Knowledge of operational patterns hinders learning and performance of equations. Child Development, 76(4), 883–899. Molina, M., & Ambrose, R. (2008). From an operational to a relational conception of the equal sign: Thirds graders’ developing algebraic thinking. Focus on Learning Problems in Mathematics, 30(1), 61–80. Musser, G. L., Peterson, B. E., & Burger, W. F. (2014). Mathematics for elementary teachers: A contemporary approach. Wiley. National Governors Association Center for Best Practices & Council of Chief State Schools Officers (2010). Reaching higher: The Common Core State Standards Validation Committee. Powell, S. R., Kearns, D. M., & Driver, M. K. (2016). Exploring the connection between arithmetic and prealgebraic reasoning at first and second grade. Journal of Educational Psychology, 108, 943–959. Schoen, R. C., LaVenia, M., Champagne, Z. M., & Farina, K. (2016a). Mathematics performance and cognition (MPAC) interview: Measuring first- and second-grade student achievement in number, operations, and equality in Spring 2014 (Research Report No. 2016–01). Learning Systems Institute, Florida State University. Schoen, R. C., LaVenia, M., Champagne, Z. M., Farina, K., & Tazaz, A. M. (2016b). Mathematics performance and cognition (MPAC) interview: Measuring first- and second-grade student achievement in number, operations, and equality in spring 2015 (Research Report No. 2016–02). Florida Center for Research in Science, Technology, Engineering, and Mathematics. Schoen, R. C., Anderson, D., & Bauduin, C. (2017). Elementary mathematics student assessment: Measuring the performance of grade K, 1, and 2 students in number, operations, and equality in spring 2016. (Research Report No. 2017-22). Learning Systems Institute, Florida State University. Schoen, R. C., Anderson, D., Riddell, C. M., & Bauduin, C. (2018a). Elementary Mathematics Student Assessment: Measuring the performance of grade 3, 4, and 5 students in number (whole numbers and fractions), operations, and algebraic thinking in fall 2015 (Research Report No. 2018-24). Learning Systems Institute, Florida State University. Schoen, R. C., LaVenia, M., & Tazaz, A. M. (2018b). Effects of the first year of a three-year CGI teacher professional development program on grades 3–5 student achievement: A multisite cluster-randomized trial (Research Report No. 2018-25). Learning Systems Institute, Florida State University. Schoen, R. C., Buntin, C., Guven, A., & Yang, X. (2021). Elementary mathematics student assessment: Measuring grade K, 1, 2, and 3 students’ performance in number (whole numbers and fractions), operations, and algebraic thinking in spring 2019. (Research Report No. 2021-03). Learning Systems Institute, Florida State University. Dimensions in Mathematics 38 Stephens, A. C., Knuth, E. J., Blanton, M. L., Isler, I., Gardiner, A. M., & Marum, T. (2013). Equation structure and the meaning of the equal sign: The impact of task selection in eliciting elementary students’ understandings. The Journal of Mathematical Behavior, 32, 173–182. Guillermo Farfan is a Graduate Research Assistant and a doctoral candidate at Florida State University, doing research on mathematics teacher education and mathematical practice. Prior to his time at FSU, Guillermo worked as a math teacher in Tampa and Orlando, FL. He is a member and has served as a reviewer for both the American Educational Research Association and the Association for Psychological Science. Dr. Robert Schoen is an associate professor of mathematics education in the School of Teacher Education and the associate director of the Florida Center for Research in Science, Technology, Engineering, and Mathematics (FCR-STEM) in the Learning Systems Institute at Florida State University. He has overseen more than one-dozen large-scale, randomized controlled trials of educational interventions designed to improve teaching and learning in mathematics. Table of Contents President’s Message 4 Completing the Incomplete: Making Sense of Completing the Square Aline Abassian, Siddhi Desai, Haley Curry and Heidi Eisenreich 5 Fun with Measurements Hui Fang Huang Su, Bhagi Phuyel and Dylan Mandolini 13 Elementary Students’ Understanding of the Equals Symbol: Do Florida Students Outperform Their Peers Guillermo Farfan and Robert C. Schoen 27 Mathematics Educator of the Year Application Kenneth P. Kidd Award 39 Grants and Awards Listings 40 VOLUME 41 NUMBER 1 SPRING 2021 Dimensions in Mathematics is published by the Florida Council of Teachers of Mathematics (FCTM) twice yearly. Subscribers have permission to reproduce any classroom activities published in Dimensions. Dimensions in Mathematics is devoted to the improvement of mathematics education at all instructional levels. Articles which appear in Dimensions in Mathematics present a variety of viewpoints which, unless otherwise noted, should not be interpreted as official positions of the Florida Council of Teachers of Mathematics. For more information about FCTM, including membership, visit the FCTM web site: www.fctm.net.
17039	https://ajc.maths.uq.edu.au/pdf/45/ajc_v45_p025.pdf	AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 45 (2009), Pages 25–35 Enumeration and dichromatic number of tame tournaments V´ ıctor Neumann-Lara∗ Mika Olsen† Instituto de Matem´ aticas Universidad Nacional Aut´ onoma de M´ exico M´ exico D. F. M´ exico Abstract The concept of molds, introduced by the authors in a recent preprint, break regular tournaments naturally into big classes: cyclic tournaments, tame tournaments and wild tournaments. We enumerate completely the tame molds, and prove that the dichromatic number of a tame tourna-ment is 3. 1 Introduction A mold is a regular tournament M such that all the paths of the domination digraph D (M) are of order at most 2. The molds were deﬁned and studied in . Two vertices u, v form a dominant pair of a regular tournament T if N−(u, T \ {u, v}) = N+ (v, T \ {u, v}) . The domination graph of a tournament T, denoted by dom(T), is the graph on the vertex set V (T) with edges between dominant pairs of T. The domination graph was deﬁned in and the domination graph of regular tournaments were characterized by Cho et al. in [4, 5]. Recently the authors have given a simpler proof of this characterization by the use of molds . The domination digraph of a tournament T, denoted by D (T), is the domination graph on the vertex set V (T) with the orientation induced by T. The domination digraph was deﬁned in . Since the paths of a domination graph dom(T) are all directed in the tournament T, the characterization of domination graphs induces a characterization of domination digraphs. ∗Passed away February 2004 † Current address: Departamento de Matem´ aticas Aplicadas y Sistemas, Universidad Aut´ onoma Metropolitana, Unidad Cuajimalpa, M´ exico D.F., M´ exico 26 V. NEUMANN-LARA,AND MIKA OLSEN The cyclic tournament → C2m+1 ⟨∅⟩is the tournament on the vertex set Z2m+1 and (u, v) ∈A → C2m+1 ⟨∅⟩ if v −u ∈Zm. We assign a mold MT and associate a weight function ϕT, to every regular non-cyclic tournament T, and every regular non-cyclic tournament T can be reconstructed from its mold MT and the weight function ϕT . We say that a mold is tame if it has a dominant pair such that the residue is isomorphic to a cyclic tournament (the residue of a subdigraph H of T, is T \V (H)). A regular tournament is tame if its mold is tame. Equivalently a regular tournament is tame if D (T) has a maximal path P such that T \ P is a cyclic tournament. We enumerate the tame molds and prove that all tame tournaments are 3-dichromatic, that is, 3 is the smallest number of colors needed to color the vertices such that no directed monochromatic cycles are formed. It has recently been proved that tame tournaments are tight ; a tournament is tight if every vertex coloring with exactly 3 colors induces a cyclic triangle with the 3 colors. The standard notation follows [1, 2, 3, 9]. 2 Preliminaries In this section, we review some basic facts about domination digraphs of regular tournaments and molds. Clearly the automorphism group of a regular tournament T acts on the domination digraph D (T) of T, so it ﬁxes the set of D-arcs. We will need the following results. Lemma 1 If T ′ is a subtournament of T, then D (T) \|T ′ is a subdigraph of D (T ′). Let T be a tournament, u, v ∈V (T) and S ⊂V (T). We say that u and v are concordant with respect to S if N−(u, S \ {u, v}) = N−(v, S \ {u, v}) . We say that u and v are discordant with respect to S if N+ (u, S \ {u, v}) = N−(v, S \ {u, v}) . Note that are u and v are discordant with respect to T if and only if (u, v) ∈ dom (T). The components of D (T) are directed paths or cycles (Lemma 2.2, ). If P is a path of D (T), then T [V (P)] is completely determined by the following lemma. Lemma 2 Let P = (u0, u1, . . . , uk) be a path of D (T), where T is a regular tournament, Pi,j = (ui, ui+1, . . . , uj) an induced subpath of P with 0 ≤i < j ≤k, and int (Pi,j) = Pi,j \ {ui, uj}. (i) If j −i is even, then ui and uj are concordant with respect to T \Pi,j, (uj, ui) ∈ A (T) and ui and uj are discordant with respect to Pi,j. NUMBER OF TAME TOURNAMENTS 27 (ii) If j −i is odd, then ui and uj are discordant with respect to T \ Pi,j, (ui, uj) ∈ A (T) and ui and uj are concordant with respect to Pi,j. The following proposition is a summary of results from [5, 7, 12] Proposition 1 A regular tournament T, of odd order, is a cyclic tournament if and only if dom (T) ∼ = Cn. Remark 1 Let m ≥2; then → C2m+1 ⟨∅⟩\ {i, i + m} ∼ = → C2m−1 ⟨∅⟩, and moreover D → C2m+1 ⟨∅⟩ = (0, m, 2m, m −1, . . . , 1, m + 1, 0) . We will use the following construction: Remark 2 Let (u, v) be a D-arc of a regular tournament T of order 2m + 3 (m ≥1) and denote by T ′ = T \ {u, v} the residual tournament of (u, v). By the deﬁnition of D (T) , there exists a natural partition of V (T ′) into the sets V −= N+ (u; T ′) and V + = N+ (v; T ′). Moreover, V −= N−(v, T ′) and V + = N−(u, T ′). Since T is regular, \|V −\| = m and \|V +\| = m + 1. Note that in Lemma 1, D (T) \|T ′ is not necessarily induced in D (T). The molds were deﬁned in . A mold is a regular tournament M such that all the paths of the domination digraph D (M) are of order at most 2. Let T be a regular non-cyclic tournament; then we assign the mold MT and associate a weight function ϕT to T. Proposition 2 Let T be a non-cyclic regular tournament T, and let MT be the mold of T and ϕT the weight function of T. Then (i) MT is a regular subtournament of T, and D (T) \|M is induced in D (T), and (ii) T can be reconstructed from its mold MT and the weight function ϕT. Note that if φ : V (T) →V (T ′) is an isomorphism, then (u, v) ∈A (D (T)) if and only if (φ (u) , φ (v)) ∈A (D (T ′)). Corollary 1 Let T, T ′ be regular non-cyclic tournaments and (M, ϕ) , (M′, ϕ′) their corresponding molds and weight functions. Then T ∼ = T ′ if and only if there is an isomorphism φ : V (M) →V (M′) such that ϕ′ ◦φ (v) = ϕ (v) for every trivial com-ponent v of D (M), and ϕ′ (φ (u) , φ (v)) = ϕ (u, v) for every non-trivial component (u, v) of D (M). We say that a D-arc is a C-arc if the residue of the D-arc is a cyclic tournament. We study some properties of the regular tournaments with C-arcs (for more de-tails, see ). 28 V. NEUMANN-LARA,AND MIKA OLSEN Proposition 3 Let T be a regular non-cyclic tournament with a C-arc (ξ−, ξ+). Then MT has an odd number of D-arcs. Recall that there are three non-isomorphic regular tournaments of order 7: a cyclic tournament, a Payley tournament and a third tournament W0 deﬁned in . The regular tournament W0 = (V, A) is deﬁned as follows: V (W0) = w− 1 , w− 2 , w− 3 , w+ 1 , w+ 2 , w+ 3 , w0 , A (W0) = w− i , w0 ∪ w0, w+ i ∪ w− i , w+ i ∪ w+ i , w− i−1 ∪ w+ i , w− i+1 ∪ w− i , w− i+1 ∪ w+ i , w+ i+1 , with i −1, i, i + 1 ∈Z3. Note that D (W0) = w+ 2 , w− 1 , w+ 3 , w− 2 , w+ 1 , w− 3 , {w0} , the D-arcs of W0 are all C-arcs and W0 is transitive in D-arcs. Proposition 4 The number of C-arcs in a mold M is at most one, except when M ∼ = W0, and in that case M has three C-arcs. Proof Suppose that the mold M has two C-arcs ξ− 0 , ξ+ 0 and ξ− 1 , ξ+ 1 . Let M0 = M \ ξ− 1 , ξ+ 1 , M1 = M \ ξ− 0 , ξ+ 0 and M′ = M \ ξ− 0 , ξ+ 0 , ξ− 1 , ξ+ 1 . By Remark 1, M′ is a cyclic tournament. Since M0 is vertex transitive, we can assume that D (M0) = 0, ξ− 0 , ξ+ 0 , m, 2m, . . . , 0 . Let D (M1) = 0, m, 2m, . . . , k, ξ− 1 , ξ+ 1 , k + m, . . . , 0 , 0 ≤ k ≤2m. Then (j, j + m) ∈D (M) if j ̸= 0, k and ξ− 0 , ξ+ 0 , ξ− 1 , ξ+ 1 ∈D (M). We know that 0, ξ− 0 , ξ+ 0 , m ∈D (M0) and k, ξ− 1 , ξ+ 1 , k + m ∈D (M1). Since d− ξ− i ; Tj = m + 1, d− ξ− i ; T = m + 2, with i ∈{1, 2}, and M0, M1, M are regular tournaments, then M ξ− 0 , ξ+ 0 , ξ− 1 , ξ+ 1 \ ξ− 0 , ξ+ 0 , ξ− 1 , ξ+ 1 ∼ = → C4. We assume that ξ+ 0 , ξ+ 1 ∈M. Let k ∈N+ ξ+ 0 . Since D (M) = 0, ξ− 0 , ξ+ 0 , m, 2m, . . . , k, ξ− 1 , ξ+ 1 , k + m, . . . , 0 , then T would be a cyclic tournament. So k / ∈N+ ξ+ 0 and k + m ∈N+ ξ+ 0 . Then D (M) is the disjoint union of the two arcs ξ− 0 , ξ+ 0 , ξ− 1 , ξ+ 1 and two paths ((m, 2m, . . . , k) and (k + m, k + 2m, . . . , 0)), one of even order and the other of odd order. Since M is a mold, then by Proposition 2, D (M) is the disjoint union of three arcs and a vertex, and then M ∼ = W0. □ As a consequence of Propositions 3 and 4, we have the following. Corollary 2 If M is a mold of order at least 9, with a C-arc (ξ−, ξ+), then (ξ−, ξ+) is the only C-arc of M, and M has an odd number (at least three) of D-arcs. A digraph is rigid if its only automorphism is the identity. Proposition 5 If T is a tournament with MT ∼ = W0, then T is rigid if and only if there are two D-paths of even order with diﬀerent length. NUMBER OF TAME TOURNAMENTS 29 Proof Let T be a tame mold. Let ϕT be the associated weight function and MT ∼ = W0. Then D(T) has four D-paths, one of odd order P1 and three of even orders P2, P3, P4. The cyclic rotation of the D-arcs of MT is an automorphism of MT . For necessity: Let l(P2) = l(P3) = l(P4) = 2k; then the weight function ϕT(a) = 2k, for all D-arcs a. Then by Corollary 1 the cyclic rotation of the D-arcs of MT induces an automorphism of T, and T is not rigid. For suﬃciency: We assume that l(P2) ̸= l(P3), and l(P2) ̸= l(P4). Then any automorphism of T ﬁxes the D-path P2 and, any automorphism of MT ﬁxes the C-arc (v0, v1), where P2 = (v0, v1, v2, . . . , v2k−1). Then MT with the weight function ϕT is rigid and by Corollary 1, so is T. □ 3 Tame molds A mold with a C-arc is called a tame mold. The C-arcs induce a natural partition of the molds in three classes: the family of cyclic tournaments → C2m+1 ⟨∅⟩, the tame tournaments and otherwise, the wild tournaments. A tame tournament is a regular tournament with a tame mold. We associate to each tame mold an even directed cycle and a weight function. We prove that there is a bijection between the tame molds and this set of even directed cycles with their weight function. Using this fact we enumerate the tame molds. Theorem 1 There exists a bijection between the tame molds and the set of even directed cycles → C2s = u− 1 , u+ 1 , u− 2 , u+ 2 , . . . , u− s , u+ s , u− 1 , s ≥1, with the weight function ρM : V → C2s →N with the following properties (i) ρM (v) ≥2 for v ∈V → C2s , and (ii) s i=1 ρM u+ i = s i=1 ρM u− i + 1. Proof Given a tame mold M we will prove that we can associate to M an even directed cycle → C2s and a weight function ρM and that this cycle with its weight function is unique. Let M be a tame mold of order 2m + 3 (m ≥2), with 2s + 1 D-arcs, where 0 < s ≤m/2. Let (ξ−, ξ+) be a C-arc, and M′ = M {ξ−, ξ+}. We denote → C2m+1 ⟨∅⟩ by M′. By Remark 1, D (M′) = (0, m, 2m, m −1, 2m −1, . . . , m + 1, 0) . We will associate an even directed cycle → C2s and a weight function ρM as follows: Let V −= N+ (ξ−, M′), V + = N+ (ξ+, M′). Since M′ is vertex-transitive and by Corollary 2, we may assume that (0, m) ∈D (M) such that m ∈V −and 0 ∈V +. 30 V. NEUMANN-LARA,AND MIKA OLSEN Let {P ε i }s i=1 be the set of maximal paths of D (M′)∩V ε, with ε ∈{−, +}. Note that P − 1 = (m, 2m, . . . , lm) and P + s = (km, . . . , m + 1, 0), for some 1 ≤l < k ≤2m. Let CM be the resulting cycle when we contract each path P − i and P + i to the vertex v− i and v+ i respectively, so CM= u− 1 , u+ 1 , u− 2 , u+ 2 , . . . , u− s , u+ s , u− 1 ; see Figure 1. The weight function ρM assigns to each vertex uε i the order of the path P ε i , ρM (uε i) = \|V (P ε i )\|. Note that ρM (uε i) ≥2, l (CM) = 2s and CM contains all the D-arcs of M except the C-arc (ξ−, ξ+). Also note that s i=1 ρM u− i = m, s i=1 ρM u+ i = m + 1 and u∈CM ρM (u) = 2m + 1 (see Figure 1). P − s P − 2 P − 1 P + 1 ε− ε+ P + s u− s u+ s u− 1 u+ 1 u− 2 u+ 2 (a) (b) Figure 1: (a) The tame tournament T, (b) the axillary cycle of T. Observe that if \|M\| ≥9, then by Corollary 2, the C-arc is unique, and the cycle CM with the associated weight function ρM is unique. Given an even cycle and a weight function with the required properties we can construct a tame mold M. Let the even cycle CM= u− 1 , u+ 1 , u− 2 , u+ 2 , . . . , u− s , u+ s , u− 1 , s ≥1, with the weight function ρM : V → C2s →N with the properties: (i) ρM (v) ≥2 for v ∈V → C2s , (ii) s i=1 ρM u− i = m, and (iii) s i=1 ρM u+ i = m + 1. We will construct a tame mold M of order s i=1 ρM u+ i + ρM u− i +2, (m ≥2), with 2s + 1 D-arcs, where 0 < s ≤m/2. Let {P ε i }s i=1 = {vε i,1, vε i,2, . . . , vε i,ρM(uε i)} be the set of directed paths such that \|V (P ε i )\| = ρM (uε i), with ε ∈{−, +}. Now we construct the cycle C as follows: for each vertex uε i ∈V → C2s , ε ∈{−, +}, take a copy of the path P ε i and connect the terminal vertex of P − i with the initial vertex of P + i and the terminal vertex of P + i with the initial vertex of P + i+1. Note that \|V (C)\| = 2m + 1. Let C = D(M′); then M′ is a cyclic tournament. We deﬁne the tame mold M as follows: NUMBER OF TAME TOURNAMENTS 31 Let V (M) = V (M′)∪{ξ−, ξ+}, let N+ (ξε) = ∪s i=1{vε i,1, vε i,2, . . . , vε i,ρM(uε i)} and let (ξ−, ξ+) ∈A(M). Note that (ξ−, ξ+) is a C-arc of M and that the tame mold M is completely determined by the cycle CM and the weight function ρM. □ Proposition 6 The tame molds of order at least 9 are rigid. Proof Let M be a tame tournament and \|V (M)\| ≥9. Let (ξ−, ξ+) be the C-arc of M and let M′ = M \ {ξ−, ξ+}. Let → C2s = u− 1 , u+ 1 , u− 2 , u+ 2 , . . . , u− s , u+ s , u− 1 be the even directed cycle associated to the tame mold M, and ρM : V → C2s →N be the weight function associated to the tame mold M. Note that an automorphism of → C2s with its weight function ρM induces an automorphism of D (M \ {ξ−, ξ+}) that ﬁxes the ex-neighborhoods of ξ−and ξ+. Since s i=1 ρM u+ i = s i=1 ρM u− i + 1, an odd rotation of the vertices of → C2s does not induce an automorphism of M (nor of T, by Corollary 1). Let ϕ be an even rotation of the vertices of → C2s. Since ϕ ﬁxes the ex-neighborhoods of ξ−and ξ+, it follows that ϕ induces a rotation of the vertices of N+ M′(ξ−) and a rotation of the vertices of N+ M′(ξ+). Since \|N+ M′(ξ−)\| = m, \|N+ M′(ξ+)\| = m+ 1 and gcd (m, m + 1) = 1, it follows that ϕ is the identity and then → C2s with its associated weight function ρM is rigid. Then M is rigid by Corollary 2 and Theorem 1. □ Proposition 7 A tame tournament T is rigid, except if MT ∼ = W0 and all the D-paths of even order of T have the same length. Proof Let T be a tame tournament. If V (MT ) ≥9, then the result is a consequence of Proposition 6 and Corollary 1. Let V (MT ) = 7. It is known that up to isomorphism there are three regular tournaments of order 7: the cyclic, the Paley tournament and W0. W0 is the only tame mold of order 7, and the result is a consequence of Proposition 5 and Corollary 1. □ Given the weight function ρM of a tame mold M (see Figure 1), we can represent ρM by alternating two partitions Π1 and Π2, with Π1 = ρM u− 1 \| ρM u− 2 \| . . . \| ρM u− s , Π2 = ρM u+ 1 \| ρM u+ 2 \| . . . \| ρM u+ s , where Π1 and Π2 are partitions of m and m + 1 respectively into s parts. Let π1 and π2 be the frequency representations of Π1 and Π2 respectively. Then πi = n ri,1 i,1 , n ri,2 i,2 , . . . , n ri,k i,k , where ni,j ≥2, for i = 1, 2 and 1 ≤j ≤k, such that j=k j=1 n1jr1j = m, j=k j=1 n2jr2j = m + 1 and j=k j=1 rij = s for each i = 1, 2. Let π = (nr1 1 , nr2 2 , . . . , nrk k ) be the frequency representation of a partition; then deﬁne ε (π)! = r1!r2! . . . rs!. 32 V. NEUMANN-LARA,AND MIKA OLSEN Lemma 3 The number of non-isomorphic molds of order 2m+3 with 2s+1 D-arcs, where 0 < s ≤m/2, and frequency representation π1, π2 is (s!)2 s · ε (π1)!ε (π2)!. Proof Let π1 and π2 be the frequency representation of the partitions Π1 and Π2 respectively. Let π1 = n r1,1 1,1 , n r1,2 1,2 , . . . , n r1,k 1,k and π2 = n r2,1 2,1 , n r2,2 2,2 , . . . , n r2,l 2,l , with ni,j ≥2 and j=k j=1 n1jr1j = m, j=k j=1 n2jr2j = m + 1 and j=k j=1 rij = s for each i = 1, 2. Using the counting argument of acyclic permutations with repetition, it follows that the number of tame molds is equal to 1 s s r1,1r1,2 . . . r1,k s r2,1r2,2 . . . r2,l = (s!)2 s · ε (π1)!ε (π2)! where n1 + n2 + . . . + nk = n and n n1n2 . . . nk = n! n1!n2! . . . nk!. □ We conclude that Theorem 2 The number of non-isomorphic tame molds of order 2m+3, with 2s+1 D-arcs, where 0 < s ≤m/2 and m ≥4, is π1,π2 (s!)2 s · ε (π1)!ε (π2)!. Proof Let π1, π2 be the frequency representation of the partitions Π1 and Π2 respectively. Then the result is a consequence of Lemma 3, summing on the frequency representations π1, π2. □ Corollary 3 There is a unique tame mold up to isomorphism, of order 2m + 3 (i) with 3 D-arcs, and (ii) m + 1 D-arcs, with m an even integer and m ≥4. Let M be a tame tournament, and let D(M) be the disjoint union of s arcs and 2t + 1 vertices. Then \|V (M)\| = 2(s + t) + 1. Let C(D) denote the set of maximal components of the digraph D. Let T ∈F(M). Let ϕ : C(D(M)) →N be the weight function that assigns to each component of D(M) the length of the corresponding path in D(T). Since the components of D(M) have order one or two, observe that \|C(D(M))\| = s + 2t + 1. Let C1(D(M)) denote the set of trivial components and let C2(D(M)) denote the set of components of order 2. Since T has odd order, we have c∈C(D(M)) ϕ(c) = 2r + 1, where r ≥s + t. NUMBER OF TAME TOURNAMENTS 33 Moreover, if c ∈C1(D(M)), then ϕ(c) is an odd positive integer and if c ∈C2(D(M)), then ϕ(c) is an even positive integer. Let k = r −s −t. Note that \|V (T)\| = V (MT ) + 2k. Let n1 and n2 be two nonnegative integers such that n1 + n2 = k, c∈C1(D(M)) ϕ(c) = 2(n1 + t) + 1 and c∈C2(D(M)) ϕ(c) = 2(n2 + s). Let P 1 and P 2 be the set of maximal paths of odd order and even order respectively of D(T); then 2n1 is the number of vertices of V (P 1) \ V (C1(D(M))) and 2n2 is the number of vertices of V (P 2) \ V (C2(D(M))). Lemma 4 Fix the partition (n1, n2) of k, ﬁx the frequency representation π1 = (nr11 11 , nr12 12 , . . . , nr1k 1k ), with n r1j 1j an odd integer, of the partition Π1 of 2(t + n1) + 1 into 2t + 1 parts and ﬁx the frequency representation π2 = (nr21 21 , nr22 22 , . . . , nr2l 2l ) of the partition Π2 of s + n2 into s parts. Then the number of non-isomorphic tame tour-naments T ∈F(M), with s D-arcs, order 2(s + t + k) + 1, \|V (P 1 i )\| = 2(t + n1) + 1 and \|V (P 2 i )\| = 2(s + n2) is exactly s!(2t + 1)! ε(π1)!ε(π2)!. Proof Fix the frequency representation π1 = (nr11 11 , nr12 12 , . . . , nr1k 1k ), with n r1j 1j an odd integer, of the partition Π1 of 2(t + n1) + 1 into 2t + 1 parts, and ﬁx the frequency representation π2 of the partition Π2 of s + n2 into s parts. Note that r11 +r12 +. . .+r1k = 2t+1 and r21 +r22+. . .+r2l = s. Using the counting argument of acyclic permutations with repetition, it follows that the number of tame molds is equal to 2t + 1 r11r12 . . . r1k s r21r22 . . . r2l = s!(2t + 1)! ε(π1)!ε(π2)!. □ Theorem 3 Let (n1, n2) be a partition of k. Let π1 = (nr11 11 , nr12 12 , . . . , nr1k 1k ) be the frequency representation of a partition Π1 of 2(t + n1) + 1 into 2t + 1 parts, with n r1j 1j an odd integer, and let π2 = (nr21 21 , nr22 22 , . . . , nr2l 2l ) be the frequency representation of a partition Π2 of s + n2 into s parts. Then the number of non-isomorphic tame tournaments T ∈F(M), with s D-arcs and order 2(s + t + k) + 1 is exactly n1,n2 π1,π2 s!(2t + 1)! ε(π1)!ε(π2)!. Proof Fix the partition (n1, n2) of k. Then the number of non-isomorphic tame tournaments T ∈F(M), such that T has order 2(s + t + k) + 1 is exactly π1,π2 s!(2t + 1)! ε(π1)!ε(π2)!. Summing on the partitions of k there are n1,n2 π1,π2 s!(2t + 1)! ε(π1)!ε(π2)! non-isomorphic tame tournaments of order 2(s + t + k) + 1 with T ∈F(M). □ 34 V. NEUMANN-LARA,AND MIKA OLSEN 4 Coloring We prove that all tame tournaments have dichromatic number 3. The dichromatic number dc (D) of a digraph D is the smallest number of colors needed to color the vertices of D such that no directed monochromatic cycles are formed . Remark 3 Let P = (u0, u1, . . . , uk) be a path of D (T), where T is a regular tournament. Let F0 = (u0, u2, . . . , u2l) and F1 = (u1, u3, . . . , u2l+1), where 2l, 2l+1 ≤ k; then T [Fi] is transitive, i = 0, 1. To put Theorem 5 below in perspective, it is convenient to recall the following: Theorem 4 Let T be a regular tournament. Then dc (T) = 2 if and only if T is a cyclic tournament. Theorem 5 Let T be a tame tournament. Then dc MT = 3 = dc (T) . Proof Let T be a tame tournament; by Theorem 4 dc(T) > 2. Let P be a path in D(T) such that T \P is a cyclic tournament. Let P = (u1, u2, . . . , u2j). We assume that T \ P = → C2m+1 ⟨∅⟩, 0 ∈N+ (u1) and m ∈N+ (u2). Let ϕ : V (T) →{c, c−, c+} be the following 3-coloring: ϕ (i) = ⎧ ⎨ ⎩ c− if i ∈({0, 1, . . . , m} ∩N+(u1)) ∪{u1, u3, . . . , u2j−1} c+ if i ∈({0, 1, . . . , m} ∩N+(u2)) ∪{u2, u4, . . . , u2j} c if i ∈{m + 1, m + 2, . . . , 2m} . Clearly {m + 1, m + 2, . . . , 2m} and {0, 1, . . . , m} are acyclic. Since u2s+1 and u2t+1 are concordant with respect to T \P by Lemma 2 (i), then by Remark 3 the chromatic class {c−} is acyclic. Analogously the chromatic class {c+} is acyclic. It follows that ϕ is an acyclic coloring with 3 colors, and then dc(T) = 3. □ Acknowledgment The authors would like to thank the referee for the valuable suggestions. References G.E. Andrews, The Theory of Partitions, (Cambridge Mathematical Library, 1998). L.W. Beineke and K.B. Reid, Tournaments, in: L.W. Beineke, R.J. Wilson (Eds.), Selected Topics in Graph Theory, Academic Press, New York, 1979, pp. 169–204. J.A. Bondy and U.S.R. Murty, Graph Theory with Applications, American El-sevier Pub. Co., 1976. NUMBER OF TAME TOURNAMENTS 35 H. Cho, F. Doherty, S.R. Kim and J. Lundgren, Domination graphs of regular tournaments II, Congr. Numer. 130 (1998), 95–111. H. Cho, S.R. Kim and J. Lundgren, Domination graphs of regular tournaments, Discrete Math. 252 (2002), 57–71. D.C. Fisher, J. Lundgren, S.K. Merz and K.B. Reid, Domination graphs of tournaments and digraphs, Congr. Numer. 108 (1995), 97–107. D.C. Fisher, J. Lundgren, S.K. Merz and K.B. Reid, The domination and com-petition graphs of a tournament, J. Graph Theory 29 (1998), 103–110. B. Llano and M. Olsen, Inﬁnite families of tight regular tournaments, Discuss. Math. Graph Theory 27 (2007), 299–311. J.W. Moon, Topics on Tournaments, Holt, Rinehart & Winston, New York, 1968. V. Neumann-Lara, The dichromatic number of a digraph, J. Combin. Theory Ser. B 33 (1982), 265–270; in Digraphs, Ann. Discrete Math. 17 (1983), 365– 370. V. Neumann-Lara, The 3 and 4-dichromatic tournaments of minimum order, Discrete Math. 135 (1984), 233–243. V. Neumann-Lara and M. Olsen, Molds of regular tournaments, Ars Comb. (accepted). V. Neumann-Lara and J. Urrutia, Vertex critical r-dichromatic tournaments, Discrete Math. 40 (1984), 83–87. (Received 15 Apr 2008; revised 4 May 2009)
17040	https://math.uchicago.edu/~dannyc/courses/minimal_surfaces_2014/minimal_surfaces_notes.pdf	CHAPTER 3: MINIMAL SURFACES DANNY CALEGARI Abstract. These are notes on minimal surfaces, which are being transformed into Chap-ter 3 of a book on 3-Manifolds. The emphasis is on the classical theory and its connection to complex analysis, and the topological applications to 3-manifold geometry/topology. These notes follow a course given at the University of Chicago in Spring 2014. Contents 1. Minimal surfaces in Euclidean space 1 2. First variation formula 8 3. Second variation formula 14 4. Existence of minimal surfaces 20 5. Embedded minimal surfaces in 3-manifolds 33 6. Acknowledgments 38 References 38 1. Minimal surfaces in Euclidean space In this section we describe the classical theory of minimal surfaces in Euclidean spaces, especially in dimension 3. We emphasize throughout the connections to complex analysis. The local theory of minimal surfaces in Riemannian manifolds is well approximated by the theory of minimal surfaces in Euclidean space, so the theory we develop in this section has applications more broadly. 1.1. Graphs. A smooth surface in R3 can be expressed locally as the graph of a real-valued function deﬁned on a domain in R2. For historical reasons, such graphs are sometimes referred to as nonparametric surfaces. Fix a compact domain Ω⊂R2 with boundary, and a smooth function f : Ω→R. Denote by Γ(f) the graph of f in R3; i.e. Γ(f) = {(x, y, f(x, y)) ∈R3 such that (x, y) ∈Ω} Ignoring the last coordinate deﬁnes a projection from Γ(f) to Ω, which is a diﬀeomorphism. An inﬁnitesimal square in Ωwith edges of length dx, dy is in the image of an inﬁnitesimal parallelogram in Γ(u) with edges (dx, 0, fxdx) and (0, dy, fydy) so area(Γ(f)) = Z Ω \|(1, 0, fx) × (0, 1, fy)\|dxdy = Z Ω p 1 + \|grad(f)\|2dxdy Date: February 10, 2019. 1 2 DANNY CALEGARI Note that in this formula we are thinking of grad(f) as a vector ﬁeld on Ωin the usual way. If g : Ω→R is smooth and vanishes on ∂Ωwe obtain a 1-parameter family of functions f(t) := f + tg and a 1-parameter family of graphs Γ(t) := Γ(f(t)). The hypothesis that g vanishes on ∂Ωensures that these graphs all have the same boundary. We are interested in how the area of Γ(t) varies as a function of t. Since grad(f + tg) = grad(f) + t grad(g) we can compute d dt t=0area(Γ(t)) = Z Ω d dt t=0 p 1 + \|grad(f) + t grad(g)\|2dxdy = Z Ω ⟨grad(f), grad(g)⟩ p 1 + \|grad(f)\|2 dxdy Integrating by parts, and using the vanishing of g on ∂Ω, this equals Z Ω −g div grad(f) p 1 + \|grad(f)\|2 ! dxdy (formally, this is the observation that −div is an adjoint for grad). Thus: f is a critical point for area (among all smooth 1-parameter variations with support in the interior of Ω) if and only if f satisﬁes the minimal surface equation in divergence form: div grad(f) p 1 + \|grad(f)\|2 ! = 0 In this case we also say that Γ is a minimal surface. Expanding the minimal surface equation, and multiplying through by the factor (1 + \|grad(f)\|2)3/2 we obtain the equation (1 + f 2 y )fxx + (1 + f 2 x)fyy −2fxfyfxy = 0 This is a second order quasi-linear elliptic PDE. We explain these terms: (1) The order of a PDE is the degree of the highest derivatives appearing in the equa-tion. In this case the order is 2. (2) A PDE is quasi-linear if it is linear in the derivatives of highest order, with coeﬃ-cients that depend on the independent variables and derivatives of strictly smaller order. In this case the coeﬃcients of the highest derivatives are (1 + f 2 y ), (1 + f 2 x) and −2fxfy which depend only on the independent variables (the domain variables x and y) and derivatives of order at most 1. (3) A PDE is elliptic if the discriminant is negative; here the discriminant is the dis-criminant of the homogeneous polynomial obtained by replacing the highest order derivatives by monomials. In this case since the PDE is second order, the discrim-inant is the polynomial B2 −4AC, which is equal to 4f 2 xf 2 y −4(1 + f 2 y )(1 + f 2 x) = −4(1 + f 2 x + f 2 y ) < 0 Solutions of elliptic PDE are as smooth as the coeﬃcients allow, within the interior of the domain. Thus minimal surfaces in R3 are real analytic in the interior. CHAPTER 3: MINIMAL SURFACES 3 If f is constant to ﬁrst order (i.e. if fx, fy ∼ϵ) then this equation approximates fxx+fyy = 0; i.e. −∆f = 0, the Dirichlet equation, whose solutions are harmonic functions. Thus, the minimal surface equation is a nonlinear generalization of the Dirichlet equation, and functions with minimal graphs are generalizations of harmonic functions. In particular, the qualitative structure of such functions — their singularities, how they behave in families, etc. — very closely resembles the theory of harmonic functions. 1.2. Second fundamental form and mean curvature. A parametric surface Σ in Rn is just a smooth map from some domain Ωin R2 to Rn. Let’s denote the map x : Ω→Rn, so Σ = x(Ω). The second fundamental form II is the perpendicular part of the Hessian of x. In terms of co-ordinates, let e1, e2 be vector ﬁelds on Ωlinearly independent at p. Then write IIp(ei, ej) := ei(ej(x))(p)⊥ i.e the component of the vector ei(ej(x))(p) ∈Rn perpendicular to Σ at x(p). It turns out that IIp is a symmetric quadratic form on TpΣ. To see this, observe ﬁrst that IIp depends only on the value of ei at p. Furthermore, since ei, ej = dx([ei, ej]) is tangent to Σ, it follows that IIp is symmetric in its arguments, and therefore it depends only on the value of ej at p. The mean curvature H is the trace of the second fundamental form; i.e. H(p) = X IIp(ei, ei) where ei runs over an orthonormal basis for Tp. Now let N be a unit normal vector ﬁeld on Σ. By abuse of notation we write ei for dx(ei) and think of it as a vector ﬁeld on Σ. Then pointwise, ⟨H, N⟩= X ⟨ei(ei), N⟩= − X ⟨ei, ei(N)⟩ where we use the fact that N is everywhere perpendicular to each ei. An inﬁnitesimal unit square in Tp spanned by e1, e2 ﬂows under the normal vector ﬁeld N to an inﬁnitesimal parallelogram, and the derivative of its area under the ﬂow is exactly P⟨ei, ei(N)⟩. Thus near any point p where H is nonzero we can change the area to ﬁrst order by a normal variation supported near p, and we see that Σ is critical for area if and only if the mean curvature H vanishes identically. This observation is due to Meusnier. For Σ a surface in R3 the unit normal ﬁeld N is determined by Σ (up to sign). We extend N smoothly to a unit length vector ﬁeld in a neighborhood of Σ. Along Σ the vectors e1, e2, N form an orthonormal basis, and the expression P⟨ei, ei(N)⟩is equal to div(N) which in particular does not depend on the choice of extension. Specializing to the case of a graph, the vector ﬁeld N := (−fx, −fy, 1) p 1 + \|grad(f)\|2 4 DANNY CALEGARI is nothing but the unit normal ﬁeld to Γ(f). We can extend N to a vector ﬁeld on Ω× R by translating it parallel to the z axis. We obtain the identity: −divR3(N) = divR2 grad(f) p 1 + \|grad(f)\|2 ! where the subscript gives the domain of the vector ﬁeld where each of the two divergences are computed. Thus Γ is minimal if and only if the divergence of N vanishes. 1.3. Conformal parameterization. We consider a parametric surface x : Ω→Rn. Let’s denote the coordinates on R2 by u and v, and the coordinates on Rn by x1, · · · , xn. The Jacobian is the matrix with column vectors ∂x ∂u and ∂x ∂v, and where this matrix has rank 2 the image is smooth, and the parameterization is locally a diﬀeomorphism to its image. The metric on Rn makes the image into a Riemannian surface, and every Riemannian metric on a surface is locally conformally equivalent to a ﬂat metric. Thus after precomposing x with a diﬀeomorphism, we may assume the parameterization is conformal (one also says that we have chosen isothermal coordinates). This means exactly that there is a smooth nowhere vanishing function λ on Ωso that ∂x ∂u 2 = ∂x ∂v 2 = λ2 and ∂x ∂u · ∂x ∂v = 0 We can identify Ωwith a domain in C with complex coordinate ζ := u + iv, and for each coordinate xj deﬁne φj := 2∂xj ∂ζ = ∂xj ∂u −i∂xj ∂v Then we have the following lemma: Lemma 1.1 (Conformal parameterization). Let Ω⊂C be a domain with coordinate ζ, and x : Ω→Rn a smooth immersion. Then x is a conformal parameterization of its image (with conformal structure inherited from Rn) if and only if P φ2 j = 0, where φj = 2∂xj/∂ζ. Furthermore, functions φj as above with P φ2 j = 0 deﬁne an immersion if and only if P \|φj\|2 > 0. Proof. By deﬁnition, X φ2 j = X j ∂xj ∂u 2 − ∂xj ∂v 2! −i X j ∂xj ∂u ∂xj ∂v ! whose real and imaginary parts vanish identically if and only if the parameterization is conformal, where it is an immersion. Furthermore, X \|φj\|2 = λ2 so the map is an immersion everywhere if and only if P \|φj\|2 > 0. □ If the parameterization x : Ω→Rn is conformal, we can consider the orthonormal basis e1 = λ−1 ∂ ∂u and e2 = λ−1 ∂ ∂v CHAPTER 3: MINIMAL SURFACES 5 Diﬀerentiating the deﬁning equations for x to be conformal, we obtain ∂2x ∂u2 · ∂x ∂u = ∂2x ∂u∂v · ∂x ∂v = −∂2x ∂v2 · ∂x ∂u and therefore ∆x is perpendicular to Σ. Thus the mean curvature H — i.e. the trace of the second fundamental form — is the vector H = λ−2 ∂2x ∂u2 + ∂2x ∂v2 = −λ2∆x Thus we obtain the following elegant characterization of minimal surfaces in terms of conformal parameterizations: Lemma 1.2 (Harmonic coordinates). Let Ω⊂C be a domain with coordinate ζ, and x : Ω→Rn a conformal parameterization of a smooth surface. Then the image is minimal if and only if the coordinate functions xj are harmonic on Ω; equivalently, if and only if the functions φj := 2∂xj ∂ζ are holomorphic functions of ζ. Proof. All that must be checked is the fact that the equation −∆xj = 0 is equivalent to the Cauchy–Riemann equations for φj: −∆xj = ∂ ∂u ∂xj ∂u −i∂xj ∂v + i ∂ ∂v ∂xj ∂u −i∂xj ∂v = 2∂φj ∂¯ ζ □ Corollary 1.3 (Convex hull). Every compact minimal surface Σ in Rn lies in the convex hull of its boundary. Proof. Every linear function on Rn is harmonic on Σ. Therefore by the maximum principle, if the boundary lies in a given half-space of Rn, so does Σ. □ A holomorphic reparameterization of Ωtransforms the coordinate ζ and the functions φj, but keeps ﬁxed the 1-form φjdζ. Combining this observation with the two lemmas, we obtain the following proposition, characterizing minimal surfaces in Rn parameterized by arbitrary Riemann surfaces: Proposition 1.4. Every minimal surface in Rn is obtained from some Riemann surface Ω together with a family of n complex valued 1-forms φj satisfying the following conditions: (1) (conformal): P φ2 j = 0; (2) (minimal): the φj are holomorphic; (3) (regular): P \|φj\|2 > 0; and (4) (period): the integral of φj over any closed loop on Ωis purely imaginary. The map x : Ω→Rn may then be obtained uniquely up to a translation by integration: xj = Re Z ζ 0 φj + cj Proof. All that remains is to observe that the period condition is both necessary and suﬃcient to let us recover the coordinates xj by integrating the real part of the φj. □ 6 DANNY CALEGARI If the φj are holomorphic and not identically zero, then P \|φj\|2 can equal zero only at isolated points in Ω. Near such points the map from Ωto its image is branched. We say that a surface is a generalized minimal surface if it is parameterized by some Ωas in Proposition 1.4, omitting the condition of regularity. 1.4. Conjugate families. Let Ωbe a Riemann surface, and φj a collection of n holomor-phic 1-forms satisfying P φ2 j = 0. Integrating the φj along loops in Ωdetermines a period map H1(Ω; Z) →Cn, and an abelian cover ¯ Ωcorresponding to the kernel of the period map. Then we obtain a further integration map Φ : ¯ Ω→Cn whose coordinates zj are given by zj = R ζ 0 φj. The standard (complex) orthogonal quadratic form has the value P z2 j on a vector z with coordinates zj; by Proposition 1.4 the image Φ(¯ Ω) is isotropic for this orthogonal form. Consequently we obtain a family of minimal surfaces in Rn parameterized by the action of the complex aﬃne group Cn ⋊(C∗× O(n, C)) where the ﬁrst factor acts on Cn by translation, and the second factor acts linearly. The Cn action just projects to translation of the minimal surface in Rn, and R∗×O(n, R) just acts by scaling and rotation; so this subgroup acts by ambient similarities of Rn. The action of S1 ⊂C∗is more interesting; the family of minimal surfaces related by this action are said to be a conjugate family. At the level of 1-forms this action is a phase shift φj →eiθφj. Lemma 1.5. Let x(θ) : Ω→Σθ be a conjugate family of generalized minimal surfaces in Rn; i.e. their coordinates are given by integration xj(θ)(p) = Re Z p 0 eiθφj for some ﬁxed family of 1-forms φj on Ωwith P φ2 j = 0. Then for any θ the composition x(0)◦x(θ)−1 : Σθ →Σ0 is a local isometry, and each ﬁxed point p ∈Ωtraces out an ellipse x(·)(p) : S1 →Rn. Proof. If we write φj = aj + ibj then P a2 j = P b2 j and P ajbj = 0; i.e. the vectors a and b are perpendicular with the same length, and this length is the length of dx(0)(∂u) in Tx(0)(Ω). But the length of dx(θ)(∂u) in Tx(θ)(Ω) is just \| cos(θ)a + sin(θ)b\| = \|a\| = \|b\| for all θ, so x(0) ◦x(θ)−1 : Σθ →Σ0 is an isometry as claimed. The second claim is immediate: xj(θ)(p) = cos(θ)Re Z p 0 φj + sin(θ)Re Z p 0 iφj □ Example 1.6 (catenoid and helicoid). If we write ζ := u + iv then φ1 = −sin(ζ), φ2 = cos(ζ), φ3 = −i satisﬁes the conditions of Proposition 1.4. The surface obtained by integrating the real part of the φj is the catenoid, with the parameterization x = cosh(v) cos(u); y = cosh(v) sin(u); z = v CHAPTER 3: MINIMAL SURFACES 7 Figure 1. A conjugate family of minimal surfaces interpolating between the catenoid and the helicoid Multipying the φj by i and integrating gives the helicoid, with the parameterization x = sinh(v) sin(u); y = −sinh(v) cos(u); z = u Interpolating between these two values gives an isometric deformation of one surface into the other. 1.5. Weierstrass–Enneper parameterization. Let φj be three holomorphic functions on a domain Ωsatisfying the conditions of Proposition 1.4. Using the condition P φ2 j = 0 we can eliminate one of the functions. If we write f = φ1 −iφ2 and g = φ3/f then φ1 = f(1 −g2)/2; φ2 = if(1 + g2)/2; φ3 = fg Conversely, any pair of functions f, g where f is holomorphic, and g is meromorphic so that f has a zero of order at least 2k wherever g has a pole of order k, deﬁne a (generalized) minimal surface. This normalization is the so-called Weierstrass–Enneper parameteriza-tion. This parameterization is particularly nice because of its connection to the Gauss map. For Σ an oriented surface in R3, the Gauss map N : Σ →S2 sends each point to its unit normal. If we take Σ = x(Ω) then the unit normal ﬁeld is given by N = (xu×xv)/\|xu×xv\|. Now, by deﬁnition xu = Re φ and xv = −Im φ so xu × xv = (Im φ × φ)/2. Likewise, since the parameterization is conformal, \|xu × xv\| = \|xu\| \|xv\| = \|φ\|2/2. It follows that N = (Im φ × φ)/\|φ\|2. Stereographic projection from the north pole sends S2 to C by St : (a, b, c) →(a + ib)/(1 −c) Thus the composition G := St ◦N ◦x : Ω→C is given by the formula G(u, v) = 2 Im φ2φ3 + 2i Im φ3φ1 \|φ\|2 −2 Im φ1φ2 = φ3 φ1 −iφ2 = g Corollary 1.7 (Gauss map conformal). If Σ is a minimal surface in R3, the Gauss map N : Σ →S2 is conformal. 1.6. Finite total curvature. Now let us suppose x : Ω→R3 with image Σ is a complete minimal surface in R3. It is natural to impose ﬁniteness conditions on x and Ω. Since Σ is minimal, we have K ≤0 everywhere. We say Σ has ﬁnite total curvature if Z Σ \|K\|darea = Z Σ −Kdarea < ∞ 8 DANNY CALEGARI The condition of ﬁnite total curvature imposes very strong constraints on Ωand x. Lemma 1.8. Suppose Σ has ﬁnite total curvature. Then Ωis of ﬁnite type; i.e. it is homeomorphic to a surface of ﬁnite genus with ﬁnitely many points removed. Proof. Fix a point p ∈Σ and let B(t) ⊂Σ be the set of q with distance at most t from p as measured intrinsically in Σ. Let ℓ(t) = length(∂B(t)). Then by Gauss-Bonnet, ℓ′(t) = 2πχ(B(t)) − R B(t) Kdarea. If Ωis not of ﬁnite type, then χ(B(t)) →−∞. But then ℓ(t) < 0 for large t which is absurd. □ Lemma 1.9. Suppose Σ has ﬁnite total curvature. Then Ωhas parabolic ends; i.e. Ωis conformally equivalent to a closed surface of ﬁnite genus with ﬁnitely many points removed. Proof. We have already seen that Ωhas ﬁnitely many cylindrical ends. For each of these ends ℓ′′(t) →0 so ℓ′(t) →constant ≥0. We estimate the modulus using the method of extremal length. For simplicity suppose there is one annular component, namely B(t) − B(s) for some ﬁxed s. Suppose lim ℓ′(t) = constant > 0. Let Γ be the system of curves from ∂(B(s)) to ∂(B(t)). Using ρ(t) = 1/ℓ(t) gives Lρ(Γ) := inf γ∈Γ Z γ ρ dlength ∼log(t) and Aρ := Z B(t)−B(s) ρ2darea ∼log(t) The modulus of the annulus B(t)−B(s) is the supremum supρ Lρ(Γ)2/Aρ over all ρ; hence as t →∞the modulus goes to inﬁnity, proving that the end is parabolic. The case lim ℓ′(t) = 0 is proved similarly using ρ = 1. □ Theorem 1.10 (Osserman). Suppose Σ has ﬁnite total curvature. Then Ωis conformally equivalent to a closed surface Ωof ﬁnite genus, with ﬁnitely many points removed. Conse-quently the Gauss map extends to a holomorphic map g : Ω→CP1, and the total curvature of Σ is an integer multiple of 4π. Proof. The curvature K is the pullback of the area form on the unit sphere under the Gauss map. Since Σ has ﬁnite total curvature, g does not have an essential singularity at each of the punctures of Ω, and therefore it extends holomorphically over each puncture, and g : Ω→CP1 is a branched cover. Thus the total curvature is the degree of g (which is an integer) times the area of the unit sphere. □ 2. First variation formula In this section we derive the ﬁrst variation formula, which characterizes those submani-folds that are critical for the volume functional among compactly supported variations: the critical submanifolds are precisely those with vanishing mean curvature. We also develop some elements of the theory of Riemannian geometry in coordinate-free language (as far as possible). 2.1. Grad, div, curl. There are many natural diﬀerential operators on functions, vector ﬁelds, and diﬀerential forms on R3 which make use of many implicit isomorphisms between various spaces. This can make it confusing to ﬁgure out the analogs of these operators on Riemannian 3-manifolds (or Riemannian manifolds of other dimensions). In this section CHAPTER 3: MINIMAL SURFACES 9 we recall the co-ordinate free deﬁnitions of some of these operators, which generalize the familiar case of Euclidean R3. 2.1.1. Gradient. On a smooth manifold M, there is a natural diﬀerential operator d, which operates on functions and forms of all orders. By deﬁnition, if f is a smooth function, d f is the 1-form such that for all vector ﬁelds X, there is an identity d f(X) = Xf Where f is nondegenerate, the kernel of d f is a hyperplane ﬁeld, which is simply the tangent space to the level set of f through the given point. If M is a Riemannian manifold with inner product denoted ⟨·, ·⟩, there are natural isomorphisms between 1-forms and vector ﬁelds (called the sharp and the ﬂat isomorphisms) deﬁned by ⟨α♯, X⟩= α(X) for a 1-form α and a vector ﬁeld X, and X♭(Y ) = ⟨X, Y ⟩ for vector ﬁelds X and Y . Using these isomorphisms, for any function f on a Riemannian manifold the gradient, denoted grad(f) or sometimes ∇f, is the vector ﬁeld deﬁned by the formula grad(f) := (d f)♯ In other words, grad(f) is the unique vector ﬁeld such that, for any other vector ﬁeld X, we have ⟨grad(f), X⟩= d f(X) For any 1-form α, the vector ﬁeld α♯is perpendicular to the hyperplane ﬁeld ker(α); thus grad(f) is perpendicular to the level sets of f, and points in the direction in which f is increasing, with size proportional to the rate at which f increases. The zeros of the gradient are the critical points of f; for instance, grad(f) vanishes at the minimum and the maximum of f. 2.1.2. Divergence. On an oriented Riemannian n-manifold there is a volume form dvol, and a Hodge star ∗taking k-forms to (n −k)-forms, satisfying α ∧∗α = ∥α∥2dvol This does not deﬁne ∗α uniquely; we must further add that ∗α is orthogonal (with respect to the pointwise inner product on (n−k)-forms) to the subspace of forms β with α∧β = 0. In other words, ∗α is the form of smallest (pointwise) norm subject to α ∧∗α = ∥α∥2ω. With this notation, ∗dvol is the constant function 1; conversely for any smooth function f, we have ∗f = fdvol. If X is a smooth vector ﬁeld, then (at least locally and for short time) ﬂow along X determines a 1-parameter family of diﬀeomorphisms φ(X)t. The Lie derivative of a (contravariant) tensor ﬁeld α, denoted LXα, is by deﬁnition LXα = d dt t=0φ(X)∗ tα For forms α it satisﬁes the Cartan formula LXα = ιXdα + dιXα 10 DANNY CALEGARI where ιXβ is the interior product of X with a form β (i.e. the form obtained by contracting β with X). The divergence of a vector ﬁeld X, denoted div(X) (or sometimes −∇∗X or ∇· X), is the function deﬁned by the formula div(X) = ∗(LXdvol) By Cartan’s formula, LXdvol = dιXdvol, because dvol is closed. Furthermore, for any vector ﬁeld X we have the identity ιXdvol = ∗(X♭) which can be veriﬁed pointwise, since both sides depend only on the values at a point. Thus we obtain the equivalent formula div(X) = ∗d ∗(X♭) The operator −∗d ∗on 1-forms is often denoted d∗; likewise we sometimes we denote the operator −div(X) by ∇∗, on the grounds that ∇∗(X) = d∗(X♭). If X is a vector ﬁeld and f is a compactly supported function (which holds automatically for instance if M is closed) then Z M ⟨X, ∇f⟩dvol = Z M d f(X)dvol = Z M d f ∧ιXdvol Now, d(fιXdvol) = d f ∧ιXdvol + fdιXdvol = d f ∧ιXdvol + fdiv(X)dvol But if f is compactly supported, R M d(fιXdvol) = 0 and we deduce that Z M ⟨X, ∇f⟩dvol = Z M −fdiv(X)dvol So that −div (i.e. ∇∗) is a “formal” adjoint to grad (i.e. ∇), justifying the notation. The divergence of a vector ﬁeld vanishes where LXdvol = 0; i.e. where the ﬂow generated by X preserves the volume. 2.1.3. Laplacian. If f is a function, we can ﬁrst apply the gradient and then the divergence to obtain another function; this composition (or rather its negative) is the Laplacian, and is denoted ∆. In other words, ∆f = −div grad(f) = d∗d f Thus formally, ∆is a non-negative self-adjoint operator, so that we expect to be able to decompose the functions on M into a direct sum of eigenspaces with non-negative eigenvalues. Indeed, if M is closed, then L2(M) decomposes into an (inﬁnite) direct sum of the eigenspaces of ∆, which are ﬁnite dimensional, and whose eigenvalues are discrete and non-negative. A function f with ∆f = 0 is harmonic; on a closed manifold, the only harmonic functions are constants. On Euclidean space, harmonic functions satisfy the mean value property: the value of f at each point is equal to the average of f over any round ball (or sphere) centered at f. In general, the value of a harmonic function f at each point is a weighted average of the values on a ball centered at that point; in particular, a harmonic function on a compact subset of any Riemannian manifold attains its maximum (or minimum) only at points on the frontier. CHAPTER 3: MINIMAL SURFACES 11 2.1.4. Curl. Now we specialize to an oriented Riemannian 3-manifold. The operator ∗d takes 1-forms to 1-forms. Using the sharp and ﬂat operators, it induces a map from vector ﬁelds to vector ﬁelds. The curl of a vector ﬁeld X, denoted curl(X) (or sometimes ∇×X), is the vector ﬁeld deﬁned by the formula curl(X) = (∗d(X♭))♯ Notice that this satisﬁes the identities div curl(X) = ∗d ∗∗d(X♭) = 0 (because ∗2 = ±1 and d2 = 0) and curl grad(f) = (∗dd f)♯= 0 On a Riemannian manifold of arbitrary dimension, it still makes sense to talk about the 2-form d(X♭), which we can identify (using the metric) with a section of the bundle of skew-symmetric endomorphisms of the tangent space. Identifying skew-symmetric endo-morphisms with elements of the Lie algebra of the orthogonal group, we can deﬁne curl(X) in general to be the ﬁeld of inﬁnitesimal rotations corresponding to d(X♭). On a 3-manifold, the vector ﬁeld curl(X) points in the direction of the axis of this inﬁnitesimal rotation, and its magnitude is the size of the rotation. 2.1.5. Flows and parallel transport. On a Riemannian manifold, there is a unique torsion-free connection for which the metric tensor is parallel, namely the Levi-Civita connection, usually denoted ∇(when we mix the connection with gradient in formulae, we will denote the gradient by grad). If X is a vector ﬁeld on M, we can generate a 1-parameter fam-ily of automorphisms of the tangent space at each point by ﬂowing by X, then parallel transporting back along the ﬂowlines of X by the connection. The derivative of this family of automorphisms is a 1-parameter family of endomorphisms of the tangent space at each point, denoted AX. In terms of diﬀerential operators, AX := LX −∇X, and one can verify that AXY is tensorial in Y . Thus, X determines a section AX of the bundle End(TM). On an oriented Riemannian manifold, the vector space End(TpM) = T ∗ p M ⊗TpM is an o(n)-module in an obvious way, and it makes sense to decompose an endomorphism into components, corresponding to the irreducible o(n)-factors. Each endomorphism decom-poses into an antisymmetric and a symmetric part, and the symmetric part decomposes further into the trace, and the trace-free part. In this language, (1) the divergence of X is the negative of the trace of AX. As a formula, this is given pointwise by div(X)(p) = trace of V →∇V X on TpM (2) the curl of X is the skew-symmetric part of AX; and (3) the strain of X (a measure of the inﬁnitesimal failure of ﬂow by X to be conformal) is the trace-free symmetric part of AX. 12 DANNY CALEGARI 2.2. First variation formula. Let M be a Riemannian n-manifold, and let Ωbe a smooth bounded domain in Rk. Let f : Ω→M be a smooth immersion with image Σ, and let F : Ω× (−ϵ, ϵ) →M be a one-parameter variation supported in the interior of Ω. Let t denote the coordinate on (−ϵ, ϵ), and let T = dF(∂t), which we think of (at least locally) as a vector ﬁeld on M generating a family of diﬀeomorphisms φ(T)t (really we should think of T as a vector ﬁeld along F; i.e. a section of the pullback of TM to Ωby F ∗). Under this ﬂow, Σ evolves to Σ(t), and at each time is equal to F(Ω, t). The ﬂow T determines an endomorphism ﬁeld AT along f. This endomorphism de-composes into a skew-symmetric part (which rotates the tangent space to Σ but preserves volume) and a symmetric part. The derivative at t = 0 of the area of an inﬁnitesimal plane tangent to TΣ(t) is the negative of the trace of AT restricted to TΣ. As in § 2.1.5 this can be expressed as the trace of V →∇V T restricted to TΣ. If ei is an orthonormal frame ﬁeld along TΣ, we obtain a formula d dt t=0volume(Σ(t)) = Z Σ X i ⟨∇eiT, ei⟩dvol The integrand on the right hand side of this formula is sometimes abbreviated to divΣ(T). If we decompose T into a normal and tangential part as T = T ⊥+ T ⊤, this can be expressed as d dt t=0volume(Σ(t)) = Z Σ div(T ⊤) + X i ⟨∇eiT ⊥, ei⟩dvol where div(T ⊤) means the divergence in the usual sense on Σ of the vector ﬁeld T ⊤, thought of as a vector ﬁeld on Σ. But ⟨∇eiT ⊥, ei⟩= ei⟨T ⊥, ei⟩−⟨T ⊥, ∇eiei⟩= −⟨T ⊥, ∇eiei⟩ by the metric property of the Levi-Civita connection, and the fact that T ⊥is orthogonal to ei. Similarly, R Σ div(T ⊤)dvol = 0 by Stokes’ formula, because T ⊤is compactly supported in the interior. The sum H := P i ∇⊥ eiei (where ∇⊥denotes the normal part of the covariant derivative) is the mean curvature vector, which is the trace of the second fundamental form, and is normal to Σ by deﬁnition; thus ⟨T ⊥, P i ∇eiei⟩= ⟨T, H⟩. Putting this together, we obtain the ﬁrst variation formula: Proposition 2.1 (First Variation Formula). Let Σ be a compact immersed submanifold of a Riemannian manifold, and let T be a compactly supported vector ﬁeld on M along Σ. If Σ(t) is a 1-parameter family of immersed manifolds tangent at t = 0 to the variation T, then d dt t=0volume(Σ(t)) = Z Σ −⟨T, H⟩dvol Consequently, Σ is a critical point for volume among compactly supported variations if and only if the mean curvature vector H vanishes identically. This motivates the following deﬁnition: Deﬁnition 2.2. A submanifold is said to be minimal if its mean curvature vector H vanishes identically. CHAPTER 3: MINIMAL SURFACES 13 The terminology “minimal” is widely established, but the reader should be warned that minimal submanifolds (in this sense) are not always even local minima for volume. Example 2.3 (Totally geodesic submanifolds). If Σ is 1-dimensional, the mean curvature is just the geodesic curvature, so a 1-manifold is minimal if and only if it is a geodesic. A totally geodesic manifold has vanishing second fundamental form, and therefore van-ishing mean curvature, and is minimal. An equatorial sphere in Sn is an example which is minimal, but not a local minimum for volume. Warning 2.4. It is more usual to deﬁne the mean curvature of a k-dimensional submanifold Σ to be equal to 1 k P i ∇⊥ eiei. With this deﬁnition, the mean curvature is the average of the principal curvatures of Σ — i.e. the eigenvalues of the second fundamental form — rather than their sum. But the convention we adhere to seems to be common in the minimal surfaces literature; see e.g. p. 5 or p. 5. 2.3. Calibrations. Now suppose that F is a codimension 1 foliation of a manifold M. Locally we can coorient F, and let X denote the unit normal vector ﬁeld to F. For each leaf λ of the foliation we can consider a compactly supported normal variation fX, and suppose that λ(t) is a 1-parameter family tangent at t = 0 to fX. Then d dt t=0volume(λ(t)) = Z λ divλ(fX)dvol and because X is normal, this simpliﬁes (by the Leibniz rule for covariant diﬀerentiation) to d dt t=0volume(λ(t)) = Z λ fdivλ(X)dvol Because X is a normal vector ﬁeld of unit length, it satisﬁes div(X) = divλ(X). Thus we obtain the lemma: Lemma 2.5 (Normal ﬁeld volume preserving). A cooriented codimension 1 foliation F has minimal leaves if and only if the unit normal vector ﬁeld X is volume-preserving. Now, suppose F is a foliation with minimal leaves, and let X be the unit normal vector ﬁeld. It follows that the (n−1)-form ω := ιXdvol is closed. On the other hand, it evidently satisﬁes the following two properties: (1) the restriction of ω to TF is equal to the volume form on leaves; and (2) the restriction of ω to any (n −1) plane not tangent to TF has norm strictly less than the volume form on that plane. Such a form ω is said to calibrate the foliation. Lemma 2.6. Let F be a foliation with minimal leaves. Then leaves of F are globally area minimizing, among all compactly supported variations in the same relative homology class. Proof. Let λ be a leaf, and let µ be obtained from λ by cutting out some submanifold and replacing it by another homologous submanifold. Then volume(µ) ≥ Z µ ω = Z λ ω = volume(λ) where the middle equality follows because ω is closed, and the inequality is strict unless µ is tangent to F. But µ agrees with λ outside a compact part; so in this case µ = λ. □ 14 DANNY CALEGARI Example 2.7. Let Σ be an immersed minimal (n −1)-manifold in Rn. Let p ∈Σ be the center of a round disk D in the tangent space TpΣ. Let C be the cylindrical region obtained by translating D normal to itself. Then C ∩Σ is a graph over D, and we can foliate C by parallel copies of C ∩Σ, translated in the normal direction. Thus there exists a calibration ω deﬁned on C, and we see that C ∩Σ is least volume among all surfaces in C obtained by a compactly supported variation. But C is convex, so the nearest point projection to C is volume non-increasing. We deduce that any immersed minimal (n −1)-manifold in Rn is locally volume minimizing. This should be compared to the fact that geodesics in any Riemannian manifold are locally distance minimizing. 2.4. Gauss equations. Recall that the second fundamental form on a submanifold Σ of a Riemannian manifold M is the symmetric vector-valued bilinear form II(X, Y ) := ∇⊥ XY and H is the trace of II; i.e. H = P i II(ei, ei) where ei is an orthonormal basis for TΣ. The Gauss equation for X, Y vector ﬁelds on Σ is the equation KΣ(X, Y )\|X ∧Y \|2 = KM(X, Y )\|X ∧Y \|2 + ⟨II(X, X), II(Y, Y )⟩−\|II(X, Y )\|2 where \|X ∧Y \|2 := \|X\|2\|Y \|2−⟨X, Y ⟩2 is the square of the area of the parallelogram spanned by X and Y , and K(X, Y ) := ⟨R(X, Y )Y, X⟩ \|X ∧Y \|2 is the sectional curvature in the plane spanned by X and Y , where R(X, Y )Z := ∇X∇Y Z− ∇Y ∇XZ −∇[X,Y ]Z is the curvature tensor. The subscripts KΣ and KM denote sectional curvature as measured in Σ and in M respectively; the diﬀerence is that in the latter case curvature is measured using the Levi-Civita connection ∇on M, whereas in the former it is measured using the Levi-Civita connection on Σ, which is ∇⊤:= ∇−∇⊥. If Σ is a 2-dimensional surface in a 3-manifold M, then we can take X and Y to be the directions of principal curvature on Σ (i.e. the unit eigenvectors for the second fundamental form). If we coorient Σ, the unit normal ﬁeld lets us express II as an R-valued quadratic form, and the principal curvatures are real eigenvalues k1 and k2. Since H = k1 + k2, for Σ a minimal surface we have k1 = −k2 and (2.1) KΣ = KM −\|II\|2 2 Where \|II\|2 := P i,j \|II(ei, ej)\|2. In other words, a minimal surface in a 3-manifold has curvature pointwise bounded above by the curvature of the ambient manifold. So for example, if M has non-positive curvature, the same is true of Σ. 3. Second variation formula The ﬁrst variation formula shows that minimal surfaces are critical for volume, among smooth variations, compactly supported in the interior. To determine the index of these critical surfaces requires the computation of the second variation. In this section, we derive the second variation formula and some of its consequences. CHAPTER 3: MINIMAL SURFACES 15 3.1. Second variation formula. We specialize to the case that Σ is a hypersurface in a Riemannian manifold. We further restrict attention to variations in the normal direction (this is reasonable, since a small variation of Σ supported in the interior will be transverse to the exponentiated normal bundle). Denote the unit normal vector ﬁeld along Σ by N, and extend N into a neighborhood along normal geodesics, so that ∇NN = 0. Let F : Ω× (−ϵ, ϵ) →M satisfy F(·, 0) : Ω→Σ, and if t parameterizes the interval (−ϵ, ϵ), there is a smooth function f on Ω×(−ϵ, ϵ) with compact support so that T := dF(∂t) = fN. Then deﬁne Σ(t) := F(Ω, t). Let ei be vector ﬁelds deﬁned locally on Ωso that dF(·, 0)(ei) are an orthonormal frame on Σ locally, and extend them to vector ﬁelds on Ω× (−ϵ, ϵ) so that they are constant in the t direction; i.e. they are tangent to Ω× t for each ﬁxed t, and satisfy [ei, ∂t] = 0 for all i. By abuse of notation we denote the pushforward of the ei by dF also as ei, and think of them as vector ﬁelds on Σ(t) for each t. For each point p ∈Σ corresponding to a point q ∈Ωthe curve F(q×(−ϵ, ϵ)) is contained in the normal geodesic to Σ through p, and is parameterized by t. Along this curve we deﬁne g(t) to be the matrix whose ij-entry is the function ⟨ei, ej⟩(where we take the inner product in M). The inﬁnitesimal parallelepiped spanned by the ei at each point in Σ(t) has volume p det(g(t)). Projecting along the ﬁbers of the variation, we can push this forward to a density on Σ(0) which may be integrated against the volume form to give the volume of Σ(t); thus volume(Σ(t)) = Z Σ(0) p det(g(t))dvol We now compute the second variation of volume. Taking second derivatives, we obtain d2 dt2 t=0volume(Σ(t)) = Z Σ d2 dt2 t=0 p det(g(t))dvol Suppose further that Σ is a minimal surface, so that det′(g)(0) = tr(g′(0)) = 0. Then since det(g(0)) = 1, we have d2 dt2 t=0 p det(g(t)) = 1 2 d2 dt2 t=0 det(g(t)) By expanding g(t) in a Taylor series, we can compute 1 2 d2 dt2 t=0 det(g(t)) = 1 2tr(g′′(0)) + σ2(g′(0)) where tr means trace, and for a matrix M with eigenvalues κi we have σ2(M) = P i 1 we can ﬁnd eigenfunctions u1, u2 for which some nonconstant linear combination changes sign somewhere, contrary to the above. □ Eigenfunctions of L with diﬀerent eigenvalues are orthogonal; consequently only the eigenfunction of lease eigenvalue does not change sign. 18 DANNY CALEGARI Corollary 3.6. If Σ is stable with trivial normal bundle, so is any cover π : ˜ Σ →Σ. Proof. If ˜ Σ is unstable, there is some compactly supported f with ⟨Lf, f⟩< 0, and by Proposition 3.5 we can take f ≥0 everywhere. Deﬁne g on Σ by g(p) = P q∈π−1(p) f(q). Since f is compactly supported, g is ﬁnite and compactly supported, and ⟨Lg, g⟩< 0 so Σ is unstable. □ 3.1.2. Stability controls area growth. Colding-Minicozzi Thm. 2.1 derive a simple upper bound on the area growth for a stable minimal surface in a Riemannian 3-manifold. They consider D an immersed disk in M of intrinsic radius r, and a non-negative operator of the form ∆Σ −ν + κ + c1KΣ where c1 > (1 + κr2)/2, and then obtain an estimate of the form (3.1) area(D) r2 + c2 2πc1 Z D ν 1 −s r 2 ds ≤c2 for c2 := 2πc1/(2c1 −1 −κr2) (here s is the intrinsic radial coordinate on the disk D). To apply this to a stable minimal surface, we can take ν = Ric(N), κ ≥2\|KM\| and c1 = 2, at least for r suﬃciently small depending on κ. For M = R3 we have ν = κ = 0 and c1 = 2 works unconditionally. In this form the estimate of Colding-Minicozzi gives: Proposition 3.7 (Colding-Minicozzi). Let D be a stable minimal disk in R3 of radius r in its intrinsic metric. Then πr2 ≤area(D) ≤4/3 πr2 Proof. For any minimal surface in R3, stable or not, we have K ≤0 pointwise, so Gauss-Bonnet gives πr2 ≤area(D). The other bound needs stability. For a stable minimal surface in R3 the operator ∆Σ + 2KΣ is non-negative; i.e. for any compactly supported test function f on Σ we have 0 ≤ Z Σ \|gradΣf\|2darea + 2 Z Σ KΣf 2darea Let f(s, θ) = η(s) be a test function on D depending only on the radius s. Since f is constant as a function of θ, we integrate out the θ direction to obtain the inequality 0 ≤ Z r 0 (η′(s))2ℓ(s)ds + 2 Z r 0 K′(s)η2(s)ds where K(s) denotes the integral of K over the disk D(s) of radius s, and ℓ(s) is the length of its boundary ∂D(s). Now, Gauss-Bonnet says that ℓ′(s) = 2π −K(s). Thus integrating by parts we get 0 ≤ Z r 0 (η′(s))2ℓ(s)ds −2 Z r 0 (2π −ℓ′(s))(η2(s))′ds Make the explicit choice η(s) := 1 −s/r so that η′ = −1/r and (η2)′ = −2/r(1 −s/r). Substituting gives −1 r2 Z r 0 ℓ(s)ds + 4 r Z r 0 ℓ′(s)(1 −s/r)ds ≤8π r Z r 0 (1 −s/r)ds = 4π CHAPTER 3: MINIMAL SURFACES 19 Whereas integration by parts shows LHS = 3 r2 Z r 0 ℓ(s)ds = 3 r2area(D) ≤4π □ 3.2. Stable complete minimal surfaces in R3 are planes. Colding-Minicozzi’s esti-mate can be used to give a short proof of a famous theorem of do Carmo and Peng : Theorem 3.8 (do Carmo-Peng). Let x : Ω→R3 be an oriented stable complete minimal immersion. Then x(Ω) is a plane. Proof. By Corollary 3.6, Stability passes to covering spaces, so we may assume Ωis topo-logically a plane. Proposition 3.7 and the coarea formula implies that ℓ′(t) ≤8/3π for big t, so the total curvature is bounded by 2/3π, and is therefore ﬁnite. But Theorem 1.10 says that the total curvature is an integer multiple of 4π, so the only possibility is that K is identically zero, so that g is constant and x(Ω) is a plane. □ Every graph is calibrated, and therefore stable. Thus this recovers in a special case a theorem of Bernstein that the only complete minimal graph in R3 is a plane. 3.3. Schoen’s curvature inequality. All minimal surfaces satisfy a pointwise upper curvature bound, by equation 2.4. In general there can be no pointwise lower curvature bounds, even for stable minimal surfaces. For example, the graph of a complex polynomial p : C →C in C2 is calibrated, and therefore is always a stable minimal surface. By scaling the graph by a homothety we can ﬁnd a sequence of such surfaces with K →−∞. However in dimension three, Schoen derived lower pointwise curvature bounds for a stable minimal surface at a point far from the boundary. Theorem 3.9 (Schoen). Let M be a closed 3-manifold, and Σ a stable minimal surface. Given r ∈(0, 1] and p ∈Σ such that Br(p)∩Σ has compact closure in Σ, there is a constant C depending only on the supremum of \|KM\| and \|∇KM\| on Br(p), and on r, such that \|II\|2(p) ≤Cr−2 Moreover, there is a constant ϵ > 0 depending on the same data so that Σ ∩Bϵr(p) is a union of embedded disks. For stable minimal surfaces in R3 this simpliﬁes as follows: Theorem 3.10 (Schoen). There exists a positive constant C > 0 such that for any stable minimal surface Σ in R3 and any point p ∈Σ with distance at least r to ∂Σ, there is an estimate KΣ(p) ≥−Cr−2 Theorem 3.9 can be derived from Colding-Minicozzi’s Equation 3.1. Theorem 3.10 follows from Theorem 3.8 as follows: a sequence of stable minimal surfaces in R3 with Kr2 →−∞ can be rescaled so that K →−1 (say) and r →∞. The limit is a proper stable minimal surface which is not a plane; this is a contradiction. Schoen’s Theorem implies a kind of compactness property for stable minimal surfaces, and has many applications to the theory of 3-manifolds. It implies that in a ﬁxed (closed) 20 DANNY CALEGARI 3-manifold M, any sequence of stable minimal surfaces Σi with points pi ∈Σi at which the Σi have injectivity radius ≥r, has a subsequence that converges on compact subsets to a stable minimal surface 4. Existence of minimal surfaces In this section we discuss techniques to demonstrate the existence of minimal surfaces satisfying certain geometric and topological conditions. 4.1. Björling’s Problem and the Schwarz surface. The following is known historically as Björling’s Problem: Problem 4.1 (Björling). Given a real analytic curve γ in R3 and a real analytic unit normal vector ﬁeld ν along γ, construct a minimal surface Σ containing γ, with normal ﬁeld extending ν. This problem was solved by Hermann Schwarz, using techniques from complex function theory. His solution is as follows. Suppose γ : I ⊂R ⊂C →R3 is parameterized to be unit speed. Then ν × γ′ is a unit normal vector ﬁeld along γ that must be tangent to such a surface Σ and perpendicular to ν. Now, since γ and ν are real analytic, they admit holomorphic extensions γC, νC : Ω→C3 for some open neighborhood Ωof I in C. If we let z denote the complex coordinate on Ω, and deﬁne x(z) := Re γC(z) −i Z z 0 νC(ω) × γ′ C(ω)dω then x has coordinates which are harmonic functions of z, and x(t) = γ(t) along I. If we deﬁne xC(z) := γC(z) −i Z z 0 νC(ω) × γ′ C(ω)dω then γ′(z) = Rex′ C(z) = x′ and Imx′ C(z) = −ν(z) × γ′(z) along γ(I). Hence ⟨x′ C, x′ C⟩ vanishes identically on I, and therefore also on Ω(because it is holomorphic). But if we deﬁne x′ C = (φ1, φ2, φ3) this implies the φj satisfy the minimal surface equation, and x solves the Björling Problem. In fact, if we run this argument in reverse, it is easy to see that x deﬁnes the most general form of a solution to the Björling Problem. From this one obtains the following corollaries: Corollary 4.2 (Schwarz). (1) Every straight line contained in a minimal surface is an axis of rotational symmetry of the surface; and (2) if a minimal surface intersects a plane perpendicularly, then that plane is a plane of reﬂectional symmetry of the surface. These corollaries follow immediately from the symmetry of the surface and the Schwarz reﬂection principle (which was invented historically in exactly this context). This corollary motivates the search for real analytic minimal surfaces Σ bounded by a Schwarz chain; i.e. a cyclic sequence of straight line segments and ﬂat planes, where we insist that ∂Σ should contain each straight line segment, and be perpendicular to each ﬂat plane. Such a surface can then be continued indeﬁnitely by repeated reﬂection and rotation in the planes and segments. CHAPTER 3: MINIMAL SURFACES 21 Example 4.3 (Schwarz surface). We describe in detail a beautiful classical example discov-ered by Schwarz. Conider the quadrilateral in R3 composed of four straight segments between the four vertices A, B, C, D in cyclic order of a regular tetrahedron. For example, we could take A = (1, 1, 1), B = (1, −1, −1), C = (−1, −1, 1), D = (−1, 1, −1) We seek a piece of minimal surface Q that is topologically a disk, and is bounded by this quadrilateral. See Figure 2. Figure 2. The quadrilateral Q At each vertex the surface Q has a corner with an angle of 60◦. The surface may be continued by rotating copies of Q around each boundary edge successively. Thus, six copies of Q ﬁt together around each vertex, three of them pointing ‘up’ and three pointing ‘down’. By repeated applications of the reﬂection principle, Q may be continued to a triply periodic properly embedded surface ˜ Σ in R3 called the D surface. To say that it is triply periodic means that there is a lattice Λ of translations in R3 leaving ˜ Σ invariant. The quotient ˜ Σ/Λ = Σ is a closed surface of genus 3, that we may think of as an embedded minimal surface in the 3-torus R3/Λ. Note that Λ is isomorphic to Z3 as an abstract group. The symmetries of ˜ Σ and Σ will let us determine the functions g, f in the Weierstrass– Enneper parameterization explicitly (see § 1.5). First we consider g, which we think of as the Gauss map g : Q →S2 under the usual stereographic identiﬁcation of S2 with CP1. Because ∂Q is the union of four straight segments AB, BC, CD, DA, the normals to Q along ∂Q are segments of great circles in S2. In fact, these great circles are precisely four of the six great circles one obtains by extending the (geodesic) edges of a spherical “cube”. Thus, the images of A, B, C, D under g are the vertices of one face of a regular cube inscribed in S2, and g(Q) is a regular spherical quadrilateral with angles of 120◦at each vertex. By extension we get g : Σ →S2, which is a double branched cover, with branch points the eight vertices of an inscribed cube. Four of these vertices are the images of A, B, C, D under g. They are the points ± √ 3 −1 √ 2 and ± √ 3 −1 √ 2 i 22 DANNY CALEGARI The other four vertices are ± √ 3 + 1 √ 2 and ± √ 3 + 1 √ 2 i We denote these eight points by ωj for j = 1, · · · , 8. To simplify matters we use the Gauss map g itself to deﬁne holomorphic coordinates on Σ, at least where it is nonsingular. By abuse of notation we write f = F(g) where f is as in the Weierstrass–Enneper parameterization. Since ∞is not a branch point for g, the 1-form F(g)dg should have a zero of order 2 at g = ∞, and should therefore be of the form F(ω) ∼ω−4 near inﬁnity. Since it has order 2 branch points at the ωj the only possibility is F(ω) := K Q j(ω −ωj)1/2 = K √ ω8 −14ω4 + 1 for some as yet undetermined constant K. To determine K we need to evaluate the periods of the one forms φ1 = K(1 −ω2) 2 √ ω8 −14ω4 + 1dω, φ2 = iK(1 + ω2) 2 √ ω8 −14ω4 + 1dω, φ3 = Kω √ ω8 −14ω4 + 1dω on the surface Σ. Taking K real ensures that the periods of the φj are pure imaginary on three suitable generators for the homology of Σ; taking K imaginary makes the period pure imaginary on the other three generators, and produces a conjugate triply periodic surface, called the P surface. In 1968, Alan Schoen , an engineer at NASA, discovered that taking K = eiθ where θ ∼0.66348 gives another triply-periodic embedded surface, known as the gyroid, or G surface, which contains neither straight lines nor planes of reﬂection symmetry. Figure 3. Associates of Q in the P and G surfaces 4.2. Plateau problem. The Plateau problem asks for the surface of least area spanned by a given closed Jordan curve Γ (in R3 if nothing more is speciﬁed). In this formulation the topology of the surface is unspeciﬁed; also left in the air is the question whether the surface should be embedded or not. Finally, unless Γ is rectiﬁable we cannot expect it to bound any surface of ﬁnite area, in which case it seems meaningless to ask for one of ‘least’ area. CHAPTER 3: MINIMAL SURFACES 23 Thus to simplify matters we insist ﬁrst that Γ should be rectiﬁable, second that the surface we are seeking should be topologically a disk, and thirdly that we do not insist that the surface is embedded or even immersed. In other words we give the Plateau problem the following formulation: Problem 4.4 (Plateau). Given a rectiﬁable Jordan curve Γ in R3, construct a map x : B →R3 from the closed unit disk to R3 so that x : ∂B →Γ is an orientation-preserving homeo-morphism, and so that x is a generalized minimal surface on B. Recall from Proposition 1.4 that x\|B is a generalized minimal surface providing it is conformal (i.e. xu ⊥xv and \|xu\|2 = \|xv\|2 everywhere) and the coordinate functions are harmonic (i.e. ∆x = 0). The solution, due to Douglas and independently to Rado , is to ﬁnd a suitable class C(Γ) of maps x : B →R3 with good compactness properties, and to show that the minimizer of a certain functional in C(Γ) solves Plateau’s problem. The trick is to decide on the correct functional. The function A(x) which assigns to a map the area of its image is no good for several reasons. First of all, there are too many symmetries: any reparame-terization of the domain produces a map with the same area. Second, the area functional is too degenerate: a map of a disk can be modiﬁed dramatically by adding ‘hair’ with very small 2-dimensional measure but very large diameter. Understanding and controlling the geometric limits of such objects is subtle. Douglas’s idea was to use a diﬀerent functional D(x), which measures energy instead of area; the minimizer of this functional at once ﬁnds the surface with the smallest area, and the most eﬃcient parameterization. A similar philosophy is familiar in the theory of Riemannian geometry, where we seek to minimize energy rather than length to deﬁne geodesics. 4.2.1. The Dirichlet integral and the class C∗(Γ). Naively we could start by deﬁning C(Γ) to be the class of maps from the closed disk to R3, smooth in the interior, which restrict to orientation-preserving homeomorphisms ∂B →Γ. However, limits of sequences of smooth maps are rarely smooth, and limits of homeomorphisms are not always homeomorphisms. Thus we extend the class C(Γ) in two ways: ﬁrst we allow maps which on B are in L2; this means the Dirichlet integral D(x) := 1 2 Z B \|xu\|2 + \|xv\|2dudv is ﬁnite (where the derivatives are taken in the sense of distribution). The quantity D(x) is called the energy of the map x, and formalizes the physical intuition of elastic energy of a membrane. Second we allow maps for which ∂B →Γ is merely monotone; this means that it is surjective, and the point preimages are connected. A monotone map between circles might degenerate the cyclic order of triples of points, but it does not reverse them. It will turn out that a generalized minimal surface whose boundary parameterization is monotone is actually a homeomorphism; we return to this in the sequel. A further source of noncompactness comes from the group of symmetries of D. The integrand of D is invariant under conformal reparameterizations of the domain; thus if ϕ : B →B is any (orientation-preserving) conformal automorphism, we have D(x) = D(x◦ϕ). 24 DANNY CALEGARI The group of conformal automorphisms of the unit disk is noncompact. Therefore it makes sense to break this symmetry by choosing three points p1, p2, p3 ∈∂B and three points q1, q2, q3 ∈Γ in the same cyclic order, and restricting to the class C∗(Γ) of maps as above for which x(pj) = qj. 4.2.2. Harmonic extension of boundary values. Now, for a given monotone map x : ∂B →Γ there is a unique extension to x : B →R3 which minimizes D(x), namely the harmonic extension; i.e. the map for which all the coordinate functions are harmonic. If we identify B with the hyperbolic plane, and ∂B with the circle at inﬁnity, any measurable function f on ∂B extends to a unique harmonic function on the interior as follows. At each point p ∈B exponentiation of geodesics deﬁnes a bijection between the unit tangent vectors UTpB to p and ∂B. Under this identiﬁcation the value of the extension at p is equal to the average of f on ∂B with respect to the angular measure on UTpB. If xi →x measurably on ∂B, their harmonic extensions converge also, and if by abuse of notation we use the same symbol to denote these extensions, then D(xi) →D(x) when these quantities are ﬁnite. Now there is one last source of potential non-compactness we must contend with. An orientation-preserving homeomorphism y : ∂B →Γ determines its graph g(y) ⊂∂B ×Γ. If we think of ∂B×Γ as a torus, the graph g(y) is a (1, 1)-curve that intersects every meridian and every longitude in exactly one point. Any sequence of homeomorphisms yj : ∂B →Γ has a subsequence for which the graphs g(yj) converge to some (1, 1)-curve g, but g might have horizontal or vertical segments; see Figure 4. Figure 4. the black curve is the graph of a homeomorphism; the red curve is the limit of a sequence of graphs but has both horizontal and vertical segments A horizontal segment is where the limiting map is monotone but not locally injective; a vertical segment indicates a discontinuity. Note that because we have normalized our boundary maps to take pj to qj, a limiting map cannot consist solely of the union of a CHAPTER 3: MINIMAL SURFACES 25 meridian and a longitude. In other words, a vertical segment in the limit graph, if it exists, has distinct endpoints. Our concern is for the possible appearance of vertical segments in g, since such a g is not the graph of a continuous function. So suppose we have a sequence of orientation-preserving homeomorphisms xi : ∂B →Γ for which the graphs g(xi) accumulate on a vertical segment. If we let x : ∂B →Γ denote the merely measurable map which is a pointwise limit of the xi then the vertical segment indicates that there is some point q ∈∂B for which the left and right limits x±(q) exist and are not equal. Since x is measurable, it still has a harmonic extension. We claim that for such an x, the harmonic extension has D(x) = ∞, and therefore D(xi) →∞. To see this, we use the conformal invariance of D. Let ϕ be a conformal automorphism of B which is a hyperbolic isometry with attracting ﬁxed point q and repelling ﬁxed point r. Consider the harmonic extensions of the sequence of maps xn := x ◦ϕn. The boundary maps converge to a step function taking the two values x± on the intervals of ∂B −{q, r}. Let y be the harmonic extension of this step function. If we let B(1/2) denote the ball of radius 1/2, then 1 2 Z B(1/2) \|yu\|2 + \|yv\|2dudv = K > 0 Choose ϕ as above so that the translates ϕn(B(1/2)) are all disjoint. We can estimate D(x) ≥ X n 1 2 Z B(1/2) \|xn u\|2 + \|xn v\|2dudv But when n is big, each of the integrals has value close to K. Hence D(x) = ∞, proving the claim. Now let xj ∈C∗(Γ) be a sequence of maps with D(xj) →infx∈C∗(Γ) D(x). By the discussion above, there is a subsequence whose restriction to ∂B converges to a map x : ∂B →Γ which is continuous and monotone. For each xj, let yj denote the harmonic extension of xj\|∂B. Then D(yj) ≤D(xj) and yj →y, the harmonic extension of x. Thus D(y) = lim D(yj) = inf x∈C∗(Γ) D(x) 4.2.3. Energy versus Area. What is the relation between D(x) and A(x), the area of the surface x(B)? We can write A(x) := 1 2 Z B \|xu × xv\|dudv Comparing integrands pointwise, we see that A(x) ≤D(x) with equality iﬀxu ⊥xv and \|xu\| = \|xv\| everywhere; i.e. precisely if x is conformal. From this it follows that any y which minimizes the Dirichlet integral in C∗(Γ) is confor-mal and, since it is also harmonic, it is minimal. For, suppose y is not conformal. Then we have A(y) < D(y). But we can pull back the conformal structure on the image y(B), and using the fact that the resulting surface is conformally equivalent to the standard disk, we can ﬁnd a (quasiconformal) homeomorphism ϕ : B →B so that y ◦ϕ is conformal, and lies in C∗(Γ). 26 DANNY CALEGARI Now, reparameterizing a surface does not change its area, so A(y ◦ϕ) = A(y). On the other hand, since y ◦ϕ is conformal, D(y ◦ϕ) = A(y ◦ϕ) < D(y) contrary to the assumption that y minimized D. We have almost solved Plateau’s problem, in that we have constructed a generalized minimal surface y with boundary on Γ, except that a priori y\|∂B might not be a home-omorphism but merely monotone. It turns out this possibility cannot occur. If y were constant on some interval I ⊂∂B we could apply the reﬂection principle to continue y across I. But this would give a minimal surface which is constant on an interval in the interior, which is impossible. Note by this discussion that the map y we obtain is also a global minimizer for A. Putting this together we deduce: Theorem 4.5 (Douglas-Rado). Any rectiﬁable Jordan curve Γ in R3 is the boundary of a generalized minimal surface with the topology of the disk, which furthermore minimizes area among all disks with boundary Γ. Figure 5. A minimal disk bounding an approximation of a Hilbert curve in the boundary of a cube 4.2.4. Morrey’s theorem. The Douglas-Rado theorem was generalized in 1948 by Morrey . A Riemannian manifold M is said to be homogeneously regular if there is a uniform upper bound on the sectional curvature KM, and a uniform lower bound on the injectiv-ity radius (note that this implies M is complete). For example: any closed Riemannian manifold, or any covering space of such a manifold, is homogeneously regular. A proof of Morrey’s theorem parallel to Douglas’ argument can be carried out by intro-ducing the energy E(f) of a map f : B →M with ∂f : ∂B →Γ, a quantity that generalizes the Dirichlet integral when M is Euclidean space. An f extending ∂f and minimizing E(f) is said to be harmonic, and a map that is harmonic and conformal is minimal. Morrey’s argument is more indirect than this sketch indicates, but a proof along these lines can be obtained by the (more general) methods outlined in § 4.4. CHAPTER 3: MINIMAL SURFACES 27 4.3. Branch points and Gulliver-Osserman’s Theorem. A generalized minimal sur-face x : Ω→R3 is and immersion away from an isolated set of branch points where dx vanishes. These branch points come in two kinds: (1) p ∈Ωis a false branch point if there is a neighborhood U of p so that x(U) is a smooth embedded surface; and (2) p ∈Ωis a true branch point otherwise. A false branch point is really an artefact of the parameterization. We say a minimal surface is locally least area if its area cannot be strictly decreased by an arbitrarily small compactly supported variation. A locally least area surface is necessarily stable, but not conversely. In high dimensions locally least area surfaces can have branch points, but for surfaces in dimension three, one has the following: Theorem 4.6 (Gulliver-Osserman). A locally least area surface in a 3-manifold has no interior branch points. The history of this theorem is that Osserman showed that a locally least area disk in R3 has no true branch points, and then Gulliver extended this result to all branch points, and to all 3-manifolds. The fundamental idea is as follows. Near a true branch point p, Osserman shows a minimal surface must have a transverse arc α of self-intersection. Let us leave the proof of this for the moment, and indicate how to modify the map along α near p. The case of a branch point of ‘order 2’ is easiest to understand, and is indicated in Figure 6. On the left is an immersed disk with an arc α of intersection emanating from a single branch point p. On the right is a pair of embedded disks which osculate along a single point on the boundary. We may cut out a neighborhood as in the left, replace it with a neighborhood as on the right, and modify a map of a surface locally so as to reduce area. One way to think of this modiﬁcation is that we have taken the branch point p and ‘pushed’ it along the arc α where the single branch sheet crosses itself, unzipping the arc of self-intersection and regluing to make two non-singular arcs on the top and bottom sheet. Figure 6. Push the branch point along an arc of self-intersection and smooth to reduce area The modiﬁcation of the domain is indicated in Figure 7. On the left we have the disk whose diameter maps to α by a two-to-one map branched over the midpoint, which maps to p. We cut open the disk along this diameter, and ‘buckle’ it outwards to form a diamond-shaped hole, then collapse the diamond along the other axis to form two disks. Note that this modiﬁcation can be performed whenever there is a branch point p that ends at an arc of intersection α with (at least) two disjoint preimages in the domain. 28 DANNY CALEGARI Figure 7. A disk is cut along a diameter, to form a diamond-shape hole which buckles out and then collapses to form two disks To demonstrate the existence of a transverse arc of self-intersection, Osserman considers the Weierstrass parameterization near a branch point in terms of holomorphic functions f and g. Let the branch point be the origin (in the domain), and suppose that the order of vanishing of f and g at the origin are l and m respectively. Then Osserman shows (, eq. 3.13 and 3.14) that if we choose a local holomorphic co-ordinate w on the domain so that g(w) = wm, the coordinates xj of the map have the form (4.1) x1 + ix2 = alwl + · · · + al+2m−1wl+2m−1 + al+2mwl+2m − l l + 2malwl+2m + · · · and (4.2) x3 = Re l l + malwl+m + · · · for certain coeﬃcients aj. From the form of this equation one can show either that x multiply covers its image (i.e. we are at a fake branch point) or there is a δ > 0 with the following property: whenever 0 < \|w1\|, \|w2\| < δ are distinct points with the same image x(w1) = x(w2), then g(w1) ̸= g(w2); i.e. the surface intersects itself transversely. It follows that suﬃciently near the branch point, the subset where the surface intersects itself consists of a discrete collection of pairs of real analytic arcs. For topological reasons there must be self-intersections arbitrarily near the branch point; thus if we analytically continue the (ﬁnitely many!) pairs of arcs of self-intersection within the disk \|w\| < δ, at least one pair can be continued all the way to 0. Thus the minimal surfaces constructed by Douglas have no interior branch points. 4.4. Minimal surfaces in Riemannian manifolds. Let M be a closed Riemannian manifold, and let f : Σg →M be a continuous map where Σg is a closed oriented surface of genus g. It is natural to ask: when is there a map h : Σg →M, homotopic to f, for which h(Σg) is a minimal surface? A rather satisfying answer was obtained independently by Schoen-Yau and Sacks-Uhlenbeck : Theorem 4.7 (Schoen-Yau, Sacks-Uhlenbeck). Let M be a closed Riemannian manifold. Suppose that f : Σg →M is a continuous map such that f∗: π1(Σg) →π1(M) has no conjugacy class corresponding to an essential simple closed curve in the kernel. Then there CHAPTER 3: MINIMAL SURFACES 29 is a map h : Σg →M for which h(Σg) is a branched minimal surface, and such that h induces the same map as f on π1. If π2(M) = 1 we may take h to be homotopic to f. Finally, if M has dimension 3 then h can be chosen to have no branch points. The basic strategy is very close to that of Douglas. If we ﬁx a marked conformal structure φ on Σg, we can look for a harmonic map fφ : Σg →M in the homotopy class of f. A harmonic map is one which minimizes energy; a harmonic map which is conformal is minimal. To construct fφ we must understand convergence properties of sequences of maps with energy bounds; a limit might ‘bubble oﬀ’ spheres; cutting oﬀthese spheres changes the homotopy class of the map by an element of π2(M). Varying the marked conformal structure leads us to confront the non-compactness of the Teichmüller space T of marked conformal structure on Σg. It is here that we make use of the hypothesis that f∗should be injective on essential simple curves. Finally we observe that the surface we obtain is locally least area; thus if M is 3-dimensional, Theorem 4.6 of Gulliver-Osserman implies that h has no branch points. 4.4.1. Harmonic maps. We discuss brieﬂy the setup for the theory of harmonic maps be-tween Riemannian manifolds, as developed by Eells-Sampson . For a smooth map f : N →M between Riemannian manifolds, the energy of f is the functional E(f) := 1 2 Z N \|d f\|2dvol The integrand \|d f\|2/2 is the energy density of f. Here we are thinking of d f as a section of the bundle T ∗N ⊗f ∗(TM), so that if ei is an orthonormal basis for TN at a point p, then \|d fp\|2 = P \|d fp(ei)\|2. Scaling the metric on N at p by λ scales \|d f\|2 by λ−2 and dvol by λd where d is the dimension of N. Thus if N is a surface, E(f) depends only on the conformal structure on the domain. Let F : N × (−ϵ, ϵ) →M be a smooth 1-parameter family of maps with ft := F(·, t) so that f0 = f and ft = f outside a compact set. Let X = dF(∂t) and if ei is an orthonormal basis of TN locally, let Yi(t) := d ft(ei). Note that X and the Yi are commuting sections of F ∗TM and therefore ∇XYi = ∇YiX for each i, where ∇is pulled back from the Levi-Civita connection on TM. By abuse of notation we use X and Yi to refer to the restrictions of these vector ﬁelds in f ∗TM over N × 0. We compute d dt t=0E(ft) = Z N X i ⟨Yi, ∇XYi⟩dvol = Z N X i ⟨Yi, ∇YiX⟩dvol Area is invariant under tangential variations, but energy is not. We deﬁne a 1-form α on N which (up to the canonical identiﬁcation of 1-forms and vector ﬁelds via the metric) is the pullback of the tangential part of X. Thus: let α be the 1-form on N deﬁned pointwise 30 DANNY CALEGARI by α(V ) := ⟨X, d f(V )⟩. Then α♯is a vector ﬁeld on N with divergence div α♯= X i ei(α(ei)) −α(∇eiei) = X i ei⟨X, Yi⟩−⟨X, d f(∇eiei)⟩ = X i ⟨∇YiX, Yi⟩+ ⟨X, ∇YiYi −d f(∇eiei)⟩ Deﬁne the tension by the formula τ(f) := X i ∇d f(ei)d f(ei) −d f(∇eiei) where the sum is taken over an orthonormal basis ei pointwise (more intrinsically we could write τ(f) = tr ∇d f). Note that this is a vector ﬁeld along f(N). Since X is compactly supported, div α♯integrates to zero, and we obtain the ﬁrst variation formula d dt t=0E(ft) = − Z N ⟨X, τ(f)⟩dvol for any smooth compactly supported variation tangent to X. In particular, f is a critical point for E if and only if τ = 0. Such a map is called harmonic. Example 4.8. If N and M are Kähler manifolds (for example, if they are nonsingular projective varieties with the induced metric), every holomorphic map f : N →M is harmonic. To see this, observe that at a point p in N, if we choose geodesic normal co-ordinates at p and f(p) the tension operator reduces (in local co-ordinates) to the usual Laplacian. But for Kähler manifolds we can choose geodesic local co-ordinates which are the same time holomorphic co-ordinates, and then use the fact that for a Kähler manifold holomorphic functions are harmonic (in the usual sense). For a conformal map f : N →M, if e′ i is an orthonormal frame on M with d f(ei) = λe′ i then τ(f) = X i λ2∇e′ ie′ i + λe′ i(λ)e′ i −d f(∇eiei) so the normal part to M of the tension ﬁeld is pointwise proportional to the mean curvature. Comparing with the ﬁrst variation formula for area (i.e. Proposition 2.1) we see that a harmonic map which is conformal is minimal. Conversely, let us suppose that N = Σ is 2-dimensional. Let f : Σ →M be harmonic but not conformal. There is another surface Σ′ with a Riemannian metric and an area-preserving diﬀeomorphism h : Σ′ →Σ so that f ◦h : Σ′ →M is conformal. It is elementary that wherever f fails to be conformal, the energy density of f ◦h is strictly smaller than that of f, and therefore E(f ◦h) < E(f). In fact, one easily sees that for any f : Σ →M the area of f is less than or equal to the energy, with equality if and only if f is conformal. We conclude that a map f : Σ →M that minimizes energy in its homotopy class over all choices of conformal structure on the domain is both harmonic and conformal, and is consequently a (locally least area) minimal surface. As a special case, any harmonic f : S2 →M is conformal and thus minimal. CHAPTER 3: MINIMAL SURFACES 31 4.4.2. Bubbling oﬀ. For maps f : N →Rn the energy is the sum of the L2 norm of the coordinate functions, and a map is harmonic if each coordinate function is harmonic in the usual sense. If fi : N →R is a sequence of smooth functions with uniformly bounded energy, a limit (if it exists) might be no more regular that L2. Rather than attempt to adapt the theory of Sobolev spaces to maps between Riemannian manifolds, it is convenient to appeal to a famous theorem of Nash that every Riemannian manifold M can be isometrically embedded in some high dimensional Rn. Then for a smooth map f : N →M the energy of f is the same as the energy of the composition N →M →Rn, and it makes sense to talk about L2 limits of sequences of maps. Note that a map f : N → M ⊂Rn is harmonic (as a map to M) if the tension ﬁeld (as a map to Rn) is normal to M. From now on we specialize to the case that the domain is Σ, a surface, mapping to a compact Riemannian manifold M. Let’s ﬁx a homotopy class of map, and look for a sequence of maps fi : Σ →M in this homotopy class whose energies converge to the inﬁmum. Since E(fi) is bounded, there is a subsequence which converges weakly and pointwise almost everywhere to an L2 map f : Σ →M for which E(f) ≤lim E(fi). To say the subsequence converges weakly means that for every L2 map g we have Z Σ ⟨d fi, dg⟩dvol → Z Σ ⟨d f, dg⟩dvol Note that at this point we do not know if f is continuous, or in case it is, that it is in the desired homotopy class. Energy can go down in a limit if a deﬁnite amount of energy of fi becomes concen-trated in a sequence of smaller and smaller regions Ui that shrink to a collection of points. Conformally rescaling fi near such a limit point gives a sequence of maps whose domains converge to C, and after reparameterization by a diﬀeomorphism we may assume (reducing the energy in the process) these maps are themselves conformal. The energy density of the resulting maps might blow up and we can iterate this process, obtaining a ‘tree’ of ﬁnite energy harmonic maps from punctured spheres to M in the limit. This situation is analyzed carefully by Sacks-Uhlenbeck. They show Thm. 3.6 that a ﬁnite area harmonic map from a punctured disk to a Riemannian manifold may be extended smoothly to a harmonic map over the puncture, and thus the limit map ﬁlls in over the punctures of each sphere to give a harmonic map S2 →M. They are able to obtain therefore a limit f deﬁned on a new domain Σ, called a bubble tree. It is built from Σ by attaching ‘depth 1’ spheres at the points where the energy density blows up in Σ, then attaching ‘depth 2’ spheres where the energy density blows up in the the depth 1 spheres, and so on. The spheres in Σ are the bubbles, and they carry the energy that has ‘bubbled oﬀ’ from Σ. The restriction of f to each sphere S2 in Σ is a (nonconstant) harmonic map, and is therefore a minimal surface. If M is compact, there is a uniform upper bound on KM and therefore by Gauss-Bonnet a positive lower bound on the area — hence the energy — of f restricted to each sphere in the tree. It follows that the number of bubbles in the tree is ﬁnite, and Σ admits the natural structure of a nodal Riemann surface. 32 DANNY CALEGARI Cutting oﬀbubbles changes the map f by elements of π2. Thus the restriction f to Σ is a harmonic map that might not be homotopic to the fi, but induces the same map on π1 (the details of the proof of this claim depend on a careful analysis of convergence to the bubble tree, ruling out the possibility that each bubble might be attached to the next by vanishingly thin ‘necks’ of positive diameter; see Parker ). 4.4.3. Variation in Teichmüller space. In the previous sections we have indicated the proof that for any continuous f : Σ →M, and for each ﬁxed marked conformal structure φ on Σ, there is a harmonic map fφ : Σ →M inducing the same map on π1 as f. We may now ask how fφ varies as a function of φ. If fφ is not conformal, we may reduce its energy by a diﬀeomorphism of the domain whose composition with fφ is conformal. Thus we will obtain a minimal surface if we can ﬁnd a φ which realizes the inﬁmum of E(fφ). Let φi be a sequence of marked conformal structures on Σ; i.e. homotopy classes of diﬀeomorphisms φi : Σ →Si where Si is a Riemann surface, and let fi : Si →M be harmonic, so that φi ◦fi induces the same map on π1 as f. Suppose further that E(fi) converges to the inﬁmum. We would like to argue that φi converges to some φ ∈T(Σ), the Teichmüller space of Σ. There are two sources of non-compactness: that coming from the conformal structures on Si, and that coming from the marking. First we show that the Si lie in a compact subset of moduli space. This means that there is a uniform upper bound on the moduli of embedded annuli in Si. Let A be an annulus in some Si conformally equivalent to a Euclidean cylinder S1 × [0, R]. We want a bound on R. Let’s denote the restriction of fi to A by h : S1 × [0, R] →M. By hypothesis, the core circle of the annulus maps to a homotopically nontrivial loop in M; therefore there is a positive constant ℓ> 0 so that the map ht : S1 →M deﬁned by ht(θ) = h(θ, t) has length at least ℓ, and therefore E(ht) ≥ℓ2/4π by Cauchy-Schwarz. Hence E(fi) ≥E(h) ≥ Z R 0 E(ht)dt ≥Rℓ2/4π so an upper bound for E(fi) gives an upper bound on R, and the claim is proved. The noncompactness of Teichmüller space is more serious, since it can easily happen that energy is not proper on T. Example 4.9. Any α ∈π1(M) which normalizes the image of π1(Σ) will induce an outer automorphism of any marking of Σ. For example, if M is a surface bundle over a circle with ﬁber homeomorphic to Σ, dragging Σ around the circle direction induces the monodromy action on the marking. It turns out that the conjugation action of the normalizer of f∗π1(Σ) in π1(M) accounts for all the non-properness of the energy functional E on T. Returning to our family of marked conformal structures φi : Σ →Si and harmonic maps fi : Si →M with E(fi) converging to the inﬁmum, we have already argued that the Si must lie in a compact subset of moduli space. Fix a subsequence with Si →S in moduli space. For each i there is a family of simple based loops Γi in Si that generate π1(Si), and we can choose such loops so that Γi →Γ ⊂S; this induces a family of homotopy equivalences ϕi : Si →S; note that we do not assume these maps are consistent with the markings φi; i.e. ϕi ◦φi need not be homotopic to ϕj ◦φj for i ̸= j. CHAPTER 3: MINIMAL SURFACES 33 As in § 4.4.2 some subsequence of the fi converges away from ﬁnitely many points (where energy might bubble oﬀin trees of spheres) to f : S →M. In particular, for large i, j the maps fi ◦ϕ−1 i and fj ◦ϕ−1 j induce the same maps on π1(S) →π1(M), and a free homotopy of the image of the basepoint under fi ◦φi to the image under fj ◦φj gives an element of π1(M) normalizing f∗(π1(Σ)). It follows that we can ﬁnd a new sequence of markings (φ′ i, Si) in a compact subset of Teichmüller space converging to (φ′, S) so that f ◦φ′ : Σ →M is a minimal surface satisfying the conclusion of Theorem 4.7. 5. Embedded minimal surfaces in 3-manifolds Theorem 4.7 allows us to construct immersed minimal surfaces in 3-manifolds under very general hypotheses on the fundamental group. However, for applications it is sometimes important for surfaces to be embedded. In this section we describe results which guarantee the existence of embedded minimal surfaces certain isotopy classes. 5.1. Embedded minimal disks with prescribed boundary. Let Γ ⊂M be an em-bedded Jordan curve in a closed 3-manifold M. If Γ is null-homotopic, Morrey’s theorem shows that there is a least area f : D →M with f : ∂D →Γ a homeomorphism. Note by Gulliver-Osserman f has no interior branch points. It is natural to ask: under what conditions is f an embedding? Even for Γ ⊂R3 this is a subtle question. One necessary condition is that Γ should be unknotted — i.e. it should bound some embedded disk. However this condition is not suﬃcient. Example 5.1 (Almgren-Thurston ). There is an unknotted Γ ⊂R3 that does not bound an embedded disk within its convex hull. Proposition 5.2 (Total curvature ≤4π). Let Γ ⊂R3 be a Jordan curve with total geodesic curvature at most 4π. Then Γ bounds an embedded minimal disk. 5.2. Mean convex boundary. Deﬁnition 5.3. Let M be a compact 3-manifold with smooth boundary ∂M. We say that M is mean convex if the mean curvature ﬁeld H along ∂M does not point out of M, and M is strictly mean convex if H is nonzero everywhere on ∂M and points in to M. For example, a convex subset of R3 is mean convex. Likewise, a small enough round ball in any Riemannian manifold is mean convex. Mean convexity is the right boundary condition to put on a manifold in order to ensure the existence of minimal surfaces. Theorem 5.4 (Meeks-Yau , Thm. 1). Let M be a mean convex 3-manifold, and let Γ ⊂∂M be an embedded Jordan curve that is null-homotopic in M. Then Γ bounds a least area disk in M, and any such disk is properly embedded. Proof. The ﬁrst step is to show that there is an immersed least area disk f : B →M with f : ∂B →Γ. To do this we embed M as a codimension 0 submanifold of a Riemannian manifold N that is homogeneously regular in the sense of Morrey (see § 4.2.4) so that there is f : B →N with f : ∂B →Γ, and then reason a posteriori that f(B) ⊂M. 34 DANNY CALEGARI By perturbing the metric on M very slightly near the boundary, we may assume ∂M is strictly convex. We let N := M ∪∂M ∂M × [0, ∞) where we put the product metric on ∂M × [0, ∞). This is a C0 Riemannian metric, and will not typically be smooth along ∂M = ∂M × 0. We give a collar neighborhood of ∂M in M the structure of a product ∂M × (−ϵ, 0] by exponentiating the unit normal ﬁeld. Then the entire ∂M × (−ϵ, ∞) ⊂N has a metric of the form g(t) ⊕dt2 where g(t) is a C0 family of metrics on ∂M, and g(t) = g(0) for t > 0. We deﬁne a new C∞family of metrics h(t) on ∂M by averaging g (pointwise as a function of t) with respect to a smoothly varying family of smooth probability measures φt centered at t and with support equal to (−ϵ, ∞). The mean curvature of ∂M × t at each point (p, t) ∈∂M × (−ϵ, ∞) with respect to the h(t) metric is (to ﬁrst order) a weighted average of the mean curvatures in the g(t) metric at nearby (p, s); in particular, the end of N is foliated by strictly mean convex surfaces. Then if f : B →N is minimal and not contained in M, it has an interior point of tangency with some strictly mean convex surface, violating minimality. We conclude f : B →M. Since our metric on M is as close to the original metric as desired, we can obtain minimal f : B →M (in the original metric) as a suitable limit. The second step is to show that any least area disk whose boundary is embedded is itself embedded. If the manifold and the map are simplicial with respect to some triangulation, one can build a tower as in Papakyriakopolos’ proof of Dehn’s Lemma, where the map of the disk lifts to each step of the tower, and where at the top step the map has only simple self-intersections; at this step one can perform cut-and-paste to build a new map with the same area but which is singular along the self-intersections — such a map cannot be minimal, thus a posteriori there could have been no self-intersections at the top of the tower, and therefore none at all: i.e. the original map was an embedding. If M and Γ are real analytic, so is the minimal surface, and everything can be taken to be simplicial, and the argument goes through. Approximating the original metric and curve by real analytic ones we obtain the desired result. □ The ﬁrst step of the argument has nothing to do with disks per se, and can be used to construct minimal surfaces in 3-manifolds with mean convex boundary. A nice application is the so-called bridge principle: Corollary 5.5 (Bridge Principle; Meeks-Yau ). If X and Y are two strictly stable two-sided minimal surfaces in R3 and α is an arc joining their boundaries, there is a new stable minimal surface Z that is close to X ∪Y attached by a ‘bridge’ along α. We indicate the idea of the proof. Suppose ﬁrst that X and Y are embedded, and that α is disjoint from both except at the endpoints. We round α at the endpoints so it is normal to ∂X, ∂Y and then build a 3-manifold M which is the union of rounded thickened neighborhoods NX and NY of X and Y with a thickened neighborhood Nα of α. Since X is minimal it has mean curvature 0. Let f be an eigenfunction for the stability operator L of least eigenvalue, and recall that f is nowhere zero. We can push X in the normal direction by a distance ±tf for small t to produce X(±t). Intuitively, since X is stable and f is an eigenfunction, the variation increases area locally to second order every-where; hence X(±ϵ) for small enough ϵ has mean curvature pointing in to X everywhere, and together they bound NX which is mean convex (NY is constructed similarly). CHAPTER 3: MINIMAL SURFACES 35 A suﬃciently thin tube Nα around α has arbitrarily large principle curvatures along the meridians into the tube, and therefore the entire boundary is mean convex. Finally, the frontier where Nα meets NX and NY can be rounded and modeled on catenoids; the result is the desired mean convex 3-manifold M. We can build a Jordan curve Γ in ∂M by attaching a 1-handle along the tube from ∂X to ∂Y , and then span this by a surface Z in the desired homotopy class; we can assume Z is least area in M, and therefore Z is stable in R3. If X and Y are not embedded we can nevertheless build M as above together with an immersion into R3. Using the bridge principle, Hall showed that one cannot replace ‘least area’ with ‘stable’ in Theorem 5.4. Example 5.6 (Hall ). There is an embedded Jordan curve on S2 that bounds an immersed stable minimal disk that is not embedded. To see this, take two copies of the equator and perturb them slightly. They bound a pair of (almost horizontal) disks; by the bridge principle there is a stable minimal disk D1 obtained by attaching a small 1-handle to these two disks. This stable minimal disk is not least area; a least area disk D2 is contained near a narrow strip around the equator. Now perturb the boundaries of D1 and D2 slightly so that D1 intersects D2 transversely, but their boundaries are disjoint in S2. By the bridge principle we can tube these disks together to make D stable and not embedded, with embedded boundary. See Figure 8. Figure 8. A self-intersecting stable disk with embedded boundary on the boundary of a convex set 5.3. Embedded incompressible surfaces and the roundoﬀtrick. Theorem 5.7 (Freedman-Hass-Scott Thm. 5.1). Let M be a closed irreducible Rie-mannian 3-manifold, and let Σ be a closed surface of non-positive Euler characteristic. Let f : Σ →M be a least area immersion which is π1-injective, and such that f is homotopic to a two-sided embedding g. Then either (1) f is an embedding; or (2) f double covers a one-sided minimal surface K embedded in M, and g(Σ) bounds a twisted I-bundle over a surface isotopic to K. 36 DANNY CALEGARI An irreducible 3-manifold has trivial π2, by the sphere theorem. If g : Σ →M is a π1-injective embedding, Theorem 4.7 guarantees the existence of a homotopic immersed minimal surface f : Σ →M in the same homotopy class. Conversely, a two-sided embed-ding g : Σ →M which is injective on essential simple curves is π1-injective. Thus we can ﬁnd f satisfying the hypothesis of Theorem 5.7 under many circumstances. In the sequel we give a brief outline of the argument, following which should be consulted for details. 5.3.1. Cut and paste and the roundoﬀtrick. Let Σ be a 2-sided oriented immersed surface in a Riemannian 3-manifold M (we do not assume Σ is connected), and suppose Σ intersects itself transversely in double curves and isolated triple points. If there are no triple points, the self-intersection locus Γ is a union of circles, and we can perform cut and paste along some circle γ. This means that we cut Σ along the preimage of γ, and reattach the local sheets by a permutation compatible with the orientations. The resulting surface is no longer immersed; it has a corner along the preimage of γ. This corner can be rounded, reducing the number of self-intersections and reducing area; this is called the roundoﬀtrick. The resulting surface Σ′ then satisﬁes area(Σ′) < area(Σ), has no triple points of self-intersection, and has fewer curves of self-intersection than Σ. Under suitable topological conditions it might happen that Σ′ as above is homotopic to Σ, so that we can deduce that any Σ as above is not least area in its homotopy class. 5.3.2. Intersections near a tangency. One of the subtleties of applying cut and paste and the roundoﬀtrick is that (self)-intersections of minimal surfaces in 3-manifolds need not be in general position. Nevertheless one can give a normal form for the intersection near points of tangency, which we now describe. Let Σ be a 2-sided oriented immersed minimal surface in a 3-manifold and let Σ1, Σ2 be two embedded sheets in Σ that intersect at a point p of common tangency. It turns out there is a choice of local C1 coordinates x, y, z with p at the origin, in which Σ1 is the x–y plane, and Σ2 looks like the graph of Re ((x + iy)n) for some n. For surfaces in R3 this can be deduced from the Weierstrass parameterization, and is analagous to (but simpler than) Osserman’s model near an interior branch point; compare § 4.3. In an arbitrary Riemannian 3-manifold one must appeal to the general theory of quasilinear second order elliptic PDE; see Lemma 1.4 and 1.5. Using this local model, we can perform cut and paste and roundoﬀfor Σ as follows. First, either Σ factors through a covering map, or the points of self-tangency are isolated (we assume we are in the second case). Let p be a point where Σ intersects itself transversely, and suppose there is a neighborhood N(p) with the property that cut and paste and roundoﬀsupported in N(p) will reduce area by at least ϵ > 0. We perturb Σ to Σ′ by a perturbation supported outside N(p) in such a way as to remove all the points of self-tangency; this can be done in such a way that area(Σ′) < area(Σ)+ϵ. Assuming there are no triple points, we can now perform cut and paste and roundoﬀon Σ′, producing a new surface Σ′′ with fewer double curves, where by hypothesis area(Σ′′) ≤area(Σ′) −ϵ < area(Σ). 5.3.3. Embedded surfaces in homotopy equivalences. If M is closed and irreducible and f : Σ →M is π1-injective, we can lift f to fΣ : Σ →MΣ, where MΣ is the covering space of M with fundamental group equal to f∗π1(Σ). Then MΣ is noncompact, but is complete CHAPTER 3: MINIMAL SURFACES 37 with respect to the metric pulled back from M, and fΣ is a homotopy equivalence (in fact, MΣ is necessarily homeomorphic to Σ × R, but that is not essential for what follows). In this context we have the following: Proposition 5.8 ( Thm. 2.1). Let M be an oriented Riemannian 3-manifold without boundary, and let f : Σ →M be a homotopy equivalence, where Σ is a closed oriented surface other than S2. If f is least area in its homotopy class, then it is an embedding. Proof. In the sequel we use homology with Z/2Z coeﬃcients. If Σ is embedded, we are done, so we suppose to the contrary that Σ is not embedded. Since f is a homotopy equivalence, it is an isomorphism on homology, and therefore if N is a regular neighborhood of f(Σ), the map Σ →N is injective on homology. Since M is homotopic to Σ and without boundary, it has two ends, which are separated by the image of Σ. Thus ∂N contains an ‘upper boundary’ A which cobounds one of the ends, and B := ∂N −A contains the ‘lower boundary’ which cobounds the other end. In particular, if g : M →Σ is a homotopy inverse to f, then A →M →Σ and B →M →Σ are both degree one, and therefore A and B both have rank H1 at least as large as Σ. We now argue separately depending on whether H1(Σ) →H1(N) is surjective or not. Case 1: H1(Σ) →H1(N) is surjective. For any compact 3-manifold, H1(∂N) →H1(N) kills a half-dimensional subspace. We conclude that A and B have the same rank of H1 as Σ, and since A →M →Σ is degree one, A is homeomorphic to A′ together with a collection of spheres, where A′ →M →Σ is a homotopy equivalence. Likewise there is B′ ⊂B where B′ →M →Σ is a homotopy equivalence. Since Σ is not embedded, by cut and paste and suitable roundoﬀwe can ﬁnd a neigh-borhood N with area(∂N) < 2 area(Σ), so that either A′ or B′ violates the hypothesis that Σ is least area in its homotopy class. Case 2: H1(Σ) →H1(N) is not surjective. In this case we apply a tower argument. We relabel M, N and f : Σ →N as M0, N0 and f0 : Σ →N0. Since f0 : H1(Σ) →H1(N0) is not surjective, there is a degree 2 cover p1 : M1 →N0 and a lift f1 : Σ →M1 with regular neighborhood N1 →M1. If H1(Σ) →H1(N1) is not surjective, we pass to another cover, and repeat; each nontrivial lift produces fj with fewer self-intersections than before, so that the tower terminates at fk : Σ →Mk with neighborhood Nk and H1(Σ) →H1(Nk) an isomorphism. Some homological algebra ( Cor. 2.3) shows that ∂Nk = Ak ∪Bk each homologous to fk(Σ). Thus, as above, there are A′ k ⊂Ak and B′ k ⊂Bk each homeomorphic to Σ and mapping down to M in the homotopy class of f. At least one has area less than Σ unless fk is an embedding. But if fk is an embedding then since pk has degree 2, the map pk ◦fk : Σ →Mk−1 has no triple points, and after a small perturbation has only embedded double curves of self-intersections. Thus as in § 5.3.2 we can do cut and paste and roundoﬀ to produce a surface in the same homotopy class with smaller area. This completes the argument. □ 5.3.4. Least area property under coverings. 38 DANNY CALEGARI 5.4. Sweepouts and index one surfaces. 5.5. Combinatorial minimal surfaces. 6. Acknowledgments Danny Calegari was supported by NSF grant DMS 1005246. References F. Almgren and W. Thurston, Examples of unknotted curves which bound only surfaces of high genus within their convex hulls, Ann. Math. 105 (1977), pp. 527–538 M. do Carmo and C. Peng, Stable complete minimal surfaces in R3 are planes, Bull. AMS 1 (1979), no. 6, pp. 903–906 T. Colding and W. Minicozzi, A course in minimal surfaces, AMS Graduate Texts in Mathematics 121, American Mathematical Society, Providence, RI, 2011 T. Colding and W. Minicozzi, Estimates for parametric elliptic integrands, IMRN 2002 (2002), pp. 291–297 J. Douglas, Solution of the problem of Plateau, Trans. AMS 33 no. 1, pp. 263–321 J. Eells and J. Sampson, Harmonic mappings of Riemannian manifolds, Amer. J. Math. 86 no. 1, (1964), pp. 109–160 M. Freedman, J. Hass and P. Scott, Lease area incompressible surfaces in 3-manifolds, Invent. Math. 71 (1983), pp. 609–642 R. Gulliver, Regularity of minimizing surfaces of prescribed mean curvature, Ann. Math. (2) 97 (1973), pp. 275–305 P. Hall, Two topological examples in minimal surface theory, J. Diﬀ. Geom. 19 (1984), no. 2, pp. 475–581 W. Meeks and S.-T. Yau, The Existence of embedded minimal surfaces and the problem of uniqueness, Math. Z. 179 (1982), pp. 151–168 C. Morrey, The problem of Plateau on a Riemannian manifold, Ann. Math. (2) 49, no. 4 (1948), pp. 807–851 R. Osserman, A survey of minimal surfaces, Second Edition. Dover Publications, Inc., NY, 1986 R. Osserman, A proof of the regularity everywhere of the classical solution to Plateau’s problem, Ann. Math. 91 no. 3, (1970), pp. 550–569 T. Parker, Bubble tree convergence for harmonic maps, J. Diﬀ. Geom. 44 (1996), pp. 595–633 T. Radó, On Plateau’s problem, Ann. Math. (2) 31 no. 3, (1930), pp. 457–469 J. Sacks and K. Uhlenbeck, The existence of minimal immersions of 2-spheres, Ann. Math. 113 no. 1, (1981), pp. 1–24 J. Sacks and K. Uhlenbeck, Minimal immersions of closed Riemann surfaces, Trans. AMS 271 (1982), no. 2, pp. 639–652 A. Schoen, Inﬁnite periodic minimal surfaces without self-intersections, NASA technical note TN D-5541, NASA, Washington D.C., May 1970 R. Schoen, Estimates for stable minimal surfaces in three-dimensional manifolds, Seminar on minimal submanifolds, 111–126, Ann. of Math. Stud. 103, Princeton Univ. Press, Princeton, NJ, 1983 R. Schoen and S.-T. Yau, Existence of incompressible minimal surfaces and the topology of three dimensional manifolds with non-negative scalar curvature, Ann. Math. 110, no. 1, (1979), pp. 127– 142 W. Thurston, The Geometry and Topology of 3-Manifolds, a.k.a “Thurston’s Notes”; available from the MSRI at B. White, Lectures on minimal surface theory, preprint; arXiv:1308.3325 University of Chicago, Chicago, Ill 60637 USA E-mail address: dannyc@math.uchicago.edu
17041	https://www.youtube.com/watch?v=jfcpnZpVeiI	How to Divide a Regular Hexagon Into 4 Triangles ExpertVillage Leaf Group 3640000 subscribers 17 likes Description 5685 views Posted: 12 Dec 2020 How to Divide a Regular Hexagon Into 4 Triangles. Part of the series: Hexagons. A regular hexagon can easily be divided into four different triangles using a basic procedure. Divide a regular hexagon into four triangles with help from an experienced math tutor in this free video clip. 2 comments Transcript: hi there my name is Ryan Malloy in this video we're going to discuss how to divide a regular hexagon into four triangles and there are two methods for doing so the first we're just going to divide our hexagon in half to get two quadrilaterals once we have a quadrilateral that's fairly easy to divide into two triangles simply pick one of our two vertices that we already drew a line on and draw the remaining line to the only other vertex that is not connected with already we have two triangles on top and then we repeat the process on the bottom pick the other vertex that we Ed to connect with our yellow line draw a line from that to the only other vertex that is not connected with voila we have one 2 3 four triangles so the second method that one can use to divide a regular hexagon into four triangles pick any vertex at all we'll just go with this one and then draw a line from that to every other vertex so these two they're already lines connecting so we don't need to do that again but these three do not have lines connecting to this vertex so we'll draw lines there like so and just like that we have four triangles 1 2 3 4 my name is Ryan Malloy and we've just discussed how to divide a regular hexagon into four triangles
17042	https://math.answers.com/other-math/How_is_the_number_of_faces_of_a_prism_related_to_the_number_of_sides_of_its_base	How is the number of faces of a prism related to the number of sides of its base? - Answers Create 0 Log in Subjects>Math>Other Math How is the number of faces of a prism related to the number of sides of its base? Anonymous ∙ 13 y ago Updated: 1/23/2025 Oh, what a happy little question! The number of faces on a prism is always two more than the number of sides on its base. You see, the base of the prism creates one face on the bottom and another face on the top, and each side of the base connects to one face on the side of the prism. Just like painting a beautiful landscape, it's all about seeing how each part comes together to create a lovely whole. BobBot ∙ 9 mo ago Copy Show More Answers (3)Add Your Answer What else can I help you with? Search Continue Learning about Other Math ### How many sides does a prism and a pyramid have? A prism has two identical polygonal bases connected by rectangular or parallelogram faces, so it has 2 sides. A pyramid has a polygonal base and triangular faces that meet at a common vertex, so it has a total of n+1 sides, where n is the number of sides of the base polygon. ### How many faces does quadillateral prism have? The number of faces in a prism is one face for each of the sides of its cross-section plus the two end faces. A quadrilateral prism has a quadrilateral as its cross-section. Therefore it has 4 + 2 = 6 faces. ### How is the total number of sides of the faces related to the number of edges? Faces + Vertices = Edges + 2 ### How many side faces and vertices does a hexagonal prism have? Sides and faces are the same thing. A hexagonal prism has 8 faces and 12 vertices. ### Can a prism have curved sides? No, a prism cannot have curved sides. A prism is a polyhedron with two parallel and congruent polygonal bases connected by lateral faces that are parallelograms. The lateral faces of a prism are always flat, planar surfaces that are perpendicular to the bases. Curved sides would not meet the definition of a prism. Related Questions Trending Questions What is another word for not broken?What is petrol measured in?Cubic meter equal to how many kgs?What is 3.499 rounded to the nearest hundredth?What times what equals 225?Who invented dice?50G in 100ml convert this to percentage?What percent is 372 of 821?What are the common factors of 133 and 161?Who created the number 2?How many times does 6 go into 180?How do you simplify aaaa?What is the sign of the product of 11 negative integers and 2 positive integers?How do you convert 25m into feet?How do you take dot product of two vectors?What number is 60 less than 345?What is the gcf of 12 36 60 84?What number is half of 1110?What are advantages of compound interest versus simple interest?What is 1 1 2 plus 1 3 8? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
17043	https://www.youtube.com/watch?v=wgn5x0Tm4CE	Pre-Calculus - Finding the maximum or minimum of a function MySecretMathTutor 242000 subscribers 731 likes Description 191652 views Posted: 16 Apr 2013 This video will show you how to use the formula -b/(2a) to find the maximum or minimum of a quadratic function. Remember to use the value of "a" to determine if you have a max or a min. For more videos please visit 50 comments Transcript: Today we're going to work on finding the maximum or minimum value of a quadratic function. Now, essentially what we're looking to do is figure out where the function takes on its greatest value and where the function takes on its absolute smallest value. Now, from looking at some of these quadratic functions, sometimes they can get quite complicated. They might have decimals, fractions, or a few other things. But finding the min or max is actually a straightforward, simple process if you know what to look for. So, let's look at the process and see how this is done. To figure out the maximum or minimum of a quadratic, the very first thing we need to do is check its a value. This is the uh coefficient in front of the x squar term. Now, what this will essentially tell us is that the quadratic is either facing up or down. Here's how it works. If the coefficient in front of x^2 is positive, then our quadratic is facing up. That means that it takes on a low value and that it will have a minimum. If however that a value is negative then our quadratic is facing down and then it has a maximum value. So in either case we're essentially just looking for the vertex. Now it's going to really help us out because for the second step of this as soon as we know which direction it's facing we simply use the formula for the ver vertex to figure out where that max or min is located. So we use x is equal tob over 2 a. Once we have that value, we plug it into the function and it'll tell us what the value of the max or min actually is. So now let's do two examples using this process. You'll see exactly how easy this is. All right, so this is my quadratic function. I want to find the maximum or the minimum depending on which one it has. Very first thing I do is go ahead and check the a value. So that again is the coefficient right in front of x^ squ. As you can see, this thing is positive. 7.5 a nice positive number. And and the information I'm getting from that is that my parabola is facing up. So I know that it will have some sort of minimum value. Now I just have to find it. All right, this is where that formula comes into play. So I know the minimum will occur at x=b all over 2 a. So I'm going to use my 7.5 for the a value and I'll use this one in front of the x for the b value. All right. So x equ= and we'll use our b value a -63 all over 2 7.5. All right. Now this will require a little bit of simplification but not too bad. So negative a negative will be a positive 63. 2 7.5 15. And looks like this just turns into a 4.2. Now be careful that 4.2 is not the minimum. This is where the minimum occurs. If we actually want to figure out what that minimum value is, we need to take this 4.2 to and put it into the function. So, let's take our function, plug in 4.2. So, everywhere you see an x, go ahead and put in 4.2. Now even though there are a lot of terms in the function you can always uh use the calculator to help you out with some of these uh calculations. When I did this I got a - 33.3. So now I can say that the minimum value is a -3.3. And in case I ever want to know where it is, remember we already found this earlier. It happens when x= 4.2. So the process is pretty straightforward, but let's see this one more time just so you have it down. Now we'll try it with a -3x^2 + 201x - 54. So again the very first thing I want to do is let's check that a value. And in this case the a value is negative. So my parabola is facing the other way which means I have a maximum. Not bad. All right. And now we will use our formula x = b / 2 a to figure out where this happens. So negative uh the value of b will be that 201. So let's grab that -201 all over 2. My value of a is the -3. All right, looking pretty good. So, looks like I have a -201 all over -6 or I can just say 2011 over 6. Now, you'll notice that this one doesn't simplify, but that's okay. Now, all we have to do is take this and plug it into our function. So f ofx, we'll plug in the 2011 all over 6. Again, everywhere we see an x, we'll put in this 2011 all over 6. Now, this one is quite a mess, so I'll definitely use my calculator to uh help me out as much as possible with this guy. Let's go ahead and grab it. All right, so let's plug this in. We'll start with a -3 2011 / 6^ 2ar + 2011 201 / 6 and then minus 54. All right, this looks like I'm getting about uh 3,00 312.75. So I can say the minimum value is 3,312.75. And again, remember if you want to know where this happens, that's the x value you found earlier. For more videos, please watch my secretmatutor.com.
17044	https://mathoverflow.net/questions/109785/what-structure-supports-division-to-a-unique-quotient-and-remainder	ra.rings and algebras - What structure supports division to a unique quotient and remainder? - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more What structure supports division to a unique quotient and remainder? Ask Question Asked 12 years, 11 months ago Modified12 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This has been bugging me for a while. According to if I divide integer a a by integer b b, I get unique t t, r r such that a=t b+r a=t b+r, 0≤r<b 0≤r<b. Furthermore, for any Euclidean domain R R, division with remainder can be defined as follows: a=t b+r a=t b+r, and either r=0 r=0 or f(r)<f(b)f(r)<f(b). where f f is the euclidean function of R R. Z fits into this classification by letting f(n)=\|n\|. My problem with this "generalization" is that t and r need not be unique. Is there a classification for structures which support division with remainder where t and r are unique? ra.rings-and-algebras division-algebras division-rings Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Oct 16, 2012 at 5:24 Benjamin BraunBenjamin Braun asked Oct 16, 2012 at 4:50 Benjamin BraunBenjamin Braun 295 2 2 silver badges 8 8 bronze badges 3 What do you mean by "t and r as defined above" for a general Euclidean domain? What does 0≤r<b mean? Qiaochu Yuan –Qiaochu Yuan 2012-10-16 05:12:25 +00:00 Commented Oct 16, 2012 at 5:12 Modified question to clarify.Benjamin Braun –Benjamin Braun 2012-10-16 05:24:50 +00:00 Commented Oct 16, 2012 at 5:24 Yes, I've seen such a classification somewhere; if memory serves, this article is referred to in "Certain Number-Theoretic Episodes In Algebra" by Sivaramakrishnan. Franz Lemmermeyer –Franz Lemmermeyer 2012-10-16 07:37:55 +00:00 Commented Oct 16, 2012 at 7:37 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. In [M. A. Jodeit, Jr., Uniqueness in the division algorithm, The American Mathematical Monthly 74 (1967), 835–836] it is shown that a Euclidean Domain in which quotient and remainder are strictly unique (for the integers there is a choice of sign) is either a field or a polynomial ring over a field. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Oct 16, 2012 at 8:31 Francis ClarkeFrancis Clarke 46 1 1 bronze badge Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. A Euclidean domain is a ring such that there is (at least one) valuation function f from the non-zero elements of R to the non-negative integers so that for a,b∈R with b≠0 R we can write a=b q+r for some q,r with either r=0 or f(r)<f(b). The valuation function is not part of the structure so I can imagine (but don't know) that there might be cases with more than one valuation and certain statements would be valid with respect to one but not the other. Maybe even a pair q,r that works with one does not with the other. But one could discuss Euclidean Domains with distinguished valuation functions. If D=4 k+1 is a positive square free integer then the field Q[√D] has ring of integers O D consisting of all α=u+v 1+√D 2 with the usual norm f(α)=\|u 2−d v 2\|. It is known when this f shows that O D a Euclidean domain, when O D is a Euclidean Domain but that f does not establish it, and when it is not a Euclidean Domain. Similar definitions exist for other algebraic extensions of Q and there are open questions. So even the question of classifying Euclidean Domains is open. All this tip-toes around your question. Let me stick for now to the Gaussian integers u+v√−1 in Q[√−1] . We can think of these as the lattice points in the plane. Given Gaussian integers a,b with A not a multiple of b, we have a b in Q[√−1] a point which falls in some lattice square. Then we can take q as any one of the corners and this gives r. To choose r uniquely we can have a rule such as q should be the corner nearest the origin or the one nearest a b with a tie-breaker rule if needed. We could even pick our favorite ordering of the Gaussian Integers and pick q (or r) as the first in our order. So uniqueness can be imposed in the countable case. I think that what you really wonder about is not specific to Euclidean Domains. In a ring a unit is an element with a multiplicative inverse. a,b are associates if a=b u for some unit u. In the integers, the non-negative integers are closed under addition and multiplication (form a semiring) and contain exactly one member from each (two element) class of associates is a member. So some starting questions are: For which rings R does the multiplicative group decompose as (R,⋅)=P U where U is the group of units and P is a set of elements closed under addition or multiplication or both. I don't know the answer to that but perhaps it is very standard. Then one could go on to ask (in the case of Euclidean domains with this property): Is the set P unique? When all this is true and (R,f) is a Euclidean Domain with distinguished valuation function f, can the remainders r always be taken to come from P and, in that case, is the choice of r∈P always unique? Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications answered Oct 16, 2012 at 8:01 Aaron MeyerowitzAaron Meyerowitz 30.3k 2 2 gold badges 50 50 silver badges 105 105 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions ra.rings-and-algebras division-algebras division-rings See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Related 2monoid ring and some structure within it - how is it called? 5Infinite dimensional division algebras with finite center, and their involutions 1Central division algebras and splitting fields 4List of Casimir elements of low dimensional Lie algebras 2Principal ideal of a non-associative magma 13Properties of finite dimensional, real division algebras that yield only R, C, H and O 2When an algebra isomorphism preserves positive involution Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today MathOverflow Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings
17045	https://www.youtube.com/watch?v=pXN6rJlNXSg	Mass Percent of a Solution Made Easy: How to Calculate Mass % or Make a Specific Concentration ketzbook 110000 subscribers 1646 likes Description 113895 views Posted: 22 Sep 2018 This is a whiteboard animation tutorial on how to solve Mass Percent calculations for solutions. Please support me on Patreon: The Mass Percent of a solution is equal to the grams of solute per 100 grams of solution. The formula is: Mass Percent = ((mass of solute) / (mass of solution)) x 100% This equation may be rearranged to solve for any one of those three variables. The mass may be in any mass units as long as the mass units are consistent with each other. It is also useful to realize that the density of water is approximately 1 g/mL, so volumes of water in mL are equal to masses in grams. In this video, I also describe how to make a solution of a certain mass percent. Please see my related video on molarity: My goal is to make chemistry easier ;) chemistry #madeeasy #solutions #ketzbook #tutorial 64 comments Transcript: Introduction Welcome to Mass Percent Made Easy, brought to you by Ketzbook In this video, we are going to learn how to calculate the mass percent of a solution and how to calculate the mass of the solute or solvent if you know the mass percent concentration. But first, what is mass percent? Mass percent is a way to measure concentration. “Per” means for each, and “cent” is Latin for 100, so percent refers to the amount of solute dissolved in 100 units of solution. “Mass” simply refers to how we measure both the solute and the solution. For example, the fat content in milk is measured in mass percent. What is 2% milk? It means that 100 grams of that milk contains 2 grams of fat. There are two grams of fat for each 100 grams of 2% milk. The equation that we will use to calculate mass percent is: mass percent equals the mass of the solute divided by the total mass of the solution all multiplied by 100 to turn the fraction into a percent. Mass Percent Example 1 Let’s try some examples to see how this works. Calculate the mass percent concentration of a 38.5 gram aqueous solution that contains 1.96 grams of sodium bromide. Before we solve this problem, let’s make some sense of it. First, the term aqueous means that water is the solvent. Next, the problem tells us that we are solving for the mass percent. 38.5 grams is the mass of the solution, and 1.96 grams is the mass of the solute. Remember that the solute is the thing that is dissolved in the solvent. Now we’re ready to plug in some numbers to calculate the mass percent. 1.96 grams goes on the top of the fraction and 38.5 grams goes on the bottom of the fraction. This fraction is then multiplied by 100%. In your calculator, type 1.96 divided by 38.5 times 100, which gives us the answer of 5.09 percent. As for the units, notice that grams on the top and bottom of the fraction cancel each other out. This is true for any kind of percent or any kind of parts per calculation. The units on the top and bottom of the fraction must be the same so that they cancel out. Percent functions as the units of our answer, but in reality percent is merely a way to represent a dimensionless ratio. Mass Percent Example 2 Okay, let’s try another problem. What is the mass percent of 125 grams of fructose dissolved in 375 mL of water? Once again, we are solving for the mass percent concentration, 125 grams is the mass of the solute, and 375 mL is the volume of the solvent. Volume? What do we do with volume? Water is the solvent, so this is relatively simple because the density of water is approximately 1 g/mL at room temperature. So for water, 1 mL has a mass of 1 gram. That means that 375 mL is the same as 375 grams. As long as we don’t need any more than 3 significant figures and as long as the water is not too warm, this approximation works fine. If we did need more precision, we would have to calculate the mass of the solvent based on its density at the given temperature. Now we can plug in some numbers. The mass percent is equal to the mass of the solute divided by the mass of the solution all times 100 percent. We put 125 grams in for the mass of the solute. You may be tempted to put 375 grams in the bottom, but that is only the mass of the solvent. The mass of the solution is the mass of the solvent PLUS the mass of the solute, that is 375 plus 125. That works out to 125 divided by 500 times 100, or 25%. Suppose, however, we wanted to solve for something different, such as the mass of the solute. Let’s try this next problem. How much sodium fluoride is in a 221 g tube of toothpaste that has 0.24% sodium fluoride? In this case, we are solving for the mass of the solute. 221 grams is the mass of the solution, and 0.24% is the mass percent concentration of sodium fluoride. We can set the problem up the same as always. The mass percent is 0.24%, which is equal to the mass of the solute divided by the mass of the solution or 221 grams, all multiplied by 100%. The only difference this time is that what we are solving for is on the right side of the equation, so we will need to do a bit of rearranging. First, let’s remove the units because percent is on both sides of the equation, and we know that the mass of the solute has the units of grams. Next, we can multiply both sides of the equation by 221, which cancels it out on the right. Finally, we can divide both sides of the equation by 100 to cancel it out on the right side as well. This leaves us with the mass of the solute being equal to 221 times 0.24 divided by 100, which calculates to be 0.53 grams. That is approximately one tenth the lethal dose of sodium fluoride for adults, but it is enough to be dangerous for small children, so caution should be exercised when allowing children to use any fluoride containing product. Mass Percent Example 3 Okay, let’s try one last problem. How could you make a 7.8% aqueous solution of glucose using 5.0 g of glucose? Like always, let’s first identify the knowns and the unknowns in the problem. 7.8% is the mass percent concentration of the solution we want to make, and 5 grams is the mass of the solute we need to use. The question doesn’t specifically say what we are solving for, but we can infer it from what is missing. Mass percent equals the mass of the solute divided by the mass of the solution all times 100 percent. The mass percent is 7.8, and the mass of the solute is 5. That leaves the mass of the solution as the only unknown, so we should solve for that. However, when you make a solution, it is more practical to weigh the solvent all by itself, so we should ultimately solve for the mass of the solvent. In order to solve this problem, we need to do a little bit of rearranging. The variable we are solving for cannot be in the denominator, so the first thing we need to do is multiply both sides of the equation by the mass of the solution. Mass of the solution then cancels out on the right side of the equation. Next, we need to get the mass of the solution all by itself, so we divide both sides by 7.8. The 7.8 on the top and bottom of the fraction cancel each other out. This leaves us with the mass of the solution equals 5 times 100 divided by 7.8, which calculates to be 64.1. Because the units for the mass of glucose are grams, the units for the mass of solution are also grams. We are almost done, but as we mentioned earlier, it is easier in the lab to weigh the solvent separately, so we need to calculate the mass of the solvent all by itself. The mass of the solution equals the mass of the solvent plus the mass of the solute, and we know that the mass of the solute is 5 grams. In other words, the mass of the solvent is 64.1 minus 5 or 59.1 grams. By the way, the reason I have included an extra significant figure for this quantity is because it is not based on a measurement but something we will make in the lab, so we want it to be as precise as possible. Now that we have the numbers we need, we can go ahead and weigh everything we need. Weigh the glucose in a weighing boat or a clean weighing container. Although we could use a graduated cylinder to measure the water, it is better to weigh the water. Once we have the correct amounts, simply combine them in a beaker or flask, and stir it a little until it is all dissolved. The resulting solution is a 7.8 percent aqueous glucose solution. Thanks for watching. If you found this video useful, please like or subscribe. Feel free to share any comments or questions you have below, and check me out at ketzbook.com.
17046	https://www.ck12.org/flexi/cbse-math/surface-area-of-cones/how-do-you-find-the-surface-area-of-a-cone/	Flexi answers - How do you find the surface area of a cone? \| CK-12 Foundation Subjects Explore Donate Sign InSign Up All Subjects CBSE Math Surface Area of Cones Question How do you find the surface area of a cone? Flexi Says: The net shows a cone, when unfolded, consists of one triangle and one circle. The circle is the base of the cone, and the triangle is the lateral face. The surface area is the sum of the areas of the two shapes. For a right circular cone of base radius π r l+π r 2 height (altitude) h and slant height π r l(l+r) we have i The formula for the total surface area of cone, TSA cone= Curved surface area + Area of the base =π r l + π r 2 =π r(l+r) Click here to learn more about the surface area of a right circular cone! Analogy / Example Try Asking: How do you find the net surface area of a cone?A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is calledHow to find the lateral area of a cone? How can Flexi help? By messaging Flexi, you agree to our Terms and Privacy Policy × Image Attribution Credit: Source: License:
17047	https://math-gpt.org/problems/question/the-base-of-a-regular-pyramid-is-a-square-and-each-of-the-other-four-sides	To find the vertical height of the regular pyramid with a sq \| MathGPT To find the vertical height of the regular pyramid with a square base and equilateral triangle sides, we can follow thes... Question: To find the vertical height of the regular pyramid with a square base and equilateral triangle sides, we can follow these steps:1. Identify dimensions of the base and sides: The base is a square with each side measuring 20 cm. Therefore, the vertices of the base are at coordinates (0, 0, 0), (20, 0, 0), (20, 20, 0), and (0, 20, 0).2. Calculate the height of the equilateral triangles: The height h of an equilateral triangle can be found using the formula: h = 32 side length For our case, the side length is 20 cm. Thus, the height of one triangle is: h = 32 20 = 103 cm 3. Find the center of the base: The center of the square base is at the point (10, 10, 0).4. Apply the Pythagorean theorem: Now, we need to find the vertical height H of the pyramid from the apex directly above the center of the base to the plane of the base. The apex forms a right triangle with the height and the distance from the center of the base to the midpoint of one side of the square base.5. Calculate that distance: The midpoint of one side of the base can be found at (10, 0, 0). The distance from the center (10, 10, 0) to the midpoint (10, 0, 0) is: d = 10 cm 6. Set up the equation using Pythagorean theorem: H^2 + d^2 = (103)^2 Substituting d = 10: H^2 + 10^2 = (103)^2 H^2 + 100 = 300 H^2 = 200 H = 200 = 102 cm Therefore, the vertical height of the pyramid is 102 cm. Answer: To find the vertical height of the pyramid, let's denote: the length of each side of the square base and the equilateral triangles as , the vertical height of the pyramid as , the slant height of the pyramid (which is also the height of the equilateral triangle) as , the length of the apothem of the base squ... Unlock with MathGPT are (half the diagonal of the square) as , where is the diagonal. Since the base is a square, we first find using the Pythagorean theorem, considering that the diagonal splits the square into two 45-45-90 triangles: Now, is also the radius of the circle circumscribed around the square base. The slant height of the pyramid can be found from the equilateral triangle side using the formula for the height of an equilateral triangle: To find the vertical height of the pyramid, we note that , , and form a right triangle, with as the hypotenuse and and as the legs. Using the Pythagorean theorem: Substituting cm and cm: Therefore, the vertical height of the pyramid is . MathGPTNext Problem ;
17048	https://www.vedantu.com/question-answer/find-the-constant-term-of-a-binomial-expa-class-12-maths-cbse-6038881cc340bb46219b69fb	Courses for Kids Free study material Offline Centres Talk to our experts Question Answer Class 12 Maths How do I find the constant ter... Question Answers for Class 12 Class 12 Biology Class 12 Chemistry Class 12 English Class 12 Maths Class 12 Physics Class 12 Social Science Class 12 Business Studies Class 12 Economics Question Answers for Class 11 Class 11 Economics Class 11 Computer Science Class 11 Biology Class 11 Chemistry Class 11 English Class 11 Maths Class 11 Physics Class 11 Social Science Class 11 Accountancy Class 11 Business Studies Question Answers for Class 10 Class 10 Science Class 10 English Class 10 Maths Class 10 Social Science Class 10 General Knowledge Question Answers for Class 9 Class 9 General Knowledge Class 9 Science Class 9 English Class 9 Maths Class 9 Social Science Question Answers for Class 8 Class 8 Science Class 8 English Class 8 Maths Class 8 Social Science Question Answers for Class 7 Class 7 Science Class 7 English Class 7 Maths Class 7 Social Science Question Answers for Class 6 Class 6 Science Class 6 English Class 6 Maths Class 6 Social Science Question Answers for Class 5 Class 5 Science Class 5 English Class 5 Maths Class 5 Social Science Question Answers for Class 4 Class 4 Science Class 4 English Class 4 Maths How do I find the constant term of a binomial expansion? Answer Verified 503.4k+views Hint: In the above question, you were asked to find the constant term of binomial expression. According to the formula of the binomial theorem that is ${{(x+y)}^{n}}$ , the term ${{y}^{n}}$ is always constant. You will see how this is a constant term. So, let us see how we can solve this problem.Complete step by step solution: We will see what do we get from the expansion of the binomial formula which is ${{(x+y)}^{n}}$ $\Rightarrow {{(x+y)}^{n}}=(\begin{matrix} n \ 0 \\end{matrix}).{{x}^{n}}+(\begin{matrix} n \ 1 \\end{matrix}).{{x}^{n-1}}.{{y}^{1}}....+(\begin{matrix} n \ k \\end{matrix}).{{x}^{n-k}}.{{y}^{k}}+....+(\begin{matrix} n \ n \\end{matrix}).{{y}^{n}}=\sum\limits_{k=0}^{n}{.(\begin{matrix} n \ k \\end{matrix}).{{x}^{n-k}}.{{y}^{k}}}$ where $x,y\in \mathbb{R},$ $k,n\in \mathbb{N}$ and $(\begin{matrix} n \ k \\end{matrix})$ denotes combinations of n things taken k at a time. So we have 2 cases1st case: When the terms of the binomial are a constant and a variable like $\Rightarrow {{(x+c)}^{n}}=(\begin{matrix} n \ 0 \\end{matrix})\cdot {{x}^{n}}+(\begin{matrix} n \ 1 \\end{matrix})\cdot {{x}^{n-1}}\cdot {{c}^{1}}+...+(\begin{matrix} n \ k \\end{matrix})\cdot {{x}^{n-k}}\cdot {{c}^{k}}+...+(\begin{matrix} n \ n \\end{matrix})\cdot {{c}^{n}}$ Here the constant term is $(\begin{matrix} n \ n \\end{matrix})\cdot {{c}^{n}}$ and its product is also constant.2nd Case: When the terms of the binomial are a variable and the ratio of that variable like $\Rightarrow (\begin{matrix} n \ k \\end{matrix})\cdot {{x}^{n-k}}.{{(\dfrac{c}{x})}^{k}}=(\begin{matrix} n \ k \\end{matrix})\cdot {{x}^{n-k}}\cdot {{c}^{k}}.\dfrac{1}{{{x}^{k}}}=((\begin{matrix} n \ n \\end{matrix})\cdot {{c}^{k}}).\dfrac{{{x}^{n-k}}}{{{x}^{k}}}=((\begin{matrix} n \ k \\end{matrix})\cdot {{c}^{k}}).{{x}^{n-2k}}$ Therefore, the middle term is constant in this case that is $k=\dfrac{n}{2}$.So, the constant term in a binomial expression which is ${{(x+y)}^{n}}$ is ${{y}^{n}}$.Note:For the above solution, there was one more case but it has no constant term. The binomial expression for the third case is ${{(x+y)}^{n}}$. We will study the details of the binomial theorem in the coming lessons. Recently Updated Pages Master Class 12 Business Studies: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 English: Engaging Questions & Answers for SuccessMaster Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Physics: Engaging Questions & Answers for Success Master Class 12 Business Studies: Engaging Questions & Answers for SuccessMaster Class 12 Economics: Engaging Questions & Answers for SuccessMaster Class 12 English: Engaging Questions & Answers for Success Master Class 12 Maths: Engaging Questions & Answers for SuccessMaster Class 12 Social Science: Engaging Questions & Answers for SuccessMaster Class 12 Physics: Engaging Questions & Answers for Success Trending doubts When was the first election held in India a 194748 class 12 sst CBSEWhich are the Top 10 Largest Countries of the World?Differentiate between homogeneous and heterogeneous class 12 chemistry CBSEWhat are the monomers and polymers of carbohydrate class 12 chemistry CBSEWhy is the cell called the structural and functional class 12 biology CBSEDifferentiate between microsporogenesis and megasporogenesis class 12 biology CBSE When was the first election held in India a 194748 class 12 sst CBSEWhich are the Top 10 Largest Countries of the World?Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE What are the monomers and polymers of carbohydrate class 12 chemistry CBSEWhy is the cell called the structural and functional class 12 biology CBSEDifferentiate between microsporogenesis and megasporogenesis class 12 biology CBSE
17049	https://mathinsight.org/local_extrema_examples_two_variables	Two variable local extrema examples - Math Insight Skip to navigation (Press Enter) Skip to main content (Press Enter) Home Threads Index About Math Insight Page Navigation Top Examples Example 1 Example 2 In threads Multivariable calculus Math 2374 Links Similar pages See also Contact us To create your own interactive content like this, check out our new web site doenet.org! Two variable local extrema examples Suggested background Introduction to local extrema of functions of two variables Example 1 Find the local extrema of f(x,y)=x 3+x 2 y−y 2−4 y f(x,y)=x 3+x 2 y−y 2−4 y. Solution Step 1: Find the critical points. The derivative of f f is D f(x,y)=[3 x 2+2 x y x 2−2 y−4].D f(x,y)=[3 x 2+2 x y x 2−2 y−4]. D f(x,y)=[0 0]D f(x,y)=[0 0] means both components must be zero simultaneously. We need x(3 x+2 y)=0(1)(1)x(3 x+2 y)=0 and x 2−2 y−4=0.(2)(2)x 2−2 y−4=0. We need to solve two equations for the two unknowns x x and y y. Equation (1)(1) is satified if either x=0 x=0 or if 3 x+2 y=0 3 x+2 y=0, i.e., if x=0 x=0 or if y=−3 x/2 y=−3 x/2. We consider these two solutions as two separate cases. For each case, we will find soultions for equation (2)(2). Case 1: Let x=0 x=0. Then we know equation (1)(1) is satisfied. We plug x=0 x=0 into equation (2)(2), which becomes 0−2 y−4=0 0−2 y−4=0, i.e., y=−2 y=−2. If x=0 x=0 and y=−2 y=−2, then both equation (1)(1) and equation (2)(2) are satisfied. Therefore the point (0,−2)(0,−2) is a critical point. Case 2: Let y=−3 x/2 y=−3 x/2. Then we know that equation (1)(1) is satisfied. We plug y=−3 x/2 y=−3 x/2 into equation (2)(2) and simplify: x 2−2(−3 x/2)−4 x 2+3 x−4(x−1)(x+4)x=1 or x=0=0=0=−4.x 2−2(−3 x/2)−4=0 x 2+3 x−4=0(x−1)(x+4)=0 x=1 or x=−4. So, we have two solutions of equation (2)(2) for case 2. The first solution is when x=1 x=1, which means y=−3 x/2=−3/2 y=−3 x/2=−3/2. If x=1 x=1 and y=−3/2 y=−3/2, then both equation (1)(1) and equation (2)(2) are satisfied. Therefore the point (1,−3/2)(1,−3/2) is a critical point. The second solution for case 2 is when x=−4 x=−4, which means y=−3 x/2=6 y=−3 x/2=6. Therefore, the point (−4,6)(−4,6) is a critical point. To summarize the results from both case 1 and case 2, we conclude that f(x,y)f(x,y) has three critical points: (0,−2)(0,−2), (1,−3/2)(1,−3/2), and (−4,6)(−4,6). You should double check that D f(x,y)=[0 0]D f(x,y)=[0 0] at each of these points. Step 2: Classify the critical points. The Hessian matrix is H f(x,y)=[6 x+2 y 2 x 2 x−2]H f(x,y)=[6 x+2 y 2 x 2 x−2] We need to check the definiteness of the H f(x,y)H f(x,y) at the critical points (0,−2)(0,−2), (1,−3/2)(1,−3/2), and (−4,6)(−4,6). For the critical point (0,−2)(0,−2), H f(0,−2)=[−4 0 0−2]H f(0,−2)=[−4 0 0−2] h 11=−4<0 h 11=−4<0 and det(H f)=8>0 det(H f)=8>0. This means H f(0,−2)H f(0,−2) is negative definite and f f has a local maximum at (0,−2)(0,−2). For the critical point (1,−3/2)(1,−3/2), H f(1,−3/2)=[3 2 2−2].H f(1,−3/2)=[3 2 2−2]. h 11=3>0 h 11=3>0 and det(H f)=−6−4=−10<0 det(H f)=−6−4=−10<0. This means H f(1,−3/2)H f(1,−3/2) is indefinite and f f has a saddle at (1,−3/2)(1,−3/2). For the critical point (−4,6)(−4,6), H f(−4,6)=[−12−8−8−2].H f(−4,6)=[−12−8−8−2]. h 11=−12<0 h 11=−12<0 and det(H f)=24−64=−40<0 det(H f)=24−64=−40<0. This means H f(−4,6)H f(−4,6) is indefinite and f f has a saddle at (−4,6)(−4,6). Example 2 Identify the local extrama of f(x,y)=(x 2+y 2)e−y f(x,y)=(x 2+y 2)e−y. Solution Step 1: Find the critical points. The derivative of f f is D f(x,y)=[2 x e−y(2 y−x 2−y 2)e−y]D f(x,y)=[2 x e−y(2 y−x 2−y 2)e−y] D f(x,y)=[0 0]D f(x,y)=[0 0] means that 2 x=0 2 x=0 and 2 y−x 2−y 2=0 2 y−x 2−y 2=0, i.e., x=0 x=0 and y(2−y)=0 y(2−y)=0. The critical points are therefore (0,0)(0,0) and (0,2)(0,2). Step 2: Classify the critical points. The Hessian matrix is H f(x,y)=[2 e−y−2 x e−y−2 x e−y(2−4 y+y 2+x 2)e−y]H f(x,y)=[2 e−y−2 x e−y−2 x e−y(2−4 y+y 2+x 2)e−y] At the critical point (0,0)(0,0) H f(0,0)=[2 0 0 2]H f(0,0)=[2 0 0 2] h 11=2>0 h 11=2>0 and det(H f)=4>0 det(H f)=4>0, so (0,0)(0,0) is a local mininum. At the critical point (0,2)(0,2) H f(0,2)=[e−2 0 0−2 e−2]H f(0,2)=[e−2 0 0−2 e−2] h 11=e−2>0 h 11=e−2>0 and det(H f)=−2 e−4<0 det(H f)=−2 e−4<0 so (0,2)(0,2) is a saddle point. Thread navigation Multivariable calculus Previous: Introduction to local extrema of functions of two variables Next: Introduction to double integrals Math 2374 Previous: Introduction to local extrema Next: The integrals of multivariable calculus Similar pages Introduction to local extrema of functions of two variables Minimization and maximization refresher Local minima and maxima (First Derivative Test) Minimization and maximization problems Solutions to minimization and maximization problems Introduction to Taylor's theorem for multivariable functions Multivariable Taylor polynomial example Critical points, monotone increase and decrease An algebra trick for finding critical points Taylor polynomials: formulas More similar pages See also Introduction to local extrema of functions of two variables Cite this as Nykamp DQ, “Two variable local extrema examples.” From Math Insight. Keywords: critical points, extrema, Taylor's theorem Send us a message about “Two variable local extrema examples” Name: Email address: Comment: If you enter anything in this field your comment will be treated as spam: Two variable local extrema examples by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. For permissions beyond the scope of this license, please contact us.
17050	https://www.lexplug.com/topics/property-law/types-of-tenancy	Outlines Constitutional Law Torts Civil Procedure Criminal Law Criminal Procedure Evidence Contracts Property Law Business Associations Wills & Trusts Family Law Practice Constitutional Law Torts Civil Procedure Criminal Law Criminal Procedure Evidence Contracts Property Law Business Associations Wills & Trusts Family Law Topics / Property Law / Concurrent Ownership / Tenancy in Common, Joint Tenancy, Tenancy by the Entirety Tenancy in Common, Joint Tenancy, Tenancy by the Entirety Premium Audio Content Subscribe to Lexplug to access audio content Start 7-Day Free Trial 0:000:00 In property law, multiple persons may hold concurrent interests in the same parcel of real property. These forms of co-ownership—Tenancy in Common, Joint Tenancy, and Tenancy by the Entirety—have distinct legal characteristics regarding possession, transferability, survivorship, and creditor rights. This section provides a detailed examination of each, illustrating their similarities, differences, and practical implications. I. Tenancy in Common A. Definition and Key Characteristics A Tenancy in Common is a form of concurrent ownership in which two or more persons hold an undivided interest in the property with no right of survivorship. The key attributes include: No Right of Survivorship: When a co-tenant dies, that individual’s fractional interest passes to their heirs or devisees, rather than automatically vesting in the other co-tenants. Freely Transferable: Each tenant in common may sell, gift, or otherwise transfer or encumber their share without the consent of the other co-tenants. Undivided Interests: All co-tenants have equal rights of possession, regardless of the size of their interests. However, the fractional shares can be unequal (e.g., 50%/50%, 70%/30%, etc.). Default Form of Co-Ownership: In most U.S. jurisdictions, if the deed is silent about the form of concurrent ownership, the law presumes a Tenancy in Common. B. Example Alice and Bob acquire a home. The deed lists “Alice and Bob, as co-owners.” Jurisdictional rules often interpret this conveyance as creating a Tenancy in Common. Alice might own a 50% interest, Bob the other 50%, and each can freely transfer that interest. If Alice dies, her interest passes under her will or by intestacy to her heirs. C. Partition If disagreements arise, a co-tenant generally has the right to seek “partition.” In a partition in kind, the court physically divides the property into separate parcels. Alternatively, a court may order a partition by sale, in which the property is sold, and proceeds are divided among the co-tenants. Courts often order partition by sale where division of the land itself is impractical. Continue reading with a 7-day free trial... Premium Content Subscribe to Lexplug to view the complete topic You're viewing a preview of this topic D. Notable Cases Gillis v. Smith, 293 P.2d 153 (Cal. 1956): Illustrates that each tenant in common may convey their share independently, and the grantee acquires the same rights as the conveying tenant. Ives v. Kimball, 196 N.E. 716 (Mass. 1935): Demonstrates partition procedures and a court’s willingness to protect each co-tenant’s right to sever co-ownership. II. Joint Tenancy A. Definition and the Right of Survivorship A Joint Tenancy is characterized by the right of survivorship: upon the death of any joint tenant, the surviving tenants automatically acquire the decedent’s interest. This ownership form avoids probate for the deceased tenant’s share. Joint Tenancy is historically disfavored by courts unless it is clearly created with the requisite formalities. B. The Four Unities Under common law, a valid Joint Tenancy requires the presence of four “unities”: Unity of Time: Each joint tenant’s interest must vest at the same moment. Unity of Title: All joint tenants must acquire title by the same instrument (e.g., the same deed, will, or other conveyance). Unity of Interest: Each joint tenant must have an identical (or substantially identical) interest in terms of duration and share. Unity of Possession: Each joint tenant has an equal right to possess and enjoy the entire property. Modern statutes and case law vary; some jurisdictions allow minor deviations from the strict common law requirements but still require explicit language reflecting an intent to create a right of survivorship (e.g., “to A and B as joint tenants with right of survivorship, and not as tenants in common”). C. Severance A Joint Tenancy can be “severed,” resulting in a Tenancy in Common among the severing party and the remaining owners, if any. Common methods of severance include: Conveyance of Interest: A joint tenant unilaterally conveys their share to a third party. The four unities are destroyed, and the grantee holds as a tenant in common with the remaining joint tenants. Mutual Agreement: All joint tenants may agree to end the joint tenancy, transforming it into a tenancy in common. Partition: A lawsuit for physical or financial partition can sever a Joint Tenancy. Severance rules vary by jurisdiction. Some states find that granting a mortgage on a joint tenant’s interest severs the joint tenancy (title theory), while others hold it does not (lien theory). D. Illustrative Cases Harms v. Sprague, 105 Ill. 2d 215 (1984): Under a lien theory jurisdiction, the Supreme Court of Illinois held that a mortgage given by one joint tenant did not sever the joint tenancy. Riddle v. Harmon, 102 Cal. App. 3d 524 (1980): Clarified that a joint tenant may effectively sever a joint tenancy by conveying the interest to oneself, so long as the jurisdiction’s statutes or rules permit such self-conveyance. E. Example Assume Sarah and Tom purchase property “as joint tenants with right of survivorship.” If Tom sells his undivided half-interest to Mindy, the Joint Tenancy is severed as to Tom’s share, leaving Sarah and Mindy as tenants in common with respect to that half. However, Sarah retains her joint tenancy interest as between herself and any remaining joint tenants who have not severed. III. Tenancy by the Entirety A. Nature of the Estate A Tenancy by the Entirety (TBE) is a specialized form of concurrent ownership recognized in many jurisdictions, available only to married couples (or, in some states, parties to a legally recognized union). Like the Joint Tenancy, a Tenancy by the Entirety features a right of survivorship, but with additional protections based on the marital relationship. B. Requirements and Unities Traditional common law viewed Tenancy by the Entirety as requiring the same four unities as a Joint Tenancy—time, title, interest, and possession—plus a fifth unity of marriage. Some states have modernized the formality requirements but still mandate a valid marriage or recognized spousal relationship at the time of conveyance. C. Transfer and Severance No Unilateral Conveyance: One spouse generally cannot unilaterally sever or convey the Tenancy by the Entirety interest. Both spouses must join in any transfer or encumbrance. Survivorship: Upon death of one spouse, the property automatically passes to the surviving spouse. Divorce: In most jurisdictions, a divorce automatically terminates the Tenancy by the Entirety, converting it into a Tenancy in Common (or in some states, a Joint Tenancy), unless otherwise provided by statute or court order. D. Creditor Protections One of the most significant features of Tenancy by the Entirety is the creditor protection it may afford in many states. Because each spouse is viewed as owning an indivisible “whole,” the interest of one spouse alone may be immune from certain creditors. This protection varies significantly by jurisdiction. Sawada v. Endo, 561 P.2d 1291 (Haw. 1977): A leading case holding that an individual spouse’s creditors cannot attach or force a sale of property held in TBE to satisfy that spouse’s separate debts. In other states, however, creditors may reach the debtor spouse’s interest or attempt a partition if permitted by local statutes or case law. E. Example Bob and Carol, married, hold their home “as tenants by the entirety.” Carol incurs significant personal debt unrelated to Bob. In many TBE jurisdictions, the creditor cannot attach Carol’s interest to force a sale without Bob’s consent, protecting the marital property from Carol’s separate liability. If Bob dies, Carol becomes the sole owner by right of survivorship. IV. Practical Considerations and Drafting Tips Clear Language Deeds should unambiguously state the parties’ intent to create a Joint Tenancy (using specific survivorship language) or a Tenancy by the Entirety (if available). Absent such clarity, courts often presume a Tenancy in Common. State Variations Always check local statutes and case law, especially regarding Tenancy by the Entirety (which is not recognized or is heavily modified in some jurisdictions) and the effect of mortgages on Joint Tenancies (title vs. lien theory). Estate Planning Joint Tenancy can be used to circumvent probate. However, estate planners should be mindful of severance risks and the potential for unintended consequences if a co-tenant conveys their share. Control and Creditor Issues Potential creditors, spousal rights, and the ability to partition or alienate interests should be carefully considered before deciding on a form of concurrent ownership. Conclusion Tenancy in Common, Joint Tenancy, and Tenancy by the Entirety each offer distinct legal relationships and consequences. They differ primarily in the presence (or absence) of survivorship rights, transferability, and protection from creditors. Mastering the nuances of these three forms of co-ownership is essential for attorneys and real estate professionals, who must draft clear instruments and advise clients on how each arrangement can meet their financial and personal goals. As always, variations in state law mandate close examination of local provisions and authorities to ensure accurate application. How can we improve this content?
17051	https://mathworld.wolfram.com/Saddle.html	Saddle -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Geometry Surfaces Miscellaneous Surfaces Saddle A surface possessing a saddle point. See also Hyperbolic Paraboloid, Monkey Saddle, Saddle Point Explore with Wolfram\|Alpha More things to try: anchor ring 6x6 latin squares d/dx(e^(ax)) Cite this as: Weisstein, Eric W. "Saddle." From MathWorld--A Wolfram Resource. Subject classifications Geometry Surfaces Miscellaneous Surfaces About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
17052	https://www.lifeskillscreations.com/products/life-skills-receipts-special-education-math-shopping-money-unit-1	Life Skills - Receipts - Special Education - Math - Shopping - Money - – Life Skills Creations Skip to content Don't forget to check out my featured products! Home Shop FREEBIES Blog Contact FAQ Shop on TpT Log in Instagram Facebook Search Home Shop FREEBIES Blog Contact FAQ Shop on TpT Search Instagram Facebook Log inCart Item added to your cart Check out Continue shopping Skip to product information Open media 1 in modal 1 / of 2 Life Skills - Receipts - Special Education - Math - Shopping - Money - Unit 1 Life Skills - Receipts - Special Education - Math - Shopping - Money - Unit 1 ----------------------------------------------------------------------------- Regular price $5.00 USD Regular price~~ $0.00 USD ~~Sale price $5.00 USD Unit price/per Sale Sold out Quantity Decrease quantity for Life Skills - Receipts - Special Education - Math - Shopping - Money - Unit 1 Increase quantity for Life Skills - Receipts - Special Education - Math - Shopping - Money - Unit 1 Add to cart Couldn't load pickup availability Refresh This is the FIRST set in my functional documents series for receipts. You will get 10 images of actual receipts your students will need to understand. Then, there are 6 follow up comprehension questions for EACH picture. That is a total of 60 real-world comprehension questions!!! Your kids will absolutely love this and you will too. I directly integrated the data into my IEP goals as well. Life skills students need exposure to vocabulary like sales tax, total vs. sub total, debit vs. credit, change, etc. Such a fun and simple activity to assess them on these areas! Purchase Includes: 10 images of receipts from various stores 60 comprehension questions (6 per image) Share Share Link Close share Copy link View full details You may also enjoy these products... ### Life Skills - Receipts - Special Education - Math - Shopping - Money - Unit 2 ### Life Skills - Receipts - Special Education - Math - Shopping - Money - Unit 2 Regular price $5.00 USD Regular price~~ $0.00 USD ~~Sale price $5.00 USD Unit price/per ### Life Skills - Receipts - Reading - Math - Special Education - Money - BUNDLE Sale ### Life Skills - Receipts - Reading - Math - Special Education - Money - BUNDLE Regular price $8.00 USD Regular price~~ $10.00 USD ~~Sale price $8.00 USD Unit price/per Sale ### Life Skills - Reading - Tickets - Real World - Reading Comprehension - Unit 1 ### Life Skills - Reading - Tickets - Real World - Reading Comprehension - Unit 1 Regular price $5.00 USD Regular price~~ $0.00 USD ~~Sale price $5.00 USD Unit price/per ### Life Skills - Reading - Tickets - Real World - Reading Comprehension - Unit 2 ### Life Skills - Reading - Tickets - Real World - Reading Comprehension - Unit 2 Regular price $5.00 USD Regular price~~ $0.00 USD ~~Sale price $5.00 USD Unit price/per ### Life Skills - Shopping - Coupons - Money - Math - Grocery Store - Unit One ### Life Skills - Shopping - Coupons - Money - Math - Grocery Store - Unit One Regular price $5.00 USD Regular price~~ $0.00 USD ~~Sale price $5.00 USD Unit price/per Payment methods © 2025, Life Skills Creations- All Rights Reserved Choosing a selection results in a full page refresh.
17053	https://math.uchicago.edu/~may/REU2017/REUPapers/Dougherty.pdf	Some applications of the Borsuk-Ulam Theorem Math REU University of Chicago Apprentice Program 2017 Jackson Dougherty September 25, 2017 1 Introduction The Borsuk-Ulam theorem with various generalizations and many proofs is one of the most useful theorems in algebraic topology. This paper will demonstrate this by ﬁrst exploring the various formulations of the Borsuk-Ulam theorem, then exploring two of its applications. The two major applications under con-sideration, the ham sandwich theorem and the Kneser conjecture, come from diﬀerent areas of mathematics and are separately interesting. Before going further, I would like to credit for inspiring the sequence of the discussion and for introducing me to many of these proofs. I have tried to emulate the clarity and simplicity of its presentation here. First, we will state the theorem in several diﬀerent ways. Theorem 1.1 (Borsuk (B1a)). For every continuous function f : Sn →Rn, there exists a point x ∈Sn with f(x) = f(−x). Theorem 1.2 (Borsuk (B1b)). For every continuous antipodal function f : Sn →Rn, there exists a point x ∈Sn with f(x) = 0. Theorem 1.3 (Borsuk (B2a)). There does not exist an antipodal mapping f : Sn →Sn−1. Theorem 1.4 (Borsuk (B2b)). There does not exist a mapping f : Bn →Sn−1 that is antipodal on the boundary ∂Bn = Sn. Theorem 1.5 (Lyusternik and Shnirel’man closed (LSc)). A covering of Sn by n + 1 closed sets, F1, . . . , Fn+1, has at least one set that contains a pair of antipodal points. Theorem 1.6 (Lyusternik and Shnirel’man open (LSo)). A covering of Sn by n + 1 open sets, U1, . . . , Un+1, has at least one set that contains a pair of antipodal points. 1 Math REU 2017 Jackson Dougherty Points x, y on Sk are said to be antipodal if y = −x. Likewise, a function f : Sk →X is antipodal if f is continuous and, for all x ∈Sk, f(−x) = −f(x) where X is any space. It also follows immediately from this deﬁnition that F ∩−F = ∅holds for a set F which does not contain a pair of antipodal points. A proof of the theorem itself would not be as illustrative as what follows, and given the scope of this paper, will therefore be excluded. However, proving the equivalence of the various forms is not particularly diﬃcult and follows imme-diately. Proof. (B1a) ⇒(B1b) Any antipodal continuous function f : Sn →Rn has a point x ∈Sn that satisﬁes f(x) = f(−x) = −f(x) by (B1a), which implies f(x) = 0. (B1b) ⇒(B1a) The converse follows by applying (B1b) to g(x) = f(x)−f(−x), since g is antipodal and continuous for any continuous f. (B1b) ⇒(B2a) An antipodal continuous function f : Sn →Sn−1 deﬁnes an antipodal function from Sn to Rn which is nowhere zero, since 0 / ∈Sn−1 and Sn−1 ⊂Rn. The existence of such a function is prohibited by (B1b). (B2a) ⇒(B1b) Similarly, a nonzero antipodal function f : Sn →Rn deﬁnes a function g : Sn →Sn−1 by g := f(x) ∥f(x)∥. Such a function contradicts (B2a). (B2b) ⇒(B2a) Note that the projection π : (x1, . . . , xn, xn+1) 7→(x1, . . . , xn) of the upper hemisphere of Sn to Bn is a homeomorphism. Thus, an antipodal map f : Sn →Sn−1 would deﬁne a function g : Bn →Sn−1 that is antipo-dal on the boundary ∂Bn according to g(x) := f(π−1(x)), contradicting (B2b). Here continuity of g is guaranteed since π is a homeomorphism, implying π−1 is continuous. (B2a) ⇒(B2b) A map g : Bn →Sn−1 that is antipodal on the boundary ∂Bn deﬁnes a function f : Sn →Sn−1 by f(x) := g(π(x)) and f(−x) := −g(π(x)) for x in the upper hemisphere. Such a function is antipodal by construction, contradicting (B2a). (LSo) ⇒(LSc) For a closed cover, F1, . . . , Fn+1, of Sn deﬁne open sets U ϵ i := {x ∈Sn : dist(x, Fi) < ϵ}, where dist(x, S) := infy∈S ∥x −y∥for a point x and a set S. If no set in the closed cover F1, . . . , Fn+1 contained a pair of antipodal points, then there is some ϵ > 0 such that the diameter of every set is less than 2−ϵ. Thus, Fi ⊂U ϵ/2 i , so U ϵ/2 1 , . . . , U ϵ/2 n+1 forms an open cover of Sn which does not contain any antipodal points, violating (LSo). (LSc) ⇒(LSo) For an open cover, U1, . . . , Un+1, we will construct F1, . . . , Fn+1, a collection of closed sets with Fi ⊂Ui for i ∈[n + 1], where [n+1] denotes the set {1, 2, . . . , n + 1}. For each x ∈Sn, let Vx be an open neighborhood of x 2 Math REU 2017 Jackson Dougherty whose closure is completely contained in one of the Ui. Applying compactness gives a cover whose closure, F1, . . . , Fn+1, doesn’t contain any pairs of antipodal points, since Fi ⊂Ui for each i, contradicting (LSc). (B1a) ⇒(LSc) For a closed cover, F1, . . . , Fn+1, let f : Sn →Rn such that x 7→(dist(x, F1), dist(x, F1), . . . , dist(x, Fn)). Because the norm is continuous, f is continuous, so there exists a point y such that y = f(x) = f(−x) by (B1a). if some component of y is zero, for example, the ith component, then dist(x, Fi) = 0 = dist(−x, Fi). Thus, Fi contains an antipodal pair. On the other hand, if no component of y is zero, then x, −x ∈Fn+1, since F1, . . . , Fn+1 covers Sn and x, −x / ∈F1, . . . , Fn. (LSc) ⇒(B2a) There exists a covering of Sn−1 by n + 1 closed sets such that no set contains a pair of antipodal points. To construct such a covering, F1, . . . , Fn+1, simply project the facets of an n-simplex containing 0 in the in-terior centrally from 0 onto Sn. Thus, an antipodal mapping f : Sn →Sn−1 gives a closed covering f −1(F1), . . . , f −1(Fn+1), no set of which contains a pair of antipodal points, contradicting (LSc). Thus, the various theorems above are equivalent statements of the Borsuk-Ulam theorem. As a quick illustration, applying (B1a) guarantees that there is some point on earth which shares a temperature and barometric pressure with its antipode. It is also interesting to observe that Borsuk-Ulam gives a quick proof of the Brouwer ﬁxed point theorem, the important result from algebraic topology which states that, for every continuous function f : Bn →Bn, there exists a point x ∈Sn such that f(x) = x. (B2b) ⇒Brouwer ﬁxed point theorem. Suppose there exists a continuous func-tion f : Bn →Bn such that f has no ﬁxed point. Deﬁne g : Bn →Sn−1 such that g(x) is the point on Sn−1 that intersects with the ray from f(x) to x. This is a well deﬁned retraction since there is no ﬁxed point at which the function would be ill-deﬁned, and it is a retraction since a ray from anywhere in Bn to a point x ∈∂Bn = Sn−1 intersects the boundary at x by construction. But, since the identity is clearly antipodal, i.e., g(−x) = −g(x) for x ∈Sn−1, such a function contradicts (B2b). Such a simple proof seems reasonable given the similarities between the Borsuk-Ulam theorem and the Brouwer ﬁxed point theorem, and, indeed, there are proofs of each theorem which share many similarities. At this point, it is worth noting that Borsuk-Ulam theorem has many generalizations and a variety of methods of proof. But we will instead focus on proving two interesting theorems, the ham sandwich Theorem and the Kneser conjecture. 3 Math REU 2017 Jackson Dougherty 2 Ham Sandwich Theorem We’ll begin by deﬁning some concepts and proceed by stating a version of the ham sandwich theorem. We then prove the theorem as well as some generaliza-tions. Finally, we apply the theorem to a simple problem. Deﬁnition 2.1. A hyperplane in Rd is a (d −1)-dimensional aﬃne subspace, or equivalently, the set {x ∈Rn : ⟨a, x⟩= b} for some a ∈Rd and b ∈R, where ⟨, ⟩is the standard inner product on Rn. A hyperplane deﬁnes two closed half-spaces of the form {x ∈Rn : ⟨a, x⟩≤b} for some a ∈Rd and b ∈R. The concept of a measure, somewhat informally described as a volume func-tion deﬁned from sets to R+ ∪∞, is also important, but we will not concern ourselves with a construction here (for such a construction, see ). As such, we will assume familiarity with measures. Deﬁnition 2.2. A ﬁnite Borel measure µ on Rd is a measure on Rd such that all open subsets of Rd are measureable and 0 < µ(Rd) < ∞. An illustrative example is that of the restriction of the Lebesgue measure to a compact subset. If A ⊂Rd is a compact subset and λd is the d-dimensional Lebesgue measure, then µ(X) := λd(X ∩A) for Lebesgue measurable sets X ⊆ Rd. Theorem 2.1 (Ham sandwich theorem). Let µ1, . . . , µd be ﬁnite Borel measures on Rd such that every hyperplane has measure 0 for each of the µi. Then there exists a hyperplane h such that µi(h+) = 1 2µi(Rd) for i = 1, . . . , d, where h+ is one of the half-spaces deﬁned by h. Informally, this theorem states that any arrangement of ham, bread, and cheese in space can be bisected by a single cut. The fact that any ham sandwich can be bisected with only one well placed cut is quite surprising. Proof. Let u = (u0, u1, . . . , ud) ∈Sn. If one of the components u1, . . . , ud is nonzero, assign the half-space h+(u) := {(x1, . . . , xd) ∈Rd : u1x1 + · · · + udxd ≤u0} to u. Antipodal points correspond to opposite half-spaces: {(x1, . . . , xd) ∈Rd : −u1x1 + · · · + −udxd ≤−u0} = h+(−x) = {(x1, . . . , xd) ∈Rd : u1x1 + · · · + udxd ≥u0} = −h+(u). We also have h+((1, 0, . . . , 0)) = {(x1, . . . , xd) ∈Rd : u1x1 + · · · + udxd = 0 ≤1} = Rd, 4 Math REU 2017 Jackson Dougherty h+((−1, 0, . . . , 0)) = {(x1, . . . , xd) ∈Rd : u1x1 + · · · + udxd = 0 ≤−1} = ∅. Then, deﬁne f : Sd →Rd by fi = µi(h+(u)). The hyperplane which deﬁnes a half-space h+(u0) such that f(u0) = f(−u0) satisﬁes the theorem. Additionally, by sub-additivity of a measure and since µ(Rd) > 0, f((−1, 0, . . . , 0)) = (µ1(∅), . . . , µd(∅)) ̸= (µ1(Rd), . . . , µd(Rd)) = f(1, 0, . . . , 0), which guarantees a well-deﬁned hyperplane. Thus, it suﬃces to prove that f is continuous in order to apply Borsuk-Ulam. We will show that µi(h+(un)) →µi(h+(u)) for a sequence (un)∞ n=1 which con-verges to u ∈Sd. For x / ∈∂h+(u) and suﬃciently large n, x ∈h+(un) if and only if x ∈h+(u). Therefore, the characteristic function gn := χh+(un) approaches the characteristic function g = χh+(u), except on ∂h+(u). By as-sumption, µi(∂h+(u)) = 0, so gn converges almost everywhere to g in the µi measure. Lebesgue’s dominated convergence theorem gives µi(h+(un)) = R gndµi → R g dµi = µi(h+(u)). This follows since 1 dominates gn and fur-ther since the assumption of a ﬁnite measure guarantees the integrability of 1. Therefore, f is continuous and Borsuk-Ulam applies, completing the proof. We next consider point sets, but the meaning of a hyperplane bisecting a point set with an odd number of points carries some ambiguity. We say that a hyperplane h bisects a point set A if each of the open half-spaces deﬁned by h contains at most 1 2 A \|⌋points in A. This deﬁnition allows at most k points from a set containing 2k + 1 points in each open half-space, thereby requiring that at least one point in an odd magnitude point set lie on the hyperplane. With this clariﬁcation, we may prove a variation of the ham sandwich theorem. Theorem 2.2 (Point set ham sandwich theorem). Let A1, . . . , Ad ⊂Rd be ﬁnite point sets. Then there exists a hyperplane h that simultaneously bisects A1, . . . , Ad. Proof. We would like to apply the ham sandwich theorem for measures to the point sets by imagining them as tiny balls. Suppose each Ai has odd cardinality and the disjoint union is in general position, i.e., Ai ∩Aj = ∅for i ̸= j and no hyperplane intersects more than d points. Then, let Aϵ i be the set Ai with each point replaced by a ball of radius ϵ centered at that point. Thus, there exists an ϵ > 0 such that no hyperplane intersects more than d balls of the union Aϵ 1 ∪· · · ∪Aϵ d. The ham sandwich theorem gives a hyperplane that bisects the sets Aϵ i. This bisection must intersect at least one of point from each Aϵ i since they each have an odd number of balls, and, since h intersects a maximum of d balls, each Aϵ i is intersected exactly once. The bisection guarantees that the hyperplane passes through the center of a ball which it intersects. Therefore, h bisects each of the original Ai. 5 Math REU 2017 Jackson Dougherty We will relax the assumptions on position. Let Ai,η for η > 0 denote a per-turbation of Ai where each point is moved by at most η such that the dis-joint union of Ai,η, in general position as before. Then, the previous argu-ment gives hη a hyperplane which bisects the Ai,η. For each hη, we may write hη = {x ∈Rd : ⟨aη, x⟩= bη} for some aη, a unit vector in Rd and some scalar bη. The bη lie in a bounded interval and the aη lie on a d −1-sphere. Thus, we may apply compactness to guarantee a cluster point (a, b) ∈Rd+1 of pairs (aη, bη) as η →0. We claim that the hyperplane h = {x ∈Sn : ⟨a, x⟩= b} bisects Ai. Take a sequence η1 > η2 > · · · which converges to 0 such that (aηj, bηj) →(a, b). Then, for x a distance δ > 0 away from h, x is at least 1 2δ away from hηj for suﬃciently large j. Thus, if one of the open half-spaces of h contains k points in Ai, then the corresponding open half space deﬁned by hηj must also contain at least k points of Ai,ηj for large enough j. Since these open half spaces cannot contain more than 1 2 A \|⌋points from Ai by construction, there can be no more than 1 2 Ai \|⌋points from Ai in the open half space deﬁned by h, so h must bisect each Ai as required. Now, we relax the condition on the parity of the Ai. Suppose some of the Ai have an even number of points. For each such Ai, remove an arbitrary point and bisect the resulting collection of point sets as before. Then, replacing the deleted points maintains the bisection because each open half space from the half-space already contained fewer than half of the points in the original sets. These two theorems give surprising results regarding the ability of hyper-planes to bisect arbitrary sets. We would like to apply these theorems to a diﬀerent problem, but we need to modify the previous theorem slightly. Corollary 2.2.1. Let A1, A2, . . . , Ad ⊂Rd be disjoint, ﬁnite point sets in gen-eral position, i.e., no hyperplane contains more than d points of the disjoint union of Ai. Then there exists a hyperplane h that bisects each Ai such that there are exactly 1 2 Ai \|⌋points from Ai in each open half-space deﬁned by h, and at most one point of Ai on the hyperplane h. Proof. Applying the previous theorem gives a hyperplane h where each of the corresponding half-spaces has at most half of the points in each Ai. However, the hyperplane could contain up to d points of a single Ai if some of the Ai have even cardinality. Fix a coordinate system so h has the equation xd = 0. Let B := h ∩(A1 ∪ · · · ∪Ad); then B consists of at most d aﬃnely independent points. We claim that it is possible to move h slightly so that any point in B is above or below or remains on h, whichever is required to achieve the desired bisection. To demonstrate this, add d−\|B\| points to B to obtain a d-point aﬃnely indepen-dent set C ⊂h. For each a ∈C, choose a point a′ such that either a′ = a, in the case where a is one of the new points and for points of B which should remain 6 Math REU 2017 Jackson Dougherty on h in order to achieve the required bisection, or a′ = a + ϵed, or a′ = a −ϵed, where ed is the standard basis vector in the d dimension. Let h′ = h′(ϵ) be the hyperplane determined by the d points a′, a ∈C. For all suﬃciently small ϵ > 0, the a′ remain aﬃnely independent, so h′(ϵ) is well deﬁned, and the motion of h′(ϵ) is continuous in ϵ. Thus, for all suﬃciently small ϵ, h′ bisects the Ai as required by the corollary. This version of the ham sandwich theorem will allow us to prove a theorem related to a necklace. We motivate this theorem with an anecdote. Two thieves steal a necklace made up of d diﬀerent types of jewels of unknown value. In order to equally divide the value of the necklace, the thieves agree to cut up the necklace so that they both get the same number of jewels. This is made easier since the necklace is open and there are an even number of gem-stones of each type. However, the jewels are set in platinum, a precious metal which the thieves don’t want to waste. So, the thieves decide to make as few cuts as possible. For a necklace with all of the stones of one type followed by all the stones of a second type, and so on, through all d diﬀerent types, it will be necessary to make d cuts to divide each type of jewel evenly. Then next theorem shows this is the maximum possible number of cuts. Theorem 2.3 (Necklace). Every open necklace with d kinds of stones can be divided between two thieves using at most d cuts. Prior to giving the proof, we introduce one more concept, the moment curve. The moment curve in Rd is the curve {γ(t) : t ∈R} given by γ(t) := (t, t2, . . . , td). We will need a lemma regarding this curve. Lemma 2.4. No hyperplane intersects the moment curve γ in Rd in more than d points. Therefore, any set of d + 1 distinct points on the moment curve is aﬃnely independent. Proof. A hyperplane has equation a1x1 + · · · + adxd = b with (a1, a2, . . . , ad) ̸= 0 ∈Rd. A point in γ(t) which intersects h satisﬁes a1t + a2t2 + · · · + adtd = b. These intersections correspond to real roots of the polynomial p(t) = (Pd i=1 aiti)− b whose degree is at most d. A polynomial of degree ≤d has at most d roots, so there are a maximum of d intersections. This will allow us to prove the necklace theorem quite easily. Proof of Necklace theorem. We embed the necklace into Rd along the moment curve. For a necklace of n stones, let Ai = {γ(k) = (k, k2, . . . , kd) : the kth stone is of the ith kind, k = 1, 2, . . . , n}. We will refer to the points of Ai as the stones of the ith kind. By the lemma, h cuts the moment curve in at most d places, so we may apply 2.2.1. Therefore, 7 Math REU 2017 Jackson Dougherty there exists a hyperplane h simultaneously bisecting each Ai. Furthermore, since every set is even, h contains no stones, and so cutting the necklace where h intersects the moment curve divides the stones as required by the theorem. This theorem is surprisingly diﬃcult to prove without topological methods, with combinatorial proofs typically using some version of Tucker’s lemma, the combinatorial version of the Borsuk-Ulam Theorem, as with . 3 Kneser’s Conjecture We will now explore a somewhat diﬀerent topic in Kneser’s conjecture. Consider the n-element subsets of a set with 2n + k elements. Then, Kneser conjectured that the minimum number of classes required to partition the subsets so that the pairwise intersection of subsets in each class is non-empty is k + 2. We can exhibit such a partition by considering the set [2n + k]. Then if Ki denotes the collection of all n-subsets whose least element is i, then K1, K2, . . . , Kk+1, and Kk+2 ∪· · · ∪Kn+k+1 generates a suitable partition with k + 2 classes. Thus, Kneser’s conjecture claims that any partition with fewer classes results in at least one class with a pair of disjoint subsets. The conjecture was ﬁrst proved by Lov´ asz nearly 20 years after Kneser pro-posed it. Lov´ asz introduced a perhaps surprising recoding in combinatorial terms which we shall introduce below. But for now, we will follow the logic of the simple proof ﬁrst given by . Thus, we desire a lemma generalizing the Lyusternik and Shnirel’man theorems introduced above. Lemma 3.1 (Generalized Lyusternik and Shnirel’man). If Sd is covered with d + 1 sets, each of which is either open or closed, then one of the sets contains a pair of antipodal points. Proof. Induct on the number t of closed sets in the cover of Sd. The base case t = 0 is a cover composed entirely of open sets. This is simply (LSo) proved above. Now assume 0 < t < d + 1 and that the conclusion holds for fewer than t closed sets as an induction hypothesis. Let C be a cover of Sd with d + 1 sets, with t closed sets and d + 1 −t open sets. Fix a closed set F in C. Suppose F did not contain a pair of antipodal points. It follows that F has diameter 2 −ϵ for some ϵ > 0. We let U be the open set of points in Sn with distance from F less than ϵ/2. Thus, F ⊂U and (C{F}) ∪U forms a cover of Sd consisting of t −1 closed sets and d + 2 −t open sets. The induction hypothesis guarantees that at least one set in this cover contains a pair of antipodal points. However, U does not contain a pair of antipodal points by construction. So some other set in C must contain such a pair, as required. We now have a slightly stronger form of the Lyusternik and Shnirel’man theorem which gives us everything required to prove Kneser’s conjecture. 8 Math REU 2017 Jackson Dougherty Theorem 3.2 (Lov´ asz-Kneser). If the n-element subsets of a 2n + k element set are partitioned into k +1 classes, then one of the classes must contain a pair of disjoint subsets. Proof. Distribute 2n + k points on Sk+1 in general position, i.e., such that no k + 2 points lie on a great k-sphere. Partition the n-element subsets of these points into k + 1 classes, call them A1, . . . , Ak+1. For i ∈[k + 1], let Ui denote the set of points a ∈Sk such that the open hemisphere centered at a, which has the form {x ∈Sk+1 : ⟨a, x⟩> 0}, contains an n-subset in the class Ai. Each Ui is clearly open, so F := Sk+1(U1 ∪· · ·∪Uk+1) is closed. The collection of all of these sets covers Sk + 1 with k + 2 sets. The lemma asserts that at least one set contains a pair of antipodal points ±a ∈Sk+1. F cannot contain ±a, since then both open hemispheres centered at a and −a would contain fewer than n points, which is impossible since then at least k+2 points must lie on the great k sphere given by {x ∈Sk+1 : ⟨a, x⟩= 0}. This contradicts the fact that the points are in general position. Thus, a and −a must lie in Ui for some i. Therefore, the open hemispheres centered at a and −a both contain an n-element subset in Ai, and these subsets must be disjoint. Kneser’s conjecture thus receives a simple proof by embedding the sets on the sphere and applying Borsuk-Ulam. However, we can encode the informa-tion diﬀerently, by imagining the sets as graphs and asking about the chromatic number of the graph. Deﬁnition 3.1. A (simple, undirected) graph G is a pair (V, E) where V is a set, known as the vertex set, and E ⊆ V 2 is the edge set, where S 2 is the collection of subsets of S with 2 elements. We say that the vertices v, v′ ∈V are adjacent if {v, v′} ∈E. Furthermore, a legal k-coloring of G is a mapping c : V →[k] such that c(u) ̸= c(v) whenever {u, v} ∈E. The chromatic number of G, denoted χ(G) is the smallest k such that G has a k-coloring. We say that edges connect vertices, giving a simple picture for graphs. Ad-ditionally, a coloring is legal if no two adjacent vertices share a color. With these concepts in mind we may recode the conjecture as follows. For a set X and F ⊂2X, where 2X is the power set of X, the Kneser graph of F, denoted KG(F), has a vertex set F and two sets F1, F2 ∈F are adjacent if and only if F1 ∩F2 = ∅. Symbolically, KG(F) = (F, {{F1, F2} : F1, F2 ∈F, F1 ∩F2 = ∅}) . If we denote the Kneser graph of F = [2k+n] n by KGn+2k,n, where [2k+n] n is the n-element subsets of [2k + n], then Kneser’s conjecture asserts χ(KGn+2k,n) = n+2. To be clear, the upper bound follows from the discussion at the beginning of the section which exhibits a coloring by k + 2 colors, while the lower bound is given by the Lov´ asz-Kneser theorem. 9 Math REU 2017 Jackson Dougherty This interpretation is interesting for several reasons. For example, if k < n, then there is no triplet of mutually disjoint n-element subsets, which means that KGn+2k,n has no triangles. Yet the chromatic number is arbitrarily large, growing with k. This is perhaps surprising since, when ﬁrst introduced to the concept of a coloring, one might expect that a high degree of interconnected-ness, for example in the form of triangles, is necessary to force a high chromatic number by reducing the size of independent sets. The Kneser graphs deny this expectation by allowing high chromatic number without triangles. Deﬁnition 3.2. Fractional chromatic number, denoted χf(G), is the inﬁmum of the fraction a b such that the vertex set is covered by a independent sets in such a way that every vertex is covered at least b times. An independent set is a subset of the vertices, V ′ ⊆V , such that, for all v, v′ ∈V ′, v ̸∼v′. This notion of coloring is slightly diﬀerent. Conceptually, it allows for mul-tiple colors of diﬀerent weights which each represent a legal coloring and whose total weight for each vertex is at least one. This reduces to the previous deﬁ-nition when each color has weight one, which gives χf(G) ≤χ(G). Returning to the Kneser graph, we ﬁnd that χf(KGn+2k,n) ≤n+2k n , since the collection of n + 2k independent sets of the form Ai = {S ∈ [n+2k] n : i ∈S} covers each point n times. Therefore, there is a large gap between the fractional chromatic number of the Kneser graph and its chromatic number. In general, giving lower bounds for χf(G) is easier than doing the same for χ(G), making the gap be-tween the two types of chromatic number quite unusual. This discussion links a theorem whose proof is highly dependent upon the topol-ogy of the sphere with a class of combinatorial objects. We thus see an unex-pected connection between algebraic topology and graph theory. 4 Conclusion This paper will hopefully provide an introduction to phrasings and applications of the Borsuk-Ulam theorem for those who are interested. Additional discussion as well as a variety of additional applications and several proofs of the theorem can be found in . Acknowledgments Thanks to Catherine Ray and Ronno Das for advising me through this process, and thanks to Professor Peter May for the opportunity to write this paper. References Ji˘ r´ i Matou˘ sek. Using the Borsuk-Ulam Theorem: Lectures On Topological Methods in Combinatorics and Geometry. Berlin, New York, Springer, 2003. 10 REFERENCES Math REU 2017 Jackson Dougherty E. M. Stein, and R. Shakarchi. Real analysis: Measure theory, integration, and Hilbert spaces. Princeton, N.J., Oxforx, Princeton University Press, 2005. Fr´ ed´ eric Meunier. Discrete Splittings of the Necklace. Mathematics of Oper-ations Research, 33(3):678-688, 2008. J. E. Greene. A new short proof of Kneser’s Conjecture. Amer. Math. Monthly, 109:918-920, 2002. K. Borsuk. Drei S¨ atz ¨ uber die n-dimensional euklidische Sph¨ are. Fundamenta Mathematicae, 20:177-190, 1933. L. Lyusternik and Shnirel’man. Topological Methods in Variational Problems (in Russian). Issledowatelski˘ i Institut Matematiki i Mechaniki pri O. M. G. U., Moscow 1930. 11
17054	https://brainly.com/question/13861734	[FREE] What pattern can you use to find the multiples of 5? - brainly.com Advertisement Search Learning Mode Cancel Log in / Join for free Log in Join for free Tutoring Session Smart guidance, rooted in what you’re studying Get Guidance Test Prep Ace exams faster, with practice that adapts to you Practice Worksheets Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified What pattern can you use to find the multiples of 5? loading See answer Explain with Learning Companion NEW Asked by banegasn90 • 11/14/2019 Advertisement Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 8799629 people 8M 5.0 2 Upload your school material for a more relevant answer Please read the answer below. Explanation To know if a number is part of the multiples of 5, we have two options:the first and the most difficult is to divide the number by 5 and if the module is 0 and the result is a whole number, then the number is a multiple of 5. For example, if we divide 30 by 5, the result is 6 exact and therefore, 30 is a multiple of 5. The second way to identify if a number is part of the multiples of 5 (and for me, the simplest) is if it ends at 0 or 5. For example 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, all these numbers, are multiples of 5 because their termination as we observe is 0 or 5. Answered by roundadworthy •1.5K answers•8.8M people helped Thanks 2 5.0 (2 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 8799629 people 8M 5.0 2 Chemical Thermodynamics - Andrea Allgood Carter Python for Informatics: Exploring Information - Charles Severance Energy and Human Ambitions on a Finite Planet - Murphy, Thomas W, Jr Upload your school material for a more relevant answer Multiples of 5 can be found by multiplying 5 by whole numbers, resulting in values like 5, 10, and 15. Additionally, if a number ends in 0 or 5, it is also a multiple of 5. This simple approach allows for easy identification of multiples of 5. Explanation To find the multiples of 5, you can follow a simple pattern. Multiples of a number are the results of multiplying that number by whole numbers. For example, to find the multiples of 5, you can multiply 5 by different whole numbers like 1, 2, 3, and so on: 5 × 1 = 5 5 × 2 = 10 5 × 3 = 15 5 × 4 = 20 5 × 5 = 25 5 × 6 = 30 5 × 7 = 35 5 × 8 = 40 5 × 9 = 45 5 × 10 = 50 These results show that the multiples of 5 are: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. Another easy way to identify multiples of 5 is to look at the last digit of the number. If a number ends in 0 or 5, it is a multiple of 5. This makes it very straightforward. For example: 25 ends in 5, so it is a multiple of 5. 40 ends in 0, so it is a multiple of 5. Therefore, remembering that multiples of 5 will either end in 0 or 5 can help you quickly identify them without needing to do multiplication. Examples & Evidence For instance, 20, 25, and 45 are multiples of 5 because they end in 0 or 5. You could also list the first ten multiples of 5 as: 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50. The definition of multiples states that they are obtained by multiplying integers by a specific number, and in this case, using 5 demonstrates that the resulting numbers meet the criteria for being multiples based on both multiplication and the last-digit rule. Thanks 2 5.0 (2 votes) Advertisement banegasn90 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them? Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216 Choose an equivalent expression for 1 0 6÷1 0 4. A. 1 0 2 B. 1 0 3 C. 1 0 10 D. 1 0 24 How would you write 1 2−3 using a positive exponent? A. 1 2 3 B. 1 2 0 C. 1 1 2 3 D. 1 2 3 1 Is this equation correct? 6 3⋅7 3=4 2 3 Previous questionNext question Learn Practice Test Company Copyright Policy Privacy Policy Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com
17055	https://my.mech.utah.edu/~pardyjak/me3700/IntroCompFlow.pdf	1 Introduction to Compressible Flow 0 ≠ Dt Dρ The density of a gas changes significantly along a streamline Compressible Flow Definition of Compressibility: the fractional change in volume of the fluid element per unit change in pressure p p p p v dp p + dp p + dp p + dp p + dv v − Compressible Flow 1. Mach Number: 2. Compressibility becomes important for High Speed Flows where M > 0.3 • M < 0.3 – Subsonic & incompressible • 0.3 <M < 0.8 – Subsonic & compressible • 0.8 <M < 1.2 – transonic flow – shock waves appear mixed subsonic and sonic flow regime • 1.2 <M < 3.0 - Supersonic – shock waves are present but NO subsonic flow • M > 3.0 – Hypersonic Flow, shock waves and other flow changes are very strong sound of speed velocity local = = c V M 2 Compressible Flow 3. Significant changes in velocity and pressure result in density variations throughout a flow field 4. Large Temperature variations result in density variations. As a result we now have two new variables we must solve for: T & ρ ρ ρ ρ We need 2 new equations. We will solve: mass, linear momentum, energy and an equation of state. Important Effects of Compressibility on Flow 1. Choked Flow – a flow rate in a duct is limited by the sonic condition 2. Sound Wave/Pressure Waves – rise and fall of pressure during the passage of an acoustic/sound wave. The magnitude of the pressure change is very small. 3. Shock Waves – nearly discontinuous property changes in supersonic flow. (Explosions, high speed flight, gun firing, nuclear explosion) 4. A pressure ratio of 2:1 will cause sonic flow Applications 1. Nozzles and Diffusers and converging diverging nozzles 2. Turbines, fans & pumps 3. Throttles – flow regulators, an obstruction in a duct that controls pressure drop. 4. One Dimensional Isentropic Flow – compressible pipe flow. 3 Approach • Control volume approach • Steady, One-dimension, Uniform Flow • Additional Thermodynamics Concepts are needed • Restrict our analysis to ideal gases Thermodynamics • Equation of State – Ideal Gas Law RT p ρ = Temperature is absolute and the specific volume is (volume per unit mass): ρ 1 = v K) J/(kg 287 .97kg/kmol 28 K) J/(kmol 8314 air of mass Molecular Constant Gas Universal ⋅ = ⋅ = = = m u M R R Thermodynamics – Internal Energy & Enthalpy • Internal Energy – individual particle kinetic energy. Summation of molecular vibrational and rotational energy. • For an ideal gas • Recall from our integral form of the Energy Equation for Enthalpy of an ideal gas: dv v u dT T u u d T v       ∂ ∂ +       ∂ ∂ = ~ ~ ~ ( ) T v u u , ~ ~ = dT c u d v = ~ ( ) T u u ~ ~ = ) (T h h = dT c dh p = pv u h + = ~ 4 Thermodynamics – Internal Energy & Enthalpy dT c dh p = RdT u d dh RT u h pv u h + = + = + = ~ ~ ~ RT p = ρ dT c u d v = ~ Substituting: const R c c R c c RdT dT c dT c RdT u d dh v p v p v p = = − + = + = + = ~ Thermodynamics – Internal Energy & Enthalpy Define the ratio of specific heats: const c c k v p = ≡ Then, 1 1 − = − = k R c k kR c v p For Air: cp = 1004 J/kg-K k = 1.4 The 2nd Law of Thermodynamics & Isentropic Processes Combining the 1st and 2nd Laws gives us Gibb’s Equation rev T Q ds       = δ We define entropy by: ∫ ∫ ∫ − = − = − = 2 1 2 1 2 1 p dp R T dT c ds dp dT c Tds dp dh Tds p p ρ ρ dT c dh p = p R T = ρ 1 5 The 2nd Law of Thermodynamics & Isentropic Processes For an Isentropic process: adiabatic and reversible We get the following power law relationship 1 2 1 2 1 2 ln ln p p R T T c s s p − = − k k k T T p p         =         = − 1 2 1 1 2 1 2 ρ ρ Control Volume Analysis of a Finite Strength Pressure Wave c 0 = V T p ρ T T p p ∆ + ∆ + ∆ + ρ ρ V ∆ Moving Wave of Frontal Area A The Speed of sound (c) is the rate of propagation of a pressure wave of infinitesimal strength through a still fluid. c V T p = ρ T T p p ∆ + ∆ + ∆ + ρ ρ V c V ∆ − = Stationary Wave Reference frame moving with wave ( ) ( )( ) ( )( ) ( ) ( ) (A) ˆ 0 2 1 ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ∆ + ∆ = ∆ ∆ + ∆ − ∆ + = ∆ − ∆ + = ∆ − ∆ + + − = • = ∫ ∫ ∫ c V V c c A V c cA dA V c dA c dA n V CS r Steady State Continuity Equation (Solve for the induced velocity ∆V): 1 2 Control Volume Analysis of a Finite Strength Pressure Wave Small Amplitude moderate frequency waves are isentropic and c V T p = ρ T T p p ∆ + ∆ + ∆ + ρ ρ V c V ∆ − = ( ) ( ) ( ) ( ) (B) ˆ 1 2 V c p c V c cA A p p pA V V m dA n V V F CS x x ∆ = ∆ − ∆ − = ∆ + − − = • = ∫ ∑ ρ ρ ρ & r Steady State Momentum Equation: (Find ∆p and c) 1 2 Now combine A & B and solve for the speed of sound: 0 of limit in the 1 2 2 → ∆ ∂ ∂ =         ∆ + ∆ ∆ = ∆ + ∆ ∆ = ρ ρ ρ ρ ρ ρ ρ ρ ρ p c p p c const p k = ρ 6 Control Volume Analysis of a Finite Strength Pressure Wave Calculating the Speed of Sound for an ideal gas: const p k = ρ ρ ρ p k p = ∂ ∂ kRT p k c = = ρ kRT c = Typical Speeds of Sound Fluid c (m/s) Gases: H2 1,294 Air 340 Liquids: Water 1,490 Ethyl Alcohol 1,200 Data From White 2003 4 . 1 ≈ = v p c c k For Air: K) J/(kg 287 ⋅ = R Example 1: Speed of sound calculation Determine the speed of sound in Argon (Ar) at 120 oC. MW = 40 kg/kmol: kRT c = 668 . 1 ≈ = v p c c k K) J/(kg 9 . 207 0kg/kmol 4 K) J/(kmol 8314 ⋅ = ⋅ = = m u M R R ( )( ) 1 -ms 8 . 318 393 J/kgK 9 . 207 668 . 1 = = K c Movement of a sound source and wave propagation V = 0 V < c V > c α Source moves to the right at a speed V M c V 1 sin = = α Zone of silence V ∆t 3 c∆t V ∆t V ∆t Mach cone 7 Example 2: a needle nose projectile traveling at a speed of M=3 passes 200m above an observer. Find the projectiles velocity and determine how far beyond the observer the projectile will first be heard 200 m α M =3 x Example 2: a needle nose projectile traveling at a speed of M=3 passes 200m above an observer. Find the projectiles velocity and determine how far beyond the observer the projectile will first be heard ( )( ) ( ) m m x x m M Mc V kRT c o 565 5 . 19 tan 200 200 tan 5 . 19 3 1 sin 1 sin m/s 6 . 1041 2 . 347 3 m/s 2 . 347 300 287 4 . 1 1 1 = = = =       =       = = = = = = = − − α α Steady Isentropic Flow – Control Volume Analysis Applications where the assumptions of steady, uniform, isentropic flow are reasonable: 1. Exhaust gasses passing through the blades of a turbine. 2. Diffuser near the front of a jet engine 3. Nozzles on a rocket engine 4. A broken natural gas line 8 Steady Isentropic Flow p T h ρ V dp p dT T d dh h + + + + ρ ρ dV V + dx ( ) ( )( )( ) dAdV d dV Ad dAdV AdV dA Vd VAd VdA AV VA dA A dV V d VA A V A V dA n V CS ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ + + + + + + + = + + + = + − = • = ∫ 2 2 2 1 1 1 ˆ 0 r Steady State Continuity Equation: 1 2 Steady Isentropic Flow p T h ρ V dp p dT T d dh h + + + + ρ ρ dV V + dx ( ) ( )( )( ) dAdV d dV Ad dAdV AdV dA Vd VAd VdA AV VA dA A dV V d VA A V A V dA n V CS ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ ρ + + + + + + + = + + + = + − = • = ∫ 2 2 2 1 1 1 ˆ 0 r Steady State Continuity Equation: Only retain 1st order differential terms & divide By ρVA V dV d A dA + + = ρ ρ 0 1 2 ~ 0 ~ 0 ~ 0 ~ 0 Steady Isentropic Flow p T h ρ V dp p dT T d dh h + + + + ρ ρ dV V + dx ( ) ( ) ( )1 2 1 2 2 1 2 2 ~ ~ 2 pv u pv u z z g V V m W Q s + − + + − + − = − & & & Steady State Energy Equation with 1 inlet & 1 exit: Neglecting potential energy and recalling: pv u h + = ~ 1 2 2 1 2 2 2 h h V V m W Q s − + − = − & & & ( ) 1 2 2 1 2 2 2 T T c V V m W Q p s − + − = − & & & Assuming and ideal gas: 1 2 9 Steady Isentropic Flow p T h ρ V dp p dT T d dh h + + + + ρ ρ dV V + dx Steady State Energy Equation with 1 inlet & 1 exit, neglecting potential energy & assuming Isentropic duct flow: 1 2 1 2 2 2 2 2 h V h V + = + Assuming and ideal gas: 1 2 1 2 2 2 2 2 T c V T c V p p + = + 1 2 1 2 2 2 1 2 1 2 RT k k V RT k k V − + = − + 1 2 Stagnation Conditions Assume the area A2is so big V2 ~ 0, then o h h V h = + = 1 2 1 2 2 1 2 Stagnation enthalpy T c V T p o + = 2 2 1 2 1 2 2 2 2 2 T c V T c V p p + = + Similarly, as we adiabatically bring a fluid parcel to zero velocity there is a corresponding increase in temperature Insolated walls Stagnation Temperature Stagnation Conditions – maximum velocity T c V T p o + = 2 2 If the temperature, T is taken taken down to absolute zero, then (+) can be solved for the maximum velocity: (+) o pT c V 2 max = No higher velocity is possible unless energy is added to the flow through heat transfer or shaft work. 10 Stagnation Conditions – Mach number relations 1 2 1 1 2 1 1 2 2 2 2 2 2 2 + − = + − = + = + = M k c V k T T c T V T T T c V T o p o p o For Ideal gases: 1 1 1 − =       − = k kRT T k kR T c p p c 2 c 1 2 1 2 + − = M k T To Recall, that the Mach number is defined as: c V M = Stagnation Conditions – Isentropic pressure & density relationships 1 1 2 1 1 1 2 1 1 2 1 1 2 1 − − − −       + − =       =       + − =       = k k o o k k k k o o M k T T M k T T p p ρ ρ 1 2 1 2 + − = M k T To Critical Values: conditions when M = 1       + = 1 2 k T T o 1 1 2 −       + = k k o k p p 1 1 1 2 −       + = k o k ρ ρ 2 1 1 2       + = k c c o 11 Critical Values: conditions when M = 1 8333 . 0 1 2 =       + = k T T o 5283 . 0 1 2 1 =       + = − k k o k p p 9129 . 0 1 2 1 1 =       + = − k o k ρ ρ For Air k = 1.4 9129 . 0 1 2 2 1 =       + = k c c o In all isentropic flow, all critical values are constant. Critical Values: conditions when M = 1 Critical Velocity: is the speed of sound c 2 1 2 1 1 2 1 2       + =       + = = = k kRT k c kRT c V o o 2 1 1 2       + = k c c o Example 3: Stagnation Conditions Air flows adiabatically through a duct. At point 1 the velocity is 240 m/s, with T1 = 320K and p1 = 170kPa. Compute (a) To (b) Po (c) ro (d) M (e) Vmax (f) V 12 Steady Isentropic Duct Flow p T h ρ V dp p dT T d dh h + + + + ρ ρ dV V + dx Recall, for Steady isentropic flow Continuity: V dV d A dA + + = ρ ρ 0 For compressible, isentropic flow the momentum equation is: VdV dp dV dp + = + = ρ ρ 2 0 2 Bernoulli’s Equation! neglecting gravity Substitute (†) into () (†) () V dV d A dA − − = ρ ρ         − = + − = dp d V dp V dp d A dA ρ ρ ρ ρ ρ 2 2 1 1 2 Steady Isentropic Duct Flow p T h ρ V dp p dT T d dh h + + + + ρ ρ dV V +         − = dp d V dp A dA ρ ρ 2 1 Recall that the speed of sound is: ρ ∂ ∂ = p c 2        − =       − = 2 2 2 2 2 1 1 1 c V V dp c V dp A dA ρ ρ Substituting the Mach number: c V M = ( ) 2 2 1 M V dp A dA − = ρ Describes how the pressure behaves in nozzles and diffusers under various flow conditions 1 2 Nozzle Flow Characteristics 1. Subsonic Flow: M < 1 and dA < 0, then dP < 0: indicating a decrease in pressure in a converging channel. 2. Supersonic Flow: M > 1 and dA < 0, then dP > 0: indicating an increase in pressure in a converging channel. 3. Subsonic Flow: M < 1 and dA > 0, then dP > 0 : indicating an increase in pressure in a diverging channel. 4. Supersonic Flow: M > 1 dA > 0, then dP < 0 : indicating a decrease in pressure in a diverging channel. ( ) 2 2 1 M V dp A dA − = ρ P P P P P P P P 13 Steady Isentropic Duct Flow – Nozzles Diffusers and Converging Diverging Nozzles ( ) 2 2 1 M V dp A dA − = ρ Describes how the pressure behaves in nozzles and diffusers under various flow conditions VdV dp + = ρ 0 Recall, the momentum equation here is: VdV dp − = ρ Now substitute () into (††) : ( ) 1 2 − = M V dV A dA ( ) 1 2 − = M V A dV dA Or, (††) () Nozzle Flow Characteristics 1. Subsonic Flow: M < 1 and dA < 0, then dV > 0: indicating an accelerating flow in a converging channel. 2. Supersonic Flow: M > 1 and dA < 0, then dV < 0: indicating an decelerating flow in a converging channel. 3. Subsonic Flow: M < 1 and dA > 0, then dV < 0 : indicating an decelerating flow in a diverging channel. 4. Supersonic Flow: M > 1 dA > 0, then dV > 0 : indicating an accelerating flow in a diverging channel. ( ) 1 2 − = M V dV A dA Converging-Diverging Nozzles Amin Subsonic Supersonic M = 1 Amax Subsonic Supersonic M < 1 Subsonic Supersonic M > 1 Flow can not be sonic 14 Choked Flow – The maximum possible mass flow through a duct occurs when it’s throat is at the sonic condition Consider a converging Nozzle: VA RT p VA m = = ρ & o o o T p ρ r p receiver plenum e e V p Mass Flow Rate (ideal gas): kRT V c V M = = MA RT k p A kRT M RT p m = = & MA RT k p m = & Choked Flow MA RT k p m = & Mass Flow Rate (ideal gas): 1 2 1 2 1 −       + − = k k o M k p p Recall, the stagnation pressure and Temperature ratio and substitute: ( ) k k o o M k MA RT k p m − +       − + = 1 2 1 2 2 1 1 & 1 2 1 2 + − = M k T To ( ) ( ) 1 2 1 2 1 1 2 1 − +         + − + = k k k M k M A A The critical area Ratio is: ( ) k k o o k RT k A p m − +       + = 1 2 1 2 1 & If the critical area (A) is where M=1:
17056	https://www.trapshooters.com/threads/probability-of-hitting-a-target.924243/	Steve W We were discussing the probability of hitting a single target with multiple shooters. If all shooters in this question all have 70% hit rate on average. We launch one target to the sky within their guns & ammo's effective range. If only one shooter fire a shot, it'll have 70% chance of hitting it. What's the probability of hitting that target with two shooters & three shooters firing at the same target at the same time (without knowing there're other shooters)? IIRC, high school math (long time ago) will be one shooter 70%, two shooters 91%, & three shooters 97.3%. But two of my friends with EE PhD @ Berkeley and ME @ USC, says: "Seems to me that the 70% is a statistical SWAG for any launch against any target. The probability of a hit would average out over a number of trials to 70% 🤔" I don't know exactly how much their education really worth, but they both got paid a lot on their jobs as rocket scientists (actually making rockets). I'm confused. Very confused. Can any of you help me with this? Probability of hitting a target Jump to Latest 12K views 98 replies 39 participants last post by DuckNut Steve W Discussion starter 8,580 posts · Joined 1998 Add to quote Only show this user #1 · (Edited) We were discussing the probability of hitting a single target with multiple shooters. If all shooters in this question all have 70% hit rate on average. We launch one target to the sky within their guns & ammo's effective range. If only one shooter fire a shot, it'll have 70% chance of hitting it. What's the probability of hitting that target with two shooters & three shooters firing at the same target at the same time (without knowing there're other shooters)? IIRC, high school math (long time ago) will be one shooter 70%, two shooters 91%, & three shooters 97.3%. But two of my friends with EE PhD @ Berkeley and ME @ USC, says: "Seems to me that the 70% is a statistical SWAG for any launch against any target. The probability of a hit would average out over a number of trials to 70% 🤔" I don't know exactly how much their education really worth, but they both got paid a lot on their jobs as rocket scientists (actually making rockets). I'm confused. Very confused. Can any of you help me with this? 4,545 posts · Joined 2019 Add to quote Only show this user #2 · Steve W said: IIRC, high school math (long time ago) will be one shooter 70%, two shooters 91%, & three shooters 97.3%. Click to expand... I think this would be right, of course your chances are going to improve if you are throwing more lead at it! Shutnlar1 477 posts · Joined 2017 Add to quote Only show this user #3 · Or if one of the shooters is jjhess son! P Pull & Mark 17,277 posts · Joined 2006 Add to quote Only show this user #4 · You wanted to know the probability. OK then, I say that one of the three shooters would probably hit the target every time. LOL. break em all jeff Ram Rod 7,527 posts · Joined 2017 Add to quote Only show this user #5 · ""Huh""" JBrooks 8,664 posts · Joined 2006 Add to quote Only show this user #6 · You buddies are correct. Think of it this way. The 3 are going to shoot 10 targets. What are the chances that all 3 shooters hit the first target? 70% for each so the average is 70%. What are the chances that all 3 shooters miss the next target? 30%. So, that average is 30% just like their individual averages. As such, the probability remains at 70% for a hit on each individual target. However, over 10 targets, the shooters are not going to shoot in unison. While they will all miss 3 targets, they all won't miss the same 3 targets. Think of this as a hat with 21 white balls and 9 black balls. the 3 guys each draw a ball simultaneously. Their individual chance of drawing a white ball is 70%. They are all white. However, the percentage of black balls in the hat has gone up compared to white balls,18-9. So, the probability of someone drawing a black ball has gone up from 30% to 33%%. Next draw there are 2 whites and one black, 66.6% white, 33.3% black. Now the ratio is 16-8. Next draw is 2 blacks and one white. We now have 21 balls left, 15 white, 6 black. Now the chance of drawing a white ball is back up to 71.4%. And so on. Not really much off of 70% As you can see, no matter how you split the draws, eventually 21 white balls will get drawn and 9 black balls will get drawn.As such for our shooters, the more often 1, 2 or 3 shooters hit the same target, the more likely it becomes that all 3 shooters will miss the same target. So, the probability is still going to be around 70% on any given target. BOXMAN76 325 posts · Joined 2021 Add to quote Only show this user #25 · (Edited) Very well stated and explained. I think you should be the statistics professor…. tomk2 10,582 posts · Joined 1998 Add to quote Only show this user #7 · (Edited) I get a 0.3 chance one shooter misses, 0.09 chance two shooters miss the same target, and a 0.027 chance that all three miss the same target. Of course they would have to be independent events, where the action of one shooter does not effect the outcome of the others. For that mathematical analysis to be correct, the target would have to remain intact with unaltered fight path and speed, and each shooter unaware what the outcome was for the other shooters. I suppose it also applies to three separate identical targets thrown for each shooter independently. As such, shooter one fires at one straight away from post 3, then shooter 2 comes out and fires at the identical target, as does shooter 3. They do not know if the other two shooters hit or missed, and you still get 0.3 chance one of them misses, 0.09 chance two of the three misses, and 0.027 chance all three of them miss. There is a 0.7 chance one of them hits, 0.49 chance two of them hit, and 0.343 chance all three of them hit. C Chichay 5,833 posts · Joined 2007 Add to quote Only show this user #8 · (Edited) The probability is independent of each other once the previous occurence(s) has become a certainty. When a guy shoots, he either hits or misses. Assuming the odds are equal, a guy takes four shots and he has already hit the first 3 targets, what is the odds of him hitting the fourth target? It is still 50%. Before he takes the 4 shots, what is the odds of him hitting all 4? It's 50% x %50% x %50% times 50%. But the odds in trapshooting isn't equal In the case of unloaded dice, what are the odds of rolling a specific number (1-6)? One sixth. What are the odds of rolling the same number twice? Before getting the result of the first roll, the odds is one sixth times one sixth. Once the result of the first roll is already known, what is odds of getting it again? Just one sixth! JBrooks 8,664 posts · Joined 2006 Add to quote Only show this user #12 · Chichay said: The probability is independent of each other once the previous occurence(s) has become a certainty. When a guy shoots, he either hits or misses. Assuming the odds are equal, a guy takes four shots and he has already hit the first 3 targets, what is the odds of him hitting the fourth target? It is still 50%. Before he takes the 4 shots, what is the odds of him hitting all 4? It's 50% x %50% x %50% times 50%. But the odds in trapshooting isn't equal In the case of unloaded dice, what is odds of rolling a specific number (1-6)? One sixth. What is odds of rolling the same number twice? Before getting the result of the first roll, the odds is one sixth times one sixth. Once the result of the first roll is already known, what is odds of getting it again? Just one sixth! Click to expand... The outcome of the shot in binary, you either break it or you don't. However that is not the probability that you break it. That is determined by your proficiency on any particular target presentation from a given post. Good shooters, 97+ average, know that they may have 3-4 targets that if they miss there is a higher probability it will be one of those presentations. As such, they have a 97% + chance of hitting most targets and maybe a 95% chance of hitting their "hard" targets. So, while the out come of an individual target is binary, the probability of a shooter hitting it is far from binary. S showmeshooter 782 posts · Joined 2017 Add to quote Only show this user #9 · If all 3 are firing at each target the probability of all 3 missing is 0.3 x 0.3 x 0.3 = 0.027. Probability of at least one hitting is therefore 1 - 0.027 = 0.973. if two are firing probability of both missing is 0.3 x 0.3 = 0.09. Probability of at least one hit is 1 - 0.09 = 0.91. Of course, this assumes independent probabilities for each shooter. If they are interacting all bets are off. JBrooks 8,664 posts · Joined 2006 Add to quote Only show this user #13 · showmeshooter said: If all 3 are firing at each target the probability of all 3 missing is 0.3 x 0.3 x 0.3 = 0.027. Probability of at least one hitting is therefore 1 - 0.027 = 0.973. if two are firing probability of both missing is 0.3 x 0.3 = 0.09. Probability of at least one hit is 1 - 0.09 = 0.91. Of course, this assumes independent probabilities for each shooter. If they are interacting all bets are off. Click to expand... So, you contention is that if you put 3 really bad, 70% average shooters on posts 2, 3 & 4 and locked the targets on straight and all shot at the same time at each target for100 targets, that they would combine for a score of 97? AusAmateur 7,148 posts · Joined 2017 Add to quote Only show this user #10 · Why, God, why???? 22hornet 2,443 posts · Joined 2007 Add to quote Only show this user #18 · I hear you, Brother! Shutnlar1 477 posts · Joined 2017 Add to quote Only show this user #11 · My head hurts, is this “new math” or the stuff my grandkids are learning? 🤯🤪 waldedw 1,828 posts · Joined 2019 Add to quote Only show this user #15 · GEEZZ the grand is over and people are really bored, I'll be just fine if I never find out the answer to this one machine121 1,344 posts · Joined 2018 Add to quote Only show this user #17 · Wait…is it winter already? o-hale 12,289 posts · Joined 1998 Add to quote Only show this user #19 · Ask V10, I was told he is actually a rocket scientist. Steve W Discussion starter 8,580 posts · Joined 1998 Add to quote Only show this user #37 · Somehow I'm under the impression he is, or at least something like that. Years ago he made a statement to correct someone else's inexcusable mistake backed by very scientific proof that impressed me. Ricko 72 posts · Joined 2022 Add to quote Only show this user #21 · (Edited) Depends on the station. If the three 70% shooters all took the shot from post 1 or 5, the statistical probability of a broken bird would be in the neighborhood of 0% to less than 71%. Shooting from post 3, the odds of at least a chip would be greater than 69%. Cornbread are square. Update: Presenting this question to my sister-in law’s - geologist half sister’s- rocket scientist husband, who works on drone mission planning for the Mars Lander, his first response was “7 1/2 or 8s ?”. Stating “eights of course”, dude let me borrow two non public pieces of hardware he borrowed from an RS assigned to the JWST that’s used for mind boggling computations just like this. The IR beam has an accuracy of 0.00 and when jury rigged to co-rectify the nonferrous linear singularity of the shooter’s 3 loaded hulls via blue tooth to the sliding scale of the (classified) parabolic statistical analyzer, the answer is 99.3%, 70% of the time using factory STS 1-1/8 #8’s. Apparently, the missing .7% is due to an anomaly involving the speed of light measured in the northern hemisphere as compared to space or down under. This equipment only works on nonferrous or NASA hulls made of unobtanium. Jim Porter 2,398 posts · Joined 1998 Add to quote Only show this user #22 · If you use 7.5 instead of 8, will the odds change?? jjhess 4,545 posts · Joined 2019 Add to quote Only show this user #24 · You beat me to it jjhess 4,545 posts · Joined 2019 Add to quote Only show this user #23 · Are they shooting 7.5s or 8s? DuckNut 13,524 posts · Joined 2016 Add to quote Only show this user #26 · This is for 2 shooters - I am not going to figure three for you! Let A denote the event that A hits Let B denote the event that B hits. Let E denote the event that the target is hit once. The shots are independant. Then to find the answer: P(A/E) = P(A∩E) / P(E) Hence: P(A∩E) = P(A∩Bc) = P(A)P(Bc) = P(A)(1−P(B)) = 0.3 × 0.3 = 0.09 and finally: 1 - 0.09 = 91% the target will be hit. You have to realize this answer is for EACH target with 2 shooters shooting at it. jjhess 4,545 posts · Joined 2019 Add to quote Only show this user #27 · If you think the odds are still 70%, go get 3 dice roll all 3 10 times, how many times will you have at least 1 even number per roll? You must be thinking 50% but I would bet it would be over 8 or 9 of the 10 rolls. JBrooks 8,664 posts · Joined 2006 Add to quote Only show this user #33 · This is an example of random results. In the case of the 3 shooters, they will miss the target 30 times out of 100 shots. So what are the odds that one each of those 30 shots will come on the same target? In effect, how many times out of 100 draws will all three draw a black ball out of the hat that starts with 90 black balls and 210 white balls? The math guys here say that will only happen 3 times. Jaburg1 3,524 posts · Joined 2017 Add to quote Only show this user #28 · Two shooters = 91% rookieshooter 587 posts · Joined 2015 Add to quote Only show this user #29 · This is bordering in the realm of Quantum mechanics. The 2 slit experiment comes to mind. It all depends on how you test it. Is the target broken or missed? Or substitute the famous ''Cat in a box'' to a target in a box. Was the target broken or missed? Then again the results collapses when observed. J Joe9499 235 posts · Joined 2013 Add to quote Only show this user #30 · If there shooting 70% they need a new hobby, maybe take up Skeet. flashmax 30,336 posts · Joined 2013 Add to quote Only show this user #31 · Sporties are satisfied at 70% B bossbasl 2,847 posts · Joined 2013 Add to quote Only show this user #32 · Similar to the question: If a couple had three natural birth children, all girls; what is the percentage probability the next child will be a boy? Obviously, the correct answer is 50%! tomk2 10,582 posts · Joined 1998 Add to quote Only show this user #34 · (Edited) uh... yeah. Kinda. Post deleted. 'the only thing I do worse than statistics is explain statistics. We really should be allowed to read our post and delete it for a short time. JBrooks 8,664 posts · Joined 2006 Add to quote Only show this user #35 · No, If the three shooters are 70% percent shooters. Each will hit the target 70 times and miss the target 30 times out of 100. As such, there will be 210 times the target will be hit by one or more of the shooters and 90 times the target will be missed by one or more of the shooters. Anytime all 3 miss on a particular target, the target won't be broken. This is not about the binary outcome. This is about the odds of all 3 shooters missing the same target and that target being one of the 30 each is going to miss. In effect, all 3 drawing a black ball out of the hat on that draw. You are confusing two different concepts. One being the simple binary outcome of either the target breaking or not breaking. The other, the statistical reality of the shooter's average. If you think a AAA shooter has the same statistical probability of breaking the next target as a D shooter, never ever go to Las Vegas. tomk2 10,582 posts · Joined 1998 Add to quote Only show this user #36 · (Edited) Since each shooters outcome is not determined by the other shooters or their own previous shots, you do not have 300 balls in a hat. By that reasoning, what the first shooter does changes the odds of what the second and third shooter can do by removing one of their possible balls. You have three separate hats, each with 30 black balls and 70 white, and each shooter draws from their own hat with 100 balls and puts the balls back into their respective hat after each draw before they take their next shot/draw. Only a small number of 100 target events will a shooter actually achieve the 70% hit and 30% miss, most will be different, demonstrating that it is not about what order you empty the hat to arrive at the predetermined odds. There will be plenty of 100 draw events that had 80 white and only 20 black, even though the hat had a 70/30 mix. I can't tell if we agree and I am misinterpreting your analogy, I figure there is a 70% chance. spitter 8,022 posts · Joined 1998 Add to quote Only show this user #38 · I'm sorry... since no one is perfect... It's always 50/50... like flipping a coin... JBrooks 8,664 posts · Joined 2006 Add to quote Only show this user #40 · You are confusing two different concepts. The target breaks or doesn't break. That is a binary. The chances of a particular shooter breaking a target is not. That is based on his proficiency in breaking that target presentation from the post he is shooting from. Another example. You need bypass surgery. The Doc tells you that given your age and medical history, you have a 99% chance of surviving the surgery. That figure is based on thousand of iterations of the surgery on people similar to you.You say those are great odds, let's do it. However, if he gave you the binary outcome it would be either you live or die. That concept disregards the outcome of those thousands of other surgeries. Same with shooting a target, The binary outcome is dead or lost. But if the shooter breaks that target presentation 98% of the time he shoots at it, he has a 98% probability of breaking it the next time he shoot at it. 3,084 posts · Joined 1998 Add to quote Only show this user #39 · When I took a Machine shop SPC (Statistical Process Control) class on how to help diagnose machine problems, I was amazed at how accurate these thigs can be. The class instructor told us to go out and ask 29 people what their Birthday date, Month and Day were, and two of those people would have the same. I did this several times around the shop, and never got disappointed. God grief, now we are doing this with shooting. LOL Related Threads ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Home Forums Trap Shooting Forums Shooting Related Discussions
17057	https://basics.sjtu.edu.cn/~yangqizhe/pdf/dm2023w/slides/DMLec2-handout-zh.pdf	《离散数学》 2-命题逻辑(II)(Proposition Logic(II)) 杨启哲上海师范大学信机学院计算机系 2023 年9 月25 日上周回顾 • 命题与命题联结词。 ◦命题的基本概念，什么是命题？简单命题和复合命题。 ◦联结词的概念，与日常用语的区别。如何将一个复杂命题符号化。 • 命题公式。 ◦命题公式的概念，如何判断一个符号串是否是命题公式。 ◦命题公式的真值，真值表的概念。三种公式类型（重言式、可满足式、矛盾式）。 2 主要内容 þ 等值演算 þ 联结词的完备集 þ 命题公式的范式 þ 可满足性问题 3 等值演算等值引入 • 在初等数学中，当我们进行一些代数式运算时，我们经常会使用一些等式，如： (a + b)2 = a2 + 2ab + b2 a2 −b2 = (a + b)(a −b) sin(x + y) = sin x cos y + sin y cos x · · · 在命题逻辑里，我们也可以建立同样的等值式，去进行相应的运算。 5 等值式回顾相等的定义，我们知道，两个数a 和b 相等，当且仅当a 和b 的值相同。类比该定义，我们可以给出等值式的定义：定义1 [等值式]. 设A 和B 是两个命题公式，若对于任何一个赋值，A 和B 的真值都相同，则称A 和B 是等值的(或等价的)，记作A ⇔B(A = B)。例2. 对于公式a →b 和¬a ∨b 来说，尽管他们的形式是不一样的(语法上不同)，但对于任何一个关于a, b 的赋值来说，他们的真值是相同的(即真值表相同)，所以这两个公式是等值的(语义上相同). 6 等值式的证明通过真值表我们可以很容易的证明两个命题公式是等值的。证明¬(p ∨q) 和¬p ∧¬q 等值只需要验证两个公式的真值相同即可。 p q 0 0 0 1 1 0 0 1 ¬(p ∨q) ¬p ∧¬q 1 1 0 0 0 0 0 0 7 关于等值式 A 和B 拥有不同的变元可以么？完全可以！事实上，假设A 的变元集合为P1,B 的变元集合为P2, 我们可以假设A, B 都是在变元集合P1 ∪P2 上的命题公式，一个不存在在A 中但存在在P1 ∪P2 的命题变元可以视作A 的哑元，即无论其真或者假，都不会影响A 的真值。 8 关于等值式 ↔与⇔的区别？尽管A ↔B 和A ⇔B 都蕴含了些A 和B 等价的意思，但两者式完全不同的概念。 • 从真假性来看，A ↔B 只有是重言式时，才能说明A 和B 等值。而A ⇔B 已经表明A 和B 等值。 • 从形式角度看，A ↔B 是一个命题公式，而A ⇔B 不是，它反映的是一种命题公式上的等价关系。定理3 [等值定理]. A ⇔B 当且仅当公式A ↔B 是重言式。 9 等值关系和数学中”=“的性质一样，等值关系具有如下三种性质： 1. 自反性(reflexivity)，即对于任何命题公式A, A ⇔A. 2. 对称性(symmetry)，即对于任何命题公式A, B, 若A ⇔B, 则B ⇔A. 3. 传递性(transitivity)，即对于任何命题公式A, B, C, 若A ⇔B 且B ⇔C, 则A ⇔C. 一般来说，满足这三个性质的关系被称为等价关系，其描绘了两者是” 等同的”。 10 等值证明的方法在前面的例子中，我们已经展示了真值表技术可以用来证明两个公式等值，但是否存在其他方法呢？等值证明的方法 • 真值表技术。 • 利用基本等值式进行等值演算。利用等值定理，我们可以构造一些相应的等值式，然后利用这些等值式进行等值演算。 11 基本等值公式(I) 等值公式 1. 双重否定律： ¬(¬A) ⇔A 2. 幂等律： A ∨A ⇔A, A ∧A ⇔A A →A ⇔1, A ↔A ⇔1 3. 交换律： A ∨B ⇔B ∨A, A ∧B ⇔B ∧A, A ↔B ⇔B ↔A 4. 结合律： A ∨(B ∨C) ⇔(A ∨B) ∨C A ∧(B ∧C) ⇔(A ∧B) ∧C A ↔(B ↔C) ⇔(A ↔B) ↔C 12 基本等值公式(II) 等值公式 5. 分配律： A ∨(B ∧C) ⇔(A ∨B) ∧(A ∨C) A ∧(B ∨C) ⇔(A ∧B) ∨(A ∧C) A →(B →C) ⇔(A →B) →(A →C) 6. 德摩根律： ¬(A ∨B) ⇔¬A ∧¬B, ¬(A ∧B) ⇔¬A ∨¬B ¬(A →B) ⇔A ∧¬B, ¬(A ↔B) ⇔A ↔¬B ⇔¬A ↔B 7. 吸收律： A ∨(A ∧B) ⇔A, A ∧(A ∨B) ⇔A 13 基本等值公式(III) 等值公式 8. 零律： A ∨1 ⇔1, A ∧0 ⇔0 A →1 ⇔1, 0 ↔A ⇔1 9. 同一律： A ∨0 ⇔A, A ∧1 ⇔A, 1 →A ⇔A 1 ↔A ⇔A, A →0 ⇔¬A, 0 ↔A ⇔¬A 10. 补余律(排中律、矛盾律) A ∨¬A ⇔1, A ∧¬A ⇔0 A →¬A ⇔¬A, A ↔¬A ⇔0 14 基本等值公式(IV) 等值公式 11. 蕴含等值式： A →B ⇔¬A ∨B 通常在运算中，¬A ∨B 要比A →B 方便。 12. 等价等值式： A ↔B ⇔(A →B) ∧(B →A) 这表明等价关系可以用双蕴含词来表达，符合两个符号形式上的联系。 13. 假言易位： (A →B) ↔(¬B →¬A) 这阐述的就像高中所说的“原命题和逆否命题等价” 。 15 基本等值公式(V) 等值公式 14. 等价否定等值式： A ↔B ⇔¬A ↔¬B 直观上说，如果A, B 是等价的，那么他们的否定也是等价的。 15. 归谬论： (A →B) ∧(A →¬B) ⇔¬A 这个直观上理解就是反证法。 16 更多的等值公式更多的等值公式 • A →(B →C) ⇔(A ∧B) →C (前提合并) • A →(B →C) ⇔B →(A →C)(前提交换) 请注意：A →(B →C) ̸= (A →B) →C! • A ↔B = (A ∧B) ∨(¬A ∧¬B) ⇔(A ∨¬B) ∧(¬A ∨B)(取值描述) • (A →C) ∧(B →C) ⇔(A ∨B) →C(蕴含合并) 17 置换规则置换规则令Φ(A) 是一个包含A 的命题公式，其中A 是任意一个命题公式，Φ(B) 是将Φ(A) 中的一些A 替换成公式B 后得到的公式, 则有：若A ⇔B, 则Φ(A) ⇔Φ(B) 回顾：代入规则若A 是一个重言式，B 是一个命题公式，若将A 中的某个命题变元p 全部用B 替换形成一个新的公式A′，则A′ 也是重言式。 18 代入规则与置换规则的区别 • 置换规则是将其中一个公式中的某个子公式替换成另一个公式，而代入规则是将其中一个公式中的某个命题变元替换成另一个公式。 • 置换规则说明了替换某个子公式的等值式不会改变公式的真值，从而可以进行等值式的证明。代入规则则是对重言式的证明。 • 置换不需要全部换掉，可以只替换其中一部分。 19 等值演算等值演算利用等值定律, 基本等值式以及替换规则进行公式推演。等值演算的一般方法一般是将⇔两边的公式推演成相同形状的公式, 从而证明等值式成立. • 尽可能的减少出现联结词的种类。 • 将否定联结词移到命题变元上。 • 可以转换成相应的范式。（后续内容） 20 等值演算的例子(I) 证明：(p ∨q) →r ⇔(p →r) ∧(q →r) 证明. 由基本等值式可得： (p ∨q) →r ⇔¬(p ∨q) ∨r (蕴含等值式) ⇔(¬p ∧¬q) ∨r (德摩根律) ⇔(¬p ∨r) ∧(¬q ∨r) (分配律) ⇔(p →r) ∧(q →r) (蕴含等值式) 也可以从右边开始进行推导 21 等值演算的例子(II) 用等值演算判断下列公式的类型 1. (P →Q) ∧P →Q. 2. ¬(P →(P ∨Q)) ∨R. 3. P ∧(((P ∨Q) ∧¬P) →Q). 证明. 只给出第一个的演算过程： (P →Q) ∧P →Q ⇔(¬P ∨Q) ∧P →Q (蕴含等值式) ⇔((¬P ∧P) ∨(Q ∧P)) →Q (分配律) ⇔(0 ∨(Q ∧P)) →Q (取补律) ⇔(Q ∧P) →Q (同一律) ⇔¬(Q ∧P) ∨Q (蕴含等值式) ⇔¬P ∨¬Q ∨Q (德摩根律，结合律) ⇔1 (同一律) 22 对偶式与内否式我们介绍一些可以简化等值公式讨论的技术，假设下面提到的公式仅使用了¬, ∧, ∨ 这三个联结词。定义4 [对偶式]. 令A 式任一命题公式，将其中的∧, ∨, 1, 0 分别用∨, ∧, 0, 1 代替形成的新公式称为A 的对偶式，记作A∗。定义5 [内否式]. 令A 式任一命题公式，将其中所有肯定形式出现的变元x 换成¬x、所有否定形式出现的变元¬x 换成x 后形成的新公式称为A 的内否式，记作A−。 23 对偶式与内否式例子 • 令A def = (¬P ∨(Q ∧R)) ∧T，则有： 1. A∗= (¬P ∧(Q ∨R)) ∨F. 2. A−= (P ∨(¬Q ∧¬R)) ∧T. 3. (A∗)−= (P ∧(¬Q ∨¬R)) ∨F 4. (A−)∗= (P ∧(¬Q ∨¬R)) ∨F 24 对偶式的一些性质定理(I) 定理6. 令A 是任一只包含¬, ∧, ∨的命题公式，则有： 1. (A∗)∗⇔A, (A−)−⇔A. 2. (A ∨B)∗⇔A∗∧B∗, (A ∨B)−⇔A−∨B−. 3. (A ∧B)∗⇔A∗∨B∗, (A ∧B)−⇔A−∧B−. 4. (¬A)∗⇔¬(A∗), (¬A)−⇔¬(A−) 5. ¬A ⇔(A∗)−.(德摩根律一般形式) 6. (A∗)−⇔(A−)∗. 25 ¬A ⇔(A∗)−的证明证明. 对A 中出现的联结词个数作归纳法证明。 BASE: n = 0，即A 没有联结词，从而A = p，从而¬A = ¬p, (A∗)−= ¬p，定理成立。 INDUCTION: 假设n ⩽k 时命题成立，则n = k + 1 时，A 中至少有一个联结词，有如下三种情形： 1. A = ¬A1, A1 的联结词个数⩽k。由归纳假设¬A1 ⇔(A∗ 1)−因此： ¬A ⇔¬(¬A1) ⇔¬((A∗ 1)−) ⇔(¬A∗ 1)−⇔(A∗)− 2. A = A1 ∨A2，A1, A2 的联结词个数⩽k。由归纳假设： ¬A ⇔¬(A1 ∨A2) ⇔¬A1 ∧¬A2 ⇔(A∗ 1)−∧(A∗ 2)−⇔((A1 ∨A2)∗)−⇔(A∗)− 3. A = A1 ∧A2，与上一种情况相同。 26 对偶式的一些性质定理(II) 定理7. 令A 是任一只包含¬, ∧, ∨的命题公式，则有： 1. A 与A−同永真，同可满足。 2. ¬A 与A∗同永真，同可满足。推论8. 若有A ⇔B, 则有A∗⇔B∗和A−⇔B−。 27 推论8的证明证明. • A ⇔B 等价于公式A ↔B 永真，也等价于¬A ↔¬B 永真。 • ¬A ⇔(A∗)−, ¬B ⇔(B∗)− • 公式(A∗)−↔(B∗)−永真 • 公式((A∗)−↔(B∗)−)−永真，即公式A∗↔B∗永真。 28 等值演算总结等值演算 • 等值式的概念、等值定理。 • 等值关系(⇔), ⇔与↔的区别。 • 等值证明的方法: 真值表技术、等值演算。 1. 基本等值式。 2. 替换规则（与代入规则的不同） 3. 一些技巧-对偶式与内否式。 29 联结词的完备集命题公式与真值表关系由命题公式写真值表是非常容易的，但是反过来呢？考察我们之前描述“异或”的概念时，用了如下的表达式： p ⊕q def = (p ∧¬q) ∨(¬p ∧q) 如果将不带¬ 的视为1，带¬ 的视为0, 上式右边的式子便可以看作由异或的取真赋值写成的公式。 31 由真值表来书写公式由取真赋值来写(由T 来写) • 核心思想：将所有取真赋值用∧表达出来，然后用∨连接起来，表示任选其中一个取真赋值便能使其为真，否则为假。 • 示例：假设p = 1, q = 1, r = 0 为其一个取真赋值，则建立公式p ∧q ∧¬r，该公式仅有在p = 1, q = 1, r = 0 时为真。由取假赋值来写(由F 来写) • 核心思想：将所有取假赋值用∨表达出来，然后用∧连接起来，表示只有不取到其中任何一个赋值时才能为真，否则为假。 • 示例：假设p = 1, q = 1, r = 0 为其一个取假赋值，则建立公式¬p ∨¬q ∨r，该公式仅在p = 1, q = 1, r = 0 时为假，意味着此时取到了取假赋值。 32 由真值表来书写公式我们通过如下一个例子来说明，假设p, q 是两个命题变元，公式A 的真值表如下： p q A 0 0 1 0 1 0 1 0 0 1 1 1 • 由T 来写：A = (¬p ∧¬q) ∨(p ∧q). • 由F 来写：A = (p ∨¬q) ∧(¬p ∨q). 当然我们也可以发现，A 其实就是p ↔q. 33 联结词的个数我们之前介绍了几种常见的联结词，一个很自然的问题是到底有多少联结词？例9. 我们可以定义很多新的联结词： • 异或(exclusiveOR): p ⊕q def = (p ∧¬q) ∨(¬p ∧q) • 与非(NAND): p ↑q def = ¬(p ∧q) • 或非(NOR): p ↓q def = ¬(p ∨q) 问题10. 对于n 个命题变元p1, . . . pn，一共可以定义出多少个不同的联结词？这其中互相独立的有多少个？ 34 真值函数一个n 元联结词实际上可以看作是由{0, 1}n 指向{0, 1} 的一个函数，我们称这样的函数为真值函数。定义11 [真值函数]. 真值函数，是指以真值为定义域和值域的函数，即{0, 1}n 到{0, 1}。例12. 一元联结词¬ 可以看成如下一个真值函数¬ : {0, 1} →{0, 1}： ¬(0) = 1, ¬(1) = 0. 因此，前面的问题变转化成：一共有多少个n 元真值函数？ 35 一元联结词的个数先从最简单的一元联结词入手。一元真值函数只有一个自变元x，它有两个取值可能0, 1。对于其每个取值，函数也有两种不同的函数值0, 1。因此一共可以定义4 个不同的一元真值函数，如下表： x f0(x) f1(x) f2(x) f3(x) 0 0 0 1 1 1 0 1 0 1 相应的，有4 个一元联结词，但f0 和f3 表示永假和永真，f1 表示不变，因此只有f2（即¬) 被经常使用。 36 二元联结词的个数同样的方式，我们可以列举出所有的二元联结词，一共222 = 16 个： x y g0(x, y) g1(x, y) g2(x, y) . . . g15(x, y) 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 1 1 0 1 0 1 显然,g1 就是我们熟知的∧. 37 n 元联结词的个数一般的，我们有：引理13. n 元真值函数一共有22n 仲，因此一共可以定义22n 种不同的联结词。例14. 我们可以顶一个一个三元联结词#: #(x, y, z) 为真当且仅当x, y, z 至少两个为真。其对应的真值函数满足: #(0, 0, 0) = 0, #(0, 0, 1) = 0, #(0, 1, 0) = 0, #(0, 1, 1) = 1 #(1, 0, 0) = 0, #(1, 0, 1) = 1, #(1, 1, 0) = 1, #(1, 1, 1) = 1 38 联结词的完备集第二个关于联结词的问题是哪些联结词是能相互表示的。例15. ¬, ∨便可以表示我们学过的其他三个联结词∧, →, ↔： • A ∧B def = ¬(¬A ∨¬B). • A →B def = ¬A ∨B. • A ↔B def = (¬(¬A ∨¬B)) ∨(¬(A ∨B)) 而仅依靠∨, ∧却不能表示所有的联结词，比如它不能表示¬. 39 联结词的完备集定义16 [完备集]. 令S 是一个联结词集合，若任何一个n 元真值函数都可以仅由S 仲的联结词构成的公式表示，则称S 是联结词完备集。例17. 由上述讨论可知，以下都是完备集： {¬, ∨}, {¬, ∧}, {¬, →}, {¬, ∨, ∧}, {¬, ∨, ∧, →, ↔} 而以下集合都不是完备集： {¬}, {∨, ∧}, {¬, ↔} 40 最小联结词完备集(基底) 定义18 [最小联结词完备集(基底)]. 完备的联结词集合的联结词是独立的，也就是说这些联结词不能相互表示；或者说，不含冗余联结词联结词集合, 称为最小联结词完备集(基底). 例19. 以下都是最小联结词完备集： {¬, ∨}, {¬, ∧}, {¬, →} 问题20. 存不存在只有一个联结词的最小完备集？ 41 与非NAND 和或非NOR 定义下述两个联结词： • 与非(NAND): A ↑B def = ¬(x ∧y) • 或非(NOR): A ↓B def = ¬(x ∨y) 定理21. {↑} 和{↓} 都是最小联结词完备集。证明. 我们只证明{↑}，事实上： • ¬A ⇔A ↑A. • A ∧B ⇔(A ↑B) ↑(A ↑B). 42 联结词总结联结词 • 由真值表书写公式. ◦由成真赋值来写。 ◦由成假赋值来写。 • 联结词的个数，真值函数。 • 联结词的完备集。 ◦完备集的概念。 ◦最小完备集。 43 命题公式的范式标准形式我们知道，尽管等值的n 元公式只有22n 个，但是公式的数量是无穷多的： x, ¬x, ¬¬x, ¬¬¬x, . . . , 能否找到一种标准形式，使得等值的不同公式都可以转换成相同的形式？ 45 范式-术语定义22 [一些术语的定义]. 我们下面介绍一些术语： • 命题变项p 和其否定¬p 都被称为文字 • 有限个文字的合取叫简单合取式。 • 有限个文字的析取叫简单析取式。 • p 与¬p 被称为互补对。定义23 [析取范式和合取范式]. • 由若干个简单合取式用析取联结词∨连接起来的公式称为析取范式。 • 由若干个简单析取式用合取联结词∧连接起来的公式称为合取范式。 46 范式举例(I) 析取范式的例子 1. (p ∧q) ∨(¬p ∧q) 2. (p ∧q ∧r) ∨(¬p ∧q) ∨(p ∧¬q) 3. (p ∧¬p) ∨(q ∧q). 合取范式的例子 1. (p ∨q) ∧(¬p ∨q) 2. (p ∨q ∨r) ∧(¬p ∨q) ∧(p ∨¬q) 3. (p ∨¬p) ∧(q ∨q). 47 范式的存在性由之前的讨论我们知道，{¬, ∨, ∧} 是一个联结词完备集，因此任何一个公式都可以转换成一个等值的只有这三个联结词的公式。再由双重否定律、德摩根律、分配律： ¬¬A ⇔A ¬(A ∧B) ⇔¬A ∨¬B ¬(A ∨B) ⇔¬A ∧¬B A ∨(B ∧C) ⇔(A ∨B) ∧(A ∨C) A ∧(B ∨C) ⇔(A ∧B) ∨(A ∧C) 可得：定理24 [范式存在定理]. 任何一个命题公式存在与其等值的的析取范式和合取范式。 48 求范式的方法上述证明也给出了求范式的方法：范式求法 1. 消去其中不是¬, ∨, ∧的联结词： ◦A →B = ¬A ∨B. ◦A ↔B = (¬A ∨B) ∧(A ∨¬B). 2. 用德摩根律将否定号移到文字上。 ◦¬(A ∨B) = ¬A ∧¬B. ◦¬(A ∧B) = ¬A ∨¬B. 3. 用分配律将公式化为合取范式或析取范式。 ◦A ∨(B ∧C) = (A ∨B) ∧(A ∨C). ◦A ∧(B ∨C) = (A ∧B) ∨(A ∧C). 4. 进行适当的化简。 ◦A ∨¬A = 1. ◦A ∧¬A = 0. 49 范式举例(II) 求(p →q) ↔r 的合取范式和析取范式 1. 先求合取范式。 (p →q) ↔r = (¬p ∨q) ↔r (消去→) = (¬(¬p ∨q) ∨r) ∧((¬p ∨q) ∨¬r) (消去↔) = ((p ∧¬q) ∨r) ∧(¬p ∨q ∨¬r) (德摩根律) = (p ∨r) ∧(¬q ∨r) ∧(¬p ∨q ∨¬r) (分配律) 2. 再求析取范式。 (p →q) ↔r = ((p ∧¬q) ∨r) ∧(¬p ∨q ∨¬r) (前面一致) = ((p ∧¬q) ∧(¬p ∨q ∨¬r)) ∨(r ∧(¬p ∨q ∨¬r)) (分配律) = (p ∧¬q ∧¬r) ∨(r ∧¬p) ∨(r ∧q) (分配律) 50 范式的作用一些具有特定形式的范式可以帮助我们判断范式的种类： • 如果合取范式里的每个简单析取式都有互补对，则该范式是重言式. ◦(p ∨¬p ∨r) ∧(q ∨¬q). • 如果析取范式里的每个简单合取式都有互补对，则该范式是矛盾式. ◦(p ∧¬p ∨q) ∨(q ∧¬q ∨r). 51 范式的作用是否可以通过范式来判断是否A ⇔B? 考察如下两个命题公式： 1. A = (¬p ∧r) ∨(q ∧r). 2. B = (¬p ∧q ∧r) ∨(¬p ∧¬q ∧r) ∨(p ∧q ∧r) ∨(¬p ∧q ∧r) 显然A, B 都是析取范式，并A ⇔B. 但是，它们的形式却完全不同。可不可以转换成一类范式，使得如果两个公式是等值的，则它们转换成的范式形式上是一摸一样的？ • 主范式！ 52 极小项定义25. 令公式只涉及n 个命题变量x1, . . . , xn，一个极小项指的是一个由n 个文字构成的合取式，其中每个命题变量xi 以xi 或者¬xi 恰好仅出现在左起第i 个文字上（即命题公式里的文字按字典序排列出现）。例26. 令p, q, r 为涉及到的所有命题变元，以下合取式都是极小项： • p ∧q ∧r. • ¬p ∧q ∧¬r. 每个极小项都包含了所有的命题变元，且每个变元恰好出现一次。 53 极小项的性质极小项的性质 • 极小项一共有2n 个。 ◦可以用n 位二进制数来记录每个极小项，即用0 来表示¬xi,1 来表示xi, 比如：m0 可以用来表示¬x1 ∧¬x2 ∧. . . ∧¬xn. ◦由上述讨论可知，n 个变元的极小项可以用m0, . . . m2n−1 来表示。 • 任何一个极小项只在一个赋值（解释）下为真，即上述表示的二进制数。 • 极小项互不等值，且任何两个极小项的合取是永假的。 • 由所有极小项的析取构成的公式是永真的。极小项的意义极小项相当于唯一确定了一成真赋值，因此对于由所有极小项的析取构成的公式来说，任何一个赋值都是使其为真的赋值，从而该公式是永真的。 54 极大项定义27 [极大项]. 令公式只涉及n 个命题变量x1, . . . , xn，一个极大项指的是一个由n 个文字构成的析取式，其中每个命题变量xi 以xi 或者¬xi 恰好仅出现在左起第i 个文字上（即命题公式里的文字按字典序排列出现）。例28. 令p, q, r 为涉及到的所有命题变元，以下析取式都是极大项： • p ∨q ∨r. • ¬p ∨q ∨¬r. 每个极大项都包含了所有的命题变元，且每个变元恰好出现一次。 55 极大项的性质极大项的性质 • 极大项一共有2n 个。 ◦可以用n 位二进制数来记录每个极大项，即用0 来表示xi,1 来表示¬xi, 比如：M0 可以用来表示x1 ∨x2 ∨. . . ∨xn. ◦由上述讨论可知，n 个变元的极大项可以用M0, . . . M2n−1 来表示。 • 任何一个极大项只在一个赋值（解释）下为假，即上述表示的二进制数。 • 极大项互不等值，且任何两个极大项的析取是永真的。 • 由所有极大项的合取构成的公式是永假的。极大项的意义极大项相当于唯一确定了一成假赋值，因此对于由所有极大项的合取构成的公式来说，任何一个赋值都是使其为假的赋值，从而该公式是永假的。 56 主范式由极小项和极大项的概念，我们可以提出唯一的表达形式-主范式的概念。定义29 [主合取范式和主析取范式]. 仅由极小项（resp. 极大项）构成的析取范式（resp. 合取范式）称为主析取范式（resp. 主合取范式）。定理30 [主范式的存在和唯一性]. • 任何一个含有n 个命题变元的公式，都有唯一的与之等值的恰含这n 个变元的主析取范式。 • 任何一个含有n 个命题变元的公式，都有唯一的与之等值的恰含这n 个变元的主合取范式。 57 主范式的求法-利用基本等值式主析取范式的求法 • 利用基本等值式求出一个析取范式。 • 如果某个简单合取式A 未出现命题变元x, 则通过： A ⇔A ∧(x ∨¬x) ⇔(A ∧x) ∨(A ∧¬x) 补足x。 • 去除重复的变量、矛盾式以及重复出现的极小项。主合取范式的求法 • 利用基本等值式求出一个合取范式。 • 如果某个简单析取式A 未出现命题变元x, 则通过： A ⇔A ∨(x ∧¬x) ⇔(A ∨x) ∧(A ∨¬x) 补足x。 • 去除重复的变量、矛盾式以及重复出现的极大项。 58 主范式的求法-举例(I) 求(p →q) ↔r 的主合取范式和主析取范式 1. 求主合取范式。 ◦由之前的讨论可得其的合取范式： (p ∨r) ∧(¬q ∨r) ∧(¬p ∨q ∨¬r) ◦对于前两个简单析取式(p ∨r) 和(¬q ∨r), 我们分别补足变元q 和p: (p ∨r) ∧(q ∨¬q) = (p ∨r ∨q) ∧(p ∨r ∨¬q). (¬q ∨r) ∧(p ∨¬p) = (¬q ∨r ∨p) ∧(¬q ∨r ∨¬p) ◦整理极大项可得： (p ∨q ∨r) ∧(p ∨¬q ∨r) ∧(¬p ∨¬q ∨r) ∧(¬p ∨q ∨¬r) 上述主合取范式也可以表示为: M0 ∧M2 ∧M5 ∧M6. 59 主范式的求法-举例(II) 求(p →q) ↔r 的主合取范式和主析取范式 1. 求主析取范式。 ◦由之前的讨论可得其的析取范式： (p ∧¬q ∧¬r) ∨(r ∧¬p) ∨(r ∧q) ◦对于后两个简单合取式(r ∧¬p) 和(r ∧q), 我们分别补足变元q 和p: (r ∧¬p) ∨(q ∧¬q) = (r ∧¬p ∧q) ∨(r ∧¬p ∧¬q) (r ∧q) ∨(p ∧¬p) = (r ∧q ∧p) ∨(r ∧q ∧¬p) ◦整理极小项可得： (¬p ∧q ∧r) ∨(¬p ∧¬q ∧r) ∨(p ∧q ∧r) ∨(p ∧¬q ∧¬r) 上述主析取范式也可以表示为: m1 ∨m3 ∨m4 ∨m7. 60 主范式的求法-利用真值表考察如下的命题公式：p →(p ∧q)，其真值表为： p q (p →(p ∧q)) 0 0 1 0 1 1 1 0 0 1 1 1 其中有3 个成真赋值00, 01, 11 和1 个成假赋值10，因此： • 主析取范式为：m0 ∨m1 ∨m3. • 主合取范式为: M2 61 主析取范式和主合取范式的转换主析取范式可以视作是将所有代表成真赋值的最小项析取而成，而主合取范式可以视作是将所有代表成假赋值的最大项合取而成。因此，我们非常容易完成其中的转换：主析取范式和主合取范式的转换令A = mi1 ∨. . . ∨mik 是一个n 个变元的主析取范式，则: 1. I = {i1, . . . , ik} 是相应的成真赋值集合。 2. J = {0, . . . , 2n −1} \ I 为其成假赋值，记为j1, . . . j2n−k。 3. Mj1 ∧. . . ∧Mj2n−k 是其主合取范式。 62 主范式运用例31. 某科研所要从3 名科研骨干A, B, C 中挑选1 ∼2 名出国进修，由于工作需要需要满足如下条件： • 若A 去，则C 也要去。 • 若B 去，则C 不能去。 • 若C 不去，则A 和B 至少要去一个。问有哪些选派方案？ 63 主范式运用解32. 令p : 派A 去 q : 派B 去 r : 派C 去则上述条件可以转换为 (p →r) ∧(q →¬r) ∧(¬r →(p ∨q)) 上述公式的主析取范式为： m1 ∨m2 ∨m5 因此一共有3 种方案： • 只派C 去(对应1(二进制形式：(001)2))。 • 只派B 去(对应2(二进制形式：(010)2))。 • 派A, C 去(对应5(二进制形式：(101)2))。 64 命题公式的范式总结命题公式范式 • 命题公式的标准形式。 ◦析取范式。 ◦合取范式。 • 求范式的方法(基本等值式、真值表)。 • 主析取范式和主合取范式 ◦极小项和极大项 ◦互相的转化 65 可满足性问题命题公式的可满足性问题定义33 [SAT 问题(Satisfiability Problem)]. 给定一个n 个变元的命题公式，可满足性问题是指，是否存在一个使其为真的赋值？我们知道对于任何一个命题公式，其可以转换成析取范式或者合取范式，哪种范式的可满足性更容易判断？析取范式！应为析取范式只需要使得其中某个简单合取式成真即可。但有时，相应的析取范式会非常的长，比如p ∨q 而言，其析取范式为： (p ∧q) ∨(p ∧¬q) ∨(¬p ∧q) 67 SAT 问题介绍一般SAT 问题考察的是一个合取范式C，n-SAT 问题则要求C 中每个析取式最多不超过n 个文字。第一个NP 完全问题 1971 年Stephen Cook和Leonid Levin提出了Cook-Levin 定理，证明了SAT 问题是第一个NP 完全问题。 SAT 问题在计算机科学中有着重要的作用，比如在程序正确性分析里，通过将所需要的性质转换成相应的合取范式，即可通过SAT 问题来判断该性质是否成立。国际上每年都有关于SAT solver 问题的竞赛，具体可以参考 68 消解法-思路令C1, C2 表示合取范式C 中两个不同的简单析取式，满足： C1 = A ∨x C2 = B ∨¬x 记Res(C1, C2) = A ∨B。我们可以发现C1 ∧C2 的可满足性与A ∨B 相同。 • 如果A ∨B 是可满足的，此时必然存在一个赋值使得A 或B 满足，不妨设为A，则令 x = 0，可以得到使C1 ∧C2 满足的一组赋值。 • 如果C1 ∧C2 是可满足的，取其中一个令其为真的解释，并不妨令x = 0，此时由定义 A 必然是可满足的，从而A ∨B 是可满足的。消解法从而我们可以得出如下的思路：对C 中的简单析取式一直进行上述的消解操作，直到不产生新的公式或者产生了矛盾式为止。 69 消解法-算法消解法输入: n 个变元的合取范式C = C1 ∧C2 . . . Cm 输出: C 是否可满足？ 1: S0 ←∅, S1 ←{C1, . . . , Cn}, S2 ←∅ 2: for any c1 ∈S0, c2 ∈S1 do 3: if c1 and c2 can be resolved then 4: c ←Res(c1, c2) 5: if c = ϵ then return False 6: else if c / ∈S0 ∪S1 then S2 ←S2 ∪{c} 7: for any c1, c2 ∈S1 do 8: if c1 and c2 can be resolved then 9: c ←Res(c1, c2) 10: if c = ϵ then return False 11: else if c / ∈S0 ∪S1 then S2 ←S2 ∪{c} 12: if S2 = ∅then return True 13: else 14: S0 ←S0 ∪S1, S1 ←S2, S2 ←∅ 15: Goto Line 2 70 消解法-例子 (¬p ∨q) ∧(p ∨q) ∧(¬q) 运用消解法的例子 1. 先找出其简单析取式： S0 = {}, S1 = {¬p ∨q, , p ∨q, ¬q}, S2 = {} 2. 对S0 和S1 之间以及S1 内部寻找可以消解的公式： ◦Res(¬p ∨q, p ∨q) = q. ◦Res(¬p ∨q, ¬q) = ¬p. ◦Res(p ∨q, ¬q) = p. 3. 更新S0, S1, S2: S0 = {¬p ∨q, , p ∨q, ¬q}, S1 = {p, ¬p, q}, S2 = {} 4. 重复上述过程，由于： Res(p, ¬p) = ϵ 我们可以得到上述公式不可满足。 71 关于SAT 问题 • 2-SAT 问题是有多项式算法的。 • 消解法并不是目前常用的SAT 问题的求解方法，其并不高效。目前比较常用的算法是基于回溯思想的DPLL 算法以及其进一步改进的CDCL 算法。这些算法在实际中有着非常好的效果。 • 此外，还有一类非完备类的算法。即该类算法只能证明一个公式是可满足的，但是不能证明一个公式是不可满足的。这类算法的代表是Stochastic Local Search 算法。 • 其衍生SMT-solver 也有着非常广泛的应用。扩展资料 • SAT 问题简介：这是中科大吉建民副教授的对SAT 问题的一个介绍，可供参考。 72 总结本章总结 • 等值演算。 • 联结词的完备集。 • 命题公式的范式。 • 可满足性问题。第二周作业本周作业已经发布到课程主页上。截至时间：2023 年9 月25 日周一晚上23:59。 73
17058	https://www.classcentral.com/course/youtube-dilutions-chemistry-khan-academy-309954	Learn SEO and GEO with Noble Desktop’s SEO Bootcamp View 0 Reviews Start learning Class Central is learner-supported. When you buy through links on our site, we may earn an affiliate commission. Science Chemistry Science Chemistry Science Chemistry Stoichiometry Dilutions - Chemistry Fundamentals and Practice Problems Khan Academy via YouTube Write review Start learning Write review Overview Coursera Plus Annual Sale: All Certificates & Courses 30% Off! Grab it Learn about the process of dilution in chemistry through this 11-minute video lecture. Explore the concept of lowering solution concentrations by adding solvent, and master the dilution equation (M1V1) = (M2V2) for calculating molarities and volumes. Begin with a practical example of diluting Kool-aid, then progress to building the dilution equation and solving practice problems. Gain valuable skills for preparing precise solution concentrations essential for various experiments, starting from concentrated stock solutions. Perfect for high school chemistry students and those seeking to understand the fundamentals of solution preparation. Syllabus Diluting Kool-aid- Building the dilution equation- Dilution practice problem 1- Dilution practice problem 2 Taught by Khan Academy Related Courses ### Chemistry 1 - Vol 2 Related Articles ### Your Formula for Chemistry: A Guide to 450+ Free Online Courses ### 7 Best Chemistry Courses for 2025 ### 10 Best Organic Chemistry Courses for 2025 ### 5 Best Free Human Physiology Courses for 2025 ### 5 Best YouTube Marketing Courses for Business in 2025 ### 9 Best Free LaTeX Courses for 2025 Reviews Select rating Start your review of Dilutions - Chemistry Fundamentals and Practice Problems Start learning Share Facebook Twitter Bluesky Email Never Stop Learning. Get personalized course recommendations, track subjects and courses with reminders, and more. Sign up for free
17059	https://jics.org.br/ojs/index.php/JICS/article/download/36/14/190	Journal of Integrated Circuits and Systems, vol. 13, n. 2, 2018 1Digital Object Identifier 10.29292/jics.v13i2.36 Abstract —The linearization of radio frequency power am-plifiers (PAs) for wireless communication systems can be per-formed by a digital baseband predistorter (DPD). The DPD design includes the parameter identification of a post-inverse or a pre-inverse model. Such process depends on measure-ments of discrete-time complex-valued envelopes. With the adoption of larger envelope bandwidths and the PA operation at stronger nonlinear regimes, the requirements on the sam-pling frequency are very stringent. To relax the specifications on the analog-to-digital and digital-to-analog converters, a band-limited memory polynomial can be employed. To reduce the amount of calculations, polynomials can be replaced by li-nearly interpolated look-up tables (LUTs). Traditionally, a non optimal two step procedure is performed to identify the values to be stored in the LUTs. This work contribution is to propose an optimal identification technique that computes directly the values to be stored in the LUTs. The reported case study uses an LTE OFDMA envelope and Matlab simulation results show that the modeling accuracy can be significantly improved by the adoption of the proposed technique instead of the tradi-tional one, quantified by reductions in normalized mean square error and adjacent channel power ratio of up to 10.8 dB and 12.4 dB, respectively. Index Terms —digital baseband predistortion; memory po-lynomial; power amplifier; radio frequency; wireless commu-nication systems. I. INTRODUCTION Digital baseband predistorters (DPDs) are included in the transmitter chain of wireless communication systems for guaranteeing the required level of linearity, at the same time keeping the most power consuming element, which is the radio frequency (RF) power amplifier (PA), operating in an efficient way . A common approach is to design the DPD block by using a discrete-time feed-forward dynamic nonli-near model with adjustable coefficients . Examples in-clude the Volterra series and its simplified versions like the memory polynomial (MP) . Two DPD identification strategies are available -. The direct and indirect learning architectures are based on measurements of discrete-time complex-valued envelope signals. When an envelope signal passes through a nonline-ar operator, its bandwidth is increased. A very high sam-pling frequency requires expensive and very power consu-ming analog-to-digital converters (ADCs) inside the measu-rement instruments . For a given number of resolution bits, to relax the specifications on the instrument ADCs, in recent years some techniques were reported in literature for the identification of DPD schemes based on measurements adopting reduced sampling frequencies that comply with the Nyquist criterion only for the undistorted envelope signal -. The success of such DPD strategies using reduced sampling frequency, or equivalently under-sampling, is theoretically justified by the works of -. The work of reports a comparison between the modeling accura-cies of the direct and indirect learning architectures in sce-narios of reduced sampling frequencies. Digital-to-analog converters (DACs) are also necessary to feed the analog PA with a digitally predistorted signal whose bandwidth is wider than the undistorted one. In , a band-limited DPD was proposed to alleviate both ADC and DAC sampling frequency specifications. Such tech-nique attenuates only the distortions located around the main channel and inside a bandwidth equal to the ADC and DAC sampling frequency. In polynomial-based models, the amount of additions and multiplications required to process a single sample throughout the DPD model increases very rapidly with the polynomial order and memory depth. To relax the computa-tional burden in real-time DPD application, in poly-nomial functions are replaced by linearly interpolated look-up tables (LUTs). In such LUT-based models, the values to be stored in the LUTs become the adjustable coefficients. Traditionally, the parameter identification is performed in a two step procedure. First, the polynomial coefficients are found assuming the continuous model description. Then, the known polynomials are sampled at a finite number of ampli-tude levels to obtain the values to be stored in the LUTs. However, the optimality of the solution provided by the tra-ditional approach can not be guaranteed. To circumvent such drawback and improve the modeling accuracy, this work proposes an identification technique that computes directly the values to be stored in the LUTs and, therefore, it is able to provide an optimum set of values. This work is organized in the following manner. Section II addresses PA linearization through the use of DPDs. Di-rect and indirect learnings are detailed in Section III. Sec-tion IV describes the LUT-based band-limited MP model adopted in this work, whereas its parameter identification is discussed in Section V. Simulation results of a comparative study are reported in Section VI. Conclusions are provided in Section VII. II. DIGITAL BASEBAND PREDISTORTION The block diagram of a DPD is shown in Fig. 1a. To co- Joel H. Chavez, Carolina L. R. Machado, Luis H. A. Lolis, and Eduardo G. Lima Department of Electrical Engineering, Federal University of Paraná, Curitiba, Brazil e-mail: elima@eletrica.ufpr.br Optimal Parameter Identification for Look-up Table based Band-limited Memory Polynomial Model using Direct and Indirect Learnings 2 CHAVEZ et al .: Optimal Identification for Look-up Table based Models rrect the nonlinearities introduced by the PA, a DPD block is put before the PA in a cascade connection. The DPD pro-vides the inverse effect with respect to the behavior of the PA. Figure 1b shows three transfer characteristics: DPD transfer characteristic (DPD output nv% as a function of DPD input nx% ), PA transfer characteristic (PA output ny% as a function of PA input nv% ) and transfer characteristic of cas-cade connection between DPD and PA (PA output ny% as a function of DPD input nx% ). As illustrated in Fig. 1b, the DPD has an expansive characteristic, in contrast with the PA compression aspect. Besides, Fig. 1c indicates that the DPD is fed with an undistorted envelope .nx% Nonlinearities inside the DPD model intentionally distort the DPD output n v% , causing spectral regrowth in the signal applied as PA input. The nonlinear distortions generated by the PA cancel those produced by the DPD, providing an undistorted PA output .ny% To design a DPD, a topology of its behavioral model must be selected. Such model must manipulate complex-valued envelopes, provide a nonlinear relationship between input and output envelopes, take into account memory ef-fects and perform as few as possible computations. Neural networks and polynomials with memory are the most com-mon choices . In common, they have adjustable coeffi-cients that must be chosen based on optimization proce-dures. In this work, polynomial approaches are employed because, in contrast to neural networks, they are linear in their parameters. Fig.1 Basic concepts in DPD: a) block diagram of a cascade connection; b) transfer characteristic curves; c) power spectral densities. III. DIRECT AND INDIRECT LEARNING ARCHITECTURES Once a DPD model is chosen, the next step is the identi-fication of its parameters. This section addresses the direct and indirect approaches for performing such task. A. Direct Learning Figure 2 shows the block diagram of the direct learning architecture . Observe that the DPD desired output is not known. In a cascade connection of a DPD followed by a PA, the purpose is that the cascade output be a linear replica (apart from a linear gain g) of the cascade input. Hence, the DPD coefficient estimation by this method must include the PA model, which manipulates in a nonlinear way the DPD output. In this way, the coefficient extraction is posed as a nonlinear optimization problem, where the nonlinear func-tion to be minimized is the mean square of an error signal n e% defined by the difference between the DPD input sig-nal nx% and the PA output signal divided by g ( ny% /g). Fig.2 Direct learning architecture. Therefore, direct learning, in addition to the need for a PA model, always requires nonlinear optimization algo-rithms, like Gauss-Newton, Levenberg-Marquardt . De-pending on the initial guess, such algorithms can be trapped in local minima that can only be corrected by the inclusion of additional algorithms (e.g. genetic algorithms, particle swarms, simulated annealing), which demand for greater computational complexity than a linear algorithm such as the least squares (LS). In the direct learning, the error signal has a bandwidth equal to the undistorted envelope signal. B. Indirect Learning Indirect learning is an architecture in which the coef-ficients of a post-distorter (PoD) – connected in cascade af-ter a PA – is first identified and then copied to a DPD – connected in cascade before a PA – of same topology. In fact, for polynomial models truncated to the same order P,the pre- and post-inverse models are the same . In a first PoD identification, as shown in Fig. 3a, the PA is excited by an undistorted input ' n v% and a distorted signal corrupted by spectral regrowth is measured at the PA output ' .ny% Since the input and output roles are exchanged in PA and PoD blocks, the PoD input ' n y% /g is distorted (where g is the PA small-signal gain), while the PoD output nw% is an undistorted signal. The polynomial parameters are obtained through a linear algorithm, such as LS, and then copied as parameters of a DPD of same topology. In a DPD, as shown in Fig. 1, the input nx% is an undistorted signal and the output n v% is a distorted signal. Hence, the DPD works in a differ-3ent configuration, concerning the input and output distortion levels, in comparison with the first identified PoD. As a consequence, degradation in linearization capability is ex-pected. To solve this problem, the PoD parameter identifi-cation process must be repeated, but now with the PA ex-cited by a predistorted signal obtained based on the DPD parameters previously identified, as shown in Fig. 3b. These new PA input " n v% and output " n y% signals are used to identify a second PoD, whose operation now resembles the opera-tion of the DPD (e.g. undistorted input and distorted out-put). Novel PoD estimations can be computed, repeating the same procedure, but using the previous DPD as the starting point for collecting PA data. From the second PoD and so on, the error signal ne% to be minimized is computed as the difference between two distorted signals. Fig.3 Indirect learning architecture: a) PoD first iteration; b) PoD second iteration. IV. LUT-BASED BAND -LIMITED MP MODEL In this work, the DPD model is described by a band-limited MP model according to: { }2( 1) ,0 1 , MPpnnmm p nmmp v x h a LPF −− −= =  = ∗   ∑ ∑ %% % (1) where nx% and nv% indicate complex-valued DPD input and output envelopes, respectively, na is the amplitude compo-nent of ,nx% ,m p h% are complex-valued parameters, P is the po-lynomial order, M is the memory length, stands for con-volution and LPF refers to a lowpass filter. The block dia-gram of the band-limited MP model is shown in Fig. 4. 2( 1) ,1 PpM p n M p h a −−= ∑%\|.\| 1 z− n x% M 1 z− 2( 1) 1, 11 Pppnp h a −−= ∑%\|.\| 2( 1) 0, 1 Pppnp h a −= ∑%\|.\| n v%f \|.\| 2 BW − 2 BW LPF Fig.4 Block diagram of band-limited MP model. As reported in , the lowpass filter has the purpose of constraining the spectral content of the predistorted signal inside the ADC/DAC sampling frequency of BW . Due to the limited bandwidth of (1), the DPD is only able to linearize the PA output at the frequency range from -BW /2 to + BW /2. For DPD purposes, the PA output must then be subject to a bandpass filter of same bandwidth. The computation of outputs from polynomials such as (1) demands for a huge quantity of additions and multiplica-tions. Besides, the amount of necessary calculation increas-es very rapidly with the polynomial order and memory length. As proposed in , to reduce the computational complexity the one-dimensional polynomials of (1) can be replaced by linearly interpolated look-up tables (LUT). To that purpose, first rewrites (1) as: { }20 ( ) , MPOL nnmmnmm v x f a LPF − −=  = ∗   ∑% % (2) where the one-dimensional polynomials POL mf (for m ranging from 0 to M) are given by: 22( 1) ,1 ( ) . PPOL pmnmm p nmp f a h a −− −= = ∑ % (3) Each POL mf takes as argument a single real-valued in-formation, given by the squared input amplitude component 2 nm a − . Each POL mf returns a complex-valued information ac-cording to (3). Instead of performing all the additions and multiplications demanded by (3) at every time a new argu-ment is applied, in the LUT-based approach each POL mf is replaced by the storage of a finite number of Q input and output values, as reported in Table I. Table I. Information stored in a LUT with Qaddressable positions. LUT real-valued input LUT complex-valued output em,1 ,1 m s% em,2 ,2 m s% MM em,Q ,m Q s% In Table I, LUT real-valued inputs are designated by em,q (with q ranging from 1 to Q) and LUT complex-valued out-puts are designated by ,m q s% (with q ranging from 1 to Q). At every time a new argument 2 nm a − is applied, the compara-tive operation 2, 1 ,m q n m m q e a e− −≤ ≤ is performed to find the integer index q, which must be on the range from 2 to Q.The outputs , 1m q s − % and ,m q s% associated to the two consecu-tive LUT inputs em,q-1 and em,q are read and, using linear in-terpolation, each POL mf is approximated by: ( ), , 12 2, 1 , 1, , 1 ( ) . m q m q POL mnmm q nmm q m q m q s sf a s a ee e −−−−−−  −≈ + −  −  % %% (4) The LUT-based technique allows for a simpler imple-mentation because it reduces the amount of necessary addi-tions and multiplications, at the expense of reading pre- Journal of Integrated Circuits and Systems, vol. 13, n. 2, 2018 4 CHAVEZ et al .: Optimal Identification for Look-up Table based Models viously stored data and then performing linear interpolation. Indeed, in the LUT-based approach, according to (4), the computation of an output is always performed by the same number of additions and multiplications, independent of the polynomial order. V. PARAMETER IDENTIFICATION OF THE LUT-BASED BAND -LIMITED MP MODEL This section addresses the identification of the band-limited LUT-based memory polynomial. Two techniques are reported here to find the LUT complex-valued output values , .m q s% All of the LUT real-valued input values em,q are predetermined as equally spaced values starting in zero. The LPF and the values for M, P and Q are also assumed known before the ,m q s% parameter identification. First, a tra-ditional technique based on a two step procedure is re-ported. Then, a proposed direct approach is introduced. A. Traditional technique The traditional parameter identification method is di-vided into two processes. The first one is to calculate all of the polynomial parameters ,m p h% of (1). Because (1) is linear in the parameters ,m p h% , such task can be performed by a li-near method such as the LS. Given a set of N input and out-put data, all polynomial parameters can be found with: 1 ( ) ( ), H HPOL POL POL − =% % % % %h X X X v (5) where the operator (.) H indicates the complex conjugate transpose, 0,1 , TM P h h =   % %% Lh is the row vector contain-ing all the polynomial parameters, [ ]1 TMN v v+=% % %Lv is the row vector containing the output data and POL %X is the regression matrix which manipulates the input samples ac-cording to: 02( 1) 111102( 1) . PMMPOL PNNNMNM x a LPF x a LPF x a LPF x a LPF −++−−−     =     % %L% M M M% %L X (6) With all polynomial parameters identified, the second phase of this approach can be done, that is, all polynomial functions of (3) are evaluated having as argument each one of the LUT input values em,q to provide the values for all , .m q s% Therefore, in the traditional approach, the order of the system to be solved by the LS is dependent on the poly-nomial order P, but not on the number of LUT addressable positions Q. B. Proposed Technique The optimality of the solution provided by the LS in the traditional approach is assured only if continuous poly-nomial functions are employed. Hence, after the sampling of polynomial functions and the application of linear inter-polation, there is no guarantee that the error is minimized. To circumvent such drawback of the traditional approach, this work proposes a technique that is able to provide an optimum set of values for the ,m q s% parameters. The pro-posed approach introduces a method to calculate all ,m q s% values directly, that is, without previously knowing any po-lynomial function. The band-limited LUT-based MP model is in fact also linear in the parameters ,m q s% . As a conse-quence, the ,m q s% parameter identification can be performed into a single step by LS. Given a set of N input and output data, all ,m q s% can be found with: 1 ( ) ( ), H HLUT LUT LUT − = % % %% %s X X X v (7) where 0,1 , TM Q s s =  % % %Ls is the row vector containing all the LUT complex-valued parameters, [ ]1 TMN v v+=% % %Lv is the row vector containing the output data and LUT %X is the regression matrix which manipulates the input samples by: 2210111220 ( ) ( ) ,( ) ( ) LUT LUT MMMLUT LUT NNNMMNM x a LPF x a LPF x a LPF x a LPF ++−−       % %LM M M% %L r rr r (8) in which 2 ( )LUT m n ma −r indicates column vectors with Q col-umns. The interpolated output of a given LUT depends only on the information of two consecutive positions of the given LUT. Hence, for each one of these ,LUT mr there are only two columns with non null values. In other words, any LUT mr is filled with ( Q-2) null values. To obtain the two values dif-ferent from zero to be inserted in each ,LUT mr first the integer index q between 2 and Q that satisfies 2, 1 ,m q n m m q e a e− −≤ ≤ must be computed. Then: 2,1, ( ) [0 0 0 0], LUT m n m m q m q a r r− −= % %L Lr (9) 2,1,1,,1 1 , nmm q m q m q m q a er e e −−−−  −= −  −   % (10) 2,1,,,1 . nmm q m q m q m q a er e e −−−  −=  −   % (11) Therefore, in the proposed approach, the order of the system to be solved by the LS is dependent on Q, but not on the polynomial order P. VI. VALIDATION In this section, Matlab simulation results of several dis-tinct DPDs are reported. The undistorted envelope signal is an LTE OFDMA with a bandwidth of 20 MHz. The PA to be linearized is described by (1), exchanging the input and output roles, with the complex-valued coefficients reported in Table II, followed by a lowpass filter with BW = 38.4 MHz described by an idealized frequency sequence of 4096 equally spaced samples, composed by ones and zeros for desired and undesired frequencies, respectively. The sam-pling frequency is set to 153.6 MHz. All the simulated DPDs share the LUT-based band-limited MP topology with M = 1, Q = 4, em,1 = 0, em,2 = 1/3, em,3 = 2/3, em,4 = 1 and using the aforementioned idealized frequency sequence to implement LPF . The DPDs differ among them by their 5training algorithms. Both direct and indirect learning archi-tectures, as well as traditional and proposed parameter iden-tifications, are employed for the DPD identification. The traditional parameter identification approach assumes P = 5. Table II. PA complex-valued coefficients. Coeff Value Coeff Value 0,1 h% -0.0515 - j0.0831 1,1 h% 1.0538 + j0.0612 0,2 h% 0.5407 + j0.0729 1,2 h% -1.6840 + j0.3680 0,3 h% -0.5777 + j0.0293 1,3 h% 2.9292 - j0.4123 0,4 h% 0.2852 - j0.1514 1,4 h% -3.2223 + j0.2051 0,5 h% -0.0218 + j0.1011 1,5 h% 1.3383 - j0.0130 The validation of the linearization capability of a DPD is assessed by the normalized mean square error (NMSE) and the adjacent channel power ratio (ACPR). The ACPR calculation employs main channel bandwidth of 18 MHz, adjacent channels with 9 MHz bandwidth and frequency separation between main and adjacent channels of 14.7 MHz. Table III presents the results obtained with the dis-tinct strategies. Five iterations of the indirect learning architecture are reported in Table III. NMSE and ACPR values improved significantly from first to second iterations. These im-provements reflect the behavior explained in Section III.B, since the second identified PoD works in a scenario closer to the DPD validation. Third to fifth iterations improve on-ly slightly the linearization capability with respect to the second iterations. For the parameter identification with the direct learning architecture, Levenberg-Marquardt algo-rithm was used as the optimization algorithm by Matlab nonlinear least squares . The initial guess uses the pa-rameters found in the fifth iterations of indirect learning. In comparison with the indirect learning, the direct learning improves the NMSE and ACPR results up to 0.9 dB and 2.4 dB, respectively. However, such improvement can only be achieved at the expense of performing a nonlinear proce-dure. Table III. NMSE and ACPR simulation results. Traditional Proposed Architecture Iter NMSE [dB] ACPR [dB] Iter NMSE [dB] ACPR [dB] Indirect Learning 1st -30.2 -42.9 1st -35.6 -49.1 2nd -31.0 -46.9 2nd -40.0 -52.8 3rd -31.1 -47.0 3rd -40.1 -52.9 4th -31.1 -47.0 4th -40.1 -52.9 5th -31.1 -47.0 5th -40.1 -52.9 Direct Learning 18 -31.5 -47.5 15 -41.0 -55.3 In a comparison with the traditional technique, indepen-dent from the adopted learning architecture, the proposed technique always shows a superior behavior. In particular, improvements in NMSE results between 4.1 dB and 10.8 dB, as well as reductions in ACPR results between 1.6 dB and 12.4 dB, are reported when the proposed approach is employed instead of the traditional one. Figure 5 shows the power spectral densities (PSDs) of linearized PA output signals. A further substantial reduction in the distortions at adjacent channels is observed when the proposed technique is employed instead of the traditional one. Figure 6 shows the output amplitude as a function of the input amplitude for the PA model, the DPD model and the cascade connection of DPD followed by PA. Notice that the gain expansion of the DPD almost completely cancels the PA gain compres-sion. -2 -1.5 -1 -0.5 00.5 11.5 2x 10 7 -40 -20 020 Frequency (Hz) PSD (dBm/Hz) traditional proposed Fig.5 Power spectral densities (PSDs) of the linearized PA output signals, when the indirect learning with the traditional and proposed identification techniques is employed. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.2 0.4 0.6 0.8 Input amplitude (V) Output amplitude (V) DPDPADPD+P A Fig.6 Transfer characteristics for: PA characteristic (in green color), DPD identified by indirect learning with proposed parameter identification (in blue color) and cascade connection of DPD followed by PA (in red color). VII. CONCLUSION An optimal parameter identification for a band-limited LUT-based MP model is proposed in this work. Using two iterations of the indirect learning, which demands for the implementation of two least squares, NMSE and ACPR re-sults of -40.0 dB and -52.8 dB, respectively, were obtained when the proposed technique is employed in the case study reported in this work. Conversely, the best NMSE and ACPR results obtained by the traditional approach, which was based on the direct learning that requires a nonlinear procedure, were equal to -31.5 dB and -47.5 dB, respective-ly. Therefore, a very significant superior performance is ob-tained by the proposed technique in comparison with the traditional one. ACKNOWLEDGEMENTS The authors would like to acknowledge the financial Journal of Integrated Circuits and Systems, vol. 13, n. 2, 2018 6 CHAVEZ et al .: Optimal Identification for Look-up Table based Models support provided by Coordenação de Aperfeiçoamento de Pessoal de Nível Superior and Organization of American States. REFERENCES P. B. Kenington, High Linearity RF Amplifier Design . Norwood, MA: Artech House, 2000. J. C. Pedro and S. A. Maas, “A comparative overview of microwave and wireless power-amplifier behavioral modeling approaches,” IEEE Trans. Microw. Theory Tech. , vol. 53, no. 4, pp. 1150–1163, Apr. 2005. J. Kim and K. Konstantinou, “Digital predistortion of wideband sig-nals based on power amplifier model with memory,” Electron. Lett. ,vol. 37, no. 23, pp. 1417–1418, Nov. 2001. R. N. Braithwaite, “General Principles and Design Overview of Digital Predistortion,” in Digital front-end in wireless communica-tions and broadcasting circuits and signal processing , Fa-Long Luo, Ed. Cambridge University Press, 2011, pp. 150–161. G. Baudoin and P. Jardin, “Adaptive polynomial pre-distortion for linearization of power amplifiers in wireless communications and WLAN,” in Proc. IEEE International Conference on Trends in Communications , Bratislava, Slovakia, Jul. 2001, pp. 157–160. C. Eun and E. J. Powers, “A new volterra predistorter based on the indirect learning architecture,” IEEE Trans. Signal Process. , vol. 45, no. 1, pp. 223–227, Jan. 1997. B. Murmann, "ADC Performance Survey 1997-2016," [Online]. Available: Y. Liu, J. J. Yan, H. -T. Dabag, and P. M. Asbeck, “Novel Tech-nique for Wideband Digital Predistortion of Power Amplifiers With an Under-Sampling ADC,” IEEE Trans. Microw. Theory Tech. , vol. 62, no. 11, pp. 2604–2617, Nov. 2014. R. N. Braithwaite, “Closed-Loop Digital Predistortion (DPD) Using an Observation Path With Limited Bandwidth,” IEEE Trans. Mi-crow. Theory Tech. , vol. 63, no. 2, pp. 726–736, Feb. 2015. Y. Liu, W. Pan, S. Shao, and Y. Tang, “A General Digital Predistor-tion Architecture Using Constrained Feedback Bandwidth for Wide-band Power Amplifiers,” IEEE Trans. Microw. Theory Tech. , vol. 63, no. 5, pp. 1544–1555, May 2015. Q. Zhang, Y. Liu, J. Zhou, S. Jin, W. Chen, and S. Zhang, “A Band-Divided Memory Polynomial for Wideband Digital Predistortion With Limited Bandwidth Feedback,” IEEE Trans. Circuits Syst. II, Exp. Briefs , vol. 62, no. 10, pp. 922–926, Oct. 2015. Y.-M. Zhu, “Generalized sampling theorem,” IEEE Trans. Circuits Syst. II , vol. 39, no. 8, pp. 587–588, Aug. 1992. W. A. Frank, “Sampling requirements for Volterra system identifica-tion,” IEEE Signal Process. Lett. , vol. 3, no. 9, pp. 266–268, Sep. 1996. J. Tsimbinos and K. Lever, “Input Nyquist sampling suffices to iden-tify and compensate nonlinear systems,” IEEE Trans. Signal Process. , vol. 46, no. 10, pp. 2833–2837, Oct. 1998. J. H. Chavez and E. G. Lima, "Direct and Indirect Learning for Pre-distorter Design Using Data with Reduced Sampling Frequency", in Proceedings of the 32nd South Symposium on Microelectronics , Rio Grande, May 2017, pp. 13–16. C. Yu, L. Guan, E. Zhu, and A. Zhu, “Band-Limited Volterra Series-Based Digital Predistortion for Wideband RF Power Amplifiers,” IEEE Trans. Microw. Theory Tech. , vol. 60, no. 12, pp. 4198–4208, Dec. 2012. A. Kwan, F. M. Ghannouchi, O. Hammi, M. Helaoui, and M. R. Smith, "Look-up table based digital predistorter implementation for field programmable gate arrays using long-term evolution signals with 60 MHz bandwidth", IET Sci. Meas. Technol. , vol. 6, no. 3, 181–188, May 2012. J. Dennis Jr., “Nonlinear least-squares,” in State of the Art in Nu-merical Analysis: Conference Proceedings , D. A. H. Jacobs, Ed. London: Academic Press, 1977, pp. 269–312. M. Schetzen, “Theory of pth-order inverses of nonlinear systems,” IEEE Trans. Circuits Syst. , vol. 23, no. 5, pp. 285–291, May 1976. M. S. Muha, C. J. Clark, A. Moulthrop, and C. P. Silva, “Validation of power amplifier nonlinear block models,” in IEEE MTT-S Int. Mi-crowave Symp. Dig. , Anaheim, Jun. 1999, pp. 759–762.
17060	https://www.mathway.com/popular-problems/Calculus/590743	Find the Average Value of the Function f(x)=x-x^2 , [0,2] \| Mathway Enter a problem... [x] Calculus Examples Popular Problems Calculus Find the Average Value of the Function f(x)=x-x^2 , [0,2] f(x)=x-x 2 f(x)=x−x 2 , [0,2][0,2] Step 1 The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined. Interval Notation: (-∞,∞)(−∞,∞) Set-Builder Notation: {x\|x∈ℝ} Step 2 f(x) is continuous on [0,2]. f(x) is continuous Step 3 The average value of function f over the interval[a,b] is defined as A(x)=1 b-a∫b a f(x)d x. A(x)=1 b-a∫b a f(x)d x Step 4 Substitute the actual values into the formula for the average value of a function. A(x)=1 2-0(∫2 0 x-x 2 d x) Step 5 Split the single integral into multiple integrals. A(x)=1 2-0(∫2 0 x d x+∫2 0-x 2 d x) Step 6 By the Power Rule, the integral of x with respect to x is 1 2 x 2. A(x)=1 2-0(1 2 x 2]2 0+∫2 0-x 2 d x) Step 7 Since -1 is constant with respect to x, move -1 out of the integral. A(x)=1 2-0(1 2 x 2]2 0-∫2 0 x 2 d x) Step 8 By the Power Rule, the integral of x 2 with respect to x is 1 3 x 3. A(x)=1 2-0(1 2 x 2]2 0-(1 3 x 3]2 0)) Step 9 Simplify the answer. Tap for more steps... Step 9.1 Combine 1 3 and x 3. A(x)=1 2-0(1 2 x 2]2 0-(x 3 3]2 0)) Step 9.2 Substitute and simplify. Tap for more steps... Step 9.2.1 Evaluate 1 2 x 2 at 2 and at 0. A(x)=1 2-0((1 2⋅2 2)-1 2⋅0 2-(x 3 3]2 0)) Step 9.2.2 Evaluate x 3 3 at 2 and at 0. A(x)=1 2-0(1 2⋅2 2-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3 Simplify. Tap for more steps... Step 9.2.3.1 Raise 2 to the power of 2. A(x)=1 2-0(1 2⋅4-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3.2 Combine 1 2 and 4. A(x)=1 2-0(4 2-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3.3 Cancel the common factor of 4 and 2. Tap for more steps... Step 9.2.3.3.1 Factor 2 out of 4. A(x)=1 2-0(2⋅2 2-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3.3.2 Cancel the common factors. Tap for more steps... Step 9.2.3.3.2.1 Factor 2 out of 2. A(x)=1 2-0(2⋅2 2(1)-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3.3.2.2 Cancel the common factor. A(x)=1 2-0(2⋅2 2⋅1-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3.3.2.3 Rewrite the expression. A(x)=1 2-0(2 1-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3.3.2.4 Divide 2 by 1. A(x)=1 2-0(2-1 2⋅0 2-(2 3 3-0 3 3)) A(x)=1 2-0(2-1 2⋅0 2-(2 3 3-0 3 3)) A(x)=1 2-0(2-1 2⋅0 2-(2 3 3-0 3 3)) Step 9.2.3.4 Raising 0 to any positive power yields 0. A(x)=1 2-0(2-1 2⋅0-(2 3 3-0 3 3)) Step 9.2.3.5 Multiply 0 by -1. A(x)=1 2-0(2+0(1 2)-(2 3 3-0 3 3)) Step 9.2.3.6 Multiply 0 by 1 2. A(x)=1 2-0(2+0-(2 3 3-0 3 3)) Step 9.2.3.7 Add 2 and 0. A(x)=1 2-0(2-(2 3 3-0 3 3)) Step 9.2.3.8 Raise 2 to the power of 3. A(x)=1 2-0(2-(8 3-0 3 3)) Step 9.2.3.9 Raising 0 to any positive power yields 0. A(x)=1 2-0(2-(8 3-0 3)) Step 9.2.3.10 Cancel the common factor of 0 and 3. Tap for more steps... Step 9.2.3.10.1 Factor 3 out of 0. A(x)=1 2-0(2-(8 3-3(0)3)) Step 9.2.3.10.2 Cancel the common factors. Tap for more steps... Step 9.2.3.10.2.1 Factor 3 out of 3. A(x)=1 2-0(2-(8 3-3⋅0 3⋅1)) Step 9.2.3.10.2.2 Cancel the common factor. A(x)=1 2-0(2-(8 3-3⋅0 3⋅1)) Step 9.2.3.10.2.3 Rewrite the expression. A(x)=1 2-0(2-(8 3-0 1)) Step 9.2.3.10.2.4 Divide 0 by 1. A(x)=1 2-0(2-(8 3-0)) A(x)=1 2-0(2-(8 3-0)) A(x)=1 2-0(2-(8 3-0)) Step 9.2.3.11 Multiply-1 by 0. A(x)=1 2-0(2-(8 3+0)) Step 9.2.3.12 Add 8 3 and 0. A(x)=1 2-0(2-8 3) Step 9.2.3.13 To write 2 as a fraction with a common denominator, multiply by 3 3. A(x)=1 2-0(2⋅3 3-8 3) Step 9.2.3.14 Combine 2 and 3 3. A(x)=1 2-0(2⋅3 3-8 3) Step 9.2.3.15 Combine the numerators over the common denominator. A(x)=1 2-0(2⋅3-8 3) Step 9.2.3.16 Simplify the numerator. Tap for more steps... Step 9.2.3.16.1 Multiply 2 by 3. A(x)=1 2-0(6-8 3) Step 9.2.3.16.2 Subtract 8 from 6. A(x)=1 2-0(-2 3) A(x)=1 2-0(-2 3) Step 9.2.3.17 Move the negative in front of the fraction. A(x)=1 2-0(-2 3) A(x)=1 2-0(-2 3) A(x)=1 2-0(-2 3) A(x)=1 2-0(-2 3) Step 10 Simplify the denominator. Tap for more steps... Step 10.1 Multiply-1 by 0. A(x)=1 2+0⋅(-2 3) Step 10.2 Add 2 and 0. A(x)=1 2⋅(-2 3) A(x)=1 2⋅(-2 3) Step 11 Reduce the expression by cancelling the common factors. Tap for more steps... Step 11.1 Cancel the common factor of 2. Tap for more steps... Step 11.1.1 Move the leading negative in -2 3 into the numerator. A(x)=1 2⋅-2 3 Step 11.1.2 Factor 2 out of -2. A(x)=1 2⋅2(-1)3 Step 11.1.3 Cancel the common factor. A(x)=1 2⋅2⋅-1 3 Step 11.1.4 Rewrite the expression. A(x)=-1 3 A(x)=-1 3 Step 11.2 Move the negative in front of the fraction. A(x)=-1 3 A(x)=-1 3 Step 12 [x 2 1 2√π∫x d x] Please ensure that your password is at least 8 characters and contains each of the following: a number a letter a special character: @$#!%?&
17061	https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality?srsltid=AfmBOoopM3nsdzZw4LM4D4kNWX310pTxDl4oN1FLichbPAV0arr71kTu	Art of Problem Solving Cauchy-Schwarz Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Cauchy-Schwarz Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Cauchy-Schwarz Inequality In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra. Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals and , with equality if and only if there exists a constant such that for all , or if one list consists of only zeroes. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests. Its vector formulation states that for any vectors and in , where is the dot product of and and is the norm of , with equality if and only if there exists a scalar such that , or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems. The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality. Contents [hide] 1 Proofs 2 Lemmas 2.1 Complex Form 2.2 A Useful Inequality 3 Real Vector Spaces 3.1 Proof 1 3.2 Proof 2 3.3 Proof 3 4 Complex Vector Spaces 4.1 Proof 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 Other Resources 6.1 Books Proofs Here is a list of proofs of Cauchy-Schwarz. Consider the vectors and . If is the angle formed by and , then the left-hand side of the inequality is equal to the square of the dot product of and , or .The right hand side of the inequality is equal to . The inequality then follows from , with equality when one of is a multiple of the other, as desired. Lemmas Complex Form The inequality sometimes appears in the following form. Let and be complex numbers. Then This appears to be more powerful, but it follows from A Useful Inequality Also known as Sedrakyan's Inequality, Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality: For any real numbers and where the following is true: Real Vector Spaces Let be a vector space, and let be an inner product. Then for any , with equality if and only if there exist constants not both zero such that . The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers. Proof 1 Consider the polynomial of This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., must be less than or equal to , with equality when or when there exists some scalar such that , as desired. Proof 2 We consider Since this is always greater than or equal to zero, we have Now, if either or is equal to , then . Otherwise, we may normalize so that , and we have with equality when and may be scaled to each other, as desired. Proof 3 Consider for some scalar . Then: (by the Trivial Inequality) . Now, let . Then, we have: . Complex Vector Spaces For any two vectors in the complex vector space , the following holds: with equality holding only when are linearly dependent. Proof The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021. Define the unit vectors , as and . Put . In other words, is the complex argument of and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let and . Importantly, we have Since and , this calculation shows that and form an orthogonal basis of the linear subspace spanned by and . Thus we can think of and as lying on the unit sphere in this subspace, which is isomorphic to . Another thing to note is that The previous two calculations established that and are orthogonal, and that the sum of their squared norms is . Now we have Equality holds when either or , or equivalently when and . Lastly, multiplying each side by , we have Problems Introductory Consider the function , where is a positive integer. Show that . (Source) (APMO 1991 #3) Let , , , , , , , be positive real numbers such that . Show that Intermediate Let be a triangle such that where and denote its semiperimeter and inradius, respectively. Prove that triangle is similar to a triangle whose side lengths are all positive integers with no common divisor and determine those integers. (Source) Olympiad is a point inside a given triangle . are the feet of the perpendiculars from to the lines , respectively. Find all for which is least. (Source) Other Resources Wikipedia entry Books The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. Problem Solving Strategies by Arthur Engel contains significant material on inequalities. Retrieved from " Categories: Algebra Inequalities Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
17062	https://www.e-magnetica.pl/doku.php/air_gap	skip to content Encyclopedia Magnetica™ User Tools Log In Site Tools Recent Changes Media Manager Sitemap Table of Contents Air gap Influence on B-H loop and permeability Importance of air gap in practical applications Rotating machines Linear motors Gapped and air-cored inductors Solenoids and linear actuators Current sensors Induction heating Energy stored in air gap Flux fringing Distributed air gap See also Air gap \| \| \| Stan Zurek, Air gap, Encyclopedia Magnetica, \| Air gap, also airgap1) or air-gap2) - is a non-magnetic part of a magnetic circuit. It is usually connected magnetically in series with the rest of the circuit, so that a substantial part of the magnetic flux (or magnetic field) flows through the gap. Diagram of an air gap in a magnetic circuit S. Zurek, E-Magnetica.pl, CC-BY-4.0 Depending on application, air gap may be filled with a non-magnetic material such as gas, water, vacuum, plastic, wood etc. and not necessarily just with air.3)4) An air gap is a practically unavoidable part of any magnetic circuit in which there is a relative movement between different parts (e.g. in motors, generators, relays, etc). Due to increased reluctance of an air gap the flux spreads into the surrounding medium causing the flux fringing effect. It is generally an unwanted phenomenon which usually increases proximity and eddy current loss in conductors located in the vicinity of the air gap. \| \| \| \| --- \| → → → Helpful page? Support us! → → → \| PayPal \| ← ← ← Help us with just $0.10 per month? Come on… ← ← ← \| Influence on B-H loop and permeability Influence of air gap ($l_g$ = 0.07 mm) on the shape of B-H loop for a cut core ($l_c$ = 250 mm), so $l_g / l_c$ = 0.00028 S. Zurek, E-Magnetica.pl, CC-BY-4.0 The B-H loop of a magnetic circuit is affected by the presence of an air gap. Permeability of non-magnetic material is low and therefore it requires greater values of $H$ to obtain the same value of $B$ as compared with magnetically soft materials. With the introduction of an air gap the B-H loop of a magnetic circuit gets "sheared" (slanted), hence the value of its slope proportional to the effective permeability is reduced. The amount of “shearing” is proportional to the length of the air gap - the larger the air gap the lower the slope. For air core coil (no magnetic material present) the B-H characteristics become by definition the same as for the non-magnetic material encircled by the winding (e.g. air). Changes of effective permeability and linearisation of the B-H loops caused by increasing air gap S. Zurek, E-Magnetica.pl, CC-BY-4.0 Conversely, if no air gap is present then the slope becomes as steep as possible, and the B-H loop will represent the closest approximation of the characteristic of the magnetic material (for a given shape of the magnetic circuit). For this reason international standards defining magnetic measurements specify procedures to ensure that the influence of non-magnetic parts of a magnetic circuit remain within negligible levels. This can be achieved for instance by careful polishing or lapping of the flat faces, in order to reduce the surface roughness and the amount of space between the magnetic surfaces. 5) Even a very small air gap can have a significant impact on effective permeability S. Zurek, E-Magnetica.pl, CC-BY-4.0 Split cores made out of ferrite are widely used in electronic transformers because they greatly simplify assembly, so that low-cost devices are possible. However, by definition, a split core comprises some small air gap, even if the mating faces are very carefully polished or lapped. Typical air gaps for such ferrite cores are as follows:6) precision-ground or lapped ~ 1 μm (0.001 mm) normally ground ~ 10 μm (0.01 mm) purposely gapped > 10 μm (0.001 mm) The effects depend on the magnetic path length of the core and the material permeability, and are more pronounced for physically smaller cores with high permeability. Even small contamination on the mating surfaces (dust, adhesive, etc.) can have significant impact on the measured values of permeability, inductance or the AL factor. Before a precise measurement is attempted, it is recommended to clean the mating phases and/or rub the two halves together in an elliptical motion.7) Importance of air gap in practical applications The air gap can take different form, shape and size depending on the type of magnetic circuit and its shape. In some circuits it might be actually an integral part ensuring correct performance of the device, but in other cases it should be as small as possible. Such requirements will be determined by the operating principle, performance, size, efficiency, and many other technological factors. Rotating machines Air gap between laminations of a rotor and stator of an induction motor S. Zurek, E-Magnetica.pl, CC-BY-4.0 In rotating machines the air gap is usually unwanted yet unavoidable due to the necessity of physical movement required between the stator and the rotor. The smallest practical air gap for industrial machines is around 0.2 mm.8) But with larger machines there are greater dimensional tolerances and the gap is sized accordingly to the diameter of the rotor to be at the order of 0.1% of the machine diameter. An air gap increases the value of magnetising current and lowers the achievable flux density.9) In rotating machines it may lead to increased magnetic loss on the surface of the rotor due to increased space harmonics. Thin laminations are thus required to reduce the effect of eddy currents in the parts of the motor which are exposed to AC flux.10) However, there are also special motors with oversized air gap required for non-magnetic reasons (electrical isolation, access to other parts, etc.)11) The air gap in this prototype of a linear motor is filled partly with Tufnol S. Zurek, E-Magnetica.pl, CC-BY-4.0 Linear motors The cylindrical shapes of rotating machines can be manufactured with tighter tolerances than it is the case for instance for linear motors. As a consequence the linear machines must operate with larger air gaps, which might impact on their efficiency.12) Trains using magnetic levitation technology are propelled by using the principle of linear motors. The air gap between the train suspension and the tracks can be as large as 13 mm for electromagnetic suspension (Germany) or 102 mm for electrodynamic suspension (Japan).13) For structural reasons the gap can be filled partly with air and partly with a solid non-magnetic material (see photo). Gapped and air-cored inductors Short-circuit current limiting air-cored choke rated at 1 kA by Stahlkocher, CC-BY-SA-3.0 Energy storing inductors Air gaps are an integral part of gapped inductors. The gap reduces effective permeability of a given magnetic circuit and allows storing much greater energy before saturation is reached. Increasing the gap reduces the inductance, so the winding must have more turns to compensate accordingly.14) For a given size of inductor the amount of stored energy versus applied air gap can be represented by a Hanna curve.15) Electronic choke with large air gap (comparable to the length of the ferrite rod) S. Zurek, E-Magnetica.pl, CC-BY-4.0 If operation with high currents is required then the air gap might be very large, so that the magnetic circuit is quite “open”. For instance, a common design for electronic chokes is to place a winding on a magnetic rod. The magnetic field lines must close through the surrounding air (outside of the winding), so the length of the air gap is comparable with the length of the rod.16) In some cases the currents are so high that it is very difficult or cost prohibitive to design the inductor with a magnetic core. In such case a so-called “air core” is used, where the windings are supported by a non-magnetic structure, and the whole magnetic circuit is effectively one big air gap. In order to reduce the flux fringing effect and the losses associated with it, in some inductors the gap is distributed into many smaller gaps.17) The distribution of air gap can be also extended even further. There are magnetic materials, which are made from small particles (mostly based on powder iron, sendust or moly permalloy powder) bound together in such a way as to contain certain percentage of non-magnetic volume in them. The resultant effective permeability is much lower, but the air gap is uniformly distributed throughout the whole material.18) The fringing effect and leakage flux is greatly reduced, which is especially important for high-frequency applications. Such cores are usually more expensive (either in initial production or further coil winding) than alternative technologies with a single gap. Variable and signal inductors Air gap is frequently used as means of tuning, or adjusting inductance to the required value. The variation can be done just once, for instance with a fixed shim introduced between parts of the core to create an air gap. But in some cases the variation must be carried out more frequently, or even cyclically. An important phenomena which necessitates tuning of an LC circuit, is to either achieve the resonance point (e.g. signal transmission through electromagnetic waves) or to be as far as possible from it (e.g. for filtering reasons). LC resonance and tuning is very important for many electromagnetic circuits, ranging from millimetre-sized low-power high-frequency signal transmission19), to high-power devices which work with parts of electricity grid (several kilometres long), as is the case for instance for Peterson coil.20) Air gap in relay between the movable armature and the core (top right corner) by R. Ruizo, CC-BY-SA-3.0 Completely air-cored inductors can also be employed in radio frequencies, where values of inductance are low, and linearity and stability is more important than other factors.21) Relays In relays the air gap is usually an integral part, as it facilitates the movement between the fixed parts (e.g. winding and magnetic core) and the active armature, which mechanically drives the main electrical contacts to be connected or disconnected. In relays, the length of the gap is a compromise between the required mechanical force, mechanical movement, and the available excitation. There are numerous constructions which allow optimisation of these different factors, according to requirement of specific application22) 23) including such that the movable armature is magnetised.24) Solenoids and linear actuators Solenoids and linear actuators work similarly to relays. The magnetic circuit comprises yoke or core and a movable armature, usually called plunger. Electromagnetic actuator (solenoid) by R. Goossens, Public Domain In the simplest form a plunger is being pulled into the space inside the coil. However, as with relays, there are multiple different designs and approaches, including those with a rotating armature, whose construction becomes therefore close to electric motors.25) Transformers In classical transformers air gap is usually avoided. The role of transformer is to deliver the energy from the primary winding to the secondary winding instantaneously, without the need for energy storage. Laminations in larger transformers are overlapped to distribute the air gap26) by United States Patent and Trademark Office, Public Domain Any air gap in the magnetic core increases leakage inductance and stores additional energy, which needs to be cyclically transferred or dissipated. All these factors impact efficiency of energy transformation. The air gap lowers the total inductance of the primary winding and causes an increase in apparent power 27) through the increase of magnetising current.28) If a transformer core is made of laminations, then they are cut and assembled in a way as to stagger the gaps, to minimise any localised effects.29) For large power transformers the laminations are cut in a specific way to accommodate such overlapping. Similarly for smaller E-I or U-I cores the assembly can be carried out such that the gaps are staggered every other lamination. Similar practice is employed with cores wound from amorphous ribbon 30) or made in the Unicore technology.31) Current and voltage transformers Current and voltage transformers of the common design follow the same principles as regular transformers - the air gap is to be minimised as it limits the measurement accuracy of both amplitude and phase. However, special designs (for instance used in protection) might include air gap to lower the effective permeability and widen the operating range of primary current (linear CTs, transactors), but also to reduce the remanence (anti-remanence CTs), tune capacitive voltage transformers, etc. 32) Flyback transformers A gapped core is used in flyback transformers S. Zurek, E-Magnetica.pl, CC-BY-4.0 “Air gap” can be made by inserting non-magnetic spacers (circle shows “gap” between the core parts) S. Zurek, E-Magnetica.pl, CC-BY-4.0 The operating principle of flyback transformers used in switched-mode power supplies is more akin to energy-storing inductors than to classical transformers. For this reason should be more correctly referred to as coupled inductor. 33) All energy must be first stored in the magnetic field in the first part of the cycle, and only passed to the secondary winding in the second part of the cycle. The energy storing capability is usually achieved by means of an air gap (or more precisely: a gap filled with non-magnetic material). However, the air gap also increases leakage inductance, which generates an unwanted back EMF at the instant of switch off, and this voltage can be high enough to damage the switching transistor if not designed correctly. So the construction of a flyback transformer must take into account a compromise between the required energy storage and any voltage spikes. Rapidly changing currents and high-frequency magnetic field “leaking” from the air gaps might cause problems with electromagnetic compatibility (EMC). Low-energy supplies might not need any EMC measures, higher energy might require filters in form of bead ferrites, but at for the highest energies a conductive (flux band) or magnetic shields can be used. 34) Current sensors Current sensor might employ a gapped core with a Hall-effect sensing element S. Zurek, E-Magnetica.pl, CC-BY-4.0 Multiple approaches are used in current sensors. The importance and influence of the gap depends on the given technology.35) 36)37) A common approach used by sensor manufacturers is the “open loop sensing” principle. a magnetic core as means of concentrating the magnetic flux, which is forced to flow through the air gap. A sensor of flux density $B$ (or magnetic field strength $H$) is placed in the gap. Since the gap non-magnetic (and hence there is a linear response) the measured value of $B$ or $H$ can be related to the value of the primary current. Another common approach is the “closed loop sensing”. This technique also uses a magnetic core, and an air gap with a sensor. However, there is a compensating winding on the core and the sensor is used as a zero detector - for a condition where the magnetic field from the primary current is compensated by the current in the winding located on the core. The value of the compensating current is proportional and hence a measure of the primary current. Closed-loop technique offers better linearity and accuracy than the open loop, but it is more costly to implement. Rogowski coils Rogowski coil is wound on a non-magnetic former - it is therefore an air-cored transformer, so that its whole magnetic circuit consist of an air gap. Because the core is non-magnetic then magnetic saturation does not occur and very high alternating currents (hundreds of kA) can be measured with high accuracy. Typical Rogowski coil is wound on a flexible non-magnetic rod S. Zurek, E-Magnetica.pl, CC-BY-4.0 For accurate measurements the Rogowski coil must fully enclose the current to be measured. But the coil must also open to enable the coil to be closed around a conductor. Therefore, there is a gap between the meeting ends, and this introduces a measurement error. There are various ways of dealing with this gap, for instance overwinding of the coil ends or positioning them in a specific way. 38) Due to low permeability of the non-magnetic core the sensitivity does not allow to measure very low currents, and coils with “high sensitivity” can typically measure currents of around 300 mA even if amplification and filtering is employed. 39) 40) For this reason for low currents Rogowski coils cannot compete with ordinary current transformers. Induction heating of a metal bar by Vector1 nz, CC-BY-SA-3.0 Induction heating Induction heating employs a coil (or set of coils) which generates alternating magnetic field, usually at kHz frequency. The coil does not need to directly touch the heated object, because the electromagnetic field induces enough power loss (e.g. through eddy currents) in the object that it is possible even to achieve melting temperatures of metals. Because of the level of current the coils are commonly water cooled from inside. Apart from the heated object, the coils often do not have any magnetic core. Popularity of the so-called flux concentrators 41) keeps growing, but applying them to already existing designs can change inductance of the coil, which might require re-tuning of the system for resonating power supplies. 42) The air gap also provides electrical insulation, so that the coil is not short-circuited by the heated object or flux concentrator. 43) Electromagnets Large electromagnet with a 200 mm long air gap S. Zurek, E-Magnetica.pl, CC-BY-4.0 A common performance expected from an electromagnet is to generate magnetic field within a given volume of an air gap. This could be done for a number of tasks, for instance: to exert mechanical force on a designed part - this operation is similar to electromagnetic actuators to exert mechanical force on inclusions or other elements suspended in non-magnetic matter - a principle used in magnetic separators44), recording of shapes on magnetic film45) and some medical applications (e.g. guiding particles inside of blood vessels)46) to provide magnetic field required for material processing47)48) Energy stored in air gap A magnetic circuit behaves like a “conductor” so that the magnetic field can be efficiently guided along desired path. If a high-permeability material is used then very little energy will be stored in the magnetic core. However, an air gap introduces a discontinuity and due to its low permeability stores significant amount of magnetic energy, as compared to the same volume of magnetic core. Using analogy to an electrical circuit - a good conductor such as thick copper wire (low resistance) does not dissipate or store much energy, but a resistor or other load (high impedance) will dissipate or store large amount of energy. This energy storing property is utilised for instance in energy storing inductors and flyback transformers, in which air gap in a pivotal design parameter. On the one hand, the air gap is used for storing the actual energy, but on the other it changes operating characteristics of the B-H curve and allows driving the inductor at higher currents hence higher magnetic field strength thus extending the range before magnetic saturation occurs. For a simple magnetic circuit with a single air gap (see the first image at the top), for which the core is made out of high-permeability material such that $μ_{material} >> μ_0$, with the air gap itself and the flux density in the air gap being uniform, and if the flux fringing can be neglected, it can be derived that the stored energy is: 49) \| \| \| --- \| \| $$E \approx \frac{B^2 ⋅ V}{2 ⋅ \mu_0}$$ \| (J) \| where: $E$ - stored energy (J), $B$ - flux density in the air gap (T), $V$ - volume of the air gap (m3), $μ_0$ - permeability of free space (H/m). Flux fringing Flux fringing (red arc) around an air gap in magnetic core S. Zurek, E-Magnetica.pl, CC-BY-4.0 Flux fringing is caused by the fact that the reluctance of the concentrated air gap is much greater than that of the core. The flux tries to spread as wide as possible in order to minimise the drop of magnetomotive force across the air gap. As a result of flux fringing the total reluctance of the circuit is somewhat lower. This has several major effects. In energy-storing inductors the inductance is related to the reluctance of the air gap. The fringing lowers the overall reluctance, so that the resulting inductance is somewhat higher. This needs to be taken into account so that the inductance value is appropriate for a given design. There are various empirical equations suggested in literature for calculating the correction of this effect. For instance McLyman suggest the following “flux fringing factor” ($F$):50) \| \| \| --- \| \| $$F = 1 + \frac{l_{gap}}{\sqrt{A}} ⋅ ln \left( \frac{2 ⋅ l_{window}}{l_{gap}} \right)$$ \| (unitless) \| where: $F$ - factor by which the inductance is increased (unitless), $l_{gap}$ - length of the air gap (m), $A$ - cross-section area of the core (m2), $l_{window}$ - length of the inside (in the window) of the core leg in which the gap is present (m). Another example is when the area of the air gap is scaled according to its length. For instance if the magnetic core cross-section is a rectangle then Kazimierczuk proposes to use the following calculation: 51) \| \| \| --- \| \| $$F = 1 + \frac{l_{gap} ⋅ (a + b + 2 ⋅ l_{gap})}{a ⋅ b}$$ \| (unitless) \| where: $a$ and $b$ are the lengths of each side of the rectangular cross-section of the magnetic core (m). Yet another approximating equation is given by Hurley and Wölfle52) \| \| \| --- \| \| $$F = \frac{(a + l_{gap} ) ⋅ (b + l_{gap}) }{a ⋅ b}$$ \| (unitless) \| However, all such equations are only approximate, and usually work only under the assumption that the length of the air gap is much smaller than any of the dimensions of the core. The second effect is additional copper loss due to the fact that fringing flux “bulges away” from the air gap. Usually most of the core window is occupied by windings and if they are exposed to fast-changing fringing flux (e.g. in flyback transformers) this causes additional eddy current losses in the windings.53) The third effect is that the fringing flux enters the core perpendicularly to the normal flow of magnetic field. In soft ferrites this is not a problem. But in laminated cores this flux does not travel along the laminations, but enters them perpendicularly to their surface, resulting in a large value of normal component, inducing elevated eddy currents and thus additional iron loss. A distributed air gap is employed in order to reduce this effect (see next section). Distributed air gap Changing one large air gap into several smaller reduces flux fringing S. Zurek, E-Magnetica.pl, CC-BY-4.0 In high-power energy storing inductors the air gaps can be quite large. This would cause for the inductance to differ by unacceptable amount and also the losses would be too large. The air gap is therefore “distributed” by introducing several smaller air gaps, whose total length is comparable to one large air gap. This reduces the fringing effect considerably. Air gaps between particles in a powder core S. Zurek, E-Magnetica.pl, CC-BY-4.0 Magnetic cores of high-frequency chokes can be made from powder cores. High-permeability magnetic particles are compressed into a core, but a certain amount of non-magnetic volume is left on purpose. This causes a very good distribution of the resulting air gap over the whole volume of such core. As a result the fringing flux greatly reduced (as compared to a core with a concentrated gap), yet a large amount of energy can be stored. In the powder cores the transition into magnetic saturation (when overexcited) is less sharp, which is also an important feature reducing severity of faults in electronic circuits using such solutions. However, such cores are usually more expensive and exhibit different magnetic loss (e.g. higher than soft ferrites), so magnetic design must take these factors into account. See also Effective magnetic permeability Flux fringing Leakage flux References 1) John R. Brauer, Magnetic Actuators and Sensors, John Wiley & Sons, 2006, ISBN 9780471777700 2) Austin Hughes, Electric Motors and Drives: Fundamentals, Types and Applications, 3rd edition, Newnes, 2005, ISBN 9780080522043 3) Multigo Vertical Centrifugal Multistage Pumps, datasheet, {accessed 16 Nov 2012} 4) Delmar R. Riffe, Richard D. Olson, David M. Edison, J. Spreadbury, Patent US4030058, Inductive coupler, 1977 5) International standard, IEC 60404-3, Magnetic materials - Part 3: Methods of measurement of the magnetic properties of magnetic sheet and strip by means of a single sheet tester 6), 7) Epcos, Ferrites and accessories, Processing notes, September 2006, {accessed 2021-11-27} 8) Juha Pyrhonen, Tapani Jokinen, Valeria Hrabovcova, Design of rotating electrical machines, Wiley, 2008, ISBN 978-0-470-69516-6, p. 298 9) Austin Hughes, Electric motors and drives - fundamentals, types and applications, 3rd edition, Newnes, 2006, ISBN 978-0-7506-4718-2, p. 197 10) S.K. Bhattacharya, Electrical Machines, 3rd edition, McGraw-Hill, 2009, ISBN 978-0-07-066921-5, p. 74 11) Dieter Gerling, Design of an Induction Motor with Multilayer Rotor Structure and Large Gap, Proceedings of the International Conference on Electrical Machines (ICEM), 2000, Vol. 1, Helsinki, Finland, p. 458 12) Hamid A. Toliyat, Gerald B. Kliman, Handbook of electric motors, 2nd edition, CRC Press, 2004, ISBN 0-8247-4105-6, p. 47 13) Vassilios A. Profillidis, Railway Management And Engineering, 3rd edition, Ashgate Publishing, 2006, ISBN 978-0-7546-4854-3, p. 33 14), 15) Alex Goldman, Magnetic components for power electronics, Kluwer Academic Publishers, 2002, ISBN 0-7923-7587-4, p. 116 16) Coilcraft, PCV-0 Series, Power chokes - vertical mount, Document 135-1, 05/01/2012, {accessed 20 Nov 2012} 17) Clifford G. Thiel, Darin Driessen, Debabrata Pal, Frank Feng, Patent US7573362, High current, multiple air gap, conduction cooled, stacked lamination inductor, 2007 18) Marian K. Kazimierczuk, High-Frequency Magnetic Components, John Wiley & Sons, 2011, ISBN 9780470714539, p. 67 19) Igor Abramov, Trimmable inductor, Patent US6005467, 1999 20) Juergen Schlabbach, Karl-Heinz Rofalski, Power System Engineering: Planning, Design, and Operation of Power Systems and Equipment, John Wiley & Sons, 2008, ISBN 978-3-527-40759-0, p. 282 21) Image of air-cored radio inductor, Wikimedia Commons, {accessed 20 Nov 2012} 22) Takashi Miura, Yoshiaki Kamiya, Electromagnetic polar relay, Patent US5150090, 1992 23) John S. Zimmer, Miniature relay with double air gap magnetic circuit, Patent US3553612, 1971 24) Masami Hori, Norimasa Kaji, Hiromi Nishimura, Yoshinobu Okada, Polarized electromagnetic relay, Patent EP0361392B1, 1994 25) Richard E. Tippner, Rotating electromagnetic solenoid motor, Patent US4214178, 1980 26) Kou C. Lin, Charles E. Burkhardt, Compact step-lap magnetic core, US Patent US4445104, 1980 27), 29) P. Marketos, T. Meydan, D. Snell, Performance of single-phase transformer cores assembled from consolidated stacks of electrical steels, Journal of Magnetism and Magnetic Materials, Vol. 254-255, 2003, p. 612-614 28) Bharat Heavy Electricals Limited , Transformers, 2nd edition, Tata McGraw Hill Publishing Company Ltd, p. 100, ISBN 978-0-07-048315-6 30) Dung A. Ngo, Kimberly M. Borgmeier, Method for manufacturing a wound, multi-cored amorphous metal transformer core, Patent US6668444, 2003 31) Unicore cores, ArcelorMittal, {accessed 19 Aug 2014} 32) Network Protection & Automation Guide, Alstom Grid, 2011, {accessed 19 Aug 2014} 33) Muhammad Rashid (editor), Power Electronics Handbook: Devices, Circuits, and Applications, Butterworth-Heinemann, 2011, ISBN 978-0-12-382036-5, p. 39 34) Power Integrations Inc., TOPSwitch (R) flyback transformer construction guide, Application note AN-18, July 1996, {accessed 28 Nov 2012} 35) Open Loop Hall Effect, Hall effect current transducers (O/L), LEM, {accessed 11 Dec 2012} 36) Hall effect current sensors, Current transformers & Toroidal cores, Telcon, {accessed 11 Dec 2012} 37) Current Sensors Line Guide, Honeywell, June 2008, {accessed 11 Dec 2012} 38) Pierre Turpin, Rogowski current sensor, Patent US20120126789, 2012 39) CWT Ultra miniature wideband current probe for ac current measurement, Powertek, {accessed 19 Aug 2014} 40) Powertek, CWT Ultra mini data sheet, {accessed 19 Aug 2014} 41), 43) Valery Rudnev, Don Loveless, Raymond Cook, Micah Black, Handbook of Induction Heating, Marcel Dekker, 2003, ISBN 0-8247-0848-2, p. 364 42) R.S. Ruffini, Jr., V.S. Nemkov, Magnetic Field Control and Concentration in Induction Heating coils, Heat Treating, Proceedings of the 16th ASM Heat Treating Society Conference & Exposition, 19-21 March 1996, Cincinatti, Ohio, US, p. 127 44) Magnetic Grate RMM 3-20-25, The Research and Development Office MAGNETO,. {accessed 8 Jan 2013} 45) Magnetic Field Indicator MFI-3, The Research and Development Office MAGNETO. {accessed 8 Jan 2013} 46) Attila Meretei, Systems and methods for removing plaque from a blood vessel, Patent US20060142632, 2004 47) A. Kozlowski, S. Zurek, R. Rygal, Influence of the number of excitation coils on uniformity of distribution of magnetic flux density in an air gap, 1&2DM Workshop, September 2012, Vienna, Austria 48) Flat (F) loops, VACUUMSCHMELZE GmbH & Co. KG, {accessed 8 Jan 2012} 49) W.G. Hurley, W.H. Wolfle, Transformers and Inductors for Power Electronics: Theory, Design and Applications, John Wiley & Sons, 2013, ISBN 9781118544679, example 2.3 50), 53) Colonel William T. McLyman, Transformer and Inductor Design Handbook, CRC Press, 2004, ISBN 9780824751159, section "Fringing flux" in chapter 8 51) Marian K. Kazimierczuk, High-Frequency Magnetic Components, John Wiley & Sons, 2011, ISBN 9781119964919, p. 38 52) W.G. Hurley, W.H. Wölfle, Transformers and Inductors for Power Electronics: Theory, Design and Applications, John Wiley & Sons, 2013, ISBN 9781118544679 Magnetic circuits, Air gap, Counter air_gap.txt · Last modified: 2023/10/18 14:15 by stan_zurek
17063	https://www.online-calculator.com/zh-CN/distance-converter/	Go 改变语言 Menu 全屏计算器 - 可以免费全屏使用的在线计算器科学计算器 - 一个伟大的科学计算器。清晰自由! 简易计算器 - 一个不错的简单免费在线计算器。易于使用和阅读. 网上珠算 - 在线算盘！教数字从 1 到 50:-) 飞镖计算器 - 忘记数学，玩飞镖吧！数学计算器 - 此在线数学计算器显示您的总和历史记录。很有帮助！体重指数计算器 - 计算出您的 BMI / 体重指数天井计算器 - 那个露台有多少块板？戒指尺寸转换器 - 您需要多大尺寸的戒指？了解差异长度转换器 - 转换长度！厘米、脚等! 百分比计算器 - 算出你需要的所有百分比！音量转换器 - 求立方体的体积！重量转换器 - 在权重之间转换长度换算 - 长度转换工具合集速度转换 - 速度转换工具合集温度转换 - 温度转换工具合集时间转换 - 时间转换工具合集重量换算 - 重量转换工具合集使用我们惊人的距离转换器工具！从英里转换为公里、米、英尺、英寸——随心所欲！非常容易使用！完美的！ :-) 距离转换我们的距离转换器工具可让您轻松地在英里、公里、米、英尺和英寸之间进行转换，让您能够快速准确地转换距离，无论您开始使用什么单位。无论您是计划公路旅行、测量建筑或工程项目的距离，还是只是需要转换学校或工作的距离，我们的工具都是完美的解决方案。凭借其简单的界面和易于使用的功能，我们的距离转换器可以在使用不同的距离单位时为您节省时间和麻烦。我们还有大量长度和距离转换工具 - 可在此处找到：长度和距离换算热门距离转换：英尺换算为厘米- 将英尺换算为厘米！厘米换算为英尺- 将厘米换算为英尺！英尺换算为公里- 将英尺换算为公里！公里换算为英尺- 将公里换算为英尺！英尺换算为英里- 将英尺换算为英里！英里换算为英尺- 将英里换算为英尺！英尺换算为毫米- 将英尺换算为毫米！毫米换算为英尺- 将毫米换算为英尺！英尺换算为码- 将英尺换算为码！码换算为英尺- 将码换算为英尺！英寸换算为厘米- 将英寸换算为厘米！厘米到英寸- 将厘米转换为英寸！英寸换算为分米- 将英寸换算为分米！英寸换算为英尺- 将英寸换算为英尺！英尺换算为英寸- 将英尺换算为英寸！英寸换算为千米- 将英寸换算为千米！千米换算为英寸- 将千米换算为英寸！英寸换算为米- 将英寸换算为米！米换算为英寸- 将米换算为英寸！英寸换算为英里- 将英寸换算为英里！英里换算为英寸- 将英里换算为英寸！英寸换算为毫米- 将英寸换算为毫米！毫米换算为英寸- 将毫米换算为英寸！英寸到码- 将英寸转换为码！码换算为英寸- 将码换算为英寸！公里换算为厘米- 将公里换算为厘米！厘米换算为公里- 将厘米换算为公里！公里换算为英尺- 将公里换算为英尺！英尺换算为公里- 将英尺换算为公里！公里到英寸- 将公里换算成英寸！英寸换算为公里- 将英寸换算为公里！公里到米- 将公里转换为米！米到公里- 将米转换为公里！公里换算为英里- 将公里换算为英里！英里换算为公里- 将英里换算为公里！公里换算为毫米- 将公里换算为毫米！毫米换算为公里- 将毫米换算为公里！公里到码- 将公里转换为码！码换算为公里- 将码换算为公里！米换算为厘米- 将米换算为厘米！厘米换算为米- 将厘米换算为米！米链- 将米转换为链条！米换算为英尺- 将米换算为英尺！英尺换算为米- 将英尺换算为米！米换算为英寸- 将米换算为英寸！英寸换算为米- 将英寸换算为米！米换算为毫米- 将米换算为毫米！毫米换算为米- 将毫米换算为米！米到码- 将米转换为码！码换算为米- 将码换算为米！英里换算为厘米- 将英里换算为厘米！厘米换算为英里- 将厘米换算为英里！英里换算为英尺- 将英里换算为英尺！英尺换算为英里- 将英尺换算为英里！英里换算为英寸- 将英里换算为英寸！英寸换算为英里- 将英寸换算为英里！英里换算为米- 将英里换算为米！米到英里- 将米转换为英里！英里换算为毫米- 将英里换算为毫米！毫米换算为英里- 将毫米换算为英里！英里换算为码- 将英里换算为码！码到英里- 将码转换为英里！英里到 KM 问题！问题：火车行驶 100 英里。它行驶了多少公里？回答：要将英里转换为公里，我们需要将英里数乘以 1.609。要找出火车行驶了多少公里，我们需要将 100 英里乘以 1.609：100 英里 x 1.609 = 160.9 公里因此，火车行驶了大约 160.9 公里。 KM 到英里问题！问题：一条船行驶50公里。它行驶了多少英里？回答：要将公里转换为英里，我们需要将公里数除以 1.609。因此，要找出船行驶了多少英里，我们需要将 50 公里除以 1.609：50 公里 ÷ 1.609 = 31.07 英里因此，船行驶了大约 31.07 英里。距离换算表 \| 转换 \| 英里 \| 公里数 \| 米 \| 脚 \| 英寸 \| --- --- --- \| \| 1英里 \| 1个 \| 1.60934 \| 1609.34 \| 5280 \| 63360 \| \| 5英里 \| 5个 \| 8.04672 \| 8046.72 \| 26400 \| 316800 \| \| 10英里 \| 10 \| 16.0934 \| 16093.4 \| 52800 \| 633600 \| \| 15英里 \| 15 \| 24.1402 \| 24140.2 \| 79200 \| 950400 \| \| 20 英里 \| 20 \| 32.1869 \| 32186.9 \| 105600 \| 1267200 \| \| 25英里 \| 25 \| 40.2336 \| 40233.6 \| 132000 \| 1584000 \| \| 30 英里 \| 30 \| 48.2803 \| 48280.3 \| 158400 \| 1900800 \| \| 35 英里 \| 35 \| 56.327 \| 56327 \| 184800 \| 2217600 \| \| 40 英里 \| 40 \| 64.3738 \| 64373.8 \| 211200 \| 2534400 \| \| 45英里 \| 45 \| 72.4205 \| 72420.5 \| 237600 \| 2851200 \| 此表包括 10 行从英里、公里、米、英尺、英寸转换的温度 Site Menu 主页 - 返回主页... :-) 全屏计算器 - 可以免费全屏使用的在线计算器科学计算器 - 一个伟大的科学计算器。清晰自由! 简易计算器 - 一个不错的简单免费在线计算器。易于使用和阅读. 网上珠算 - 在线算盘！教数字从 1 到 50:-) 飞镖计算器 - 忘记数学，玩飞镖吧！数学计算器 - 此在线数学计算器显示您的总和历史记录。很有帮助！体重指数计算器 - 计算出您的 BMI / 体重指数天井计算器 - 那个露台有多少块板？戒指尺寸转换器 - 您需要多大尺寸的戒指？了解差异长度转换器 - 转换长度！厘米、脚等! 百分比计算器 - 算出你需要的所有百分比！音量转换器 - 求立方体的体积！重量转换器 - 在权重之间转换长度换算 - 长度转换工具合集速度转换 - 速度转换工具合集温度转换 - 温度转换工具合集时间转换 - 时间转换工具合集重量换算 - 重量转换工具合集跑表 - 在线秒表，全屏秒表。非常适合会议、教室、大会、学校等任何地方……:-) 随机名称选择器 - 随机名称选择器轮和更多随机生成器！距离转换器 - 转换距离！英里数、公里数等! 温度转换器 - 太热/太冷？ / 正好！圆形工具 - 圆半径、圆直径或圆周长转化计算器使用我们种类齐全、简单易用的换算计算器，轻松换算单位、货币等！这里只是其中的一部分，请务必检查我们提供的所有其他内容！我们的体重转换工具准确、易于使用且提供范围广泛的单位。体重换算与我们的用户- 友好而全面的长度转换计算器。长度转换我们的温度转换计算器准确、易于使用，并且涵盖各种温度单位。温度转换大量其他易于使用的转换工具！更多转化... KG 换算为磅石头到公斤盎司到磅天到小时分钟到天小时到秒英尺到 CM 英里换算为 KM 英尺到英寸摄氏到华氏 MM 换算为米码到英寸特色工具：这些是网站上一些最流行的计算器和工具！飞镖计算器我们的飞镖计算器可帮助玩家快速准确地计算分数和结算，使游戏更加有趣和更具竞争力！飞镖计算器体重指数计算器我们的 BMI 计算器可以快速计算身体质量指数，帮助用户了解他们的体重状况并做出明智的健康决定。体重指数计算器在线珠算在线珠算科学计算器科学计算器 Online-Calculator.com 自 2007 年以来一直存在！那是一段很长的时间！最初的计算器是在 17 世纪由一位名叫 Blaise Pascal 的法国人发明的！他才 18 岁，想帮父亲算税。和我们一起了解计算器的迷人历史！计算器的历史！ About Us Contact us Languages Terms & FAQ Privacy Policy Stopwatch
17064	https://chem.libretexts.org/Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/11%3A_Stoichiometry-_Quantities_in_Chemical_Reactions/11.01%3A_General_Stoichiometry/11.1.5%3A_AcidBase_Titration	11.1.5: Acid–Base Titration - Chemistry LibreTexts Skip to main content Table of Contents menu search Search build_circle Toolbar fact_check Homework cancel Exit Reader Mode school Campus Bookshelves menu_book Bookshelves perm_media Learning Objects login Login how_to_reg Request Instructor Account hub Instructor Commons Search Search this book Submit Search x Text Color Reset Bright Blues Gray Inverted Text Size Reset +- Margin Size Reset +- Font Type Enable Dyslexic Font - [x] Downloads expand_more Download Page (PDF) Download Full Book (PDF) Resources expand_more Periodic Table Physics Constants Scientific Calculator Reference expand_more Reference & Cite Tools expand_more Help expand_more Get Help Feedback Readability x selected template will load here Error This action is not available. chrome_reader_mode Enter Reader Mode 11.1: General Stoichiometry 11: Stoichiometry- Quantities in Chemical Reactions { } { "11.1.01:_Mass_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.1.02:_Solution_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.1.03:_Stoichiometry_and_the_Ideal_Gas_Law" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.1.4:_Gas_Stoichiometry_at_STP" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.1.5:_AcidBase_Titration" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } { "11.01:General_Stoichiometry" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.02:_Limiting_Reactant" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.03:_Theoretical_Yield_and_Percent_Yield" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.04:_Enthalpy_Change_is_a_Measure_of_the_Heat_Evolved_or_Absorbed" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1", "11.E:_Stoichiometry_Applications(Exercises)" : "property get Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1" } Mon, 19 Dec 2022 19:22:45 GMT 11.1.5: Acid–Base Titration 409060 409060 Joshua Halpern { } Anonymous Anonymous User 2 false false [ "article:topic", "acid\u2013base titration", "showtoc:no", "authorname:anonymous", "source-chem-47562", "license:mixed", "source-chem-367852" ] [ "article:topic", "acid\u2013base titration", "showtoc:no", "authorname:anonymous", "source-chem-47562", "license:mixed", "source-chem-367852" ] Search site Search Search Go back to previous article Sign in Username Password Sign in Sign in Sign in Forgot password Contents 1. Home 2. Campus Bookshelves 3. Fresno City College 4. Introductory Chemistry, Atoms First for FCC 5. 11: Stoichiometry- Quantities in Chemical Reactions 6. 11.1: General Stoichiometry 7. 11.1.5: Acid–Base Titration Expand/collapse global location Introductory Chemistry, Atoms First for FCC Front Matter 1: Basics of Measurement 2: Atomic Structure 3: Light, Electrons, and the Periodic Table 4: Compounds and Chemical Bonds 5: The Mole and Chemical Formulas 6: Lewis Structures, Shapes, and Intermolecular Forces 7: Solids, Liquids, and Phase Changes 8: Gases 9: Aqueous Solutions 10: Chemical Reactions 11: Stoichiometry- Quantities in Chemical Reactions 12: Acids and Bases 13: Chemical Equilibrium 14: Radioactivity and Nuclear Chemistry Back Matter xx delete-test 11.1.5: Acid–Base Titration Last updated Dec 19, 2022 Save as PDF 11.1.4: Gas Stoichiometry at STP 11.2: Limiting Reactant picture_as_pdf Full Book Page Downloads Full PDF Import into LMS Individual ZIP Buy Print Copy Print Book Files Buy Print CopyReview / Adopt Submit Adoption Report Submit a Peer Review View on CommonsDonate Page ID 409060 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Table of contents 1. Learning Objectives 2. Example 11.1.5.1: Equivalence Point 1. Solution Exercise 11.1.5.1 Exercise 11.1.5.2 Exercise 11.1.5.3 Indicator Selection for Titrations Summary Contributions & Attributions Learning Objectives Understand the basics of acid-base titrations. Understand the use of indicators. Perform a titration calculation correctly. The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because so many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a titration. A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions. During an acid-base titration, an acid with a known concentration (a standard solution) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base. The indicator will signal, by color change, when the base has been neutralized (when [H+] = [OH-]). At that point—called the equivalence point, or end point—the titration is stopped. By knowing the volumes of acid and base used, and the concentration of the standard solution, calculations allow us to determine the concentration of the other solution. It is important to accurately measure volumes when doing titrations. The instrument you would use is called a burette(or buret). Figure 11.1.5.1: Equipment for Titrations. A burette is a type of liquid dispensing system that can accurately indicate the volume of liquid dispensed. For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted: mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl We also have the balanced chemical reaction between HCl and NaOH: HCl+NaOHNaCl+H⁡A 2⁢O So we can construct a conversion factor to convert to number of moles of NaOH reacted: 0.002766⁢m⁢o⁢l⁢H⁡C⁢l×1 m⁢o⁢l N⁢a⁢O⁢H 1⁢m⁢o⁢l⁢H⁡C⁢l=0.002766 m⁢o⁢l N⁢a⁢O⁢H Then we convert this amount to mass, using the molar mass of NaOH (40.00 g/mol): 0.002766⁢m⁢o⁢l⁢H⁡C⁢l×40.00⁢g⁡N⁢a⁢O⁢H 1⁢m⁢o⁢l⁢H⁡C⁢l=0.1106⁢g⁡N⁢a⁢O⁢H This type of calculation is performed as part of a titration. Example 11.1.5.1: Equivalence Point What mass of Ca(OH)2 is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO 3? The balanced chemical equation is as follows: 2 HNO⁢A 3+Ca⁡(OH)⁢A 2Ca⁡(NO⁢A 3)⁢A 2+2⁢H⁡A 2⁢O Solution In liters, the volume is 0.04402 L. We calculate the number of moles of titrant: moles HNO 3 = (0.04402 L)(0.0885 M) = 0.00390 mol HNO 3 Using the balanced chemical equation, we can determine the number of moles of Ca(OH)2 present in the analyte: 0.00390⁢m⁢o⁢l⁢H⁡N⁢O 3×1 m⁢o⁢l C⁢a⁢(O⁢H)2 2⁢m⁢o⁢l⁢H⁡N⁢O 3=0.00195 m⁢o⁢l C⁢a⁢(O⁢H)2 Then we convert this to a mass using the molar mass of Ca(OH)2: 0.00195⁢m⁢o⁢l C⁢a⁢(O⁢H)2×74.1⁢g⁡C⁢a⁢(O⁢H)2 m⁢o⁢l C⁢a⁢(O⁢H)2=0.144⁢g⁡C⁢a⁢(O⁢H)2 Exercise 11.1.5.1 What mass of H 2 C 2 O 4 is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows: H⁡A 2⁢C⁢A 2⁢O⁢A 4+2 NaOHNa⁢A 2⁢C⁢A 2⁢O⁢A 4+2⁢H⁡A 2⁢O Answer0.182 g Exercise 11.1.5.2 If 25.00 mL of HCl solution with a concentration of 0.1234 M is neutralized by 23.45 mL of NaOH, what is the concentration of the base? Answer0.1316 M NaOH Exercise 11.1.5.3 A 20.0 mL solution of strontium hydroxide, Sr(OH)2, is placed in a flask and a drop of indicator is added. The solution turns color after 25.0 mL of a standard 0.0500 M HCl solution is added. What was the original concentration of the Sr(OH)2 solution? Answer3.12×10−2 M Sr(OH)2 Indicator Selection for Titrations The indicator used depends on the type of titration performed. The indicator of choice should change color when enough of one substance (acid or base) has been added to exactly use up the other substance. Only when a strong acid and a strong base are produced will the resulting solution be neutral. The three main types of acid-base titrations, and suggested indicators, are: The three main types of acid-base titrations, suggested indicators, and explanations\| Titration between . . . \| Indicator \| Explanation \| --- \| strong acid and strong base \| any \| \| \| strong acid and weak base \| methyl orange \| changes color in the acidic range (3.2 - 4.4) \| \| weak acid and strong base \| phenolphthalein \| changes color in the basic range (8.2 - 10.6) \| Summary A titration is the quantitative reaction of an acid and a base. Indicators are used to show that all the analyte has reacted with the titrant. Contributions & Attributions Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. This page titled 11.1.5: Acid–Base Titration is shared under a mixed license and was authored, remixed, and/or curated by Anonymous. 14.6: Acid–Base Titration by Henry Agnew, Marisa Alviar-Agnew is licensed CK-12. Original source: Toggle block-level attributions Back to top 11.1.4: Gas Stoichiometry at STP 11.2: Limiting Reactant Was this article helpful? Yes No Recommended articles 11.1: General StoichiometryWe have used balanced equations to set up ratios, in terms of moles of materials, that we can use as conversion factors to answer stoichiometric quest... Article typeSection or PageAuthorAnonymousLicenseMixed LicensesShow Page TOCno on page Tags acid–base titration source-chem-367852 source-chem-47562 © Copyright 2025 Chemistry LibreTexts Powered by CXone Expert ® ? The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Privacy Policy. Terms & Conditions. Accessibility Statement.For more information contact us atinfo@libretexts.org. Support Center How can we help? Contact Support Search the Insight Knowledge Base Check System Status× contents readability resources tools ☰ 11.1.4: Gas Stoichiometry at STP 11.2: Limiting Reactant Complete your gift to make an impact
17065	https://openoregon.pressbooks.pub/ctetechwriting/chapter/grammar-lesson-apostrophes-possessive-s/	Skip to content Grammar Lesson – Apostrophes/Possessive S There are two main ways to show ownership in writing: using a possessive apostrophe or using a possessive pronoun. This section will define and provide examples of each. Possessive Apostrophes Apostrophes are signals telling the reader that a word is either possessive or a contraction. As a technical communicator, it’s important to understand the difference between the two. Apostrophes are used to form contractions to indicate omitted letters, such as couldn’t (the apostrophe indicates the missing letter o ). Apostrophes are also used to signal omitted numbers, such as The ‘80s (the apostrophe indicates the missing numbers 19). But this has nothing to do with apostrophes used to show possession. To use an apostrophe to show ownership, you simply add apostrophe s or s apostrophe to a noun, depending on whether it’s singular or plural. Singular Possessive Apostrophe: to indicate singular ownership, add apostrophe s: EXAMPLES: The car’s new tires were next to John’s workstation. (there is only one car and one John, so we simply add an apostrophe s to indicate singular ownership). The woman’s home needed refurnishing, so she used last week’s pay to go furniture shopping. Plural Possessive Apostrophe: to indicate plural ownership, add s apostrophe. EXAMPLES: The cars’ new tires were stacked up next to the mechanics’ workstations (in this case there is more than one car and more than one mechanic, so we would use s apostrophe). The roommates’ house needed repairs, so they all agreed to use some of the extra months’ rent money they’d saved to go furniture shopping. Joint and Individual Ownership: to show joint ownership, only the last noun/name has the apostrophe s. To show individual ownership, each noun/name has an apostrophe s. EXAMPLES: Joint: Mary, Beth, Phil, and Bill’s house. Individual: Mary’s, Beth’s, Phil’s, and Bill’s houses. Nouns Ending is S: When making a possessive of a singular noun that already ends in s, writers can make the possessive by adding ’s to the word; however, some writers and editors argue that there’s no need to include an s after the apostrophe, since the apostrophe already tells readers that the word is possessive. Others argue that you should drop the final s only on words of several syllables but retain it on short words. Since there is no agreement on this, must make your own choice. Regardless of which option you choose, be consistent. EXAMPLES: Table 2 shows three proper nouns that end in s, each of which is correct: TABLE 2: Proper Nouns Ending in S \| NAME \| APOSTROPHE S \| S APOSTROPHE \| \| James \| James’s \| James’ \| \| Jones \| Jones’s \| Jones’ \| \| Jesus \| Jesus’s \| Jesus’ \| NOTE: There are irregular nouns like fish (one fish, two fish) and goose (one goose, two geese), but we won’t worry about those right now. Possessive Pronouns Pronouns, such as him, her, they, and them are stand-ins for proper nouns; in other words, they refer to someone or something specific without using the proper noun or name. Possessive pronouns show ownership. Some are used alone, while others are used to modify or describe a noun. Used alone: mine, yours, his, hers, ours, theirs, whose EXAMPLE: That computer is hers. That car is mine. Used as modifier: my, your, his, her, its, ours, their, whose EXAMPLE: That is her computer. The car needs its clutch replaced. Note that none of the possessive pronouns uses an apostrophe to show ownership. Pay extra attention to your use of possessive pronouns, as several of them sound like some commonly-used contractions. For example, watch your use of the following commonly confused possessive pronouns and contractions: Your/You’re, Its/It’s, Their/They’re, and Whose/Who’s. Additional Resources “Basic Rules of Punctuation,” a resource on general punctuation rules from Professional Communication “Apostrophes” from OER Service’s Technical Writing
17066	https://artofproblemsolving.com/wiki/index.php/AM-GM_Inequality?srsltid=AfmBOoogfofw5svRSGGYlyVVyoaBJY_nnbHnBc9-b3Rea2cPvXDcBeaU	Art of Problem Solving AM-GM Inequality - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki AM-GM Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search AM-GM Inequality In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same. In symbols, the inequality states that for any real numbers , with equality if and only if . The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests. Contents [hide] 1 Proofs 2 Generalizations 2.1 Weighted AM-GM Inequality 2.2 Mean Inequality Chain 2.3 Power Mean Inequality 3 Problems 3.1 Introductory 3.2 Intermediate 3.3 Olympiad 4 See Also Proofs Main article: Proofs of AM-GM All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for . Generalizations The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM. Weighted AM-GM Inequality The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article. Mean Inequality Chain Main article: Mean Inequality Chain The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain. Power Mean Inequality Main article: Power Mean Inequality The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality. Problems Introductory For nonnegative real numbers , demonstrate that if then . (Solution) Find the maximum of for all positive . (Solution) Intermediate Find the minimum value of for . (Source) Olympiad Let , , and be positive real numbers. Prove that (Source) See Also Proofs of AM-GM Mean Inequality Chain Power Mean Inequality Cauchy-Schwarz Inequality Inequality Retrieved from " Categories: Algebra Inequalities Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
17067	https://math.stackexchange.com/questions/2928272/why-does-dividing-a-polynomial-by-x-a-give-the-same-quotient-as-evaluating-it	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Why does dividing a polynomial by $x-a$ give the same quotient as evaluating it at $x=a$ using synthetic division? Ask Question Asked Modified 2 years, 6 months ago Viewed 2k times 4 $\begingroup$ I know synthetic division is a table representation of the calculations that occur when evaluating a polynomial in Horner's form at some $x$ value ($x=a$). For example, if there's some polynomial in standard form say $x^3 + 4x^2 -5x + 5$ it can be transformed into Horner's form through successive groupings and factoring out $x$. $x^3 + 4x^2 -5x + 5$ $=(x^2+4x-5)x+5$ $=((x+4)x-5)x+5$ If this polynomial is evaluated at $x=3$ the result is $p(3) =(((3)+4)(3)-5)(3)+5$ $=((7)(3)-5)(3)+5$ $=(21-5)(3)+5$ $=(16)(3)+5$ $=48+5$ $=53$ Which is also what we get using synthetic division Interestingly, synthetic division also gives the same quotient as dividing $x^3 + 4x^2 -5x + 5$ by $x-3$ Which of course is $x^2 + 7x + 16 +\frac{53}{x-3}$ My question are: 1) Why would dividing a polynomial by $x-a$ (using long division) give the same quotient as evaluating it at $x = a$ when using synthetic division? 2) Furthermore why are only the coefficients and $a$ term used in synthetic division? What happened to the variables and their respective powers? They seem to "disappear" from the calculation when using synthetic division yet the same quotient is derived just as using long division. algebra-precalculus polynomials Share edited Sep 24, 2018 at 0:15 SleckerSlecker asked Sep 23, 2018 at 22:36 SleckerSlecker 1,1191212 silver badges3030 bronze badges $\endgroup$ 5 3 $\begingroup$ The techniques are essentially identical. The only major difference is the technique on the right (synthetic division) provides the information in a much more compact easy to read form, avoiding getting bogged down by so many parentheses or variables. All of the same information however can be extracted however and all the same steps are taken (though perhaps phrased slightly differently). $\endgroup$ JMoravitz – JMoravitz 2018-09-23 22:38:32 +00:00 Commented Sep 23, 2018 at 22:38 2 $\begingroup$ JMoravitz is right, though if you are asking why you get $53$ in two different ways (one by plugging in $2$ to compute $p(x)$ and one by looking at the remainder when dividing $p(x)$ by $x-2$), consider that by division (synthetic or long), you showed that $p(x)=x^3+4x^2-5x+5=(x-2)(x^2+7x+16) + 53$. If you set $x=2$ on the left, you get $p(2)$, and if you set $x=2$ on the right, you get $0+53$, or the remainder from division. $\endgroup$ Steve Kass – Steve Kass 2018-09-23 23:06:06 +00:00 Commented Sep 23, 2018 at 23:06 $\begingroup$ @SteveKass Yes, I understand why the two give the same remainder as its a result of the Remainder Theorem. I'm wondering why the two methods give the same answer: quotient, $x^2 + 7x + 16$, and remainder $\frac{53}{x-3}$ $\endgroup$ Slecker – Slecker 2018-09-23 23:22:12 +00:00 Commented Sep 23, 2018 at 23:22 1 $\begingroup$ Why wouldn't they? Both methods are the same, of course they will give you the same answer. I am trying to understand your question, but I think it has been answered in the 2 comments above $\endgroup$ Villa – Villa 2018-09-23 23:35:01 +00:00 Commented Sep 23, 2018 at 23:35 $\begingroup$ All the commenters above have said it better than I do, but what I say is: 1. Forget Synthetic Division, its just a technique for doing Genuine Division. 2. Now look at what Division says: if you start with $P(x)$ and divide by $x-a$, you get a quotient and a remainder of degree less than the degree of the divisor $x-a$, and thus a constant. So you have $P(x)=Q(x)(x-a) + R$, where $R$ is a constant. Now substitute $x=a$, and get $P(a)=R$. If you dont grasp this, maybe you havent grasped what the meaning of substitution is. Lots of high-school students seem to have trouble with just this. $\endgroup$ Lubin – Lubin 2018-09-24 01:33:44 +00:00 Commented Sep 24, 2018 at 1:33 Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ I'm responding separately to several questions about "how" or "why" synthetic division works. This topic is often not presented very elegantly in high school math classes, which makes it seem more mysterious (and simultaneously more mundane) than it actually is. It's easy enough to see how this algorithm mirrors the steps of polynomial long division, and because it uses the root $a$, instead of the divisor $(x-a)$, the algorithm involves addition rather than subtraction to find each intermediate remainder. But things get more interesting when the steps are interpreted as repeated application of the Remainder Theorem. Using Horner's form, as shown in the question, it's quite straightforward to see that the end result is $p(a)$, and by the Remainder Theorem, that this would be the remainder for $\frac{p(x)}{x-a}$. What's not so clear is why the intermediate results are the coefficients of the quotient polynomial. Turns out, it's just repeated application of the Remainder Theorem, taken one step at a time. There are two critical keys to understanding how synthetic division builds upon the Remainder Theorem. The first is Horner's form of the polynomial (a.k.a. the nested form), which is absolutely essential. It is instructive to consider the efficiency of computation ($n$ multiplications and $n$ additions to evaluate a polynomial of degree $n$) for synthetic substitution, versus direct evaluation of the polynomial at $x=a$ using the order of operations. It is an efficient algorithm. The second key to understanding why this works is to be absolutely clear on what is meant by "synthetic substitution." Substitution is performed one "$x$" at a time, moving from left to right, in the nested form of the polynomial. So at each step, the first two remaining terms are regrouped as the product of a linear factor $(Ax + B)$ and a power of $x$. Then, the Remainder Theorem is applied to just the linear factor, by substituting $a$ for that one (and only that one) $x$. Other instances of $x$ remain unresolved, and that's how each result becomes a coefficient in the quotient polynomial. When $(Ax + B)$ is divided by $(x-a)$, the quotient is $A$ (because $x-a$ is monic), and the remainder is $Aa + B$ (by the Remainder Theorem). For each successive term, the remainder from the prior operation becomes the leading coefficient for the next linear factor, which then undergoes the same treatment. Each successive linear factor is multiplied with a power of $x$, with the power decreasing each step. Division of each new linear factor by $x-a$ means its leading coefficient goes to the next term of the quotient, and substitution of $x=a$ in that linear factor produces the next remainder coefficient. That number will become the leading coefficient and hence the coefficient for the next quotient term in the following step, producing yet another remainder, and so on, until the final remainder is reached. Consider $p(x)=x^3+4x^25x+5$, as given in the question. The first step involves $(x+4)x^2$ (leaving off the other terms), and we apply the Remainder Theorem on the factor $x+4$. So, $\frac{x+4}{x-a}$ yields quotient $1$ and remainder $a+4$. Multiplying back the $x^2$, the full quotient term becomes $x^2$, and the full remainder term is $(a+4)x^2$. Next, the $-5x$ term is incorporated, and the prior remainder becomes part of a refactored term $((a+4)x-5)x$. Dividing the left factor by $x-a$ produces quotient $a+4$ and remainder $(a+4)a-5$. Multiplying back the factor $x$ on the right gives us the actual quotient term $(a+4)x$ and remainder term $((a+4)a-5)x$. Finally, incorporating the constant term $5$, we have remaining $\frac{((a+4)a-5)x+5}{x-a}$, which gives us the trailing quotient term $(a+4)a-5$ and the final remainder $((a+4)a-5)a+5$. Putting all of this together, the complete quotient is $x^2+(a+4)x+((a+4)a-5)$. Notice how $a$ was substituted for $x$ one degree at a time, as we proceeded through each term of the original polynomial. When $a=3$, this quotient becomes $x^2+7x+16$ with remainder $53$. Share edited Mar 15, 2023 at 10:59 answered Mar 14, 2023 at 21:28 Michael MalioneMichael Malione 15144 bronze badges $\endgroup$ 1 $\begingroup$ This is the missing piece of the puzzle that most explanations never properly explain. Well done. $\endgroup$ Sam Keays – Sam Keays 2023-08-09 11:43:15 +00:00 Commented Aug 9, 2023 at 11:43 Add a comment \| 1 $\begingroup$ It is easy to understand why the remainder obtained upon 'dividing' by $(x-a)$ and substituting for $x=a$ for finding the value polynomial gives you the same answer. I'm sure you must have proved the remainder theorem earlier: $$f(x)=q(x)d(x)+r(x)$$ Where $q(x)\ \&\ r(x)$ are the quotient and remainder respectively obtained when $f(x)$ is divided by divisor $d(x)$ (knowing here that $\deg r(x) < \deg d(x)$ and $d(x)$ is non-zero polynomial). So if $f(x)$ is to be divided by some $x-a$ we get: $$f(x)=(x-a)q(x)+r(x)$$ And it is not difficult to conclude that for the case $x=a$ we get: $$f(a)=(a-a)q(a)+r(a)=0 \cdot q(a)+r(a)=0+r(a)=r(a)$$ It is important, and to maintain standards of rigor, that we have assumed at the beginning of the proof that $q(x)$ and $r(x)$ are obtained uniquely for division of $f(x)$ by $d(x)$, which is actually an assertion of Euclid's division algorithm. So I don't see why you wouldn't understand why you get the result. Perhaps, you could read about Hornet's synthetic division. It is nothing but an optimized version in the case of divisor being a linear polynomial of the standard long division algorithm we use. You can check this video and this one. Share answered Oct 24, 2018 at 17:53 Suhrid SahaSuhrid Saha 34911 silver badge1010 bronze badges $\endgroup$ Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algebra-precalculus polynomials See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked Why does synthetic division work? How does synthetic division work? 2 Proof of Remainder Theorem for polynomials 1 proving a statement about Horner's method Related Division of polynomial with multiple variable using synthetic division Find the remainder of the polynomial division $p(x)/(x^2-1)$ for some $p$ 9 How does synthetic division work? 0 Finding solution of an algebraic equation by Horner's synthetic division method using a complex root 1 Can this method that makes use of Horner's Method(Synthetic Division) be used to obtain the zeroes of any polynomial? 1 I have a small problem with long division when dividing polynomials of the same degree Hot Network Questions Determine which are P-cores/E-cores (Intel CPU) How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Checking model assumptions at cluster level vs global level? Origin of Australian slang exclamation "struth" meaning greatly surprised Do sum of natural numbers and sum of their squares represent uniquely the summands? RTC battery and VCC switching circuit Interpret G-code Exchange a file in a zip file quickly Is there a specific term to describe someone who is religious but does not necessarily believe everything that their religion teaches, and uses logic? What happens if you miss cruise ship deadline at private island? Why is the fiber product in the definition of a Segal spaces a homotopy fiber product? Copy command with cs names What is the meaning of 率 in this report? Who is the target audience of Netanyahu's speech at the United Nations? Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator I'm having a hard time intuiting throttle position to engine rpm consistency between gears -- why do cars behave in this observed way? Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation Why is the definite article used in “Mi deporte favorito es el fútbol”? Passengers on a flight vote on the destination, "It's democracy!" I have a lot of PTO to take, which will make the deadline impossible Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth? What "real mistakes" exist in the Messier catalog? Storing a session token in localstorage more hot questions Question feed
17068	https://math.stackexchange.com/questions/3519640/find-the-remainder-when-px-is-divided-by-x2-1	polynomials - Find the remainder when $p(x)$ is divided by $x^2 -1$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find the remainder when p(x)p(x) is divided by x 2−1 x 2−1 Ask Question Asked 5 years, 8 months ago Modified5 years, 8 months ago Viewed 3k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Polynomial p(x)p(x) leaves a remainder of 4 4 when divided by x−1 x−1 and a remainder of −2−2 when divided by x+1 x+1. Find the remainder when p(x)p(x) is divided by x 2−1 x 2−1 . According to Remainder Theorem, when a polynomial p(x)p(x) is divided by (a x+b)(a x+b), the remainder is p(−b a)p(−b a) . So, I did the following: p(1)p(−1)=4=−2 p(1)=4 p(−1)=−2 p(x)p(x)=(x 2−1)q(x)+A x+B=(x−1)(x+1)q(x)+A x+B p(x)=(x 2−1)q(x)+A x+B p(x)=(x−1)(x+1)q(x)+A x+B When p(1)A+B=A(1)+B=4(1)p(1)=A(1)+B(1)A+B=4 When p(−1)−A+B=−A+B=−2(2)p(−1)=−A+B(2)−A+B=−2 Doing (1)+(2)(1)+(2) gives: 2 B B=2=1 2 B=2 B=1 Substitute B=1 B=1 into (1)(1) gives A=3 A=3 So, I got the remainder as 3 x+1 3 x+1 But, the answer in the book is x+3 x+3 , which means my values of A A and B B have been mixed up. Please tell me where I went wrong polynomials Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jan 23, 2020 at 11:18 Nick 1,259 10 10 silver badges 22 22 bronze badges asked Jan 23, 2020 at 10:53 gc3941dgc3941d 487 1 1 gold badge 5 5 silver badges 15 15 bronze badges 3 If we divide x+3 x+3 by x+1 x+1 , the remainder is clearly 2 2. Weird !Peter –Peter 2020-01-23 11:00:52 +00:00 Commented Jan 23, 2020 at 11:00 @Peter Why would you divide the remainder by the factor?gc3941d –gc3941d 2020-01-23 11:05:22 +00:00 Commented Jan 23, 2020 at 11:05 Apparently, I mixed the answer (which gives the remainder) with the polyomial, sorry.Peter –Peter 2020-01-23 11:13:04 +00:00 Commented Jan 23, 2020 at 11:13 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. 3 x+1=3(x−1)+4 3 x+1=3(x+1)−2 3 x+1=3(x−1)+4 3 x+1=3(x+1)−2 So your answer 3 x+1 3 x+1 is correct. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 23, 2020 at 11:35 gandalf61gandalf61 15.6k 15 15 silver badges 32 32 bronze badges 3 I see, thank you gc3941d –gc3941d 2020-01-23 12:04:59 +00:00 Commented Jan 23, 2020 at 12:04 What is missing : The proof that this is the only polynomial modulo x 2−1 x 2−1 that does the job. Otherwise we could not conclude the remainder.Peter –Peter 2020-01-23 12:09:51 +00:00 Commented Jan 23, 2020 at 12:09 1 @Peter Remainder is unique because if there are two remainders r r and r′r′ then their difference must be divisible by x 2−1 x 2−1. But since the degrees of both r r and r′r′ are less than the degree of x 2−1 x 2−1, we must have r−r′=0 r−r′=0, so r=r′r=r′.gandalf61 –gandalf61 2020-01-23 12:41:25 +00:00 Commented Jan 23, 2020 at 12:41 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions polynomials See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 2Find undetermined coefficients in polynomial quotient and remainder 6What is the remainder when x 100−8 x 99+12 x 98−3 x 10+24 x 9−36 x 8+3 x 2−29 x+41 x 100−8 x 99+12 x 98−3 x 10+24 x 9−36 x 8+3 x 2−29 x+41 is divided by: x 2−8 x+12 x 2−8 x+12. 1how to find the remainder when a polynomial p(x)p(x) is divided my another polynomial q(x)q(x) 0Find remainder when P(x) is divided by x²-3x+2 4Polynomial remainder theorem 15Why do we automatically assume that when we divide a polynomial by a second degree polynomial the remainder is linear? 1A polynomial is exactly divided by x+1 x+1, and when it is divided by 3 x−1 3 x−1, the remainder is 4 4... (continued in post) 4Finding remainder when unknown f(x)f(x) is divided to g(x)g(x) 1Finding a polynomial f(x)f(x) that when divided by x+3 x+3 yields quotient 2 x 2−x+7 2 x 2−x+7 and remainder 10 10 0Deriving remainder of the division of a polynomial using two methods Hot Network Questions Should I let a player go because of their inability to handle setbacks? Identifying a movie where a man relives the same day What’s the usual way to apply for a Saudi business visa from the UAE? Why do universities push for high impact journal publications? Origin of Australian slang exclamation "struth" meaning greatly surprised Alternatives to Test-Driven Grading in an LLM world "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results Does the curvature engine's wake really last forever? If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church? Checking model assumptions at cluster level vs global level? Is it safe to route top layer traces under header pins, SMD IC? Numbers Interpreted in Smallest Valid Base I have a lot of PTO to take, which will make the deadline impossible Does the Mishna or Gemara ever explicitly mention the second day of Shavuot? Suspicious of theorem 36.2 in Munkres “Analysis on Manifolds” Passengers on a flight vote on the destination, "It's democracy!" What is a "non-reversible filter"? Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Determine which are P-cores/E-cores (Intel CPU) Is it ok to place components "inside" the PCB How can blood fuel space travel? What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? Can you formalize the definition of infinitely divisible in FOL? more hot questions Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
17069	http://math.bu.edu/keio2019/talks/Nguyen.pdf	EQUIDISTRIBUTION ESTIMATES FOR THE k-FOLD DIVISOR FUNCTION TO LARGE MODULI DAVID T. NGUYEN Abstract. For each positive integer k, let τk(n) denote the k-fold divisor function, the coeﬃcient of n−s in the Dirichlet series for ζ(s)k. For instance τ2(n) is the usual divisor function τ(n). The distribution of τk(n) in arithmetic progression is closely related to distribution of the primes. Aside from the trivial case k = 1 their distributions are only fairly well understood for k = 2 (Selberg, Linnik, Hooley) and k = 3 (Friedlander and Iwaniec; Heath-Brown; Fouvry, Kowalski, and Michel). In this note I’ll survey and present some new distribution estimates for τk(n) in arithmetic progression to special moduli applicable for any k ≥4. This work is a reﬁnement of the recent result of F. Wei, B. Xue, and Y. Zhang in 2016, in particular, leads to some eﬀective estimates with power saving error terms. 1. Introduction and survey of known results Let n ≥1 and k ≥1 be integers. Let τk(n) denote the k-fold divisor function τk(n) = X n1n2···nk=n 1, where the sum runs over ordered k-tuples (n1, n2, . . . , nk) of positive integers for which n1n2 · · · nk = n. Thus τk(n) is the coeﬃcient of n−s in the Dirichlet series ζ(s)k = ∞ X n=1 τk(n)n−s. This note is concerned with the distribution of τk(n) in arithmetic progressions to moduli d that exceed the square-root of length of the sum, in particular, provides a sharpening of the result in . 1.1. Distribution of arithmetic functions. C. F. Gauss, towards the end of the 18th century, conjectured the celebrated Prime Number Theorem concerning the sum X p≤X 1 as X approaches inﬁnity, where p denotes a prime. It is more convenience to count primes with weight log p instead of weight 1, c.f. Chebyshev; this leads to consideration of the sum X p≤X log p. To access the Riemann zeta function more conveniently we also count powers of primes, leading to the sum X pα≤X α≥1 log p, which is equal to the unconstrained sum over n X n≤X Λ(n) 1 2 DAVID T. NGUYEN where Λ(n) is the von Mangoldt function–the coeﬃcient of n−s in the series −ζ′(s)/ζ(s). In 1837, P. G. L. Dirichlet considered the deep question of primes in arithmetic progression, leading him to consider sums of the form X n≤X n≡a(mod d) Λ(n) for (d, a) = 1. More generally, the function Λ(n) is replaced by an arithmetic function f(n), satisfying certain growth conditions, and we arrive at the study of the congruence sum (1.1) X n≤X n≡a(mod d) f(n). This sum (1.1) is our main object of study. For most f appearing in applications, it is expected that f is distributed equally among the reduced residue classes a(mod d) with (a, d) = 1, e.g., that the sum (1.1) is well approximated by the average (1.2) 1 ϕ(d) X n≤X (n,d)=1 f(n) since there are ϕ(d) reduced residue classes modulo d, where ϕ(n) is the Euler’s totient function. The quantity (1.2) is often thought of as the ‘main term’. Diﬀerent main terms are also considered. Thus, the study of (1.1) is reduced to studying the ‘error term’ (1.3) ∆(f; X, d, a) := X n≤X n≡a(mod d) f(n) − 1 ϕ(d) X n≤X (n,d)=1 f(n), for (a, d) = 1. measuring the discrepancy between the the sum (1.1) and the expected value (1.2). If f satisﬁes f(n) ≤Cτ B(n) logB X for some constants B, C > 0, which is often the case for most f in applications, then a trivial bound for the discrepancy ∆(f; X, d, a) is ∆(f; X, d, a) ≤C′ 1 ϕ(d)X log−B′ X, for some constants B′, C′ > 0. The objective is then to obtain an upper bound for ∆(f; X, d, a) as small as possible for d as large as possible–the smaller the discrepancy and the larger the range of d, the better the distribution estimates for f is. We consider the k-fold divisor function f(n) = τk(n), k ≥1. It is well known that the function τk is closely related to prime numbers; see Remark 1 below. Let us next survey known results on the distribution of τk(n). 1.2. Individual estimates for each modulus d. For (a, d) = 1 deﬁne Tk(X, d, a) = X n≤X n≡a(mod d) τk(n). For k = 1 we have T1(X, d, a) = X n≤X n≡a(mod d) 1 = X d + O(1), EQUIDISTRIBUTION ESTIMATES FOR THE k-FOLD DIVISOR FUNCTION TO LARGE MODULI 3 and this is valid for all d < X. We wish to ﬁnd real numbers θk > 0, as large as possible, such that the following statement holds. (S1) For each ϵ > 0 there exists δ > 0 such that (1.4) Tk(X, d, a) − X ϕ(d)Pk(log X) ≪X1−δ ϕ(d) uniformly for all d ≤Xθk−ϵ. Here Pk(log X) is a polynomial in log X of degree k −1 given by Cauchy’s theorem as Pk(log X) = Res(s−1Lk(s, χ0)Xs−1; s = 1), where χ0 is the principal character of modulus d; for instance, see and . The number θk is called the level of distribution for τk. It is widely believed that (S1) is valid for all θk < 1 for each k ≥1; though, the only known case is for k = 1. For ease of referencing, we record this as Conjecture 1. For each k ≥2 statement (S1) holds for any θk < 1. The Generalized Riemann Hypothesis implies that statement (S1) holds for all θk < 1/2 for any k. In Table 1 we summarize known unconditional results towards Conjecture 1. We now give a brief survey of the known results. Table 1. Known results towards Conjecture 1 for individual estimates and refer-ences. Only for k = 1, 2, 3 is the exponent of distribution θk for τk known to hold for a value larger than 1/2. k θk References k = 2 θ2 = 2/3 Selberg, Linnik, Hooley (independently, unpublished, 1950’s); Heath-Brown (1979) [27, Corollary 1, p. 409]. k = 3 θ3 = 1/2 + 1/230 Friedlander and Iwaniec (1985) [24, Theorem 5, p. 338]. θ3 = 1/2 + 1/82 Heath-Brown (1986) [28, Theorem 1, p. 31]. θ3 = 1/2 + 1/46 Fouvry, Kowalski, and Michel (2015) [21, Theorem 1.1, p. 122], (for prime moduli, polylog saving). k = 4 θ4 = 1/2 Linnik (1961) [40, Lemma 5, p. 197]. k ≥4 θk = 8/(3k + 4) Lavrik (1965) [39, Teopema 1, p. 1232]. k = 5 θ5 = 9/20 Friedlander and Iwaniec (1985) [23, Theorem I, p. 273]. k = 6 θ6 = 5/12 Friedlander and Iwaniec (1985) [23, Theorem II, p. 273]. k ≥7 θk = 8/3k Friedlander and Iwaniec (1985) [23, Theorem II, p. 273]. k ≥5 θk ≥1/2 Open. The classical result for k = 2 giving θ2 = 2/3 depends crucially on the Weil bound for Klooster-man’s sum S(a, b; d): (1.5) S(a, b; d) := d X n=1 (n,d)=1 ed (an + bn) ≪τ(d)(a, b, d)1/2d1/2. This important result for θ2 is an unpublished result of Selberg, Hooley, and Linnik obtained independently in the 1950’s, though none of them seem to have formally written it down. They discovered that Weil bound (1.5) for Kloosterman sums implies that for every ϵ > 0, there exists δ > 0 such that T2(X, d, a) = X ϕ(d)Pd(log X) + O X1−δ ϕ(d) , 4 DAVID T. NGUYEN uniformly for all d < X2/3−ϵ, where Pd is the linear polynomial given by Pd(log X) = Res   ζ2(s) Y p\|d (1 −p−s Xs−1 s ; s = 1    (1.6) = ϕ(d)2 d2 (log X + 2γ −1) −2ϕ(d) d X δ\|d µ(δ) log δ δ . (Note that the main term here is diﬀered from that in (1.3) by an admissible quantity X n≤X (n,d)=1 τ(n) −XPd(log X) ≪X1/2dϵ; see, e.g., [1, Lemma 2.5, p. 7].) As noted in , the work of Hooley in 1957 essentially gives this result for θ2. A formal proof of this result can be found in [27, Corollary 1, p. 409], the work of Heath-Brown in 1979. The divisor problem for arithmetic progressions (c.f. ) then amounts to extending θ2 beyond 2/3. This diﬃcult question has seen no improvement since the 1950’s. In one aspect, the problem is asking for a better uniform estimates for Kloosterman sums beyond those that are immediately available from the Weil bound (1.5); see, e.g., the discussion in . The next, and only known, case where θk > 1/2 is for k = 3. Friedlander and Iwaniec’s spectacular breakthrough work in 1985 yields, in particular, θ3 = 58/115 > 1/2 for k = 3. More precisely, they proved in [24, Theorem 5, p. 338] that, for any ϵ > 0, X92/185 < d < X58/115, (a, d) = 1, we have T3(X, d, a) = X ϕ(d)P(log X) + O(XA+ϵd−B), where A = 271/300, B = 97/120, and P is the quadratic polynomial for which P(log X) is the residue at s = 1 of (Q p\|d(1 −p−sζ(s))3(Xs−1/s). The proof uses multiple exponential sums which rests on Deligne’s deep work on the Riemann Hypothesis over ﬁnite ﬁelds together with Burgess’ bounds on character sums. Heath-Brown’s improvement [28, Theorem 1, p. 31] in 1986 of this exponent of distribution θ3 to 1/2 + 1/82 gives a diﬀerent proof and removes the condition (a, d) = 1. Consequently, he replaced the expected main term in (1.4) by Mk(X, d, a) = X ϕ(d/δ)Res        ( ∞ X m=1 (m,d)=δ τk(m)m−s)Xs−1 s ; s = 1        where δ = (a, d). Heath-Brown showed in [28, Theorem 1, p. 31] that for any ϵ > 0, if q ≤X21/41, then T3(X, d, a) = M3(X, d, a) + O(X86/107+ϵd−66/107). His proof uses deep estimates for multiple Kloosterman sums also powered by Deligne’s Riemann Hypothesis. Most recently, in 2015, Fouvry, Kowalski, and Michel’s result for k = 3 to prime moduli uses the spectral theory of modular forms and estimates for exponential sums using Frobenius trace functions. They showed in [21, Theorem 1.1, p. 122] that for every non-zero integer a, every ϵ, A > 0, every X ≥2, and every prime q, coprime with a, satisfying q ≤X1/2+1/46+ϵ, EQUIDISTRIBUTION ESTIMATES FOR THE k-FOLD DIVISOR FUNCTION TO LARGE MODULI 5 we have ∆(τ3; X, q, a) ≪ X q logA X , where the implied constant depends on ϵ and A and not on a. The estimates for k = 1, 2, and 3 are the only known cases where (S1) holds with θk > 1/2. The exponent of distribution θ4 = 1/2 for k = 4 is explicit in Linnik’s work of 1961. Lavrik’s result in 1965 for k ≥4 uses Burgess’ estimates for character sums and the fourth power moment estimate for L(s, χ) averaged over characters χ modulo q. Friedlander and Iwaniec’s improvement in 1985 of Lavrik’s result for k ≥5 uses Burgess’ estimates for character sums and Heath-Brown and Iwaniec’s work on the diﬀerence between consecutive primes. Remark 1. The k-fold divisor problem for arithmetic progressions, which asks whether the range of d for which statement (S1) holds can be extended beyond X1/2 for k ≥4, has important implica-tions to the distribution of prime numbers. For instance, the estimate (1.4) has application to the asymptotic for the divisor problem X n≤X τk(n)τℓ(n + h). See, for instance, the recent works [42,43], or [56, p. 4] for discussion towards application to power moments of the Riemann zeta function. More concretely, the distribution of the ternary divisor function τ3(n) in arithmetic progression has shown to play an important rˆ ole in the sensational work of Y. Zhang towards the problem of bounded gaps between primes, bringing the gap from inﬁnity to a ﬁnite number. The distribution of τ4 to large moduli will undoubtedly have important consequence to prime numbers. See also [17, Th´ eoreme 4] for precise connection between distribution of τk and distribution of prime numbers. 1.3. Average estimates over d. To obtain further progress on Conjecture 1, a larger range for d can be obtained by averaging over d. This type of result is of Bombieri-Vinogradov type. More precisely, let (S2) be the following statement. (S2) For each ϵ > 0 there exists δ > 0 such that (1.7) X d<Xθk max (a,d)=1 Tk(X, d, a) − X ϕ(d)Pk(log X) ≪X1−δ, provided that d ≤Xθk−ϵ. We have an analogous conjecture for θk. Conjecture 2. For each k ≥2 statement (S2) holds for any θk < 1. Of course, Conjecture 1 implies Conjecture 2. In Table 2 we list known results towards Conjecture 2. Averaging over d, Fouvry in 1985 was able to break through the X2/3-barrier for θ2 and showed in [18, Corollaire 5, p. 74] that for any ϵ > 0, there exists a constant c = c(ϵ) > 0 such that X (q,a)=1 d≤X1−ϵ d/ ∈[X2/3−ϵ,X2/3+ϵ] \|∆(τ2; X, d, a)\| ≪X exp(−c log1/2 x) uniformly for all \|a\| ≤exp(c log1/2 x). His proof uses estimates for Kloosterman sums together with Poisson’s formula and the dispersion method. 6 DAVID T. NGUYEN Table 2. Known results towards Conjecture 2 for average over all moduli d, and references. k θk References k = 2 θ2 = 1 Fouvry (1985) [18, Corollaire 5, p. 74] (exponential saving); Fouvry and Iwaniec (1992) [20, Theorem 1, p. 272]. k = 3 θ3 = 1/2 + 1/42 Heath-Brown (1986) [28, Theorem 2, p. 32]. θ3 = 1/2 + 1/34 Fouvry, Kowalski, and Michel (2015) [21, Theorem 1.2, p. 123], (for prime moduli, polylog saving). k ≥4 θk < 1/2 Follows from the general version of Bombieri-Vinogradov theorem, see, e.g., or , (polylog saving). k ≥4 θk ≥1/2 Open. The gap X2/3−ϵ < d < X2/3+ϵ is closed by Fouvry and Iwaniec for the case of squarefree moduli d which have a square-free factor r of a certain size. They showed in [20, Theorem 1, p. 272] that if r is square-free with (a, r) = 1, r ≤X3/8, then we have X rs2<X1−6ϵ (s,ar)=1 \|∆(τ2; X, d, a)\| ≪r−1X1−ϵ, where the implied constant depends on ϵ alone. The proof depends on an estimate for sums of Kloosterman sums [20, Theorem 2, p. 272]. In the proof of Theorem 2, some estimates of sums of Laurent polynomials in ﬁve variables over a ﬁnite ﬁeld were required; an estimate for these sums is proved in the appendix of by Katz. A larger range for θ3 obtained by Heath-Brown is a result of a sharper average bound for #{n1, n2, n3, n4 : 1 ≤ni ≤N, (ni, d) = 1, n1 + n2 ≡n3 + n4 (mod d)}. He proved in [28, Theorem 2, p. 32] that for any ϵ > 0, we have X d≤D max x≤X max a(mod d) \|T3(x, d, a) −M3(x, d, a)\| ≪X40/51+ϵD7/17, and this is non-trivial for D ≤X11/21−3ϵ. With a ﬁxed, Fouvry, Kowalski, and Michel’s second result in shows that for every non-zero integer a, for every ϵ > 0, and every A > 0, we have, X q≤X9/17−ϵ q prime,q̸\|a \|∆(τ3; X, d, a)\| ≪ X logA X , where the implied constant depends on ϵ, A, and a. This is the best result on the numerical value of the exponent of distribution θk for τk presently. Their proof combines estimates for divisor twists of trace functions together with “Kloostermaniac” techniques through the seminal work of Deshouillers and Iwaniec in . Notably, in this type, the statement (S2) holds for any θk < 1/2 for all k ≥5. This follows from the general version of the Bombieri-Vinogradov theorem; see, e.g., or . 1.4. Average estimates over a. Instead of averaging over the modulus d, a diﬀerent average over the reduced residue classes a(mod d) has also been considered by several authors. For instance, in 2005, Banks, Heath-Brown, and Shparlinski showed in [1, Theorem 3.1, p. 9] that for every ϵ > 0, EQUIDISTRIBUTION ESTIMATES FOR THE k-FOLD DIVISOR FUNCTION TO LARGE MODULI 7 we have the mean value estimate (1.8) d X a=1 (a,d)=1 T2(X, d, a) − X ϕ(d)Pd(log X) ≪          d1/5X4/5+ϵ, if d > X1/2, d2/5X7/10+ϵ, if X1/3 < d ≤X1/2, d1/2X2/3+ϵ, if X1/6 < d ≤X1/3, X3/4+ϵ, if d ≤X1/6, where Pd(log X) is given as in (1.6), and the implied constant depends only on ϵ. In particular, their result shows that the left side of the above is ≪X1−ϵ/6 uniformly for all d < X1−ϵ. Their method relies on average bounds for incomplete Kloosterman sums–they showed and crucially used that the Weil-type bound for certain incomplete Kloosterman sums can be sharpened when averaged over all reduced residue classes modulo d. The upper bounds in (1.8) are substantially sharpened by Blomer in 2008 where the condition (a, d) = 1 is removed and the summand is squared instead. More precisely, letting (1.9) ˜ Pd(log X) = X δ\|d rδ(a) δ (P(log X) + 2γ −1 −2 log δ), Blomer showed in [4, Theorem 1.1, p. 277] that for X a large real number and d ≤X a positive integer, then for any ϵ > 0, we have (1.10) d X a=1 T2(X, d, a) −X d ˜ Pd(log X) 2 ≪X1+ϵ, where rc(h) is the Ramanujan’s sum. When (a, d) = 1, the above main term matches that of (1.8). In particular, by Cauchy’s inequality, (1.10) gives, for all d > X1/3 d X a=1 T2(X, d, a) −X d ˜ Pd(log X) ≪(dX)1/2+ϵ. The proof of (1.10) uses Voronoi summation (the functional equation of twists of the corresponding L-function). Blomer’s proof also works for other arithmetic functions such as Fourier coeﬃcients of primitive (eigenform of all Hecke operators) holomorphic cusp forms [4, Theorem 1.2] and the square-free numbers µ2 [4, Theorem 1.3]. Even though τ(n) > 0 for all n while Hecke eigenvalues can be negative, it is well known that their mean values share analogous oscillatory properties; see, e.g., the detailed exposition of Jutila. Lau and Zhao in 2012 obtained asymptotic results for the variance of τ(n) for d and X going to inﬁnity at diﬀerent rates. Fouvry, Ganguly, Kowalski, and Michel in 2014 show, in a restricted range, that the divisor function τ(n) in residue classes modulo a prime follows a Gaussian distribution, and a similar result for Hecke eigenvalues of classical holomorphic cusp forms. There is also a conjecture for general k. In a function ﬁeld variant, the work of Keating, Rodgers, Roditty-Gershon, and Rudnick in leads to the following conjecture over the integers for the variance of τk (c.f. [47, Conjecture 1]). Conjecture 3 (Keating–Rodgers–Roditty-Gershon–Rudnick ). For X, d →∞such that log X/ log d → c ∈(0, k), we have d X a=1 (a,d)=1 ∆(τk; X, d, a)2 ∼ak(d)γk(c)X(log d)k2−1, 8 DAVID T. NGUYEN where ak(d) is the arithmetic constant ak(d) = lim s→1+(s −1)k2 ∞ X n=1 (n,d)=1 τk(n)2 ns , and γk(c) is a piecewise polynomial of degree k2 −1 deﬁned by γk(c) = 1 k!G(k + 1)2 Z [0,1]k δc(w1 + · · · wk)∆(w)2dkw, where δc(x) = δ(x −c) is a Dirac delta function centered at c, ∆(w) = Q i<j(wi −wj) is a Vandermonde determinant, and G is the Barnes G-function, so that in particular G(k + 1) = (k −1)!(k −2)! · · · 1!. This conjecture is closely related to the problem of moments of Dirichlet L-functions; see, e.g, the works [10–13] of Conrey and Keating on moments of the Riemann zeta function and correlations of divisor sums. 1.5. Average estimates over both a and d. The estimate of the form (1.7) in (S2) is of Bombieri-Vinogradov type where an average over moduli d is taken. In addition to this average, taking an additional average over all primitive residue classes a in each modulus d yields better result for the range for d. These results are of Barban-Davenport-Halberstam type (see, e.g., [14, §29] and have a long history, starting from the initial works [2,3] of Barban in 1963 on primes in arithmetic progressions, and of Davenport and Halberstam a few years latter. Hooley has written a series of papers on the Barban-Davenport-Halberstam sums, dating from 1975, which numbers nineteen, as of currently. In 1976 Motohashi obtained an asymptotic formula for τ(n) averaged over both a and d. More speciﬁcally, there are explicit numerical constants Sj, 0 ≤j ≤3, such that X d<X d X a=1 T2(X, d, a) −˜ Pd(log X) 2 = X2(S3 log3 X + S2 log2 X + S1 log X + S0) + O(X15/8 log2 X), where ˜ Pd(log X) is as in (1.9) Very recently, Rodgers and Soundararajan in 2018 consider smoothed sums of τk averaging over both a and d, and conﬁrm an averaged version of Conjecture 3 for a restricted range. More precisely, for k ≥2, let ∆k(D; X) = X d Vk(d; X)Φ d D , where Vk(d; X) = d X a=1 (a,d)=1  X n≡a(d) τk(n)Ψ n N − 1 ϕ(d) X (n,d)=1 τk(n)Ψ n N   2 , with Φ and Ψ ﬁxed smooth non-negative functions compactly supported in the positive reals nor-malized so that R Φ = 1 and R Ψ2 = 1. Then letting c = log x/ log D, Rodgers and Soundarara-jan in [47, Theorem 1] obtained asymptotics for ∆k(D; X) as X, D →∞uniformly in c for all δ ≤c ≤(k + 2)/k −δ for some δ > 0 suﬃciently small. Under GRH, a larger range δ ≤c ≤2 −δ independent of k is obtained in [47, Theorem 2]. Their results are closely related to moments of Dirichlet L-functions as discussed above, and their proof relies on the asymptotic large sieve. EQUIDISTRIBUTION ESTIMATES FOR THE k-FOLD DIVISOR FUNCTION TO LARGE MODULI 9 The latter work of de la Bret eche and Fiorilli in considers a related variance in the range 1 ≤c < 1+O(1/k) using an arithmetic approximation motivated by work of Vaughan. Interestingly, the asymptotic for their arithmetic variance in [7, Theorem 1.2] matches those in [47, Theorem 1]. In our ﬁrst result, we prove a distribution estimate for τk averaging over both a and d to moduli d as large as X, similar to that of Barban-Davenport-Halberstam type theorems for the von Mangoldt function. Theorem 1. For k ≥4 we have X d≤D d X a=1 (a,d)=1 ∆(τk; X, d, a)2 ≪(D + X1−1/6(k+2))X(log X)k2−1. Motivated by the recent work of Heath-Brown and Li in 2017, we also prove analogous estimates for pairs of τk(n)’s and τk(n)Λ(n) to moduli d that can taken to be almost as large as X2. Theorem 2. For k ≥4 and any ϵ > 0 there holds (1.11) X d≤D d X a=1 (a,d)=1     X m,n≤X m≡an(mod d) τk(m)τk(n) − 1 ϕ(d)     X n≤X (n,d)=1 τk(n)     2    2 ≪X4−1/3(k+4) for any D ≤X2−1/3(k+2). In particular, the above estimate is valid if one of the τk is replaced by the von Mangoldt function Λ. We have (1.12) X d≤D d X a=1 (a,d)=1     X m,n≤X m≡an(mod d) τk(m)Λ(n) − X ϕ(d) X n≤X (n,d)=1 τk(n)     2 ≪X4−1/3(k+4) for any D ≤X2−1/3(k+2). It might look surprising at ﬁrst that the moduli in Theorems 2 can be taken almost as large as X2, but proof is in fact rather simple; the proof of Theorem 2 follows essentially from the multiplicative large sieve inequality. Remark 2. Assuming the Generalized Riemann Hypothesis, it might be possible to show that the estimates (1.11) and (1.12) hold in a larger range for d with right side replaced by ( X2−δ, for 1 ≤D ≤X1+ϵ, X2D(log X)k2, for X1+ϵ < D ≤X2, for some constant δ > 0. 1.6. Average estimates over smooth moduli. Recently progress has been made breaking the X2/3-barrier towards the divisor problem for arithmetic progression for special moduli. It has been discovered that if, in addition to averaging over moduli d, we also restrict d to those that have good factorization properties, we can obtain results for θk ≥1/2 beyond the 1/2-barrier. A natural number m is called Xδ-smooth if all proper prime divisors of m are less than or equal to Xδ for some δ > 0. Let (S3) be the following statement. 10 DAVID T. NGUYEN (S3) For each ϵ > 0 there exists δ > 0 such that X d 0, provided that d ≤Xθk−ϵ. Conjecture 4. For each k ≥2 statement (S3) holds for any θk < 1. Table 3. Known results towards Conjecture 4 for average over smooth moduli d, and references. k θk References k = 2 θ2 = 2/3 + 1/246 Irving (2015) [32, Theorem 1.2, p. 6677] (square-free moduli). θ2 = 2/3 + ϵ Khan (2016) [36, Theorem 1.1, p. 899] (prime powers moludi). k = 3 θ3 = 1/2 + 1/34 Xi (2018) [55, Theorem 1.1, p. 702], (square-free moduli, polylog saving). k ≥4 θk = 1/2 + 1/584 Wei, Xue, and Y. Zhang (2016) [53, Theorem 1.1, p. 1664] (square-free moduli, exponential saving). In 2015, Irving succeeded in broken through the Weil bound for smooth moduli and obtained sharp individual estimate for each d for d almost as large as X55/82, thus showing θ2 > 2/3. More precisely, Irving proved in [32, Theorem 1.2, p. 6677] that for any ϖ, η > 0 satisfying 246ϖ + 18η < 1, there exists a δ > 0, depending on ϖ and η, such that for any Xη-smooth, squarefree d ≤X2/3+ϖ and any (a, d) = 1, we have (1.13) ∆(τ2; X, d, a) ≪d−1X1−δ. The proof is based on a q-analog of van der Corput’s method and bounds on complete Kloosterman sums of Fouvry, Kowalski, and Michel [22, Corollary 3.3]. Khan in 2016 considered an important case of prime power moduli d = pm and proved in [36, Theorem 1.1, p. 899] that for a ﬁxed integer m ≥7, there exists some constants δ > 0 and ρ > 0, depending only on m, such that (1.13) holds uniformly for X2/3−ρ < d < X2/3+ρ. Khan’s method is diﬀerent from that of Irving. The proof uses cancellation of a sum of Kloosterman sums to prime power moduli via the method of Weyl diﬀerencing. Khan’s result is made uniform in m in . For the ternary divisor function, in an extension of the work of Fouvry, Kowalski, and Michel on θ3 for prime moduli, Xi in 2018 obtained individual estimates for smooth, square-free d to the same level of distribution θ3 = 9/17. More precisely, for each ϵ > 0 and A > 0, there exists a constant η > 0 such that if X9/17−ϵ ≪d ≪X9/17−ϵ, d is Xη-smooth and square-free, then ∆(τ3; X, d, a) ≪X d (log X)−A holds uniformly for (a, d) = 1. In , F. Wei, B. Xue, and Y. Zhang showed that the methods of Zhang in in the problem of bounded gaps between primes applies not only to the von Mangoldt function Λ, but also equally to the k-fold divisor function τk. They proved in [53, Theorem 1.1, p. 1664] that for any k ≥4 and a ̸= 0, we have (1.14) X d∈D d<X1/2+1/584 µ(d)2 \|∆(τk; X, d, a)\| ≪X exp(−log1/2 X), EQUIDISTRIBUTION ESTIMATES FOR THE k-FOLD DIVISOR FUNCTION TO LARGE MODULI 11 where (1.15) D = {d ≥1 : (d, a) = 1, (d, Y p X71/584}, and the implied constant depends on k and a. The condition on the moduli d in (1.15) slightly relaxes the constraint on d being smooth, allowing for d to have some, but not too many, prime factors larger than X1/1168. The error term and, more importantly, the exponent of distribution θk = 293/584 = 1/2 + 1/548 in (1.14) hold uniformly in k. The proof uses the Cauchy-Schwartz inequality, combinatorial arguments, the dispersion method, the Weil bound on Kloosterman sums, together with an estimate of Birch and Bombieri for a variant of a three-variable Kloosterman sum proved in the Appendix to powered by Deligne’s Riemann Hypothesis, and, crucially, the factorization d = qr to Weyl shift a certain incomplete Kloosterman sum to the modulus r, thus gaining over Deligne’s Riemann Hypothesis by a power of r, hence saving a power of d, since d is a multiple of r. Our main result improves upon the error term obtained by F. Wei, B. Xue, and Y. Zhang in 2016 for the distribution of family of the k-fold divisor function τk for k ≥4, and is the following: Theorem 3 (Main theorem). Let ϖ = 1 1168 and θk = min 1 12(k + 2), ϖ2 . For a ̸= 0, let D = {d ≥1 : (d, a) = 1, \|µ(d)\| = 1, (d, Y p≤Xϖ2 p) < Xϖ, and (d, Y p≤Xϖ p) > X71/584}, where µ is the M¨ obius function. Then for each k ≥4 we have (1.16) X d∈D d<X293/584 X n≤X n≡a(mod d) τk(n) − 1 ϕ(d) X n≤X (n,d)=1 τk(n) ≪X1−θk. The implied constant is eﬀective, and depends at most on a and k. Remark 3. Theorem 3 admits several reﬁnements. The particular choice of ϖ = 1/1168 is not optimal and there are certain ways to improve the numerics in Theorem 3, for instance using the extensive work of the Polymath 8 project. Though we do not focus on this aspect here, it is an open problem to replace ϖ and θk on the right side of (1.16) by values that are as large as possible. Conditionally, if we assume the Generalized Riemann Hypothesis, or the weaker Generalized Lindel¨ of Hypothesis, for Dirichlet L-functions we can obtain a stronger result. Theorem 4. On the Generalized Lindel¨ of Hypothesis, the estimate (1.16) holds with the right side replaced by X1−ϖ2, where the θk power saving is replaced by a positive constant independent of k. This uniform power saving is the result of sharper estimates of L(s, χ)k on the critical line that are independent of k. 12 DAVID T. NGUYEN 2. Sketch of proof The actual details are quite lengthy. We summarize here the main steps. To prove (1.16) we follow standard practice and split the summation over moduli d in into two sums: one over d < X 1 2 −δ which are called small moduli and the other over X 1 2 −δ ≤d < X 1 2 +2ϖ which are called large moduli. For small moduli, we estimate (1.16) directly using the large sieve inequality together with a direct substitute for the Siegel-Walﬁsz condition. For the von Mangoldt function Λ(n), the M¨ obius function µ(n) is involved and, hence, the Siegel-Walﬁsz theorem is needed to handle very small moduli. For us, fortunately, τk is simpler than Λ in that µ is absent–this feature of τk allows us to get a sharper bound in place of the Siegel-Walﬁsz theorem. The constant here is eﬀective. For large moduli, we adapt the methods of Zhang in to bound the error term which goes as follows. After applying suitable combinatorial arguments, we split τk into appropriate convolutions as Type I, II, and III, as modeled in . We treat the Type I and II in our Case (b), Type III in our Case (c), and Case (a) corresponds to a trivial case which we treat directly. The main ingredients in Case (b) are the dispersion method and Weil bound on Kloosterman sums. The Case (c) depends crucially on the factorization d = qr of the moduli to Weil shift a certain incomplete Kloosterman sum to the modulus r. The shift modulo this r then induces a Ramanujan sum. This allows for a saving of a power of r, and since d is a multiple of r, this saves a small power of X from the trivial bound. References W. D. Banks, R. Heath-Brown, and I. E. Shparlinski, On the average value of divisor sums in arithmetic progres-sions, Int. Math. Res. Not. 2005, no. 1, 1-25. MR2130051 M. B. Barban, ¨ Uber die Verteilung der Primzahlen in arithmetischen Progressionen ‘im Mittel’, (On the distribu-tion of primes in arithmetic progressions ‘on average’, in Russian), Dokl. Akad. Nauk USSR, no. 5 (1964), 5-7. Google Scholar M. B. Barban, The “large sieve” method and its application to number theory, Uspehi Mat. Nauk 21 no. 1 (1966), 51-102. MR0199171 V. Blomer, The average value of divisor sums in arithmetic progressions, Q. J. Math., 59 (2008), 275-286. MR2444061 E. Bombieri, On the large sieve, Mathematika 12 (1965), 201-225. MR0197425 E. Bombieri, J. B. Friedlander, and H. Iwaniec, Primes in arithmetic progressions to large moduli, Acta Math. 156 (1986), 203-251. MR0834613 R. de la Breteche and D. Fiorilli, Major arcs and moments of arithmetical sequences, arXiv:1611.08312. D. A. Burgess, On character sums and L-series, II, Proc. London Math. Soc. (3), 13 (1963), 524-536. MR0148626 J. B. Conrey and A. Ghosh, Remarks on the generalized Lindel¨ of Hypothesis, Funct. Approx. Comment. Math., Volume 36 (2006), 71-78. MR2296639 J. B. Conrey, J. P. Keating, Moments of zeta and correlations of divisor-sums: I, Phil. Trans. R. Soc. A 373:20140313 (2015). MR3338122 J. B. Conrey, J. P. Keating, Moments of zeta and correlations of divisor-sums: II, In Advances in the Theory of Numbers Proceedings of the Thirteenth Conference of the Canadian Number Theory Association, Fields Institute Communications (Editors: A. Alaca, S. Alaca & K.S. Williams), 75 85, 2015. MR3409324 3 J. B. Conrey, J. P. Keating, Moments of zeta and correlations of divisor-sums: III, Indagationes Mathematicae, 26(5): 736-747 (2015). MR3425874 J. B. Conrey, J. P. Keating, Moments of zeta and correlations of divisor-sums: IV, Research in Number Theory, 2(24): (2016). MR3574307 H. Davenport, Multiplicative Number Theory, Graduate Texts in Mathematics 74, third edition, 2000. MR1790423 H. Davenport and H. Halberstam, Primes in arithmetic progressions, Michigan Math. J. 13 1966 485-489. MR0200257 EQUIDISTRIBUTION ESTIMATES FOR THE k-FOLD DIVISOR FUNCTION TO LARGE MODULI 13 J.-M. Deshouillers and H. Iwaniec, Kloosterman sums and Fourier coeﬃcients of cusp forms, Invent. Math. 70 (1982/1983), 219-288. MR0684172 ´ E. Fouvry, Autour du th´ eor eme de Bombieri-Vinogradov (French) , Acta Math. 152 (1984), no. 3-4, 219-244. J. Reine Angew. Math. 357 (1985) 51-76. MR0741055 ´ E. Fouvry, Sur le prob` eme des diviseurs de Titchmarsh (French) , J. Reine Angew. Math. 357 (1985) 51-76. MR0783533 ´ E. Fouvry, S. Ganguly, E. Kowalski, and P. Michel, Gaussian distribution for the divisor function and Hecke eigenvalues in arithmetic progressions, Comment. Math. Helv., 89(4): 979-1014, (2014). MR3284303 ´ E. Fouvry and H. Iwaniec, The divisor function over arithmetic progressions, Acta Arithmetica 61, no. 3 (1992): 271-87. With an appendix by Nicholas Katz. MR1161479 ´ E. Fouvry, E. Kowalski, and P. Michel, On the exponent of distribution of the ternary divisor function, Mathe-matika 61 (2015), no. 1, 121-144. MR3333965 ´ E. Fouvry, E. Kowalski, and P. Michel, A study in sums of products, Philos. Trans. Roy. Soc. A 373 (2015), no. 2040, 20140309, 1-26. MR3338119 J. B. Friedlander and H. Iwaniec, The divisor problem for arithmetic progressions, Acta Arith. 45 (1985), 273-277. MR0808026 J. B. Friedlander and H. Iwaniec, Incomplete Kloosterman sums and a divisor problem, Ann. of Math. (2) 121 (1985), no. 2, 319-350. MR0786351 J. B. Friedlander and H. Iwaniec, Close encounters among the primes, Preprint arXiv:1312.2926. D. Goldston, J. Pintz, and C. Yıldırım, Primes in tuples. I, Ann. of Math. 170 (2009), no. 2, 819-862. MR2552109 D. R. Heath-Brown, The fourth power moment of the Riemann Zeta-function, Proc. London Math. Soc. (3) 38 (1979), 385-422. MR0532980 D. R. Heath-Brown, The divisor function d3(n) in arithmetic progressions, Acta Arith. 47 (1986), 29-56. MR0866901 D. R. Heath-Brown and H. Iwaniec, On the diﬀerence between consecutive primes, Invent. Math. 55 (1979), 49-69. MR0553995 D. R. Heath-Brown and X. Li, Prime values of a2 + p4, Invent. Math. 208 (2017), no. 2, 441-499. MR3639597 C. Hooley, An asymptotic formula in the theory of numbers, Proc. London Math. Soc. (3) 7 (1957), 396-413. MR0090613 A. Irving, The divisor function in arithmetic progressions to smooth moduli, Int. Math. Res. Not. IMRN, 15 (2015), pp. 6675-6698. MR3384495 H. Iwaniec, E. Kowalski, Analytic Number Theory, American Mathematical Society Colloquium Publications Vol. 53, 2004. M. Jutila, Lectures on a Method in the Theory of Exponential Sums, in: Lectures on mathematics and physics (Bombay: Tata Institute of Fundamental Research) Vol. 80, Springer-Verlag (1987). MR0910497 J. Keating, B. Rodgers, E. Roditty-Gershon, and Z. Rudnick, Sums of divisor functions in Fq[t] and matrix integrals Math. Z. (2018) 288: 167-198. MR3774409 R. Khan, The divisor function in arithmetic progressions modulo prime powers, Mathematika 62 (2016), no. 3, 898-908. MR3521360 N. M. Korobov, Estimates of trigonometric sums and their applications (Russian), Usp. Mat. Nauk, 13 (1958), no. 4, 185-192. MR0106205 Y.-K. Lau and L. Zhao, On a variance of Hecke eigenvalues in arithmetic progressions, J. Number Theory, 132(5): 869-887, (2012). MR2890517 A. F. Lavrik, On the problem of divisors in segments of arithmetical progressions, (in Russian), Dokl. Akad. Nauk SSSR 164:6 (1965), 1232-1234. MR0205953 Yu. V. Linnik, All large numbers are sums of a prime and two squares (A problem of Hardy and Little-wood) II, Mat. Sb. (N. S.) 53 (95) (1961), 3-38; Amer. Math. Soc. Transl., 37 (1964), 197-240. MR0120207 K. Liu, I. E. Shparlinski, T. Zhang, Divisor problem in arithmetic progressions modulo a prime power, Adv. Math. 325 (2018), 459-481. MR3742595 K. Matom¨ aki, M. Radziwi l l, and T. Tao, Correlations of the von Mangoldt and higher divisor functions I. Long shift ranges, Proc. London Math. Soc.. MR3909235 K. Matom¨ aki, M. Radziwi l l, and T. Tao, Correlations of the von Mangoldt and higher divisor functions II. Divisor correlations in short ranges, arXiv:1712.08840 14 DAVID T. NGUYEN Y. Motohashi, An induction principle for the generalization of Bombieri’s prime number theorem, Proc. Japan Acad. 52(6) (1976), 273-275. MR0422179 M. R. Murty, Problems in Analytic Number Theory, Graduate Texts in Mathematics (Book 206), Springer; 2nd edition (2007). MR1803093 D.H.J. Polymath, New equidistribution estimates of Zhang type, Algebra Number Theory 8 (2014), no. 9, 2067-2199. MR0422179 B. Rodgers and K. Soundararajan, The variance of divisor sums in arithmetic progressions, Forum Math. 30 (2018), no. 2, 269-293. MR3769996 P. Shiu, A Brun-Titchmarsh theorem for multiplicative functions, J. Reine Angew. Math. 313 (1980), 161-170. MR0552470 P. Tur´ an, On the remainder-term of the prime-number formula I,II, Acta Math. Acad. Sci. Hung., 11, (1950) 48-63; 155-166. [Collected Papers, 1, 515-530, 541-551. Akademiai Kiado, Budapest 1990.] MR0043121 A. I. Vinogradov, The density Hypothesis for Dirichlet L-series, Izv. Akad. Nauk SSSR Ser. Mat. 29 (1965), 903-934. MR0194397 I. M. Vinogradov, A new estimate of the function ζ(1+it), lzv. Akad. Nauk SSSR, 22 (1958), 161-164. MR0103861 A. Weil, Sur les courbes alg´ ebriques et les vari´ et´ es qui s´ en d´ eduisent (in French), Actualit´ es Sci. Ind. 1041, Hermann, 1948. MR0027151 F. Wei, B. Xue, and Y. Zhang, General divisor functions in arithmetic progressions to large moduli, Sci. China Math. 59 (2016), no. 9, 1663-1668. MR3536029 D. Wolke, ¨ Uber die mittlere Verteilung der Werte zahlentheoretischer Funktionen auf Restklassen, I. (in German), Math. Ann. 202 (1973), 125. MR0327688 P. Xi, Ternary divisor functions in arithmetic progressions to smooth moduli, Mathematika 64 (2018), no. 3, 701-729. MR3826483 M. Young, The fourth moment of Dirichlet L-functions, Ann. of Math. (2) 173 (2011), no. 1, 1-50. MR2753598 Y. Zhang, Bounded gaps between primes, Ann. of Math. 179 (2014), no. 3, 1121-1174. MR3171761 Department of Mathematics, South Hall, University of California, Santa Barbara, CA 93106. E-mail address: David.Nguyen@math.ucsb.edu
17070	https://www.ahdictionary.com/word/search.html?q=wander	svg {visibility: hidden !important; opacity: 0 !important} img {visibility: hidden !important; opacity: 0 !important} video {visibility: hidden !important; opacity: 0 !important} canvas {visibility: hidden !important; opacity: 0 !important} iframe [type="application/x-shockwave-flash"] {visibility: hidden !important; opacity: 0 !important} [style="background-image"] {visibility: hidden !important; opacity: 0 !important} .hide {visibility: hidden !important; opacity: 0 !important} American Heritage Dictionary Entry: wander AMERICAN HERITAGE DICTIONARY APP The new American Heritage Dictionary app is now available for iOS and Android. THE AMERICAN HERITAGE DICTIONARY BLOG The articles in our blog examine new words, revised definitions, interesting images from the fifth edition, discussions of usage, and more. See word lists from the best-selling 100 Words Series! Find out more! INTERESTED IN DICTIONARIES? Check out the Dictionary Society of North America at \| \| \| wan·der (wŏndər) Share: Tweet v. wan·dered, wan·der·ing, wan·ders v.intr. 1. To move about without a definite destination or purpose. 2. To go by an indirect route or at no set pace; amble: We wandered toward town. 3. To proceed in an irregular course; meander: The path wanders through the park. 4. To behave in a manner that does not conform to morality or norms: wander from the path of righteousness. 5. To turn the attention from one subject to another with little clarity or coherence of thought: I had a point to make, but my mind started wandering. 6. To be directed without an object or in various directions: His eyes wandered to the balcony. v.tr. 1. To wander across or through: wander the forests and fields. 2. To be directed around or over: Her gaze wandered the docks. n. The act or an instance of wandering. --- [Middle English wanderen, from Old English wandrian.] --- wander·er n. wander·ing·ly adv. --- Synonyms: wander, ramble, roam, rove1, range, meander, stray, gallivant, gad1 These verbs mean to move about at random or without destination or purpose. Wander and ramble stress the absence of a fixed course or goal: The professor wandered down the hall lost in thought. "They would go off together, rambling along the river" (John Galsworthy). Roam and rove emphasize freedom of movement, often over a wide area: "Herds of horses and cattle roamed at will over the plain" (George W. Cable). "For ten long years I roved about, living first in one capital, then another" (Charlotte Brontë). Range suggests wandering in all directions: "a large hunting party known to be ranging the prairie" (Francis Parkman). Meander suggests leisurely wandering over an irregular or winding course: "He meandered to and fro ... observing the manners and customs of Hillport society" (Arnold Bennett). Stray refers to deviation from a proper course or area: "The camels strayed to graze on the branches of distant acacias" (Jeffrey Tayler). Gallivant refers to wandering in search of pleasure: gallivanted all over the city during our visit. Gad suggests restlessness: gadded about unaccompanied in foreign places. \| The American Heritage® Dictionary of the English Language, Fifth Edition copyright ©2022 by HarperCollins Publishers. All rights reserved. Indo-European & Semitic Roots Appendices Thousands of entries in the dictionary include etymologies that trace their origins back to reconstructed proto-languages. You can obtain more information about these forms in our online appendices: Indo-European Roots Semitic Roots The Indo-European appendix covers nearly half of the Indo-European roots that have left their mark on English words. A more complete treatment of Indo-European roots and the English words derived from them is available in our Dictionary of Indo-European Roots. American Heritage Dictionary Products The American Heritage Dictionary, 5th Edition The American Heritage Dictionary of Idioms The American Heritage Roget's Thesaurus Curious George's Dictionary The American Heritage Children's Dictionary
17071	https://science.howstuffworks.com/how-old-is-earth.htm	Published Time: 2025-04-08T18:38:20-04:00 How Old Is Earth and How Did Scientists Figure It Out? \| HowStuffWorks ___ Science Tech Home & Garden Auto Culture More Health Money Animals Lifestyle Entertainment Quizzes Coupons 1 Submit Search Search Close Search Science Tech Home & Garden Auto Culture Health Money Animals Lifestyle Entertainment Quizzes Coupons Submit Search Sign up for our Newsletter! HowStuffWorks Newsletter Advertisement HowStuffWorks Science Space Astronomy The Solar System How Old Is Earth and How Did Scientists Figure It Out? By: Patrick J. Kiger \| Updated: Apr 8, 2025 How old is planet Earth? People have been trying to figure that out since, well, since the beginning of time. Roberto Machado Noa / Getty Images If you look up the age of Earth on science websites and in publications, you'll generally find an estimate of 4.54 billion years, plus or minus 50 million years. What you may be surprised to discover is the accepted estimate dates back to the 1950s and has remained pretty much the same since then, even though scientific knowledge has progressed so dramatically since then in other areas. So what's up with that? Advertisement Contents Historical Theories of the Earth's Age Radiometric Dating Clair Patterson's Discovery Historical Theories of the Earth's Age Glacial drift helps us understand the age of the Earth. Artur Debat / Getty Images Efforts to figure out the Earth's age go back many centuries. The classical Greek philosopher Aristotle, who thought time had no beginning or end, also believed that Earth was infinitely old. Meanwhile, religious scholars in ancient India, who envisioned a universe that perpetually exploded, expanded, and collapsed only to begin anew, calculated that Earth had existed for 1.97 billion years. During the medieval era, various Christian theologians scrutinized the Bible for clues, and came up with estimates of between 5,471 and 7,519 years, according to G. Brent Dalrymple's book "The Age of the Earth." From the 1700s and 1800s, an assortment of scientists came up with various figures based on clues ranging from Earth's rate of cooling and the accumulation of sediment, to the chemical evolution of the oceans. Advertisement Radiometric Dating Carbon dating, used to determine the age of dinosaur fossils, is a type of radiometric dating. Oliver Helbig / Getty Images Radiometric dating is a powerful scientific method used to determine the age of the Earth and various geological materials with remarkable precision. The fundamental principle behind radiometric dating lies in the decay of radioactive elements found in rocks and minerals. Certain naturally occurring radioactive elements, known as parent isotopes, undergo spontaneous decay over time into stable isotopes, referred to as daughter isotopes. By measuring the ratio of parent to daughter isotopes in a sample, scientists can calculate the time elapsed since the material's initial formation. The breakthrough in radiometric dating came in the early 20th century when researchers, such as Bertram Boltwood, pioneered the use of this technique to estimate the age of rocks and minerals. Subsequently, radiometric dating techniques were further refined, and their application extended beyond Earth to other celestial bodies, including the moon. In fact, the age of the moon was determined by analyzing moon rocks brought back by the Apollo missions in the late 1960s and early 1970s, providing crucial evidence for the validity and reliability of radiometric ages. Today, radiometric dating remains an essential tool in unraveling the mysteries of Earth's history and the broader universe. Advertisement Clair Patterson's Discovery In the early 1950s, a California Institute of Technology geochemist named Clair C. Patterson, who had worked on the Manhattan Project to develop the atomic bomb during World War II, measured the isotopic composition of lead from the Canyon Diablo meteorite and several other pieces of space rock, which were believed to date back to the disc of material from which Earth also formed. In 1953, Patterson came up with an estimate of 4.5 billion years. Since then, research on meteorites and lunar rocks has refined that number only slightly. Although Patterson's breakthrough made him famous in the world of science, he didn't see it as a big deal. At the time, "No one cared about it," he recalled in an oral history interview conducted shortly before his death in 1995. "Even today, people don't care how old the earth is. In fact, less today than 40 years ago, when I measured it." Advertisement Now That's Interesting Patterson had an even bigger impact by publishing a 1965 article on the extent of toxic lead pollution on Earth, which helped to start a movement to get rid of dangers such as lead paint and leaded gasoline, according to this 2015 article from the CalTech website. Earth Age FAQ How old is the Earth according to the Bible? According to the Bible, the Earth is not the same age as what science says. It's often interpreted to be significantly younger than what scientific evidence suggests, with some religious interpretations proposing a shorter timeline for Earth's history. In this case, genealogical records combined with Genesis 1 of creation are used to guess the Earth's and Universe's age of about 6000 years. When did life start on Earth and how do they know? Life on Earth is believed to have originated around 3.5 billion years ago, as evidenced by the discovery of microfossils and biomarkers in some of the oldest rocks on the planet. These rocks, found in places like Western Australia, provide valuable clues about the existence of ancient microbial life and the gradual development of complex organisms over time, offering insights into the early stages of our planet's formation and evolution. How was the sun born? According to NASA, the sun is a yellow dwarf star. It's a hot ball of glowing gases and its gravity is responsible for holding the solar system together. How many more years will Earth survive? Scientists suggest that in about 7.5 billion years, the sun will use up its hydrogen and switch to helium, which will turn it into an even bigger ball of hot gas. It will burn up both Mars and Earth, notes NBC MACH. How old is the Earth in human years? If you look up the age of Earth on science websites and in publications, you'll generally find an estimate of 4.54 billion years, plus or minus 50 million years. How old is the sun? The exact age of the Sun is estimated to be about 4.5 billion years, According to NASA,, making it a relatively young sun compared to many other stars in the universe. Its age is determined through various scientific methods, including the study of radioactive isotopes in meteorites and the analysis of its internal processes. How old is the Earth? In the early 1950s, a California Institute of Technology geochemist named Clair C. Patterson measured the isotopic composition of lead from the Canyon Diablo meteorite and several other pieces of space rock, which were believed to date back to the disc of material from which Earth also formed. In 1953, Patterson came up with an estimate of 4.5 billion years. Cite This! Please copy/paste the following text to properly cite this HowStuffWorks.com article: Copy Patrick J. Kiger "How Old Is Earth and How Did Scientists Figure It Out?" 1 January 1970. HowStuffWorks.com. 12 May 2025 Close Citation Related Links Up Next Who Named Planet Earth? Explore More Environmental Science How Do Scientists Determine the Age of Dinosaur Bones? Physical Science What Is an Isotope? People How Creationism Works Environmental Science How Carbon Dating Works Physical Science Scientists Unlock Secrets of H-bomb Element Einsteinium Physical Science Half-Life Formula: Components and Applications You May Like What Did the Big Bang Sound Like? Explore More Environmental Science Tibetan Glacial Ice Core May Hold Clues About Climate Change Environmental Science Earthquakes on the Mississippi: The New Madrid Seismic Zone Space Earth: A Primer on the Third Rock From the Sun Environmental Science How did scientists find soft tissue in dinosaur fossils? Physical Science How Nuclear Radiation Works Extinct Animals The Mastodon Boneyard That Stole Thomas Jefferson's Heart Keep Reading What Are the Solar System Planets in Order? Explore More Advertisement Loading... Advertisement Advertisement Advertisement Advertisement Advertisement We're as obsessed with finding answers as you are. Send me the latest news in science, culture and tech! Success You are now subscribed. Continue Newsletter Sign Up Send me the latest news in science, culture and tech! Learn How Everything Works! Company About Careers Contact Us Help Sitemap Information Terms Reprints Privacy Policy For Advertisers Do Not Sell My Info Explore Quizzes Coupons Video Privacy Policy Terms Do Not Sell My Info Copyright © 2025 HowStuffWorks, a System1 Property Image 33Image 34Image 35
17072	https://mathwithfriends.org/easy-activities-for-solving-equations/	Easy Activities for Solving Equations - Math With Friends Skip to content Math With Friends Home Blog Teachers Pay Teachers Youtube Pinterest About Me Contact Privacy Policy Math With Friends Navigation Menu Navigation Menu Home Blog Teachers Pay Teachers Youtube Pinterest About Me Contact Privacy Policy Easy Activities for Solving Equations by mathwithfriends@yahoo.com There are countless important topics covered in High School Algebra, but one of the most important by far is solving equations. Just because it’s important doesn’t mean it can’t be fun! Today we will look at three easy, low-prep activities for Solving Equations. I’m confident you’ll find one that you can use in your classroom. Easy Activities for Solving Equations Activity #1 – Partner Check Activity If you’ve been on this blog before, you know that I love a good partner activity, and there are none easier than the classic partner check activity. For those of you that haven’t used one before, here’s what they look like. Students will be partnered up and will either be “Partner A” or “Partner B”. They each will have ten different questions to answer. Although they have different questions, they should get the same answer! I always tell the students that if they disagree on their answers, then they should switch and work out the other partners problem to see if they can find out what went wrong. Only when they are still stumped can they raise their hand and ask for help. This kind of activity really allows the students to work on their math discourse skills as well as their solving skills. Having to explain how they got an answer to their partner is invaluable! Plus, since the students will largely be working autonomously it will free you up to help the students that truly need the help. It’s a win-win! You can find this partner assignment here. Easy Activities for Solving Equations Activity #2 – Desmos Solving Equations Adventure! This activity is definitely one of the most fun and creative Desmos Activities I’ve seen. Students will first be tasked with a little combining like terms review, as seen here. However, the twist is that if you get the question right, Link from The Legend of Zelda will cross the screen and continue his adventure! If you get it wrong, he will lose some health and you’ll have to try again. Students will repeat this task one more time by combining some more like terms to continue on the adventure. Finally the main event happens. Students will now need to solve equations to continue on the journey. I truly love this activity and it’s always a hit with the students. It reminds me of just how creative some Desmos Activities can be, like the one featured in this post. Desmos is really just a gift that keeps giving! Easy Activities for Solving Equations Activity #3 – Blooket Tower Defense Blookets are always a go-to activity in my classroom. Students always enjoy them and they’re easy to assign and monitor. If you are unfamiliar with the different modes, I would read this first. The main decision to make when assigning a Blooket is which mode is best for the assignment. In the case of Solving Equations, you don’t want to rush your students. The speed at which students can solve equations can vary wildly, so making sure that your students feel comfortable working slowly is important. Because of this, I will recommend using the Tower Defense mode. First of all, you can find the Blooket here! With questions like the ones below, I think you’d agree that some students will require some time to accurately solve. Now, let’s talk about the Tower Defense mode. With this mode students will answer questions in groups of three. For each question they get right, they will get some coins that they can use to build or upgrade towers to help defend themselves from the attackers. You can assign this as either Homework (HW) or as a live game (Host). Since we want to give our students time to work out the problems, I recommend assigning it as homework. The default setting is the Round Goal of 20, but that means students will need to answer 60 questions! I would drop the goal down to 10. If the students can survive through 30 equations, that’s definitely a good sign! Well, that’s all for this article. I hope you found one or more activities that you can use for your lesson. As always, email me at MathWithFriends@Yahoo.com if you have any questions! Neve \| Powered by WordPress We are using cookies to give you the best experience on our website. You can find out more about which cookies we are using or switch them off in settings. Accept Close GDPR Cookie Settings Privacy Overview Strictly Necessary Cookies Powered by GDPR Cookie Compliance Privacy Overview This website uses cookies so that we can provide you with the best user experience possible. Cookie information is stored in your browser and performs functions such as recognising you when you return to our website and helping our team to understand which sections of the website you find most interesting and useful. Strictly Necessary Cookies Strictly Necessary Cookie should be enabled at all times so that we can save your preferences for cookie settings. Enable or Disable Cookies- [x] Enabled Disabled If you disable this cookie, we will not be able to save your preferences. This means that every time you visit this website you will need to enable or disable cookies again. Enable All Save Settings
17073	https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/hydrostatic-pressure	Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Tissue Engineering and Regenerative Medicine: Applications 6.7.4.1.2.1 Hydrostatic pressure Hydrostatic pressure is a physiological factor that regulates cartilage biosynthesis independently of other factors such as cell/matrix deformation, fluid flow, or electro-kinetic effects. Continuous or intermittent pressures of <3 MPa may (Kimura et al., 1985; Hall et al., 1991; Lafeber et al., 1992) or may not (Lippiello et al., 1985) affect PG synthesis in normal AC. Higher pressures (5–50 MPa) modulate cartilage biosynthesis to various degrees depending on the magnitude, frequency, and duration of pressurization. Hydrostatic pressure in the presumably physiological range (~5–15 MPa) in the short term (up to 2 h) increased aggrecan and COL biosynthesis, but higher pressures (20–50 MPa) had no effect (Hall et al., 1991). Cartilage biosynthesis was inhibited with longer exposure (2 h) to pressures of 30–50 MPa. PG synthesis was upregulated in a subsequent 2 h following a single application of 5–15 MPa pressure for 20 s or for 5 min, but was not altered with a single application of higher pressures (20–50 MPa) (Hall et al., 1991). The frequency of the applied pressure is also important, as a 0.5 Hz cyclic 5 MPa pressure was shown to stimulate biosynthesis, while the application of the pressure with same magnitude at 0.0167–0.25 Hz did not have any effect on synthesis (Parkkinen et al., 1993). View chapterExplore book Read full chapter URL: Reference work2017, Comprehensive Biomaterials IIA.R. Raleigh, ... R.L. Sah Chapter The Microcirculation and Lymphatics 2013, Cardiovascular Physiology (Tenth Edition)Achilles J. Pappano PhD, Withrow Gil Wier PhD Hydrostatic Pressure is the Principal Force in Capillary Filtration Changes in the venous resistance affect capillary hydrostatic pressure more than do changes in arteriolar resistance. A given change in venous pressure produces a greater effect on the capillary hydrostatic pressure than does the same change in arterial pressure, and about 80% of an increase in venous pressure is transmitted back to the capillaries. Despite the fact that capillary hydrostatic pressure (Pc) varies from tissue to tissue (even within the same tissue), average values, obtained from many direct measurements in human skin, are about 32 mm Hg at the arterial end of the capillaries and 15 mm Hg at the venous end of the capillaries at the level of the heart (Figure 8-9). When a person stands, the hydrostatic pressure is higher in the legs and lower in the head. Tissue pressure, or more specifically ISF pressure (Pi) outside the capillaries, opposes capillary filtration. It is Pc − Pi that constitutes the hydrostatic driving force for filtration. In the normal (nonedematous) state of the subcutaneous tissue, Pi is close to zero or slightly negative (−1 to −4 mm Hg). Hence Pc represents essentially the hydrostatic driving force. View chapterExplore book Read full chapter URL: Book2013, Cardiovascular Physiology (Tenth Edition)Achilles J. Pappano PhD, Withrow Gil Wier PhD Review article Protein Dynamics: Experimental and Computational Approaches 2011, Biochimica et Biophysica Acta (BBA) - Proteins and ProteomicsPatrizia Cioni, Edi Gabellieri 1 Introduction Hydrostatic pressure is a modulator of biochemical processes: it inhibits bacterial growth [1,2], affects the viability of viral particles [3,4], alters protein structure, and activates/inactivates enzymatic reactions . Above 3–4 kbar it induces protein denaturation , whereas in the range of 1–2 kbar compression promotes the dissociation of oligomers [6,7]. Recently it has become clear that along with other parameters such as temperature and solvent composition, pressure can be used for providing a more detailed thermodynamic and kinetic descriptions of bioprocesses and biosystems and for regulating their behavior. At present the majority of biophysical techniques have been adapted to high pressure studies to foster basic and applied research with numerous applications in the industrial, medical, and pharmaceutical fields [8–13]. Le Chatelier's principle states that at equilibrium a system tends to minimize the effect of an external perturbation. Consequently, in a chemical reaction, an increase in pressure shifts the equilibrium in favor of the reactants that occupy a smaller volume. A simple thermodynamic analysis shows that the effect of pressure P on the equilibrium constant K of a process is given by where K(p) and K(0) are the equilibrium constants governing the process at pressure p and atmospheric, respectively, T is the temperature, and R is the universal gas constant. Pressure measurements therefore enable ΔV of the process under investigation to be evaluated, which is a fundamental parameter for characterizing the nature of the process and its mechanism. As it only affects its volume, pressure represents a mild perturbation of the system. In regards to proteins, pressure perturbs their structure in a continuous, controlled way by changing only intermolecular distances as in the pressure range of biotechnological interest (0.001–10 kbar) it has generally no effect on covalent bonds. This is in contrast with other perturbations, such as a temperature change or the addition of denaturation agents, both of which have multiple effects on the system. A temperature variation produces simultaneous changes in the total energy and volume, whereas the effects of a denaturant are dependent on its binding properties. High pressure has been widely applied to study protein stability [15,16], folding [14,17] as well as aggregation/dissociation processes [18,19]. Furthermore, since protein compressibility is directly related to structural and conformational fluctuations of proteins , studies on the molecular mechanisms that account for the effect of high pressure on the structure of biological macromolecules help us to understand the dynamical behavior of proteins at atmospheric pressure. In fact, during the past years the use of pressure perturbation to study protein dynamics has increased by employing different techniques such as tryptophan fluorescence and phosphorescence [21,22], Fourier transformed infrared spectroscopy [23,24], NMR spectroscopy and, more recently X-ray crystallography . In this paper, we first report on recent results on protein flexibility provided by NMR and X-ray crystallography along with high pressure. Then, we describe the application of tryptophan phosphorescence emission for monitoring changes in the flexibility of proteins under pressure. View article Read full article URL: Journal2011, Biochimica et Biophysica Acta (BBA) - Proteins and ProteomicsPatrizia Cioni, Edi Gabellieri Review article Proteins Under High Pressure 2006, Biochimica et Biophysica Acta (BBA) - Proteins and ProteomicsDietrich Knorr, ... Roman Buckow Hydrostatic pressure may be generated by the addition of free energy, e.g., heating at constant volume or mechanical volume reduction. It is now technically feasible to reach pressures up to several gigapascals and to keep it constant for a comparably long time in specially designed vessels made from highly alloyed steel. View article Read full article URL: Journal2006, Biochimica et Biophysica Acta (BBA) - Proteins and ProteomicsDietrich Knorr, ... Roman Buckow Review article Articular cartilage functional histomorphology and mechanobiology: a research perspective 2003, BoneM Wong, D.R Carter —Hydrostatic pressure, either applied statically or cyclically, generally involves the pressurization of a fluid-filled chamber. Here the compressive strains are nearly zero due to the near incompressibility of cartilage. The pressure is carried primarily by the fluid phase (Fig. 4 right). View article Read full article URL: Journal2003, BoneM Wong, D.R Carter Chapter Subject-Specific Computational Prediction of the Effects of Elastic Compression in the Calf 2017, Biomechanics of Living OrgansFanny Frauziols, ... Stéphane Avril 3 Results The patient-specific FE model was implemented for the calf of four volunteers. The response of their lower leg to elastic compression was calculated. Here we present the results of these simulations. 3.1 Distribution Hydrostatic Pressure Without muscle contraction. Fig. 12 shows hydrostatic pressure maps for the four subjects. The hydrostatic pressure is a meaningful metric of the effects of elastic compression, as it does not depend on the choice of coordinate system, and it can be applied to analyze pressure gradients that may indicate possible fluid flows. With muscle contraction. Muscle contraction induces a significant change of the geometry of the calf, and also a change of material properties. The geometry of the calf was acquired using US imaging for an active state of muscles for each subject. This active state corresponded to a voluntary contraction equal to about 30% of the maximum voluntary contraction. The material properties of the fascia cruris and of the skin and fat tissues were also measured using USE in the active state of muscles. Only the material properties of deep soft tissues could not be identified for the active state of muscles because of the discomfort induced by the compression test on contracted muscles. Moreover, to our knowledge, the literature does not report material properties of muscles for a contraction equal to 30% of the maximum voluntary contraction. It was therefore chosen to run a parametric study by varying the elastic parameter C10 of deep soft tissues in a range defined as follows, from 18 to 360 kPa. Fig. 13 shows hydrostatic pressure maps obtained by these simulations for the four subjects. It is observed that the increase in the elastic modulus of the deep soft tissues has an impact on the distribution of the hydrostatic pressure by increasing the tension and compression between the skin and the fascia cruris. Effects of elastic compression. For the four volunteers, the distributions of hydrostatic pressure appear heterogeneous (Figs. 12 and 13). The maximum pressure is located where the radius of curvature is lower (near the tibia); the minimum pressures were observed in flat areas. Substantial variability between subjects is observed in the distribution of these pressures. The maximum pressure range is from 4.1 to 5.3 kPa. Local maxima and minima are not located at the same place depending on the subjects. For example, for subjects two and three (Fig. 13B and C), a very low pressure is observed at the small saphenous vein located on the posterior side of the leg whereas the pressure on the small saphenous vein for subject four appears much larger (Fig. 13D). 3.2 Closure of Veins The percentage of closing for the greater and the small saphenous veins is calculated. The small saphenous vein is located on the posterior side of the leg and the greater saphenous vein on the anteromedial side of the leg (Fig. 1). To derive the closure ratio, the coordinates of nodes of the venous lumen are taken at the initial state and at the end of the FE simulation. Then, the luminal areas of the veins in the initial and final state are calculated to derive the closing percentage. For each vein of each subject, the closing ratio values are given in Table 2. In this table, results are given for the inactive muscle state and for the active muscle state. For each vein of each subject in the active muscle state, the closing ratio is also shown in Fig. 15. Table 2. Closure Ratios of Veins Under Elastic Compression for an Inactive Muscle Condition and for an Active Muscle Condition \| Subject \| Muscular State \| Case Studied \| Closure of Saphenous Veins Under Elastic Compression (%) \| \| --- --- \| Small \| Greater \| \| 1 \| Inactive \| \| 7.2 \| 17.6 \| \| Active \| i \| 4.5 \| 14.4 \| \| i \| 3.6 \| 11.4 \| \| iii \| 3.1 \| 10.3 \| \| iv \| 2.9 \| 9.9 \| \| v \| 2.8 \| 9.8 \| \| 2 \| Inactive \| \| 1.9 \| 18.8 \| \| Active \| i \| 2.8 \| 11.5 \| \| ii \| 2.1 \| 8.3 \| \| iii \| 1.9 \| 7.5 \| \| iv \| 1.8 \| 7.2 \| \| v \| 1.7 \| 6.9 \| \| 3 \| Inactive \| \| 7.3 \| 17.7 \| \| Active \| i \| 3.3 \| 11.3 \| \| ii \| 2.1 \| 8.8 \| \| iii \| 1.7 \| 8.0 \| \| iv \| 1.6 \| 7.7 \| \| v \| 1.4 \| 7.6 \| \| 4 \| Inactive \| \| 1.6 \| 10.8 \| \| Active \| i \| 2.9 \| 9.4 \| \| ii \| 1.7 \| 5.2 \| \| iii \| 1.3 \| 3.9 \| \| iv \| 1.1 \| 3.5 \| \| v \| 1.1 \| 3.3 \| For the latter, the studied cases correspond to C10 values set as follows: (i) 18 kPa; (ii) 72 kPa; (iii) 180 kPa; (iv) 288 kPa; (v) 360 kPa. Influence of muscle contraction. In Table 2, we note that the overall vein closure ratios are larger in the inactive state than in the active one. Indeed, in the inactive muscle condition, material properties of superficial soft tissue measured using USE were found smaller, as shown in Fig. 14, where we report all the results obtained with the USE technique that were previously published (Frauziols et al., 2015). More compliant material properties facilitate the closure of the veins by the elastic compression. For the active muscle condition, we see in Fig. 15 for all subjects and for both veins that the larger the elastic modulus in deep soft tissues, the smaller the closure ratio. Influence the vein location. In Fig. 15, the closure ratio of the small saphenous vein is smaller than the one of the greater saphenous vein because the small saphenous vein is located on the posterior side of the leg where the curvature is low. In this area, which is mainly composed of soft tissues, the CS applies a smaller pressure. The larger closing ratio of the greater saphenous vein is also because of its position, specifically to its proximity with the tibia bone. Indeed, the greater saphenous vein is located on the anteromedial part of the leg along the tibia. Therefore, even though the area is slightly curved, low pressure applied by the CS results in a large closure ratio of the vein. View chapterExplore book Read full chapter URL: Book2017, Biomechanics of Living OrgansFanny Frauziols, ... Stéphane Avril Chapter HYDROSTATIC EFFECTS ON CELLULAR FUNCTION 1971, Underwater PhysiologyJ.V. Landau Publisher Summary This chapter focuses on hydrostatic pressure, which has been an extremely useful tool in the analysis of various biological phenomena. Hydrostatic pressure experimentation falls under two broad areas: (1) one concerning morphological and other visibly measurable effects, and (2) the other concerning biochemical analysis of the fundamental processes involved. Depending upon temperature, the application of a specific pressure may result in either stimulation or inhibition of 14C-amino acid incorporation into protein. The effects of relatively low increments of hydrostatic pressure generated by increased inert gas pressure merit major consideration by those concerned with the fundamental dynamics of living cells. The application of hydrostatic pressure reversibly solates the gel component of living cytoplasm by opposing the volume increase accompanying gelation. This solation occurs in a precise, predictable pattern, and the degree of solation is directly proportional to the magnitude of the pressure applied at a given temperature over a given time period. View chapterExplore book Read full chapter URL: Book1971, Underwater PhysiologyJ.V. Landau Chapter CIRCULATION \| Circulatory System Design: Roles and Principles 2011, Encyclopedia of Fish PhysiologyK.R. Olson Fluids at Rest: Hemostatics Unlike many molecules that have specialized proteins to help them move across membranes or epithelia, there are no known carriers for water and the only way water moves from one location to another is down a pressure gradient. There are three forms of pressure – hydrostatic, hydraulic, and osmotic. Hydrostatic pressure Hydrostatic pressure is determined by the weight of fluid directly above a point of reference; the latter depends on the distance between the top of the water and the reference point and the density of the fluid (Figure 1(a)). The pressure at the reference point (P) is then P = ρgh, where ρ is the density of the fluid (in g L−1), g is the gravitational acceleration constant (9.8 m s−2), and h is the height (in m) from the reference point to the top of the water. There is an interesting dichotomy here, however, because as fish descend deeper into water, the hydrostatic pressure on the outside of the fish is offset by the hydrostatic pressure in the fish’s cardiovascular system (see also DESIGN AND PHYSIOLOGY OF ARTERIES AND VEINS \| Physiology of Capacitance Vessels). Thus, actual blood pressure in a fish (and in physiological experiments) is always measured relative to the surface of the water. Furthermore, the only time hydrostatic pressure is important in fish is when they are out of water (except at depths of greater than 1000 m; (see also DEEP-SEA FISHES). Hydraulic pressure Hydraulic pressure is pressure supplied by a pump, in this case the heart. Hydraulic pressure is exerted in all directions in a blood vessel (Figure 1(b)). Hydraulic pressure will be examined in greater detail in subsequent sections. Osmotic and oncotic pressure Osmotic pressure is the pressure caused by water at different concentrations due to the dilution of water by dissolved molecules (solute), notably salts and nutrients. Osmotic pressure is very important in bony fish where the difference in osmotic pressure between the body fluids (typically 300 milliosmoles per liter; mosm L−1) and the environment (< 1 mosm L−1 in freshwater and 1000 mosm L−1 in saltwater) can create substantial forces that, if otherwise uncontrolled, can cause volume overload in the former and dehydration in the latter. For instance, a difference of 10 mosm across the membrane can create a pressure difference of 190 mmHg, which is approximately 6–8 times the arterial blood pressure of a fish. A unique form of osmotic pressure called oncotic or colloid osmotic pressure is presumed to exist across the capillaries. This pressure is due solely to the difference in protein concentration between plasma and interstitial fluid because proteins are presumed to be the only solute molecules that do not freely cross capillaries. While oncotic pressure is important in maintaining fluid balance across mammalian capillaries, its importance in trans-capillary fluid balance in fish is controversial. Blood pressure definitions Blood pressure can be measured anywhere in the circulation and usually the location is specified, that is, arterial pressure, venous pressure, etc. Arterial pressure is not constant but varies with the cardiac cycle (see also DESIGN AND PHYSIOLOGY OF THE HEART \| Physiology of Cardiac Pumping). Systolic pressure is the maximum pressure and occurs with ejection of blood from the heart (systole). Diastolic pressure is the minimum pressure and occurs when the heart is at rest (diastole). Pulse pressure is the difference between systolic and diastolic pressure, and mean blood pressure is the average pressure. In many fish the oscillations in arterial pressure take on a sawtooth pattern and the mean pressure is the arithmetic average of systolic and diastolic pressure (i.e., mean pressure = (systolic + diastolic) 2−1). In other fish the diastolic period is longer than systolic and mean pressure is the time-integrated average pressure weighted toward the diastolic. The formula derived for mammals, mean = (systolic + 2 diastolic) 3−1, is often used in these instances. View chapterExplore book Read full chapter URL: Reference work2011, Encyclopedia of Fish PhysiologyK.R. Olson Review article Special Issue: Environmental Biotechnology 2017, Trends in BiotechnologyFrancesca Mapelli, ... Daniele Daffonchio Hydrostatic pressure (HP) is an intrinsic feature of the deep sea. HP linearly increases with depth (about 0.1 MPa every 10 m of seawater column) and affects cell structures and functions, with a genome-based response proposed for piezophilic adaptation. Le Chtelier’s principle predicts that processes entailing volume reduction are favoured under HP. HP can affect chemical reactions (both for volume change at equilibrium and for activation) and macromolecule structures, particularly the weak noncovalent bonds and protein complexes with several subunits. This is the case for ribosome assembly or for nucleic acid/protein complexes with volume increases [76–78], where effects are generally observed at sublethal pressures. Another recognized HP effect is on cell membranes, whose relative abundance in mono- (and to a lower extent poly-) unsaturated fatty acids, necessary to maintain membrane fluidity and cell homeostasis, can be affected. A high degree of fatty acid unsaturation does not favour fatty acid acyl group packing due to the physical encumbrance of the lateral chains, with positive effects on the cell membrane’s homeoviscous properties and curvature elastic stress . Increased unsaturation of fatty acids on the membrane may result from another typical deep-sea condition, that is, low temperature, whose effects are additive to HP . HP affects membrane transport systems and transmembrane enzymes, such as ATPases and cytochromes, due to direct effects on enzyme folding or the lipid environment on the membrane. Adaptation to HP potentially involves energy generation in the cell, likely to counteract HP-related stressing effects. The model piezophile Photobacterium profundum SS9 possesses two complete operons for the F0F1 ATPase and multiple cytochrome sets, supporting the pivotal importance of electron and proton transport under high HP . Similarly, Shewanella piezophila possesses two respiratory chains, active either under low or high HP . The capacity to offset HP impact on cell turgor pressure has been little studied. This would involve the intracellular accumulation of piezolytes, as occurring for osmolytes and salinity. While their mechanism of action remains unclear, HP increase is consistent with the accumulation of N-trimethylamine oxide , β-hydroxybutyrate and ectoine . In Alcanivorax borkumensis, intracellular ectoine accumulation and gene upregulation under HP were correlated with decreased cell damage and higher cell number, but culture activity was not increased. Experiments on the synergistic effects of osmotic and HP increase suggest that ectoine water-reclamation capacity might also explain its function at high HP . View article Read full article URL: Journal2017, Trends in BiotechnologyFrancesca Mapelli, ... Daniele Daffonchio Chapter REGULATION OF WATER FLUX ACROSS ISOLATED GASTRIC MUCOSA 1981, Gastrointestinal Defence MechanismsLeopoldo Villegas Hydrostatic pressure The use of externally applied hydrostatic pressure difference has no effect on the net ion fluxes, the transmucosal resistances or the fine structure of the intercellular space (Villegas 1978). These facts make possible the calculation of a hydrostatic induced water flux from the slope of the regression line between the net water flux and the hydrostatic pressure difference. A water flux of 42.9 ± 3.2 μl/cm2h atm of hydrostatic pressure was obtained in histamine stimulated mucosae incubated in isosmotic solution (Villegas 1978). View chapterExplore book Read full chapter URL: Book1981, Gastrointestinal Defence MechanismsLeopoldo Villegas Related terms: Enzymatic Hydrolysis Actin Enzyme Pressure Gradient Cell Membrane Glomerulus Filtration Tension Oncotic Pressure Solution and Solubility Microorganism View all Topics We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy
17074	https://en.wikipedia.org/wiki/Harmonic_measure	Harmonic measure - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk Contents move to sidebar hide (Top) 1 Definition 2 Properties 3 Examples 4 The harmonic measure of a diffusion 5 General references 6 References 7 External links [x] Toggle the table of contents Harmonic measure [x] 2 languages Español 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikimedia Commons Wikidata item Appearance move to sidebar hide From Wikipedia, the free encyclopedia In mathematics, especially potential theory, harmonic measure is a concept related to the theory of harmonic functions that arises from the solution of the classical Dirichlet problem. Harmonic measure is the exit distribution of Brownian motion In probability theory, the harmonic measure of a subset of the boundary of a bounded domain in Euclidean spaceR n{\displaystyle R^{n}}, n≥2{\displaystyle n\geq 2} is the probability that a Brownian motion started inside a domain hits that subset of the boundary. More generally, harmonic measure of an Itō diffusionX describes the distribution of X as it hits the boundary of D. In the complex plane, harmonic measure can be used to estimate the modulus of an analytic function inside a domain D given bounds on the modulus on the boundary of the domain; a special case of this principle is Hadamard's three-circle theorem. On simply connected planar domains, there is a close connection between harmonic measure and the theory of conformal maps. The term harmonic measure was introduced by Rolf Nevanlinna in 1928 for planar domains, although Nevanlinna notes the idea appeared implicitly in earlier work by Johansson, F. Riesz, M. Riesz, Carleman, Ostrowski and Julia (original order cited). The connection between harmonic measure and Brownian motion was first identified by Kakutani ten years later in 1944. Definition [edit] Let D be a bounded, open domain in n-dimensionalEuclidean spaceRn, n≥2, and let ∂D denote the boundary of D. Any continuous functionf:∂D→R determines a unique harmonic functionH f that solves the Dirichlet problem {−Δ H f(x)=0,x∈D;H f(x)=f(x),x∈∂D.{\displaystyle {\begin{cases}-\Delta H_{f}(x)=0,&x\in D;\H_{f}(x)=f(x),&x\in \partial D.\end{cases}}} If a point x∈D is fixed, by the Riesz–Markov–Kakutani representation theorem and the maximum principleH f(x) determines a probability measureω(x,D) on ∂D by H f(x)=∫∂D f(y)d ω(x,D)(y).{\displaystyle H_{f}(x)=\int {\partial D}f(y)\,\mathrm {d} \omega (x,D)(y).} The measure _ω(x,D) is called the harmonic measure (of the domain D with pole at x). Properties [edit] For any Borel subset E of ∂D, the harmonic measure ω(x,D)(E) is equal to the value at x of the solution to the Dirichlet problem with boundary data equal to the indicator function of E. For fixed D and E⊆∂D, ω(x,D)(E) is a harmonic function of x∈D and 0≤ω(x,D)(E)≤1;{\displaystyle 0\leq \omega (x,D)(E)\leq 1;}1−ω(x,D)(E)=ω(x,D)(∂D∖E);{\displaystyle 1-\omega (x,D)(E)=\omega (x,D)(\partial D\setminus E);}Hence, for each x and D, ω(x,D) is a probability measure on ∂D. If ω(x,D)(E)=0 at even a single point x of D, then y↦ω(y,D)(E){\displaystyle y\mapsto \omega (y,D)(E)} is identically zero, in which case E is said to be a set of harmonic measure zero. This is a consequence of Harnack's inequality. Since explicit formulas for harmonic measure are not typically available, we are interested in determining conditions which guarantee a set has harmonic measure zero. F. and M. Riesz Theorem: If D⊂R 2{\displaystyle D\subset \mathbb {R} ^{2}} is a simply connected planar domain bounded by a rectifiable curve (i.e. if H 1(∂D)<∞{\displaystyle H^{1}(\partial D)<\infty }), then harmonic measure is mutually absolutely continuous with respect to arc length: for all E⊂∂D{\displaystyle E\subset \partial D}, ω(X,D)(E)=0{\displaystyle \omega (X,D)(E)=0} if and only if H 1(E)=0{\displaystyle H^{1}(E)=0}. Makarov's theorem: Let D⊂R 2{\displaystyle D\subset \mathbb {R} ^{2}} be a simply connected planar domain. If E⊂∂D{\displaystyle E\subset \partial D} and H s(E)=0{\displaystyle H^{s}(E)=0} for some s<1{\displaystyle s<1}, then ω(x,D)(E)=0{\displaystyle \omega (x,D)(E)=0}. Moreover, harmonic measure on D is mutually singular with respect to t-dimensional Hausdorff measure for all t>1. Dahlberg's theorem: If D⊂R n{\displaystyle D\subset \mathbb {R} ^{n}} is a bounded Lipschitz domain, then harmonic measure and (n−1)-dimensional Hausdorff measure are mutually absolutely continuous: for all E⊂∂D{\displaystyle E\subset \partial D}, ω(X,D)(E)=0{\displaystyle \omega (X,D)(E)=0} if and only if H n−1(E)=0{\displaystyle H^{n-1}(E)=0}. Examples [edit] If D={X∈R 2:\|X\|<1}{\displaystyle \mathbb {D} ={X\in \mathbb {R} ^{2}:\|X\|<1}} is the unit disk, then harmonic measure of D{\displaystyle \mathbb {D} } with pole at the origin is length measure on the unit circle normalized to be a probability, i.e. ω(0,D)(E)=\|E\|/2 π{\displaystyle \omega (0,\mathbb {D} )(E)=\|E\|/2\pi } for all E⊂S 1{\displaystyle E\subset S^{1}} where \|E\|{\displaystyle \|E\|} denotes the length of E{\displaystyle E}. If D{\displaystyle \mathbb {D} } is the unit disk and X∈D{\displaystyle X\in \mathbb {D} }, then ω(X,D)(E)=∫E 1−\|X\|2\|X−Q\|2 d H 1(Q)2 π{\displaystyle \omega (X,\mathbb {D} )(E)=\int _{E}{\frac {1-\|X\|^{2}}{\|X-Q\|^{2}}}{\frac {dH^{1}(Q)}{2\pi }}} for all E⊂S 1{\displaystyle E\subset S^{1}} where H 1{\displaystyle H^{1}} denotes length measure on the unit circle. The Radon–Nikodym derivatived ω(X,D)/d H 1{\displaystyle d\omega (X,\mathbb {D} )/dH^{1}} is called the Poisson kernel. More generally, if n≥2{\displaystyle n\geq 2} and B n={X∈R n:\|X\|<1}{\displaystyle \mathbb {B} ^{n}={X\in \mathbb {R} ^{n}:\|X\|<1}} is the n-dimensional unit ball, then harmonic measure with pole at X∈B n{\displaystyle X\in \mathbb {B} ^{n}} is ω(X,B n)(E)=∫E 1−\|X\|2\|X−Q\|n d H n−1(Q)σ n−1{\displaystyle \omega (X,\mathbb {B} ^{n})(E)=\int {E}{\frac {1-\|X\|^{2}}{\|X-Q\|^{n}}}{\frac {dH^{n-1}(Q)}{\sigma {n-1}}}} for all E⊂S n−1{\displaystyle E\subset S^{n-1}} where H n−1{\displaystyle H^{n-1}} denotes surface measure ((n−1)-dimensional Hausdorff measure) on the unit sphere S n−1{\displaystyle S^{n-1}} and H n−1(S n−1)=σ n−1{\displaystyle H^{n-1}(S^{n-1})=\sigma _{n-1}}. Harmonic Measure on Simply Connected Planar Domains If D⊂R 2{\displaystyle D\subset \mathbb {R} ^{2}} is a simply connected planar domain bounded by a Jordan curve and X∈{\displaystyle \in }D, then ω(X,D)(E)=\|f−1(E)\|/2 π{\displaystyle \omega (X,D)(E)=\|f^{-1}(E)\|/2\pi } for all E⊂∂D{\displaystyle E\subset \partial D} where f:D→D{\displaystyle f:\mathbb {D} \rightarrow D} is the unique Riemann map which sends the origin to X, i.e. f(0)=X{\displaystyle f(0)=X}. See Carathéodory's theorem. If D⊂R 2{\displaystyle D\subset \mathbb {R} ^{2}} is the domain bounded by the Koch snowflake, then there exists a subset E⊂∂D{\displaystyle E\subset \partial D} of the Koch snowflake such that E{\displaystyle E} has zero length (H 1(E)=0{\displaystyle H^{1}(E)=0}) and full harmonic measure ω(X,D)(E)=1{\displaystyle \omega (X,D)(E)=1}. The harmonic measure of a diffusion [edit] Consider an Rn-valued Itō diffusion X starting at some point x in the interior of a domain D, with law Px. Suppose that one wishes to know the distribution of the points at which X exits D. For example, canonical Brownian motion B on the real line starting at 0 exits the interval (−1,+1) at −1 with probability ⁠1/2⁠ and at +1 with probability ⁠1/2⁠, so B τ(−1,+1) is uniformly distributed on the set {−1,+1}. In general, if G is compactly embedded within Rn, then the harmonic measure (or hitting distribution) of X on the boundary ∂G of G is the measure μ G x defined by μ G x(F)=P x[X τ G∈F]{\displaystyle \mu {G}^{x}(F)=\mathbf {P} ^{x}{\big [}X{\tau {G}}\in F{\big ]}} for _x∈G and F⊆∂G. Returning to the earlier example of Brownian motion, one can show that if B is a Brownian motion in Rn starting at x∈Rn and D⊂Rn is an open ball centred on x, then the harmonic measure of B on ∂D is invariant under all rotations of D about x and coincides with the normalized surface measure on ∂D General references [edit] Garnett, John B.; Marshall, Donald E. (2005). Harmonic Measure. Cambridge: Cambridge University Press. ISBN978-0-521-47018-6. Øksendal, Bernt K. (2003). Stochastic Differential Equations: An Introduction with Applications (Sixth ed.). Berlin: Springer. ISBN3-540-04758-1.MR2001996 (See Sections 7, 8 and 9) Capogna, Luca; Kenig, Carlos E.; Lanzani, Loredana (2005). Harmonic Measure: Geometric and Analytic Points of View. University Lecture Series. Vol.ULECT/35. American Mathematical Society. p.155. ISBN978-0-8218-2728-4. References [edit] ^R. Nevanlinna (1970), "Analytic Functions", Springer-Verlag, Berlin, Heidelberg, cf. Introduction p. 3 ^R. Nevanlinna (1934), "Das harmonische Mass von Punktmengen und seine Anwendung in der Funktionentheorie", Comptes rendus du huitème congrès des mathématiciens scandinaves, Stockholm, pp. 116–133. ^Kakutani, S. (1944). "On Brownian motion in n-space". Proc. Imp. Acad. Tokyo. 20 (9): 648–652. doi:10.3792/pia/1195572742. ^F. and M. Riesz (1916), "Über die Randwerte einer analytischen Funktion", Quatrième Congrès des Mathématiciens Scandinaves, Stockholm, pp. 27–44. ^Makarov, N. G. (1985). "On the Distortion of Boundary Sets Under Conformal Maps". Proc. London Math. Soc. 3. 52 (2): 369–384. doi:10.1112/plms/s3-51.2.369. ^Dahlberg, Björn E. J. (1977). "Estimates of harmonic measure". Arch. Rat. Mech. Anal. 65 (3): 275–288. Bibcode:1977ArRMA..65..275D. doi:10.1007/BF00280445. S2CID120614580. P. Jones and T. Wolff, Hausdorff dimension of Harmonic Measure in the plane, Acta. Math. 161 (1988) 131-144 (MR962097)(90j:31001) C. Kenig and T. Toro, Free Boundary regularity for Harmonic Measores and Poisson Kernels, Ann. of Math. 150 (1999)369-454MR 172669992001d:31004) C. Kenig, D. Preissand, T. Toro, Boundary Structure and Size in terms of Interior and Exterior Harmonic Measures in Higher Dimensions, Jour. of Amer. Math. Soc. vol 22 July 2009, no3,771-796 S. G. Krantz, The Theory and Practice of Conformal Geometry, Dover Publ. Mineola New York (2016) esp. Ch 6 classical case External links [edit] Solomentsev, E.D. (2001) , "Harmonic measure", Encyclopedia of Mathematics, EMS Press \| v t e Measure theory \| \| Basic concepts \| Absolute continuityof measures Lebesgue integration L p spaces Measure Measure space Probability space Measurable space/function \| \| Sets \| Almost everywhere Atom Baire set Borel set equivalence relation Borel space Carathéodory's criterion Cylindrical σ-algebra Cylinder set 𝜆-system Essential range infimum/supremum Locally measurable π-system σ-algebra Non-measurable set Vitali set Null set Support Transverse measure Universally measurable \| \| Types of measures \| Atomic Baire Banach Besov Borel Brown Complex Complete Content (Logarithmically)Convex Decomposable Discrete Equivalent Finite Inner (Quasi-)Invariant Locally finite Maximising Metric outer Outer Perfect Pre-measure (Sub-)Probability Projection-valued Radon Random Regular Borel regular Inner regular Outer regular Saturated Set function σ-finite s-finite Signed Singular Spectral Strictly positive Tight Vector \| \| Particular measures \| Counting Dirac Euler Gaussian Haar Harmonic Hausdorff Intensity Lebesgue Infinite-dimensional Logarithmic Product Projections Pushforward Spherical measure Tangent Trivial Young \| \| Maps \| Measurable function Bochner Strongly Weakly Convergence: almost everywhere of measures in measure of random variables in distribution in probability Cylinder set measure Random: compact set element measure process variable vector Projection-valued measure \| \| Main results \| Carathéodory's extension theorem Convergence theorems Dominated Monotone Vitali Decomposition theorems Hahn Jordan Maharam's Egorov's Fatou's lemma Fubini's Fubini–Tonelli Hölder's inequality Minkowski inequality Radon–Nikodym Riesz–Markov–Kakutani representation theorem \| \| Other results \| Disintegration theorem Lifting theory Lebesgue's density theorem Lebesgue differentiation theorem Sard's theorem Vitali–Hahn–Saks theorem For Lebesgue measure Isoperimetric inequality Brunn–Minkowski theorem Milman's reverse Minkowski–Steiner formula Prékopa–Leindler inequality Vitale's random Brunn–Minkowski inequality \| \| Applications&related \| Convex analysis Descriptive set theory Probability theory Real analysis Spectral theory \| Retrieved from " Categories: Measures (measure theory) Potential theory Hidden category: CS1: long volume value This page was last edited on 20 June 2024, at 02:42(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Search Search [x] Toggle the table of contents Harmonic measure 2 languagesAdd topic
17075	https://www.numerade.com/ask/question/write-number-in-scientific-notation-00000003-10288/	Write number in scientific notation. 0.0000003 Log inSign up This problem has been solved by verified expert Jennifer Stoner 100% free to try - get full access instantly Start Free Trial Guest user Add your school New Chat Textbooks Files More FlashcardsNotes & ExamsBootcamps Numerade Plus Expert Educators Textbook Videos Unlimited Ace chats AI Study Tools… Sign up for free Download App Question Write number in scientific notation. 0.0000003 Write number in scientific notation. 0.0000003 Added by Steven B. Grade 6 Mathematics: Open Up Resources, Common Core State Standards Edition N/A 6th Edition Instant Answer Solved by Expert Jennifer Stoner 05/11/2022 Step 1 In this case, we move it 7 places to the right, so we get 3.Show more… Show all steps View the full answer Close Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration 0:00 Loaded: 0% Progress: 0% Stream Type LIVE Remaining Time-0:00 Playback Rate 2x 1.5x 1x, selected 0.5x 1x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected English Auto Captions Audio Track Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration 0:00 Loaded: 0% Progress: 0% Stream Type LIVE Remaining Time-0:00 Playback Rate 2x 1.5x 1x, selected 0.5x 1x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected Audio Track Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. Please give Ace some feedback Your feedback will help us improve your experience Submit Thanks for your feedback! Write number in scientific notation. 0.0000003 Hi ! Need help with this question? I know these concepts can sometimes be confusing, but I’ll make it simple. Just tell me what’s on your mind! Play audio Play audio Feedback Powered by NumerAI Jennifer Stoner and 95 other Prealgebra educators are ready to help you. Ask a new question Key Concepts Key Concept Premium Feature Explore the core concept behind this problem. View Video View Video Key Concept Premium Feature Explore the core concept behind this problem. Your browser does not support the video tag. Close Try it Now Recommended Videos Write number in scientific notation. 0.000000567 Exponents and Polynomials Scientific Notation Write the number in scientific notation. −0.0087 Introduction to Functions and Graphs Numbers, Data, and Problem Solving Write the number in scientific notation. the number 0.000459 Exponents and Exponential Functions Scientific Notation Recommended Textbooks Grade 6 Mathematics: Open Up Resources, Common Core State Standards Edition N/A 6th Edition 1,510 solutions Grade 7 Mathematics: Open Up Resources, Common Core State Standards Edition N/A 7th Edition 1,901 solutions Grade 8 Mathematics: Open Up Resources, Common Core State Standards Edition N/A 8th Edition 1,670 solutions Transcript 00:01 We're going to write 0 .000000 -03 in scientific notation. 00:07 So we're going to get a number times 10 to some exponent. 00:12 My number is going to be 3 .0 because we always have a whole number, one whole number in front of the decimal... Need help? Use Ace Ace is your personal tutor. It breaks down any question with clear steps so you can learn. Start Using Ace Ace is your personal tutor for learning Step-by-step explanations Instant summaries Summarize YouTube videos Understand textbook images or PDFs Study tools like quizzes and flashcards Listen to your notes as a podcast GET THE ANSWER Get Answer to This and Millions of Other Questions Question Write number in scientific notation. 0.0000003 Create your account Start a free 7-day trial to get step-by-step solutions. Sign in with Google Signed in with Google Continue with Facebook Continue with Apple Continue with Apple Continue with Clever or Name Email Password Create Account By creating an account, you agree to the Terms of Service and Privacy Policy Already have an account? Log In A free answer just for you Watch the video solution with this free unlock. View the Answer Log in to watch this video ...and 100,000,000 more! EMAIL PASSWORD Log in OR Sign in with Google Signed in with Google Continue with Facebook Continue with Apple Continue with Apple Continue with Apple Continue with Clever Don't have an account? Sign Up About Our Story Careers Our Educators Numerade Blog Newsroom Browse Bootcamps Books Directory Notes & Exams Ask our Educators Topics More Test Prep Ask Directory Notes Directory Online Tutors AP Content Subjects NEW Math Chemistry Physics Biology Economics Support Help Privacy Policy Terms of Service Get the app 680 E Colorado Blvd, Suite 180 - Pasadena, CA 91101
17076	https://www.youtube.com/watch?v=kz2suvi6mEw	Given cost and demand functions, maximize profit Math Tutorials 1880 subscribers 9 likes Description 2933 views Posted: 15 Mar 2022 In this example problem, we are given a cost function and a constant demand function and asked the what production level will maximize profit. We create a profit function by using that profit is revenue - cost. To create the revenue function, we use that revenue is price multiplied by quantity. After we create the profit function, we find its 1st derivative and set it equal to zero to find critical numbers (critical values). We identify that we will have a maximum at this value because our profit function is a quadratic with a negative leading coefficient. This means that the graph will be a downward facing parabola. We then use the critical number to find the maximum profit by evaluating the profit function at that value. This video contains examples that are from Business Calculus, 1st ed, by Calaway, Hoffman, Lippman. from the Open Course Library, remixed from Dale Hoffman's Contemporary Calculus text. It was extended by David Lippman to add several additional topics. The text is licensed under the Creative Commons Attribution license. Transcript: for this problem we're given a cost function and a demand function now the demand function always connects together our quantity and a price so being that this is given to us as p of x that's going to indicate that x represents a quantity however many are being produced or sold okay so we're asked to find the production level that will maximize our profit so let's try to set this up and represent our profit function now a profit function is going to be however much money you bring in we call that the revenue minus any of the costs that you have to expect okay so our profit function we already have the cost but we need to represent the revenue so revenue is going to be whatever you're charging price multiplied by whatever however many you sell now our price we're given that price function so i can say that's the same thing as p of x multiplied by our quantity which we said was going to be x minus our cost function c of x i'm going to go ahead and call this capital p of x for profit not to be confused with the lowercase p of x which was our demand function representing the price from here let's go ahead and fill in the information we know so p of x the little p of x was 1740 multiplied by x minus and here's a big key that we want to fill in the whole cost function but the whole thing has to be subtracted away so i'm using a big set of parentheses and i'm going to go ahead and fill in 5550 plus 580 x plus 1.6 x squared all in a big set of parentheses because all of that cost function needs to be subtracted away from here let's go ahead and do a little bit of distributing still representing our profit function so 1740x minus 5550 minus 580x minus 1.6 x squared from here i'm going to go ahead and combine some like terms and put this in descending order so negative 1.6 x squared plus 1160 x minus five five five zero okay so we've represented our profit function from here we would like to go ahead and find any critical values or critical numbers to do so let's find the first derivative so power rule a couple times we'll do negative 3.2 x plus 1160 and the derivative of that constant is going to be zero to find the critical values or critical numbers we need to set this equal to zero and do a little bit of solving down so we'll go ahead and subtract the 1160 to the other side and divide both sides by negative 3.2 so that'll get us x on one side all by itself and we'll get 3 162.5 and we can represent that as a critical number all right now to determine whether this is going to be a maximum or minimum we could do either the first or second derivative test but instead i'm just going to go back to our profit function and this version we can see that it's a quadratic and the leading term is going to be negative so the pre it's going to be a graph of a parabola and this one opens down because that leading term is going to be negative that tells us that we're going to get a maximum at this x value all right the first second derivative test would do the exact same thing for us the last thing that wasn't asked for us uh asked of us but we could definitely answer is we could figure out what is that maximum profit actually going to be now we know it occurs at an x value of 362.5 to find what that actual profit would be we could simply plug back into the profit function 362.5 gets filled back into the profit function replace each one of the x's with it negative 1.6 times 362.5 squared plus 1160 times 362.5 five minus five hundred fifty and our very maximum profit we could have with this situation is two hundred four thousand seven hundred dollars all right so i hope this helps out as you're given a price function and a cost function and trying to maximize profit good luck
17077	https://www.doubtnut.com/qna/11312787	In X-ray tube , when the accelerating voltage V is halved, the difference between the wavelength of Kα line and minimum wavelength of continuous X-ray spectrum remain constant because more than two times because half because less than two times More from this Exercise The correct Answer is:D Δλ=λKα−λmin Where V is haved, λmin becomes two times but λKα remains the same. ∴Δλ'=λKα−2λmin =2(Δλ)−λKα ∴Δλ'<2(Δλ) Topper's Solved these Questions Explore 13 Videos Explore 62 Videos Explore 17 Videos Explore 65 Videos Explore 40 Videos Similar Questions Find the voltage applied to an X-rays tube with nickel anticathode if the wavelength difference between the kα line and the short-wave cut-off of the continuous X-rays spectrum is equal to 84pm. When the voltage applied to an X-ray tube increased from V1=15.5kV to V2=31kV the wavelength interval between the Kα line and the cut-off wavelength of te continuous X-ray spectrum increases by a factor of 1.3. If te atomic number of the element of the target is z. Then the value of z13 will be: (take hc=1240eVnm and R=1×107m) Knowledge Check What is the minimum wavelength of X–rays In a Coolidge tube, the potential difference used to accelerate the electrons is increased from 24. 8 kV to 49.6 kV . As a result, the difference between the wavelength of Kα -line and minimum wavelength becomes two times. The initial wavelength of the Kα line is [Take hce=12.4kVÅ] When the voltage applied to and X-ray tube is increases from V1=10KVtoV2=20kV, the wavelength difference between the Kα line and short wavelength limit of the continuous X-ray spectrum increases by a factor 3. The atomic number of the element of which the tube anticathode is made will be An X-ray tube has nickel as target. The wavelength difference between the ka X-ray and the cut-off wavelength of continuous X-ray spectrum is 84 pm. Find the accelerating voltage applied in the X-ray tube. [Take hc=12400eVÅ] In x-ray tube experiment accelerating voltage is given by 1.24 MV. Find cutoff wavelength. If the accelerating voltage across an x-ray tube is doubled When the voltage applied to an X-ray tube increased from V1=15.5kV to V2=31kV the wavelength interval between the Kα line and the cut-off wavelength of te continuous X-ray spectrum increases by a factor of 1.3. If te atomic number of the element of the target is z. Then the value of z13 will be: (take hc=1240eVnm and R=1×107m) When an X-ray tube is operated with an accelerating potential difference V, the cutoff wavelength is proportional to CENGAGE PHYSICS-ATOMIC PHYSICS-Single Correct According to Bohr's theory of hydrogen atom , the product of the bindi... An electtron and a photon have same wavelength . If p is the moment of... In X-ray tube , when the accelerating voltage V is halved, the differe... In an excited state of hydrogen like atom an electron has total energy... The circumference of the second Bohr orbit of electron in hydrogen ato... X-ray emitted from a copper target and a molybdenum target are found t... An electron in a Bohr orbit of hydrogen atom with the quantum number N... In the Bohr model of a pi-mesic atom , a pi-mesic of mass m(pi) and of... In the spectrum of singly ionized helium , the wavelength of a line ab... AK(alpha) X-ray emitted from a sample has an energy of 7.46 ke V. Of w... Hydrogen atom absorbs radiations of wavelength lambda0 and consequentl... Energy liberated in the de-excitation of hydrogen atom from the third ... Which of the following products, in a hydrogen atom , are independent ... According to Bohr's theory of hydrogen atom , for the electron in the... When a hydrogen atom is excited from ground state to first excited st... In an X-ray tube, the voltage applied is 20 kV. The energy required to... Suppose the potential energy between an electron and a proton at a dis... Let A(0) be the area enclined by the orbit in a hydrogen atom .The gra... Mark out the correct statement(s). A hydrogen atom having kinetic energy E collides with a stationary hyd... Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation Contact Us
17078	https://www.youtube.com/watch?v=xr_j5eRc8X8	Circle (Coordinate Geometry) - Full Chapter \| Concept and Solutions \| Class 11/12/JEE Maths Magnet Brains 14000000 subscribers 339 likes Description 20036 views Posted: 25 Dec 2022 👉Previous Video : 👉Next Video : ✔️📚👉 Watch Full Free Course: ✔️📚👉 Get Any Class & Subject's Topic Video Here:- ✔️📚👉 Get All Subjects Playlists: ✔️📚👉 Grab Notes by Expert Teachers Here: ✔️📚👉 Place Your Queries For Books Here: ✔️📚👉 Get Books Prepared by Our Expert Teachers at Your Door Step: 📢 Full Playlist Link: ✅ In this video, ✔️ Class: 11th/12th/JEE (Mains & Advanced) ✔️ Subject: Maths ✔️ Chapter: Circle ✔️ Topic Name: Circle (Coordinate Geometry) - Full Chapter \| Concept and Solutions \| Class 11/12/JEE Maths ✔️ Topics Covered In This Video : This video provides students with an overview of the full chapter circle from Deependra Sir's Mathematics Book for Class 11, 12 and JEE. It starts by introducing the different classifications of circles and goes on to explain the various properties and equations related to circles. Deependra Sir provides practical examples and illustrations to help students understand and master the full chapter. This video is ideal for those preparing for Class 11, 12 and JEE and wanting to gain a better understanding of circles. 00:00 Introduction: Circle 01:18 Definition 05:16 Equation of Circle in Centre-Radius Form: 44:45 Equation of Circle Touching One or Both the Coordinate Axes: 01:11:10 General Form of Equation of Circle: 01:41:46 Equation of Circle Passing Through Three Points: 02:06:14 Equation of Circle in Parametric Form: 02:24:32 Question & Solutions: Assignment Questions: Circle: Que. Find the radius of the circle (x-5)(x-1) + (y-7)(y-4) = 0. 02:30:14 Geometry with Circle 02:34:19 Question & Solutions: Important Questions: Circle: 03:11:11 Intersection of Line and Circle: 04:28:45 Secant of Circle: 04:54:02 Equation of Chord of Circle Bisected at Given Point: 05:35:00 Tangent and Normal to Circle: 06:23:51 Equation of Tangent to Circle Having Given Slope: 06:43:53 Equation of Tangents to Circle Drawn from External Point: 06:58:13 Equation of Pair of Tangents: 07:27:52 Chord of Contact 08:54:00 Different Cases of Two Circles: 09:16:57 Radical Axes of Three Circles Taken Two at a Time and Radical Centre: 09:27:20 Two Disjoint Circles 10:09:52 Two Circles Touching Externally: 11:24:06 Two Intersecting Circles 12:02:19 Radical Axis and Orthogonal Intersecting of Circles: 12:34:26 One Circle Contained in the Other Circle Without Touching: 13:16:30 Question & Solutions: Exercise: (Single Correct Answer Type Questions): 14:32:05 Question & Solutions: Exercise: (Multiple Correct Answer Type Questions): 15:38:56 Question & Solutions: Exercise: (Linked Comprehension Type Questions): 17:59:53 Question & Solutions: Exercise: (Matrix Match Type Questions): 18:25:07 Question & Solutions: Exercise: (Numerical Value Type Questions): 19:00:45 Archives (JEE Advanced): 24:03:18 Website Overview Why study from Magnet Brains? Magnet Brains is an online education platform that helps gives you NCERT/CBSE curriculum-based free full courses from Kindergarten to Class 12th so that you can perform well in any and all exams you give in your academic career. 👉 Contact us 🤑🤑 ➡️ Connect with us : magnetbrainsbhopal@gmail.com ➡️ Website : ➡️ Subscribe to us on YouTube: ➡️ Subscribe to Magnet Brains Hindi Medium : ➡️Facebook-: ➡️Telegram-: ➡️Instagram:- cengagemaths #class11maths #jeemains #jeeadvanced #class12maths #coordinategeometry circle coordinate geometry pdf coordinate geometry questions and answers pdf general equation of circle coordinate geometry formula what is coordinate geometry types of circle Disclaimer: "This video is for educational and informational purposes only and is not intended to infringe on any copyrights. If you believe that this video has used any copyrighted material in a way that constitutes copyright infringement, please contact us at contact@magnetbrains.com and we will take appropriate action." 65 comments Transcript: हेलो स्टूडेंट्स वेलकम यू ऑल तू मैग्नेट ब्रेंस सर्कल्स आपका फोर्थ चैप्टर कॉर्डिनेट ज्यामिति का जो आपको आईआईटी जी मांस और एडवांस की तैयारी करवा रहे हैं हम singhgaj बुक के थ्रू आज शुरू हो रहा है आज एक और नई चीज हम पढ़ना शुरू करेंगे और नहीं चीज एक्सटेंड करेंगे वो सारी चीजों के साथ जो अभी तक आप पढ़ते हुए आए हैं ये बुक एक्चुअली पंच पार्ट्स में डिवाइडेड है इसी के साथ-साथ आपको कैलकुलस algebrain 3D एंड ट्रिगोनोमेट्री हर एक सेक्शन को हम कंप्लीट करेंगे और हर एक बुक अच्छे से पढ़ेंगे आपको अगर नोट्स की एक्सेस चाहिए स्टूडेंट्स तो हमारी टीम से आप कॉन्टैक्ट कर सकते हैं इसी वीडियो के डिस्क्रिप्शन में आपको गेट नोट्स एक लिंक दिखाई देगा हम किंडरगार्टन से 12th स्टैंडर्ड तक अलग-अलग बोर्ड अलग-अलग मीडियम और कई सारे कॉम्पिटेटिव एग्जाम्स की तैयारी आपको करवाते हैं अब्सोल्युटली फ्री ऑफ कॉस्ट कोई हिडन कॉस्ट नहीं होती है कोई यू नो चार्जेस नहीं होते हैं कोई ऐसा सब्सक्रिप्शन बेस्ड हमारा मॉडल नहीं है कुछ भी ऐसा नहीं है की ट्रायल वर्जन के बाद पेट वर्जन शुरू हो जाए हमारा पूरा कोर्स पूरा पैकेज पूरा मॉड्यूल आपको सकता है अब्सोल्युटली फ्री ऑफ कॉस्ट विद डट स्टूडेंट्स आज बात करेंगे सर्कल्स की और सर्कल में आज हम कौन सा टॉपिक शुरू करेंगे सर्कल का जो आज हमारा फर्स्ट टॉपिक होगा वो है सर इक्वेशन ऑफ अन सर्कल इन सेंटर रेडियो फॉर्म मतलब अगर मुझे किसी सर्कल की सेंटर की इक्वेशन दे दी जाए मतलब सर्कल की इक्वेशन फिगर आउट करनी है और मुझे दो क्या-क्या एक तो उसके सेंटर के कोऑर्डिनेट्स और उसकी रेडियस तो कैसे उसे इक्वेशन को फिगर आउट किया जाता है यह हम निकलना सीखेंगे और ध्यान से suniyega स्टूडेंट्स काफी सारी इंटरेस्टिंग सी चीज है कैसे समझते हैं अन सर्कल आई होप आप सर्कल की परिभाषा जानते हैं सर्कल की परिभाषा अब सब समझते हैं सर्कल को लेकर बहुत सारी अलग-अलग डेफिनिशन आती हैं जैसे एक डेफिनेशन आती है सर आप पॉलीगों बनाओ अब ट्रायंगल बनाओ क्वॉड्रिलैटरल बनाओ पेंटागन बनाओ आप डेका ये सारे पॉलिगंस बनाते चले जाओ अब जैसे-जैसे नंबर ऑफ साइट्स इंक्रीस करते जाओगे वो सिमेट्री की तरफ बढ़ता जाएगा आई एम टॉकिंग अबाउट अन रेगुलर पॉलीगों यानी की सर अगर एक मैं ऐसा पॉलीगों इमेजिन करूं जिसकी infainight साइड्स honicis के बीच की डिस्टेंस को एकदम जीरो कर दिया या ये जो थोड़ी सी कॉम्प्लिकेटेड से डेफिनेशन हो गई इसके बजाय जो एक बेसिक से डेफिनेशन की अगर हम बात करें तो लोकस के कॉन्टेक्स्ट में जो हम बात करना सीखें हैं की सर एक पॉइंट का लॉक इस तरीके से की उसे पॉइंट की किसी एक फिक्स्ड पॉइंट से डिस्टेंस हमेशा कांस्टेंट रहे आप समझ रहे हो यह जो पॉइंट है मैं इस पॉइंट के बाद की बात कर रहा हूं यह जो पॉइंट है यह पॉइंट हमेशा इस पॉइंट से जैसे आप कहते हो क्या सेंटर उससे हमेशा इस पॉइंट की जो डिस्टेंस है वह अगर फिक्स जो रहे उसे हम क्या का लेते हैं भाई रेडियस तो ये पॉइंट ये जो पॉइंट पथ ट्रेवल करेगा ना वो क्लीयरली क्या होगा भाई एक सर्कल आई थिंक यह परिभाषा बड़ी आसान सी है और इसी पर आज हम बात करना चाह रहे हैं की सर अगर यही लोकस की अंडरस्टैंडिंग से मैं कॉन्सेप्ट फिगर आउट करना चाहूंगा तो मैन लो ये एक पॉइंट है जैसे आप क्या मानते हो सर आप मैन लेते हो ह कॉम के इसके सेंटर के कोऑर्डिनेट्स में मैन लेता हूं जैसे X1 कॉम y1 और इसकी जो वो फिक्स्ड डिस्टेंस है उसे मैं मैन लेता हूं स्मॉल आर है तो डिस्टेंस फॉर्मूला मेरी क्या हेल्प करता है डिस्टेंस फॉर्मूला मुझे कहता है सर की आप सीधी-सीधी सी बात जानते हो ह कॉम के की जो इस सेंटर से डिस्टेंस है वो फिक्स्ड है और वो क्या है रेडियस तो टेक्निकल मैं क्या कहूंगा मैं कहूंगा सर ह एक ही X1 y1 से डिस्टेंस हो ह - X1 होल स्क्वायर का अंडर रूट विल बी इक्वल तू आर और √ हटा के अगर मैं यहां लगा हूं तो ये हो जाएगा r² अब एक छोटा सा कम स्टूडेंट्स हम आखिर में फाइनली क्या करते हैं हमारी हैबिट है की हम ह कमा के को एक्स ए से रिप्लेस करते हैं तो आप लिखते हैं एक्स - X1 का होल स्क्वायर प्लस ए माइंस ए वैन का होल स्क्वायर इस इक्वल तू आर स्क्वायर का जो इक्वेशन होगा जो मैथमेटिकल एक्सप्रेशन होगा या मैं इसकी लैंग्वेज में हम जब उसे डिनोट करना चाहेंगे तो इस बात को ट्रेवल करेगा इससे इक्वेशन को सेटिस्फाई करेगा आई होप यह बात बहुत defectile नहीं हो रही बात आसानी से आप डाइजेस्ट कर का रहे हो और यही बात एक्चुअली यहां पर लिख कर दी जा रही है अन सर्कल इसे दी लोकस ऑफ अन पॉइंट विच मूव्स इन अन प्लेन सच डेट इट इस इट्स डिस्टेंस फ्रॉम अन फिक्स्ड पॉइंट इन डेट से प्लेन इस ऑलवेज कांस्टेंट फ्रॉम अन फिक्स्ड पॉइंट मतलब ये क्या हो जाएगा सर उसका सेंटर इस कांस्टेंट मतलब ये वो जो डिस्टेंस है वो क्या हो जाएगी सर उसकी रेडियस यही बात हम करना चाह रहे हैं देखो का रहा है फिक्स्ड पॉइंट का क्या कहते हो सेंटर और जो कांस्टेंट डिस्टेंस है उसे आप कहते हो रेडियस ऑफ डी सर्कल आई थिंक बहुत डिफिकल्ट कॉन्सेप्ट नहीं बहुत बहुत अजीब सा कॉन्सेप्ट नहीं है और यह बेसिक सी बातें हैं जो मेरे ख्याल से अब बहुत आसानी से डाइजेस्ट कर पाया और यही बात है सर यहां पर लिखी गई हैं की उसे एक्स कमा ए की X1 y1 से जो डिस्टेंस होगी वो क्या हो जाएगी सर वो हो जाएगी ऑफ कोर्स आपकी क्या लुकास या आपके सर्कल की इक्वेशन सर मैं आपसे एक छोटी सी बात पूछना चाह रहा हूं थोड़ी देर के लिए मैन लेते हैं की सर्कल का सेंटर लेट से ओरिजिन पर है सर्कल का सेंटर लेट से ओरिजिन पर है और इस सर्कल की रेडियस है लेट से आर तो यहां पर जो पॉइंट होगा वो क्या होगा ह से हम मैन लेते हैं एक्स को तो अब अगर मैं इसकी इक्वेशन निकलना चाहूं तो क्या आएगी आप खुद सोचिए एक्स - 0 का होल स्क्वायर प्लस ए माइंस जीरो का होल स्क्वायर यानी क्या Y2 = आई थिंक यह बात आपको समझ ए रही है मतलब अगर किसी सर्कल का सेंटर ओरिजिन पर हो तो उसे सर्कल की इक्वेशन bikams दिस ऑफ कोर्स इट इस गिवन की उसकी रेडियस क्या है आर कोई डाउट स्टूडेंट्स कोई परेशानी यहां तक चीज समझ ए रही है क्या और यही बात यहां पर लिखकर आपको कहने की कोशिश की गई है की अगर आपके X1 y1 क्या बन जाए ओरिजिन तो उसे केस में आपके सर्कल की इक्वेशन हो जाने वाली है x² + y² = 0 सर्कल्स हेविंग से सेंटर बट डिफरेंट रेड आय आर कॉल्ड centrif सर्कल्स मतलब अगर कोई दो सर्कल हैं बट कोई इंसीडेंटली उनकी जो सेंटर है वो से है तो इन दोनों सर्कस को जैसे C1 और C2 से मैं दिनो करू तो ये दोनों सर्कल C1 और C2 मेरे लिए क्या होंगे कोस एट्रिक्स सरफेस यह लिंगो याद रखिएगा स्टूडेंट्स अगर कभी भी मैं मेंशन करता हूं की देखो सेंट्रिक सर्कल्स हैं तो आपको रिकॉल होना चाहिए आपको recallect होना चाहिए की मैं एक या दो ऐसे या दो से ज्यादा भी अगर कोर्स है सर्कल हुए तो मैं उन सर्कल्स की बात कर रहा हूं जो सबका सेंटर से है और क्या सर और suniyega हर डी कॉमन सेंटर ऑफ तू सर्कल से अगर X1 कमा ए वैन और उनके रेडियस है R1 एंड r2 है ना अगर आपके पास एक सर्कल है बेसिकली दो सर्कल्स हैं जो को सेंट्रिक सर्कल्स हैं उनकी अगर सेंटर के cardinate सेक्शन V1 बट उनकी इंडिविजुअल जो रेडियस है वो है R1 और r2 तो क्लीयरली में जानता हूं सर दोनों की इंडिविजुअल इक्वेशंस में लिख सकता हूं मेरा आपसे ये कहना है की जब आप इक्वेशंस को एक्सपेंड करोगे जब आप लिखोगे x² + x1² - 2x1 y² + y1² - 2y1 = r² और इसी तरीके से जब आप इसे एक पार्ट करोगे तो आपको ये मिलेगा ध्यान से suniyega डेट इक्वेशंस ऑफ कोर्स एंड ट्रिक सर्कल्स डिफर बाय अन कांस्टेंट ओनली थॉट्स अन वेरी इंपॉर्टेंट रिमार्क विच यू आर मास्टर नोट टेस्ट और आई एम ट्राईंग तू मेक यू अंडरस्टैंड अब बात समझो ये पार्ट और ये पार्ट तो एग्जैक्ट से है ना अगर ये कोर्स सेंटर का सर्कल तो डी ओनली पार्ट विच इस गेटिंग चेंज्ड विच इस मेकिंग अन डिफरेंस हर इस दिस पार्ट विच इस नथिंग बट अन कांस्टेंट वैल्यू और हम ट्राईंग तू मेक यू अंडरस्टैंड दिस इस बात को समझ गए बहुत कृष्ण स्टेटमेंट है जैसे अगर एक सर्कल है जिसके सेंटर के कार्ड से 2 3 और उसकी रेडियस है लेट्स से तू सिमिलरली एक और सर्कल है जिसके सेंटर के अकॉर्डिंग तू कमा थ्री बट उसकी रेडियस लेट्स से वैन ये आपका है सर्कल वैन और ये आपका है सर्कल तू मुझे आप बता सकते हो क्या की सर्कल वैन की इक्वेशन क्या बनेगी सर सोच सकते हैं सर्कल वैन की जो इक्वेशन बनेगी आप बोलोगे एक्स - 2 का होल कोई बात समझ ए रही है तो मैं लिखूंगा एक्स - 2 का स्क्वायर मैंने ये एक्स की लिखा सॉरी इट शुड हैव बिन ए है ना यहां पर मैं क्या लिखूंगा ए - 3² = 1² डेट्स वैन नौ माइंड युवर वैन व्हेन यू आर विशिंग तू सिंपलीफाई आपको रिलाइज क्यों नहीं हो रहा है ये जो भी देगा ये वाला पार्ट ये एग्जैक्ट आईडेंटिकल होगा से होगा की आप ये बात इमेजिन या फिर रिलैक्स कर का रहे हो इसे ध्यान से देखो ये क्या बनेगा फिर ये बनेगा x² है ना प्लस ऑफ कोर्स किया फोर और माइंस क्या फॉरेक्स यह क्या बनेगा सर यह बनेगा ए स्क्वायर सिंपलीफाई करें तो ध्यान से देखिए स्टूडेंट्स आपको क्या मिलेगा आपको जो मिलेगी इक्वेशन होगी x² + y² फिर suniyega -4x - 6x फिर क्या मिलेगा स्टूडेंट्स - 4X - 6x फिर क्या मिलेगा सर नाइन प्लस फोर कितना 13 और 13 में से वैन सूत्र किया 13 में से वैन सब्सट्रैक्ट किया तो जल्दी बताओ स्टूडेंट्स कितना हो जाएगा ऐसा ही कर रहा हूं क्या मैंने ये वाली इक्वेशन फोर लिख देता हूं पहले मैं C1 लिख ले रहा हूं कोई दिक्कत तो नहीं है फिर से रिपीट करता हूं फोर से फोर कैंसिल तो ये बच्चा कितना नाइन तो यहां पर ए जाएगा कितना + 9 = 0 ये आपके सर्कल वैन के क्वेश्चंस हैं सर्कल तू की इक्वेशन भी आप सिंपलीफाई कर लीजिए मैं डायरेक्ट लिख दे रहा हूं x² + y² - 4X - 6 अब ध्यान से सुना सर्कल तू में जो आएगा यहां पर भी फोर और नाइन ऐसे कैसे रहेंगे बट उसे तरफ क्या रहेगा वैन अब समझ का रहे हो तो कितना हो जाएगा सही हो जाएगा 12 आप प्लीज इस बात को ध्यान से देखिए दोनों चीजों को नोटिस करिए और ऑब्जर्व करिए ये दोनों ही इक्वेशंस आईडेंटिकल है एक्सेप्ट क्या बस इन दोनों में क्या चेंज है सर इनकी ये जो कांस्टेंट वैल्यू है क्योंकि क्लियर इनके सेंटर से है ना बस डिफरेंस का एरिया का तो cosrick सर्कल विल ऑलवेज रेमैन से उनकी इक्वेशंस वैसे ही रहेंगे सर एक बात बताओ हमने एक चैप्टर पढ़ा था पैर ऑफ स्ट्रेट लाइंस क्या वहां की कोई लर्निंग यहां पर कम ए रही है थोड़ा ध्यान से देखो x² + y² - 2X + 2y मतलब प्लस माइंस 6 प्लस 9 काफी कुछ दिख रहा है सर वो एक्सी वाली टर्म मिसिंग है उसे बारे में भी बात करेंगे और सर एक और बात नोटिस करने में ए रही है x² और Y2 का कॉएफिशिएंट से है या आदर वैन है इन सारी बातों पर हम बात करेंगे इन सारी बातों पर हम बात करेंगे पर मेरा ये कहना है आपको रिलाइज हो रहा है जब पैर ऑफ स्ट्रेट लाइंस चैप्टर हम पढ़ रहे द और हम एक्स और ए में एक क्वाड्रेटिक इक्वेशन की जब बात कर रहे द तो हमने डिस्कस किया था की ये वो एक इक्वेशन है जो चाहे तो पैर ऑफ स्ट्रेट लाइंस या किसी कर्व जैसे की सर कल पैराबोला एलएफसी हाइपरबोला किसी को भी रिप्रेजेंट करती है और उनके बारे में हम डिटेल डिस्कशन रिस्पेक्टिव चैप्टर में करते चले जाएंगे आई होप आपको ये सारी बातें यादें तो अगर इस बात पर बात की जाए तो ये स्टेटमेंट याद रखिएगा की अगर आपको कभी भी ऐसा मिलता है तो हमेशा आप ऐसे सर्कल्स में ये चीज निकल पाएंगे अन्य लाइन पासिंग थ्रू डी सेंटर ऑफ डी सर्कल इस कॉल्ड डायमीटर ऑफ डी सर्कल और सर्कल के infainight डायमीटर्स होते हैं ये बड़ी बेसिक सी कॉमन सी अंडरस्टैंड सी बात है जो हम हमेशा से पढ़ते आए हैं इन सारी बातों से अब फाइनली हम बढ़ते हैं एक क्वेश्चन की तरफ जहां पर हम बात करेंगे बहुत डिटेल में जो सारी चीज अभी तक हमने सीखी हैं उनके इंप्लीमेंटेशन को लेकर कैसे सोचते हैं स्टूडेंट्स ध्यान से देखिएगा तू और टैक्स तू vertises ऑफ़ एन इक्विलैटरल ट्रायंगल आर - 1 कमा जीरो एंड वैन कमा जीरो है ना एंड इट्स थर्ड वर्टेक्स लिस अबोव डी एक्स एक्सिस को क्रैक करने का हमेशा एक अच्छा तरीका होगा की आप इन्हें विजुलाइज करें तो मैं विजुलाइज करूंगा सर मैं क्वेश्चन सोचने की कोशिश करता हूं की चीज कैसे बन रही होंगी है ना तो टेक्निकल आपके पर दो पॉइंट्स है कौन-कौन से सर एक तो है -1 0 और इससे इतनी ही दूर कुछ है जो की हैव कोर्स क्या वैन कॉमर्स है की जो थर्ड वर्टेक्स होगी सर वो यहां कहीं जाकर कुछ इस तरीके से निकल जाएगी आय होप ये बात आप समझ का रहे हो कोई दिक्कत तो नहीं है सर जो थर्ड वर्टेक्स होगी उसके कोऑर्डिनेट्स निकल लूं क्या मैं पहले निकल लेंगे थोड़ा पेशेंस रखिए फाइंड डी इक्वेशन ऑफ डी सरकम सरकम सर्कल ऑफ डी ट्रायंगल मतलब इस ट्रायंगल का सरकम सर्कल हमें ढूंढना है मैं ढूंढ लूंगा सर मैं ट्रायंगल से सरकमसर्किल ढूंढ लूंगा मुझे सरकमसर्किल निकलना है ना तो मुझे तो दो बातें हमेशा पति ही होनी चाहिए की एक तो उसके क्या सेंटर के cardinate और एक क्या उसकी रेडियस तो इन दो चीजों तक मुझे पहुंचना है जो मैं शायद पहुंच जाऊंगा लेकिन उससे पहुंचने से पहले क्या मैं यहां पर कुछ सोच सकता हूं 2 मिनट में दे रहा हूं और उसके बाद आप यह क्वेश्चन ट्राई करेंगे क्योंकि यहां पर जो भी चीज करनी है ये सब आपको पढ़ गई है क्या आप डिस्टेंस फॉर्मूला लगा के यहां तक पहुंच सकते हो पहुंच सकते हो पहुंच सकते हो और चूंकि ये इक्विलैटरल ट्रायंगल है तो इसका सेट्रॉयड circumsar निकल सकते हो सरकमसर्किल का सरकम सेंटर मिल गया ऑर्थो सेंटर या इन सेंटर कुछ भी हो एक ट्रायंगल में सबसे होता है अगर वह मिल गया तो क्या मैं सरकम रेडियस नहीं निकल सकता निकल सकते हो आप वह निकल गई तो यह क्वेश्चन खत्म हो जाएगा मेरा यकीन करिए बहुत आसानी से हिंट आपको दे दी गई है एक क्वेश्चंस जरा आप ट्राई कर लीजिए और मुझे बता दीजिए इस क्वेश्चन का आंसर और फिर बढ़ेंगे हम आगे हाजी तो आपने निकल लिया क्या इस क्वेश्चन का सॉल्यूशन और आंसर लिख रहे हो आप लोग कमेंट्स में नहीं हुआ तो साथ में देखते हैं और देखते हैं क्या करना होगा बात करिएगा ध्यान से देखो सर डिस्टेंस फॉर्मूला अगर उसे किया जाए तो देखो ये कितना है -1 0 ये है वैन कमा जीरो इन दोनों के बीच के डिस्टेंस 1 - 1 यानी 1 + 1 2 का स्क्वायर का अंडर रूट थॉट्स तू इतना सोचने की जरूरत नहीं है ना ये है वैन और ये डिस्टेंस भी वैन वैन प्लस वैन तू तो shailise एक ही सर ये जो डिस्टेंस है ये कितनी है ये डिस्टेंस है तू यूनिट्स अच्छा सर अपनी सहूलियत के लिए क्या मैं इसे ए बी और सी का सकता हूं बिल्कुल का लीजिए एक और बात हम सब जानते हैं सर की यह जो है यह होता है 160 डिग्री एंगल बिल्कुल जानते हैं सर और अगर इसमें मैं यह जानता हूं की ये जो डिस्टेंस है वैन 60 डिग्री नाभि पता हो तो है ना और अगर ये जो डिस्टेंस है ये है तू अब सुनेगा ध्यान से तो अगेन अगर मैं पाइथागोरस थ्योरम उसे करूं हाइपोटेन्यूज बेस परपेंडिकुलर आपको दिख रहा है ना तो मैं क्या कहूंगा भाई मैं कहूंगा तू का स्क्वायर 4 - 1 3 और 3 का अंडर रूट अंडर रूट थ्री तो यहां पर ये जो डिस्टेंस है यहां से ये कितनी है √3 तो इसके अकॉर्डिंग तो जाएंगे जीरो कमा अंडर रूट थ्री क्या ये क्वेश्चन यहां तक समझ आया आई थिंक 60° एंगल की वैसे भी जरूरत नहीं पड़ी है मेरा आपसे बस ये कहना है की सर सारी बातों में सबसे ज्यादा जरूरी बात की ये ये और ये तीनों साइड्स इक्वल है यानी ये एक इक्विलैटरल ट्रायंगल है उसने खुद कहा तो अगर ये एक इक्विलैटरल ट्रायंगल है तो क्या इसका सेंटर राइटिंग निकल सकता हूं याद है स्टूडेंट सेट्रॉयड के कोऑर्डिनेट्स जीरो प्लस वैन माइंस थ्री ये आते हैं आपके किस के सेट्रॉयड के अकॉर्डिंग और सर अगर बात कर रही जा रही है एक इक्विलैटरल ट्रायंगल में सेट्रॉयड की तो वही ना सिर्फ एंड्रॉयड होगा बल्कि आपका ऑर्थोस मतलब ये हो जाएगा आपका सरकम सेंटर अलसो सेंटर या इन सेंटर भी यही होगा मुझे तो सर सर कम सेंटर की जरूरत है तो वो हमें मिल गया क्या ये हो जाता है सर आपका ये वाला पॉइंट जिसे आप कहते हो जीरो कमा 1 / √3 अब तो एक क्वेश्चन हो गया सर ऐसे कैसे हो गया बिल्कुल हो गया जनाब आप ध्यान से देखिए यह इसलिए हो गया अब आप ऐसे सर्कल की इक्वेशन ढूंढ रहे हो आप एक ऐसे सर्कल की इक्वेशन ढूंढ रहे हो जिसकी vertitis से पास होता है उसे हम कहते हैं सरकमसर्किल उसके मैंने सेंटर के अकॉर्डिंग निकल लिए हैं तो 0 1 / √3 सेंटर हुआ और इससे या तो इसकी या इसकी या इसके डिस्टेंस निकल लेना वो रेडियस हो जाएगी तो क्या अब आप इस सर्कल की क्वेश्चन लिख सकते हैं बिल्कुल लिख सकते हैं सर ध्यान से देखना पहले तो सेंटर है ना तो एक्स - 0 का होल स्क्वायर यानी x² + ए - 1 / √3 का स्क्वायर इस इक्वल तू रेडियस का स्क्वायर अब रेडियस का स्क्वायर मतलब ये रेडियस निकलने के स्टूडेंट्स रेडियस होगी जीरो माइंस जीरो कोई फर्क नहीं पड़ता वैन बाय अंडर रूट 3 में से √3 से अट्रैक्ट क्या है ना तो यहां पर ए जाएगा 1 / √3 - √3 का होल स्क्वायर क्या ये बात आपको समझ आई थिंक सर जैसे ही इसे आप सिंपलीफाई कर लेंगे आप शायद अपना आंसर ढूंढ लेंगे है सर x² यह हो जाता है ए स्क्वायर और ये कितना दिखेगा -2y/√3 सो राइट हैंड साइड पर अगर आप देखें कोई तकलीफ तो नहीं स्टूडेंट जो अभी हमने किया उसे बात से अगर राइट हैंड साइड पर अगर आप देखें तो वैन बाय रूट थ्री का स्क्वायर 1/3√3 का स्क्वायर ऑफ कॉस्ट थ्री और ध्यान से देखना सीधी सी बात सर - 2ab तो ये कितना बचेगा सिर्फ -2 क्या ये फैक्ट आप सभी को समझ ए रहा है सर थोड़ा अगर सिंपलीफिकेशन करने की कोशिश की जाए तो एक कम करते हैं बहुत ध्यान से suniyega आ यहां पे तो 3 - 2 कितना बच रहा है वैन तो क्या मैं इसे सीधे-सीधे वैन लिख डन लिख दो सर है ना अब मैं कम करता हूं एलएस और एलएस दोनों को 3 से मल्टीप्लाई करता हूं वैसे तो करने से कोई खास सा फायदा नहीं होगा आप ऐसे ही रख लो एक छोटा सा कम और आप देख का रहे होंगे की सर ये वैन बाय थ्री से ये वैन बाय थ्री है जा रहा है तो ज्यादा चीज मत करो सर एडजस्ट ये आपका सॉल्यूशन है क्या आप ये जो तू बाय तू बाय बाय एंड रूट थ्री है जीरो और यही होगा आपका सॉल्यूशन लिखना है मतलब कोशिश करो की x² को कॉएफिशिएंट वैन हो इससे बहुत हेल्प मिलेगी आपको ऐसे कैसे हेल्प सर बात करेंगे अभी बात करेंगे की इसे वैन रखना कब प्रेफर्ड है है ना पर मेरे ख्याल से आपको ये सॉल्यूशन समझ आया की ये जो आपका इक्विलैटरल ट्रायंगल है जो की आपका इस तरीके से बन रहा है उसके सरकमसर्किल की इक्वेशन क्या होगी एक आसान सा बेहतर तरीके से सिंपल सा शॉर्ट ट्रिक्स डायरेक्ट क्वेश्चन था आई होप ये सारी चीज आपको समझ ए रही हैं इसी के साथ-साथ अब बढ़ते हैं हमारे आज के नेक्स्ट क्वेश्चन की तरफ क्या लिखा हुआ है समझ गया ध्यान से स्टूडेंट का मिड पॉइंट का लॉकर्स कब सर जब जो वह जो कोड है वह कोड एक राइट एंगल सब्सटेंड करती है सी कमा जीरो पर सिर्फ सारी बातें बहुत अजीब अजीब सी लग रही है मुझे तो सबसे पहले ये देख के कुछ जो-जो पढ़ के समझ ए जाए उसे बात को उसे कर लो फिर कहीं-कहीं बहुत जाओगे तो मैं यहां से कहानी शुरू करता हूं x²+y² = a² कुछ हेल्प मिल रही है क्या सर याद करिए हमने पढ़ा था ये एक सेंटर इसका जो सेंटर है ओरिजिन पर लाइक करने वाले एक सर्कल की इक्वेशन है जिसकी रेडियस है ए बहुत बढ़िया सर ये तो अच्छी बात है तो अगर हम इसी से शुरुआत करें तो पहले तो मैं इसे प्लॉट करता हूं suniyega ध्यान से ये आपका ए एक्सिस ये आपका एक्स एक्सिस है ना एक सर्कल एक सर्कल विच सेंटर डेट ओरिजिन सब लोग ये बात जानते हैं की सर्कल के सेंटर के cardinate है जीरो कमा जीरो अब suniyega कम की बात है ना अब मैं मैन लेता हूं सर इस सर्कल पर इस सर्कल पर लेट से ये वो कार्ड है इस सर्कल पहले से ये वो कॉर्ड है है ना अगर आपको लगता है की सर तो सर्कल के अंदर ही होती है बाहर नहीं जाती है तो ये वो कार्ड है वह यह का रहा है की यह जो सी कमा जीरो है सी कमा जीरो कहीं भी हो सकता है पर मुझे इतना तो पता है वो कहां होगा एक्स एक्सिस पर तुम्हें मैन लेता हूं यहां कहीं है आपका सी कॉमर्स जीरो तो जो सी कमा 0 पर ये एंगल सब्सटेंड कर रही है ना जो सी कमा जीरो पर ये एंगल सब्सटेंड कर रही है ना वो एक 90° है अब कोई बातें समझ ए रही है स्टूडेंट्स क्या ये बातें बहुत अच्छे से आपको समझ ए रही है अब अगर मैं सीधे कम की बात पर हूं वो आपसे क्या पूछ रहा है वो ये जो कार्ड है ना स्टूडेंट्स इस कॉल्ड का मिड पॉइंट आपसे पूछ रहा है अच्छा सर ठीक है हम निकलेंगे और सारी बातों से सबसे ज्यादा जरूरी बात सर अगर मैं थोड़ी चीज ढूंढना चाहूं तो कुछ-कुछ चीज तो आपको पता चल रही है एक बात सोच के बताओ स्टूडेंट जब आपसे कोई कहे की एक ऐसी कार्ड है मतलब इस सर्कल को भूल जाओ अभी अभी मैं का रहा हूं एक ऐसी कॉर्ड है एक ऐसी कार्ड है जो की कहीं पर किसी पॉइंट पर 90° एंगल से प्रिंट कर रही है तो कैन आई से की सर थोड़ा सा कुछ तो याद ए रहा है हमें क्या याद ए रहा है सर मुझे एक बहुत जरूरी बात ये याद ए रही है सर की इस लाइन का अच्छा सारी बातें छोड़ो मैं इसको एक अलग तरीके से बताता हूं ताकि आप इसे इस तरीके से हमेशा याद रखें अगर कभी भी किसी राइटिंग में आप देखोगे तो जो हाइपोटेन्यूज का मिड पॉइंट होता है ना जो हाइपोटेन्यूज का मिढ्वाइंट होता है वो सारी वर्टेक्स है इससे इससे और इससे इक्वल डिस्टेंस पर होता है और अन्य राइट एंगल ट्रायंगल राइट एंगल ट्रायंगल है तो वो इस हाइपोटेन्यूज का मिढ्वाइंट होगा और उसकी किस-किस से सुना उसे पॉइंट की इससे और इससे डिस्टेंस से होगी और मुझे चूंकि इसका लोकस निकलना है तो इस मिड पॉइंट को क्या मैं थोड़ी देर के लिए ह कमा के से दिनो कर सकता हूं मेरे ख्याल से मैंने एक अच्छी डीसेंट अमाउंट ऑफ आपकी हेल्प कर दिए चलो मैं एक और क्लूज दे देता हूं उसके बाद मैं ये छोड़ दूंगा क्वेश्चन आपके ऊपर क्या अगर मैं इसे इससे कनेक्ट कर डन और अगर मैं इससे कनेक्ट कर डन तो क्या है दो चीज आप जानते हैं मैंने पॉइंट्स को थोड़ा ना डिनोट कर देता हूं जैसे ये जीरो ये है ना तो इसे हम सी लगा लेंगे यहां पे मैं ऐसे बोल देता हूं बी और इसे मैं बोल देता हूं सी पॉइंट मुझे आप ओए और ओ भी भी बता सकते हो स्टूडेंट्स अगर आप थोड़ा सोचो तो अगर आपको थोड़ी सी भी दिक्कत हो रही है तो मैं इसे थोड़ा और कॉम्प्लिकेटेड करना चाहूंगा अगर मैं इस पॉइंट को कहता हूं पी तो आपको पीसी की डिस्टेंस एक भी काफी ज्यादा हेल्प मिल जाएगी मेरे ख्याल से आपके पास काफी कुछ है अगर अब आप इस क्वेश्चन को fikraout करना चाहें तो मैंने काफी ज्यादा रेलीवेंट इनफॉरमेशन आपको बता दिए है और अब बस एक क्वेश्चन ट्राई करिए क्वेश्चन बहुत ही साधारण सिंपल और डायरेक्ट सा क्वेश्चन है जो की आप आसानी से कर सकते हो स्टूडेंट्स मुझे बस एक बात का जवाब दे दो आप मुझे बस फाइनली एक बात का जवाब दे दो की क्या हम इस बात से आगे बढ़ाएं कहानी सुनेगा ध्यान से ये जो पी है ये जो पी है क्या मैं ये जानता हूं सर की पी से ए की डिस्टेंस और पी से सी की डिस्टेंस या पी से बी की डिस्टेंस से होगी बिल्कुल होगी सर अरे होगी की नहीं सर होगी बढ़िया अब सुनते हैं ध्यान से सुनना कम की बात सर इन सारी बातों में भी एक और बहुत जरूरी बात निकल के ए रही है की सर अगर मैं फाइनली इस कंक्लुजन पर बात करूंगा की का और पीसी की डिस्टेंस से होंगे आपसे क्या कहना चाह रहा हूं मैं आपसे कहना चाह रहा हूं की का और पीसी के डिस्टेंस से होगी तो इस बात से आप कहां ले जाना चाह रहे हो सर इस बात से आप क्या कहना चाह रहे हो मैं आपसे कहना चाह रहा हूं की क्या आप फाइनली ये ब की लेंथ या फिर एटलिस्ट पीसी की लेंथ कुछ तो निकल सकते हो सर बहुत दूर नहीं जाना होगा अगर मुझसे कोई पूछे पीसी की लेंथ तो कितनी बड़ी बात है सर पी के कोऑर्डिनेट्स बताएं सी के अकॉर्डिंग निकल सकते हैं तो चलो भाई पीसी की लेंथ निकलती हैं सर में पीसी के लेंथ निकलना चाहूं तो मैं क्या कहूंगा कम की बातें स्टूडेंट्स ध्यान से suniyega अगर पीसी की हम लेते हैं अप्लाई करता हूं स्टूडेंट्स तो पीसी की वैल्यू क्या कहूंगा मैं सुनना √k - 0 का होल स्क्वायर इनसाइड एंड अंडर रूट ये आपकी आती है पीसी की वैल्यू कोई दिक्कत नहीं है सर अच्छा सर सारी बातें छोड़ो क्या आप हमारी हेल्प कर सकते हो यहां पर ये चीज फिगर आउट करने में की अगर मैं ओए और ओ बी को जानता हूं आई नो ओए एन ओ बी तो क्या इनसे इस ट्रायंगल का कोई रिलेशन निकाला जा सकता है अब मैं जो हिंट आपको बोलना चाह रहा हूं वो ये की ये जो आपके पास ओए और ओ भी है क्या इनको देखकर आप कुछ हेल्प ले सकते हो जरा इस चीज को अच्छे से नोटिस करो स्टूडेंट्स आप ये क्वेश्चन खत्म कर दोगे अच्छा अगर इन सारी बातों में एक छोटी सी बात और नोटिस की जाए एक छोटी सी बात और नोटिस की जाए तो सर एक राइट एंगल ट्रायंगल को अगर आप देखो कौन सा राइट एंगल ट्रायंगल आपका एक होगी रिपीट माय स्टेटमेंट ओए पर मैं ए रहा हूं लेकिन अगर आप इस राइट एंगल ट्रायंगल एबीसी में भी कुछ-कुछ पता है क्या हान सर कुछ कुछ चीजों तो पता है देखना कैसे इस राइट एंगल ट्रायंगल एबीसी में अगर पूछूं आपसे तो देखो हम ई निकल सकते हैं क्या सर कुछ चीज अगर दिमाग लगाएंगे तो निकल ही जा सकती हैं बट ज्यादा दूर जाने की जरूरत मेरे ख्याल से है नहीं क्या मैं इस एक की लेंथ निकल सकता हूं वो भी निकल सकता हूं पर वहां तक अभी आपको जाने की जरूरत है नहीं सर इसके अलावा ये भी भी निकाला जा सकता है अगर ह कॉम के उसके मिढ्वॉइंट्स पता है तो पर उन सारी बातों से पहले इस पॉइंट पर आते हैं की ओरिजिन से ए और पी की डिस्टेंस जब हम निकलेंगे तो वो क्या निकल कर आएगी बात वहीं आकर boyaldown हो रही है की आप ए और बी के डिस्टेंस निकल लीजिए आपका ये क्वेश्चन खत्म हो जाएगा स्टूडेंट्स एक और हिंट मेरे दिमाग में जो ए रही है जिससे शायद बात बन जाए आपको मैं याद दिलाता हूं स्टूडेंट्स की प्रॉपर्टी हमने कभी ना कभी पड़ी है की सर अगर कोई सर्कल है सर्कल और सर्कल के पास कोई भी ऑडियो सर्कल कल से सेंटर है तो अगर सर्कल के सेंटर से कॉर्ड पर परपेंडिकुलर ड्रॉप करें तो उसे कट को बायसेक्स करती है ये प्रॉपर्टी याद है क्या अगर इस प्रॉपर्टी को यहां सोचें तो मेरा कहना है सर ओ से पी को कनेक्ट करिए ओ से पी को कनेक्ट करिए तो कहना है सर की ये जो पॉइंट है आप और ये बीपी या तो इक्वल लेंथ के होंगे पर ये जो आप होगा ये लाइन अब पर परपेंडिकुलर होगा आई होप आप ये बात समझ का रहे हो और मैं इसे किसी और चीज से शो करने की कोशिश करूं तो ये परपेंडिकुलर होगा अब मैं बार-बार आपसे जब का रहा हूं की आप ओए अब से डील करो तो मैं बार-बार आपसे कहना चाह रहा हूं की देखो यहां ओपा एक राइट एंगल ट्रायंगल बन रहा है बिल्कुल बन रहा है तो ध्यान से देखना क्या मैं यहां पर ये का सकता हूं सुनेगा ध्यान से की सर oa² जो की है हाइपोट ए न्यूज़ विल बी इक्वल तू बिल्कुल सही बात अगर यहां से मैं चीज सोचना चाहूं यहां से मैं चीज सोचना चाहूं तो ये मैं आप तक या का तक पहुंचना चाह रहा हूं तो देखना आप को मैं पी ए लिखूं तो कोई बुराई तो नहीं है तो देखना स्टूडेंट्स हम का के स्क्वायर को क्या लिखेंगे हम का के स्क्वायर को लिखेंगे ओए का स्क्वायर अब ओए का स्क्वायर मतलब इस सर्कल की क्या रेट है ये ओ और ये ए तो सर्कल की रेडियस मतलब एक का स्क्वायर तो ये हो जाएगा a² - ओ p² क्या ओरिजिन से पी यानी ह कमा के की डिस्टेंस ह कमा के की डिस्टेंस क्या होगी सर वो भी अंडर रूट ओवर क्या h2 + के स्क्वायर कोई तकलीफ नहीं है सर तो ये हो गया आपका का स्क्वायर लेकिन सर ये तो है का स्क्वायर आई थिंक यह बात आपको आसानी से समझ ए रही है सर का निकल कर आया यह टच निकल कर आया यह आई थिंक कुछ तो कर सकता हूं सर का और पी सी इक्वल तो यही तो आपने प्रिडिक्ट किया था तो बस इसी बात पर बात कर लेते हैं अगर का और पी सी इक्वल है अगर का और पी सी इक्वल है तो क्या कहूंगा ध्यान से देखो यहां पर है ह-सी का होल स्क्वायर देखो अब बात समझना ये है पीसी जिसमें अंडर रूट है और ये है का जिसमें भी अंडर रूट है तो क्या अंडर रूट अंडर रूट कैंसिल बिल्कुल सर तो अंदर वाली बात देखो ह-सी का होल स्क्वायर प्लस के स्क्वायर यहां पर क्या लिखा है सर यहां पर लिखा है ह-सी का होल स्क्वायर प्लस के स्क्वायर ये किसके इक्वल होगा सर ये जो एक्सप्रेशन ये इक्वल होगा इसके जो की क्या है a² - एक्स स्क्वायर प्लस के स्क्वायर इन साइड एंड अंडर रूट तो ये क्या है a² - K2 जो की है अंडर रूट में और मेरे ख्याल से तो ये क्वेश्चन खत्म हो चुका है सर क्योंकि आप ह और के में कोई ना कोई इक्वेशन बना पाए हो जहां पर आपके पास ये ए और सी जो द वो नॉन द वो उन्होंने दिए द राइट तो मैं इस इक्वेशन को सिंपलीफाई करके लिखूं तो ये हो जाएगा क्या ह को आप रिप्लेस करते हो एक्सेस माइंस सी का होल स्क्वायर प्लस ए स्क्वायर माइंस कितना अंडर रूट ओवर x² + y² ये उसे कार्ड की उसे चोर्ड की मिड पॉइंट का लॉस है जो की इस सर्कल पर ड्रा की है x² + y² a² जो की c0 पर राइट एंगल ड्रॉप करती है क्या कहीं से कहीं तक भी इस क्वेश्चन में कोई डाउट आई थिंक क्वेश्चन आसान था बस आपको यहां पर चीज थोड़ी ठीक से विजुलाइज कर लेनी थी और हमने वही सर्कल्स की कोड्स की प्रॉपर्टीज उसे की जो अभी तक शायद हम सीखते हुए आए हैं अगर यह क्वेश्चन समझ आए तो थोड़ा और आगे बढ़े क्या स्टूडेंट्स कुछ और चीज एक्सप्लोर करें इसके बाद आते हैं एक और क्वेश्चन पर है जहां पर इसे थोड़ा और इनसाइटफुली हम देख पाएंगे और इसे और बेहतरीन तरीके से हम समझने की कोशिश करेंगे इस क्वेश्चन में क्या लिखा है वह देखिएगा स्टूडेंट्स फाइंड डी इक्वेशन ऑफ डी सर्कल विच इसे टच हो सकता है शायद में से आपको क्योंकि मैं ऑलमोस्ट हर वो चीज जरूरी है बता दूंगा पर मैं चाहूंगा की क्वेश्चन आप खुद से ट्राई करें फिर आगे क्या लिख रहा है देखिएगा ऊपर शायद मैं रोक लो और का डन की नहीं इसके बाद आप कर लेना क्योंकि इतना बता दिया की ये क्वेश्चन अब आप कर सकते हो फिर भी हम कोशिश करते हैं देखते हैं चीज क्या है सेंटर ऑन डी पॉजिटिव डायरेक्शन ऑफ डी एक्स एक्सिस सेंटर है पॉजिटिव डायरेक्शन ऑफ एक्स एक्सेस पर ठीक है आपसे बात समझने की मैं कोशिश करना चाह रहा हूं की देखो ये आपका आईएसआईएस रहा होगा ये आपका एक्स एक्स रहा होगा है ना आपका एक्स-एक्स जो है वो मैं थोड़ा यहां बना लेता हूं सर वही इसे इक्वल तू एक्स एक्स स्ट्रेट लाइन है तो ए = एक्स जो स्ट्रेट लाइन है वो मैं बना देता हूं ये कुछ ऐसी जाती है तो ये है आपकी ए = 2X स्ट्रेट लाइन है ना अब बात समझना सर्कल का सेंटर अगर पॉजिटिव एक्स-एक्स स्क्वायर यानी यहां कहीं है और अगर ये टच करती है तो सर्कल यहां यहां तो नहीं बनेगा सर्कल यही कहीं बनेगा तो सर एक ऐसा सर्कल बना लेते हैं एक ऐसा सर्कल बना लेते हैं जो y= कहीं ना कहीं टच कर रहा है क्या किसी भी स्टूडेंट को इस बात से कोई परेशानी मेरे ख्याल से तो नहीं होनी चाहिए अब आपका क्या कहना है सर इस सर्कल का कहीं ना कहीं सेंटर है वो सेंटर जो है ना वो सेंटर जो है मेरे से थोड़ा सा और अच्छे से बना देता हूं क्योंकि मुझे इस सर्कल ना बहुत ज्यादा सही दिख नहीं रहा है रिलीज सॉरी इस सर्कल को मैं यहां से चलेगा क्या हान सही है ना अब हम बात करते हैं किसकी ये जो स्ट्रेट लाइन है आपकी ये कौन सी है ये ए = ए = एक्स है ना ये जो की सर्कल की यह जो स्ट्रेट लाइन यहां पर दिख रही है वो क्या है सर वो है ए = एक्स फिर से बना देता हूं यह है ए = एक्स बहुत बढ़िया अब क्या सर इस सर्कल का सेंटर जो है वो एक्स एक्सिस पर कहीं ना कहीं सिचुएटेड है होगा समझ लेंगे अगर आपको कन्फ्यूजन है तो मैन लेते हैं क्या लेट से ए कॉमर्स जीरो कोई दिक्कत तो नहीं है सर नहीं है अब सुनना अब कम की बात है स्टूडेंट्स एक स्ट्रेट लाइन है एक्स - √3y = 0 अच्छा से मैं थोड़ा और लिखूं ना तो समझना बात को मैं ऐसे ऐसे लिख सकता हूं क्या √3y=x या फिर ए = 1 / √3 सेक्स देखो ए = एमएक्स ये क्लीयरली ओरिजिन से पास होने वाली स्ट्रेट लाइन है ये ओरिजिन से पास होने वाली स्ट्रेट लाइन है और 1 / √3 है 30 डिग्री ये 45 डिग्री है याद ए रहा है ना तो 30° मतलब एक ऐसी लाइन जा रही होगी एक ऐसी लाइन जा रही होगी और ये क्या हुई आपकी गोद जो की कौन सी लाइन है यही एक्स - √3y = ऑफ कोर्स जीरो किसी भी स्टूडेंट को यहां तक कोई डाउट है तो पूछो भाई ऑलरेडी एक्सप्लेन दिस पार्ट अब क्या सर अब सुनना ध्यान से वो ये का रहा है suniyega ध्यान से डी कोड ऑफ लेंथ तू यूनिट्स तो यह जो कोड है ना सर ये जो कॉर्ड है ये यहां से यहां तक की ये जो कट है इसकी लेंथ है तू यूनिट्स सर एक बात बताओ सारी बातें छोड़ो आप अगर मैंने इस सर्कल के सेंटर से इस कॉर्ड पर परपेंडिकुलर ड्रॉप किया तो shailise की ये कट बायसेक्स हो जाएगी बिल्कुल हो जाएगी एक और बात बताओ सर अगर मैंने सर्कल के सेंटर से सर्कल की टैसेंट पर परपेंडिकुलर ड्रॉप किया तो कहना इसे वो रेडियस हो जाएगी बिल्कुल हो जाएगी मैंने काफी सारी हिंट दी हैं काफी सारी हिंट दी है अगर आप सोच का रहे हो जो बाद में करना था सर एक और बात बताओ सोच के बताना अगर यह कार्ड की लेंथ जो थी वो थी तू यूनिट्स तो बायसेक्स होके डिस्टेंस कितनी होगी वैन और अगर यहां से मैं इससे यहां ज्वाइन करूं सर्कल के सेंटर से सर्कल की सर का फ्रेश तो क्या वो भी सर्कल की रेडियस होगी अगर आपको कन्फ्यूजन हो रहा है तो मैंने कुछ पॉइंट्स को नाम दे देता हूं जैसे की मैं इसे का देता हूं लेट्स से सेंटर है ना तो हम इसे कुछ भी का देते हैं हिंट जो मैं देना चाह रहा हूं जो शायद आपका क्वेश्चन सॉल्व कर ही देगी उससे क्या अब जो है वह एड के इक्वल होगी बिल्कुल होगी सर क्योंकि अब जो है वो आपके सर्कल के रेडियस है और एड जो है वो भी तो आपके सर्कल के रेडियस है आप बात समझ पाए क्या आई होप ये फिगर बहुत कॉम्प्लिकेटेड नहीं हो रहा है हम जान का रहे हैं सर की सर्कल की जो रेडियस है इस सर्कल की जो रेडियो से वो क्या है सर आप कहोगे उसे या तो ए बी या फिर आप उसे कहोगे भाई रेडी कोई डाउट तो नहीं है सर नहीं है अब आते हैं कम की बात पर कम की बात क्या है suniyega ध्यान से सर एड कैसे निकलोगे एड कुछ नहीं है सर एक्स माइंस यस परपेंडिकुलर ड्रॉप किया गया है है तो वह परपेंडिकुलर लेंथ निकल लेंगे यह हमने स्ट्रेट लाइन में सिखा है एक्स - ए = 0 पर परपेंडिकुलर लेंथ ये क्या होगी एड है ना तो पहले तो मैं क्या निकलता हूं एड है ना ए कमा जीरो से अब सीधी सीधी सी बात है सर ए 1 - 0 कुछ बच नहीं रहा है अपॉन में क्या वैन प्लस वैन अंडर रूट में तो ये हो जाएगा ए/√2 मैं डिस्टेंस फॉर्मूला लगा रहा हूं सुनना ए 1 + 0 -1 + 0 मतलब न्यूमैरेटर में क्या बचा ए डिनॉमिनेटर में इसका कॉएफिशिएंट यानी वैन इसका कॉफिशिएंट माइंस वैन वैन का स्क्वायर माइंस वैन का स्क्वायर वैन वैन प्लस वैन तू तू इनसाइड इन अंडर रूट तो इसकी जो आपकी ये आएगी भाई क्या इसकी जो रेडियस आएगी वो क्या होगी ए / √2 और ये हमने क्लीयरली पॉजिटिव वैल्यू मणि है पॉजिटिव x² है ना तो एक लिलियम ने यहां पर एक पॉजिटिव वैल्यू है तो ए / √2 भी तो रेडियस है वो डिस्टेंस पॉजिटिव ही आएगी तो सर इसकी रेडियस ए रही है ए / √2 एड वाली एप्रोच से वो कितनी ए रही है ए/√ को आपने कहा अब के इक्वल्स सर आप एड को कहते हो किसके इक्वल मैं कहता हूं एड जो है सर वो अब के इक्वल होगा अब ए बी क्या है अब समझ लेते हैं सर कैन आई से की सर एबीसी एक राइट एंगल ट्रायंगल है उसमें मुझे ई पता है क्या एक पता कर सकता हूं कर सकता हूं क्या परपेंडिकुलर के लेंथ निकल सकता हूं एक घबराना मत हम क्या निकलने वाले हैं स्टूडेंट्स हम निकलने वाले हैं अभी फिलहाल तो एक तो एक क्या होगा देखना ए 1 यानी कितना ए फिर वही बात 0 10 नोट मैकेनाइज्ड बाकी सब चीज जीरो और मोड ए लिखने से अच्छा सर यही लिख दो क्योंकि कोई फर्क नहीं पड़ता क्योंकि ए पॉजिटिव वैल्यू है पॉजिटिव एक्स-एक्सिस पर ए एक पॉजिटिव वैल्यू हमने मणि है और क्या और मेरा आपसे एक कहना है कम की बात है suniyega की सर डिनॉमिनेटर में जो आएगा वो होगा वैन का स्क्वायर और √3 का स्क्वायर इसका सैम कर देंगे तो क्या अब का स्क्वायर ए जाएगा सर तो ये कितना है तो ई का स्क्वायर कितना वैन का स्क्वायर और अगर मैं इसका अंडर रूट ले लूं अगर मैं इसका अंडर रूट ले लूं तो साहब आपको क्या मिल जाएगा तो सर हमको क्लीयरली जो वैल्यू मिलेगी वो होगी चल रही है सारी बातें क्या आसानी से पूरा सिंपलीफिकेशन समझ ए रहा है आई थिंक क्वेश्चन आसान है बस चीजें ठीक से विजुलाइज कर ली जाए टू नहीं है सर मैं इसको ए के लिए सॉल्व करने में इंटरेस्टेड हूं पर स्क्वायर कर दे तो अच्छा रहेगा बिल्कुल कर दो तो हो जाएगा a² / 2 यहां पर हो जाएगा a² / 4 और ये हो जाएगा +1 a² / 4 को यहां लेकर आए a² / 4 को यहां लेकर आए तो आई थिंक चीज कुछ ऐसी हो जाएंगी बिल्कुल हो जाएंगे सर a² कॉमन ले लो 1 / 2 में से 1 / 4 सब्सट्रैक्ट किया तो वैन बाय फोर मिलेगा नहीं समझ का रहे हो तो ऐसे सोच लो ध्यान से देखो स्टूडेंट ये वैन बाय तू फाइव पॉइंट फाइव फोर बचेगा तो ये बचेगा a² / 4 ए स्क्वायर तो ए की वैल्यू कितनी आएगी प्लस माइंस तू मैं प्लस तू लूंगा दिख रहा है आप सभी को तो एक ही वैल्यू क्या लूंगा सर मैं ए की वैल्यू लूंगा +2 और जैसे ही मैं लेता हूं एक ही वैल्यू प्लस तू जैसे ही मैं लेता हूं ए की वैल्यू प्लस तू हमसे क्या पूछा जा रहा था इक्वेशन ऑफ डी सर्कल इक्वेशन ऑफ डी सर्कल इस गोइंग तू बी नथिंग ये जो cardinate है ये जो cardinate है सर ये आए आपके 2 0 और सर्कल की रेडियस कितनी आई है सर सर या तो का लो एड या एड ये दोनों ही क्या द सर याद करो या तो का लो एड या एड या दोनों ही क्या था रेडियस तो किसी में भी रिप्लेस कर दो आप किसी में भी एक ही वैल्यू रिप्लेस कर दो यहां भी करोगे तो भी बात वही बनेगी यहां भी करोगे तो भी बात वही बनेंगे अब करके देख लीजिए अब बात समझ का रहे हैं स्टूडेंट्स अगर आप बात समझ का रहे हैं फाइनली तो करते हैं स्टूडेंट्स देखो अगर मैं ए की वैल्यू तू यहां सब्सट्रैक्ट यू नो रिप्लेस करता हूं जैसे मैं क्या लिख दूंगा फाइनली रूट तू कोई दिक्कत तो नहीं यह होगी इसकी रेडियस यहां रखना है क्या तो इस सर्कल की रेडियस कितनी होगी सर इस सर्कल की रेडियस हो जाती है रूट तू इस सर्कल की जो रेडियस निकल कर आती है वो होती है सर्कल के सेंटर के कोऑर्डिनेटर सर्कल की रेडियस है क्या एक्शन नहीं लिखे जा सकती सर यही तो पढ़ रहे हैं आज हम तो निकल लो भाई तू कमा जीरो √2 और अगर हम बात करें तो एक्स - 2 का होल स्क्वायर प्लस ए माइंस कोई तकलीफ क्या यह क्वेश्चन इसका कंप्लीट सॉल्यूशन के साथ आपको समझ आया कॉन्सेप्ट है यह क्या एक नई सी चीज है बात करते हैं सुने का ध्यान से आप एक सेकंड के लिए थोड़ा सा इस कॉन्सेप्ट को इमेजिन करो जैसे की सर मैन लो ये है विक्स है और यहां पर मैन लो एक खास तरीका सर्कल मुझे दिखाई दिया इस सर्कल की खास बात मैन लो यह रही सर की सर्कल ने दोनों ही एक्सेस को कहीं ना कहीं किसी पॉइंट पे टच किया मुझे एक बात बताओ सर आप अगर यह सर्कल टच कर रहा है क्योंकि ना ऐसे अगर मैंने इसकी ये जो टैसेंट है यहां से एक परपेंडिकुलर ड्रॉप किया होप आपको दिख रहा है जो मैं कहना चाह रहा हूं सिमिलरली इसकी इस इंजन पर भी एक परपेंडिकुलर ड्रॉप किया है तो कहना इसे की सर यह जो पॉइंट निकल के आएगा सर्कल का सेंटर होगा भाई बेसिक ज्यामिति हम ट्राईंग तू मेक अंडरस्टैंड डेट एंड डी होल पॉइंट हेयर स्पाइट सैंपल की सर देखो आप ध्यान से यह जो है ये डिस्टेंस भी रेडियस है और उसको घुमा कर ले तो ये डिस्टेंस भी रेडियस है मतलब अगर ये डिस्टेंस है डस्ट ये डिस्टेंस भी है अब बात समझ का रहे हो तो इसकी जो क्वाड्रेट्स होंगे वो क्या होंगे सर वो भी तो आर कॉम ए रही होंगे बट ऑफ कोर्स मैं यहां पर ये मैन के चल रहा हूं की सब कुछ हो रहा है फर्स्ट क्वाड्रेंट में तो ये आर पॉजिटिव है तो रेडियस ही है और ये भी आर पॉजिटिव है तो उसके क्वाड्रांट्स मैं फर्स्ट क्वाड्रेंट में जो क्वाड्रांट्स होंगे वो भी क्या होंगे पॉजिटिव इस पुरी बात के थ्रू जो पॉइंट मैं आपको समझाना चाह रहा हूं वो ये स्टूडेंट्स की अगर कोई सर्कल कोई सर्कल अगर दोनों ही एक्सेस को टच करें तो उसके जो एक्स और ए cardinate होते हैं सेंटर के वो उसकी रेडियस जितने ही होते हैं मेग्नीट्यूड की टर्म में साइन वाइस ये डिपेंड करता है की वो किस क्वाड्रेंट में है फर्स्ट में दोनों पॉजिटिव सेकंड में एक्स नेगेटिव ए पॉजिटिव थर्ड में दोनों negghtive फोर्थ में ए नेगेटिव और एक्स पॉजिटिव बस यह ध्यान रखना है क्या यह बात आपको समझ आई सर ये तो ए गई लेकिन और क्या कहना चाह रहे हो और आई होप इसको देखकर आप सर्कल की इक्वेशन लिख सकते हो कोई बहुत बड़ी बात नहीं है इसको देख के सर्कल की इक्वेशन क्या होगी सर एक्स - आर का स्क्वायर प्लस ए माइंस आर स्क्वायर इस इक्वल्स तू रेडियस का स्क्वायर ये आपके सर्कल की इक्वेशन ए जाएगी क्या ये बहुत टू टास्क है सर ये तो बड़ी आसान सी बात है पर इन आसान से ही बातों में दो बातें थोड़ी सी ट्रिक की हो जाती है एक बात ही आप उसे का कर देख लो ये सर अगर ये सर्कल दोनों एक्सिस को टच ना करते हुए किसी एक ही एक्सेस को टच करें तो जैसे मैन लो यहां पर ये सिर्फ इसी को टच करें अगर ये एक्स ए कोई टच करें तो क्या तो समझते हैं क्या होगा समझो बात को यहां पर इस सर्कल का सेंटर होगा इस सर्कल के सेंटर से जब आप इसकी टैसेंट पर यानी अभी फिलहाल एक्स एक्सेस को टच कर रहे हैं तो उसे पर जब आप परपेंडिकुलर ड्रॉप करेंगे तो ऑफ कोर्स क्या होगी वो रेडियस सर एक बात बताओ सिंपल सी बात है इसका जो भी excordinate रहा हो मुझे नहीं पता सेंटर का लेकिन ए कोऑर्डिनेट्स क्या रहेगा आप समझो ना ये सर्कल के सेंटर के व्हाइट कोऑर्डिनेट्स के डिस्टेंस या एक्सेस पर ऑप्शन जो बात को ये जो डिस्टेंस है वही तो है तो ये डिस्टेंस और ये डिस्टेंस से है अगर ये डिस्टेंस ए रहे हो तो डिस्टेंस भी आर होगी तो इसका ए कार्ड नहीं है तो आर अभी भी फिक्स्ड है आप बात समझ का रहे हो एक्स कोऑर्डिनेट्स कुछ ना कुछ होगा मुझे नहीं पता एक्स koardinate कितना होगा बट ऐसा कुछ होगा और लेट्स से वो डिस्टेंस से अल्फा तो मैं उससे का देता हूं अल्फा बिल्कुल का सकते हो सर यानी मैं आपसे ये कहना चाह रहा हूं की इस केस में अभी भी मैं सारे हूं फर्स्ट क्वाड्रेंट के कॉन्टेक्स्ट में ले रहा हूं ताकि आपको कन्फ्यूजन ना हो तो सर इस बार क्या हो रहा होगा सर इस बार अगर आप लिखोगे तो ये हो जाएगी एक्स - अल्फा का स्क्वायर प्लस ए माइंस आर का स्क्वायर इस इक्वल्स तू आर स्क्वायर अल्फा मतलब सेंटर का एक्स कॉर्डिनेट आर मतलब रेडियस और सेंटर का ऑर्डर कब जब ये एक्स एक्सिस को टच करेगा क्या इसी बेस पर अब आप एक और प्रिडिक्शन दे सकते हो कैसा असर जब बात हो रही हो सर अगर ये ए एक्सिस टच करें तो क्या आसान है अब सोच का रहे होंगे की सर अगर सर्कल आपका ए-एक्सिस को टच करें फिर सर्कल हैपेंस तू टच डी ए एक्सिस व्हाट विल बी हैपनिंग इन डैटसन आई वुड से इट्स क्यूट सिंपल तू प्रोटेक्ट क्योंकि अब हमने एक सिनेरियो तो देख लिया है सर इफेक्ट हैपेंस तू बी सी सेंटर ऑफ डी सर्कल आई थिंक मेरे लिए चीज आसान है क्योंकि ये उतनी ही डिस्टेंस है जितनी सर्कल की रेडियस है अगर ये है आर तो ये भी कितना हुआ आर तो सर्कल के सेंटर का जो एक्स कोऑर्डिनेट्स होगा वो होगा मैं अभी सारी चीज फर्स्ट क्वाड्रेंट में ही विजुलाइज कर रहा हूं सर अगर सर्कल के सेंटर का व्हाइट कोऑर्डिनेट्स हुआ बी या बिता तो ये डिस्टेंस क्या होगी सर ये टेक्निकल डिस्टेंस होगी बिता सर्कल के सेंटर का एक्स कोऑर्डिनेट्स आर है जो की सर्कल की रेडियस भी है इस तरीके से अगर आप चीज देख पाएं तो क्या मैं सर्कल की इक्वेशन लिख सकता हूं सर फिर वही बात एक्स - क्या आर का स्क्वायर प्लस ए - बी बेटा का स्क्वायर इस इक्वल्स तू सर्कल से ग्रास कर लिया जाए और यही सारी बातें यहां पर लिखी गई हैं देखो अगर सर्कल जो है आपका वो एक्स एक्सिस का अगर टच करता है तो ऑफ कोर्स मैं मैन लेता हूं की सर्कल का सेंटर अगर अल्फा कमा बिता है तो सर हम बिल्कुल का सकते हैं की सर्कल की जो रेडियस होगी वो क्या होगी आपकी ये डिस्टेंस बट ऑफ कोर्स इसका क्या ले लीजिएगा मोड इसका आप क्या ले लीजिएगा मोड ये बिता जो भी कोऑर्डिनेट्स हो ये नेगेटिव पॉजिटिव कुछ भी हो सकता है तो आप क्या लेंगे इसका बोट आय होप याद रखेंगे उसे तरीके से आप सर्कल की इक्वेशन लिख सकते हो वही बात है जो हमने घुमा फिर कर किसी और तरीके से सोचिए अगर वो बाइक्स इसको टच कर रहा है तो ए एक्सिस को टच करें तो वही बात स्टूडेंट्स की जो उसका एक्स कोऑर्डिनेट्स है वो उसकी क्या हो जाएगी रेडियस ऑफ कोर्स उसका मोड ले लीजिएगा उससे क्या होगा सर उससे फिर मैं सर्कल की इक्वेशन लिख पाऊंगा वही बात है वापस बट इस बार जो रेडियस है वो क्या है आपका एक्स कट है उल्टा एक एक्स एक्सिस को टच कर रहा है तो ए koardinate आपके रेडियस है जब ए एक्सिस को टच कर रहा है तो आपका एक्स कोऑर्डिनेट्स उसकी रेडियस है बहुत सिंपल सी बात है बिल्कुल सर अच्छा सर अगर इन सारी बातों से भी कनक्लूड करके एक और बात मैं करूं तो क्या ये भी याद रखिएगा की अगर सर्कल दोनों koardinate को टच करें तो वो कुछ इस तरीके से होगा ऑफ़ कोर्स वो अल्फा अल्फा हो सकता है वो अल्फा माइंस अल्फा हो सकता है वो माइंस अल्फा कॉम α हो सकता है या माइंस अल्फा माइंस अल्फा हो सकता है डिपेंडिंग ये वो कौन से क्वाड्रेंट में pleashed है पर एक बात तो तय है ना की अब तो उसकी रेडियस फिक्स्ड है मॉडल फॉर और अगर मॉडल फर्स्ट की रेडियस फिक्स्ड है तो सर क्वेश्चन खत्म हो गया की सर्कल की इक्वेशन क्या हो जाएगी कुछ ऐसी जो हम डिस्कस कर चुके हैं आई थिंक एक बेहतरीन से अंडरस्टैंडिंग आप बिल्ड कर का रहे हो सर्कल के इस कॉन्सेप्ट पर लेकर की अगर सर्कल के अगर टच करने पर उसकी कौन से cardinate से वो टच हो रहा है इक्वेशन निकलना बहुत टू कम नहीं है हमारे भी और अगर ये कम आपके लिए बहुत टू नहीं रहा तो आप मुझे प्रूफ करके बता दीजिए इस क्वेश्चन को सॉल्व करके दो ही बातें स्टूडेंट्स या तो आप स्क्रीन पॉज करके अभी ट्राई कर लीजिए जो की हमेशा से हाईली रिकमेंड रहा है इनके ऐसा करने पर क्वेश्चंस सॉल्व नहीं होता है तो बेशक प्ले करिए और अब वापस करिए जब मैं क्वेश्चंस समझा डन उसके बाद फिर से ट्राई करना स्क्रीन पॉज करके फिर भी ना हो तो बीच-बीच में स्क्रीन पीओएस करके कोशिश करते रहिए आप खुद करिए क्या लिखा है फाइंड डी इक्वेशंस ऑफ डी सर्कल्स कुछ तो हिंट उसने भी दी है इक्वेशंस ऑफ डी सर्कल्स टचिंग वही एक्सिस आते डी पॉइंट जीरो कमा फोर एंड कटिंग्स सर कुछ समझ नहीं ए रहा है थोड़ा सा पतियों सर को स्टूडेंट्स अगर यह आपका ए से मैन लो और ये अगर आपका एक्स एक्सिस है ना एक सर्कल है मेक सर्कल इमेजिन करता हूं जो की वह एक्सेस को जीरो कमा फोर पे टच कर रहा है ठीक है सर वो एक्स इसको जीरो कमा फोर पे टच कर रहा है जैसे पॉइंट है जिसका क्वाड्रेंट है जीरो कमा फोर समझ रहे हो क्या तो ए एक्सिस को अगर ये सर्कल सेल्स 0 कमा 4 पे टच कर रहा है अगर ये पॉइंट है लेट से 0 4 तो आपसे ये का रहा है ये का रहा है की जो इसकी कॉर्ड की लेंथ है ना सर जो उसकी कॉर्ड की लेंथ है जो इसकी कोर्ट की लेंथ है वो 6 यूनिट्स हैं आप बात समझ रहे हो जो एक्स के हेल्प से वो कॉर्ड बन रही है वो कितनी यूनिट की है सिक्स यूनिट्स की कोई तकलीफ तो नहीं है ठीक है सर अच्छी बात है सारी बातें समझ आई तो सारी बातें समझ आई और इस दौरान मुझे ये भी समझ आए की अगर सर्कल के सेंटर के कोऑर्डिनेट्स ढूंढो सर्कल के सेंटर्स के अकॉर्डिंग ढूंढो तो एक्स कोऑर्डिनेट्स की ज्यादा जानकारी मुझे नहीं है पर क्या हमें ए koardinate बता सकता हूं बिल्कुल बता सकते हो सर सर्कल के सेंटर का जो बाय cardinate होगा वो क्या होगा फोर क्या किसी भी स्टूडेंट को इस बात से परेशानी क्या किसी भी स्टूडेंट को इस बात से कोई परेशानी सर अभी तक तो नहीं है मैं आपको अगली हिंट जो देना चाहूंगा वो ये होगी स्टूडेंट्स की एक बात बताओ अगर सर्कल के सेंटर से अगर सर्कल के सेंटर से मैंने यहां पर एक परपेंडिकुलर ड्रॉप किया आप समझ का रहे हो जो मैं कहना चाह रहा हूं अगर सर्कल के सेंटर्स मैंने यहां पर एक परपेंडिकुलर ड्रॉप किया और अगर मैंने इस परपेंडिकुलर से इन दोनों लांस की बात आपसे की तो आप क्या कहोगे सर कार्ड पे सेंटर से ड्रॉप किया गया है परपेंडिकुलर कॉर्ड को बाय सेट करता है तो अगर इसकी पुरी लेंथ कितनी हो जाएगी 3 इंच क्या किसी भी शख्स को इस बात से कोई परेशानी बस मैंने कुछ गड़बड़ तो नहीं की नहीं किया सर एक बात अगर आप निकल लें तो यह क्वेश्चन खत्म हो जाएगा और वो बात यह है सर की इस पॉइंट से इस पॉइंट तक की ये डिस्टेंस आपको कुछ बनता हो रहा है अच्छा सारी बातें छोड़ो मुझे बताओ ये जो फोर है ये 4 है इस फोर का मतलब क्या है इस फोर का मतलब है सर ये जो डिस्टेंस है 4 है मतलब यह जो डिस्टेंस है ना सर ये फोर है अगर मैं कुछ पॉइंट्स को आपसे कहूं ना तो अगर मैं इसे कहूं लेट से थोड़ी देर के लिए ए इसे मैं आपसे कहूं बी और वैसे मैं आपसे कहूं सी तो आपको दिख रहा है ट्रायंगल एबीसी एक राइट एंगल ट्रायंगल है हान सर दिख तो रहा अगर ट्रायंगल एबीसी एक राइट एंगल ट्रायंगल है तो आप प्लीज इस बात को सुनिए अगर ये राइट एंगल ट्रायंगल है तो क्या सर ये जो एक है ये है थ्री है फोर तो एक कितना हो जाएगा फाइव pythagorain ट्रिपलेट 9 16 25 का अंडर रूट फाइव बताने की जरूरत तो नहीं है स्टूडेंट्स सर एक आया फाइव बात मैन लिया आपकी बात बहुत अच्छे से मैन ली हमने लेकिन अभी भी एक समस्या तो है क्या सर इसका एक्स कॉर्डिनेट कैसे निकले मेरे ख्याल से तो ये क्वेश्चन खत्म होने को अगर आप चीज सोच का रहे हो तो हरि पेट में स्टेटमेंट मेरे ख्याल से तो ये क्वेश्चन खत्म होने को है अगर आप ये चीज सोच का रहे हो तो क्योंकि बात समझो स्टूडेंट्स ये क्वेश्चन में थोड़ा सा चीजे ध्यान से देखोगे तो आप इसका एक cardinate निकल सकते हो बहुत टू टास्क नहीं है और जैसे ही उसके सेंटर का एक्स कोऑर्डिनेट्स आपको मिल जाएगा ये क्वेश्चन करते ही डिफिकल्ट क्वेश्चन नहीं है मेरा यकीन करिए और सर एक्स कोऑर्डिनेट्स निकलेगा कैसे एक्स कोऑर्डिनेटर तक पहुंचेंगे कैसे और एक्स कोऑर्डिनेट्स ढूंढेंगे कैसे जोर्नेट मिल गया ना आपको सी ये जो मिला है ओवरऑल ये जो आपको कब जो आपको मिल रहा है आई एम रियली सॉरी ये जो मिल रहा है वो फाइव है ना तो फाइव का क्या मतलब स्टूडेंट्स आप प्लीज इस बात को समझो ये आपकी रेडियस हुई ना भाई तो अगर सर ये एक आपकी रेडियस है और इसमें मैंने रोते किया इससे मैंने रोते किया आप देख का रहे हो तो अगर मैं इसे थोड़ी देर के लिए कहूं तो आपकी एडीबी तो फाइव यूनिट्स ये होगी क्योंकि रेडियस है ना मतलब ये जो डिस्टेंस है यही ये डिस्टेंस होगी तो एड की डिस्टेंस अगर 5 यूनिट्स होती है मतलब यह बेसिकली यह वाली डिस्टेंस आपकी 5 मिनिट्स होती है मतलब इसका जो एक्स कोऑर्डिनेट्स है वो कितना है फाइव मेरे ख्याल से तो एक क्वेश्चन खत्म हो गया है पर सर इसमें कुछ परेशानी है क्या मुझे जल्दी से कमेंट में लिख के बताओ की अभी भी इसमें क्या गड़बड़ है सर एक बात आप शायद मिस कर रहे हो उसने हिंट दी थी आप ठीक से पढ़ते नहीं हो क्वेश्चन उसने एक हिंट दी थी आप क्वेश्चन ठीक से नहीं पढ़ते हो सर यही की यही बातें इधर भी तो हो सकती थी सर यही का यही से सर्कल यहां भी तो रिप्लिकेट हो सकता था यह जो सर्कल है सर आपका यह इधर भी तो बन सकता था यही सर्कल से सर्कल इस तरीके से अगर बन रहा होता तो आई थिंक अगर मैं इसे बनाऊं थोड़ा और बड़ा करना पड़ेगा शायद मेरी ख्याल से अगर सर ये सर्कल यहां बन रहा होता तो यहां पर भी तो से बातें हो सकती थी यहां पर भी तो यह सर्कल इसे ए एक्सिस को जीरो कमा फोर पे इंटरसेक्ट कर रहा है मतलब टच कर रहा है और यहां भी तो एक्स एक्सेस पे वही डिस्टेंस है तो सर पॉसिबिलिटी थी की सर्कल के सेंटर के क्वाड्रेंट कितने हो जाएं सर एक्स-एक्सिस का वही मेरे इमेज मिल जाएगी और वो व्हाइट कार्डिनल तो वही रहेगा और अभी भी रेडियस क्या रहने वाली है सर रेडियस अभी भी आपकी वही 5 रहने वाली है तो दो सर्कल बन सकते हैं सर आप क्यों नहीं सोच का रहे हो ये बात कौन से दो सर्कल्स पहला सर्कल ये वाला जिसके सेंटर के अकॉर्डिंग 5 दूसरा सर्कल ये वाला जो इसका सेंटर के अकॉर्डिंग मेरे ख्याल से थोड़ा सा इसके लिए क्या हो जाएगा 5 हो तो क्या लिखोगे एक्स - 5 होल स्क्वायर इस इक्वल तू फाइव स्क्वायर विच इस 25 आई होप ऐसे आप सिंपलीफाई करके बना दोगे सर इसको अगर बनाना चाहे तो एक्स - 5 तो एक्स - 5 यानी एक्स प्लस फाइव होल स्क्वायर प्लस ए माइंस ए गई हैं तो आगे बढ़ जाते हैं और नेक्स्ट क्वेश्चन डिस्कस करते हैं यहां क्या लिखा हुआ है सर अन सर्कल पासिंग थ्रू डी वर्टेक्स सी ऑफ ए रेक्टेंगल ए बी सी दी एंड टचिंग इट्स साइड्स अब एंड एन बहुत कुछ दिया उसने अगर मुझे कोई रेक्टेंगल दिया है जब रेक्टेंगल पिक्चर में आता है ना तो मैं कोशिश करूंगा इस तरीके सिनेरियो की हेल्प लेने की सनग्लासेस एक एक्टिंग हमें दिया हुआ है ना तो लेट्स इमेजिन दिस रेक्टेंगल हो गया तो एक एक्टिंग पिक्चर में ए रहा है सर ठीक है अब वो आपसे ये का रहा है की जो सर्कल है यह वर्टेक्स सी से तो पास होता है और अब और टच करते हैं सुना ध्यान से बहुत ध्यान से मिक्स सर्कल बनाता हूं ना पहले तो ए बी सी दी लिख लेता हूं तो अब और एड को टच करता है इट पासेस थ्रू सी एंड अलांग विथ टच सर्कल प्लॉट करने की कोशिश करते हैं हम जॉब कोर्स क्या करें सर जो की डेफिनेटली सी से तो पास हो जो की डेफिनेटली सी से तो पास हो मेरे ख्याल से एक अच्छा एग्जांपल हो सकता है काफी हद तक तो हम बोल सकते हैं सेकंड में थोड़ा सा ऐसे शिफ्ट कर डन तो है ना बहुत अच्छे से ड्रोन नहीं कर का रहा हूं बट आई होप आप मेरी बात है समझ सन और सिख का रहे होंगे है ना मैन लेते हैं यह वह सर्कल है कोई तकलीफ किस स्टूडेंट को सर क्लियर सी बात है बड़ा हिस्सा शानदार सा सिंपल सा सिंपलीफिकेशन है यह जो सर्कल है यह इस रेक्टेंगल की एक वर्टेक्स से तो पास हो रहा है और अब और एड को टच कर रहे हैं और एक और बात है अब को कहां एम पर और एड को कहां एन पॉइंट पर टच कर रहा है इफ डी डिस्टेंस इफ डी डिस्टेंस फ्रॉम सी फ्रॉम सी तू डी लाइन सेगमेंट म = 5 मिनिट्स अगर मैं यह एम और एन को ज्वाइन करने वाली लाइन एक बनाऊं ये एन यहां टच हो रहा है ये एन यहां टच हो रहा है रिपीट में स्टेटमेंट यहां टच हो रहा है यहां टच हो रहा है तो एम और एम को ज्वाइन करने वाली स्ट्रेट लाइन अगर मैं बनाऊं है ना अगर मैंने ज्वाइन करने वाली स्ट्रेट लाइन बना हूं तो मैं एक स्ट्रेट लाइन भी बना देता हूं चीज अच्छे से ही बनाई जाए तो ज्यादा बेहतर है मेरे ख्याल से तो अगर यह लाइन हमने प्लॉट की देखेगा ध्यान से इस लाइन को अगर मैं ड्रा करूं तो यह स्ट्रेट लाइन हो जाएगी क्या इस बात से किसी भी स्टूडेंट को स्ट्रेट लाइन अब वह यह कहना चाह रहा है आपसे बहुत कृष्ण सी बात है की सर आप अगर सी से इस में पर परपेंडिकुलर ड्रॉप करोगे तो उसे परपेंडिकुलर की लेंथ कितनी होगी फाइव मिनिट्स एबीसी तो ए बी सी दी रेक्टेंगल का एरिया क्या होगा अब कोई स्टूडेंट से का सकता है की सर ये तो आईआईटी जी एडवांस्ड लेवल का क्वेश्चन है कहा जा सकता है अगर आपको बारीकियां ना पता हो और अगर आपको चीज अच्छे से सीखी गई हैं पता है तो ये बहुत ही आसान क्वेश्चन यह सच में यकीनन तौर पर एक अच्छा आसान और मजेदार क्वेश्चन है क्यों suniyega सर सबसे पहले तो है तो कोऑर्डिनेट्स की हेल्प पता है कौन करेगा कोऑर्डिनेट्स की हेल्प आपकी कर देगा एड को थोड़ा एक्सटेंड कर डन और अगर मैं अब को थोड़ा एक्सटेंड कर डन तो क्या है आपको एक्स एक्सिस और ए एक्सिस दिख रहा है कुछ समझ आया मैं क्या कहना चाह रहा हूं सर अगर यह आपको एक्स-एक्सिस और ए एक्सिस दिख रहे हैं तो बात मजेदार से ये हो जा रही है बात बड़ी मजेदार से ये हो जा रही है सर की आप कुछ तो बहुत ऐसा करने वाले हो जो शायद आज हम पढ़ रहे हैं बिल्कुल वही तो करेंगे हम जो पढ़ते आए हैं मुझे एक बात बताओ मुझे एक बड़ी बेसिक सी बात बताओ अब की कुछ ना कुछ डिस्टेंस होगी है ना अब की कुछ ना कुछ डिस्टेंस होगी लेट्स कॉल इट फॉर अन विले अभी हमारी अंडरस्टैंडिंग के लिए बी के कोऑर्डिनेटर क्या होंगे पी 0 दी के कोऑर्डिनेटर क्या होंगे जीरो की एक रेक्टेंगल है तो कहना है कॉल की सर सी के क्वाड्रेंट होंगे पी के बहुत बढ़िया यहां तक कोई परेशानी किसी भी स्टूडेंट को अभी तक तो नहीं है भाई अच्छा सा सारी बातें समझ आई अब क्या करें अब एक बात बताओ अब एक बड़ी बेसिक सी बात बताओ क्या मैं इस इस सर्कल के सेंटर के कोऑर्डिनेट्स निकल सकता हूं सर मुझे जो दिमाग में ए रही है वो बात ये की अगर ये इस सर्कल को यहां पर टच कर रही है तो इसके पैरेलल अगर इसके पास होती है कैसे लाइट और इसके भी पास होती है परपेंडिकुलर लाइन ड्रॉ की तो ये जो डिस्टेंस है कुछ नहीं है सर्कल की रेडियस है आई होप आपको बात समझ ए रही है आपके एक्स एक्सिस और ए एक्सिस दोनों को ये टच कर रहा है और अगर ये एक्स एक्सिस और ए एक्सिस दोनों को टच कर रहा है सर्कल तो इसके सेंटर के जो cardinate होंगे वो से होंगे और फर्स्ट क्वाड्रेंट भी माना है तो चीज हमने मैन ली मैन लो आर कॉम आर अल्फा कॉमन अल्फा कुछ भी मैन लो दो हिंट आपको मैं दे रहा हूं पहली हैंड सर्कल की इक्वेशन पहली है सर्कल की इक्वेशन और दूसरी हिंट यहां पे आप चीज सोचने वाले हो की सर इसकी रेडियस और इसके एक्स और ए koardinate सेंटर के से होंगे क्योंकि ये दोनों कोऑर्डिनेट्स को टच करता है बस ये दोनों हिंट सप्लाई कर लो ये क्वेश्चन कहीं नहीं गया हाथों-हाथ सॉल्व हो जाएगा अगर आप चीज सोच पाए तो पर सर अभी तो चीज बहुत कन्फ्यूजन सी लग रही हैं पता नहीं क्या करना है हमें कुछ क्लियर नहीं हो रहा है मेरा कहना है की आप पहले सबसे पहले सर्कल की इक्वेशन लिख सकते हो सर बिल्कुल लिख सकते हैं अच्छा सर्कल की इक्वेशन लिखने के साथ-साथ क्या आप यह जो लाइन है में इसकी इक्वेशन लिख सकते हो सर दोनों लिख सकते हैं आपको दोनों के स्टूडेंट्स अभी भी वक्त स्क्रीन पॉज करके ट्राई करिए आप एक क्वेश्चन कर लेंगे एक बात बताओ अगर इस सर्कल के सेंटर के अकॉर्डिंग तो क्या मुझे सीखने की जरूरत है की एम के जो कोऑर्डिनेट्स होंगे वह होंगे आरकॉम 0 और एन के जो कोऑर्डिनेट्स होंगे वो होंगे जीरो कमा आर मैंने आपसे दो बातें पूछी थी स्टूडेंट्स मुझे आप बताओ की सर्कल की इक्वेशन क्या होगी और इस लाइन में की इक्वेशन क्या होगी आप कहेंगे सर सर्कल की जो इक्वेशन होगी सेंटर के अकॉर्डिंग आर कॉम आर रेडियस भी आर तो सर्कल की इक्वेशन होगी एक्स - r² + ए - r² = r² इसे अगर आप सिंपलीफाई हो जाता है x² + r² - 2xr ये कितना हो जाता है ए स्क्वायर प्लस आर स्क्वायर इसे आप थोड़ा सिंपलीफाई कर ले तो सर्कल की इक्वेशन आपको क्या मिलती है suniyega सर एक आर स्क्वायर तो कैंसिल फिर क्या दिख रहा है मुझे एक मुझे दिख रहा है 2rx और एक मुझे दिख रहा है तू रे और एक मुझे दिख रहा है प्लस r² = 0 ये आपकी सर्कल की इक्वेशन आई है कहीं से कहीं तक कोई दिक्कत एक्स स्क्वायर + y² + 2X + 2y + r² बिल्कुल करेंगे इस बारे में इस बारे में बात करेंगे पर इससे भी पहले क्या मैं इस लाइन में की इक्वेशन निकल सकता हूं सर म तो आसान है एक सेंटर से आर हुआ इंटरसेप्ट आर इंटरसेक्ट फॉर्म तो में की इक्वेशन अगर मैं पूछूं आपसे है ना इक्वेशन ऑफ में अगर मैं पूछूं आपसे है ना बचपन में आप बचपन में लिखेंगे ऐसा ऐसा एग्जाम में नहीं लिखना है आपको बिल्कुल नहीं तो में की इक्वेशन को आप क्या कहोगे सर इंटरसेप्ट फॉर्म ये भी आर डिस्टेंस ये भी आर डिस्टेंस तो ये क्या हो जाएगा सभी हो जाएगा एक्स / आर + ए / आर = 1 तो आप इसे क्या लिखोगे एक्स + ए - आर = 0 स्ट्रेट लाइन की इक्वेशन मुझे बस यह बताओ स्टूडेंट्स मुझे बस यह बताओ अगर हमने सी के कोऑर्डिनेट्स माने द पीके अगर हमने माने से सीखे कोऑर्डिनेट्स पीके तो सर पीके से अगर मैंने परपेंडिकुलर ड्रॉप किया स्ट्रेट लाइन पर तो वो परपेंडिकुलर डिस्टेंस निकल जा सकती है अगर मैंने सी से इस स्ट्रेट लाइन पर परपेंडिकुलर ड्रॉप किया तो परपेंडिकुलर डिस्टेंस निकल जा सकती है बिल्कुल निकल जा सकती है तो हम वो निकलते हैं पीके से इस स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस हम निकलने वाले हैं सुन्नी ध्यान से सो पी 1 थॉट्स पी के सही यह जो मोड है यह फाइव टाइम्स अंडर रूट 2 अब क्या अब एक बात बताओ क्या सर हम यह जानते हैं की सर जो पी के है जो पी कमा क्यों है वह सर आपके इस सर्कल पर लाइक करता है और अगर वह सर्कल पर लाइक करते उसे सेटिस्फाई करेगा तो अगर पी कमा के रिप्लेस किया एक्स की जगह पी तो हो जाता है कुछ चीज शायद बैलेंस आउट करने के लिए क्या इस बात का कोई लॉजिकल कंक्लुजन आप सोच का रहे हो जो मैं कहना चाह रहा हूं आपसे क्या इस बात का कोई लॉजिकल क्वेश्चन आप देख का रहे हो जो मैं कहना चाह रहा हूं आपसे और सर ऐसे नहीं सोच का रहे हो तो एक कम करो क्योंकि हमारे पास जरूरत किसकी हो समझना हमें जरूरत है पी + के - आर की हमें जरूरत है आप प्लस तू पी के ही ऐड और सब्सट्रैक्ट करो मेरी बात फिर से suniyega अगर आप यहां पर कंफ्यूज हो रहे हैं तो मैंने एक्स और ए की जगह क्या रिप्लेस किया तो एक इक्वेशन बन रही थी बनते-बनते मुझे रिलाइज हुआ की सर p² + q² + r² - 2pq -2 मतलब अगर वो मिल जाता तो प्यार और कर भी मिल रहा है और शायद इससे मैं ये बना पाऊं शायद इससे बना पाऊं बिल्कुल बना पाओगे स्टूडेंट्स अब यही बात हम करना चाहे कम की बात है suniyega यही क्रैक से इस क्वेश्चन का ध्यान से देखो सोच के ये बताओ ध्यान से देखो और सोच के बस ये बताओ देखो जरा ध्यान से a² + b² + c² अगर आपका सी प्लस आर है तो सुनना का रहे हो तो ये टेक्निकल ये टेक्निकल किस एक्सप्रेशन का एक्सपेंशन है सर ये टेक्निकल पी + के - आर के स्क्वायर का एक्सपेंशन है ए + बी + सी का होल स्क्वायर लेकिन सी को मैं माइंस आर से ट्रीट कर रहा हूं तो उसने इन दो टर्म्स को नेगेटिव बना दिया बाकी सब पॉजिटिव रहेगी क्योंकि जहां-जहां ये मल्टीप्लाई वो पॉजिटिव ही रहा जहां जहां जाए इनके स्क्वायर हुए वो पॉजिटिव ही रहा बस जहां-जहां ए रहा है वहां वहां नेगेटिव हो गया बिल्कुल सही बात है सर और ये जो स्क्वायर है किसके इक्वल है सर ये इक्वल है तू पी के के सर आप क्या ज़ाहिर करना चाह रहे हो क्या फिगर आउट करना चाह रहे हो आप क्या निकलना चाह रहे हो हो क्या रहा है इससे ये हो रहा है की पी + के - आर मैं जानता हूं सर फाइव रूट तू है एल्डोस का मोड तो यहां पे स्क्वायर ही तो है तो स्क्वायर से मोड को हम सॉल्व कर सकते हैं मोड और स्क्वायर बहुत सॉर्ट इट से फंक्शंस हैं तो मैं यहां पर फाइव रूट तू रख सकता हूं क्योंकि चाहे इसका स्क्वायर मोड का स्क्वायर है एक ही बात होगी क्योंकि वो स्क्वायर है वो एनीवे पॉजिटिव बनाएगा तो मैं यहां पर क्या रखा हूं फाइव रूट तू यहां पर क्या रख रहा हूं लेफ्ट हैंड साइड पर ले आता हूं तू पी के जो की राइट हैंड साइड पर है और यहां रख रहा हूं पी + के - आर की जगह 5√2 जिसका हम कर रहे हैं स्क्वायर क्या किसी भी स्टूडेंट को यहां तक कोई परेशानी 2 पी के को एस इट इस रखा 5 का स्क्वायर 25 और √2 का स्क्वायर तू अब देख का रहे होंगे स्टूडेंट्स के तू से तू कैंसिल तो पी के आया 25 आप का रहे हो सर पिंटू के आया 25 तो आपने क्या कर लिया मेरा कहना है क्वेश्चन खत्म हो गया क्या बात कर रहे हो सर अभी कहां से क्वेश्चन खत्म हो गया हमें तो कुछ समझ ही नहीं ए रहा 1 मिनट थोड़ी शांति रखें इस रेक्टेंगल की लेंथ कितनी थी के और अगर रेक्टेंगल की लेंथ और विथ पी और क्यों है तो रेक्टेंगल का एरिया कितना होगा सर पिंटू जो की आया है 25 तो क्या माफ करेंगे ऑफ कोर्स एडवांस में इस तरह के क्वेश्चंस आएंगे पर उन्हें कैसे सोचना होगा क्या थॉट प्रक्रिया होगी क्या माइंडसेट होगा ऐसे इन चीजों को वर्कआउट हम करेंगे कोई तकलीफ किसी भी स्टूडेंट को यहां तक कोई भी डाउट है तो पूछ लो भाई वर्ण हम आगे बढ़ते हैं इसे एक बार अच्छे से समझ लो सर्कल इसका क्या मतलब होता है जरा वो समझने की कोशिश करते हैं suniyega ध्यान से अगर मुझे एक स्टैंडर्ड इक्वेशन दे दी जाए है ना जैसे मैं इसे हटा के होप आपको समझ ए रहा है एक स्टैंडर्ड इक्वेशन क्या है x² + y² + 2G एक्स + 2fy प्लस सी इसे इक्वल तू जीरो आपको इसको देख के कुछ याद ए रहा है है इससे जस्ट पहले वाला चैप्टर लाइंस के बारे में मैं आपको यह वहीं से लाकर देना चाह रहा हूं स्टूडेंट जब हम बात करते हैं बस इसमें एक चीज miscenge वो ह वाली टर्म जो तू ह एक्स वे दिया होता था ये अभी भी क्वाड्रेटिक है एक्स और ए में लेकिन यहां पर 2h5 नहीं दिख रहा है अब याद करो अगर आपके पास एक क्वेश्चन होती है जहां पर आप लिखते हो एक्स - X1 का होल्स स्क्वायर प्लस ए माइंस y1 का होल स्क्वायर इस इक्वल्स तू आर स्क्वायर अगर आप इस इक्वेशन को तोड़ मरोड़कर सिंपलीफाई करके देखो तो आपको कुछ इस तरह से टर्म्स देखेंगे आपको कुछ इस तरीके से टर्म्स दिखेंगे मतलब मैं टेक्निकल ये कनक्लूड करना चाह रहा हूं की ये जो इक्वेशन बन कर ए रही है ये इसी का एक सिंपलीफाइड या स्टैंडर्डाइज्ड वर्जन है जिसे हम अपनी लैंग्वेज में कहते हैं जनरल फॉर्म ऑफ सर्कल जनरल इक्वेशन ऑफ अन सर्कल इसका मतलब क्या है एक सेकंड जब आप इसे तोड़ के ठीक से लिखोगे ना जब आप इसे तोड़ के ठीक से लिखोगे ना तो आप इसे लिख पाओगे जब आप इसे परफेक्ट करने की कोशिश करोगे जब ये आप x² को देखोगे 2gx को देखोगे तो यहां पे आप g² भी लगा दोगे सिमिलरली जब आप y² को देखोगे और जब आप 2 फी को देखोगे तो आप f² भी लगा दोगे मैं जो कहना चाह रहा हूं समझता मैं फिर से इसे लिखता हूं देखना बहुत मजेदार सी बात है ये लिखा है x² है ना फिर लिखा था y² में यहां पे लिख देता हूं 2gx बस सुनना ऐसा क्यों कर रहा हूं फिर आपके पास क्या लिखा होता है y² फिर लिखा होगा क्या 2fy और प्लस ऑफ कोर्स सी लिखा हुआ था उसे सी को मैं राइट हैंड साइड पर लग जाता हूं इक्वेशन यहां पे लिख देता हूं माइंस सी ऐसा क्यों कर रहे हो थोड़ा शांति से सुनिए यहां पर मैं लिख देता हूं एक जी स्क्वायर ऐड कर देता हूं तो क्या यहां पर भी मुझे एक G2 करना होगा बिल्कुल करना होगा सर यहां पर मैं एक स्क्वायर ऐड कर देता करता हूं तो ऑफ कोर्स मुझे यहां पर थोड़ा सा शांत रहिए थोड़ा सा ध्यान से सुनिए एक बात बताओ स्टूडेंट्स अगर मैं इस एक्सप्रेशन को ध्यान से देखूं a² + b² + 2ab तो shailise ही ये सर एक्स + जी के होल स्क्वायर जैसा दिख रहा है a² + b² + 2ab तो सर ये ए + एफ के होल स्क्वायर जैसा दिख रहा है और अगर यहां पर मैं देखूं तो क्या में से लिख सकता हूं g² + x² - सी के अंडर रूट का स्क्वायर कैसे ही बातें कर रहे हो घुमा घुमा के बात वही कर रहे हो अंडर रूट ला के स्क्वायर लगा रहे हो कैंसिल हो जाएगा 1 मिनट थोड़ा सी की है सर्कल की की सर अगर सर्कल के सेंटर के कोऑर्डिनेटर X1 y1 और उसकी रेडियस आर तो सर्कल के इक्वेशन हम हमारी भाषा में कुछ ऐसी लिखते हैं हमने एक इक्वेशन उठाई हमने क्वेश्चन उठाई ऐसी कुछ हमने इस इक्वेशन को तोड़ मरोड़कर कुछ ऐसा बनाया और उसे कंपेयर किया हमारी सर्कल की स्टैंडर्ड इक्वेशन से और इस जड़ों ज़हर में हम यह निकल पाए किसे देखो भाई कंपेयर करो - X1 आया जीके बनाया एफ के इक्वल और r² आया इसके इक्वल यानी आप फाइनली अगर कंपेयर करें अगर आप फाइनली कंपेयर करें तो X1 y1 X1 और y1 जैसे आप कुछ नहीं सर्कल के सेंटर भी कहते द वो ए जाएंगे क्या वो ए जाएंगे सर माइंस जी कमा माइंस एफ ऑफ कोर्स माइंस एक्स वैन जी था -51 एफ था तो X1 क्या होगा -3 और y1 क्या होगा माइंस है सर जो आप बोला करते द आर सर्कल की क्या रेट है जो आप माना करते द आर सर्कल की रेडियस वो यहां पर क्या ए रही है सर वो ये ए रही है r² राइट तो आर क्या ए रहा है सर आर जो आपका ए रहा है अंडर रूट ओवर जी स्क्वायर प्लस एक्स स्क्वायर माइंस सी मतलब मैं बस इतनी सी बात कहना चाह रहा हूं अगर मुझे कोई सर्कल की स्टैंडर्ड जनरल एक्शन लिख कर दे दे क्योंकि जस्ट अभी हमने डिस्कस की x² + y² + कोस किया 2gx + 2x5 + सी = 0 तो मैं इसे देख कर ये आसानी से प्रिडिक्शंस दे सकता हूं सर की जो एक्स का कॉएफिशिएंट है और जो ए का कॉएफिशिएंट है उनके नेगेटिव हाफ जो होंगे वो होंगे आपके सेंटर के अकॉर्डिंग और उन टर्म्स के स्क्वायर उन जी और एफ के स्क्वायर के सब में से अगर आप सी सब्सट्रैक्ट कर देंगे तो आपको क्या मिल जाएगी रेडियो इस स्टैंडर्ड इक्वेशन से यह डिडक्शंस निकले जा सकते हैं क्या यहां तक किसी भी स्टेप में कोई दिक्कत आई होप एक बेसिक सा कंक्लुजन अभी तक हमने निकाला है जिसके लेके जिसको लेकर आपको कोई डाउट या परेशानी नहीं है सर इन सारी बातों को तो हम समझ पाए पर आप बातें क्या करना चाह रहे हो आई होप ये बात आप समझ का रहे हो की ऑफ कोर्स हम समझ पाए सर की ये जो है रेडियस है और अगर ये रेडियस है तो हम ये बात हमेशा कहेंगे की सर जी स्क्वायर प्लस x² - सी आपका हमेशा ग्रेजुएशन जीरो होना चाहिए तभी वो अंडर रूट के अंदर से कोई पॉजिटिव वैल्यू देगा आप बात समझ का रहे हो है और तभी जाकर वह क्या होगा सर तभी जाकर वह होगा रियल सर्कल और एक और कंक्लुजन की सर x² और Y2 के जो कॉफिशिएंट हैं वो कैसे होने चाहिए से और राजा री वुड से यूनिटी से भी रहे तो एक उसी वैल्यू से डिवाइड कर दूंगा तो x²y² के क्वेश्चन क्या हो जाएंगे वैन वैन तो इसलिए मैं का रहा हूं से आई होप यहां तक कोई डाउट नहीं यहां तक कोई परेशानी नहीं है अगर कोई परेशानी नहीं है तो कुछ कंक्लुजंस पर बात करते हैं suniyega ध्यान से बहुत कम की बातें हैं स्टूडेंट्स सुनते जाएगा एक जरूरी कंक्लुजन तो ये ऑफ कोर्स की हम जानते हैं की अगर आपकी रेडियस हो गई जीरो अगर आपके सर्कल की रेडियस होगी जीरो तो सर्कल की रेडियस जीरो मतलब एक पॉइंट सर्कल रेडियस जीरो मतलब लेंथ है ही नहीं ना मतलब रेडियस है ही नहीं मतलब सर्कल का सेंटर तो है पर रेडियस नहीं है तो बस वही सर्कल है मतलब वो सेंटर समझ का रहे हो आप जो मैं कहना चाह रहा हूं अच्छा सर इसके बाद क्या अगर इसके अलावा अगर सर्कल की रेडियस नेगेटिव आई तो मतलब वो अंडर रूट के अंदर वाला पार्ट नेगेटिव रेडियस तो नेगेटिव होगी नहीं अंडर रूट के अंदर वाला अंडर रूट g² + x² - सी तो √ जो होता है G2 + x² - अगर वो नेगेटिव हुआ तो उसका सीधी-सीधी सी बात है उसे केस में आपकी पास कोई रियल रेडियस ही नहीं वो इमेजिनरी रेडियस है और इमेजिनरी मतलब देयर इस नो रियल सर्कल ऐसा कोई रियल सर्कल है ही नहीं वहां पर फाइनल कंक्लुजन के सर हमने पैर ऑफ स्ट्रेट लाइंस के दौरान इस बारे में बात की थी की एक सेकंड ऑर्डर की यार आदर एक्स और ए के क्वाड्रेटिक इक्वेशन है इसमें पहली बात तो आपको ये टर्म्स का ना होना जरूरी है अगर एक सर्कल है तो दूसरी बात सर ये जो x²y² के कॉएफिशिएंट से वो से होना जरूरी है और तीसरी बात ऑफ कोर्स यह जब जीरो हो जाता था तो ये पैर ऑफ स्ट्रेट लाइन सरफेस करता था तो इसका जीरो नहीं होना जरूरी है लेस का जो डिटर्मिननेंट्स होता था कौन सा डिटर्मिननेंट्स स्टूडेंट्स क्या भूल गए हो हम बात करते आए हैं एबीसी और क्या ऑफ कोर्स आपका एफजी ये जो होते द आपके एफ जी और ह और यहां पे आप लिखते द फग और ह इसका जो डिटर्मिननेंट्स होगा वो जीरो नहीं होगा इन ऑर्डर तू मेक इट anequation ऑफ अन सर्कल बहुत कन्ज्यूरिंग तो नहीं हो रही है स्टूडेंट्स यह बातें आप इन्हें नोट कर रहे हो या नहीं इनसे चीज समझ का रहे हो या नहीं आई होप बड़ी बेसिक सॉल्यूशन बातें हैं सब अगर यहां से मैं आगे बढूं और आऊं अब सीधे एक पॉइंट पर तो ये रहा आज का आपके सामने पहला क्वेश्चन यह इसी कॉन्सेप्ट को कंसोलिडेटेड करने के लिए आपके सामने रखा गया है आप इसे अच्छे से ध्यान से देखेंगे पढ़ेंगे समझेंगे और खुद से ट्राई करेंगे आप कर सकते हैं मेरा यकीन करिए सर कैसे करेंगे बात समझा दो हमें अगर आपने कोशिश की आपसे नहीं हुआ तो इस क्वेश्चन को मैं समझा दे रहा हूं ध्यान से suniyega वो का रहा है की सर ये जो इक्वेशन अन रियल सर्कल ठीक है सर ये एक सर्कल रिप्रेजेंट करता है विद अन नॉन जीरो रेडियस दें फाइंड डी वैल्यू ऑफ ए सर मेरा यह कहना है की मैं इसे कंपेयर करूं है ना तो जी कितना आएगा सर एक्स के कॉएफिशिएंट का हाफ तो एक्स के कॉएफिशिएंट का नेगेटिव हाफ है ना ये है -2 -2 का नेगेटिव तू तू कहा वैन तो ये हो जाएगा वैन सर एफ कितना आएगा एफ होगा सर ए के कॉएफिशिएंट का नेगेटिव - ए और सर सी क्या आएगा वो कांस्टेंट टर्म जो बच जाती है तो सी क्या होगा सर सी जो होगा वो होगा ए + 3 अब उसने आपसे कहा उसने आपसे कहा की ये एक रियल सर्कल है जिसकी नॉन जीरो रेडियस है मैंने तो सर एक छोटी सी बात ये सीखी है की सर जो रेडियस होती है ना जो अंडर रूट के अंदर वाला पार्ट होता है उसमें G2 + x² - सी वो ग्रेटर दें जीरो होना चाहिए क्या बात आप समझ का रहे हो सर जी की वैल्यू वैन है तो यह हो जाएगा सर आपका a² है ना ये हो जाएगा -ए -3+1 कितना -2 ग्रेटर दें जीरो क्या हम स्प्लिटिंग नाम मिडिल टर्म उसे कर सकते हैं सर कोशिश कर सकते हैं तो अगर स्प्लिटिंग डी मिडिल टर्म की मैं बात करूं आपसे तो आप क्या लेकर ए रहे हो सॉल्यूशन सर प्रोडक्ट जो चाहिए वो हो जाना चाहिए वैन और माइंस तू यानी कितना -2 और जो सैम चाहिए वो हो जाना चाहिए -1 -2 प्लस वैन आई थिंक बिल्कुल सही तो आप इसे लिखोगे क्या बिल्कुल नहीं है सर सर यहां से अगर ए कॉमन लिया तो ये बन जाएगा ए - 2 यहां भी ए - 2 हो जाएगा ए + 1 > 0 सिर्फ सबसे अच्छा कम जीवन में जो हमें लगता है वो है व कर्व मेथड अभिले तो नहीं हो भाई ये हो जाता है आपका माइंस इंफिनिटी हो जाता है प्लस इंफिनिटी इनके बीच में जो नंबर्स मिलेंगे आपको माइंस वैन ऑफ कोर्स और एक क्या तू माइंस वैन और तू के बीच कौन आता है सर उनके बीच आता है जीरो इनके बीच आता है जीरो रखो तो नेगेटिव और ये पॉजिटिव जीरो माइंस तू माइंस तू जीरो प्लस वैन प्लस वैन नेगेटिव पॉजिटिव और 2 से इंफिनिटी तक कोई दिक्कत है suniyega ध्यान से हमें चाहिए एक्सप्रेशन पॉजिटिव नहीं था ना नहीं तो मैं फाइनल कंक्लुजन जो दूंगा की सर जो आपका ए होगा वो क्या होगा सर वो या तो हो माइंस इंफिनिटी से माइंस वैन तक या फिर यानी यूनियन कहां से कहां तक तू से इंफिनिटी तक क्या यह पूरा क्वेश्चन ये पूरा सॉल्यूशन ये सॉर्टेड सा एक्सप्लेनेशन आप सभी को समझ आया और एक अच्छे अंडरस्टैंडिंग आपको क्लियर हो रही है की किस तरीके से चीज सोचनी होगी आपको एग्जामिनेशन हॉल में या ऑफ कोर्स एक जी मांस लेवल पे पूछा जा सकता है किसने किसी फॉर्म में किसी ना किसी तरीके में किसी क्वेश्चन में इंपोर्ट करके इस तरह के कॉन्सेप्ट्स को जरूर डाला जाता है जहां पे ऑफ कोर्स आपकी क्या हो रहे होंगे आपका सर्कल का भी नॉलेज उसे हो गया आपकी क्वाड्रेटिक की भी नॉलेज उसे हो गई और आपकी ऑफ कोर्स इन इक्वेशंस की भी मेथड उसे होगी तो यहां पे तीन चैप्टर में समाहित के लिए सर्कल क्वाड्रेटिक और inquations और यही आईआईटी एडवांस एग्जाम है जहां पर चैप्टर नहीं आएंगे क्योंकि इससे चैप्टर इसका एक क्वेश्चन वो कंप्रिहेंसिव नॉलेज आपका चेक करेगा वो इंडिप्थ आपका नॉलेज चेक करेगा जैसे अगर मैं इस क्वेश्चन की बात करूं तो वो क्या लिख रहा है suniyega ध्यान फाइंड डी इमेज ऑफ डी सर्कल इमेज ऑफ डी सर्कल इन दिस लाइन तो टेक्निकल कहने को एक क्वेश्चन बहुत अजीब लग सकता है की सर देयर इस अन सर्कल देयर सर्कल और सर इस सर्कल की इमेज बनानी है सर्कल की इमेज के क्लाइंट में चाहिए स्लैंग के रिस्पेक्ट में तो कैसे ढूंढेंगे सर एक-एक पॉइंट की इमेज निकालो क्या नहीं नहीं साहब ऐसा नहीं करना है एक बात बताओ सारी बातों से पर एक बहुत जरूरी बात आप सबको दिख रही होगी की सर अगर इस सर्कल का कोई सेंटर है और सर्कल का कोई सेंटर है तो कहना है इसे सेंटर भी तो मिरर होगा एक बात जो मुझे बात समझ ए रही है की सर ऐसा नहीं होगा क्या की इसका जो सेंटर है वो सेंटर यहां पे मेरा रो रहा होगा मतलब सेंटर की जो इससे परपेंडिकुलर डिस्टेंस है और इसकी जो इससे परपेंडिकुलर डिस्टेंस है वो से होगी ना आई होप आपको ये बात समझ आई दूसरी बात जो फिक्स होगी सर की अगर सर्कल मिरर हो रहा है तो सर्कल की जो रेडियस है वो भी इसी सर्कल की रेडियस भी तो होगी यह क्वेश्चन यहीं पर खत्म हो जाता है कैसे सर सुनना ध्यान से एक्स स्क्वायर प्लस ए स्क्वायर -2 का हाफ -1 उसका नेगेटिव वैन फोर का हाफ तू तू का नेगेटिव -2 सो वैन कमा तू आपके क्या हो जाएंगे ये हो जाएंगे आपके सर्कल के सेंटर के कोड है सर्कल की रेडियस निकल सकता हूं क्या सर निकलती हैं कोशिश करते हैं सर्कल की अगर मैंने रेडियस निकालनी चाहे तो वो क्या होती है सर वो होती है जो रेडियस होती है सर अभी जस्ट हमने पढ़ा अंडर रूट ओवर G2 मतलब वैन का स्क्वायर माइंस वैन का स्क्वायर एक ही बात है प्लस एक्स स्क्वायर जो किसी का स्क्वायर जो की क्या हो जाएगा फोर माइंस सी माइंस माइंस फोर यानी प्लस ये हो जाएगा +4 है ना और ऐसे में ध्यान से देखो तो ये हो जा रहा है 99 का रूट थ्री तो ये हमारा ये वाला सर्कल है पर मुझे तो इस सर्कल के क्वेश्चन चाहिए जो की इसकी इमेज है मेरा बस आपसे कहना है सर क्या वैन कमा - 2 की इस लाइन के रिस्पेक्ट में मिरर इमेज निकल जा सकती है याद करो भाई अगर मैं इमेज निकलना चाह रहा हूं तो मैं क्या लिखूंगा जो इसके सेंटर के कोऑर्डिनेटर जिन्हें मैं कहता हूं ह कॉम के तो क्या कहूंगा भाई ह- डिवाइडेड बाय इस लाइन में एक्स का कॉएफिशिएंट तू इसे इक्वल तू है ना सिमिलरली के माइंस माइंस थ्री और क्या करेंगे सर अब आप पास कर देंगे वैल्यूज यानी की क्या 1 2 + - 2 -3 थॉट्स कितना 6 + ऑफकोर्स क्या फाइव और इसका आप नेगेटिव डबल लेंगे क्योंकि आप मिरर इमेज निकल रहे हो तो इसका आप नेगेटिव डबल हो गया राइट तो मैं इसको थोड़ा सा यहां लिख देता हूं ह - 1 / 2 को मैं थोड़ा देर शिफ्ट कर देता हूं स्टूडेंट्स है ना ये आपका ह माइंस 1 / 2 है और ये इसके नेगेटिव डबल के इक्वल है और डिवीज़न में डिनॉमिनेटर में अंडर रूट नहीं लेना है और डिस्टेंस फॉर्मूला होता है जब आप मिरर इमेज या फीट के अकॉर्डिंग रहते हो जैसे स्क्वायर इसका सैम करते हैं तू का स्क्वायर नाइन प्लस फोर कितना थर्टीन तो यह ए जाएगा कितना डायरेक्ट पंच 11 और दो 13 30 से 13 डिवाइड हुआ कैंसिल तो यहां पे क्या बच जा रहा है -2 अब सुनना ध्यान से आपको क्या मिल रहा है आपको मिल रहा है सर ह माइंस वैन डिवाइडेड बाय तू इसे इक्वल तू -2 = व्हाट इसे इक्वल तू के प्लस तू डिवाइडेड बाय -3 अब आप यहां से ह कमा के निकालिए कैसे निकलेंगे सर ह कमा कज अब निकालो तो ह निकलने के लिए -4 - 4 + 1 कितना -3 सिमिलरली ध्यान से देखना स्टूडेंट्स ये कितना हो रहा है प्लस सिक्स में से तू सब्सट्रैक्ट किया तो कितना फोर तो के की वैल्यू कितनी ए रही है 4 तो इसके जो cardinate हैं वो कितने हैं -3 कमा 4 और रेडियस तो वापस वही थ्री रहेगी ना रेडियस के परफेक्ट होगी तो रेडियस तो अभी भी थ्री रहेगी तो सर आपको एक ऐसे सर्कल की इक्वेशन बनानी है जिसके सेंटर के कोऑर्डिनेट्स और -3 कमा फोर और उसकी जो रेडियस है वो थ्री आई थिंक बहुत टास्क नहीं है सर बना सकते हैं sochiyega एक्स - 3 यानी एक्स + 3 तो होल स्क्वायर प्लस ए - 4 = 3² आप चाहें तो इसे सिंपलीफाई कर लीजिए अगर ऑप्शंस में ये ना दिया हो तो है ना तो आप क्या लिखेंगे आप लिखेंगे x² + 9 + कोस 6x प्लस क्या लिखेंगे y² + 16 - 8y = 9 और जो आपने इसे सिंपलीफाई करना चाहा तो आप देखोगे की सर देखो भाई ये नाइन से नाइन कैंसिल तो सर फाइनली जो इक्वेशन बन के ए रही है वो क्या होगी सर वो होगी x² + y² + 6 है एन - 8y + 16 = 0 एंड डेट यू आर करेक्ट आंसर मतलब ऐसा नहीं है की एक-एक पॉइंट की आपको मिरर इमेज निकालनी है इतना नहीं सोचता है जो बातें हैं वो लॉजिकल सोचनी है की सर इस सर्कल का अगर कुछ मिरर हो रहा है तो वो सेंटर भी तो होगा रेडियस वही रहने वाली है बस ये लॉजिक है और ये लॉजिक बहुत मजेदार होते हैं इन्हीं से क्वेश्चंस सॉल्व थोड़ा आगे बढ़ते हैं कुछ और चीज डिस्कस करते हैं और मूव करते हैं आज के नेक्स्ट क्वेश्चन की तरफ इस क्वेश्चन को अच्छे से देखिए अच्छे से समझ लीजिए और अच्छे से स्टडी करिए क्योंकि यह क्वेश्चन बहुत मजेदार है मैं चाह रहा हूं की इसे आप एक बार खुद से ट्राई कर ले पहले ट्राई किया स्टूडेंट्स आपने छोड़ो मुझे आप एक बात बताओ सर्कल की इक्वेशन क्या पड़ी है सर वो है स्क्वायर मैन लेता हूं मैं एक पॉइंट मैन लेता हूं लेट्स सर एम कमा 1 / एम मेक पॉइंट मैन लेता हूं लेट्स से एम वैन बाय एम एक पॉइंट है जो इस सर्कल पर लाइक करते हैं तो इसको सेटिस्फाई करेगा तो वो क्या हो जाएगा सर तो हो जाएगा m² + 1/2gm है ना प्लस ऑफ कोर्स 2f / एम + सी = 0 सेट डिनॉमिनेटर में m² अच्छा नहीं लग रहा है तो क्या पुरी इक्वेशन को m² से मल्टीप्लाई कर दो कर दो सर तो जैसे ही किया तो ये हो जाएगा एन की पावर 4 प्लस अब कोस कितना वैन प्लस कितना है स्टूडेंट्स प्लस ये हो जाएगा कितना m² से मल्टीप्लाई किया ना तो ये हो जाएगा 2G m³ यहां पर मल्टीप्लाई किया तो कुछ गड़बड़ कर रहा हूं क्या नहीं है ना नहीं कर रहा हूं तो फिर ये कितना हो जाएगा 2fm और ये हो जाएगा सी m² क्या कोई डाउट देखो फिर से बोलता हूं एम स्क्वायर बिल्कुल सही सर वैसे थोड़ा री राइट करना चाह रहा हूं ऐसे एक पॉलिनॉमियल की तरह रिवाइट करना चाह रहा हूं तो सुनना शुरुआत होगी एम तू डी पावर फोर्सेस शुरुआत होगी एम तू डी पावर फॉर से फिर आपको क्या देखा मैं क्यूबिक टर्म ले रहा हूं एम की पावर में एक्स प्रेस कर रहा हूं फिर आता है सर 2G m³ है ना फिर क्या आता है सर 2gm³ के बाद cm² और 2fm तो ये आता है cm² और फिर क्या आता है 2fm और फिर क्या आता है सर फिर आता है आपका आखिर में प्लस वैन फिर क्या आता है आखिर में +1=0 आप बस एक बात का जवाब दो अगर आपको एक बेसिक अंडरस्टैंडिंग है अगर आपको एक बेसिक अंडरस्टैंडिंग है इस चीज की सर देखो अगर ये एम में पॉलिनॉमियल है जिसकी डिग्री है 4 तो इसके मैक्सिमम चार रूट्स हो सकते हैं जिम मैन लेता हूं m1 M2 M3 में एक बात सीखी है कहीं ना कहीं की सैम ऑफ डी रूट्स टेकन वैन टाइम इसका क्या मतलब होगा इसका मतलब होगा m1 M2 फिर क्या m1 M3 फिर क्या सर m1 M4 फिर आप मल्टीप्लाई करो M2 को तो M2 M3 फिर क्या सर M2 M4 आई हो बिल्कुल सही फिर आप मल्टीप्लाई करो M3 को तो M3 को M4 से मल्टीप्लाई किया यह सारे रूट्स इसके अलावा मेरे ख्याल से कोई और पेड़ नहीं बन सकता m1 के बने तू थ्री फोर तू के बने थ्री और फोर थ्री का वैन फोर जो भी पैर बन्ना है बन गया तो सैम ऑफ डी रूट्स टेकन इन डी टाइम ये जो बनेगा स्टूडेंट्स ये बनेगा यहां से शुरू करते हैं हम दिस अपॉन दिस विद डी नेगेटिव साइन फिर यहां शिफ्ट कर तो सी / 1 विद्या पॉजिटिव साइन ये नेगेटिव था तो ये वाला पॉजिटिव अब अगर मैं बात करूं सैम ऑफ डी रूट्स टेकन 38 ए टाइम टेकन 38 ए टाइम मतलब m1 का M2 और M3 के साथ प्रोडक्ट फिर क्या सर फिर सर m1 का m1 का M2 फिक्स रखा और क्या कर दिया M4 लिया है कोई दिक्कत तो नहीं स्टूडेंट्स आई होप ये बातें आपको समझ ए रही है मैंने फिर क्या किया m1 और M2 को फिक्स किया थ्री और फोर ले आए ऐसे करते-करते सारे रूट्स जितने भी थ्री थ्री के पैर बन सकते हैं जितने भी बन सकते हैं उन सबके मतलब सैम ऑफ डी रूट्स टेकन 3 टाइम्स जैसे यहां दो बने द ऐसे थ्री वाले इन सब का जो समय किया था दिस अपॉन दिस नेगेटिव साइन दिस अपॉन दिस पॉजिटिव साइन तो ये जो होगा वापस क्या होगा नेगेटिव साइन के साथ तो यहां पर ए जाएगा आपका -2f अपॉन ऑफ कोर्स वैन फाइनली मैं बात करूं तो सैम ऑफ डी रूट्स टेकन फोर आते टाइम तो चार ही तो रूट्स हैं सर कौन-कौन से m1 M2 M3 और M4 तो सैम ऑफ डी रूट्स टेकन 4 ए टाइम क्या होगा स्टूडेंट्स वो होगा 2f के बाद अगली टाइम क्या बचेगा कांस्टेंट हम तो 1 / 1 ये नेगेटिव साइन था तो वैन के आगे क्या आएगा पॉजिटिव साइन तो आप कनक्लूड करोगे सर की अगर यहां पर इस इक्वेशन के इस इक्वेशन के कितने रूट्स होते हैं चार रूट्स तो इन चारों का प्रोडक्ट कितना होता है वैन मेरा ये कहना है ये आपके कम की बात नहीं थी ये तो बस आपको सिखाई है की ऐसा पता होना चाहिए ऐसा होता है है ना ये हम बात करेंगे कब जब हम क्वाड्रेटिक पढ़ेंगे तब भी हम डिटेल में डिस्कस करेंगे बट याद रखिएगा इस polynom इस तरीके से बिहेवियर शो होता है क्या बात याद रहेगी स्टूडेंट्स अच्छे से बात बस इतनी सी है बात समझने की कोशिश करना अगर आप इस पार्ट में कंफ्यूज हो रहे हो तो ऐसा मैन लो मुझे कोई कहे ए एक्स कब प्लस bx² + बेटा + गामा भी बता सकते हो आप सैम ऑफ डी रूट्स टेकन 2 आते ए टाइम अल्फा बेटा गामा और गामा अल्फा भी बता सकते हो आप सैम ऑफ डी रूट्स टेकन ऑल थ्री टाइम भी बता सकते हो सर ऐसे कैसे बता सकते हैं suniyega अब शुरू करिए इससे तो नेगेटिव साइन के साथ शुरू करिए तो ये होता है -बी/ए आप जब उसे पर आते हैं तो स्पष्ट हो जाइए बट सिन चेंज कर दीजिए नेगेटिव का तो पॉजिटिव जब आप इन पर आए हैं तो आप यहां पर ए जाइए और ये क्या हो जाएगा अगर ये पॉजिटिव दी अपॉन है इस तरीके से चीज चलती हैं बस यही लर्निंग मैंने यहां पर इसमें अप्लाई की है और अगर इसमें मैंने अप्लाई की है तो आप ध्यान से देखो वो आपसे यही बोल रहा था की m1 M2 M3 और M4 की वैल्यू निकालो जो की कितनी निकलकर ए रही है वैन डिड यू आर अंडरस्टैंड हो इजी थी और कैसे इन्हें सोचा जा सकता है इफ यू नो डी बेसिक अगर आपने हर एक चीज अच्छे से पढ़िए और क्लीयरली ये क्वेश्चन ना सिर्फ ना सिर्फ सर्कल का है बल्कि इस में क्या-क्या हो गया सर सर्कल तो उसे हुआ ही हुआ इसमें वापस क्या उसे हो गया क्वाड्रेटिक जहां पे आप इक्वेशंस का नॉलेज उसे करते हो जहां पे आप पॉलिनॉमियल्स की अंडरस्टैंडिंग उसे करते हो जब आप रूट से खेलना सीखते हो इक्वेशन के रूट्स में क्या रिलेशन होता है सैम ऑफ डी रूट्स टेकन इट ए टाइम्स सैम ऑफ डी रूट्स टेकन तू आते ए टाइम ऐसे इनमें क्या क्या रिलेशन होता है उसके बारे में हमने बातें की तो एक बेहतरीन एक्सेस और ए एक्सिस पर एक सर्कल इंटरसेप्ट बना रहा है तो क्या होता है बात समझना जैसे की ऑफ कोर्स मैन लो ये आपका एक्सेस है और ये लेट से आपका एक्स एक्सिस है बहुत ध्यान से suniyega कम की बात है और लेट से आपके सर्कल ने एक्स और ए एक्सिस पर कुछ इंटरेस्ट है जैसे मैन लो ये आपको समझने के लिए मैं कुछ ऐसा बनाता हूं आपको रिलाइज हो रहा है की आप दिख रहा है स्टूडेंट्स में किसकी बात कर रहा हूं मैं बात कर रहा हूं आपसे ये वाला जोन जो है ये ऑफ कोर्स आपका कौन सा इंटरसेप्ट है ये आपका एक्स इंटरसेप्ट ये जो डिस्टेंस है सिमिलरली यह और यह पॉइंट्स के बीच की जो डिस्टेंस है यह कंसंट्रेशन सर क्लीयरली यह ए-एक्सिस वाला इंटरसेप्ट है मेरा कहना है अगर आपको याद होगा स्टूडेंट्स हम जनरल इक्वेशन ऑफ सर्कल क्या लिखते हैं सर हम लिखते हैं x² + y² + 2gx + 2fy + सी = 0 बड़ी सिंपल सी बात है की सर अगर मैं इसे एक्स-एक्सिस के साथ सॉल्व करो एक्स एक्सिस के साथ सॉल्व करना मतलब एक्स एक्सेस की इक्वेशन क्या होती है सर एक्स-एक्सिस हमारे लिए है एक्स स्ट्रेट लाइन और इसकी इक्वेशन क्या होती है सर इसकी इक्वेशन होती है ए = 0 सो ए = 0 इन दिस इक्वेशन आप खुद इमेजिन करो या सोचो या विजुलाइज करो की ये एलिमिनेट हो जाएगा और ये एलिमिनेट हो जाएगा तो आपके पास टेक्निकल टेक्निकल एक्स में क्वाड्रेटिक बचेगी उससे आपकी एक्स की दो वैल्यूज हो जाएंगी वो जो दो वैल्यूज आएंगे वो क्लीयरली ये वाली दो वैल्यूज आएंगे डू यू ऑल गेट डेट क्या आपको ये बात समझ ए रही है सर ऐसा भी तो हो सकता है क्वाड्रेटिक के रियल रूट्स ना हो मतलब नहीं बन रहा है सर ऐसा भी तो हो सकता है क्वाड्रेटिक के इक्वल और रियल रूट्स मतलब वो एक्स एक्सिस को एक ही पॉइंट पर टच कर रहा है और ऑफ कोर्स अगर दो पॉइंट्स मिले मतलब इस तरीके से कुछ है और जैसे ही यह दो पॉइंट्स मिलेंगे एक्स इंटरसेक्ट की लेंथ निकल सकता हूं बिल्कुल निकल सकता हूं क्या यही प्रक्रिया उसे करके आप सोच के देखो ए एक्सिस भी तो एक स्ट्रेट लाइन है जिसकी इक्वेशन हमारे लिए क्या होती है एक्स = 0 जैसे ही एक्स = 0 रखेंगे तो ये ये दोनों टर्म्स कैंसिल तो ये फिर ए में एक क्वाड्रेटिक बन जाएगा और उसको क्वाड्रेटिक को सॉल्व करेंगे तो आपको ये दोनों पॉइंट्स के इसके और इसके यू नो koardinate मिल जाएंगे और जैसे यह मिल जाएंगे डिस्टेंस फॉर्मूला लगाकर आप इंटरसेक्ट भी निकल सकते हो यह था मॉडल ऑपरेटिंग हो क्या चीज कैसे हो रही होगी अगर मैं एक्सप्रेशन विजय algebraicly बात करूं तो हम सर समझते आए हैं की अगर आपके पास एक सर्कल है तो सर्कल की इक्वेशन आप ये कहते हो आई होप आपको ये बातें याद है अब मेरा कहना है की सर एक बार एक्स की जगह जीरो और एक बार ए की जगह मैं जीरो रखता हूं तो ए की जगह जीरो रखने पर एक्स में क्वाड्रेटिक बनती है क्वाड्रेटिक सॉल्व करते हैं तो यह हमें दो रूट्स मिलते हैं प्लस साइन के साथ एक माइंस साइन के साथ और जब आप इसके साथ डिस्टेंस फॉर्मूला लगाते हो तो फाइनली आपको जो एक्स इंटरसेक्ट की लेंथ मिलती है वह होती है तू टाइम्स अंडर रूट ओवर जी स्क्वायर माइंस सी इस फॉर्मूले को नोट डाउन कर लीजिएगा स्टूडेंट्स इसी पर बात बनेगी सिमिलरली जब आप ए एक्सिस के इंटरसेप्ट की बात करते हो से वही बात एक्स की जगह जीरो रखना वही में क्वाड्रेटिक बन्ना फिर वही आपके दो पॉइंट्स आना फिर डिस्टेंस फॉर्मूला लगाना चीज बिल्कुल सिमिलर होंगी और उसे केस में जो वही इंटरसेक्ट आपको लेंथ देगा वो होगा तू टाइम्स अंडर रूट ओवर f² - सी वीडियो ऑल गेट डेट क्या यहां तक कोई भी डाउट किसी भी चीज को लेकर कोई परेशानी स्टूडेंट्स आई थिंक अभी तक तो नहीं है और अगर अभी तक नहीं है तो क्या आप सीधे हम बात करें क्वेश्चन की बहुत जल्दी क्वेश्चन नहीं लिया है बिल्कुल ज्यादा कुछ समझने जैसा है नहीं और अगर आपको समझना भी है तो इस क्वेश्चंस से आप बहुत अच्छे से सुनेंगे फॉर्मूले दो बड़े कृष्ण से हैं जो याद रखने हैं आपको ये की याद रखिएगा जो आपका एक सेंटर शिफ्ट होता है उसकी लेंथ ये और वही इंटरसेप्ट जो होता है उसकी लेंथ ये दोनों फॉर्मूले काफी क्रोशिया हैं जिन्हें आपको नहीं भूलना है तो ठीक है सर एक क्वेश्चन ही ट्राई कर लेते हैं क्या लिखा है पढ़ते हैं फाइंड डी लेंथ ऑफ इंटरसेक्ट विद डी सर्कल विच दिस मैक्स ऑन डी एक्सेस नौ देर इस डी फॉर्मूला जो रत्न की आपको बिल्कुल भी आवश्यकता नहीं है वो क्या पूछ रहा है लेंथ ऑफ इंटरसेक्ट विद डी सर्कल मैक्स ऑन डी एक्सेस फॉर्मूला क्या है सर फॉर्मूला है तू टाइम्स अंडर रूट हो जाएगा तू टाइम्स अंडर रूट ओवर x² की लेंथ हो जाएगी जो की आप उसे कर सकें मैन लो अगर मैं एग्जाम में फॉर्मूला भूल जाऊं तो मैं कैसे करूंगा इन केस अगर आप भूल जाओ तो वर्ण ये फॉर्मूला आपके पास हमेशा है सर अगर मैं फॉर्मूला भूल जाता हूं तो मैं क्या करूंगा उसे suniyega सर्कल की इक्वेशन पहले मैं ए = 0 रखता हूं अगर ए = 0 पास किया तो देखो इक्वेशन क्या मिल जाएगी सर इक्वेशन मिलेगी x² ध्यान से suniyega प्लस तन एक्स आई होप आपको समझ ए रहा है प्लस 9 आई होप कोई डाउट नहीं सर अगर ऐसे में क्वाड्रेटिक के तरह ट्रीट करूं तो स्प्लिटिंग डी मिडिल टर्म पे बात की जाए तो सर ये क्या हो जाएगा x² + 9x + एक्स + 9 = 0 कोई तकलीफ तो नहीं है स्टूडेंट्स और जरा थोड़ा ध्यान से देखिएगा सर यहां से आप फैक्ट्रीज करोगे तो ये मिलेगा एक्स + 9X आप कॉमन ले लोग वहां कॉमन लोग तो ये हो जाएगा एक्स + 1 तो जो आपके दो रूट्स आएंगे रूट्स आएंगे वो क्या होंगे -1 ए जाएगा मतलब -1 मिलेगा दूसरा पॉइंट क्या मिलेगा -9 कमा जीरो क्योंकि दूसरा रूट क्या मिलेगा -9 और चूंकि एक्स एक्सिस पर ही है तो इनके बीच के डिस्टेंस क्या होगी आई थिंक डिस्टेंस फॉर्मूला भी उसे मत करिए स्टूडेंट सीधी-सीधी सी बात है -9 -1 यानी -8 डिस्टेंस है तो क्या हो जाएगा तो जो एक्स इंटरसेक्ट बनेगा ना वह क्या होगा प्लस आते इतनी सी बात है मेरा यकीन करिए इतनी सी बात है और अगर आपको यह नहीं करना है इतनी बुरी प्रक्रिया से नहीं गुजरना है तो फॉर्मूला याद कर लीजिए फॉर्मूला क्या है सर ध्यान से देखो तू टाइम्स √ ओवर g² नौ व्हाट इसे दी ये होता है आपका 2gx ये होता है 2gx तो देखो ना आप 10 का हाफ कितना होता है 5 तो 5 का स्क्वायर 5 का स्क्वायर कितना 25 - सी - सी यानी -9 और 25 - 9 कितना होगा ज़रा सोच के होता है स्टूडेंट्स सो 25 - 9 होता है 16 सॉरी बोल रहा हूं क्या 25 - 9 होता है 16 का अंडर रूट फोर फोर इन तू वापस कितना आते कोई बात समझ आई कोई डाउट स्टूडेंट्स कोई परेशानी यहां तक अच्छा सर ये तो रहा आपका एक्स इंटरसेक्ट क्या ऐसे ही हुआ इंटरसेप्ट वही बात ना वापस आप निकल लो अगर विंड निकलना है तो क्या होगा सर तू टाइम्स अंडर रूट ओवर x² बताओ है -3 -3 का स्क्वायर कितना सर वो हो जाएगा 9 क्या आपको ये बात समझ ए रही है और नाइन में से अगर नाइन सूत्र क्या है रिपीट माय स्टेटमेंट नाइन मिस नाइन सब्सट्रैक्ट है तो कितना हो रहा है जीरो और जीरो इन तू जीरो तो इसका वही इंटरसेक्ट वैन ही नहीं रहा है इसके वही इंटरसेक्ट की जो लेंथ है वो है जीरो क्यों हैं इसको समझने की कोशिश करते हैं देखो ध्यान से अगर मैं इसमें अगर मैं आपको इसकी इंटरनल मैं मेकैनिज्म समझने की कोशिश करो तो इस इक्वेशन में कम करते हैं एक्स की जगह जीरो रखते हैं जैसे ही एक्स की जगह जीरो रखा तो क्या बन जा रहा है suniyega गौर से टूल्स एक्स की जगह जीरो रखते ही क्वाड्रेटिक बनती है Y2 है ना -6y+9=0 अब गौर फरमाइए स्टूडेंट्स अगर मैं आपसे एक बात पूछूं ये तो है y² है ना प्लस नाइन को क्या मैं प्लस थ्री स्क्वायर लिख सकता हूं और ये जो -6y है कैसे मैं -2 3 ए = 0 लिख सकता हूं ध्यान से देखो स्टूडेंट्स ए स्क्वायर प्लस बी स्क्वायर माइंस बी का स्क्वायर तो ये हो जाता है ए - 3² मतलब ए इंटरसेक्ट के लिए जो आपको हेल्प मिल रही है वो क्या मिल रही है ए - 3² = 0 मतलब ए की वैल्यू कितनी है थ्री मतलब इसके दो अलग-अलग रूट्स नहीं ए रहे हैं मतलब ये सर्कल टेक्निकल जो कर रहा है ना ये कुछ ऐसा सर्कल बन रहा है अगर मैं आपको ये सर्कल प्लॉट करके ड्रॉ करके डन तो सर्कल वायरस इसको टच कर रहे हैं और एक बात बताओ अगर ये वैक्स को टच कर रहा है तो ए इंटरसेक्ट नहीं बन रहा है इसलिए उसकी लेंथ क्या ए रही थी जीरो बहुत दिक्कत बहुत परेशानी आप चाहे तो फॉर्मूले से निकल लीजिए आप चाहे तो इस तरीके से निकल लीजिए दोनों तरीके आपके पास है हमेशा मेरा मोटिव यही रहेगा की मैं आपको इस तरीके से चीज सिखाऊ की आपको कन्वेंशनल वो ट्रेडिशनल एप्रोच याद रखें की ओरिजिनल कहां से वाइफ फॉर्मूला क्योंकि आईआईटी जी एग्जाम नहीं है जो फॉर्मूला ratvakar आपसे आपके रत्न की काबिलियत चेक करें यह एग्जाम है जो देखेंगे की आपने कॉन्सेप्ट्स कितने अच्छे से समझे आपको चीज बारीकियां में पता है या नहीं अपने हर कॉन्सेप्ट को सिखाया नहीं उसके पीछे के क्यों को आप जानते हो या नहीं ये बातें याद रखेंगे चलिए थोड़ा आगे बढ़ते हैं स्टूडेंट्स और अगले कॉन्सेप्ट की तरफ बढ़ते हैं जहां पर देखते हैं एक और क्वेश्चन पर यह जरा इस क्वेश्चन को क्या लिखा हुआ है सर अगर इस क्वेश्चन को समझे इफ डी इंटरसेक्ट ऑफ डी वेरिएबल सर्कल ऑन व्हाट ऑन एक्स एक्सिस एंड ए एक्सिस आर तू यूनिट्स एंड फोर यूनिट suspectivali फाइंड डी लोकस ऑफ डी सेंटर ऑफ डी वेरिएबल सर्कल मेरा यकीन करिए एक बहुत ही आसान बहुत ही बेसिक बहुत ही सिंपल सा एक क्वेश्चन है जरा सोच के देखो आप तो एक बात बताओ क्या हमने एक इक्वेशन पड़ी है सर इक्वेशन क्या पड़ी है x² + y² + 2G एक्स + 2fy प्लस तू जीरो बिल्कुल पड़ी है सर अच्छा मुझे बस इतना बताओ आपके एक सेंटर सेट की लेंथ क्या होती है सर वो होती है तू टाइम्स अंडर रूट कितनी है सर एक्स सेंटर सबकी लेंथ है तू से तू कैंसिल कितना वैन तो यहां से क्या मैं का सकता हूं की G2 - सी कितना हो जाएगा वैन दोनों तरफ स्क्वायर कर दिया सिमिलरली हमने सर यह भी सिखा है की तू टाइम्स अंडर रूट ओवर x² - सी ये कितना होता है सर ये होता है 4 यूनिट ठीक है सर 24 कितना तू टाइम्स स्क्वायर किया तो कितना ए जाएगा तो ए जाएगा सर f2 - सी आई होप समझ रहे हो 2 का 2 टाइम्स और तू का स्क्वायर 4 तो ये चीज कुछ ऐसी ए जाएगी इन सब से क्या अचीव कर लेना चाह रहे हो अब सर अब क्या चाह रहे हो मुझे कुछ समझ नहीं ए रहा मेरा बस आपसे ये कहना है थोड़ा सा पेशेंस रखिए और मुझे ये बताइए अगर मैं इसमें से इसको सब्सट्रैक्ट करता हूं तो देखो क्या मिलेगा मुझे मिलेगा सर ध्यान से देखना अगर मैं इसमें से इसको सब्सट्रैक्ट करता हूं तो मुझे मिलेगा G2 अब बात समझ का रहे हो ऐसा क्यों सर ऐसा इसलिए होगा क्योंकि माइंस सी से माइंस सी कैंसिल और 1 - 4 कितना हो जाएगा -3 कोई डाउट्स स्टूडेंट्स इनके अगर आप कोई -3 अच्छा नहीं लग रहा है थोड़ी देर के लिए मैन लेते हैं तो आप इसे ऐसा लिख लो x² - g² = 3 कोई डाउट छोटा सा कम नहीं कर सकते हैं क्या बस ऐसे ही मेरा मैन हो रहा है सर इस सर्कल के सेंटर के कार्ड जो है मिनर चाहिए मैन लेते हैं और आखिर में उन्हें एक्स रिप्लेस कर देते हैं तो इसके सेंटर के कोऑर्डिनेट्स होते हैं तो आपके लॉकर्स की इक्वेशन जो होगी y² - एक्स स्क्वायर कॉन्सेप्ट की लेंथ का फॉर्मूला है क्योंकि जितना जल्दी से याद कर लोग उतना जल्दी आप क्वेश्चंस कर पाओगे और इसी बात को हम कहते हैं एफिशिएंसी या स्मार्ट वर्क या टाइम और एफर्ट्स से करना अच्छी बात है सर यह सारी बातें आपके समझ आई अब अगर हम सीधे मूव करें नेक्स्ट कॉन्सेप्ट की तरफ तो मैं कहूंगा इतनी सारी बातों से भी ज्यादा जरूरी बात की इन केस कभी भी कोई आपसे कहे की सर एक सर्कल है जो की X1 कमा y1 इसकी जगह कुछ लिख के देगा क्वेश्चन में माइंस फोर कमा 5 या 2 3 ऐसी एक दे देगा वो X2 कमा Y2 है ना ऐसे एक और पॉइंट दे देगा लेट से क्या X3 कमा थ्री अब देख कर ये बहुत कॉम्प्लिकेटेड सी बात रख सकती है पर जैसे मैन लो मैं एक एग्जांपल लेके बताता हूं जैसे वो लिख देगा क्या माइंस वैन कमा तू फिर वो क्या लिख देगा लेट से 1 कॉमन लेट से 1 ऐसे तीन पॉइंट्स देगा और पूछेगा की इन तीनों पॉइंट से पास होने वाले सर्कल की इक्वेशन बताओ मैं क्या करूंगा मैं कहूंगा सर मेरे लिए तो सर्कल की इक्वेशन होती है x² + y² + 2G x+y+=0 बिल्कुल सही कहा सर आपने अब क्या अब चूंकि ये सर्कल इस पॉइंट से पास होता है तो ये पॉइंट है सर्कल के क्वेश्चन को सेटिस्फाई करेगा तो एक बार आप ये रख देंगे तो आपके रिलाइज हो रहा है एक्स और ए तो है जाएंगे और मुझे क्या मिलेगा मुझे जीएफ और सी में एक क्वेश्चन मिलेगी से इसको पास करेंगे तो जीएफ और सी में एक और एक क्वेश्चन मिलेगी से इसको पास करेंगे तो जीएफ और सी में एक और एक क्वेश्चन मिलेगी सो तीन वैरियेबल्स में तीन इक्वेशंस मिल जाएंगे और उसकी हेल्प से मैं जीएफ और सी को क्रैक कर लूंगा और जीएफ और सी जो मेरे क्रैक हो गए जो जीएफ और सी पता चल गए तो मैं सर्कल की इक्वेशन लिख दूंगा बस इतना आसान सा तरीका है दिस इस दी स्टैंडर्ड ऑपरेटिंग प्रोसीजर जो आप फॉलो करेंगे जब आपको तीन पॉइंट्स दे दिया जाए और पूछा जाए इन तीन पॉइंट से पास होने वाले सर्कल की इक्वेशन बताइए आई होप ये बात आपको समझ ए रही है की आपको कैसे सोचना है आपको क्या कर लेना है सर आपके पास सर्कल की इक्वेशन होगी उसमें आप वो गिवन जो जो पॉइंट्स हैं उनको पास कर दोगे उसे सर्कल की इक्वेशन मिल जाएगी उन इक्वेशंस को सॉल्व इक्वेशन की लीनियर इक्वेशन मिल जाएगी क्योंकि आपके एक्स ए और ये एक्स और ए है जाएंगे ना आपके पास सिर्फ जीएफ और सी में एक तीन इक्वेशंस बचेगी तो जीएफ और सी तीन वैरियेबल्स है तीन वैरियेबल्स में तीन इक्वेशंस को सॉल्व कर लेंगे इनकी वैल्यूज हो जाएंगे उन्हें प्लेस कर देंगे तो हमें क्या मिल जाएगी वापस सर्कल की इक्वेशन आपको ये बात समझ ए रही है जो जनरल इक्वेशन ऑफ अन सर्कल है जो ये स्टैंडर्ड इक्वेशन ऑफ अन सर्कल है ये कितने कम की है सर्कल के बारे में कई सारी जरूरी बातों को अगर आउट करने के लिए बहुत बढ़िया अगर यहां तक चीजें क्लियर है तो आई लाइक तू गिव यू अन असाइनमेंट यह आज का आपका पहला असाइनमेंट क्वेश्चन है जैसे आप का सकते हो आपका होमवर्क क्वेश्चन जैसे स्टूडेंट्स कहते हैं dppap कई सारे स्टूडेंट्स का रहे हैं सर डीपी का प्रोवाइड करोगे इन्हीं क्वेश्चन को हम डीपी पे कहेंगे स्टूडेंट्स जैसे देखो क्या लिखा है इफ अन सर्कल हु सेंटर इस वैन कमा -3 टचेस डी लाइन दिस दें फाइंड इट्स रेडियस मेरा आइकन करिए इससे आसान क्वेश्चन नहीं होगा आप थोड़ा इमेजिन करो अगर एक सर्कल है और अगर ये सर्कल किसी 9 को टच करता है तो क्या आप इसकी रेडियस नहीं निकल सकते क्या आपको क्लिक हुआ अगर अभी भी नहीं हो रहा है तो मैं फिर से थोड़ी इंटरेस्ट देता हूं इसका सेंटर है वैन कमा -3 इस लाइन की इक्वेशन पर क्या होगी भाई सर वो होगी रेडियस है और क्या मैं आपको यह डिस्टेंस फॉर्मूला सिखाऊ या आप मुझे सिखा देंगे हेल्प आपको पता है 1 3 - -3 4 - 5 / √ ओवर 3² + 4² + - 4 का स्क्वायर जो की एक ही बात है और उसका आप मोड ले लेंगे वही आपके सर्कल की क्या होगी रेडियस आई होप आप ये क्वेश्चंस सॉल्व करके ये आंसर तू ले आएंगे तो जल्दी से इस क्वेश्चन का स्क्रीनशॉट ले लीजिए चलो इसके बाद क्या सर फाइंड डी इक्वेशन ऑफ सर्कल विच टचेस बोथ डी एक्सेस एंड डी लाइन एक्स = 2 क्या का रहा है वह जो दोनों एक्सेस को टच करता है सर अगर यह दोनों एक्सेस को टच करता है तो बहुत मजेदार सी बात यह होती है अगर कोई भी सर्कल दोनों एक्सेस को टच करें अब मैं यह नहीं का रहा हूं हमेशा फर्स्ट क्वाड्रेंट में ऐसा होता है बस आपको समझने के लिए सब बनाते रहो अगर वो इस तरीके से एक्सेस को टच करें और लाइन एक्सेस इक्वल्स तू एक्स = 2 मतलब क्या एक्स = 2 मतलब ए एक्सिस के पैरेलल एक लाइन है है ना तो ये एक्स = 2 तो आई थिंक सर बहुत आसानी से आप निकल सकते हो बहुत डिफिकल्ट बात नहीं है मेरा यकीन करिए बहुत इजीली आप का सकते हो अच्छा एक पॉसिबिलिटी और सर हो सकती है अगर ये सर्कल यहां बन रहा है तो यही सर्कल यहां थर्ड क्वाड्रेंट में भी बन सकता है तो दो आंसर्स आएंगे अगर आप ध्यान से देख का रहे हो तो अब ये क्वेश्चन कहां पूछा जा सकता है ये क्वेश्चन आपका मल्टीपल करेक्ट ऑप्शंस एक से ज्यादा करेक्ट ऑप्शंस पूछा जा सकता है जहां पे ये और ये दोनों ही आपके आंसर्स हो सकते हैं क्या ये बात आपको समझ आई अगर ये क्वेश्चन आप ट्राई करेंगे ऐसे क्वेश्चंस किए हैं हमने इसलिए बातें ए रही है है ना यहां पर अगर मैं आपसे पूछूं देखो इस क्वेश्चन में कुछ नहीं है मतलब यह टच कर रहा है उसकी रेडियस आपको पता है आर इसकी रेडियस अगर आपको पता है आरकॉम आर तो इसके सेंटर के कोऑर्डिनेटर को मार के साथ रेडियस भी ए रहे हैं और अगर आप देख का रहे हो तो उसका डायमीटर भी आपको पता है अगर मैं कुछ हिंट देने की कोशिश करूं तो बस इन हिंट्स पर कम करिएगा आपका एक क्वेश्चन खत्म हो जाएगा आज का आपका जो अगला होमवर्क क्वेश्चन है वो क्या है इसे ध्यान से पढ़िए स्टूडेंट्स आई होप आप स्क्रीनशॉट ले ले रहे हो आपने लास्ट क्वेश्चन का स्क्रीनशॉट लिया है या नहीं इस क्वेश्चन में देखो क्या लिखा हुआ है फाइंड डी इक्वेशन ऑफ सर्कल विद सेंटर्स आते थ्री कमा - 1 एंड विच कट ऑफ एन इंटरसेक्ट ऑफ लेंथ सिक्स एंड इट्स ये आज ही के कॉन्सेप्ट पर बेस्ड क्वेश्चन है देखो जरा पहली बात तो सर्कल जिसका सेंटर है थ्री कमा -1 पर ठीक है सर जो की एक इंटरसेक्ट मेरा हमेशा से मस्त लग रहा है क्वेश्चंस समझने का अच्छा तरीका है विजुलाइजेशन कोऑर्डिनेटर ज्यामिति में तो विजुलाइज अच्छी कोई बात ही नहीं है तो सर सेंटर है थ्री कमा - 1 पे ठीक है सर थ्री बनाता है सिक्स यूनिट्स का इस लाइन पर सर्कल का सेंटर देता हूं पहले तो इस लाइन से परपेंडिकुलर लेंथ आप निकल लेंगे तो उससे ये जो लाइन है ये बायसेक्स हो जाएगी अगर ये लाइन बायसेक्स हुई है तो इसकी लेंथ कितनी है इसे इस कॉल्ड की लेंथ है सिक्स यूनिट्स तो बायसेक्स होने की वजह से हो जाएगा थ्री जब आप ये डिस्टेंस निकलोगे तो ये डिस्टेंस भी ए जाएगी उससे क्या हो जाएगा आपके पास परपेंडिकुलर और बेस ए गए की आप हाइपोटेन्यूज नहीं निकल सकते अरे पाइथागोरस थ्योरम और हाइपोटेन्यूज निकल लोग मतलब रेडियस ए जाएगी सेंटर के अकॉर्डिंग बता सकते आई थिंक बहुत इजीली आप ये भी आसानी से निकल लेंगे मेरा यकीन करिए क्या ये क्वेश्चन भी आपको समझ आया आज का आपका अगला होमवर्क क्वेश्चन जरा ऐसे ध्यान से देखिए इस क्वेश्चन का भी स्क्रीनशॉट ले लीजिएगा और मेरा यकीन है आपने इस क्वेश्चन स्क्रीनशॉट ले लिया है अच्छा इस क्वेश्चन में क्या दिया है देखो भाई 4 तू डी सर्कल टचेस इट अप्वॉइंट ए इन डी फर्स्ट क्वाड्रेंट फाइंड डी कोऑर्डिनेट्स ऑफ आदर पॉइंट बी ऑन डी सर्कल सच डेट अब = 4 अगेन क्वेश्चन विजुलाइज कारी स्टूडेंट्स हो जाएगा क्या का रहा है वो एक सर्कल है इस सर्कल पे फोर कमा जीरो सो देयर इसे अन पॉइंट लेट्स से फोर कमा जीरो ऑफ कोर्स वो एक्स एक्सिस पर ही होगा फोर कमा जीरो से अगर मैंने एक टांगें ड्रॉ की किस पर इस सर्कल पर क्या इस सर्कल को आप देख का रहे हो ये सर्कल सर्कल के लिए कैसा सर्कल है जो जीरो कमा जीरो पर है अगर मैं इसको बहुत अच्छे से आपके सामने प्लॉट करवाओ ना तो देखना ध्यान से सुनना ये आपका एक सर्कल है जिसके सेंटर के कोऑर्डिनेट्स से 0 और जिसकी रेडियस है 8 का अंडर रूट यानी 2√2 अब बात समझ का रहे हो तो यहां कहीं होगा आपका तू रूट तू और यहां कहीं होगा यहां भी टेस्ट कर सकता था जो की होता आपका फोर्थ क्वाड्रेंट अब वो क्या कहना चाह रहा है फाइंड डी कोऑर्डिनेटर सर्कल और उसे तरीके से चीज निकल लेना है कंडीशन अप्लाई करके की अब की वैल्यू 4 है बहुत बेसिक सा बहुत डायरेक्शन स्ट्रेट पावर क्वेश्चन है जो आप यकीनन तौर पर कर लेंगे अगर आपके बेसिक्स सही हैं तो आई होप आप ये क्वेश्चन भी ट्राई करेंगे बहुत अच्छे से बहुत आराम से कर लेंगे बिना ज्यादा परेशान हो गए तो ये आज का आपका अगला असाइनमेंट क्वेश्चन जिसका स्क्रीनशॉट आपको ले लेना है एक और असाइनमेंट क्वेश्चंस ध्यान से देखिए क्या लिखा हुआ है फाइंड डी लेंथ ऑफ डी इंटरसेक्ट डेट डी सर्कल मैक्स ऑन डी एक्स एक्सिस क्या का रहा है वह का रहा है फाइंड डी लेंथ ऑफ डी इंटरसेप्ट था द सर्कल मैक्स ऑन डी एक्स एक्सिस मेरे ख्याल से आज का इससे आसान कोई क्वेश्चन नहीं होगा क्योंकि वही कोई ऑप्शन यू नो एप्रोच पर बेस्ड है जो हमने जस्ट अभी सिखा है आई थिंक ये वैसा ही क्वेश्चन तो ये बेसिकली आपकी डीपी शॉर्ट ऑफ थिंग जो की आपके लिए असाइनमेंट क्वेश्चंस के तौर पर आपको दी जाती है तो जरूर आप इन क्वेश्चंस को करिए और अपने सारे कॉन्सेप्ट्स कंसोलिडेटेड करिए अगला असाइनमेंट क्वेश्चन ध्यान से देखिएगा प्रूफ डेट डी लोकस ऑफ डी पॉइंट डेट मूव्स सच डेट डी सैम ऑफ डी स्क्वायर ऑफ इट्स डिस्टेंस आपको करना है आते हैं इस पॉइंट पर केसर मैन लो अगर मुझे कोई एक्लिप्स सर्कल दे दिया लेट्स इमेजिन की सर हमारे पास है बेसिकली लेट्स से एक सर्कल है ना सर्कल देने से उसका मतलब है की मैन लो मुझे उसे सर्कल की क्या दे दी जाए एक कोड दे दी जाए है ना उसे कॉर्ड के exfremes दे दिया जाए स्कॉट के एक्सपीरियंस मतलब की सर स्कॉट का एक एक्सट्रीम है और इस कोड का एक एक्सट्रीम है इसका यह एक्सट्रीम के कोऑर्डिनेट्स है लेट से X1 y1 और स्ट्रीम के कार्ड हैं X2 कमा व्हाइट और ये बात जरूर बशर्ते बताई जाए की ये जो कॉर्ड है ये सर्कल की सरकम्फ्रेंसेस पर कहीं ना कहीं लत से एक देता एंगल सब्सटेंड करती है मैं मैन लेता हूं की इस कॉल्ड का यह पॉइंट हालत कर रही है मैं क्या ऐसा का सकता हूं की सर मैं बड़ी बेसिक सी बात समझता हूं या जानता हूं की ये जो कोड है ये सर्कल की सरकम्फ्रेंसेस पे कहीं पर भी थीटा एंगल ही सब्सटेंड करेगी इस सेगमेंट में बिल्कुल सर तो क्या मैं इस सी का लोकस अगर निकल लूं तो मैं सर्कल की इक्वेशन निकलने के काफी करीब रहूंगा बिल्कुल सर आप सही का रहे हो आपको ये बातें समझ ए रही है स्टूडेंट्स जो मैं कहना चाह रहा हूं पर ये निकलेंगे कैसे मुझे एक बात बताओ क्या आप कब की स्लोप निकल सकते हो आई थिंक सर वही माइंस Y2 / एक्स - X2 क्या आप का की स्लोप निकल सकते हो बिल्कुल सर वही माइंस फाइव वैन अपॉन एक्स - X1 और अगर मुझे इन दोनों की स्लोप मिल जाती हैं अगर मुझे इन दोनों की स्लोप मिल जाती हैं तो इसकी स्लोप को अगर मैं का दूंगा m1 और इसकी स्लोप को अगर मैं का दूंगा M2 तो क्या मैं दोनों की स्लोप की हेल्प से उनके बीच का एंगल फिगर आउट कर सकता हूं तन थीटा के लिए तो मैं फॉर्मूला जानता हूं जो की होता है m1 - m2/कोस कितना है स्टूडेंट्स वैन प्लस m1 M2 बस यही सारी चीज जब आप करेंगे बस सब ये सारी चीज जब आप करेंगे किन हालातो में जब आपको ये तीन बातें पता हो कौन सी सर एक तो ये की उसे कोड के एक्सट्रीम कोऑर्डिनेट्स क्या है और वो कॉर्ड सर्कल की सरकम्फ्रेंसेस पे कितना एंगल सब्सटेंड कर रही है अगर ये तीन बातें पता हो तो उसे केस में भी आप सर्कल की इक्वेशन इस फॉर्मूले से लिख सकते हैं ये फॉर्मूला कहां से आया ओरिजिन वापस वही है आप m1 निकल लोग यहां से आप M2 निकल लोग यहां से आई होप आप Y2 - y1 अपॉन एक्स - X1 ये फॉर्मूले जानते हैं स्लोप निकलने के वही वैल्यूज यहां रिप्लेस करेंगे जब तो फाइनली सिंपलीफिकेशन से आप इस रिजल्ट पर pahunchoge मेरा आपसे यह कहना है सोच के देखो की सर अगर यह थोड़ी देर के लिए मैं मैन लूं की ये जो कार्ड हमें दी गई है ये जो कॉर्ड हमें दी गई है सर मैन लो वो थोड़ी देर के लिए अगर एक डायमीटर हो तो सर अगर किसी सर्कल की कोई क्वाड डायमीटर हो तो मैं ये जानता हूं की उसकी जो कॉर्ड होगी ना वो सर्कल की सरकम्फ्रेंसेस पे कहीं भी कितना डिग्री एंगल सब्सटेंड करेगी सो इट विल सब ट्रेंड अन 90° एंगल मेरा यह कहना है की सर सर कल की कॉर्ड के बजाय सर्कल के डायमीटर के एक्सट्रीम कोऑर्डिनेट्स दे दिए जाएं मगर इसकी वैल्यू सेक्स वैन ए वैन और ये है आपका X2 Y2 तो यह बात तो बाय डिफॉल्ट नॉन है की ये थीटा डिग्री नहीं ये एक 90° एंगल बनेगा और एक इसके एक स्क्रीन koardinate से स्टूडेंट्स सर ये इसके डायमीटर के ये इसकी किसके कोऑर्डिनेट्स हैं सर ये इसके डायमीटर के एक्सट्रीम कोऑर्डिनेट्स हैं मतलब ये यहां पर जहां सरकम्फ्रेंसेस पर पास हो रहा होगा मतलब अगर डायमीटर के कोऑर्डिनेट्स पता चल गए तो क्या मैं जानता हूं θ कितना होता है 90 अगर मैं इस एक्सप्रेशन में cot90 रखो तो वो कितना हो जाएगा सर वो हो जाएगा इंफिनिटी एंड जीरो अपॉन वैन तो जीरो इन दिस विल बी जीरो अपॉन वैन तो वैन इन दिस विल बी दिस यानी की अगर मुझे सर्कल के डायमीटर के कोऑर्डिनेट्स दे दिए जाए अगर मुझे सर्कल के आय होप आपको ये बात समझ ए रही है डायमीटर के कोऑर्डिनेट्स दे दिया जाए तो इसे देखकर क्या आपके नहीं लिख सकते की सर सर्कल की इक्वेशन क्या होगी आई हो बड़ी बेसिक सी बात मैं आपसे कहना चाह रहा हूं बस यहां पर एक छोटा सा करेक्शन जरूर करिएगा स्टूडेंट जैसे ए - y1 एक्स - X2 है और ये है यहां पर बस इससे आप लिखेंगे ए - Y2 ना ऐसे आप लिखेंगे ए - Y2 आई रिपीट माय स्टेटमेंट बहुत ध्यान से सुनना है इसका कम की बात को अगर मुझसे कोई कहे किस किसी सर्कल के डायमीटर के एक्सट्रीम कोऑर्डिनेट्स हैं कौन-कौन X1 y1 और ऑफ कोर्स X2 Y2 तो आप सर्कल की इक्वेशन डायमीटर के एक्सट्रीम से भी बता सकते हो एक्स - X1 टाइम सेक्स - X2 + ए - y1 ए - Y2 = 0 ऐसा कैसे आई होप ये प्रूफ आपको याद रहेगा मैं डायरेक्टली रिजल्ट लिख रहा हूं स्टूडेंट्स क्या लिखोगे आप एक्स - X1 टाइम्स एक्स माइंस Y2 = 0 दिस गिव्स यू डी इक्वेशन ऑफ अन सर्कल व्होस कोऑर्डिनेट्स ऑफ डी डायमीटर्स आर y1 X1 y1 रिलीज सॉरी y1 नहीं X1 y1 एंड X2 Y2 पहला कंक्लुजन तो ये आज के लेक्चर का जो आपको याद रखना है तो नोट्स ये वाला बना लो स्टूडेंट्स की अगर आपको कभी भी किसी सर्कल के डायमीटर के एक्सट्रीम दे दिए जाएं तो सर्कल की इक्वेशन ऐसे भी लिखी जा सकती है मैं इसे कंसोलिडेटेड करने के लिए आपको एक क्वेश्चन देता हूं इस इस एक्सप्रेशन को ध्यान से देखो और थोड़ी सी चीज समझने की कोशिश करो मैन लो इस एक्सप्रेशन में मैं आपसे कहूं की ये जो बी पॉइंट आपको दिया गया है होप आपको दिख रहा है जिसके cardinate हैं वैन कमा वैन इस बी पॉइंट से इस बी पॉइंट से मैं चार परपेंडिकुलर ड्रॉप करता हूं इस बी पॉइंट से मैं चार परपेंडिकुलर ड्रॉप करता हूं कहां-कहां सर पहला तो मैं ड्रॉप करता हूं इस लाइन पर दूसरा लेट्स में ड्रॉप करता हूं इस लाइन पर तीसरा और चौथा मैं ड्रॉप करता हूं इन दोनों लाइंस के परपेंडिकुलर बाईसेक्टर्स पर क्या मैं कंफ्यूज तो नहीं कर रहा हूं स्टूडेंट्स तो मैंने आपसे चार बातें कही कौन सी चार बातें स्टूडेंट्स की इस बी पॉइंट से हमने चार परपेंडिकुलर ड्रॉप किए एक तो ए = 3X + 4 पर दूसरा ए = - 3X + 6 पर और दो इनके परपेंडिकुलर बाईसेक्टर्स बाईसेक्टर्स पर क्या यहां तक कोई कन्फ्यूजन स्टूडेंट्स कोई डाउट अब मेरा आपसे यह पूछना है स्टूडेंट्स मेरा आपसे ये पूछना है की आप एक ऐसे सर्कल की इक्वेशन बताओ आप एक ऐसे सर्कल की इक्वेशन बताओ जो की जो की ध्यान से सुनना पहले से मैं परपेंडिकुलर ड्रॉप कर देता हूं ताकि आपको समझ आए जब इस बी पॉइंट से इस पर परपेंडिकुलर ड्रॉप किया तो shailise है वो परपेंडिकुलर यहां आकर कहीं गिरेगा आई होप आपको दिख रहा है जब भी इसे इस लाइन पर परपेंडिकुलर ड्रॉप किया तो आई थिंक वो परपेंडिकुलर यहां आकर कहीं गिरा आई होप ये बात भी आप समझ पाए दूसरी जरूरी बात जिसे ध्यान से सुनना मैन लो मैं कहूं की सर बी से जब इस पर परपेंडिकुलर ड्रॉप किया तो यहां आकर गिरा और जब बी से इस पर परपेंडिकुलर ड्रॉप किया तो वो यहां करके कोई दिक्कत तो नहीं स्टूडेंट्स कोई परेशानी या कोई तकलीफ तो नहीं मतलब आपने 20 से किस पर ड्रॉप किया परपेंडिकुलर एक तो इस स्ट्रेट लाइन पर जो थी ए = 3X + 4 आपने इस पर परपेंडिकुलर ड्रॉप किया जो की थी ए = - 3X + 6 और हम एक बार बहुत अच्छे से जानते हैं सर की जो किन्ही भी दो स्ट्रेट लाइंस के परपेंडिकुलर बाईसेक्टर्स होते हैं वो आपस में एक दूसरे पर परपेंडिकुलर होते हैं पैर विजय आई होप ये बात समझते हो यानी की अगर इन दोनों लाइंस का एक परपेंडिकुलर बाईसेक्टर यह है तो दूसरा वाला इस पर परपेंडिकुलर होगा क्या यहां तक कोई परेशानी अब सवाल suniyega अब सवाल suniyega क्वेश्चंस suniyega बहुत मजेदार क्वेश्चन है बहुत इनसाइड फुल क्वेश्चन है सवाल ये है क्या आप बता सकते हो क्या बता सकते हो उसे सर्कल की इक्वेशन इक्वेशन ऑफ अन सर्कल किस खास सर्कल के क्वेश्चन जिसके डायमीटर के कोऑर्डिनेट्स हैं जिसके डायमीटर के क्वाड्रांट्स हैं इन एंगल बाईसेक्टर पर ड्रॉप किए गए परपेंडिकुलर के बेस यानी की अगर मैं इसे थोड़ी देर के लिए का डन लेट से ए और लेट्स से मैं इसे थोड़ी देर के लिए का डन लेट से बी इससे अलग से मैं का डन थोड़ी देर के लिए सी और इस समय थोड़ी देर के लिए का डन दी और उसे मैं का डन थोड़ी देर के लिए आई तो आपसे सवाल यह पूछा जा रहा है की आप बताइए उसे सर्कल की इक्वेशन आप बताइए उसे सर्कल की इक्वेशन जिसके डायमीटर के कोऑर्डिनेट्स हैं ऐसी क्या आपको क्वेश्चन समझ ए रहा है जिसका जो डायमीटर है जिसका जो डायमीटर है वह एक अब एक सवाल तो यह सर की क्वेश्चन कितना घुमा फिर के पूछा जा सकता है क्वेश्चन कितना घुमा फिर के पूछा जा सकता है अगर मैं ढूंढ की सर पहले तो इसके और इसके एंगल बाईसेक्टर की इक्वेशन निकालो फिर इससे ड्रॉप किए गए परपेंडिकुलर के फीट निकालो यहां पर और फिर इससे ड्रॉप किया गया फिट निकालो यहां पर फिर वो cardinate ए जाएंगे और फिर निकालो डायमीटर की इक्वेशन बेशक मॉडल ऑपरेटिंग यही होनी चाहिए टेक्निकल पर मैं का रहा हूं एक इंटेलिजेंट स्टूडेंट तो ऐसे नहीं सोचेगा एक इंटेलिजेंट एप्रोच तो ये नहीं कहेगी बिल्कुल नहीं जी मांस और एडवांस में ये सवाल आता है तो बिल्कुल आप ऐसे नहीं सोचेंगे आप और आसान तरीका बना सकते हो जिसका मेरा आपसे कहने का तरीका होगा suniyega ध्यान से की सर आप बेवकूफ अगर आप ठीक से नहीं देख का रहे हो क्यों क्योंकि देखो ए बी अच्छा बाय डी वे यह भी बी है तो इसको मैं कुछ और का देता हूं लेट से वैसे आपको का देता हूं एफ है ना फिर से suniyega एबीसीडी मोर स्पेसिफिक यहां पर परपेंडिकुलर और दी जो पॉइंट है वो तो दो बाईसेक्टर्स के बीच का एंगल है तो 90 डिग्री होता ही है तो क्लीयरली सर दिस इस अलसो माउंट तू बी अन 90° एंगल अगर कैन आई कनक्लूड ए बी सी दी इस आर एक्टिंग सर सारी बातें छोड़ो आप ये सब बातें क्यों कर रहे हो इतनी घुमा फिर के बस ये बताओ आप मुझे ये बताओ अगर एक ए बी सी दी रेक्टेंगल है मैन लो ये एक ए मैं बस ऐसे ही कुछ भी एक रैंडम सर एक्टिंग कर ले ले रहा हूं तो क्या आप ये नहीं जानते हैं मतलब अगर आपसे कोई ऐसे सर्कल की क्वेश्चन बताओ जो बीड़ी से पास होती हो जिसका डायग्नल बीबी हो जिसका डायमीटर बी हो तो मैं कहूंगा सर वो उसी सर्कल की क्वेश्चन पूछ रहा है जिसका डायमीटर एक है क्योंकि वीडियो और ऐसी से वही डायमीटर है मतलब ये ऐसा वाला है और ये ऐसा वाला है पर ये दोनों डायमीटर है आप मेरी बात समझ का रहे हो मेरी बात बस इतनी सी है स्टूडेंट्स ये क्वेश्चन इतना आसान बन गया है इतना घुमा फिर के की ये क्वेश्चन है ये क्वेश्चन है एक ऐसे सर्कल की इक्वेशन जिसका डायमीटर हो एक मैं कहूंगा नहीं की एक क्वेश्चन है एक ऐसे सर्कल की इक्वेशन जिसका डायमीटर हो बी तो सर उससे क्या फायदा आपने कौन सा हमारा कम आसान कर दिया एक मिनट सब्र रखिए आपका कम आसान हो गया है ऐसे कैसे सर सुनिए भी पता है दी निकल सकते हो बी पता है दी निकल सकते हो सर कैसे निकल सकते हैं अरे साहब इन दोनों लाइंस का पॉइंट ऑफ इंटरसेक्शन ही तो है दी क्वेश्चन समझ ए रहा है क्वेश्चन समझ ए रहा है की नहीं तो सर ट्राई करते हैं इक्वेशन सॉल्व करते हैं ए = 3X + 4 एक तो इक्वेशन है आपकी ए = 3 एक्स + 4 और ए = -3 दोनों को एड कर लो ये हो जाएगा 2y = कोस कितना है कैंसिल हो जाएगा 10 तो ए की वैल्यू कितनी 5 किसी भी स्टूडेंट को अभी तकलीफ तो नहीं है मैंने क्या किया इन दोनों को ऐड किया तो ये कैंसिल 2y = 10 तो ए कितना आया फाइव अगर ए आता है 5 ए की वैल्यू यहां रख दो यार रख दो एक्स की वैल्यू कितनी आती है वैन बाय थ्री फाइव मुझसे जब कोई पूछेगा तो मैं कहूंगा सर मुझे तो ज्यामिति के अंडरस्टैंडिंग है और आपने भली घुमा फिराक के क्वेश्चन दिया जिसमें वो जो सर्कल की क्वेश्चन चाहिए जिसका डायमीटर हो एक मैं कहूंगा नहीं सर आप घुमा रहे हो आप तो एक ऐसे सर्कल के क्वेश्चन बताओ जिसका डायमीटर है बी दी क्वेश्चन समझ ए रहा है और यही एक इंटेलिजेंट एप्रोच है लगते complicateds हैं पर रियलिटी में मैं बार-बार यही बात का रहा हूं अगर आपको बारीकी या पता है तो क्वेश्चंस आसान होते हैं और हमारे डिटेल्स सीखने पर फोकस कर रहे हैं की देखो स चीज पता होनी चाहिए ऐसे घुमा फिर के क्वेश्चंस बनते हैं तो मैं कहूंगा की अब तो सर क्वेश्चन आसान है क्वेश्चन क्यों आसान है क्योंकि आपको एक ऐसे सर्कल की इक्वेशन बतानी है जिसके डायमीटर के कोऑर्डिनेट्स हो बी डायमीटर के एक्सट्रीमली इसके कोऑर्डिनेट्स को बी और दी और बी और दी इसलिए क्योंकि वो मुझे पता है आई होप आप निकल सकते हो मुझे सीखने की जरूरत नहीं है कौन-कौन से सर एक तो क्या है 1 और एक है 1 / 3 क्या करते हैं सर एक्स - X1 टाइम्स एक्स माइंस एक्स तू प्लस ए माइंस y1 ए - Y2 = 0 कोई डाउट स्टूडेंट्स कोई परेशानी कोई दिक्कत आई थिंक सर सिंपलीफाई किया जा सकता है बिल्कुल कर सकते हैं आप चाहें तो आपको अगर सिंपलीफिकेशन चाहिए तो थ्री से मल्टीप्लाई कर लो पर अभी मत करो बाद में कर लेना सर देखो करना ही है आखिर में बाद में अभी कभी जब जैसे करना है कर लेना पर मेरे ख्याल से अभी सिंपलीफाई कर लो और मल्टीप्लाई कर दो देखो क्या हो रहा है सर एक्स तू एक्स स्क्वायर आई थिंक सारी क्वेश्चन खत्म हो जाएगा बस आप सिंपलीफाई कर डन आंसर आपके सामने है क्या लिखना है सर देखो भाई x² हो Y2 = लिख दो सर आप x² + y² और क्या लिख दो सर देखोगे कितना हो जाएगा 3 एक्स - एक्स - 4X / 3 तो ये हो जाएगा -4x/3 - ए - 5 / हो जाएगा -6y और क्या बचा है सर फाइव प्लस वैन डेट सिक्स कोई तकलीफ देखो ऐसे नहीं करना है ना तो एक और तरीका बताता हूं इन दोनों का मिड पॉइंट निकल लोग तो वो क्या हो जाएगा सेंटर और सेंटर से इसकी डिस्टेंस निकालो तो हो जाएगी रेडियस वैसे भी लिख सकते हो पर उतना घुमा फिर के करने से अच्छा है डायरेक्ट ये तीन लाइन में आपका आंसर ए जाएगा क्या आपको यह पूरा क्वेश्चन यह पूरा एक्सप्लेनेशन समझ आया और आपको रिलाइज हुआ की चीज कैसे सोचनी है और अगर ये बात आपको समझ आई तो क्या अब हम मूव करें हमारे आज के नेक्स्ट कॉन्सेप्ट की तरफ जहां पर हम बात करेंगे की सर किसी सर्कल की इक्वेशन पैरामेट्रिक फॉर्म में भी लिखी जा सकती है क्या उसके लिए मैं आपसे चाहूंगा की आप याद करो जो हमने cardinate सिस्टम जो चैप्टर पढ़ा था ना फर्स्ट चैप्टर वहां पे बहुत सारी कम की चीज डिस्कस की थी और मैंने बार-बार कही थी शायद की सर सर्कल में भी हम ये सारी बातें आपसे करेंगे आप याद करो पोलर फॉर्म हमने सिखा था आपको याद ए रहा है पोलर फॉर्म के बारे में हमने बात की थी अगर याद नहीं कर का रहे हो तो थोड़ा याद दिलाता हूं जैसे मैन लो आपके एक्स एक्सिस बाइक्स के cardinate सिस्टम जो है आपका प्लेन है इस पर कोई cardinate है कोई पॉइंट है ये दिख रहा है क्या और इस पॉइंट की खास बात ये है सर ये जो पॉइंट है ये ओरिजिन से कुछ एक सर्टेन डिस्टेंस पर प्लेस पॉइंट को ज्वाइन करने वाली लाइन पॉजिटिव एक्स के साथ कोई एंगल बनाती है बड़ी आसान तरीके से आप उसे कहते हो आर कोस थीटा आर सिन थीटा आई होप आप ये भूले तो नहीं हो आय होप आप यह भूले तो नहीं हो की अगर किसी पॉइंट की ओरिजिन से डिस्टेंस आर है और अगर उसे पॉइंट को ओरिजिन से मिलने वाली स्ट्रेट लाइन पॉजिटिव एक्स-एक्सिस के साथ एंटीक्लाकवाइज डायरेक्शन में थीटा डिग्री एंगल बनाती है तो हम उसे कहते हैं उसे पॉइंट के कोऑर्डिनेट्स को पोलर फॉर्म में rcosθ मेरा आपसे बस इतना कहना है की जरा ध्यान से देखो स्टूडेंट्स प्लीज ध्यान से देखो ये हमेशा कोई ना कोई एंगल बना रहा होगा और टेक्निकल क्या एंगल बना रहा होगा सर जीरो से 360 इस सर्कल की सरकम्फ्रेंसेस पर लाइक करने वाले जो भी सारे पॉइंट्स होंगे वो जीरो से 9180 तक फिर 270 तक आते आते 360 तक अलग-अलग एंगल्स बना रहे होंगे ना और जब वो अलग-अलग एंगल्स बना रहे होंगे तो उसे दौरान वो अपनी डिस्टेंस ओरिजिन से फिक्स रखेंगे और यह कौन सा सर्कल है अगर मैं आपसे एक अलग तरीके से पूछूं तो आप कहोगे सर ये इक्वेशन है x² + y² की छोड़ो सारी बातें आप बोलते हो ना की ये पी पॉइंट है इसके कोऑर्डिनेट्स आप क्या कहोगे सर इसका लॉकर्स में ह कॉम के मैन लूं तो मैं एक्स को क्या कहूंगा सर मैं एक्स को कहूंगा rcostheta और ऑफकोर्स ए को क्या कहूंगा ए को कहूंगा रस साइन थीटा और हमने पैरामीटर्स में आना सिखा है सर अगर इन दोनों को स्क्वायर करके ऐड करें तो लेफ्ट हैंड साइड पर क्या ए जाएगा आपसे यही तो बात मैं कहना चाह रहा हूं आपसे स्टूडेंट्स की यह जो सर्कल की इक्वेशन है यह पैरामेट्रिक फॉर्म में इस तरीके से सूजी जाती है पर सर जीवन हमेशा इतना आसान थोड़ी रहता है जीवन तो तकलीफ में ए जाता है कब जब आपका सर्कल का सेंटर ओरिजिन की वजह कहीं और शिफ्ट हो जाए तो क्या करें तो जब सर्कल का सेंटर ओरिजिन से कहीं और शिफ्ट हो जाए तो हमने नहीं सिखा है क्या की सर ओरिजिन को कहीं और शिफ्ट कर दें तो आपके नए कोऑर्डिनेट्स क्या हो जाते हैं अगर ओरिजिन को हम कहीं और शिफ्ट कर दे तो आपके नए कोऑर्डिनेट्स क्या हो जाते हैं क्या हमने नहीं सिखा है और अगर नहीं सिखा है तो आज उसे बारे में भी बात करेंगे इस क्वेश्चन पर आने से पहले उसे बारे में भी बात कर लेंगे इसके थ्रू बड़ी सिंपल सी दो बातें हैं suniyega की अगर सर आपका सर्कल है कौन सा ओरिजिन पर सेंटर जिसकी रेडियस है तो उसकी पैरामीटर के क्वेश्चन आप जानते हो एक और θ और ए साइन थीटा होता है वो इस सर्कल की इक्वेशन है कोई तकलीफ नहीं होनी चाहिए लेकिन सर अगर कोई सर्कल है जिसका एंगल ऑफ कॉस्ट थीटा है वो पूरा पैरामीटर फॉर्म में जीरो से 360 डिग्री तक लेकिन इस बार उसका जो सेंटर है वो मैन लो सर ह कमा पेपर है तो सर्कल में कोई बड़े बदलाव नहीं हुए रेडियस जो रहा करती थी ये वही है उसका सब कुछ वही बस ओरिजिन की जगह जीरो कमा जीरो की जगह सेंटर के cardinate ए गए ह कॉम ए के तो कुछ नहीं बदला है सर एक्स जो होगा आपका जो अभी आर कोस थीटा या ए कोस थीटा रहा करता था उसे आप लिखोगे h+ ए कोस थीटा और जो ए koardinate आप लिखोगे उसे लिखोगे आप के + ए सिन थीटा क्योंकि याद करो स्टूडेंट्स कुछ नहीं होना है बस यहां से थोड़ी डिस्टेंस इससे ह और इससे के डिस्टेंस शीट किया तो जो नए cardinate जाएंगे जो नए कोऑर्डिनेट्स आएंगे अगर मैं आपसे उनकी बात करूं अगर आपसे बात करूं नए दो कोऑर्डिनेट्स की जहां पे सर्कल को मैन लो यहां शिफ्ट कर दिया तो कुछ नहीं बदला है कुछ नहीं बदला है बस इस बार क्या बदल गया है ह कॉम के जो आपका ओरिजिन पर रहा करता था सर्कल अब उसके ओरिजिन की बजाय सेंटर के cardinate हैं ह कॉम के बट अभी भी नहीं सन्डे है उसकी जो रेडियस है वो आर या ए जो भी आप कहना चाहो वही है तो इसके अब पैरामेट्रिक फॉर्म में जॉब कार्ड की आप एक्स के इक्वल रख दोगे ए के इक्वल और उससे भी आंसर ए जाएगा कैसे करेंगे सर इन क्वेश्चंस को सॉल्व तो उसके लिए ये रहा आपके सामने क्वेश्चन वैसे तो एक समझदार एक इंटेलिजेंस स्टूडेंट इस क्वेश्चन को देख कर आंसर कर देगा इस क्वेश्चन को देख कर आंसर कर देगा पर अगर नहीं दिख रहा है तो वो स्टैंडर्ड एप्रोच उसे करेगा तो मैं दोनों से बता देता हूं देख कर हम आंसर करेंगे पर पहले एक स्टैंडर्ड एप्रोच देखते हैं फाइंड डी सेंटर क्या लिख रहा है सुनना यह लिख रहा है एक्स = -1 कोस थीटा लिखूं तो आपको कोई आपत्ति नहीं होनी चाहिए सर यहां पे ये लिखा है ए = 3 + 2 सिन थीटा अगर थ्री को इधर ले आऊं तो क्या मैं इसे ऐसा लिख सकता हूं ए - 3 = 2sin थीटा और 2 को भी इधर अगर ले आऊं तू को भी इधर ले आओ तो क्या मैं इसे ऐसा लिख सकता हूं मेरे ख्याल से यहां पर भी तू होना चाहिए था क्योंकि उसे केस में ये पैरामेट्रिक फॉर्म नहीं बनेगी अगर यहां तू नहीं होता तो इस क्वेश्चन में टाइप हो है स्टूडेंट्स की वहां पर तू आना चाहिए था मेरे ख्याल से तो टेक्निकल तो इस टाइप को प्लीज करेक्ट कर लीजिएगा देर शुड हैव बिन अन तू आय रिपीट माय स्टेटमेंट मैं इसे फिर से ठीक करता हूं suniyega ध्यान से कम की बात है पहले इस क्वेश्चन को ठीक कर लीजिए स्टूडेंट्स यहां पर इट शुड हैव बिन अन तू है ना ये तू है अब क्या सर अगर आप इसे राइट करोगे तो आप इसे क्या लिखोगे एक्स प्लस वैन थीटा आप चाहे तो यही से यही से इसे देखकर आंसर बता सकते द बट कोई बात नहीं हम अगर यहां पर इस क्वेश्चन को देखें तो मेरा आपसे कहना है की देखो दोनों तरफ एलएस और एलएस के स्क्वायर करके सैम किया तो ये हो जाएगा एक्स + 1 का होल स्क्वायर प्लस ए माइंस थ्री का होल स्क्वायर अब्बास समझना तू का स्क्वायर तू का स्क्वायर वही बात cos² थीटा + sin² थीटा 1 और क्या बन जाएगा 2² क्या आपको कुछ चीज मैं दिखाऊंगी आप खुद का देंगे सर बड़ी आसान सी बात है इसको अगर आप लिखना हो तो एक्स - 1 का होल स्क्वायर प्लस ए माइंस थ्री का होल स्क्वायर इस इक्वल तू स्क्वायर क्या आपको चीज है आसान सी डायरेक्ट सिख लिया ऐसे नहीं दिख रही है की आपका हो जाएगा सेंटर के कोऑर्डिनेट्स जैसे मैं थोड़ी देर के लिए अल्फा बेटा या ह के जो भी आप कहना है का लीजिए और ये आपकी क्या हो जाएगी रेडियस अब दूसरा गौर फरमाइए जरा गौर फरमाइए अभी जस्ट हमने डिस्कस किया स्टूडेंट्स की सर अगर आप पैरामेट्रिक फॉर्म में सर्कल की इक्वेशन लिखते हो तो आप लिखते हो h+ ए कोस थीटा + साइन थीटा एक्स और ए cardinate होते हैं तो उसकी सरगम पेरेंट्स पे लाइक करने वाले लोकस करो प्रेजेंट करते हैं इसमें हम ये जान का रहे हैं सर जो ह और के हैं ये जो ह और के हैं क्लीयरली आपके सर्कल के सेंटर के कॉर्ड से और ये जो ए है ये जो ए है सर दिस इस नथिंग बट व्हाट सर्कल का रेडियस इसी बात को अगर मैं आपको थोड़ा ठीक तरीके से समझने की कोशिश करूं तो क्या आपको ये नज़र नहीं ए रहा है की सर देखो एक्स = ए = एक्स = h+ ए कोस थीटा के + ए सिन थीटा तो ह कॉम के ह के माइंस वैन कमा 3 - 1 3 क्या है सर्कल के सेंटर के कोऑर्डिनेट्स और ए कोस थीटा ए साइन थीटा तो ए क्या हो गया सर ए हो गया सर्कल तू इतना कुछ करने की जरूरत नहीं थी अगर आप इसे देख के मानते हो तो ये टू था क्या है क्वेश्चन इसमें क्या था जो आपको नहीं पता था आई थिंक चीजें आसान है चीज अगर आसान है तो एक और क्वेश्चन आपके सामने स्क्रीन पर इस क्वेश्चन को अच्छे से पढ़ो अच्छे से समझो क्योंकि ये क्वेश्चन आपसे होना चाहिए एक बेहतरीन सा ये क्वेश्चन है एक अच्छा सा एक क्वेश्चन है मैं समझा जरूर देता हूं पर ट्राई आप ही करेंगे फिर हम साथ में करेंगे पी इस डी वेरिएबल पॉइंट ऑन डी सर्कल विथ सेंटर आते सी ठीक है का एंड सीबीआर परपेंडिकुलर्स फ्रॉम सी ऑन एक्स-एक्सिस इन ए एक्सिस ठीक है विजुलाइज करते जाना मैं भी हेल्प करूंगा पब इसे अन सर्कल विद सेंटर आते डी सेट्रॉयड ऑफ ट्रायंगल सब बहुत बड़ा क्वेश्चन हो गया नहीं समझ ए रहा है एक सेकंड अब तो पहले बताओ पी इस डी वेरिएबल पॉइंट ऑन डी सर्कल विथ सेंटर आते सी और का और कब जो द वो परपेंडिकुलर द सी से किस पर एक्स-एक्सिस और ए एक्सिस पर पहले तो ये बात आपको समझ आई क्या बिल्कुल आगे सर तो पहले इसी मुद्दे पर इसी मसाले पर थोड़ी बात कर लेते यहां से चीजे हम कैसे सोचेंगे suniyega ध्यान से सर चीज थोड़ी प्लॉट करने की कोशिश करते हैं देखो चलेगा क्या आई थिंक यहां बना लेते हैं देखो मैन लो ये आपका सर्कल है कोई तकलीफ किसी भी स्टूडेंट को और सर इस सर्कल के लिए आपका ए एक्सिस हो गया क्योंकि आपका एक्सेस हो गया सिर्फ फर्स्ट क्वाड्रेंट में ही क्यों बनाते हो मेरे ख्याल से कहीं भी बना लो बस बनाना है ना इस सर्कल की एक स्टैंडर्ड एक क्वेश्चन रही होगी अब मुझे एक बात बताओ सर्कल का कोई ना कोई प्यारा होगा सेंटर इस सर्कल के क्वेश्चन में हमारी वाली मानता हूं क्या एक्स स्क्वायर प्लस ए स्क्वायर प्लस 2G एक्सप्रेस 12 प्लस सर्कल माइंस जी से अभी हमें क्वेश्चन के अकॉर्डिंग डिनोट करना है किस सी से है ना कोई तकलीफ वेरिएबल पॉइंट है जिसके बारे में बात करेंगे उसने कहा का और कब आपके एक्स एक्सिस और ए एक्सिस पर ड्रॉप किए गए परपेंडिकुलर है तो यहां से अगर मैंने ड्रॉप किया तो यह है आपका बी सर यह तो मैं बता सकता हूं एक आर्डिनेंस कुछ नहीं होंगे एक-एक cardinate होंगे माइंस के कमा जीरो और बी के क्वाड्रेंट होंगे जीरो कमा - एफ किसी भी स्टूडेंट को यहां तक कोई तकलीफ कोई परेशानी नहीं होनी चाहिए एक और बात बताओ स्टूडेंट्स अगर मैं यहां पर कोई भी एक रैंडम पॉइंट मैन लो लेट से पी तो पी के akordinates आपके अकॉर्डिंग क्या होंगे सर पता है हमें कुछ तो याद करो स्टूडेंट जल्दी से ह कमा ए कोस्थेटा सॉरी h+ ए कोस थीटा के + ए सिन थीटा रेडियस का आप क्या करू अरे भाई इस इक्वेशन को देखकर आप मुझे सर्कल की रेडियस नहीं बता सकते क्या अंडर रूट ओवर जी स्क्वायर प्लस एक्स स्क्वायर माइंस सी एंड रूट ओवर जी स्क्वायर प्लस एक्स स्क्वायर माइंस ये बता दो की पीके कोऑर्डिनेट्स क्या होंगे क्योंकि ये निकलना तो आपको सिखाया गया है जैसे मैन लो अगर तुम्हें ऐसे नहीं लेना है तो मैन लो इसकी सर रेडियस क्या है आर अगर इसकी रेडियस ए रहे हैं तो पीके कार्ड के होंगे सर पीके आर्डिनेंस होंगे माइंस जी प्लस ऑफ कोर्स आर कोस थीटा और कमा माइंस है पैरामेट्रिक फॉर्म आई होप पैरामेट्रिक फॉर्म समझते हो सेंटर के कार्ड सिमिलरली रेडियस टाइम्स आर मतलब वही आर रेडियस टाइम्स सिन थीटा लोकस ऑफ डी सेट्रॉयड ऑफ ट्रायंगल पी अब सारी बातें छोड़ो मैं पी ए और बी के सेट्रॉयड को निकलने की कोशिश करता हूं सेट्रॉयड क्या होता है और उसी का मैन लेता हूं ह के तो मेरे ह और के जो आएंगे उनके बारे में बात करते हैं जो ह आएगा वो होगा -g+rcosθ+-जी + 0 तो -3 - 0 - 2G तो ये हो जाएगा -2g+r कोस थीटा / एफ कोस 3 तो क्या मैं थ्री यहां लिख सकता हूं सिमिलरली की जो वैल्यू आएगी आपको ध्यान से देखो माइंस एफ माइंस साइन थीटा क्योंकि यहां तो जीरो है ना तो कितना हो जाएगा सही हो जाएगा -2f+ ऑफ कोर्स आर सिन थीटा अब मैं बस आपसे कहना चाह रहा हूं इस सेट्रॉयड के कोऑर्डिनेट्स को ध्यान से देखो इस सेट्रॉयड के कोऑर्डिनेट्स को ध्यान से देखो अगर अभी भी आप नहीं देख का रहे हो तो मैं आपकी और हेल्प करता हूं ध्यान से देखो चलो बताओ ये 2G को उधर ले तो कहना ऐसे ही हो जाएगा 3h+2g = आर कोस थीटा सिमिलरली यहां भी ले तो अमरेली सॉरी दिस डिवाइडेड बाय थ्री होता तो थ्री यहां पे ले आता हूं तो ये कितना हो जाएगा 3k + ऑफकोर्स 2f = 4G स्क्वायर अगर मैं चाहता तो इसमें से थ्री को ना यहीं रख लेता उसे ना बात यह बन जाती है की मुझे चीज दिख जाती है बस 1 सेकंड स्टूडेंट्स थोड़ा सा पेशेंस रखो मैं एक छोटा सा कम करना चाह रहा हूं छोटा सा कम ही करना चाह रहा हूं की स्क्वायर करके ऐड करने से पहले मैं इस थ्री को उधर ही रखना चाह रहा हूं थ्री को उधर ही रखने से मेरा मतलब क्या है समझना बात को यहां से थ्री से पूरा डिवाइड किया तो थ्री से जब डिवाइड किया तो यहां भी ए जाएगा 3 एम रियली सॉरी यहां भी ए जाएगा रियली सॉरी आई एम उसको ठीक कर रहा हूं एक सेकंड दीजिए यहां भी ए जाएगा थ्री यहां भी ए जाएगा थ्री यहां भी ए जाएगा थ्री और यहां भी ए जाएगा 3 मैंने क्या किया वो थ्री को यहीं रखा कोई दिक्कत तो नहीं है ऐसा क्यों कर रहे हो सर दो मिनट suniyega अब अगर मैं स्क्वायर करके ऐड करता हूं अब अगर मैं स्क्वायर करके ऐड करता हूं तो एलएस पर क्या आएगा सर एलएस पे आएगा h+ 2G / 3 क्या ह को मैं फाइनली एक्स रिप्लेस करूंगा लोक से तो ह को फाइनली एक्सक्यूज और रिप्लेस करेंगे और प्लस को मैं लिख लेता हूं - -2g/3 का होल स्क्वायर प्लस करेंगे ए माइंस कितना माइंस टफ बाई थ्री इक्वल तू अब कम की बात देखो आर/3 का स्क्वायर आर / 3 का स्क्वायर वही बात cos² थीटा + sin² थीटा 1 तो यहां पर क्या मिलेगा आपको यहां पर मिलेगा आर/3 का स्क्वायर क्योंकि मेरा आपसे ये कहना है बस 2 मिनट suniyega क्या आप मुझे कब के सेट्रॉयड के कार्ड बता सकते हो अभी हमने किसके सेट्रॉयड निकले अभी हमने निकले सर्कल माइंस तू अब बात समझ का रहे हो - 2G / 3 और -2f/3 है और एक बात बस इतनी सी बताओ स्टूडेंट्स इस सर्कल की रेडियस कितनी थी सर इस सर्कल की रेडियस हमने मणि थी आर इस सर्कल की रेडियस कितनी थी हमने मणि थी आर अब गौर से suniyega अब बहुत गौर से सन शो डेट डी लोकस ऑफ डी सेट्रॉयड ऑफ ट्रायंगल पी अब जो हमने निकाला है सेट्रॉयड ऑफ दी ट्रायंगल पी अब santroid ऑफ दी ट्रायंगल ऑफ पब ये आपके पब के सेट्रॉयड का लोकस है जिसे मैं देख का रहा हूं जैसे मैं बहुत आसानी से देख का रहा हूं एक्स - alpha² + ए - b² = रेडियस का स्क्वायर ये सर्कल की इक्वेशन - 2 एफ/3 जिसकी रेडियस है वह सर्कल है तो हम प्रूफ कर चुके है ना वहीं पर आपके इस सर्कल के सेंटर के कार्ड ऑफ डी रेडियस ऑफ डी गिवन सर्कल इसकी रेडियस थी आर और इसकी रेडियस है 1/3 कोई परेशानी आई थिंक एक आसान सा क्वेश्चन था पर यह बहुत आसानी से सॉल्व किया जा सकता है जब हमें क्या पता हो पैरामीटर बहुत सिंपल सा क्वेश्चन है जिसे हम आसानी से सॉल्व कर सकते हैं अगर हमें चीज पता हो तो अभी तक जो सारी बातें हमने पड़ी उन पर बेस्ड आज का ये आपका पहला असाइनमेंट क्वेश्चन जैसे आप चाहे तो कहेंगे क्या होमवर्क क्वेश्चन या फिर आप इसे कहेंगे असाइनमेंट क्वेश्चन या फिर आप इसे का लीजिए अपनी आज की डीपी क्वेश्चन खत्म हो गया है याद करो भाई एक्स - X1 एक्स - X2 ए - y1 ए - Y2 जिसके डायमीटर के कार्ड उसका हाफ कर डन तो मुझे क्या मिल जाएगी क्वेश्चन खत्म हो जाएगा क्या आई थिंक आप इसे करके मुझे आंसर बता देंगे वैसे आंसर आपके सामने है अगला सवाल फाइंड डी इक्वेशन या उसका आपका कौन सा आज का आपका दूसरा असाइनमेंट क्वेश्चन देख लीजिए इसे ध्यान से आई होप आपने इस क्वेश्चन का स्क्रीनशॉट ले लिया था अगले क्वेश्चन का स्क्रीनशॉट ले लीजिएगा स्टूडेंट्स जल्दी से लीजिए अब क्या लिख रहा है वो सुनिए फाइंड डी इक्वेशन ऑफ डी सर्कल्स विच पासेस विच पास थ्रू व्हाट ओरिजिन एंड कट ऑफ कार्ड ऑफ लेंथ ऑफ अन फ्रॉम एच आदर ऑफ डी लाइंस इक्वल्स तू एक्स एंड ए = - एक्स मतलब वो ए लेंथ ये ए लेंथ की कॉर्ड बनाता है किस किस से एक तो ए = एक्स वाली लाइन से भी और एक ए = - एक्स वाली लाइन से इस सर्कल की आपको इक्वेशन बनानी है जो ऑफ कोर्स कहां से पास होता है जो ऑफ कोर्स आपका ओरिजिन से पास होता है तो आप जब इसे सॉल्व करेंगे तो बेसिक आपके ये जो दो इक्वेशंस ए रहे हैं यही आपके आंसर्स आएंगे अब ये क्वेश्चन एक्चुअली मस्क में ए सकता है मल्टीपल सिलेक्ट मतलब एक से ज्यादा करेक्ट ऑप्शन क्योंकि यहां से दो आंसर्स ए रहे हैं यहां से दो आंसर्स आए टोटल चार आंसर्स आए और हो सकता है वो चारों के चारों आप को opshns में दे दें और आप शायद या तो ये निकालो या निकालो यस का एक ही निकल लो जब तक आप चारों नहीं ढूंढते तब तक आपका परफेक्ट करेक्ट आंसर नहीं आएगा और इसलिए एग्जाम आसान नहीं है इसलिए एग्जाम इंटरेस्टिंग है इसलिए एग्जाम इनसाइटफुल है इसलिए एग्जाम मजेदार है चलो भाई अगर ये क्वेश्चन आप ट्राई करेंगे एक होप आपने स्क्रीनशॉट लिया तो आज का आपका अगला असाइनमेंट क्वेश्चन ये आज का थर्ड आपका असाइनमेंट क्वेश्चन है जो आपको डीपी हम देते हैं बेसिकली आप का सकते हैं डीपी हम इन्हें कहेंगे प्रैक्टिस क्वेश्चंस के असाइनमेंट क्वेश्चंस फाइंड डी इक्वेशन ऑफ डी सर्कल पासिंग थ्रू डी ओरिजिन सो ये भी ओरिजिन से पास हो रहा है एंड कटिंग इंटरसेप्ट्स ऑफ लेंथ थ्री यूनिट्स एंड फोर यूनिट्स फ्रॉम डी पॉजिटिव एक्सेस पॉजिटिव एक्सेस से मतलब सर वही बात तू टाइम्स अंडर रूट ओवर G2 टाइम्स इन दोनों से तो मैं कुछ कुछ निकल लूंगा और फिर कुछ-कुछ करके निकल लूंगा क्योंकि उसमें सी नहीं होगा सी इसलिए नहीं होगा क्योंकि ये ओरिजिन से पास होता है तो अगर सी होगा तो फिर इक्वेशन कैसे सेटिस्फाई होगी साहब जीरो कमा जीरो कैसे सेटिस्फाई होगा हर बात का अपना एक मीनिंग है हर बात का अपना एक साइनिफिकेंस है बस यह बातें अगर ठीक से पढ़ी जाए सोची जाए और अप्लाई की जाए अप्लाई तब होगी जब आपने हर चीज सीखी होगी पड़ी होगी समझी होगी आपको पता होगी चलो भाई अगर आप यह क्वेश्चन कर लेंगे तो क्या हम अगले क्वेश्चन की तरफ बढ़े नेक्स्ट आज का आपका होमवर्क क्वेश्चन जो है जैसे आप चाहे तो का लीजिए असाइनमेंट क्वेश्चन जिसका स्क्रीनशॉट जल्दी से ले लीजिए आय होप आपने इस क्वेश्चन का स्क्रीनशॉट लिया इस क्वेश्चन में क्या बोल रहा है वो फाइंड डी वैल्यूज ऑफ के फॉर विच डी पॉइंट डी पॉइंट ले ऑन ए सर्कल पहली बात तो वो जीरो कमा जीरो दे दे रहा है 0 मतलब सी नहीं होगा अब आप एक स्टैंडर्ड जनरल इक्वेशन मैन लीजिए उसको 0 1 0 0 1 से सेटिस्फाई करवा दीजिए आपकी जी और एफ की वैल्यू हो जाएगी सी वैसे भी नहीं है क्योंकि जीरो कमा जीरो है की वो सर्कल इसे भी सेटिस्फाई करेगा जैसे ही सेटिस्फाई करेगा रख दीजिएगा आप ट्राई कर लेंगे एक आसान क्वेश्चन है आपके सारे बेसिक आपको रिकॉल करवाने के लिए आई होप आपने इस क्वेश्चन का भी स्क्रीनशॉट ले लिया है मेरे ख्याल से यहां पर अगर कुछ है क्योंकि आज की पैरामीटर फॉर्म ऑफ मैन सकते हो जल्दी बोलो भाई पैरामेट्रिक फॉर्म क्या होती है सर अरे मैं क्यों बताऊं आपको आप मुझे बता दीजिए पैरामेट्रिक फॉर्म h+r कोस थीटा और के प्लस rintheta तो इसमें ह और के क्या होते हैं सर उसके सेंटर के कोऑर्डिनेट्स सर आर क्या होती है उसकी रेडियस तो आई थिंक जब आप इसे देखोगे तो आप क्या बोलोगे भाई सर हम बोलेंगे की इसके सेंटर है जीरो कमा जीरो तो ह और के जीरो और रेडियस 7 तो आर कोस थीटा आर सिन थीटा एक पॉइंट ए जाएगा स्टूडेंट्स की आपको याद होना चाहिए नहीं याद है तो फिर गड़बड़ है भाई है ना अब बहुत दिक्कत हो जाएगी अगर अब भी आपको नहीं याद ए रहा है की राइट एंगल ट्रायंगल मुझे कुछ स्टूडेंट्स लिख के जरूर बताएं ताकि जिन स्टूडेंट्स को नहीं पता है वो कमेंट्स में पढ़ ले सेकंड बात स्टूडेंट्स याद करो अगर वह औरतों सेंटर की बात पूछ रहा है तो दो cardinate है एक ए जाएगा कोस थीटा सिन थीटा कुछ तो कर ही लोग आप और उससे शायद आप लोग कस बता दोगे बिल्कुल बता दोगे मैं बड़ी ही बेसिक से सैंपल से बात है सर की अगर कोई पॉइंट मुझे दिया जाए लेट्स से मुझे कोई भी कभी भी पॉइंट दे दिया जाए और उसे पॉइंट के बारे में पूछा जाए की वो पॉइंट है सर्कल के अंदर लाइक करता है सर्कल पर लाइक करता है या सर्कल के बाहर लाइक करता है सर अगर कोई पॉइंट सर्कल पर लाइक करेगा तो क्लीयरली सर्कल अगर वह सर्कल के अंदर लाइक करेगा तो वह सर्कल की इक्वेशन में जब मैं वैल्यूज पास करूंगा ना तो वो आपको नेगेटिव वैल्यू देगा मतलब लेस दें जीरो कोई वैल्यू देगा और अगर वह पॉइंट सर्कल के बाहर कहीं लाइक करेगा तो वो पॉइंट के cardinate जब आप सर्कल की इक्वेशन में रखोगे ना तो वो आपको पॉजिटिव वैल्यू देगा मतलब जैसे की मैं मैन लेता हूं एक पॉइंट है X1 y1 लेट से देयर इसे अन पॉइंट X1 अगर वह सर्कल पर लाइक करता तो क्या ए जाता भाई तो ए जाता इक्वल तू जीरो अगर वह सर्कल के अंदर लाइक करता तो X1 y1 जब आप सर्कल की इक्वेशन में पास करोगे तो उसकी वैल्यू जो आती है वो ए जाती है नेगेटिव और अगर वह सर्कल के बाहर लाइक करेगा तो सर्कल की इक्वेशन में उसे पास करने से उसे वैल्यू को जो आप पाएंगे रखने पर वो होगी पॉजिटिव सर्कल जो कॉन्सेप्ट आज हम पहले एक्सप्लोर करेंगे क्या यहां तक कोई कन्फ्यूजन suniyega ध्यान से अगर एक सर्कल है हमें आइडिया लग रहा है की सर्कल के सेंटर के कार्ड से माइंस के माइंस सी मेरा बस आपसे यह कहना है की सर क्लियर सी बात है मतलब बहुत ही बेसिक सी बात है हम सब समझ का रहे हैं की अगर एक पॉइंट है जो की सर्कल के बाहर लाई कर रहे हैं मतलब इस तरीके से भी सोचा जा सकता है जिस बात को मैं आपको डाइजेस्ट करवाना चाह रहा हूं अगर एक पॉइंट है जो सर्कल के बाहर लाई कर रहे हैं तो क्लीयरली उसे पॉइंट की सर्कल्स के सेंटर से जो डिस्टेंस होगी और उसे पॉइंट सर्कल के सेंटर से सर्कल पर लाइक करने वाले पॉइंट के जो डिस्टेंस होगी और सर्कल के अंदर अगर कोई पॉइंट है उसकी सर्कल के सेंटर से डिस्टेंस होंगी उनके बारे में क्लीयरली मेरा आपसे बस इतना सा कहना है सर्कल से बाहर कोई पॉइंट है तो उसकी डिस्टेंस रेडियस से ज्यादा होगी सर्कल पर कोई पॉइंट है तो उसकी डिस्टेंस सर्कल की रेडियस जितनी ही होगी सर्कल के अंदर कोई पॉइंट है तो उसकी डिस्टेंस सर्कल के सेंटर से सर्कल की रेडियस से कम होगी इसी बात को थोड़ा और एक्सटेंड किया जाए तो आप बस ये मैन के चलिए की सर अगर सर्कल की इक्वेशन में उसे पॉइंट के कोऑर्डिनेट्स में X1 और y1 पास कर डन सर्कल की इक्वेशन में उसे पॉइंट के कोऑर्डिनेट्स अगर मैं X1 और y1 पास कर डन तो सर बिल्कुल आप निश्चित होकर यह का दीजिए अगर उसकी वैल्यू ग्रेटर दें जीरो आई मतलब सर्कल के बाहर है उसकी वैल्यू अगर सर्कल पर है और उसकी वैल्यू लेस दें जीरो आई मतलब वो सर्कल के अंदर है क्या बड़ा बेसिक सा डायरेक्ट सा क्लियर सा सिंपल सा ऑब्जर्वेशन आप याद रखेंगे तो क्या एक कम करें इससे पढ़ते हैं सुनेगा रिप्रेजेंट्स डी रीजन इन साइड डी सर्कल डी रीजन आउटसाइड डी सर्कल और अगर इसमें इक्वल तू जीरो लगा दे तो डी रीजन ऑन डी सर्कल यही क्रश है इसे नोट डाउन कर लीजिए यही आज के सेशन का ट्रक से जस्ट है पूरा सर पूरा निचोड़ वैसे ही पहले और इसी से आज हम सारी बातें करेंगे ऑफ कोर्स यह की लर्निंग जरूर याद रखना स्टूडेंट्स की अगर कोई सर्कल में कोई पॉइंट सर्कल के अंदर है बाहर या सर्कल पर है तो रेडियस के मुकाबला उसकी डिस्टेंस क्या उसकी दूरी में क्या रिलेशन होगा वो जरूर हमको सोचना होगा कहीं ना कहीं क्वेश्चन की रिटायरमेंट के अकॉर्डिंग उसको बात कर लेते हैं कुछ और चीजों की यहां तक अगर कोई परेशानी नहीं है तो मैं सीधे आता हूं एक क्वेश्चन पर और इस क्वेश्चन को हम करते हैं बहुत ध्यान से देखना वो क्या का रहा है एक सर्कल है सर x² + y² - 10y + के सर्कल की रेडियस निकलने में तो थोड़ी तकलीफ जरूर आएगी पर सर्कल की रेडियस के बजाय सर्कल का सेंटर निकल सकता हूं क्या देखो भाई -6 - 10 - 3 - 5 इस सर्कल का जो सेंटर एस को टच करता है और ना ही इंटरसेक्ट करता है मतलब इससे दूर है मतलब इससे यह दूर है आपको समझ ए रही है और क्या सुनेगा ध्यान से इनसाइड डी सर्कल सर जो वैन कमा 4 यहां कहीं हैं ये पॉइंट है वैन कमा फोर ये सर सर्कल के अंदर है तो आप मुझे बता दो की के की वैल्यूज की क्या रेंज होगी के कहां से कहां तक हो सकता है सर क्वेश्चन तो मजेदार सा है पर करेंगे क्या कुछ समझ नहीं ए रहा है कुछ आइडिया नहीं लग रहा है आप मुझे सारी बात छोड़कर बस ये बताओ की आप सर्कल की रेडियस निकल सकते हो रिपीट माय क्वेश्चन क्या आप सर्कल की रेडियस निकल सकते हो अच्छा मैं जो पूछ रहा हूं ऐसे सी मिल जाएगा तो क्या मैं सर्कल की रेडियस निकल सकता हूं सर अगर मैं सीधा सीधा थॉट आपसे मांगूं तो रेडियस क्या होती है सर जो आपकी रेडियस होगी वो होगी अंडर रूट ओवर जी स्क्वायर एनएफ कोर्स माइंस थ्री का स्क्वायर कितना 3 सॉरी -3 का स्क्वायर मतलब माइंस फाइव का स्क्वायर कितना है 25 विदाउट तो नहीं है स्टूडेंट्स - सी तो माइंस कितना के सर ये ए जाती है तो 25 प्लस 9 आई थिंक 34 हो जाता है अंडर रूट 34 माइंस के एक बात ज़रा सोच के बताओ सर अगर यह जो सर्कल है यह जो सर्कल है ये इन सारे सरकमस्टेंसस में फुलफिल करता है चीजों को तो कहना इसे सर की इस सर्कल के जो एक्स कोऑर्डिनेट्स के डिस्टेंस है ए एक्सिस से देखो भाई इस सर्कल का जो सेंटर का एक्स कॉर्डिनेट है वो यहां कहीं है जिसके डिस्टेंस है थ्री और जरा बताओ सर्कल की रेडियस क्या है सर सर्कल की रेडियस है सो कैन आई कनक्लूड की सर सर्कल की जो रेडियस है वो इस थ्री से तो छोटी होगी देखो ना थ्री इतना बड़ा है रेडियस इतना सा है क्लियर सर्विस है तो पहला कंक्लुजन सर्कल की रेडियस जो है सर वो थ्री से छोटी होगी इससे मैं कुछ तो पता कर लूंगा कुछ तो पता कर ही लूंगा ना और क्या और थोड़ा और ध्यान से देखो यह रही इस सर्कल की रेडियस इस नजरिया से और यह रही इस सर्कल के सेंटर के सर्कल के सेंटर की एक्स एक्सिस से डिस्टेंस जो की क्लीयरली कितनी है भाई जो की क्लीयरली फाइव तो सर ये डिस्टेंस है फाइव और ये है आपकी रेडियस सो कैन आई कनक्लूड की सर इससे अच्छी एक और परिभाषा होगी की रेडियस जो है वो फाइल से छोटी होनी चाहिए अब इन दोनों में ज्यादा अच्छी कौन सी बात है एक का रहा है की रेडियस 5 से छोटी होनी चाहिए एक कैरियर एक का रहा है की रेडियस जो है वो 3 से छोटी होनी चाहिए तो मैं कहूंगा सर की ज्यादा स्पेसिफिक कौन का रहा है तीन छोटा होगा और करीब ले ए रहा है यही और फिल्टर कर दे रहा है प्रेफर हेविंग दिस तो एक हेल्प एक आइडिया तो मुझे यहां से मिल जाएगा सर एक हेल्प एक आइडिया तो मुझे यहां से मिल जाएगा मैं इससे जरूर उसे करूंगा इस क्वेश्चन में बिल्कुल सही बात है लेकिन क्या सर इतने ब्लू इतनी हिट काफी है क्या सच में बस यही कहानी खत्म हो जाएगी एक मिनट क्या आप भूल गए स्टूडेंट्स बड़ी ही साधारण सी बात की सर आपने ये क्यों नहीं सोचा की थ्री कमा फाइव से वैन कमा फोर की भी तो कुछ डिस्टेंस होगी और 3 कमा 5 की जो वैन कमा फोर से डिस्टेंस होगी वो रेडियस से छोटी होगी इस बार रेडियस इससे ज्यादा होगी क्योंकि वैन कमा फोर इस सर्कल के अंदर ले करता है तो एक और बात स्टूडेंट्स की जो डिस्टेंस होगी थ्री फाइव और वैन कमा फोर के बीच डिस्टेंस निकलने का फटाफट देखो 3 और 1 का डिफरेंस तू तू का स्क्वायर 4 का डिफरेंस वैन वैन का स्क्वायर 1 4 + 15 यानी √5 यानी एक और कंक्लुजन सर की जो आपका आर है जो आपकी रेडियस है वो ज्यादा होनी चाहिए अंडर रूट 5 से ये आपके इस क्वेश्चन को सॉल्व करने का दूसरा ग्रुप मेरे ख्याल से जितना भी डाटा जितनी भी इनफॉरमेशन आपके लिए मुझे इस क्वेश्चन में देनी चाहिए थी मैं दे चुका अब जरा आप मुझे इस क्वेश्चन को आंसर कर दीजिए और मुझे बता दीजिए की इस क्वेश्चन का आप क्या आंसर मार्क करेंगे आई थिंक बहुत टू बहुत डिफिकल्टी क्वेश्चन है नहीं आप सीधे सीधे सोच के बता दो की आपके आंसर मार्क करोगे क्योंकि अब तो आना चाहिए भाई अब तो अब तो चीज कुछ बच ही नहीं है सर ऐसे कैसे आंसर ए जाएगा पहले तो देखो हमने रेडियस निकल ली थी ना रेडियस क्या थी अंडर रूट 34 - ठीक है सर रेडियस में बात करते हैं √34 - पहली कंडीशन क्या है देखो भाई सर पहली कंडीशन है की रेडियस जो है वो होना चाहिए थ्री से छोटी हो जाएगा 34 माइंस के के उधर भेजते हैं तो के जो होगा सर वो ग्रेटर दें होगा 34 - 9 3 4 - 30 और 30 - 25 तो सर पहला कंक्लुजन तो ये की ये जो है वो 25 से बड़ा होना चाहिए एंड आई होप आप समझते हो इन सारे शब्दों के मतलब की इसके साथ-साथ मतलब ये तो होना ही चाहिए एंड ये भी होना चाहिए ये तो होगा ही होगा और ये भी होना चाहिए अब ये जनाब क्या का रहे हैं सर आर जो है वो अंडर रूट फाइव से बड़ा होना चाहिए और जो है आर क्या है सर आर जो है आपका क्या माइंस के और ये किस बड़ा होना चाहिए सर आर जो है √5 से बड़ा होना चाहिए तो बात क्या निकल कर आती है की 34 - के से बड़ा होना चाहिए तो के को मैं कहूंगा लेस दें होगा 34 -5 जो की क्लीयरली कितना है स्टूडेंट्स प्लीज इस बात को suniyega खैर यही देख कर आप आंसर मार्क करें तो ज्यादा खुशी होगी यह कहता है की कोई वैल्यू है 25 उससे आपका के बड़ा होना चाहिए यह कहता है की सर कोई ना कोई वैल्यू है 29 उससे आपका के छोटा होना चाहिए तो दोनों कहां सेटिस्फाई हो रहे हैं दोनों को कौन सी वैल्यू सेटिस्फाई कर रही हैं तो मैं कहूंगा सर आपका आंसर होगा आपका आंसर होगा की के जो होगा सर बिलॉन्ग करेगा 25 से 29 के बीच की हर एक वैल्यू को 25 से 29 एक्सक्लूडिंग दी तू पॉइंट्स अगर आप कोई भी वैल्यू 25.1.25.2 26 27.8 28.9 तक कुछ भी ले लें तो बेशक वो आपकी ये कंडीशन इन सर करेंगे की यह सर्कल एक्स को ना तो टच करें और ना ही इंटरसेक्ट करें और ऑफ कोर्स यह वैन कमा फोर को अपने अंदर कंटेन करें जिसमें आपको अपनी सर्कल की सारी अंडरस्टैंडिंग की बेसिक्स यहां पर अप्लाई करने द आई थिंक क्वेश्चन आप सभी को समझ आया नेक्स्ट क्वेश्चन ट्राई करते हैं स्टूडेंट्स नेक्स्ट क्वेश्चन देखिएगा और नेक्स्ट क्वेश्चन पर अगर हम बात करें तो ये आज का आपका अगला क्वेश्चन जरा इसे ध्यान से देखिए पढ़िए और समझने की क्या पूछ रहा है अच्छा क्वेश्चन है मजेदार क्वेश्चन है और बेहतरीन सा क्वेश्चन है जो काफी कुछ आपको सिखाएगा फाइंड डी एरिया ऑफ डी रीजन तो सर एरिया ऑफ डी रीजन निकलना है इन विच डी पॉइंट सेटिस्फाई डी इनिक्वालिटीज एक तो ये और एक ही है सर पहले इस पे बात करते हैं जब भी ऐसी inqualities की बात आएं हैं तो पहले इक्वल साइन ले लो जैसे मैं x² + y² = 4 की अगर आपसे बात करूं तो suniyega ध्यान से अगर कोई मुझसे कहे x² + y² = 4 तो shailise है की सर ऐसा तो मैंने कुछ पढ़ा है ऐसा तो मैंने कुछ पढ़ा है आप समझना स्टूडेंट्स इस बात को बहुत ध्यान से समझना की सर अगर मुझसे कभी भी कोई कहे x² + y² = 4 तो मैं कहूंगा एक्स स्क्वायर प्लस ए स्क्वायर समझ ए रहा है की नहीं उससे यह जो एक्स स्क्वायर प्लस ए स्क्वायर ग्रेटर दें ग्रेटर दें मतलब क्या मतलब मैं उन सारे पॉइंट्स में इंटरेस्टेड हूं जो इस सर्कल के बाहर है रिपीट माय स्टेटमेंट जब आपसे कोई का रहा है x² + y² - 4 = 0 ये वही inquality है इसका ये मतलब है की सर्कल से इस सर्कल से बाहर सारे पॉइंट्स आपने निकल लो कोई तकलीफ तो नहीं हो रही है स्टूडेंट्स आई थिंक यह बात आपको समझ ए रही होगी अगर यहां तक चीज समझ ए रही हैं तो मैं इस जोन में ये सारे जो पॉइंट्स लाइक कर रहे होंगे ये सब जो पॉइंट्स लाइक कर रहे होंगे हम इनकी बात कर रहे हैं अब क्या करना चाह रहे हो सर एक मिनट इसकी बात करते हैं यहां जब लिखा है ये वाला तो ये क्या लिखा है सर x² + y² = 4² है ना पर इस बार भी वो वही सर्कल है जिसके सेंटर के cardinate से जीरो कमा जीरो आई होप आप मेरी बात समझ का रहे हो इस बार भी वैसा ही कोई सर्कल है जिसके सेंटर के कोऑर्डिनेट्स जीरो कमा जीरो पर इस बार लिखा है x² + y² माइंस सर्कल के अंदर मतलब इस सर्कल के अंदर का सारा हिस्सा अगर इन दोनों को सेटिस्फाई करना है मतलब सर इससे बाहर लेकिन इससे अंदर तो किसकी बात करें वह बात कर रहा है इस जोन की वो बात कर रहा है इस जोन की वो बात इस जॉन की कर रहे हैं मतलब मुझे इस सेक्शन का एरिया निकलना है सर मैं निकल लूंगा बहुत ही सारा सा क्वेश्चन हो जाएगा तो और एक मिनट क्वेश्चन खत्म नहीं हुआ है ना अभी अगर यही क्वेश्चन होता तो मैं कैसे करता सर ये तो बचपन में हमने किया है सर हम पहले क्या निकलती द इस बड़े वाले सर्कल का रही है उसमें से अंदर वाले सर्कल के एरिया को माइंस कर देते द तो रीजन का एरिया ए जाता था और एक क्वेश्चन खत्म हो जाता था पर यह क्वेश्चन यहां तक नहीं होगा क्योंकि आईआईटी जी मांस का या एडवांस का लेवल का क्वेश्चन है उसको करने के लिए क्या करेंगे अभी पूरा पढ़ेंगे और पूरा पढ़ने के बाद फिर आगे बढ़ेंगे तो आगे घर में भरू सर तो आगे क्या बोलता है वो बोलता है 3x² - y² इस ग्रेटर दैन जीरो सॉरी तो बहुत अजीब सी कोई बात लिखी है ये तो बहुत अजीब सी कोई बात लिखी है यह बहुत अजीब सी हो सकती थी आपके लिए पर मैं कहूंगा अभी आपके लिए बहुत अजीब से नहीं होनी चाहिए क्योंकि आपने पैर ऑफ स्ट्रेट लाइंस पढ़ लिया जनाब मतलब क्या कहना चाह रहे हो सर मैं आपसे ये कहना चाह रहा हूं अगर आप इसे ध्यान से देखें सोच के देखो अगर मैं इसे ऐसा लिखूं suniyega ध्यान से अंडर रूट 3X का होल स्क्वायर माइंस ए स्क्वायर है ना इस ग्रेटर दैन जीरो क्या मैं इसे ऐसा लिख सकता हूं प्लीज ध्यान से सुनना अंडर रूट 3X - ए ग्रेटर दें जीरो जरा इस एक्सप्रेशन को ध्यान से देखो और रिकॉल या री कलेक्ट करो वो जो आपको पढ़ाया गया है है वह जो आपको पढ़ाया गया अभी आप मुझे नहीं बता का रहे हैं की मैं आपसे क्या पूछना चाह रहा हूं अच्छा सर सारी बातें छोड़ो आप तो पॉइंट पर आओ मुद्दे पर आओ मुद्दा यह व्हाट दस इट रिप्रेजेंट अगर यह ग्रेटर दें का साइन में भूल जाऊं और अगर मैं यहां पर थोड़ी देर के लिए मैन लेता हूं इक्वल का साइन हान तो मेरे लिए क्या है ये सुनना गया यहां से एक तो है आपके पास √3x+y और एक है अंडर रूट 3X - ए = 0 अगर मैं एकल तू जीरो रख लूं तो ये क्या है सो दिस इस नथिंग बट अन पैर ऑफ स्ट्रेट लाइंस ये आपकी पहली लाइन है दूसरी लाइन तो मैं बना सकता हूं मैं अगर इसे प्लॉट करता हूं तो कैसे करता सुनना ध्यान से सर सबसे पहले तो अगर मैं इसे लिखता है ऐसे लिखता तो ये क्या हो जाता है सही हो जाता है ए = - √3 6 और ये कितना हो जाता है सर ये हो जाता है ए = + √3x √3 कितना होता है सर √3 होता है 60° तो बेसिकली ओरिजिनल से पास होने वाली 60 डिग्री की स्लोप वाली स्ट्रेट लाइन की इक्वेशन है ये अरे हान या ना और आप इसे देखोगे तो ये आपकी वही ओरिजिनल से पास होने वाली स्ट्रेट लाइन की इक्वेशन है बट ये ये वाली स्ट्रेट लाइन है ये जो है मतलब मैंने बहुत अच्छे से नहीं बनाई है बट मैं अगर आपको बहुत ही प्रॉपर्ली इसको ड्रॉ करके डन अगर मैं बहुत ही अच्छे से आपको समझने की कोशिश करूं तो सुनना कितनी मजेदार सी बात है एक तो आपकी स्ट्रेट लाइन यह है जो ओरिजिन से पास हो रही है मैं इसे थोड़ा सा टिल्ट करूंगा इन्हें रोते करूंगा 60° के लिए नेशन दूंगा और एक जो दूसरी स्ट्रेट लाइन है आपकी वह आपकी यह वाली है जो की ऑफ कोर्स पास तो ओरिजिन से ही हो रही है बट इसका भी इंक्लिनेशन इस वाले ऐसा कुछ ऐसा है पर सर इन दोनों ने तो इस रीजन को आपके प्लेन को चार हसन में बांट दिया एक दो तीन चार अब यह इनिक्वालिटी के सिर्फ रीजन को रिप्रेजेंट करेगी आप कैसे पता करोगे 2 सेकंड मेरे पास एक तरीका है मेरे पास एक तरीका है सीधी-सीधी सी बात मुझे क्या चाहिए मुझे चाहिए सर अंडर रूट 3X + ए टाइम्स जैसे ही इसके कोऑर्डिनेट्स लिए मैंने अंडर रूट थ्री कमा जीरो इसमें रख कर देखते हैं सुनना √3 यानी कितना 3 प्लस जीरो से बड़ा होता है क्या होता है तो सर आपको ये वाला रीजन पूरा लेना है समझ आया क्या सुनना अब यहां से मेक और कम करता हूं सर ये वाले रीजन का ये वाले रीजन और ये वाले रीजन का पता करना है सिमेट्री से आपका हो गया तो आपको पता चल गया है लेकिन फिर भी अगर आपको अपने मैन को शांति देनी है तो आप कहोगे सर यहां कहीं पॉइंट होगा माइंस अंडर रूट थ्री कमा जीरो बिल्कुल होगा सर रखकर देखते हैं तो -3 + 0 तो -3 - 0 - 3 + 9 और प्लस 90 वाला रीजन भी पूरा आपको ले लेना है तो कौन नहीं लेना है आपको ये और ये रीजन नहीं लेना है सर एक ऐसे कैसे शो करो आप नहीं लेना है चेक तो किया ही नहीं कर लो आपकी तसल्ली कर लेना यहां पे कुछ होगा एनालिसिस जीरो कमा 1 अरे होगा की नहीं 0 कमा वैन होगा जीरो वैन जीरो माइंस वैन इन माइंस वैन माइंस वैन माइंस वैन जीरो से बड़ा नहीं होता है इंक्लूड नहीं करना है ये करना है क्या लेके देख लो 00 - 1 ही लेके देख लो वापस माइंस वैन आएगा आप मेरी बात समझ का रहे हो मतलब ये ये जो ये वाला आपका कर्व है कौन सा वाला ये जो इनिक्वालिटी है ये किस रीजन को रिप्रेजेंट कर रही है सर ये जो inquality है ये इस रीजन को रिप्रेजेंट कर रही है ध्यान से सुनना ये आपके एक तो स्ट्रेट लाइन ये होगी सर साहब की जो की कितने डिग्री पर इंक्लाइंड होगी जो की ऑफ कोर्स इंक्लाइंड होगी 60 डिग्री पर और सिमिलरली आपकी एक और स्ट्रेट लाइन होगी सर और वह भी ऑफ कोर्स आपके 60 डिग्री पर इंक्लिन होगी है ना वो भी आपके 60 डिग्री पर इंक्लाइंड होगी और इन दोनों के बीच का रीजन अब सुनना बात को सुनना अब ये क्वेश्चन यहां पर एक अच्छा मोड लेता है इन दोनों सर्कल ने कहा था इन दोनों सर्कल ने कहा था की हमारे इस इस स्ट्रिप पर आने वाले हर चीज का एरिया ले लेना रीजन का लेकिन इन दोनों लाइंस ने और उसे रिस्टिक कर दिया तो अब जो आपको रीजन लेना है वो कौन सा लेना होगा भाई अब जो आपको रीजन लेना है सर वो ये वाला रीजन है अब आपको इस रीजन का एरिया लेना है क्या यह क्वेश्चन बहुत डिफिकल्ट है स्टूडेंट्स मेरा यकीन करिए नहीं है सर रीजन का एरिया आप कैसे निकलेंगे टू नहीं लग रहा आपको नहीं है टू ऐसे कैसे नहीं है बहुत बेसिक सी बात है यह क्वेश्चन इसलिए खत्म हो जाता है यहां पर यह कितना डिग्री एंगल 60 डिग्री एंगल अब सारी बातें छोड़ो अब सारी बातें छोड़ो मुझे ये बताओ बस किसी भी तरीके से निकल लो जैसे आपको ठीक लगे की आप मुझे इस रीजन का क्या आप मुझे इस रीजन का एरिया निकलने में हेल्प कर सकते हो कर तो सकते हैं सर कैसे बस ये बताओ आप तो सुनना क्या यह पूरा 90 डिग्री एंगल था मैं अगर आपसे इस सर्कल की इस इस सर्कल का एरिया पूछता इस सर्कल का एरिया पूरे सर्कल में से इस हिस्से का एरिया पूछता तो सही होता है 14th होता की नहीं होता सर सुनना अभी बस दो मिनट सुनना पर इस समय से अगर मुझे इस वाली पट्टी का एरिया चाहिए होता इस वाली पट्टी का एरिया चाहिए होता तो मैं कहता सर बड़े वाले सर्कल में से छोटे वाले सर्कल को सब्र करो ऐसा बड़े वाले सर्कल की रेडियस कितनी थी सर बड़े वाले सर्कल की रेडियस थी 4 अंदर वाले सर्कल की रेडियस थी तू तो पाई r² याद है ना तो सर पाई r² यानी पाई r² यानी 4² तो 4 का स्क्वायर मतलब 16 पी तो 16π में से पाई r² वो अपना अंदर वाला जो सर्कल ठाणे 4 पाई को सब्सट्रैक्ट करता यह आपकी इस पुरी पट्टी की एरिया ए जाता है आता की नहीं भाई ए जाता सर मेरा आपसे यह कहना है प्लीज इस बात को सुनना ध्यान से यह पूरा एंगल कितना होता है 90 डिग्री पर आपको 90 डिग्री का एरिया नहीं चाहिए होगा इसका 23 क्योंकि 60 / 90 चाहिए ना ये पूरा 98 डिग्री का ये एरिया है 90° का ये एरिया है तो 60° का कितना होगा 60° 90° के रिस्पेक्ट में तू थर्ड है ना तो 60 / 90 रेश्यो इतना प्रोपोर्शन आपको चाहिए तो 65 90 को इन ए बटोर लैंग्वेज कैन आई राइट s2/3 यस यू कैन कैन आय राइट दिस s2/3 एस यू ऑल कैन यस ओर नो स्टूडेंट्स और ये तो इस वाले पत्ते का इस वाले रीजन का आया ना ऐसा एक दो तीन चार टोटल चार रीजंस हैं तो मैं इसके लिए इसको मल्टीप्लाई कर दूंगा फोर से टेक्निकल ये आपका क्या होगा आंसर इसमें थोड़े से बदलाव बदलाव मतलब इसको थोड़ा सिंपलीफाई करेंगे कैसे देखो भाई 16π -4π कितना 12 कितना आता है ये आता है तो जो रीजन पूछ रहा है उसका एरिया कितना होगा क्यों सोचने की जरूरत है ये तो बड़ी मजेदार सी बात है सर क्यों मजेदार सी बात है आप समझो बात को देयर इसे डी सर्कल लेट्स देयर इस अन सर्कल अब सर कोई पॉइंट है लेट्स से पॉइंट है आप खुद बताओ अगर मैं इस पॉइंट से इस पॉइंट से अगर एक ऐसी लाइन ड्रॉ करता हूं अगर इस पॉइंट से कैसे लाइन ड्रॉ करता हूं जो ऑफ कोर्स इसके सेंटर से पास होती है जो ऑफ कोर्स इसके सेंटर से पास होती है जैसे मैं दिनो करता हूं सी और ये जब सर्कल से मैं एक ऐसी लाइन ड्रॉ करता हूं जिसके सेंटर से पास होती है और इस पॉइंट से जो लाइन आपने ड्रा किया वो इस सर्कल को ए और बी पर टच करती है तो क्या बड़ी बेसिक सी बात समझ का रहे हो पहली बात तो ये की अगर आपने ए पर टांगें ड्रॉ की होती तो पी से बेसिकली आपने एक नॉर्मल ड्रॉप किया है सर्कल पर ये सारी बातें हम समझेंगे जब हम टैसेंट और नॉर्मल की बातें करेंगे अभी फिलहाल मेरा उद्देश्य आपको सिर्फ ये समझाना है इस बात को सुनना की सर इस पॉइंट से इस ए तक की जो डिस्टेंस है ना ये इस सर्कल से इस पॉइंट की मिनिमम डिस्टेंस है और इस पॉइंट से जो बी तक डिस्टेंस है ना वो इस सर्कल से इस पॉइंट की मैक्सिमम डिस्टेंस है कॉन्सेप्ट खत्म मतलब हान ज्यामिति के लिए कॉन्सेप्ट खत्म बट अलजेब्राइक ली तो अब शुरू हुआ है हम समझेंगे की चीज कैसे होती है कैसे निकल जाती क्या-क्या करना होता है सारी बातें पर मैं का रहा हूं ज्यादा दूर नहीं है कॉन्सेप्ट समझो बात को ये कॉन्सेप्ट कितना आसान है पीके कुछ ना कुछ कोऑर्डिनेट्स होंगे जिन्हें आप कहोगे सर X1 कमा y1 सर्कल के सेंटर के कुछ ना कुछ कोऑर्डिनेट्स होंगे जिन्हें मैं कुछ भी का लेता हूं लेट्स से एक्स तू कमा Y2 क्या आप मेरी बात समझ का रहे हो निकल देंगे ना हम सर्कल की इक्वेशन से और ये पॉइंट दिया होगा खैर इतने आसान क्वेश्चंस आएंगे नहीं बट फिर भी मैं आपको ये आइडिया देना चाह रहा हूं की चीज कैसी होंगी पर ये कॉन्सेप्ट उसे हो रहा होगा इस पर बेस्ड क्वेश्चंस जरूर आएंगे पर कॉन्सेप्ट जरूर समझो अब क्या मैं ए से सी तक की डिस्टेंस बता सकता हूं सर ये रेडियस है क्या मैं सी से बी तक की डिस्टेंस बता सकता हूं सर की रेडियस है आप मेरी बात समझ रहे हो क्या आप सन नहीं रहे हो ध्यान से सुनो सर कल की जो मिनिमम डिस्टेंस है जो मिनिमम डिस्टेंस है उसे क्या कहेंगे सर उसे हम कहेंगे का सर पिए तो निकलते नहीं ए रही है मुझे अरे आती है आपको कैसे आती है सर पी से सी तक की डिस्टेंस निकल सकते हैं ना सर डिस्टेंस फॉर्मूला से बिल्कुल निकल सकते हैं तो का को मैं कैसे निकलूंगा भाई मैं उसे कहूंगा पीसी माइंस सर्कल की रेडियस गाने खत्म कहानी कैसे खत्म सर सर कल सेंटर से उसे पॉइंट की डिस्टेंस निकल और उसमें से सर्कल की रेडियस का सूत्र कर दिया तो मुझे शॉर्टेस्ट या मिनिमम डिस्टेंस मिल गई अब पूछ रहा है मैक्सिमम डिस्टेंस क्या होगी तो सर मैक्सिमम डिस्टेंस क्या होगी पी से बी तक के डिस्टेंस अरे होगी या नहीं तो मैं क्या कहूंगा सर जब आप मुझसे पूछोगे मैक्सिमम डिस्टेंस तो मैं क्या कहूंगा सर मैक्सिमम डिस्टेंस में कहूंगा पीवी लेकिन सर पी भी तो मुझे निकलती नहीं आती तो मत निकालो तो पी भी निकलने के लिए क्या मैं पीसी प्लस रेडियस ऐड कर दो सेंटर से उसे पॉइंट के डिस्टेंस में एक बार रेडियस अट्रैक्ट कर दीजिए मिनिमम डिस्टेंस मिल जाएगी सेंटर से उसे पॉइंट के बीच के डिस्टेंस में एक बार रेडियस ऐड कर दीजिए मैक्सिमम डिस्टेंस मिल जाएगी बस इतना सा ही एक कॉन्सेप्ट है बस इतना सा ही कॉन्सेप्ट है मतलब यह को हमने कैसे निकाला सर का को निकलने के लिए जो हमने किया वो था पीसी-एक और सर ब को निकलने के लिए जो हमने किया वो हमने किया पीसी + व बस इतनी सी साधारण सी सॉल्यूशन बातें आपसे कहना चाह रहा हूं इससे ज्यादा हम कुछ नहीं करने वाले हैं इस टॉपिक में पर ये टॉपिक सच में इतना आसान है क्या सर मतलब पहले तो आई होप आप ये समझ का रहे हो चीजें सिंपल सी होती है की सर अगर सर्कल की इक्वेशन दे दी जाए ये तो सर्कल के सेंटर के अकॉर्डिंग हम जानते हैं सर माइंस जी कमा माइंस है और सर्कल की रेडियस हम निकल लेते हैं अंडर रूट G2 + x² - सी अब किसी एक्सटर्नल पॉइंट से मैं इसके सेंटर के डिस्टेंस निकल सकता हूं निकल सकता हूं उससे जो ये डिस्टेंस होगी ये डिस्टेंस जो होगी उसमें से सर अगर आप रेडियस अट्रैक्ट कर दोगे तो वो मिनिमम डिस्टेंस मिल जाएगी और सर यह जो डिस्टेंस है उसमें अगर आप रेडियस ऐड कर दोगे तो मैक्सिमम डिस्टेंस मिल जाएगी बस इतना सा ही है कॉन्सेप्ट है प्लीज इस बात को समझ लीजिए मेरा तो बस आपसे यह कहना है की इस चीज को नोट डाउन कर लीजिए आपके लिए बस इतनी ही कम की बात है इसे अच्छे से नोट डाउन कर लो स्टोरेज यही आपके आज के इस लेक्चर का क्रश है इस कॉन्सेप्ट का क्रस्ट क्रस्ट है जस्ट जैसे आप कहते हैं जैसे आप निचोड़ सर या सारांश जो कहते हैं तो हमारा फंडा तो साफ बड़ा सिंपल सा है किसी भी कॉन्सेप्ट्स को अगर हमें सीखना है तो हम तो उसके क्वेश्चंस करेंगे एप्लीकेशंस करेंगे और न्यूमेरिकल से अच्छा कोई तरीका नहीं है किसी कॉन्सेप्ट को सीखने का तो क्या ये क्वेश्चन आप मुझे करके बताएंगे बेशक मैं करवाऊंगी और मैं ही करवाऊंगी पर पहले आपको ट्राई करना है और आपको बताना है मुझे इस क्वेश्चन का आंसर स्टूडेंट्स ये यकीनन तौर पर बहुत अच्छा और आसान क्वेश्चन है वो क्या लिख रहे हैं पड़ी है फाइंड डी पॉजिटिव वैल्यूज ऑफ ए सच डेट डी पॉइंट इनसाइड डी रीजन बॉन्डेड बाय दी सर्कल्स कुछ समझ नहीं ए रहा है सर कुछ समझ ही नहीं ए रहा है सर एक सेकंड बात समझना स्टूडेंट्स 1 सेकंड बात समझना मैं सब कुछ कर लूंगा लेकिन मुझे थोड़ा यह सर्कल की इक्वेशन देख के कुछ तो क्लिक हो रहा है आपको क्या क्लिक होना चाहिए की मुझे क्या इस सर्कल के सेंटर के कोऑर्डिनेट्स पता है अब देखो ए वाली टर्म तो है नहीं क्या मुझे इस सर्कल के सेंटर के कोऑर्डिनेट्स बताएं अब बताओ 2gx तो एक्स का कॉएफिशिएंट क्या -2 -2 का हाफ - 1 - 1 का नेगेटिव वैन और ए वाली टर्म है ही नहीं तो ए का कॉएफिशिएंट जीरो का हाफ जीरो और जीरो का नेगेटिव जीरो से वही -2x यहां लिखा है तो इसके भी सेंटर के कोऑर्डिनेट्स वैन कमा जीरो डेट मैक्स अस अंडरस्टैंड डेट डीज तू सर्कल्स आर को सेंट्रिक सर्कल्स इन दोनों के सेंटर्स से हैं कोई साइड कर रहे हैं मतलब थोड़ा सा विजुलाइज करते हैं सर प्रॉब्लम को फिर समझेंगे दो सर्कल है मेरे पास मेरे पास दो सर्कल है सर है ना अब दो सर्कल में एक छोटी सी बात और मैं आपसे पूछूं तो बता दोगे क्या सर अगर मैं सर्कल की रेडियस निकलना चाहूं तो देखो भाई रेडियस कैसे निकलती हैं सर्कल रेडियस क्या होती है है तो वैन का स्क्वायर प्लस एक्स स्क्वायर 16√4 तो इसकी रेडियस कितनी ए जाएगी 4 में से r2 का देता हूं इसे R1 क्या देता हूं क्या इस बात से आपको कोई आपत्ति ठीक है सर आप जैसा कहें एक सर्कल है जिसकी रेडियस वैन है एक सर्कल है जिसकी रेडियस फोर है अब suniyega यहीं पर इस क्वेश्चन का करेक्ट छुपा हुआ है अगर मैं आपसे बात करूं तो suniyega ध्यान से ये आपका सर्कल है और दोनों सर्कल यहीं पर है बहुत कम की बात है स्टूडेंट्स suniyega और सर इस सर्कल का सेंटर जो है वो कहां है सर वो एक्स एक्सिस पर है क्योंकि सेंटर के कोऑर्डिनेट्स से वैन कमा जीरो इस सर्कल का जो सेंटर है उसके कोऑर्डिनेट्स हैं वैन कमा जीरो होनी चाहिए अब सर सारी बातें भूल जाते हैं बातें हैं इस पॉइंट के पास अब आते हैं इस पॉइंट के पास इस पर आएंगे वापस पर इस पॉइंट के पास आते हैं क्या मैं यहां से क्या मैं यहां से एक ही वैल्यू निकल सकता हूं देखो यहां से एक ही वैल्यू क्या आएगी सर एक ही वैल्यू यहां से आएगी एक्स - 1 क्या यह की वैल्यू यहां रख सकता हूं तो सर यहां से ए की वैल्यू आएगी ध्यान से देखना के बात आपको समझ आई क्या यह बात आपको समझ आई अब सुनना ध्यान से बहुत ध्यान से सुनना स्टूडेंट्स अगर मैं आपसे सीधे सीधे पॉइंट की मुद्दे की बात करूं बहुत ध्यान से सुनना बहुत कृष्ण पार्ट है ये वाला अगर मैं सीधे पॉइंट की मुद्दे की आपसे बात करूं तो सर यह एक स्ट्रेट लाइन की इक्वेशन है यह एक स्ट्रेट लाइन की इक्वेशन है जो की कहां से पास होती है जो की अगर मैं बात करूं तो इंटरसेक्ट फॉर्म में भी निकल सकते हो वर्ण आप सीधे-सीधे भी सोच सकते हो सर्कल में एक्स की जगह जीरो रख डन तो ये हो जाएगा कितना -√3 तो ये हो जाएगा 0 - √3 और suniyega ध्यान से अगर मैं ए की जगह जीरो रख दो अगर मैं ए की जगह जीरो रख डन तो जीरो रखा तो अंडर रूट 3 रह जाएगा तो एक्स की वैल्यू हो जाएगी वैन तो एक पॉइंट हो जाएगा क्या वैन कमा जीरो तो सर ये स्ट्रेट लाइन इन दोनों पॉइंट से पास होती है यह स्ट्रेट लाइन इन दोनों पॉइंट से पास होती हम बात करेंगे मैं वैन कमा जीरो जीरो माइंस अंडर रूट थ्री से पास होने वाली एक स्ट्रेट लाइन बनाना चाह रहा हूं जो की ऑफ कोर्स हम यहां बनाएंगे सुनेगा ध्यान से अगर मैं एक स्ट्रेट लाइन बनाना चाह रहा हूं तो मैं स्ट्रेट लाइन बनाता हूं ध्यान से सुनना है स्टूडेंट्स ट्रक से क्वेश्चन है ना एक स्ट्रेट लाइन मैंने बनाई जो की वैन कमा जीरो और माइंस अंडर रूट थ्री कमा जीरो से तो मैंने 1 कमा जीरो से जाती है ऐसा क्यों किया सर सुनना ध्यान से इसमें एक बात आप देख रहे हो सर अगर ए = एमएक्स + सी फॉर्म में देखे ना तो इसकी स्लोप है अंडर रूट 3 यानी 1060 यानी कितना डिग्री एंगल बना रही है थ्री एंगल बना रही है 60° मतलब यहां पर ये जो एंगल है यहां पर ये जो एंगल है ये है 60° बहुत कम की बात मैं आपसे का रहा हूं स्टूडेंट्स बहुत कम की बस इससे विजुलाइजेशन से जो क्वेश्चन खत्म हो जाएगा कैसे सर समझाइए जरा अब बात सन मुझे एक ही वह सारी वैल्यूज चाहिए मुझे वो सारी पॉजिटिव वैल्यूज ऑफ ए चाहिए सो था अच्छा सो था आपका जो पॉइंट है वो किस रीजन में रहे वो इन दोनों सर्कल से बॉन्डेड रीजन में रहे इन दोनों सार्वजनिक का बॉन्डेड रीजन क्या था सर इन दोनों सर्कल का बाउंड्री रीजन ये था इनके अंदर अगर आप ए रखना चाहते हो इस तरीके से की आपका जो ए है वो इस कंडीशन को फुलफिल करें की ये इस तरीके का एक cardinate है तो सर ए बेसिकली सत स्ट्रेट लाइन पर लाइक करता है तो इन दोनों रीजन के अंदर ए अगर होगा तो या तो यहां इस लाइन पर लाइक करेगा या यहां इस लाइन पर लाइक करेगा लेकिन एक मिनट सर उसने एक और कंडीशन लगाई थी पॉजिटिव वैल्यूज ऑफ ए ये एक्स एक्सिस है इससे नीचे अगर यहां पर ए देखोगे तो वो नेगेटिव होगा ए-एक्सिस पर ए जाएगा ना नेगेटिव साइड पर तो मुझे ये वाले ए चाहिए मुझे ये वाले सारे पॉइंट्स के कोऑर्डिनेट्स चाहिए इनको एक कॉमन तरीका है क्या बताने का सर शायद हो सकता है ढूंढने की कोशिश करते हैं कैसे बहुत ध्यान से सुनना कैसे इसे सुनना मैं आपसे कहना चाह रहा हूं सर अगर मैं इसके कोऑर्डिनेट्स निकल लूं और अगर मैं इसके कोऑर्डिनेट्स निकल लूं तो मैं कहूंगा इन दोनों के बीच में जो पॉइंट्स हैं उन्हें की हम बात कर रहे हैं अब मेरी बात समझ का रहे हो क्या फाइंड डी पॉजिटिव वैल्यूज ऑफ ए है ना तो इसके और इसके बीच की मैं बात कर रहा हूं अगर मैं पैरामेडिक फॉर्म याद करूं कैसी होती थी सर याद करो स्टूडेंट्स भुला मत करो पैरामेट्रिक फॉर्म पड़ी थी ना हमने कैसे पढ़ी थी सर अगर किसी सर्कल के सेंटर के कोऑर्डिनेट्स कॉम के उसकी रेडियस लेते फॉर्म में आप क्या लिखते द आप लिखते द h+ ए कोस थीटा वापस आता हूं आपसे इस बात को करने मुझे जरा बताओ इस पॉइंट के कोऑर्डिनेटर क्या होंगे बस ऐसे ही बता दो यह बताओ इस पॉइंट के अकॉर्डिंग निकलूं तो सुनना h+ ए कितना सर इस अंदर वाले सर्कल की रेडियस A1 और उसे बाहर वाले सर्कल की रेडियस कितनी है 4 आई होप ये नहीं बोले हो इस पॉइंट की अगर मैं कोऑर्डिनेट्स सबसे पूछूं तो आप क्या कहोगे h+ यानी वैन प्लस अगर मैं इसे थोड़ी देर के लिए का डन पी और इसे मैं का डन क्यों तो पहले मैं बात करता हूं आपसे पी की तो पी क्या होगा सुनना सर पी जो होगा वो होगा ह यानी 1 + 1 + रेडियस कितनी है वैन कोस 60 तो कितना हो जाएगा 1 + 1 टाइम्स कोस 60 और ऑफ कोर्स जीरो प्लस वैन टाइम्स की बात है स्टूडेंट्स बहुत कम की बात है ये क्या निकल कर आया है सर ये निकल कर आता है आपका पी क्या इसी तरीके से मैं क्यों भी निकल सकता हूं लेकिन रेडियस क्या है लेकिन रेडियस क्या है फोर मतलब मैं क्या कहूंगा भाई के के कोऑर्डिनेट्स वैन प्लस फोर नहीं जीरो प्लस फोर टाइम्स साइन 60 आई होप वही बातें आपको दिख रही होंगी सीधी सीधी से जो समझ आणि चाहिए अब अगर मैं ऐसे निकलूंगा तो ये कितना आएगा स्टूडेंट्स सोच के देखो कब जो आएगा वो होगा 1 + कोस 60 यानी 1 / 2 ये हो जाएगा 4 / 2 एन है तू कोई तकलीफ तो नहीं है भाई यहां तक अच्छा सर यहां पे बता दीजिए sin60√3/2 तू से फोर तू टाइम्स तो ये कितना हो जाएगा सही हो जाएगा 2√3 तो के के कोऑर्डिनेट्स कितने आएंगे ए जाएंगे थ्री कमा तू रूट थ्री मेरा बस आपसे ये कहना है या तो यह और यह ले लीजिए या फिर यह और यह ले लीजिए आपका आंसर ए जाएगा ए प्लस वैन या अंडर रूट थ्री मैं मैन लेता हूं थोड़ी देर के लिए आपको समझने के लिए लेट से ए प्लस वैन सर ये जो आपका ए + 1 है ये जो आपका ए + 1 है जो एक्स कोऑर्डिनेट्स है वो यहां से यहां तक जा रहा है यहां से यहां तक जा रहा है मतलब 3 / 2 से 3 तक जा रहा है अरे हान या ना 3/2 से 3 तक जा रहा है तो मिनिमम वैल्यू 3 / 2 होनी चाहिए मैक्सिमम वैल्यू थ्री होनी चाहिए समझ ए रहा है की नहीं तो सर इसे तो मैं सॉल्व कर सकता हूं 1 / 2 ये हो जाएगा ए और ये हो जाएगा तू तो सर ए की जो वैल्यूज होंगी वो होंगी 1 / 2 से 2 थॉट्स एंड थॉट्स योर आंसर क्या किसी भी स्टेप में कहीं भी कोई डाउट है इस स्टूडेंट्स कहीं पर भी कोई बात नहीं समझ आई हो तो पूछो मेरा कहना है अगर आपको ये ठीक नहीं लग रहा था तो आप इस पर भी बात कर सकते द आपको सर ये अच्छा नहीं लग रहा है तो आप का सकते द अंडर रूट 3 ये क्या था वही कॉर्डिनेट यह भी पी और के के ए koardinate के बीच में लाइक करेगा मतलब अंडर रूट 3/2 से 2√3 तक क्या भाई √3 / 2 से तू रूट थ्री तक क्या √3 कैंसिल करते हैं मतलब मैं आपसे कहना चाह रहा हूं तीनों को 2/√3 से मल्टीप्लाई करता हूं तो यहां भी 2 / √3 से मल्टीप्लाई किया और यहां भी 2 / √3 से मल्टीप्लाई किया इससे क्या होगा सर आप खुद देख लो भाई आपको दिख नहीं रहा है ऐसा देखो ये तो पूरा वैन हो जाएगा काट रहा हूं समझ में तो ए रहा है ये है जाएगा और ये है जाएगा तो दिख ही रहना तू ए और ये फोर तो ए की वैल्यू कितनी ए जाएगी सर रे की वैल्यू हो जाएगी 1 / 2 और 2 वही बात वापस आई थिंक ये क्वेश्चन और इसका सॉल्यूशन आपको समझ आया और अच्छे से क्लेरिटी डिवेलप हो रही है की जब मुझसे कोई का रहा है की कोई पॉइंट जो इस कंडीशन को सेटिस्फाई करता है टेक्निकल होगा कहना चाह रहा था इस स्ट्रेट लाइन पर लाइक करता है तो वो चाह रहा था की इन दोनों सर्कल्स के रीजन में वो बॉन्डेड हो लेकिन एक ही पॉजिटिव वैल्यू चाहिए मतलब मैं यहां नहीं आऊंगा क्या बात समझ का रहे हो क्या आपको अच्छे से चीज क्लियर हो रही है बस यही बात यहां पर पूछी समझी या जानी गई थी आई थिंक सर्कल को अगर आपने एक सर्कल को एक स्ट्रेट लाइन से इंटरसेक्ट करवाया तो आप खुद सोचो या तो स्ट्रेट लाइन ऐसे निकल जाएगी या फिर हो सकता है क्या बन जाए स्ट्रेट लाइन दस नॉट इंटरसेक्ट विद डी सर्कल इट ऑल क्या यह बात आप सभी को समझ आई हंस रही तीन बातें होंगी तो इस बात को थोड़ा और डिटेल में डिस्कस करते हैं आज का लेक्चर आपका इसी पॉइंट इसी स्टेटमेंट इसी कंक्लुजन पर बेस्ड है मीत डी सर्कल आते इदर वैन ओर तू पॉइंट्स ऑफ कार्स सॉरी हम जानते हैं मैन लो ये एक सर्कल है जिसे हम शुरुआत से डिस्कस करते आए हैं x² + y² + 2gx + 12 + सी अब क्या अब आपसे ये कहना है की सर ये जो लाइन है मैन लो आपकी एमएक्स + सी अब से यहां पर ले चुके हैं तो उसकी जगह मैं क्या ले लेता हूं एमएक्स + अल्फा नौ लेट्स इमेजिन डेट ये जो स्ट्रेट लाइंस है ये सर्कल को इंटरसेक्ट करती है या नहीं करती है कैसा करती है सब बातें अगर मुझे पता करनी है तो क्या सर ज्यादा लंबी चौड़ी बातें मत करो आप तो ये बताओ मैन लो एक स्ट्रेट लाइन है एक स्ट्रेट लाइन है ना और मैन लो यह आपका सर्कल है एक बात बताओ अगर यह स्ट्रेट लाइन अगर यह स्ट्रेट लाइन सर्कल को इंटरसेक्ट नहीं कर रही है अगर यह स्ट्रेट लाइन सर्कल को इंटरसेक्ट नहीं कर रही तो एक बात बताओ इस सर्कल का कोई ना कोई सेंटर होगा जिसे आप क्या कहोगे माइंस के माइंस होगी जैसे आप का रहे हो एमएक्स + अल्फा तो सर मेरा जितना नॉलेज कहता है अब बड़ा खुद सोच के बता देना की आपको नहीं लग रहा की अगर मैं सर्कल के सेंटर से इस स्ट्रेट लाइन पर एक परपेंडिकुलर अगर ड्रॉप करूं तो यह जो परपेंडिकुलर की लेंथ होगी यह ऑफ कोर्स इस सर्कल की रेडियस से ज्यादा होगी अरे हान या ना बिल्कुल सर तो क्या आपको कुछ पैरामीटर समझ ए रहा है जिसके बेसिस पर हम यह डिसाइड करने वाले हैं की वह लाइन इसे टच करती है इससे टच करती ही नहीं है या ऐसे दो पॉइंट पर अट्रैक्ट करती है वो कैसे पता करेंगे कैसे पता करेंगे थोड़े समझा दो उसको समझ ए जाएंगे एक इंटेलिजेंट ब्रेन यहां से खुद से सोचेगा पर देखो कैसे मैन लो सर ये लाइन एक ऐसी है जो इसे टच करती है तो आप खुद बताओ अगर एक स्ट्रेट लाइन से टच करती हुई जाती है तो क्या हो रहा होगा सर अगर ये टच करती हुई जा रही है ना तो फिर तो सीधी सीधी सी बात है सर अगर आपने उससे उसे पर परपेंडिकुलर ड्रॉप किया अगर आपने उसे सर्कल के सेंटर से उसे टैसेंट पर परपेंडिकुलर ड्रॉप किया तो वो क्या होगा रेडियस तो उसे केस में जी कमा माइंस है उसे स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस इक्वल होगी आप समझ का रहे हो जो बातें मैं का रहा हूं और थर्ड कंक्लुजन बड़ा आसान सा ये रहा भाई की सर अगर जो उसने assistrate लाइन को दो पॉइंट्स पर इंटरसेक्ट किया होता अगर 2 पॉइंट पर इंटरसेक्ट किया होता तो क्या हो रहा था तो खुद आप मुझे बता दो भाई क्या हो रहा होता अब तो सर बड़ी सिंपल सी बात है की अब अगर जो मैंने इस स्ट्रेट लाइन पर परपेंडिकुलर ड्रॉप किया तो सर उसे परपेंडिकुलर की लेंथ रेडियस से कम होगी दो बातें दो बातें क्या यार तीन बातें परपेंडिकुलर की लेंथ है वो जो परपेंडिकुलर की लेंथ है वो या तो क्या होगी ग्रेटर दें रेडियस या फिर इक्वल तू रेडियस या फिर लेस दें रेड्स अगर वह परपेंडिकुलर के लेंथ ग्रेटर दें रेडियस है मतलब लाइन इंटरसेक्ट नहीं कर रही है अगर वो परपेंडिकुलर लेंथ रेडियस के इक्वल है मतलब टच कर रही है लेकिन एक्जेक्टली एक पॉइंट पर और अगर उसे परपेंडिकुलर की लेंथ रेडियस से कम है मतलब उसे वो इंटरसेक्ट करके हुए जा रही है दो पॉइंट्स पर आई होप लिंग को यह परिभाषा आपको क्लियर हो रहा है चीज कैसे सोचनी है बस यही बातें आपको मैं समझने की कोशिश कर रहा हूं की सर हम जानते हैं सर्कल का सेंटर हो जाएगा -जी कमा एफ पर ये बात हम शुरुआत से डिस्कस करते आए हैं और सर्कल की रेडियस हम डिस्कस कर चुके हैं g² + x² - सी इनसाइड एंड अंडर रूट अब बड़ी सिंपल सी बात है या तो आप सॉल्व कर लो या तो आप सॉल्व कर लो स्ट्रेट लाइन और सर्कल की इक्वेशन को जब आप सॉल्व करोगे ए की जगह एमएक्स प्लेयर से रिप्लेस कर दोगे तो एक्स में क्वाड्रेटिक बन जाएगी अब जैसे ही एक्स में क्वाड्रेटिक बनेगी तो दो बातें सॉल्व होंगे या तो सॉल्यूशन निकलेगा ही नहीं सॉल्व हो गई या तो एक सॉल्यूशन निकलेगा सॉल्व होकर या फिर दो सॉल्यूशन निकलेंगे अब सीधी सीधी सी बात है कौन डिसाइड करवाइए की सॉल्यूशन नहीं है एक सॉल्यूशन है या दो सॉल्यूशन क्वाड्रेटिक स्टूडेंट्स याद करिए अगर मैं क्वाड्रेटिक इक्वेशन रखूं आपके सामने x² + बीस + सी तो हम जानते हैं सर क्वाड्रेटिक का डिस्क्रिमिनेंट सारा खेल खेलता है discribinant ग्रेटर दैन जीरो मतलब ये केस डिस्क्रिमिनेंट इक्वल तू जीरो मतलब ये केस डिस्क्रिमिनेंट नेगेटिव मतलब नो रियल रूट्स मतलब इंटरसेक्ट कर ही नहीं रही है इक्वल तू जीरो मतलब वैन रूट या इक्वल रूट्स रियल रूट्स ग्रेटर दैन जीरो मतलब तू डिस्टिंक्ट डिफरेंट रियल रूट्स ऐसे भी सोच सकते हो क्या मेरी बातें आपको समझ ए रही है मतलब मैं आपको एक सिंपल सा एग्जांपल लेना चाह रहा हूं जैसे की मैन लो आपके पास सर्कल की इक्वेशन है ना सर्कल की इक्वेशन कैसी होगी सर जैसे लिखा होगा x² + ए स्क्वायर जीरो और अब आपसे कोई पूछे की सर एक स्ट्रेट लाइन एक्सट्रैक्ट लाइन है जैसे ए = 3X - 1 अब वो पूछ रहा है ये स्ट्रेट लाइन ऐसे इंटरसेक्ट करती है या नहीं कैसे पता करेंगे या करती भी है तो कितने करती है करती है तो टच करती है और कितने पॉइंट्स पर करती है नहीं करती जो भी ये बताओ सर तो मैं क्या करूंगा इस पुरी इक्वेशन में जहां-जहां ए है उसकी जगह रख दूंगा तो देखो मैंने इन दोनों को सॉल्व किया तो x² आगे इट इसे है ना फिर क्या ए की जगह रख दिया फिर कितना ए जाएगा 3X - 1 का क्या होल स्क्वायर प्लस ऑफ कोस कितना 2X - ए की जगह रख दिया तो माइंस थ्री एक्स प्लस वैन प्लस थ्री आपको रिलाइज हो रहा है इसे आप सिंपलीफाई करके क्या बनाओगे सर ये एक्स में क्वाड्रेटिक बन जाएगी क्योंकि पावर क्या मिलेगी तू अब उसे क्वाड्रेटिक का आपको क्या निकलना है डिस्क्रिमिनेट की तीन बातें होंगी या तो ग्रेटर दें जीरो होगा या इक्वल तू जीरो होगा ग्रेटर दें जीरो मतलब 2 पॉइंट्स पे इंटरसेक्ट कर रही है इक्वल तू जीरो मतलब टच करके निकल जा रही है 0.1 < 0 मतलब टच ही नहीं कर रही है आते ऑल बस ये आपके इस क्वेश्चन को ये इस तरह के क्वेश्चंस को सॉल्व करने की स्टैंडर्ड ऑपरेटिंग प्रोसीजर रहने वाली है कोई डाउट कोई परेशानी अगर यहां तक चीज है क्लियर है अगर यहां तक चीज क्लियर है यही बात मैं यहां पर आपको कहने की कोशिश कर रहे हैं अगर आप समझ पाए तो है ना अगर यहां तक चीज क्लियर है तो क्या थोड़ा आगे बढ़े थोड़ी और चीज समझे यही बात कहना है नौ देर आर थ्री पॉसिबल केसेस फॉर ए लाइन इन ए सर्कल है ना प्लेन की प्लेन में एक लाइन और एक सर्कल के बीच तीन किस हो सकता है पहला किस्सा लाइन सर्कल से दो अलग-अलग डिस्कस किया था की सर सर्कल के सेंटर से उसे लाइन की जो परपेंडिकुलर डिस्टेंस है वो रेडियस से कम होगी आप विजुलाइज कर का रहे हो ना आई होप से नारियों में बार-बार समझने की कोशिश कर रहा हूं दूसरा केस दूसरा केस क्या होगा सर अगर लाइन सर्कल से एक ही पॉइंट पे मीत कर रही है मतलब वो लाइन सर्कल के लिए किस तरह बिहेव कर रही है सर वो लाइन सर्कल के लिए टैसेंट की तरह बिहेव कर रही है वो लाइन सर्कल के लिए कैसे बिहेव कर रही है टैसेंट की तरह तो उसे केस में क्लीयरली सर सर्कल के सेंटर से उसे पर पब्लिक ड्रॉप करो यानी वो किसके इक्वल होगा रेडियस के या फिर उसे क्वाड्रेटिक के एक ही रूट होंगे इक्वल रूट क्या हो जाएगा तो डिस्क्रिमिनेंट होगा आपका क्या जीरो यही बात वो कहना चाह रहा है अगर आप ठीक से पढ़ ले तो थर्ड बात वह क्या कहना चाह रहा है सन लीजिए ध्यान से अगर लाइन दस नॉट मीत डी सर्कल आते ऑल अगर लाइन सर्कल से मीठी नहीं करती है तो पहला मतलब यह की वो जो क्वाड्रेटिक बन रही है उसके रियल रूट्स नहीं होंगे मतलब डिस्क्रिमिनेंट क्या होगा नेगेटिव या ज्यामिति के लिए सर अगर मैं सर्कल के सेंटर से उसे लाइन की परपेंडिकुलर डिस्टेंस निकलूं तो उसकी रेडियस सिर्फ कोर्स ज्यादा आएगी क्या इन सारी बातों में कहीं पर भी किसी भी चीज को समझने को लेकर आपको कोई परेशानी आई थिंक या आप अच्छे से डाइजेस्ट हो चुका है और ये कॉन्सेप्ट अगर आपने अच्छे से डाइजेस्ट कर लिया है तो हमारा तो तरीका आपको क्लियर है साहब हम तो क्वेश्चंस करेंगे क्वेश्चंस के थ्रू कॉन्सेप्ट समझेंगे और उनके एप्लीकेशन सीखेंगे और यही तरीका वज़ीर और एक वैलिड और एक ऑथेंटिक तरीका है मेरा ऐसा पर्सनली बात है आप इस क्वेश्चन को ट्राई करिए स्क्रीन पॉज करके और मुझे बताइए जरा की आपके अकॉर्डिंग के आंसर ए रहा है देखो भाई फाइंड डी रेंज ऑफ पैरामीटर फॉर विच डी वेरिएबल लाइन डी वेरिएबल लाइन लिस बिटवीन डी सर्कल डी वेरिएबल लाइन लिस बिटवीन डी सर्कल्स विदाउट इंटरसेप्टिंग और टचिंग इदर ऑफ दिस सर्कल है मैन लो ये आपकी स्ट्रेट लाइन है स्ट्रेट लाइन के एक क्वेश्चन क्या है सर ये है वही इस इक्वल्स तू 2X + suniyega बहुत कम की बात है अब बस ध्यान से सन नहीं रहे हो वो ये का रहा है की जो आपकी स्ट्रेट लाइन है ना वो आपके दोनों सर्कल्स को एक सर्कल ये होगा मैन लो है ना और एक और सर्कल होगा सर एक और सर्कल होगा जो की मैन लो यह रहा तो वो ये कहना चाह रहा है की वो स्ट्रेट लाइन कुछ इस तरीके से लाइक करना चाहिए की वो आपके दोनों ही सर्कल्स को किसी भी हाल में किसी भी हाल में क्या ना करें भाई टच करना तो दूर की बात इंटरसेक्ट करना तो दूर की बात उनसे अलग ही रहे और अलग में ऐसा नहीं की दोनों के या तो लेफ्ट में रहेगा दोनों के राइट में रहे नहीं नहीं वो दोनों के बीच में से गुजरे उसे हालत के लिए आपको एक ही वैल्यू निकालनी है क्वेश्चंस समझ ए रहा है क्वेश्चन समझ आते आते पता है आपको हिंट पता चल गई होगी इस क्वेश्चन की कैसे करना है आपको चलिए या नहीं इसकी हिंट पता भाई अब ध्यान से देखो ये क्वेश्चन सर्कल्स का है ही नहीं यह क्वेश्चन तो कोऑर्डिनेट्स सिस्टम जो फर्स्ट चैप्टर में पढ़ा था ना वहां का है ऐसे कैसे बोल रहे हो सर कुछ भी एक कम करो बस सर्कल की नॉलेज से बता दो की यह जो दोनों सर्कल्स हैं इनकी सेंटर हो जाएगा 81 क्या आपको दिख रहा है इसके सेंटर क्या बताएंगे सर इसके सेंटर वैन आपको अभी भी नहीं समझ ए रहा है मैं जो कहना चाह रहा हूं अब तो मैंने काफी कुछ लिख दिया है और अगर इसके आगे अगर मैंने अगला स्टेटमेंट भी का दिया तो फिर यह क्वेश्चन हो जाएगा फिर नहीं रहेगा मेरा कोई भी रोल तो यहां पर मैं चाहूंगा की अब आप मुझे जल्दी से बता दो की इस क्वेश्चन को कैसे करना है बातें बड़ी आसान सी थी स्टूडेंट्स ये क्वेश्चन थोड़ी देर के लिए भूल जाओ आप याद करो जीवन में हमने एक बहुत ही मजेदार सी किस्सा सिखा था किस्सा यह था की सर मैन लो ये एक स्ट्रेट लाइन है यह स्ट्रेट लाइन जीरो की है X2 कुछ याद आया नहीं आया और सोचते हैं मेरा तो बस आपसे यह कहना है भाई की सर आपने यह क्यों नहीं पढ़ा अब भूल क्यों रहे हो की एक रीजन पूरा जो भी यहां से वैल्यूज उठाओगे इस लाइन के ऊपर से वो पॉजिटिव देगा और उसके नीचे से नेगेटिव देगा कौन सा क्या देगा मुझे वहां नहीं पता चल रहा है क्योंकि उन्हें ज्यामिति के लिए प्लॉट नहीं किया उसको बिल्कुल cardinate ज्यामिति बिल्कुल कहां से क्या होगा और मुझे भी पता है एक रीजन आपको पॉजिटिव वैल्यू दे गया है क्रिएशन नेगेटिव देगा मतलब आप समझ रहे हो क्या मैं जो कहना चाह रहा हूं कोई एक रीजन पॉजिटिव देगा कोई एक रीजन नेगेटिव देगा मतलब सुनना अगर मैं आपसे कहूं ax1 प्लस सी मैन लेते हैं पॉजिटिव होगा तो ax2 + सी नेगेटिव होगा और पॉजिटिव इन नेगेटिव क्या होगा नेगेटिव मैन लो यह पॉजिटिव देगा तो यह नेगेटिव देगा तो भी तो प्रोडक्ट क्या होगा नेगेटिव बस यही तो बात हो रही है की आप मेरी बात समझ का रहे हैं अगर आप मेरी बात समझ का रहे हैं तो वापस आते हैं इस क्वेश्चन पर इस स्ट्रेट लाइन की इक्वेशन को हम रराइटिंग है ना बटोर मनोर सर ए को उसे तरफ शिफ्ट करते हैं ए को जैसे ही उसे तरफ शिफ्ट किया तो ये इक्वेशन क्या हो जाएगी suniyega ध्यान से ये हो जाएगा 2X - ए आई थिंक - ए और क्या प्लस ए = 0 पहली कंडीशन की सर दोनों सर्कल्स के सेंटर्स अलग-अलग हो एक इधर हो और एक इधर हो और बातें करेंगे और पहले ये तो होना ही चाहिए बिल्कुल सर तो एक से पॉजिटिव आएगी एक्शन नेगेटिव मतलब दोनों का प्रोडक्ट नेगेटिव तो सुनेगा कैसे 2X - 5 + ए है ना 2X - ए + 1 आपके लाइन क्या है सर 2X - ए + रेट है ना तो suniyega ध्यान से अगर मैं कंडीशन अप्लाई करूं वैन रखा तो हो जाएगा तू है ना माइंस वैन प्लस तू कितना 16 - 1 + ए और इन दोनों का प्रोडक्ट नेगेटिव होना चाहिए इसे सॉल्व किया सर थोड़ा आगे जाकर तो क्या मिलेगा 2 - 1 1 + ए तो इसमें लिख लेता हूं ए + 1 16 - 15 तो ये हो जाएगा 15 प्लस लेता हूं ए + 15 और अगर मैं यहां पर आपसे पूछूं तो क्या आप बोलोगे की सर्वे कर मेथड उसे कर लो याद ए रहा है क्या आई थिंक यह सब तो आपको समझने के लिए मैं बनाता हूं पर अब आपको याद हो जाना चाहिए की सर एक पॉइंट होगा क्या -15 और एक पॉइंट होगा क्या -1 अब बड़ी सिंपल सी बात है सर आपको बता सकते हो एक कम करो\| जीरो रखते हैं जीरो प्लस वैन पॉजिटिव जीरो प्लस 15 पॉजिटिव तो ये दोनों होगा पॉजिटिव ये होगा नेगेटिव और ये होगा पॉजिटिव यहां से ये कंक्लुजन मिल रहा है की आपको ये एक्सप्रेशन नेगेटिव चाहिए तो एक इसको बिलॉन्ग करेगा सर ए बिलॉन्ग करेगा -15 से माइंस वैन के बीच की वैल्यू उसको लेकिन ये क्वेश्चन यहां खत्म नहीं हुआ है यह क्वेश्चन यहां खत्म क्यों नहीं हुआ सर यह क्वेश्चन यहां इसलिए खत नहीं हुआ है क्योंकि अभी तो एक और कंडीशन लगना बाकी है सर सर एक और कौन सी कंडीशन लगेगी आप सब तो कर लिया आपने और क्या बच रहा है आप थोड़ा गैस करके बता सकते हो क्या की अगली कंडीशन अब क्या होनी चाहिए कोई भी स्टूडेंट मुझे दिखा सकता है बता सकता है समझा सकता है की अगली कंडीशन क्या होगी सर ये जो लाइन है ना ये टच भी ना करें ऐसा कुछ करना है आपको ये जो लाइन है ये टच भी ना करें इन सर्कल्स को ऐसा कुछ करना है आपको कुछ पढ़ा है क्या है सामने कुछ ऐसा पड़ा है की आज हमने सर मुझे कुछ तो याद ए रहा है तो याद करो भाई सर इस सर्कल की जो रेडियस है ना इस सर्कल की जो रेडियस है उससे ज्यादा होनी चाहिए इस लाइन की परपेंडिकुलर डिस्टेंस सिमिलरली यह लाइन इस सर्कल को भी टच ना करें मतलब सर इस सर्कल की जो रेडियस है ना उससे ज्यादा होनी चाहिए इस पॉइंट से इस लाइन की परपेंडिकुलर डिस्टेंस हान या ना तो यह दोनों कंडीशंस भी लगानी है पहले सर्कल की रेडियस निकलती हैं तो होता है g² + x² तो 64 + 1 आपका खास है यहां से तो 64 प्लस वैन 65 65 - 61 कितना हो जाएगा 4 और 4√ कितना 2 तो इसकी रेडियस आती है तू आई होप ये बात आपको समझ ए रही है इसकी रेडियस कितनी आती है भाई इसकी रेडियस आती है तू है ना सिमिलरली यहां पे भी निकल लेते हैं वैन स्क्वायर प्लस वैन स्क्वायर वैन कमा वैन से इस लाइन की परपेंडिकुलर डिस्टेंस तू एक्स माइंस फाइव प्लस वैन से क्या हो गया सर देख के बताओ तो यह कितना हो जाएगा 2 - 1 + a² + 1 यानी कितना √5 और इस पे आप ऑफ कोर्स क्या लगाते हो मोड और आपका कहना है की सर ये जो परपेंडिकुलर डिस्टेंस है इस पॉइंट से स्ट्रेट लाइन को इसकी रेडियस से ज्यादा होनी चाहिए मतलब किस से ज्यादा होनी चाहिए सर मतलब वैन से ज्यादा होनी चाहिए ठीक है सर इसे हम सॉल्व कर लेंगे इसी के साथ साथ एक और बात आपको निकालनी है सर ये जो 2X - 5 + है आपकी स्ट्रेट लाइन है 2X - 5 + है आपकी जो स्ट्रेट लाइन है इसकी परपेंडिकुलर डिस्टेंस एट कमा से कैसे निकले टाइप होने से बचाने के लिए क्या लगा लो मोड और यह भी बड़ी होनी चाहिए ये भी ज्यादा होनी चाहिए इसकी रेडियस है जो की कितना है तू कैन दोनों बातों में कहीं पर भी कोई परेशानी थोड़ा सिंपलीफाई कर लेते हैं तो ये क्या लिखा है सर ये लिखा है 3 + A3 प्लस ए और इसका जो मोड है वो होना चाहिए अंडर रूट 5 से बड़ा किसी को भी कोई परेशानी सिमिलरली सर यहां पर अगर मैं देखूं तो ये लिखा है 17 प्लस ए और इसका जो मोड है वो होना चाहिए 2√5 से बड़ा कोई परेशानी सर मोड ए गया है इनिक्वालिटी ए गई है यह कैसे सॉल्व करेंगे 1 मिनट आप कर सकते हो सुनो सारी बातें भूल जाओ मुझे बस यह बताओ की आपने यह पढ़ रखा है क्या आपको कभी यह पता था की सर अगर कभी भी मुझसे कहीं कहे की मोडेक्स जो है वो थ्री से बड़ा होना चाहिए तो इसका ये मतलब होता है की सर जो है वो या तो थ्री से बड़ा हो या फिर एक्स जो है वो -3 से छोटा हो आप खुद सोच के देखो ना माइंस थ्री से छोटी कोई भी वैल्यू लेस से माइंस सेवन माइंस सेवन का नोट 7 7 3 से बड़ा होता है थ्री से बड़ी कोई भी वैल्यू से 10 तो 10 का मोड 3 से बड़ा होता है क्या आपकी बात समझ पाए क्योंकि यही बात यहां पर सोचनी है जब कोई का रहा है की 3 + ए का मोड √5 से बड़ा क्या का रहे द वो का रहा है 3 + ए का मोड √5 से बड़ा हो मतलब 3 + ए या तो √5 से बड़ा हो या फिर इसी के साथ-साथ आपको एक और कंडीशन अप्लाई करनी है की सर 17 प्लस ए का मोड जो है वो 2√5 से बड़ा हो मतलब मैं क्या कहूंगा सर मैं कहूंगा की 17 प्लस ए जो है वो 2√5 से या तो बड़ा होगा या फिर 17 प्लस ए जो है वो नेगेटिव ऑफ तू रूट फाइव से छोटा होगा आसानी से आप डाइजेस्ट कर का रहे हो अब फाइनली हम उसे पाठ की तरह डिस्कस करेंगे बहुत डिटेल में सुनना यहां पर यह क्या का रहा है यहां पर यह का रहा है सर ये जो है यह जो बड़ा होना चाहिए -√5 -3 से -√3 -3 से कोई दिक्कत अच्छा आगे suniyega बहुत कम की बात है suniyega यहां पर यह का रहा है सर ये जो है वो बड़ा होना चाहिए किस तू रूट फाइव माइंस 17 इन सारी बातों में एक बात मत भूलना की हमने एक और कंक्लुजन निकाला था की ए जो है वो माइंस 15 और -1 के बीच में होना चाहिए तो इस कंडीशन वो कंडीशन और तीसरी कंडीशन के साथ आपको सोचना है क्या फाइनली किस कंक्लुजन पर पहुंचेंगे मेरे ख्याल से यहां तक तो कोई परेशानी नहीं है किसी भी स्टूडेंट को किसी भी स्टूडेंट को कोई तकलीफ अगर हो तो देख लीजिए और बता दीजिए वर्ण फिर हम आगे बढ़ेंगे और आगे की कहानी डिटेल में डिस्कस करेंगे यहां तक कोई परेशानी हो तो बता दो भाई बस एक बार थोड़ा क्रॉस चेक कर लेते हैं अगर हमने कोई चीज नहीं की है गलत की हैं तो क्योंकि देखो भाई यह जो cardinate ए रहे हैं वैन कमा वैन ए रहे हैं ना वो जो cardinate ए रहा है वो ए रहा है 8 1 रेडियस हम निकल ही चुके हैं इसकी इक्वेशन ए जा रही है 2X - 5 + ए है ना अब जो हमने सॉल्व 2X - 5 + 1 को किस वैन कमा वैन से है ना तो कितना आता है ये आता है तू माइंस वैन प्लस 3 प्लस एक के लिखा बिल्कुल सही मैंने ये गलती की है भाई हम रियली सॉरी आप में से कई सारे स्टूडेंट्स नोटिस किया होगा की यार आना चाहिए था वनप्लस है ऐसा ही मैं यहां भी देख लेता हूं मुझे क्रॉस चेक कर लेने दीजिए स्टूडेंट्स 8 2 16 - 1 था 15 है ना तो इससे 17 पता नहीं क्यों मैंने इसे है के आई एम रियली सॉरी फॉर डेट अपने सबने नोटिस किया होगा की सर कैसे बेवकूफी कर रहे हो आप है ना तो ये 15 प्लस ए होना चाहिए था ये हो जाएगा वैन प्लस ये है वनप्लस है और ये भी है वैन प्लस 15 बेसिकली ये जो है ये 15 है और ये भी 15 है तो 15 के कारण यहां भी 15 आएगा और यहां भी 15 ही आएगा रियली सॉरी फॉर दिस आई होप यहां तक कोई कन्फ्यूजन आपको नहीं है एंड आय रियली सॉरी फॉर दिस पार्ट अगर इसने कोई कन्फ्यूजन क्रिएट किए हो तो अब कम की बात सुनना स्टूडेंट्स अब बहुत कम की बात है और यही सबसे कृष्ण पार्ट है इस क्वेश्चन का जो आपको समझ लेना है सुनना बहुत कम का पार्ट है यहीं पर स्टूडेंट्स कंफ्यूज होते हैं सुनो हमें यह भी चाहिए और यह भी चाहिए दोनों चाहिए तो पहले इससे पूछते हैं इसका क्या मतलब है इसने देखो इसने क्या कहा सुना मैं नंबर स्केल पे प्लॉट करूंगा सारी चीज नंबर स्किल पर अगर मैंने प्लॉट किया तो देखो अंडर रूट 5 - 1 कितना होता है 2.25 और उसमें से वानस्पतिक है तो 1.2 के आसपास कुछ तो लेट्स से ये यहां आता है ये मैं यहां पे बना लेता हूं कौन अंडर रूट 5 - सुनते जाएगा कम की बात है ना अच्छा दो बातें एक और बात तू टाइम्स अंडर रूट 5 - 15 पॉजिटिव होगा क्या सर मुश्किल लग रहा है क्यों क्योंकि √5 समथिंग तू पॉइंट 2 पॉइंट में तू से मल्टीप्लाई भी किया तो कितना हो जाएगा 4.3 लेट्स से 4.5 फॉर अन ऑयल है ना 4.5 को भी सहूलियत के लिए मैन लेता हूं थोड़ी देर के लिए फाइव फाइव माइंस 15 माइंस 10 तो कितना ए रहा है -10 मतलब यहां कहीं है ना ये एप्रोक्सीमेटली -10 है मैं अपनी सहूलियत के लिए से लिख लेता हूं तू टाइम्स अंडर रूट 5 - 15 संवारे अराउंड -10 ऐसा नहीं का रहा हूं -10 अपनी सहूलियत के लिए प्लॉट करने के लिए नंबर कम की बात है सुनते चले जाएगा हमने इसे एप्रोक्सीमेटली माना है -5 -5 - 15 - 20 तो ये और पीछे चला जाएगा सर तो इससे थोड़ा इधर और खींच लीजिए क्योंकि आपको इधर थोड़ी जरूरत पड़ने वाली है तो लेट्स से मैं इसे यहां पर मैन लेता हूं है ना ये कितना है आपका -2 √5 - 15 जो की कितना है क्योंकि एप्रोक्सीमेटली है -20 खैर इस तक आने की जरूरत ही नहीं पड़ेगी के क्योंकि मैं अभी बस बताने वाला हूं आपको ऐसी क्यों नहीं पड़ेगी अच्छा एक छोटा सा एग्जांपल लो ये वाला पार्ट और हमें सॉल्व कर लेना है देखना ध्यान से माइंस एंड √5 वही 2.5 2.25 मतलब तू में वैन और माइंस तू में माइंस वैन ऐड किया तो माइंस थ्री तो ऐसे -3 यहां कहीं होगा तो टेक्निकल ये जो नंबर है कौन सा - √5 -1 यहां देखिएगा अब कम की बात सुनो बहुत कम की बात मुझे इसका और इसका इंटरसेक्शन लेना है क्योंकि बीच में क्या लगा है अभी बीच में लगा है एंड मतलब मुझे इसका और इसका इंटरसेक्शन है पहले देखते हैं ये क्या का रहा था सर ये बोल रहा था सुनना ध्यान से ये का रहा था सुना √5 -1 से बड़ा होना चाहिए मतलब यह जो है वह कुछ ऐसा होगा यह का रहा था की अंडर रूट 5 - 1 से छोटा होना चाहिए 2√5-15 से बड़ा होना चाहिए तो 2√5-15 से बड़ा होना चाहिए मतलब कुछ ऐसा और इसने कहा की 2√5 - 15 से छोटा आना चाहिए मतलब कुछ ऐसा कम की बातें सुनना अभी एक बात आप भूल रहे हो एक छोटी सी बात आप बोल रहे हो सर क्योंकि आपने एक और बात निकल थी की ये जो है वो माइंस 15 से माइंस वैन के बीच होना चाहिए आपकी एक और बात पर एंड लगेगा की सर ए जो है ए जो है वो माइंस 15 से -1 के बीच होना चाहिए सोच के बोलो ध्यान से देखना यहां पर कुछ ए रहा था कोई नेगेटिव वैल्यू यहां पर ए रही है कोई पॉजिटिव वैल्यू है ना तो कहना है से माइंस वैन ये कितना ए रहा है सर ये माइंस थ्री के आसपास कुछ ए रहा था बस ऐसे ही कहने के लिए दे रहा हूं क्योंकि ऐसे यहां कहीं होगा -1 आई होप आप मेरी बात समझ का रहे हो ये माइंस 10 एप्रोक्सीमेटली -20 तो यहां कहीं होगा आपका -15 तो ये का रहा है की आप माइंस 15 से -1 के बीच रखें - 15 माइंस वैन के बीच मतलब मैं यहां लिख डन तो चलेगा क्या अब इन सब में जो सबसे ज्यादा कॉमन पार्ट है जिसमें ये लाइन ये लाइन और ये लाइन तीनों इंटरसेक्ट कर रही है तीनों का कॉमन पार्ट तीनों का इंटरसेक्शन तो सुनना यहां पे दो तो है पर तीसरी नहीं है यहां पर तो अकेली है यहां पर भी ये दोनों तो है लेकिन तीसरी नहीं है लेकिन सर ये जोन कौन सा जोन सुनना ध्यान से यहां से यहां तक ये तीनों हैं दिख रही हैं क्या कहां से कहां तक देखना मैं हाईलाइट करता हूं यहां से यहां से यहां तक ये जो ग्रीन पार्ट मैंने आपके लिए बनाया है ये जो जोन है इसमें ये भी है ये भी है और ये भी उसके आगे फिर दिक्कत है यहां पर ये है और ये है यहां पर तो सिर्फ यह और इसके बाद सिर्फ यह दोनों है तो मेरा फाइनल कंक्लुजन जो होगा तीनों का इंटरसेक्शन तीनों का इंटरसेक्शन क्या स्टूडेंट्स तू अंडर रूट 5 - 15 से -√5- तो मैं कनक्लूड करूंगा की सर जो ए बिलॉन्ग करेगा सर जो ए बिलॉन्ग करेगा ये जो बिलॉन्ग करेगा वो किसको बिलॉन्ग करेगा 2√5 - 15 ये आपका बिगिनिंग होगी और यहां से कहां तक जाएगा सर यहां से जाएगा ये माइंस एंड रूट 5 - 1 तक इस पार्ट को जरा देख लो इससे अच्छे से समझ लो क्योंकि यही आपका इस क्वेश्चन का है भाई आंसर मुझे सच में बता दो एक बार अच्छे से देख के कहीं कोई डाउट तो नहीं स्टूडेंट्स पूरा पार्ट अच्छे से समझ लो क्या ये पार्ट आपको समझ आया की कैसे सोचा पहली कंडीशन हमने क्या सोची थी की सर आपका जो ए है जो आपका ए है वो इस तरीके से आएगा दोनों सर्कल के सेंटर्स अलग-अलग लाइक करेंगे तो उनकी वैल्यूज पास करने पर ये पॉजिटिव नेगेटिव होगा और दूसरी कंडीशन हमने ये अप्लाई कारी कहीं पर भी इस क्वेश्चन में किसी भी स्टेप में कोई डाउट है तो पूछ लो आई विल reclarify डी होल पार्ट सैम हो सैम वे कहीं ना कहीं किसी ना किसी लेक्चर के दौरान चलिए यह क्वेश्चन अगर समझ आया तो मूव करते हैं नेक्स्ट क्वेश्चन की तरफ और एक और क्वेश्चन देखते हैं इसी तरीके के मिलते-जुलते कॉन्सेप्ट पर बेस्ड पढ़िए भाई क्या लिखा है सुनना बहुत कम की बात है ठीक है सर पीक्यूब अन्य लाइन पीक्यूब अन्य लाइन पासिंग थ्रू दिस ठीक है सर ठीक है यह समझ ए रहा है राइट एंगल अब अगर यह आईटीजे मैन से एडवांस में क्वेश्चन आया ना तो एक स्टूडेंट पहले तो ये सोचेगा की ये सर्कल का क्वेश्चन है क्या तो इंटेलिजेंट स्टूडेंट सोच लेगा यह पढ़ते ही यह पढ़ते ही की ये सर एक सर्कल का क्वेश्चन है ऐसा कैसे सुनो एक बात बताओ एक बात बताओ सोच के देखो अगर अब आपकी एक लाइन है है ना अब आपकी लाइन है जो की किस से गुजर रही है सर जो की ऑफ कोर्स आपकी ए जो है वो है माइंस वैन कमा जीरो ये आपका है ए और बी जो है वो क्या है सर वो है आपका थ्री कमा जीरो suniyega कम की बात है है ना एक सर्कल बना लेता हूं मैन लेता हूं मैं एक सर्कल है ना एक सर्कल है एक सर्कल अगर मैंने बनाना चाहा तो पहले मुझे कायदे से बनाना था सर्कल एक समझदार टीचर की तरह ताकि वह मीटर में फटाफट बना देता अब सुनना ध्यान अगर एक सर्कल है जिसका डायमीटर है अब सर्कल हीसिटेटिंग्ली मैं कहूंगा की अब इस सर्कल की सरकम्फ्रेंसेस पे कहीं पर भी एक 90° एंगल ही सब करेगी सन रहे हो क्या अच्छा एक और बात मैं आपसे जानना या बोलना चाह रहा हूं मैन लो एक लाइन है एक लाइन है ना एक स्ट्रेट लाइन ऐसी और वह इस सर्कल को 2 पॉइंट्स पी और के पर इंटरसेक्ट कर रहे मैन लेते हैं और यह लाइन में जानता हूं सर फोर कमा 1 से पास हो रही है अब आप सन लीजिएगा और मुझे जवाब दे दीजिएगा अगर आपको बात समझ ए गई है तो हान सर एक बात तो बड़ी मजेदार से पता चली ये जो लाइन अब है यह जो लाइन अब है यह पीपल करेगी क्योंकि इसका डायमीटर है ना आई होप आप समझ का रहे हो और सिमिलरली लाइन यह जो लाइन अब है ये पॉइंट के पर भी 90 डिग्री एंगल करेगी क्योंकि ऑफ कोर्स अब जो है वो आपका डायमीटर है डिड योर रिलाइज हो इजी दिस क्वेश्चन bikams अन क्वेश्चन ऑन सर्कल्स डी मोमेंट यू क्रैक दिस पार्ट ऑफ दिस क्वेश्चन डी मोमेंट यू क्रैक दिस पार्टिकुलर पार्ट ऑफ डी क्वेश्चन बट हो अवर गोइंग तू सॉल्व दिस क्वेश्चन डेट्स अंदर बिगड़ डफर क्वेश्चन नंबर suniyega क्वेश्चन आसान है यह इसी कॉन्सेप्ट पर बेस्ड जो आज पढ़ाया गया है मेरा कहना है सर आप मुझे पीके लाइन की इक्वेशन लिखने में हेल्प कर सकते हो क्या उसने खुद तो बोला स्लोप एम है और वो फोर कमा वैन से पास होती है तो पीके की क्वेश्चन क्या होगी सर ए - y1 इस इक्वल्स तू एम एक्स माइंस नहीं है अब बस मुझे बता दो अगर यह लाइन अब इस पीके के किन्हीं दो पॉइंट्स पर अगर 90° सब्सटेंड कर रही है दो पॉइंट्स पर तो यह तब ही संभव है जब पीके ना तो इसे टच करें और ना ही इसे बाहर हो जब ये पीके टच करेगा तो एक ही पॉइंट पे होगा पीके एक ही पॉइंट हो जाएगा वही 90° एंगल बनेगा ये तब ही संभव है तब ही पॉसिबल है जब पी के इस सर्कल को दो डिस्टिंग पॉइंट्स पर इंटरसेक्ट कर रही हो आप यह हिंट समझ का रहे हो क्या यह क्वेश्चन खत्म हो जाएगा बस अब आप मुझे आंसर बता दो तो क्योंकि अब जो भी इस क्वेश्चन के लिए नीडेड नेसेसरी इनफॉरमेशन या डाटा था आपको दे दिया गया है इससे ज्यादा मेरे ख्याल से बताने का कोई सेंस ही नहीं है मेरा यकीन करिए पर कैसे करेंगे सर ये क्वेश्चन कैसे सोचेंगे इसमें क्या करना होगा मुझे तो कोई आइडिया दिख नहीं रहा है सर इस क्वेश्चन को करने का मैं का रहा हूं एक एक छोटा सा थॉट लाओ पहले तो मुझे बताओ सारी बातें छोड़ो की आप इस सर्कल की इक्वेशन लिख सकते हो अरे भाई आप कैसी बातें करते हो कभी कभी सर्कल की इक्वेशन भाई यह सर्कल इस सर्कल की क्वेश्चन कैसे लिखेंगे इस सर्कल की इक्वेशन तो बड़ी सिंपल सी है ना की डायमीटर है आपको डायमीटर के कोऑर्डिनेट्स नहीं दिख रहा है माइंस वैन कमा जीरो थ्री कमा जीरो एक क्वेश्चन खत्म हो गया है सर्कल की इक्वेशन तो निकल चुकी है सर एक्स माइंस एक्स वैन तो एक्स - 1 या नहीं एक्स + 1 एक्स - X2 तो ये कितना हो जाएगा एक्स - 3 + ए - y1 यानी कितना ए - 0 ए - Y2 तो ए - 0 वही माइंस जीरो और इसे इक्वल तू जीरो और इसे आप थोड़ा सिंपलीफाई कर लो तो ये दिखेगा x² + y² - 3X + एक्स कितना -2x और कितना -3 = 0 ये इस सर्कल के क्वेश्चन और मैं बचका का सकता था सर इसका जो मिड पॉइंट होगा वो पॉइंट ऑफ इंटरसेक्शन होगा पॉइंट ऑफ इंटर मिड पॉइंट जो होगा वो इस सर्कल का सेंटर होगा इनके अगर पूछा जाता तो वो कितना होगा बस ऐसे ही पूछ रहा हूं रेडियस कितनी होगी बाय डी वे सो रेडियस भी दिख ही रही है डायमीटर की लेंथ कितनी है 3 - -14 जीरो माइंस जीरो जीरो सर्कल की रेडियस कितनी आती है सर इस सर्कल की रेडियस आती है तू क्यों निकल रहे हो क्योंकि शायद मुझे इसकी जरूरत पड़ेगी सारी बातें छोड़ो क्या आप मुझे सर्कल के सेंटर के अकॉर्डिंग बता सकते हो शायद चाहो तो यहां से बता देंगे यहां से बता दो थ्री माइंस वैन तू तू डिवाइडेड बाय तू वैन जीरो बाय जीरो जीरो मतलब जीरो प्लस जीरो जीरो बाय तू इसे जीरो बात समझ जाइए आपको तो इस सर्कल के सेंटर के कोऑर्डिनेट्स से वैन कमा जीरो अब सन लीजिएगा बस कहानी यही खत्म हो जाएगी किस कॉन्सेप्ट से सर ज्यामिति के लिए हमने पढ़ा है अगर ये लाइन में चाहता हूं सर्कल को दो जगह इंटरसेक्ट करें मैं चाहता हूं की यह लाइन इस सर्कल को दो जगह इंटरसेक्ट करें यह तब ही पॉसिबल है यह सिर्फ और सिर्फ पॉसिबल है जब इस सर्कल के सेंटर से इस सर्कल के सेंटर से इस लाइन पर ड्रॉप किए गए परपेंडिकुलर की लेंथ इस सर्कल की रेडियस से कम हो इक्वल भी नहीं चलेगा उसे जगह में तो एक ही जगह इंटरसेक्ट करेगा यानी टच करेगा और वो एक ही पॉइंट पे 90 डिग्री celtate करेगा मुझे कहां चाहिए मुझे एक्जिस्ट दो पॉइंट्स पर चाहिए एक्जेक्टली प्रॉब्लम समझ ए रही है और शायद उसे रद्द हो जाए मैं एम की रेंज निकल लूं बिल्कुल सर पहले तो इस लाइन को थोड़ा फिर से ही राइट करते हैं देखो ये लाइन की इक्वेशन क्या ए जाएगी सुनना इस लाइन की क्वेश्चन ए जाएगी देखो ये ए जाएगी सर एमएक्स है ना एमएक्स ए इधर आए तो क्या हो जाएगा -ए फिर क्या हो जाएगा -4m और ये क्या हो जाएगा प्लस वैन इसे इक्वल्स तू जीरो ये आपकी स्ट्रेट लाइन है और इसकी डिस्टेंस चाहिए वैन कमा जीरो से तो suniyega आपकी स्ट्रेट लाइन क्या है सर आपकी स्ट्रेट लाइन की इक्वेशन आई होप बहुत डिफिकल्ट नहीं हो रहे हैं लेक्चरर्स आपको एक अच्छी अंडरस्टैंडिंग बिल्ड हो रही है की किस तरह के क्वेश्चंस आपको डील करना है और ऑफ कोर्स सेंटर के cardinates के सर सेंटर के cardinate से वैन कमा जीरो तो इसकी परपेंडिकुलर डिस्टेंस निकालनी है मुझे वैन कमा जीरो से निकल सकते हो क्या हान सर ये तो फॉर्मूला पड़ा है 1 एम तो कितना म 0 -1 0 फिर ये कितना ए जाएगा -4m+1 और इस पर आप क्या लगाते हो मोड डिवाइडेड बाय अंडर रूट ओवर m² + - 1² यानी 1 + m² + 1 जो भी आप लिखना चाहें सर ये जो डिस्टेंस है यह जो इस सेंटर से इस लाइन की परपेंडिकुलर डिस्टेंस है इसकी रेडियस रेडियस कितनी है सर इसके रेडियस तू से कम होनी चाहिए इसकी रेडियस तू से कम होनी चाहिए क्या इस क्वेश्चन को सॉल्व करना बहुत डिफिकल्टी आपके लिए करते हैं इससे वहां शिफ्ट किया तो ये कितना हो जाएगा एन - 4m 3m -3 हो जाएगा 1 - 3m का मोड लेस दें तू टाइम्स अंडर रूट ओवर 1+m2 सर एक छोटा सा कम करते हैं दोनों तरफ स्क्वायर कर लेते हैं क्योंकि देखो इधर अंडर रूट वाला पार्ट है सो डी बेस्ट थिंग विच आई एम सजेस्टिंग और थिंकिंग ऑफ डूइंग इसे इट्स स्क्वायर बोथ डी साइड्स सो सी कैन एड ऑफ डी स्मॉल आगे वेल आगे स्क्वायर थिंग तो 1 - 3 एम का स्क्वायर है ना तो मैं यहां पे क्या लिखूंगा 1 - 3 एम का स्क्वायर एंड दिस इस लेस दें तो ये कितना हो जाएगा 4 टाइम्स वनप्लस m² इसे आपने अगर सिंपलीफाई करना चाहा तो देखो भाई क्या दिखेगा चलिए ए जाएगा वैन प्लस नहीं होनी चाहिए ये क्या है सर 4 कर्सिव राइटिंग में क्यों लिख रहा हूं मैं है ना ये हो जाएगा 4 + 4m² कोई दिक्कत स्टूडेंट्स नहीं होनी चाहिए चीज अगर शिफ्ट करते हैं 9 m² - 4m² तो कितना हो जाएगा बोलो भाई 5n² यह लिखा हुआ है -6 कितना हो जाएगा स्टूडेंट्स की सर यहां पर अब मुझे क्या चाहिए 5 3 15 और 15 के लिए तो मुझे 5 और 3 ही दिखाई दे रहे हैं कहीं कुछ गड़बड़ तो नहीं की कहीं कुछ गलती तो नहीं की इसे एक बार देख लो भाई क्योंकि मुझे शायद कुछ अजीब सा दिख रहा है क्या करें कहीं कुछ दिक्कत लग रही है क्या क्योंकि स्प्लिटिंग द मिडिल टर्म कम नहीं कर रही है तो कुछ गड़बड़ तो नहीं की ना हमने कुछ गड़बड़ की होती तो नहीं किए मुझे जो लग रहा है वैसे मैं आगे बढ़ता हूं देखते हैं जो भी आएगा नहीं किया है ना जहां तक मैं शो करूं नहीं किया तो अब अगर मैं स्प्लिटिंग डी मिडिल टर्म उसे करता हूं नहीं कर का रहा हूं तो श्रीधर आचार्य मेथड तो 5 m² - 6x 6m - 3 इस < 0 अब सारी बातें छोड़ो आप तो ये समझो अगर मुझे कभी भी ऐसा दिखे लिखा हुआ दिखे की सर इसके दो रूट्स हैं एक्स - अल्फा एक्स माइंस बिता लेस दा वलम गिवन विले आई एम गिवन के सर अल्फा जो है वो बिता से छोटा है तो मुझे कोई कहे की ये सर आपका बेबी कर मेथड यहां पे अल्फा है और ये बिता है और ऑफ कोर्स ये आपका माइंस इंफिनिटी और इंफिनिटी है आपको चाहिए था की एक्सप्रेशन लेस दें जीरो हो तो यहां पॉजिटिव आता है नेगेटिव आता है पॉजिटिव आता है ये आपको याद हो जाना चाहिए आप बचका ट्री करके देख लेना अल्फा से बिता की किसी भी वैल्यू के लिए एक्सप्रेशन नेगेटिव होगा यह हमेशा ट्रू अगर ऐसा कोई एक्सप्रेशन बनकर ए रहा है तो मैं बात करना चाह रहा हूं इस क्वेश्चन के कॉन्टैक्ट में कैसे सुना अगर मैं इसके रूट्स निकलता हूं तो अल्फा बेटा कहेंगे सुनना तो जब आप इसके रूट्स निकलोगे तो अल्फा बेटा जो आएंगे माइंस भी यानी सिक्स का माइंस सिक्स का नेगेटिव सिक्स प्लस माइंस अंडर रूट ओवर b² यानी 36 - 4ac तो माइंस माइंस प्लस 4 3 12 12 5 60 अपॉन 2 फैक्ट्रीज किया जा सकता है 96 4 से डिवाइड होता है तो 96 को मैं लिखूंगा 24 4 लिख सकता हूं क्या 24 भी क्या होता है 6 4 तो लिख सकता हूं 6 4 4 है ना तो मुझे दिख रहा है 6 4² देखो आप मेरी बात समझ का रहे हो 24 गए 4 टाइम्स होता है ना तो 96 को मैं लिख का रहा हूं 6 4² 6 4 स्क्वायर ए जाएगा फोर बाहर रहेगा तो ये हो जाएगा 4 टाइम्स अंडर रूट सिक्स डिवाइडेड बाय 10 आई थिंक सर सिंपलीफिकेशन पॉसिबल है ये किसकी वैल्यू है सर अल्फा और बिता की है ना तो तू से आई थिंक अगर मैंने इसे सिंपलीफाई किया तू से अगर सिंपलीफाई किया तो देखो कितना हो जाएगा ये हो जाएगा 3 + - 2 टाइम्स √6/5 आई थिंक वे सी कर मेथड क्या कहती है सर सी सी कर मेथड कहती है की यहां पे कहीं होगा - इंफिनिटी मैं लिख रहा हूं ऐसा करने की जरूरत आपको नहीं होनी चाहिए ये होगा सर थ्री माइंस तू सिक्स डिवाइडेड बाय फाइव ये बिता पॉइंट होगा आपका क्या थ्री प्लस तू रूट 6/5 और यहां कहीं जाकर इंफिनिटी आएगा अब आपको चाहिए था क्या सर आपको चाहिए था की आपका एक्सप्रेशन कैसा हो याद करो आपका एक्सप्रेशन आपको चाहिए था नेगेटिव होता है अल्फा और बिता के बीच में मतलब इन दोनों के बीच में यहां पे ये पॉजिटिव होगा यहां नेगेटिव होगा और ये पॉजिटिव होगा नहीं नेगेटिव साइन नहीं है नेगेटिव जॉन है क्या बात समझ का रहे हो तो आपका आंसर होगा आपका आंसर होगा सर की 3 - 2√6/5 से लेकर 3 + 2 रूट 6/5 तक के बीच की सारी वैल्यूज क्या होंगी एम की रेंज अगर इस लाइन की स्लोप इन दोनों वैल्यूज के बीच में रही तो डेफिनेटली स्ट्रेट लाइन डेफिनेटली स्ट्रेट लाइन इस सर्कल को दो डिस्टिंक्ट पॉइंट्स पर इंटरसेक्ट करेगी उससे क्या होगा उससे ये होगा सर की आपके ये जो दो पॉइंट्स हैं ये जो दो पॉइंट्स हैं बनने वाला जो बेसिकली डायमीटर है वो इस लाइन पर पी और के पर 90° एंगल सब्सटेंड करेगा व्हाट टू तो नहीं था एक क्वेश्चन क्या आपको पूरा क्वेश्चन समझ आया और आपको रिलाइज हो रहा है यही जी मांस और एडवांस एग्जाम का नेचर है पहली बात तो आपको क्वेश्चन देख के रिलाइज ही नहीं होगा की ये किस चैप्टर का क्वेश्चन है आपको रिलायंस होगा की सर ये सिर्फ सर्कल्स का नहीं है सर्कल्स के बजाय स्ट्रेट लाइन भी लग गया डिस्टेंस फॉर्मूला भी लग गया cardinate सिस्टम आते आते सर थोड़ा अगर मैं आगे गया तो मुझे तो अलजेब्रा भी लगने लगा मॉड्यूल से डील करना पद गया अलजेब्रा में सर मोड और इनिक्वालिटी से डील करते-करते एहसास हुआ की हमें तो क्वाड्रेटिक से भी डील करना है और अभी तो आप रुकिए थोड़ा पेशेंस रखिए जनाब अभी सीक्वेंस इन सीरीज डिटर्मिननेंट्स मेट्रिक्स पीएनसी प्रोबेबिलिटी स्टैटिसटिक्स सब दल देंगे वो वो ट्रिगो दल देंगे लोगरिथम आएगा सब कुछ मार्च करके मंगल करके पूछेंगे और मजेदार क्वेश्चंस बनेंगे जिनको सॉल्व करके आपके दिमाग के सारे तार खुल जाएंगे और आपको ऐसा लगेगा की हान कुछ किया आप समझ रहे हो और ऐसे ही चीजों को करने का अलग ही मजा है चैलेंज जस्ट नीचे है ना अच्छे क्वेश्चंस करिए वो ना सिर्फ आपके दिमाग के दार खोलेंगे आपको बहुत एक्सप्लोर करने में हेल्प करेंगे बल्कि बस शर्तें आपको नए नई चीज भी सिखाएंगे बहुत ध्यान से सुनना है स्टूडेंट्स मैन लो की मेरे पास एक सर्कल है है ना मैन लो मेरे पास एक सर्कल है और इस सर्कल पर मैं किसी पॉइंट से किसी पॉइंट से ड्रॉ करता हूं मैंने किसी पॉइंट से ड्रॉ किया तो इस पॉइंट से आई ड्रा अन सैम व्हाट लाइक दिस किसी भी स्टूडेंट को कोई डाउट आई कूद हैव एक्सटेंडेड इट फॉर डी बट मैं यहीं पर ऐसे रोकना चाह रहा हूं कोई रीजन होगा सर ये सर्कल का कोई ना कोई सेंटर होगा ऑफ कोर्स सर्कल का सेंटर होगा अगर मैं मैन लेता हूं सर्कल है आपका वही स्टैंडर्ड एक्स स्क्वायर + y² + 2G एक्स + 2 अब suniyega ध्यान से सर अगर आप सर्कल के सेंटर जिसके कोऑर्डिनेट्स हैं -3 - एफ इस पॉइंट के जिसके कोऑर्डिनेट्स में मैन लेता हूं y1 से tagent ड्रॉ करता हूं क्या आप एक बात जानते हो या समझ रहे हो स्टूडेंट की सर आप सीधा-सीधा एक कम क्यों नहीं करते की अगर इस पीस है इस सर्कल के सेंटर को कनेक्ट किया जाए इस पी से सर्कल के सेंटर को कनेक्ट किया जाए और सर्कल के सेंटर से लेट से यहां पर एक परपेंडिकुलर ड्रॉप किया जाए तो आपको ये क्यों नहीं दिख रहा है की ये जो ट्रायंगल है पीटीसी ये एक राइट एंगल ट्रायंगल है ये जो ट्रायंगल है पीटीसी ये राइट एंगल ट्रायंगल विच इस राइट एंगल बनाता है टी पर किसी भी स्टूडेंट को यहां तक कोई परेशानी मेरा ये कहना है सर मुझे दो बातें पता है मुझे कौन सी दो बातें पता है पर मुझे एक बात तो ये पता है की पी से सी तक की डिस्टेंस क्या है कैसे पता है डिस्टेंस फॉर्मूला क्या समझ का रहे हो सर मुझे सी से टी तक कैसे पता चल जाता है सर रेडियस है ना भाई तो सर ये क्या हो गई रेडियस या तो इसे रेडियस लिख लो या फिर क्या लिख लो अंडर रूट ओवर जी स्क्वायर प्लस एक्स स्क्वायर पता है ये बात क्या हो जाएगी सर ये हो जाएगी अंडर रूट ओवर डिस्टेंस फॉर्मूला लगाया तो X1 - - न X1 + कोस क्या जी का होल स्क्वायर प्लस y1 - एफ यानी y1 + एफ का होल स्क्वायर इन साइड एंड अंडर रूट मुझे ये डिस्टेंस और डिस्टेंस पता है सर राइट एंगल में हाइपोटेन्यूज पता है परपेंडिकुलर पता है तो क्या बेस पता नहीं कर सकते पाइथागोरस थ्योरम तो मैं पॉइंट को क्या कहूंगा सर मैं पॉइंट अगर लेंथ निकलने में इंटरेस्टेड हूं तो मैं पॉइंट के बारे में बड़ी साधारण से टिप्पणी करूंगा जो pt² होगा है ना जो pt² होगा ही तो आप समझ रहे हो बेस का स्क्वायर क्या होगा हाइपोटेन्यूज का स्क्वायर माइंस परपेंडिकुलर का स्क्वायर hypotaneous है सर हाइपोटेन्यूज है इसका स्क्वायर मतलब दिस एंड दिस एंड्रयू कैंसिल सो X1 + जी का होल स्क्वायर प्लस y1 + एफ का स्क्वायर है ना तो X1 + जी का स्क्वायर मतलब सर x1² y1 + एफ है ना तो एफ का स्क्वायर प्लस तू fy1 मिनट से थोड़ी और चीज आप करने वाले द क्या बिल्कुल इसके साथ में इसमें से आर का स्क्वायर हाइपोटेन्यूज स्क्वायर माइंस परपेंडिकुलर स्क्वायर ए होप आप समझ का रहे हो मुझे कहना चाह रहा हूं मैं ये कहना चाह रहा हूं की पाइथागोरस थ्योरम से मैं क्या कहूंगा मैं कहूंगा सर की pt² + tc² जो होगा वह पीसी स्क्वायर को क्या कहूंगा pc² - tc² और tc² मैंने इसका स्क्वायर अंडर रूट से स्क्वायर कैंसिल तो g² + x² - सी आपको सब्सट्रैक्ट करना है तो क्या सब्सट्रैक्ट करना है सर आपको आपको सब्सट्रैक्ट करना है g² f² और -सी यानी माइंस ऑफ माइंस यानी प्लस सी कोई दिक्कत मेरे ख्याल से नहीं होनी चाहिए सर एक बात देखो आप पता नहीं क्यों ध्यान से देखते नहीं हो ये और ये और ये कैंसिल हो जा रहा है अब चीज जरा ध्यान से देखना ये pt² मतलब सही हो जाता है x1² + में लिख लेता हूं y1 स्क्वायर प्लस में लिख लेता हूं 2G X1 प्लस में लिख लेता हूं अगर ये है तो पॉइंट की लेंथ कितनी हो गई भाई श्री कुछ रीसेंबल होता हुआ नहीं लग रहा है आर यू नॉट एबल तू अंडरस्टैंड की कुछ तो रीजनेबल करें इट इस रीसेंबलिंग विथ समथिंग समथिंग विच सी ऑल स्टेड एक्स स्क्वायर + ए स्क्वायर है तो सर्कल की इक्वेशन में आपने X1 y1 रख दिया है बिल्कुल सही बात बिल्कुल सही बात मतलब इतना आसान सा ये कॉन्सेप्ट है किसी भी पॉइंट से किसी भी पॉइंट से किसी सर्कल पर ड्रा की गई टैसेंट की लेंथ इस गिवन बाय उसे सर्कल की इक्वेशन में जिस पॉइंट से आपने टैसेंट ड्रॉ की है वो cardinate रख दीजिए और उसका अंडर रूट ले लीजिए थॉट्स इट थॉट्स ऑल डी पॉइंट सेस आज का कॉन्सेप्ट बस इतना सा ही है क्या कहीं पर भी कोई परेशानी सुनते जाइए ध्यान से आज की बात मैन लेते हैं एक टैसेंट आपने ड्रा किया किसी पॉइंट से हम जानते हैं सर किसी भी पॉइंट से आप सर्कल पर अगर ड्रॉ कर पाते हैं तो इंजन ड्रा कर पाएंगे अगर वो बाहर ए पॉइंट तो और वो दोनों ही पॉइंट्स की लेंथ से होती है मतलब पी से एक ही लेंथ और पी से बी की लेंथ दोनों ही से होगी अच्छा सर सीधी सीधी सी बात है वही सारी बातें जो यहां पर लिखी हैं तो फाइनली मैं इस इस स्टेटमेंट को क्या कहूंगा एस वैन एस वैन का मतलब क्या होता है ज्यामिति अगर आपके पास एक सर्कल है जैसे मैं कहता हूं x² + y² होगा अगर आपका X1 y1 तो ये आपने यहां पास कर दिया क्या बात समझ पाए और आपको निकलना है अंडर रूट S1 क्योंकि यही आपकी उसे पॉइंट से उसे पर टच होने वाले मतलब बेसिकली tagent की डिस्टेंस होगी जो की क्या हो जाएगी जो की होगी x1² + y1² + 2gx1 + 2fy वैन प्लस सी तो सर इतने आसान क्वेश्चंस आएंगे क्या बिल्कुल नहीं सर एडवांस में इतने आसान क्वेश्चंस तो आएंगे नहीं की आपको X1 y1 रख देना है और अंडर रूट रखकर उसकी डिस्टेंस निकल देनी है पर हान इस पर बेस्ड क्वेश्चंस आएंगे यह एक एप्रोच है अब इसका आइडिया और डेवलपमेंट करेंगे की आपको ये पता है या नहीं कुछ अच्छे शानदार से बेहतरीन से क्वेश्चंस के थ्रू वही बात वापस क्वेश्चंस ट्राई करेंगे सर्कल मतलब इन दोनों सर्कल्स के सेंटर से है ठीक है सर बीइंग टिस डेट ऑफ सी वैन सी वैन की जो भी रेडियस उसका डबल है C2 के रेडियस सब चीज थोड़ी बना ही ले तो अच्छा बिल्कुल बना लो भाई तो दो को सेंट्रिक सर्कल है ये एक सर्कल है और एक और सर्कल है आपका जिसकी रेडियस इस सर्कल का डबल है जिसकी रेडियस इस सर्कल का देखना चाह रहा हूं बिल्कुल डबल है है ना अब क्या सर suniyega यह कोसिन का मतलब दोनों का सेंटर से जगह मतलब अगर मैं इस अंदर वाले रेडियस को अंदर वाले रेडियस को कहूं आर तो बाहर वाली रेडियस कितनी है सर बाहर वाले जो सर्कल की रेडियस है डेट इसे 2r यहां तक तो बात है उसने बात करते हैं एक और बात करें C2 जो है डेट इसे ठस ऑफ सी वैन तो C2 आपका बाहर वाला सर्कल है मैं इसे लिख देता हूं यहां पर ये है आपका C2 और ये आपका अंदर वाला सर्कल है C1 ये पॉइंट नहीं है सर्कल्स के नाम है ना एवं और C2 अब क्या सर अब सुनते जाएगा फ्रॉम अन पॉइंट पी ऑन सी तू तो सर C2 पर एक पॉइंट है पी सर C2 पर मैन लेता हूं मैं C2 पर दिस इस अन पॉइंट पी है ना सो दिस इस अन पॉइंट पी एक तो का है ना लेट्स क्वालिटी कर रहा हूं इसको थोड़ा और ठीक से बना लेते हैं ये हो जाएगी आपकी क्या ब आई थिंक हो गई आपकी बीवी तो ये क्या हो गया सर ये हो गया का और ये हो जाता है ब कोई दिक्कत ये इन दोनों का सेंटर है ना कोई तकलीफ किसी भी स्टूडेंट को प्रूफ डेट डी सेट्रॉयड ऑफ डी ट्रायंगल पब लिस ऑन C1 वो क्या कहना चाह रहा है suniyega क्वेश्चन खत्म हो जाएगा अगर आप चीज ध्यान से देख पाते हो पब ये जो ट्रायंगल है इसका जो सेट्रॉयड है वो ट्रायंगल सी वैन पर लाइक करते हैं मतलब आप से कहना चाह रहे हैं अगर आपने इस ट्रायंगल को कंप्लीट किया अगर आपने ए और बी को मिलाया है ना तो इस पब का जो सेट्रॉयड है ना वो यहां कहीं लाइक करें ऐसा वो आपसे बोलना पूछना है समझाना ये बताना चाह रहा है मतलब वो इस पर यहां कहीं लाइक करेगा कुछ ऐसी वो बात आपसे कर रहा है इन बातों को लेकर कोई कन्फ्यूजन तो नहीं हो रही है स्टूडेंट्स कोई दिक्कत तो नहीं हो रही है क्या एक क्वेश्चन बहुत डिफिकल्ट तो नहीं हो गया आई थिंक सर कर तो सकते हैं बट करना क्या है कुछ समझ नहीं ए रहा है अपने बना तो दिया है बट करेंगे कैसे थोड़ी सी hintlo थोड़ा सा पिक्चर आपके सामने है आपको बस ये प्रूफ करना है की ये जो पीईबी का centralide है वो C1 पर लाइक करेगा कोई आइडिया कोई थॉट कोई कोई भी दिमाग में बात ए रही है तो बताओ स्टूडेंट्स एक बार ट्राई कर लो एक बार खुद से तो देखो क्योंकि होना चाहिए इस क्वेश्चन में ज्यादा ऐसी कोई बहुत डिफिकल्ट या मुश्किल बातें हैं नहीं अगर मैं सच कहूं तो क्योंकि बात समझो स्टूडेंट्स आप कैसे निकलोगे सर दो बातें सुनना मुझे बस यह बताओ सबसे पहले तो स्टूडेंट्स आप देख का रहे हो की क्या मैं इस एंगल को थोड़ी देर के लिए थीटा कहूं तो चलेगा क्या का सकता हूं क्या सुनना ध्यान से इसको कहने का थीटा रीजन क्या है अच्छा मैं इस पॉइंट को लेट से थोड़ी देर के लिए का देता हूं ओ कोई तकलीफ तो नहीं है अच्छा मैं ओ से सर ए को ज्वाइन करता हूं और मैं ओ से बी को भी ज्वाइन करता हूं तो सर जब आपने ओ से एक ज्वाइन किया ना आई होप आप देख का रहे हो मैं इसको थोड़ा और अच्छे से बनाता हूं की सर जब आप ओ से ए को ज्वाइन कर रहे हो मैं थोड़ा कुछ अलग बना देता हूं ताकि आप चीज है कंफ्यूज ना हो और चीज अच्छे से समझ में और सिमिलरली सर ओ से बी को भी ज्वाइन कर लेते हैं और इसको भी हम कुछ बना लेते हैं कुछ ऐसा बन रहा है ना अब suniyega ध्यान से बहुत ध्यान से सुनना है मैं आपसे बात करना चाह रहा हूं अभी फिलहाल ट्रायंगल पाओ में सर एक और बात सुनना ध्यान से क्या मैं जानता हूं ये जो ए है ये इसकी रेडियस है जो की मैं का दूंगा स्मॉल आर है और सर आप क्या है अरे आप बाहर वाले सर्कल की रेडियस है आपको मैं अगर यहां कन्फ्यूजन हुई तो अलग से लिखता हूं अच्छा बताओ ओए क्या है सर ओए अंदर वाले सर्कल की रेडियस है जैसे हम क्या कहेंगे स्मॉल आर अच्छा आप क्या है सर आप बाहर वाले सर्कल की रेडियस है C2 की जो की है तू आर सुनना ध्यान से अगर ये राइट एंगल ट्रायंगल है कौन सा सर अगर आपका राइट एंगल ट्रायंगल है ओ का तो क्या ये एस राइटिंग थीटा निकल सकता हूं क्यों निकल रहे हो बस थोड़ा सा पेशेंस रखो अगर मैं सिन थीटा निकलूं तो सिन थीटा क्या होता है सर ये थीटा है परपेंडिकुलर क्या होता है तो परपेंडिकुलर अपॉन हाइपोटेन्यूज तो ये क्या हो जाएगा ओए अपॉन ओ पी आई होप अगर कंफ्यूज हो गया तो मैं फिर से का देता हूं सिन थीटा क्या होगा परपेंडिकुलर अपॉन हाइपोटेन्यूज ये बेस है ना तो ओए अपॉन ओ पी ओ ए कितना है सर ओए है आर और आप कितना है सर ओ पी है तू आर तो ये कितना ए जाता है 1 / 2 1 / 2 का क्या मतलब सर 1 / 2 का मतलब अगर साइन थीटा आपका आया 1 / 2 तो थीटा कितना होगा थीटा होगा 30 डिग्री के बाद आपको समझाइए देता हो गया 30 डिग्री अब कोई आइडिया कोई थॉट कोई दिमाग में एप्लीकेशन जो आप सोच का रहे हो की हान सर अब बात बन जाएगी मुझे क्या प्रूफ करना है मुझे प्रूफ करना है की सर इसका जो सेट्रॉयड है इसका जो सेट्रॉयड है वो बेसिकली इस C1 सर्कल पर लाइक करता है अच्छा एक बात एक मैन लेता हूं जैसे की ये जो सेट्रॉयड है वो यहां पर है मेरे अकॉर्डिंग ही होना चाहिए जैसे मैं अभी फिलहाल के लिए वैसे जी से दिनो करता हूं मेरा बस इंटरेस्ट इतना सा ए रहा है मेरा बस कंसर्न इतना सा ए रहा है क्या मैं इस पीजी तक पहुंच सकता हूं किसी तरीके से क्या मैं इस पीजी तक पहुंच सकता हूं क्वेश्चन है इस हिंट को उसे करके बस मुझे बता दो क्या मैं किसी भी आइडिया से किसी भी हिंट से किसी भी थॉट से पीजी तक पहुंच सकता हूं क्या जल्दी से स्टूडेंट सोचकर बस इतना बता दो हमारा क्वेश्चन जो है खत्म हो जाएगा हमारा क्वेश्चन बहुत अच्छे से आराम से खत्म हो जाएगा अगर आप इतना सोच का रहे हो तो मैं क्या पूछना चाह रहा हूं अच्छा इस पीजी से पहले एक छोटी सी बात का जवाब दो ध्यान से सुनना बहुत ध्यान से सुनना सर अगर मैं इस पॉइंट को कुछ कहूं तो का लो भाई लेट्स अस पॉइंट को मैं कहता हूं सी इस पॉइंट को मैं कहता हूं सी अब आप देखना ट्रायंगल ओए सी में कौन से ट्रायंगल में ओक में कौन सी ट्रायंगल में बात कर रहा हूं भाई अलग से बनाता हूं देखना है ना बस घबराना मत ट्रायंगल आपका ओक चीज थोड़ी सी सिंक में रखना देखो सर ये अगर एंगल थीटा है अगर ये एंगल थीटा है तो हम जानते हैं सर सीधी सी बात है ये आपका कार्ड है ये अब आपकी कॉर्ड है और ये भी जो कॉर्ड है और आप क्या है आपकी रेडियस तो क्लियर सी बात है सर मतलब ओ जी क्या है आपकी रेडियस आप देख का रहे हो ना ओ जी क्या है आपकी रेडियस बहुत ध्यान दो ना ओ जी आपकी रेडियस है तो इस रेडियस ने इस ए बी कट को क्लीयरली परपेंडिकुलरली बायसेक्स किया होगा ये बात हम जानते हैं मतलब कैन आई सी की सर ये जो एंगल होगा ये 90 डिग्री एंगल होगा कम की बात से सुनना बहुत ध्यान से सुनना अगर ये 90 डिग्री एंगल है तो ये भी एक नाइटी दिख रही है और अगर ये थीटा है अगर यह थीटा है तो ये एंगल 90 - θ होगा 180° वाले प्रॉपर्टी से अगर यह एंगल 90 - थीटा है और यह पूरा 90 तो यह थीटा होगा क्या मतलब मैं फिर से आता हूं कम की बात पर suniyega ध्यान से यह 90 यह थीटा तो ये 90 - θ तो ये कितना थीटा रेट थीटा होगा ना आई होप आप मेरी बात समझ का रहे हो बिना किसी कन्फ्यूजन के बिना किसी दिक्कत के आई होप ये बात अच्छे से देख रहे हो मैं फिर से रिपीट करता हूं ये जो दिख रहा है ये एंगल 90 है और ये थीटा है तो यह एंगल कितना होगा सर ये होगा 90 - θ तो यह कितना होगा सर ये थीटा होगा ऐसे नहीं ए रहा है 90 है ना मैन लो ये एंगल मुझे नहीं पता ये एंगल मुझे नहीं पता है तो मैं इसको मैन लेता हूं अल्फा और ये कितना है 90 - θ ये कितना है अल्फा और ये है 90 रन दोनों का सैम कितना है सर इन दोनों का सैम है आप मेरी बात समझ पाए क्या आप खुद देखो 90 से 90 के आंसर सॉरी वो इक्वल आना था ना हम रिलीज सॉरी ये जो सैम है ये किसके इक्वल है ये पूरा सैम 90 के इक्वल मैं फिर से रिपीट करता हूं ये एंगल अगर अल्फा है और ये अगर 90 - अल्फा है तो ये पूरा सैम किसके इक्वल है 90 के 90 से 90 कैंसिल तो देता किसके इक्वल आता है अल्फा के मैं आपसे यही कहना चाह रहा हूं की ये जो एंगल है ये जो एंगल यहां पर है ये भी कितना है थीटा आई होप ये बात आप विजुलाइज कर का रहे हो मैं वापस यहीं पर प्लॉट कर देता हूं आपको बिना कंफ्यूज किए की यहां पर ये जो एंगल है ये है कंफ्यूज नहीं हो रहे हो अगर ये एंगल थीटा है तो सुना अगर ये एंगल थीटा है तो क्या मैं कोस थीटा निकल सकता हूं सुना कौन से राइट एंगल में ओ सी ए में है ना हम किस्म बात कर रहे हैं हम ओक या ओक जो भी आप का रहे हो यहां पर मैंने इंटरेस्ट हूं कोस थीटा कम की बातें सुनना इससे आपको बहुत क्लेरिटी मिलेगी सर ओ सी ए में जब आप कोस थीटा निकल रहे हो तो कोस थीटा क्या होता है सर बेस अपॉन हाइपोटेन्यूज बेस क्या है सर बेस है आपका एक बेस क्या है भाई एक और हाइपोटेन्यूज क्या है आओ बेस है आपका एक और हाइपोटेन्यूज क्या है आओ ही होप आप बात समझ का रहे हो थीटा के नीचे वाली साइड बेस और उधर आपका हाइपरटेंशन होता है तो क्लीयरली सर एक / आओ है ना तो मैं यहां पे क्या लिखने जा रहा हूं मैं यहां पर लिखने जा रहा हूं एक / ए बस यही इस क्वेश्चन का करेक्ट से अगर आप ध्यान से सोच पाओ तो अब एक/ए क्या जानता हूं सर मैं आओ को जानता हूं कितना आर अरे हम जानते हैं ना बस यही इस क्वेश्चन का करेक्ट है की अगर मैं आपसे पूछूं की जो कोस्थेटा है तो एक अभी हम वर्क आउट करने वाले हैं जिसकी मैं आपसे बात करूंगा लेकिन क्या मैं और लिख सकता हूं अब मुझे बस इतना बता दो क्या कोई और तरीका है स्टूडेंट्स एक को लिखने का क्या एक को लिखने का कोई और तरीका आपके दिमाग में ए रहा है कोई भी तरीका अन्य वैन हेविंग अन्य थॉट क्योंकि उसका करेंगे क्या उसको कैसे निकलेंगे आई थिंक सर आसान सा तरीका है क्योंकि आप बात नहीं समझ रहे हो आप सिंपल सी बात बताओ सोच के बताओ मेरे ख्याल से अगर मैं एक को निकलना चाह रहा हूं तो क्या मैं कोस थीटा जानता हूं आई थिंक कोस θ जानता हूं सर आप बात समझो ना सिथेंटा कितना है सो देता है 30° और अगर थीटा है आपका 30 ध्यान से देखना अगर थीटा है आपका 30° तो बात समझना यहां पर कोस 30 निकल सकते बिल्कुल निकल सकता हूं सर सो जैसे ही कोस 30 निकाला कोस 30 कितना होता है सर √3 / 2 तो कितना ए जाएगा सर ये ए जाएगा एक / आर कोई दिक्कत तो नहीं अच्छा इसके साथ एक छोटा सा कम और मैं करना चाह रहा हूं मुझे शायद इसके साथ एक छोटा सा कम और मैं करना चाह रहा हूं क्या यहां पर भी मैं सिन थीटा सोच सकता हूं रिपीट मैं स्टेटमेंट क्या यहां पर भी मैं सिथेंटा सोच सकता हूं देखो सिथेंटा क्या होगा सुनना सो सिन थीटा होगा परपेंडिकुलर अपॉन हाइपोटेन्यूज साइन थीटा साइन थीटा क्या होगा यहां सिन थीटा होगा आपका ओक / ओए होप आप सन का रहे हो सिन थीटा कितना होता है सर 1 / 2 उसी कितना होता है क्या है रेडियस suniyega ध्यान से ओक हमें निकलना है और ओए हमें पता है रेडियस है ना तो ओक कोयल की आर / 2 अब सुनना अब सुनना इस बात को स्टूडेंट्स ओक अगर आया आर/2 अगर ओक आया आर/2 तो क्या आप मुझे ओ जी बता सकते हो आई होप आप कंफ्यूज नहीं हो रहे हो ना ध्यान से मैं यहां पर इसे अलग से कर दे रहा हूं मुझे पता था सर बहुत ध्यान से सुनना सर ओ जी जो आपका था ओ जी कितना था आर है ना सर ओ सी कितना है सर ओ सी आया आर/2 बहुत कम की बात है इस बात को सुनो तो क्या आप मुझे जीसी बता सकते हो क्यों निकल रहे हो सर इतनी सारी चीज बस दो लाइन और ये क्वेश्चन खत्म हो जाएगा की आप मुझे जीसी बता सकते हो बिल्कुल सर ध्यान से देखना अगर मैं ओ जी में से ओक सब्सट्रैक्ट कर डन तो जीसी ए जाएगा क्या तो जीसी कितना होगा सर ध्यान से देखो ओ जी में से उसी सब्सट्रैक्ट किया तो कितना ए गया आर/2 बस एक क्वेश्चन अब खत्म होने को बस पेशेंट के लिए इस लाइन को सुनो बस इस लाइन को सुनना ध्यान से देखो जीसी कितना आया आर/2c कितना आया आर/2 अब सुनना क्या आप मुझे बता सकते हो बड़ी ही बेसिक सी बात आय होप जिसे हमारा आर/2 ही आया है क्योंकि ओ जी कितना था सर ओ जी था आर और ओसीआर / 2 तो जीसीआई आर / 2 अब सुनना क्या आप मुझे ये इतनी सी बात बता सकते हो suniyega ध्यान से की ये जो आप था ये जो आप था इसकी लेंथ कितनी थी सर आप बड़े वाले सर्कल की रेडियस थी रिपीट मैं स्टेटमेंट जो आप था वो लेंथ कितनी थी सर बड़े वाले सर्कल की रेडियस जो की कितनी थी तू आर या हमने निकल थी याद ए रहा है अब अगर आप में से मैं ओक सब्सट्रैक्ट कर डन तो क्या मुझे पीसी मिल जाएगा suniyega बस यहां से तो पीसी क्या आएगा सुनना सर यह जो पीसी है यह क्या होगा आप - ओक ये कितना होगा आप - ओक आप कितना है सर आप है 2r और ओक कितना है ओक है ओर / 2 सर दो में से आधा घटना है यानी 1 1/2 बचा ना यानी 3/2 तो पी सी कितना ए रहा है ना सर जो पीसी ए रहा है पीसी जो ए रहा है वो ए रहा है थ्री आर / 2 बस यही आपके इस क्वेश्चन का आंसर है या प्रूफ है के सुनना जीसी क्या है ये और पीसी क्या है पीसी क्या है ये क्यों सर सुनना कम की बात कम की बात ये की सर अगर मैं एक ट्रायंगल पर आपसे बात करूं अगर मैं ट्रायंगल पर आपसे बात करूं ये ट्रायंगल है ना इस ट्रायंगल में आपके पास क्या है सिर्फ कोर्स ट्रायंगल में आपके पास है एक मेडियन है अगर इस ट्रायंगल में मेडियन के पास एक सेट्रॉयड है अगर सेट्रॉयड है तो सोच कर बताओ तो जीसी की लेंथ है आर / 2 जैसे की कितनी भाई आर / 2 तो यहां पर ये जो जीसी की लेंथ है ये है आर / 2 ऐसे प्रूफ हो चुका है वो पर फिर भी अगर आपको नहीं दिख रहा है तो सुनना सर थ्री आर / 2 में से आर / 2 है आप बात समझो ये कितना है डेढ़ में से आधा यहां गया तो ऑफ कोर्स एक यहां ए जाएगा तो यह डिस्टेंस कितनी होगी सर ये डिस्टेंस होगी आर डिस्टेंस कितनी होगी आर अब खुद सोचो आप मुझे इसका और इसका रेश्यो बता सकते हो क्या आप रेशों बताओ स्टूडेंट्स आर / आर / 2 तो हो जाएगा 1 / 2 यानी कितना रेश्यो हो जाएगा चलिए ए जाएगा रेशों 2 1 और हम जानते हैं सर centeroide मीडियम को 2:1 में डिवाइड करता है तो क्या हम ये इनडायरेक्टली डायरेक्टली या सेटल इंप्रूव कर का रहे हैं की हान सर ये syndroid ही है क्योंकि अगर ये जी नहीं होता यहां तो फिर ये 2:1 में डिवाइड ही नहीं कर रहा था फिर ये यहां पर इस सर्कल पर क्यों लाइक कर रहा था तो अगर ये सर्कल पर लाइक करेगा तभी जाकर वो 2:1 में डिवाइड करेगा तभी जाकर वो सेट्रॉयड होगा जो की मेडियन पर लाइव तरीके से कर रहा होगा क्या ये पुरी ज्योमैट्रिकल प्रॉब्लम आपको क्लियर है अंडरस्टैंड थोड़ा सा ट्रिकी हो गया ये वाला पार्ट थोड़ा सा इनविन हो गया कौन सा ये वाला विच वास नॉट नीडेड कौन सा ये कोस थीटा वाला पार्ट आई थिंक लेकिन इसके अलावा ये पार्ट हमारे लिए कृष्ण रहा आई थिंक बहुत डिफिकल्ट बहुत टू ये क्वेश्चन नहीं था क्या बेसिक सा कॉन्सेप्ट आप समझ पाए की प्रूफ कैसे हमने किया मेरा बस कहना सर इतना सा है की सर अगर जी है तो वो 2:1 में डिवाइड करेगा बात कर और ये रेश्यो हमने निकल लिया और जी कहां पर लाइक कर रहा है हमारे azmption के अकॉर्डिंग जी यहां पर लाइक करेगा तभी हम इसको 2:1 में ला का रहे हैं बात बस इतनी सी है और अगर बात बस इतनी सी है तो क्या अब हम आज के हमारे अगले क्वेश्चन की तरफ बने इस क्वेश्चन को जरा ध्यान से देखो स्टूडेंट्स और बहुत बेसिक सा बहुत सिंपल सा शॉर्ट ट्रिक्स सा क्वेश्चन है अगर आप समझ पाओ तो वो का रहा है इफ फ्रॉम अन्य पॉइंट पी इफ फ्रॉम अन्य पॉइंट पी ऑन दिस सर्कल ऑन दिस सर्कल है ना इस सर्कल पर एक पॉइंट है बाय डी वे यहां पर आपका यह इक्वल नहीं होना चाहिए x² + y² + 2X आई थिंक ये प्लस साइन ही होगा ऑफ कोर्स यहां पे ये प्लस साइन नहीं प्लस साइन नहीं होगा ना तो दे शुड बी प्लस बेसिकली स्टूडेंट इस शुड हैव बिन अन प्लस साइन है ना ये आपका प्लस है वो ये का रहा है की इस सर्कल पर एक पॉइंट है उससे मैंने टांगें ड्रॉ की है तो आपसे ये पूछ रहा है की उन दोनों टांगें के बीच एंगल क्या होगा की आपको एक क्वेश्चन समझ आया मेरा यकीन करिए ये कीनन तौर पर एक बहुत ही सिंपल बहुत ही बेसिक और बहुत ही आसान सा क्वेश्चन है अगर आप इस क्वेश्चन को विजुलाइज कर पाएं तो सर कैसे करेंगे देखो दो चीज आपको इस क्वेश्चन को देख के ऑब्जर्व करनी चाहिए कौन सी दो चीज पहली तो ये सर की अगर इस सर्कल की मैं बात करूं इस सर्कल की मैं बात करूं तो इसका सेंटर क्या है सर इसका सेंटर मुझे दिख रहा है माइंस जी कॉम माइंस है सिमिलरली इसकी यह कांस्टेंट टर्म बड़ी अजीब सी है पर इस सर्कल का भी सेंटर तो -डी कमा - एफसी अगर एक बात आपको नज़र आई ये दोनों को सेंट्रिक सर्कल है वापस जिनकी रेडियस में ऑफ कॉस्ट रेफरेंस हैं जो की हम बात करेंगे लेकिन अगर इन दोनों सर्कल्स में मैं बात करूं तो ये दोनों सर्कल cosentric सर्कल्स हैं है ना आपके पास एक सर्कल तो ये रहा जो आपका बाहर वाला सर्कल बना लेते हैं और एक अंदर वाला सर्कल से ये रहा ये आपका अंदर वाला सर्कल क्यों नहीं बनाने दे रहे हो मुझे ये आपका एक और सर्कल है अंदर वाला जो की आपका छोटा वाला सर्कल है जिसकी रेडियस इससे कम है आई थिंक यह कोर्स एंड ट्रिक देख रहे होंगे आपको अब बात करते हैं क्या वो कहना चाह रहा है वो कहना चाह रहा है की यह वाला जो सर्कल है ना ये कौन सा आपका x² + y² + 2G + 2 + सी है ना इस पर एक पॉइंट है ड्रा की है तो ये भी 90 डिग्री एंगल होगा और ये भी एक 90° एंगल होगा और वो ये जो दोनों अपने tenjent ड्रॉप किया है ना इनके बीच का एंगल जानना चाह रहे हैं पर उससे पहले मैं इस कहानी को ऐसे खत्म करना चाह रहा हूं क्योंकि मैं मैन लेता हूं ये जो एंगल है थोड़ी देर के लिए थीटा है मैं मैन लेता हूं एंगल बाईसेक्टर होगा क्योंकि टांगें सीक्वल लेंथ की है कॉमन साइड है तो एंगल बाईसेक्टर होगा तो थीटा अगर निकल लूं मैं थीटा निकल लूं मैं तो θ का डबल कर दूंगा वही आपका दोनों टांगें के बीच का एंगल होगा तो हमारा क्वेश्चन है थीटा निकलना लेकिन अभी कहानी बाकी है साहब कहानी में क्या-क्या बाकी है सर ये इसकी रेडियस है अंदर वाले ट्रायंगल अंदर वाले सर्कल की और ये जो है ये इसकी रेडियस है किसकी बाहर वाले सर्कल की मतलब अगर मैं ऐसा का डन थोड़ी सी ओ है ना इसे मैं का देता हूं इससे ए इसमें मैं का देता हूं भी कोई तकलीफ तो नहीं यहां तक हम निकलने चले हैं सर सारी बातें करते हैं आई थिंक ओए आप समझ का रहे हो ना आपके अंदर वाले सर्कल की रेडियस और आपकी आप जो है वो बाहर वाले सर्कल के लिए आय होप ये बात आप समझ का रहे हो ना सर ठीक है सारी बातें कर लिया आपसे अब हम समझना चाह रहे हैं की हम θ निकलेंगे कैसे तो बस मेरी हेल्प करो पहले तो इस सर्कल की रेडियस निकलने में सर इसकी रेडियस तो हमको आती है निकलना क्या है हमने पढ़ रखा है ना इसकी रेडियस क्या होगी सर इसकी जो रेडियस होगी अगर मैं दिनो भी कर डन आपके लिए तो ये बाहर वाला सर्कल है बाहर वाला सर्कल मतलब ओपी तो ये क्या है सर आप कितना आएगा सर √ ओवर G2 + एक्स स्क्वायर वाले सर्कल की रेडियस क्या इसे देखकर अंदर वाले सर्कल की रेडियस बता सकते हैं जो की कितनी होगी सर अंदर वाले सर्कल की रेडियस है ओए ओ बी अंदर वाले सर्कल के रेडिएशन है ओए या अब जो भी आपके आना चाहे वो कितनी होगी इसे ध्यान से देखना इसे ध्यान से g² + x² ऐसे क्या लिखेंगे जी स्क्वायर प्लस एक्स स्क्वायर माइंस कितना स्टोरेज देखना सी टाइम्स माइंस सी टाइम्स क्या साइन स्क्वायर माइंस जी स्क्वायर प्लस स्क्वायर माइंस ये वाली टर्म का नेगेटिव तो मैं पुरी टर्म नेगेटिव साइन के साथ लिख देता हूं है ना नेगेटिव साइन बाद में अंदर मल्टीप्लाई करेंगे तो ये हो जाता है सी टाइम्स स्क्वायर अल्फा प्लस जी स्क्वायर कर सकते हो क्योंकि मुझे कुछ कुछ चीज आसान होती हुई दिख रहे हैं कैसे दिख रही है तो बन जाएगा 1 - sin² + sin² अल्फा हो तो सिंपलीफाई कर लो मेरा कहना है सिंपलीफाई करने से चीज शायद आसान दिख जाए तो यही जो ए या वो भी हम निकल रहे हैं वो क्या बनेगा सुनना बात सुनना ये G2 कोस स्क्वायर जी स्क्वायर कॉमन ले लेंगे तो बचेगा 1 - cos² कर लेना sin² अल्फा तो क्या देखेगा सर आपको दिखेगा जी सिमिलरली f2 तो आपको ऐसा ही क्या दिखेगा f2 सिन स्क्वायर अल्फा तो ये कितना हो गया माइंस सी sin² अल्फा और ये आपकी क्या बन गई इसकी रेडियस सर एक छोटा सा कम और कर लेते हैं sin² अल्फा कॉमन ले लिया क्या जैसे ही साइन स्क्वायर α कॉमन लिया तो अंडर रूट में क्या बचा g² + x² - सी और sin² अल्फा जो कॉमन लिया वो अंडर रूट से कैंसिल हो की क्या बन जाएगा सर वो बच जाएगा आपका ऑफ कोर्स क्या साइन स्क्वायर तो ये आपकी ए गई ओए ओ बी की वैल्यू कोई तकलीफ अब मेरा बस आपसे यह कहना है suniyega ध्यान से अगर मैं थीटा निकलना चाह रहा हूं तो मैं बात कर रहा हूं आपसे ट्रायंगल आप में सुनेगा ट्रायंगल आप में यह जो देता है यह परपेंडिकुलर हाइपोटेन्यूज अगर मैं आपसे पूछूं सिथेंटा निकलने को तो सिन थीटा क्या होगा सर साइन थीटा कुछ नहीं होगा ध्यान से देखो परपेंडिकुलर अपॉन हाइपोटेन्यूज यानी ओए अपॉन आप ये क्या हो गया क्या है सर ओए क्या है सर सर ओए मैंने अभी निकाला ये √g स्क्वायर प्लस एक्स स्क्वायर माइंस सी टाइम्स साइन α तो ओए कितना है सर ओए आपका है क्या आपने आई थिंक वो निकाला है दिख रहा है क्या कुछ सर आप हमने निकाला है जो की है अंडर रूट ओवर कितना G2 एसएससी हर बार आप ये नहीं कहते की थीटा ही अल्फा के इक्वल होगा प्रिडिक फंक्शन है सब कुछ आपका 90 डिग्री और 180° के अंदर ही है तो आप कहोगे θ किसके इक्वल है अल्फा के इन tenjins के बीच जो एंगल होगा वह थीटा प्लस थीटा तो 2 थीटा अगर इनके बीच का एंगल है हमारे अकॉर्डिंग तो उनके अकॉर्डिंग कितना होगा 2 अल्फा तो इस क्वेश्चन में जब आपसे कोई पूछे की इन दोनों के बीच का एंगल कितना होगा उसने खुद का दिया है वो ये है तो आपका जो दोनों टांगेंट्स के बीच का एंगल होगा वो होगा 2 अल्फा इन डी ईयर फाइनल आंसर क्या ये पूरा क्वेश्चन और इसका एक्सप्लेनेशन बिल्कुल क्लियर है किसी भी डाउट या परेशानी के बिना आई होप आपको समझ ए रहा है की सर्कल पे चीज कैसे-कैसे घुमा कर पूछी जा सकते हैं रहिए तो उसका क्या मतलब है इस बात को बहुत ध्यान से सुनिए स्टूडेंट्स मैन लो इमेजिन देयर इस अन सर्कल किसी एक्सटर्नल पॉइंट से लेट्स में किसी एक्सटर्नल पॉइंट से इंटरनल पॉइंट से भी बनाई जा सकती है एक्सटर्नल पॉइंट से भी बनाई जा सकती है देस नॉट मेक मच ऑफ डी डिफरेंस जैसे किसी एक्सटर्नल पॉइंट से मैं सिकन तू ड्रा कर रहा हूं आई होप ये बात आप समझ का रहे हो तो इस तरीके से अब कोस किसी एक एक्सटर्नल पॉइंट पीस है मैंने कोई सेकंट दरो की है लेट्स से जो की इसे ए बी सी और दी ऐसे चार पॉइंट्स पर इंटरसेक्ट करती हैं अब इसका मतलब क्या है इससे क्या फायदा होता है इससे क्या चीज निकल कर आते हैं उसे बारे में हम बात करेंगे क्या जरूरी है सर ये किसी एक्सटर्नल पॉइंट से ही ड्रॉ हो बिल्कुल नहीं यानी क्या ऐसा नहीं हो सकता की एक सर्कल है और उसके लिए कोई सीटेट इस तरीके से जा रही हो ऑफ कोर्स दिस इस अन सीटेट क्योंकि ये ट्रायंगल को सॉरी सर्कल को दो हसन में मतलब दो बेसिकली पॉइंट्स पर बायसेक्स कर रही है जैसे ये पी से निकल ही सीक्वेंट है जो की ऑफ कोर्स सर्कल को अगेन अब और सीडी ऐसे चार पॉइंट्स पर इंटरसेक्ट कर रही है तो सेकंट हमारे फिलहाल इसने मैं दो केसेस याद दो त्रिकोण से देख सकती है कैसे एक तो किसी एक्सटर्नल पॉइंट से ड्रा की गई हो या फिर ऑफ कोर्स एक किसी इंटरनल पॉइंट से ड्रॉ की गई अब इन पर क्या प्रॉपर्टीज बनती है उनके बारे में जरा बात करते हैं suniyega ध्यान से अगर इंटरनल किसी पॉइंट से बनाई गई है तो कुछ ऐसी एक्सटर्नल किसी पॉइंट से बनाई गई है तो कुछ ऐसी suniyega बहुत ध्यान से स्टूडेंट्स बहुत कम की बात है ये ऑफ कोर्स आपकी एक सेकंड जिसमें पॉइंट से ए और बी पॉइंट ऑफ इंटरसेक्शन एक और सेकंड जिसके पॉइंट ऑफ इंटरसेक्शन है सी आर दी वही से बात यहां पर हो रही है अब क्या suniyega ध्यान से आप बेसिकली बड़ी आसानी से ये प्रूफ कर सकते हो जैसे की देखो मैं अगर इन दोनों ट्रायंगल को सिमिलर प्रूफ करना चाहता हूं जैसे यहां पे एक एग्जांपल लेते हैं क्या मैं ये का सकता हूं सोच के बोलना स्टूडेंट्स की सर अगर आप बी को एक कॉर्ड की तरह ट्रीट करो इफ यू ट्रीट बीड़ी क्या आपको यह बात सीधी-सीधी सी नज़र ए रही है की जो बी आपकी कार्ड है जो वीडियो है वो आपका ये एक एंगल क्रिएट कर रही है जो की आपका यहां पर एक एंगल है की अपनी बात समझ का रहे हो और सिमिलरली जो बी आपकी कार्ड है वो आपका एक एंगल यहां पर क्रिएट कर रही है जो की आपका ये है तो आप इस तरीके से इन ट्रायंगल में सिमिलरिटीज प्रूफ कर सकते हो अभी आप जानते हो इक्वल लेंथ सब्सटेंड इक्वल एंगल आते ऑन दिस सरकम्फ्रेंसेस ऑफ डी सर्कल बस यही बात मैं आपको याद दिलाना चाह रहा हूं की इस बेसिस पे मैं इन दोनों ट्रायंगल को यहां पर इसे और इसे सिमिलर प्रूफ कर सकता हूं और जैसे ही हम सिमिलर प्रूफ कर देंगे मैं कुछ मोलेरिटी के थ्रू बड़े इंपॉर्टेंट से कृष्ण से रिजल्ट्स डेरिव कर सकता हूं आई होप आप ये बात जानते हो की सर सिमिलर प्रूफ करने के लिए क्या एक एंगल काफी है नहीं अगर आप दूसरा एंगल देखोगे तो ये वर्टिकल अपोजिट एंगल से तो ये भी से हो गए और सीधी सीधी सी बात है सर ये से हो गए तो ये तो से होना यही है ना या फिर इसको ऐसे सोच लो की एक कट थी मैन लो ए दी ए सी तो ए सी कार्ड ने एक बार ये एंगल बनाया और एक कौर ने एक बार ये एंगल बनाया था ऑफ कोर्स ये भी से होना है तो एंगल एंगल से milaiti से मैं सिमिलर ट्राएंगल्स पर जो हमारी प्रोपोर्शन वाली प्रॉपर्टीज होती है वो लगा सकता हूं और जैसे ही आप proporation वाली प्रॉपर्टी अप्लाई करेंगे तो इस तरह के किसी रिजल्ट की तरफ ए पाएंगे suniyega ध्यान से आप सीधे-सीधे प्रूफ कर सकते हो ये जो की अभी जस्ट हमने डिस्कस किया अब इस निष्कर्ष के बेसिस पर इस रिजल्ट के बेसिस पर आप प्रूफ कर का रहे हो की एडीपी और एबीपी suniyega ध्यान से क्लियर सी बात है एडीपी आपका एक तो ये वाला ट्रायंगल और एक ये वाला ट्रायंगल ये आप प्रूफ कर का रहे हो की सिमिलर है और ये सिमिलर है तो हम जानते हैं सर सीधी सीधी सी बात सोच के देखना आप / क है ना कौन कौन ए पी / क किसके इक्वल होगा डीपी / डीपी के थर्ड साइड भी रेश्यो में इक्वल ए जाती अगर आप एड अपॉन सी भी लिखते बट डेट इस नॉट नीडेड हर के क्योंकि जो प्रॉपर्टी हम प्रूफ करना चाह रहे हैं वो कैसे का ध्यान से आप/क विल बी इक्वल तू डीपी/बीपी जिसमें अगर आप बीपी को इधर ले आएं क को इधर ले आएं तो क्या कंक्लुजन आता है कंक्लुजन आता है आप बी पी जैसे मैं कैसे याद रखता हूं सुनेगा ध्यान से मैं ए सकता हूं आप बी पी विल बी इक्वल तू क पद या डीपी क्या कहना चाह रहा हूं स्टूडेंट सोच के देखना बड़ी सिंपल सी बात है बड़ी ही सिंपल सी बात है पहले आप उठा लो ये वाली सेकंड के इस रेड लाइन के पॉइंट्स इस स्ट्रेट लाइन के पॉइंट्स उठा तो कौन कौन एक तो आप और ब तो पहले मैं उठा लेता हूं इन दोनों को जिसकी लेंथ है क्या आप और पी और सिमिलरली जब इस वाली seacate में आप चीजे उठाओगे तो यहां पर कौन-कौन है पद और पीसी तो इन्हें ले लीजिए कौन-कौन स्टूडेंट आपके पास कौन-कौन है क और बीपी डिस्टेंस याद रहेगा सेकंट के लिए बना है क्या सिर्फ ये इसी के लिए है नहीं ये प्रॉपर्टी आप इस वाले सीटेट के लिए भी होल्ड करते हो कैसे suniyega ध्यान से आप बी पी विल बी इक्वल तू पीसी पद क्या आप मेरी बात समझ पाए मैं फिर से दोहराता हूं suniyega ध्यान से मैं कैसे याद रखता हूं देखो यहां पे अब बड़ी बेसिक सी बात है अच्छे से गौर से सुनना स्टूडेंट्स आप पहले तो लिखोगे का ये आपका डिस्टेंस ये जो डिस्टेंस है तो पहले आप लिखोगे का इन ऑफ कोर्स क्या इन ऑफ कोर्स ब ये आपका जो पॉइंट है बी है है ना इन पीवी डेट विल बी इक्वल तू डेट विल बी इक्वल तू दिस पीसी ये क्या लेंथ है स्टूडेंट्स ये लेंथ है पीसी ऑफ कोर्स पद सो दिस विल बी इक्वल तू पीसी पद क्या ये प्रॉपर्टी याद रहेगी आपको सेकंड को लेकर मतलब जिस पॉइंट से वो सेकंट ड्रॉन है ना उससे जो पॉइंट इंटरसेक्ट हो रहे हैं तो इस सीक्वेंट से कहां का इंटरसेक्ट हो रहे हैं ए और बी पर तो ए था की डिस्टेंस और बी तक की डिस्टेंस सिमिलरली यहां पर कहां का हो रहे हैं सी और दी पर तो पी से उसी पॉइंट से सी तक की डिस्टेंस और उसी पॉइंट से डी तक के डिस्टेंस इनके डिस्टेंस इसके प्रोडक्ट्स इक्वल होते हैं क्या ये एक्सटर्नल सीक्रेट के केस में और इंटरनल पॉइंट से ड्रॉन सिकुड़ के केस में दोनों केस में आपको अंडरस्टूड हैं आई थिंक एक आसान सी ये बात है सुलझी हुई सी बात है पर ये बात हम कर क्यों रहे हैं सर मेरा यकीन करिए ये बहुत कृष्ण प्रॉपर्टी ये तो इस पॉइंट को समझने के लिए आप नोट डाउन दो चीज तो कर ही लीजिए एक तो ये फिगर एक ये फिगर और ऑफ कोर्स यह प्रॉपर्टी कम से कम ये तीन चीज याद रखें इन पर बेस्ड क्वेश्चंस आईआईटी जी मांस और एडवांस में आते हैं और इन पर बेस सवाल बनते हैं और इन से कई सारी चीज निकल कर आती हैं तो ये प्रॉपर्टी और ये दोनों तो याद रखना ही रखना है भले तभी समझ आएगा फिर से सन लीजिए अगर बोल रहे हो तो सर इसकी डिस्टेंस और इसकी डिस्टेंस का प्रोडक्ट यह डिस्टेंस और यह डिस्टेंस की इक्वल होगा एक्सटर्नल पॉइंट से ड्रॉ की किसी एंड के केस में सर ये डिस्टेंस और ये डिस्टेंस का प्रोडक्ट इस डिस्टेंस और इस डिस्टेंस के प्रोडक्ट के इक्वल होगा भूलोगे तो नहीं अच्छे से याद रहेगा आई होप ये बातें आप याद रखेंगे और यहीं से कहानी आगे बढ़ते हैं इस कॉन्सेप्ट को कंसोलिडेटेड करने के लिए अच्छा एक और छोटी सी बात निकल कर आती है एक बहुत जरूरी बात निकल कर आती है की सेकंट के साथ सर अगर मैंने कुछ ऐसा सिनेरियो बनाया होता ये भी बहुत मजेदार कॉन्सेप्ट है इससे भी आप काफी कुछ ऐसी हो गए सर वही मैन लो की एक्सटर्नल पॉइंट पी है लेकिन इस एक्सटर्नल पॉइंट से एक तो हमने बनाई मतलब क्या सर यहां पर इसने टच किया होकर टांगें बन गए मतलब उसने दो पॉइंट्स पर इंटरसेक्ट नहीं किया इससे किसी एक पॉइंट पर टच किया तो क्या एक और मजेदार से प्रॉपर्टी पिक्चर में आती है मैं इसके पूरे प्रूफ पर नहीं जा रहा हूं मैं डायरेक्ट आपसे कंक्लुजन आपसे रिजल्ट शेयर कर रहा हूं जैसे आप याद रखिए इनके रिव्यू ऑन तू लुक आते डी प्रूफ प्रूफ बड़ा आसान सा है फिर से ध्यान से देखो आप ये दो चीज प्रूफ करोगे पहले तो की सर जब आपने tenjent ड्रॉ की हर जब आपने सेकंट ड्रॉप की है तो क्लियर सी बात है जो एशिया आपकी है ये यहां बी पर कोई एक एंगल सब्सटेंड करेगी और ये प्रॉपर्टी है की यही कोड अल्टरनेट सेगमेंट में अगर इससे कोई टांगें गुजर रही है इसी पॉइंट पी से जिससे सेकंट बनी थी जैसे ये ए और सी आपके पॉइंट निकल रहे हैं मतलब मैं सिंपल सी बात आपसे जो कहना चाह रहा हूं वो ये suniyega ध्यान से ये प्रॉपर्टीज सीधे-सीधे याद रखना की आपने क्या किया आपने एक एक्सटर्नल पॉइंट पी से एक एक्सटर्नल पॉइंट पी से आपने एक सेकंड और एक टांगें ड्रॉ की क्या बात आपको समझ आई एक एक्सटर्नल पॉइंट पी से आप नेक्स्ट वीकेंड और एक टांगें ड्रॉ की क्या बात आपको समझ आई अब जो आपने सेकंट ड्रॉ की है ना जो आपने सिकन ड्रॉ की है suniyega ध्यान से इस सीटेट का जो पॉइंट ऑफ इंटरसेक्शन का जो पॉइंट ऑफ इंटरसेक्शन है उससे आपने कार्ड बनाए कोई दिक्कत तो नहीं है स्टूडेंट्स अब जो यह आपने कोर्ट बनाई है ना इसमें आप कुछ मत करो आप इसमें इस टी पॉइंट को इस बी से भी कनेक्ट कर दो तो क्लियर सी बात है सर ये जो कॉर्ड है 80 या जिसे आप इस क्वेश्चन के अकॉर्डिंग अगर कहना चाहो जनरली हम ठीक कहते हैं बट अगर आप इसे कहना चाहो तो सी का लो तो इस कॉल्ड एक ने जो ये एंगल सब्सटेंड किया है यहां पर वही एंगल वही एंगल वही एंगल ये जो बी और सी को ज्वाइन करने वाली है ना वही एंगल इस टैसेंट के साथ भी बनाएगी इस टैसेंट के साथ भी बनाएंगे यानी अगर ये लेट्स से थीटा है तो ये भी थीटा होगा ये प्रॉपर्टी है इसे प्रूफ किया जा सकता है ज्यामिति के लिए बट मैं डायरेक्टली आपसे रिजल्ट डिस्कस करना चाह रहा हूं जिस मुद्दे पर मैं आना चाह रहा हूं की कंक्लुजन ये होगा बहुत कम की बातें suniyega की आप इस जादू जाम है ये प्रूफ कर देंगे वापस की ट्रायंगल पीसीबी और ट्रायंगल पीएसी कौन-कौन से ट्रायंगल स्टूडेंट्स आप याद रखिएगा पीसीबी ये आपका बड़ा वाला ट्रायंगल और ये छोटा वाला ट्रायंगल ये वापस सिमिलर होंगे क्यों सिमिलर होंगे क्योंकि देखो ये एंगल कॉमन है ना पीसीबी में भी एंगल है और पीएसी में भी एंगल है और इन दोनों में एक-एक एंगल और कॉमन है पीएसी में ये एंगल थीटा है और पीसीबी में ये एक एंगल थीटा है तो बचा हुआ एंगल कौन सा पीएसी में ये वाला एंगल और पीसीबी में ये वाला एंगल ये भी से होगा डिड यू ऑल गेट माय पॉइंट क्या आप सभी को यह बात समझ आई और जैसे ही आप अप्रूव कर देते हो क्लियर ही सर वापस यह दोनों ट्राएंगल्स कौन-कौन से पीएसी और पीसीबी सिमिलर है और उसे कारण आप निकल पाओगे पीसी / पी ए विल बी इक्वल तू ब / पीसी क्या क्या कहना चाह रहा हूं पीसी / पी ए विल बी इक्वल तू ब/पीसी कोई तकलीफ मेरा ये कहना है अगर आपने वो सेकंट वाली प्रॉपर्टी पढ़ी थी ना यही इसी प्रॉपर्टीज आप इसे भी आसानी से प्रूफ कर सकते हो देखो मैंने इसे बचपन में कैसे याद किया था मैं शेयर करना चाह रहा हूं वो तकनीक सुनना ध्यान से बहुत कम की बात है मैंने सेकंड वाली तकनीक की याद की थी और मैं सोचा करता था की सर यहां पर भी वैसी सोच सकते ना मैन लो अगर ये सी केंट होती है तो ये दो पॉइंट्स में इंटरसेक्ट कर रही होती एक पॉइंट पे इंटरसेक्ट या टच करने के बजाय तो हम क्या करते हैं वो सेकंड वाली थ्योरम के फॉर्म बोलते हैं का ब वुड हैव बिन इक्वल तू पीस इन तू पी दी फिर दी की जगह पीसी है तो मैं मैन लेता हूं तो आप इसको लिखते पद लेकिन वापस वह पीढ़ी की जगह एक ही पॉइंट है ना तो वापस क्या है पीसी अब समझ का रहे हो मैं कहना चाह रहा हूं मैं बस ऐसे ही शॉर्टकट के थ्रू याद रखता था तो मैं क्या करता था स्पेसी का क्या कर लेता था अभी स्क्वायर तो डायरेक्टली मुझे ज्यादा थ्योरम से आती रहनी पड़ी अगर सेकंट थ्योरम कोई मैं थोड़ा एक्सटेंड करूं और उसको मैन लो की ये सेकंट ही है और ये सिर्फ दो पॉइंट सी और सी पर इंटरसेक्ट कर रही है जो एक्चुअली में एक ही पॉइंट है तो सेकंट वाली हमारी थ्योरम क्या कहती है सर वो कहती है काब होता है पीसी पद बट दी है ही नहीं तो पीसी पीसी जो की क्या हो जाएगा टेक्निकल पीसी स्क्वायर आई होप ये बात आप आसानी से समझ का रहे हो बिना ज्यादा कंफ्यूज करो तो दूसरा आप जो चीज नोट डाउन करेंगे वो होगा ये वाला पॉइंट इससे अच्छे से नोट डाउन कर लीजिए स्टूडेंट्स bhuliyega मत क्या है हमारा कंक्लुजन हमारा कंक्लुजन है की सर आपका यह जो पिक्चर है कौन सा पिक्चर ये वाला जो पिक्चर है ये वाला जो सिनेरियो है इस क्वेश्चन पर मैं ए रहा हूं ये वाला जो सिनेरियो है इसमें याद रखिएगा सर की आपके ये कौन-कौन का ये वाला लेंथ ब ये वाला लेंथ स्पाइस के स्क्वायर के इक्वल होता है क्या ये प्रॉपर्टी आप याद रखेंगे अगली प्रॉपर्टी जो कृष्ण आपको याद रखनी है वो है आज के लिए ये इससे नोट डाउन कर लीजिए और इस इमेज को ड्रॉ कर लीजिए ताकि आप जब भी रिवाइज करें तो आपको ये बात समझ ए जाए क्या आप यह प्रॉपर्टी याद रखेंगे तो नहीं भूलना तो नहीं चाहिए पर अगर आप भूल भी रहे होंगे तो जरा इस क्वेश्चन को देख लीजिए और कर लीजिए आपको कोई प्रॉपर्टी अच्छे से डाइजेस्ट हो जाएगी अगर यह प्रॉपर्टी में पढ़ो तो प्रॉपर्टी में क्या लिखा हुआ ऐसा देखना इफ डी लाइंस दिस एंड दिस कट डी कोऑर्डिनेटर accessically वह किसी सर्कल पर लाइक करते दो पॉइंट्स पर इंटरसेक्ट कर रही होगी कहीं ना कहीं ये भी एक्स और ए एक्सिस पर की नहीं दो पॉइंट्स पर इंटरसेक्ट कर रही होगी कहीं ना कहीं पहले हम निकलती हैं उन पॉइंट्स के बारे में इनफॉरमेशन suniyega ध्यान से मैं सीधे आता हूं मैन लो अगर हमारा ये रहा सर लेट से एक्सेप्ट मैन लेते हैं बस वैसे ये रहा लेट से आपका सर्कल क्या आपको बातें समझ ए रही है आई होप आपको ये बातें समझ ए रही हैं अब मेरा आपसे ये कहना है सर की मैं सीधी-सीधी सी बात आपसे करता हूं की ये जो आपकी पहली इक्वेशन है ये जो आपकी फर्स्ट इक्वेशन है ये एक्स एक्सिस और ए एक्सिस पर कहां है रिसेट करेगी हमने पढ़ा है ना स्ट्रेट लाइंस में अगर मुझे पता करना है इसका एक्स इंटरसेप्ट तो मैं ए की जगह जीरो रख दूंगा जैसे ही ए की जगह जीरो रखा तो एक्स कितना ए जाएगा सर एक्स ए जाएगा -c1/a1 क्या बात समझ पाए ये हो जाएगा इसका एक्स इंटरसेक्ट सिमिलरली इसी लाइन का अगर मैं वही इंटरसेक्ट निकलना चाहूं तो मैं एक्सक्यूज जगह जीरो रख दूंगा तो ए इंटरसेक्ट कितना ए जाएगा सर वो ए जाएगा -c1/b1 आई थिंक सर इसी नॉलेज इसी इनफॉरमेशन के बेसिस पर मैं चाहूं तो मैं इस लाइन के भी एक्स ओर विंड एसेट निकल सकता हूं इसका एक्स इंटरसेप्ट निकलने के लिए ए जीरो रख देंगे तो वो कितना हो जाएगा सर वो हो जाएगा कितना भाई - C2 / ऑफ कोर्स A2 और इसी तरीके से अगर इसका वही इंटरसेप्ट निकलना चाहो तो इसे जीरो कर दिया तो वो कितना ए जाएगा भाई वो ए जाएगा -c2 / B2 कोई डाउट कोई डाउट कोई परेशानी तो सर इस लाइन के जो इंटरसेक्ट होंगे इस लाइन के जो इंटरसेप्ट आपको समझ ए रही है सुनना ध्यान से मैं मैन लेता हूं ये आपकी ये वाली लाइन थी ये जो आपकी लाइन थी वो आपकी ये वाली लाइन थी ना वो मैन लेते हैं वह आपकी मतलब मैं बहुत अच्छी लाइंस बताने की ड्राइंग कर पता हूं तो मैं ले लूंगा इसमें मौजूद एक टूल की हेल्प और मैं उससे आपको बताने की कोशिश करूंगा की देखो यह स्ट्रेट लाइंस एक और स्ट्रेट लाइन आपके पास है कौन सी ये वाली है अच्छा थोड़ा सा रोते करने की जरूरत अब नहीं है ना अब सुनना ध्यान से इस लाइन ने यह हमारी लाइन है मैन लो तो इसके जो एक्स और ए इंटरसेप्ट किया है -c1/a1 0 तो ये जो पॉइंट है इसके कोऑर्डिनेट्स क्या होंगे सर ये होंगे -c1 / ए1 कॉमर्स जीरो ये बात आपको समझ आई कोई दिक्कत तो नहीं है भाई अच्छा इसका वही कॉर्डिनेट क्या था सर इसका ए koardinate था -c1/b1 है ना तो इसका जो ए कार्ड ये कितना होगा सर ये होगा जीरो अपॉन 0 -1 / b1 सिमिलरली सिमिलरली क्या इस दूसरी वाली स्ट्रेट लाइन के भी मैं एक्स और ए इंटरसेप्ट निकल सकता हूं बिल्कुल निकल सकती हो सर हमने निकले ही द तो इसका एक सेंटर सेट कितना ए रहा था वो ए रहा था -c2 / A2 कमा जीरो और इसका ए इंटरसेप्ट कितना ए रहा था सो जीरो कमा - c2/b2 किसी भी स्टूडेंट को कोई परेशानी किसी भी स्टूडेंट को कोई दिक्कत अब आप सोच के देखो स्टूडेंट्स अगर मैं ओरिजिन को ओरिजिन को अगर एक पॉइंट मैन लो क्योंकि ऑफ कोर्स ओरिजिन है इसे मैं थोड़ी देर के लिए का लूं इसे मैं थोड़ी देर के लिए का लो लेट से बी आप सन रहे हो एसएससी और इससे मैं थोड़ी देर के लिए का डन दी तो क्या आपको जस्ट अभी हमने सीखी आपको अप्लाई करते ए रही है सर आप कहोगे ओए अब कहोगे की नहीं सर आप कहोगे प्रॉपर्टी से याद आया की नहीं आई रिपीट माय स्टेटमेंट मैं क्या कर रहा हूं मैं ये लेंथ ओए इन अब दिस लेंथ अब मल्टीप्लाइंग विद ओक इन ऑडी बट 98 एवरीवन सी आर मल्टीप्लाइंग डी लेंथ हम लेंथ मल्टीप्लाई कर रहे हैं और अभी जो मुझे पता है वो परली koardinate से वो भी इमेजिन की ये इधर ए रहा होगा इधर ए रहा होगा इधर ए रहा होगा इधर आओ मैंने चारों जगह लिए ताकि आप कंफ्यूज ना हो मतलब माइंस साइन का मतलब नहीं होता ना की वो माइंस पर ये जा रहा होगा मुझे डिसाइड करना पड़ेगा किसी वैन और A1 की वैल्यू क्या है उसके सी वैन और बी वैन की वैल्यू क्या है उसके अकॉर्डिंग वो नेगेटिव है पॉजिटिव साइड होगा इन सारी बातों के अलावा जो मेरा फाइनल कंक्लुजन मुझे नज़र आता है वो ये की सर वो ए जो लेंथ होगी वो ओ से एक ही लेंथ होगी और इस एक्सिस पे ये लेंथ ये होगी C1 b1 लेकिन क्या होगा मोड तो नेगेटिव साइन तो हटा देता हूं तो मैं क्या लिख लेता हूं मोड में C1 / b1 आई होप आपको सिग्नेचर से कोई फर्क नहीं पड़ता क्योंकि आपको डिस्टेंस चाहिए और डिस्टेंस के लिए मैंने उसे पर क्या लगा दिया है मोड की अगर C1 b1 नेगेटिव हुआ भी तो डिस्टेंस में कन्वर्ट करने का कम कौन कर देगा मॉल कोई तकलीफ कोई परेशानी आप समझो ना ये जीरो कमा जीरो है यहां से यहां तक की डिस्टेंस क्या होगी C1 V1 बट अगेन मोड में क्योंकि अभी नहीं पता है की C1 और b1 के साइन क्या है कोई तकलीफ सिमिलरली सर ओ बी की लेंथ कितनी होगी C2 B2 हान या ना तो सर इससे क्या लिख देंगे हम इसे लिख देंगे C2 / V2 सिमिलरली सर आप ओक और ऑडी क्या लिखोगे भाई ओक कितना हो जाएगा सर ओ सी हो जाएगा C2 / A2 हान या ना और ओडी कितना हो जाएगा सर c1/a1 तो ये हो जाएगा कोई तकलीफ किसी भी स्टूडेंट को आई थिंक नहीं होनी चाहिए आई थिंक किसी भी स्टूडेंट को कोई परेशानी तो यहां तक नहीं होनी चाहिए राइट C1 अब क्या करना है शायद मैं प्रूफ कर सकता हूं क्योंकि सर मोड अगर ऐसे मल्टीप्लाई हो रहा है तो आप एक कॉमन मोड भी लगा सकते हो ज्यादा सोचने की जरूरत नहीं है तो मैं लिख सकता हूं सी वैन सी तू अपॉन b1 b2c12 से सी वैन सी तू कैंसिल मोड में भी है तो कैंसिल हो ही रहा है ना तो A1 A2 इधर ए जाएगा b1 B2 वहां चला जाएगा तो फाइनल कन्फ्यूजन क्या लिखोगे आप लिखोगे सर A1 A2 = यही मेरे ख्याल से हमें प्रूफ करने को कहा गया था की आप A1 A2 और b1 B2 के प्रोडक्ट के मोड्स को इक्वल प्रूफ करिए कब बस शर्त है बस शर्तें ये कब है सर ये कब है जब आपके चारों पॉइंट्स क्या है कंसीडर्ड conscikali का मतलब क्या conscitelic का मतलब इस स्ट्रेट लाइन में जो इंटरसेप्ट्स बनाए और इस स्ट्रेट लाइन में जो इंटरसेप्ट्स बनाए जहां पर एक्स-एक्सिस और ए-एक्सिस पर इंटरसेक्ट किया उन चारों पॉइंट से एक सर्कल गुजरता है एक सर्कल पर वो चारों पॉइंट्स लाइक करते हैं तो हमने हमारी सीक्वेंट वाली थ्योरम अप्लाई कर दी की सर ओ से स्ट्रेच की गई ये दो सीक्वेंट है बेसिकली और हम जानते हैं सर ओए इन अब विल बी इक्वल तू ओक ओडी ये प्रॉपर्टी तो जस्ट अभी हमने पढ़ी है क्या आपको एक क्वेश्चन और इसका एक्सप्लेनेशन समझ आया अब एग्जामिनेशन में ऐसा नहीं दिया होगा एग्जामिनेशन में कैसा आएगा से में से एडवांस कैसे पूछेगा ये प्रूफ डेट वाला पार्ट वो नहीं देगा वो क्या देगा की एक लाइन यह है और एक लाइन ये है ये एक क्वाड्रेंट एक्सेस पे इंटरसेप्ट करती है कंसाई के लिए पॉइंट पे और इनकी जो अभी कोई न्यूमेरिकल वैल्यू दे देगा और फिर पूछ लेगा दें करके क्वेश्चन मार्क करो ऑप्शन में कुछ वैल्यूज दे देगा तो जहां पर ऐसा कुछ दिख रहा होगा वह आपका आंसर होगा मैंने एक जनरल आपको एप्रोच दिखाई की देखो अगर ऐसा कुछ दिया वहां पे वैल्यू दे देगा 3X + 4y+7 ऐसा कुछ करके और इस तरीके से चीज निकल कर ए रही होंगी आप बात समझ का रहे हैं स्टूडेंट या मैन लो इसमें कहीं ना कहीं कोई एक वेरिएबल दे दिया की लामबीडीए और उसकी वैल्यू पूछ लेगा तो यहां ये प्रॉपर्टी आपको उसे करनी पड़ेगी और ये तभी उसे कर पाओगे जब ये बात याद होगी तो बेशक हमने इसे ट्राई किया क्वेश्चन की तरह पर ये है तो एक बजे से प्रॉपर्टी जिस पर बेस्ड क्वेश्चंस आपको करने होंगे क्या आप ये बात याद रखेंगे अगर आप सब यह बात याद रखेंगे तो लेट्स मूव तू शब्द डी नेक्स्ट क्वेश्चन फॉर टुडे ये रहा आज का आपका अगला सवाल इस क्वेश्चन को ध्यान से देखिए वो क्या लिख रहा है क्या पूछ रहा है पहले इसे पढ़ते हैं और फिर हम आगे बढ़ते हैं ट्राई की आप लोगों ने क्वेश्चन स्टूडेंट सर कैसे सोचेंगे देखो भाई बड़ी सिंपल सी बात है सर मैन लो यहां पर एक सर्कल है सर्कल और सर्कल बड़ा मजेदार क्यों बड़ा मजेदार सर आपको देखो ना ये कैसा सर्कल है सर ये लिखा है x² + y² = a² मतलब यहां से एक लाइन ड्रॉ की गई है जो की सर्कल को ए और बी पर टच करती है जो की सर्कल को ए और बी पर इंटरसेक्ट करती है ये मैन लो की एक पॉइंट है और यहां से मैंने एक लाइन ड्रॉ की ऐसी और यह लाइन से इंटर सेट करती है और वो पूछ रहा है का और ब की वैल्यू कैसे निकलेंगे छोटा सा कम यह कर सकते हैं क्या सर आप जो सोच रहे हो बिल्कुल अच्छी बात है लेकिन क्या मैं पी से एक टैसेंट ड्रा कर सकता हूं क्या सकता हूं जो इस सर्कल को सी पर कनेक्ट कर रही है टच कर रही है और जैसे ही मैंने ऐसा किया की आपके दिमाग में कोई थॉट ए रहा है हमसे जब वह पूछ रहा है तो वैल्यू ऑफ पीएसी भी तो मेरा गणित कहता है सर ये जो होता है मेरी थ्योरम के अकॉर्डिंग पीसी स्क्वायर के इक्वल होता है याद आया पर ये पीसी निकलेंगे कैसे एक मिनट थोड़ी शांति रखें पीसी निकलने के लिए मेरे पास जो विचार ए रहा है जो थॉट मेरे दिमाग में ए रहे हो suniyega क्या हम ऐसा नहीं कर सकते सर की इसके सेंटर से इसकी यू नो पॉइंट ऑफ कॉन्टैक्ट पर इस टांगें के पॉइंट ऑफ कॉन्टैक्ट पर एक परपेंडिकुलर ड्रॉप करें सर कर लो और इसके सेंटर से इसके सेंटर से इसको पी को मैं कनेक्ट कर डन और मैन लो मैं इसे का दो कनक्लूड था दिस ट्रायंगल सर बड़ा मजेदार ऑब्जर्वेशन है की अगर मुझे पीसी निकलना है तो अगर मैं पो और ओक पता कर लूं तो पाइथागोरस थ्योरम को ढाल नहीं आउट सर कैसे ध्यान से बात करेंगे उसे बारे में सुनना क्या आप देख का रहे हो पो इस नथिंग बट डी हाइपोटेन्यूज यह परपेंडिकुलर या बेस है ना तो मैं क्या कनक्लूड करूंगा मैं कहूंगा सर की पो का स्क्वायर विल बी इक्वल तू ओ सी का स्क्वायर प्लस पीसी का स्क्वायर सही बात है क्या हमें निकलना है पीसी का स्क्वायर जैसे क्या मैं का सकता हूं वो होगा पीसी का स्क्वायर माइंस ओक स्क्वायर बिल्कुल सही बात है लेकिन पो होगा क्या अरे भाई पो क्या होगा स्टूडेंट्स सोच के बताओ सर पीओएस नथिंग बट डी डिस्टेंस बिटवीन डीज तू पॉइंट्स कौन-कौन से निकल सकता हूं मैं आता है मुझे क्या होगा सर अल्फा माइंस जीरो का होल स्क्वायर प्लस बिता माइंस जीरो का स्क्वायर कैंसिल ड्रॉप करते हैं तो वो रेडियस होती है तो उसी क्या है सर उसी इस नथिंग बट डी रेडियस ऑफ डी सर्कल तो उसी का स्क्वायर मतलब स्क्वायर एग्जाम में भी ऐसे ही क्वेश्चंस आएंगे क्या नहीं है एग्जाम में यहां पर कोई ना कोई वो वैल्यू दे दे देगा एक ही यहां पर पी की जगह पे कोई ना कोई वैल्यू दे देगा अब आपसे जब पी और पी भी पूछेगा तो इंटरनल आप ये मेकैनिज्म उसे करोगे जो आपको सिखाई गई है तो अभी हम इन्हें डेरिवेशंस या थ्योरम या फिर कुछ कॉन्सेप्ट्स की तरह क्वेश्चंस के फॉर्म में सिख रहे हैं पर इन्हें आपको याद रखना है और एग्जाम में अगर कोई क्वेश्चन ऐसा आता है तो इसे अप्लाई करना है मैं बस न्यूमेरिकल्स नहीं रख रहा हूं यहां पर मतलब मैं न्यूमेरिकल वैल्यू में उसे जनरल अलजेब्राइक फॉर्म में लिख रहा हूं पर आपको ये बात याद रहनी चाहिए की चीज कुछ ऐसी बनती है आई होप आप ये बात याद रखोगे और बहुत अच्छे से याद रखोगे भूलोगे नहीं इस बात को गौर से सुनना की देखो जैसे मैन लो आपके पास एक क्या है सर्कल है इट्स अन देयर इस अन सर्कल और इस सर्कल का ऑफ कोर्स कोई ना कोई इक्वेशन दी होगी इक्वेशन क्या होगी भाई एक स्टैंडर्ड एक क्वेश्चन तो मैन लेते हैं x² + y² + 2gx + 2fy प्लस सी इसे इक्वल तू जीरो सर इसको देख के क्या दिमाग में आता है मैं ये कनक्लूड कर सकता हूं की इसके सेंटर के कोऑर्डिनेट्स क्या होंगे सर इसके सेंटर के कोऑर्डिनेट्स होंगे माइंस जी कमा माइंस एफ अब इसको छोड़कर पहले छोटी सी बात और करना चाह रहा हूं जैसे मैन लो आपके पास मैंने एक सर्कल दिया आपको मैंने एक सर्कल दिया और मुझे कहा जाए की सर्कल के अंदर कहीं ना कोई पॉइंट है तो अब बता सकते हो की इस पॉइंट से कितनी कोड्स पास हो सकती है सर infainight कार्ड्स कैन पास कैसे सर infainight देखो हम ड्रॉ करने की कोशिश करते हैं है ना ऑफ कोर्स में मैन लेता हूं ये सर्कल का जो है ये है सेंटर है ना अब infainight चॉर्ड्स कैसे सर एक कार्ड हो सकती है सर ऐसे जा रही हो है ना एक कोड सर हो सकती है जो ऐसी जा रही हो है ना सर एक कार्ड हो सकती है जो ऐसे जा रही हो आई होप आपको बातें समझ ए रहे हैं एक कार्ड हो सकती है सर ऐसे जा रही हो क्या आपको बातें समझ ए रही है हो सकता है सर एक कार्ड हो सकती है जो थोड़ा इसे रोते कर दूंगा जो की शायद ऐसे जा रही हो आपको कन्फ्यूजन ना समझ लें इस तरीके से infainights कैन पास अब आप खुद समझ का रहे हो जितनी ज्यादा सर्कल के सेंटर के पास आती जाएगी उसकी लेंथ उतनी ही बढ़ती चली जाएगी है तो सबसे छोटी यानी शॉर्टेस्ट कोट कौन सी होगी सर मेरा कहना है जो सबसे छोटी कट होगी ना डी शॉर्टेस्ट पर अगर मैं सर्कल के सेंटर से परपेंडिकुलर ड्रॉप करूं तो वो बायसेक्स हो जाए की आप मेरी बात सुनो वैसे तो क्लीयरली सारी कोट्स बाईसेक्टर होते द मतलब इस कॉर्ड पर अगर मैंने परपेंडिकुलर ड्रॉप किया तो यहां पर बायसेक्स हो जाएगी हर कोड सर्कल के सेंटर से परपेंडिकुलर ड्रॉप करने पर बायसेक्स हो जाएगी पर सुनो ना इस लाइन को जिस पॉइंट से गुजरने की बात हो रही थी ना सर जिस पॉइंट से गुजरने की बात हो रही थी उसे पॉइंट को और सर्कल के सेंटर को मैं ज्वाइन करूं उसे पॉइंट को और सर्कल के सेंटर को अगर मैं ज्वाइन करूं तो उसे ज्वाइन करने वाली लाइन से परपेंडिकुलर क्रिएट हो जाए जिस कॉर्ड पर वो कॉर्ड हमारी शॉर्टेस्ट लेंथ की कॉर्ड होगी रिटायरमेंट आप समझ का रहे हो मतलब सर नहीं हो रही होगी ना उसे लाइन के थ्रू कौन सी लाइन के थ्रू मैं बात कर रहा हूं जिस लाइन के थ्रू ये येनो ये ये सारी कोट्स पास हो रही है और सर्कल के सेंटर को मैंने ज्वाइन किया तो आप देख का रहे हो सर कोई ये तो 90 डिग्री एंगल नहीं बना रही है ये तो 90° एंगल नहीं बना रही है और ऑफ कोर्स ये वाली कट को बायसेक्स तो नहीं हो रही है आपको समझ ए रहा है सिर्फ एक ही कोई खास कार्ड होगी जो बायसेक्स होगी और वो कौन सी कॉर्ड है सर अगर आप ध्यान से देखोगे तो ये वाली जो कॉर्ड है ये वाली जो कार्ड है ये क्लीयरली सर बायसेक्स हो रही है क्योंकि देखो इसकी ये लेंथ और रियल लेंथ इक्वल है सारी चॉर्ड्स नहीं हो रही है बस सिर्फ और सिर्फ हम रिलीज और ये इक्वल साइज मैंने यहां बनाया सिर्फ ये और ये कॉर्ड है ना मैं इसे थोड़ा और ठीक करता हूं क्योंकि मैंने शायद कंफ्यूज कर दिया आपको ध्यान से सुनना ये कौन सी कोर्ट की बात कर रहा हूं मैं इसकी ये जो कार्ड आपकी जा रही है है ना ये जो कार्ड आपकी जा रही है जो की यहां पर जा रही है है ना मतलब कौन सी स्टूडेंट्स ये जो एंगल है ये और ऑफ कोर्स में बात करूं तो ये ये एक 90° एंगल बन रहा है की आपको बात समझ ए रही है और ऑफ कोर्स के अगर मैं पॉइंट्स निकलूं जैसे मैन लो ये है पॉइंट ए और ये है पॉइंट बी अरे जो ड्रॉप किया है अपने परपेंडिकुलर यानी जिस पॉइंट से वो पास होती थी तो आपका एक और कब = लेंथ के होंगे है ना सेंटर के cardinate कुछ भी मैन लो जैसे की हम अभी फिलहाल मैन लेते हैं ओ क्या यह कंक्लुजन समझ ए रहा है की सर किसी पॉइंट से infainight कार्ड कैन पास पर मैं उसे ढूंढना चाह रहा हूं जो की बेसिकली जो की बेसिकली इस पॉइंट की वजह से मिड पॉइंट क्रिएट कर रहा हूं यानी ये जो पॉइंट है ये पॉइंट का मिड पॉइंट बन रहा हो और ये स्क्वाड का मिढ्वाइंट कब बनेगा पहली बात तो ये जो कॉर्ड होगी ये शॉर्टेस्ट लेंथ की होगी याद रखिएगा और ये मिढ्वाइंट तब बनेगा जब सर सर्कल के सेंटर से उसे पॉइंट को कनेक्ट करेंगे तो कार्ड क्या और रही होगी बायसेक्स यानी सर्कल के सेंटर से इस पॉइंट को कनेक्ट करने पर यह जो लाइन है यह जो लाइन है सेंटर से इस पॉइंट पर परपेंडिकुलर होनी चाहिए बहुत कन्ज्यूरिंग तो नहीं है स्टूडेंट से स्टेटमेंट अगर ये नहीं है स्टेटमेंट कन्ज्यूरिंग तो वापस आते हैं इस बात पर सर्कल का सेंटर है -जी कमा माइंस है और सर्कल पर ऑफ कोर्स कहीं ना कहीं कोई पॉइंट है लेट्स से ये एक पॉइंट है है ना ये एक पॉइंट है अब मुझसे कोई कहता है की कुछ भी हो सकता है ना लेट्स से अगर आपको समझना है तो मैं का देता हूं X1 मोबाइल अब मुझे आप खुद सोच के बताओ ने एक ऐसी कॉर्ड की इक्वेशन ढूंढने की तलाश में हूं जिसका मिड पॉइंट है X1 तो वह कोर्ट कौन सी होगी सर वो कोर्ट बड़ी बेसिक सी होगी की पहली बात तो ये की आप इससे इस पॉइंट को कनेक्ट करना इससे इस पॉइंट को कनेक्ट करना और आप एक ऐसी लाइन बनाना आप एक ऐसी लाइन बनाना जो की ऑफ कोर्स उसे पर क्या हो परपेंडिकुलर तो यह जो कोर्ट बनी ना सर यह जो कोड बनी है आपकी जो की ऑफ कोर्स कहां टच हो रही है एक तो आप कई पॉइंट हो और ये आपका पॉइंट है जैसे आप का सकते हो ए और बी तो ये जो ए बी कट बनी है ना ये वो कॉर्ड है जिसका मिड पॉइंट है आई होप ये अपेरेंट है के aparimet है क्योंकि ऑफ कोर्स ये बन गया 90° और ये बन गया क्या आपके इक्वल सेक्शंस बाय सेट हो गया और भी कार्ड्स पास हो सकती थी मैं बार-बार यही बात कहना चाह रहा हूं इस पॉइंट से और भी क्वैड्स पास हो सकती थी पर क्या इस कॉल्ड का मिढ्वाइंट होता है नहीं सर ये तो छोटी डिस्टेंस तो ये बहुत ज्यादा डिस्टेंस है तो ये वो कंडीशन फुलफिल नहीं करता तो आपको बातें समझ ए रही है मैं उसे कोड की इक्वेशन ढूंढने की तलाश में हूं जिसका मिड पॉइंट है X1 कमा y1 कैसे सोचेंगे सर एक बात बताओ अगर मुझे अब की क्वेश्चन चाहिए मैं दो-तीन चीज आपसे डिस्कस करना चाह रहा हूं जैसे मैं सर्कल का सेंटर कहता हूं सी और मैं इसे कहते हैं तो क्या आप मुझे क की स्लोप बता सकते हो हान से निकल सकते हैं क की स्लोप निकलना कौन सी बड़ी बात है अगर आप मुझसे पूछते हो क की स्लोप तो वही बात Y2 - y1 अपॉन X2 - X1 निकल लोग क्या सर जैसे ही क की स्लोप ए जाएगी मैं जानता हूं की अब और क क्या है परपेंडिकुलर अरे हम जानते हैं ना सर की अब और क क्या है एक दूसरे पर परपेंडिकुलर तो अगर क की स्लोप पता चल गई तो अब की स्लोप पता चल जाएगी क्या सर अब के स्लोप कुछ नहीं होगी नेगेटिव रस्सी पर्पल होगी किसका क की स्लोप का कोई तकलीफ अब सर मुझे अगर अब के स्लो पता चल गई और मुझे ये पता है मुझे यह पता है की अब किस पॉइंट से पास होता है तो क्या मैं अब के क्वेश्चन नहीं लिख सकता ये तो एक स्ट्रेट लाइन की इक्वेशन लिखना होगा ना जो किसी एक सर्टेन पॉइंट से पास होती है और जिसकी स्लोप हम जानते हैं बिल्कुल सर और जब आप ऐसा कुछ करेंगे तो एक बहुत ही मजेदार सा रिजल्ट निकल कर आएगा की आपको ये पुरी मॉडेस्ट ऑपरेटिंग नहीं समझ ए रही है की हम कैसे निकलने वाले हैं अब मैं सीधे कंक्लुजन की तरह बड़ा हूं जो की आपको समझना चाहिए जैसे मैन लो सीधी-सीधी सी बात ये आपका एक सर्कल है x² + y² वनप्लस 2G एक्सप्रेस 285 + सी = 0 इसके सेंटर के cardinate हम जानते हैं सर क्या होंगे -जी कमा माइंस एफ अब मुझसे कोई कहे मुझसे कोई कहे की सर X1 कमा y1 X1 कमा y1 आपका एक पॉइंट है जिससे ओबवियसली मल्टीपल और रदर infainight कार्ड्स के पास मैडम लुकिंग फॉर डेट स्पेशल कोड विच हज X1 एस इट्स मिड पॉइंट मतलब X1 कमा y1 उसे कॉल्ड का मिढ्वाइंट है हमें उसे खास कार्ड की तलाश में हूं कैसे निकलेंगे सर आई होप ये बड़ा बेसिक सा ऑपरेशन आपको क्लियर हो रहा है की सर उसे X1 से अगर सर्कल के सेंटर को कनेक्ट किया जाए तो सर्कल का सेंटर सर्कल के सेंटर से कनेक्ट होने वाली वो लाइन उसे कट पर अगर परपेंडिकुलर है तो क्लीयरली सर वो कट बायसेक्स हो रही होगी तो हमें अब कार्ड की तलाश में मेरा कहना है जो स्टैंडर्ड ऑपरेटिंग प्रोसीजर हम फॉलो करेंगे वह क्या होगी सर वो ये होगी की सर्कल के सेंटर हम जानते हैं -जी कमा माइंस है और जो मिड पॉइंट है आपकी कोर्ट का वो है X1 y1 तो क्या मैं ये कम लाइन की स्लोप निकल सकता हूं बिल्कुल सर मैं कम लाइन के स्लोप निकल लूंगा क्या y1 - एफ/x1-जी जो की ये हो जाएगी अगर यह है स्लोप सी एम की तो अब की स्लोप क्या हो जाएगी सर इसका नेगेटिव रिसिप्रोकल जिसका रिसिप्रोकल करके नेगेटिव निकल दिया सर मैं जानता हूं अब X1 y1 से पास होती है तो अब की स्लोप पता है और अब किस पॉइंट से पास होती है तो क्या अब की इक्वेशन निकल सकता हूं जब आप फाइनली अब की इक्वेशन निकलेंगे और उसे सिंपलीफाई करेंगे तो ये बहुत ही मजेदार सा ऑब्जर्वेशन आपको दिखाई देगा जो की फाइनली कुछ ऐसा होगा सुनना ध्यान से जो की फाइनली कुछ ऐसा होगा एक्स एक्स वैन प्लस ए ए वैन प्लस जीएक्स + X1 + फी प्लस 51 + सी इस इक्वल तो x1² + 1² + 2G X1 + 2a 51 + सी इसका क्या मतलब है इस बात पर अब मैं आपसे बात करने वाला हूं सुनेगा ध्यान से जब भी आपके पास कोई ऐसी इक्वेशन आती है बहुत इंपॉर्टेंट बात है इसे suniyega x² + y² + 2gx + 2fy प्लस सी इसे इक्वल तू जीरो अगर मैं इस इक्वेशन की टांगें सॉरी इस इक्वेशन की क्वाड की इक्वेशन निकलना चाह रहा हूं टेंशन की भी हम बात करेंगे तो अगर कभी भी मैं कहूं की आप निकालिए टी अगर मैं कभी भी आपसे कहूं आप निकल लिए टी तो टी का मतलब क्या होगा जब आपसे कहूं मैं की आप निकालिए 80 तो आप क्या करेंगे आप x² को एक्स एक्स वैन से रिप्लेस करेंगे आपको पॉइंट दिया होगा तीन निकलने के लिए आपको एक पॉइंट दिया होगा अच्छा आई होप आपको ये कंक्लुजन समझ ए रहा है की ये आपकी इक्वेशन निकल कर ए रही है एक्स एक्स वैन प्लस ए ए वैन प्लस ऑफ कोर्स जी टाइम से एक्स + X1 + एफ ए + 51 + सी = ये कुछ इसी क्वेश्चन तक लाने का एक शॉर्टकट हम बना रहे हैं एक-एक तकनीक डिवेलप करें क्योंकि यह जो तरीका अब हम बनाने वाले हैं ना ये ना सिर्फ सर्कल की टैसेंट या नॉर्मल की इक्वेशन या जो हम अभी बात करें इस इक्वेशन में बल्कि हाइपरबोला पैराबोला एलिप्स इन हर चैप्टर में कम आने वाली है तो इस बात को बहुत ध्यान से सुना ये बात उसे चीज की शुरुआत है क्या जब भी मैं बात करूं आपसे टी की अब टी एक स्टैंडर्ड कन्वेंशन है उसे टीका मतलब क्या होता है आज हम उसे बारे में बात करने वाले हैं जैसे अगर आपको एक कर्व की क्वेश्चन दे दी जाएगी क्योंकि फिलहाल अभी हमारे केस में सर्कल है और एक पॉइंट दे दिया जाएगा और इससे हमें टी निकलना होगा तो टी क्या होगा एक जनरल कन्वेंशन है टी जो होगा उसके लिए हम क्या करेंगे सर हम x² को एक्स एक्स वैन से रिप्लेस कर देंगे सिमिलरली हम y² को हम ए स्क्वायर और एक बात तो शायद आपको याद होगी हम डिस्कस कर चुके हैं अगर मैं कभी भी आपसे पूछूं ये की डी वे आपकी क्या निकल करेगा ये सब आपके निकले में हेल्प करेंगे टी इसी तरीके से एक चीज आप भूले नहीं हो की अगर मैं आपसे पूछूं S1 क्या होता है तो S1 कुछ नहीं होता ऐसा ये जो आपकी सर्कल की इक्वेशन है इसमें एक्स और ए की जगह आप क्या रख दो X1 y1 एंड S1 क्या होगा सर S1 कुछ नहीं होगा एक्स और ए की जगह आप क्या रख दो आपके X1 और y1 क्या आप मेरी बातें समझ का रहे हैं क्या आप मेरी बातें बहुत अच्छे से समझ का रहे हैं स्टूडेंट्स जो कहना चाह रहा हूं उसे सुनना जब आप यह निकलोगे टी और S1 और अगर आपने अगर आपने टीको S1 की इक्वल रख दिया तो आपको सर्कल की उसे कोड की क्वेश्चन मिल जाएगी जिसके मिढ्वाइंट के कॉइन मिड पॉइंट के कोऑर्डिनेट्स हैं X1 कॉम y1 मैं थोड़ी और क्लेरिटी देने की कोशिश करता हूं सोच के देखो जैसे मैन लो हमारे पास एक सर्कल है x² + ए स्क्वायर कर देता हूं जैसे की माइंस है ए + 3 = 0 एक सर्कल है ना मैं कोई भी पॉइंट लेता हूं इसके अंदर इसके अंदर कौन सा पॉइंट होगा जैसे इसके सेंटर के अकॉर्डिंग - मतलब 2 - 1 / 2 2 रेडियस भी कुछ ना कुछ हो जाएगी तो ऑफ कोर्स इसके अगर मैं सेंटर के अकॉर्डिंग लेना चाहूंगा मैं रेडियस कितनी हो जाएगी जैसे मैन लो ये है फोर फोर का स्क्वायर कितना मतलब ये हो जाएगा 22 का स्क्वायर कितना 4 और 4 में से अगर 1 / 2 तो ये बात थोड़ी बिगाड़ देगा तो इस 3 को हमें मैन लेता हूं थोड़ी देर के लिए लेते -3 उससे हमारी हेल्प हो जाएगी बस मैं एक सर्कल की इक्वेशन बना रहा हूं कोई क्वेश्चंस बनाने के लिए मैन लो आपकी ये सर्कल की इक्वेशन है अब suniyega कम की बात ध्यान से कम की बात ये है जैसे मैं इस ए को भी थोड़ी देर के लिए 2y ही लिख देता हूं ताकि हम चीज अच्छे से सोच पाए है ना suniyega ध्यान से कम की बात है बहुत कम की बात है suniyega ध्यान से मैं क्या कहना चाह रहा हूं आपसे ये एक सर्कल्स इस सर्कल का सेंटर है सर इस सर्कल का सेंटर है तू कमा -1 है ना और इस सर्कल के लिए रेडियस क्या है सर्कल की रेडियस अगर आप निकलोगे तो रेडियस आएगी तू स्क्वायर फोर 4 प्लस वैन कितना फाइव प्लस थ्री करेंगे तो 5 + 3 कितना हो जाएगा 8 है ना मैं इसमें लिख देता हूं फोर फोर क्यों लिख रहे हो सर बस ऐसे ही मेरा मैन हो रहा है अगर मैंने लिखा होता 5 - 4 या 5 + 49√ कितना होता है तो 3 क्या होती है इसकी रेडियस अब बात समझ का रहे हो मैं कैसा सर्कल visilise है मैं इमेजिन करना चाह रहा हूं जिसका सेंटर है तू कमा -1 पर अब तू कमा - 1 मतलब सर ये होगा तू और ये होगा माइंस में तो यहां कहीं और इसकी रेडियस है थ्री कुछ ना कुछ एक ऐसा सर्कल बन रहा होगा आई होप आप मेरी बात समझ का रहे हो मैन लेता हूं वैन कमा जीरो एक पॉइंट आप खुद सोचो वैन कमा जीरो यहां पे रख के देखो वैन जीरो माइंस फोर जीरो माइंस फोर माइंस फोर माइंस फोर माइंस एट माइंस एट प्लस वैन माइंस सेवन तो इसमें 1 0 जब मैंने रखा तो वैन कमा जीरो मुझे इसमें रखने पर क्या दे रहा है एक नेगेटिव वैल्यू मतलब वैन कमा जीरो सर्कल के अंदर लाइक करता है जैसे मैन लो वैन कमा जीरो ऑफ कोर्स यहां कहीं होगा जो सर्कल के अंदर लाइक करेगा अब मेरा आपसे सवाल है की आप मुझे उसे कार्ड की क्वेश्चन बताओ उसे कोड की क्वेश्चन बताओ जिसका मिड पॉइंट है वैन कमा जीरो जिसका मिड पॉइंट है वैन कमा जीरो क्या आपको क्वेश्चन समझ आया की मुझे आपसे उसे कोड की क्वेश्चन चाहिए जिसका मिड पॉइंट है वैन कॉमर्स जीरो तो आप बोलोगे सर टी = S1 आपके हेल्प करेगा अब सुनना ये है आपके सर्कल की इक्वेशन और ये है आपका उसे पॉइंट का वो मिड पॉइंट उसे कार्ड का मिड पॉइंट जिसकी आपको एक क्वेश्चन निकालनी है तो टी क्या होता है सर टी क्या होता है suniyega ध्यान से x² को एक्स एक्स वैन से रिप्लेस करना तो एक्स तो एक्स है X1 कौन है वैन तो 1 एक्स तो आप क्या लिखोगे एक्स क्या बात आपको समझ आई ए स्क्वायर कितना हो जाएगा जीरो कोई दिक्कत तो नहीं है स्टूडेंट्स फिर एक्स को एक्स + x1/2 से रिप्लेस करना तो एक्स को एक्स + X1 यानी एक्स + 1 / 2 2 से -4 कैंसिल हो तो -2 टाइम से एक्स + 1 ठीक है हो जाएगा सही हो जाएगा -2 टाइम्स x+1 आई होप ये बाते समझ ए रही है ए को ए + y1 / 2 से रिप्लेस करना तू से ये तू कैंसिल हो जाएगा ए + y1 ए + 0 तो ये कितना बचेगा सर ये बचेगा आपका और सिर्फ 5 और कांस्टेंट हमको कांस्टेंट ही रहने दीजिए ये तो आपका आयत अब आप निकालिए S1 S1 मतलब इस इक्वेशन में स्क्रब की चूंकि अभी फिलहाल इस इक्वेशन में आपके X1 और y1 पास कर दीजिए तो फिर से वही बात 100 पीएम में रख रहा हूं -4 -4 -4 - 8 - 8 + 1 - 7 तो यहां आप क्या लिखेंगे -7 ये जो लीनियर इक्वेशन बनकर ए रही है ये इसी कोड की इक्वेशन है जिसके मिड पॉइंट है वैन कमा जीरो क्या यह बातें आसानी से आपको समझ ए रही हैं सर अगर आप इसे लिखना चाह रहे हो तो आप इसे क्या लिखोगे देखो ये है एक्स और ये है -2x तो कितना हो जाएगा सही हो जाएगा -एक्स कोई दिक्कत तो नहीं सर ये है प्लस ए तो इसे मैं लिख लेता हूं प्लस फाइव सो यहां से मिलेगा -2 ये है -4 ये हो जाएगा -6 - 7 है तो -6 + 7 जो की कितना हो जाएगा +1=0 और इसे आप थोड़ा सिंपलीफाई करें तो मेरे ख्याल से हो जाएगा एक्स - ए = 1 ये आपकी सर्कल की क्वाड की इक्वेशन होगी जिसका मिड पॉइंट है वैन कमा जीरो क्या ये पुरी मेथड समझ आई सर ऐसा ही क्यों कर रहे हैं टी = S1 ही क्यों कर रहे हैं क्योंकि जब आपने आपकी थ्योरम के अकॉर्डिंग निकल ना जब आपने आपकी थ्योरम के अकॉर्डिंग क्वेश्चन निकल तो वही तो वही देखो टी एक्स एक्स वैन ए ए वैन एक्स स्क्वायर की जगह होता है ना 2G एक्स तो तू से तू कैंसिल एक्स की जगह एक्स + X1 फी प्लस ए तो ए + y1 आपकी ए की जगह आपने रखा और तू रखा था तो तू से ही ये तू कैंसिल प्लस सी इक्वल तू x1² + y1² + 2G X1 +2f y1 + सी ये तो S1 है सर तो यही तो मैं का रहा हूं जब आपने पुरी इक्वेशन को सॉल्व किया क्वाड की तो आपको फाइनली टी = h1 जैसी एक फॉर्म मिली इसलिए आप कनक्लूड कर पाए की आप इस तरीके से चीज निकलोगे याद रखोगे क्या क्या आप ये बात याद रखोगे अच्छा सर एक और बात मैं जानना चाह रहा हूं आपसे की सर उसे पॉइंट से पास होने वाली सबसे बड़ी और सबसे छोटी क्वाड कौन सी होगी इस बात पे थोड़ी बात वापस करते हैं आप अब तो समझ ही गए हो ना सर की ये तो बड़ी आसान सी बात है इतना सोच कर पहले आपके सर्कल का सेंटर है है और मैं इस पॉइंट को का देता हूं थोड़ी देर के लिए क्या वही X1 y1 तो क्या सर आप मुझे बता सकते हो इस पॉइंट से पास होने वाली सबसे बड़ी क्वाड अरे भाई सर सबसे बड़ी कोर्ट तो आसान है आपको क्यों नहीं दिख रही है सर सबसे बड़ी कोर्ट मतलब मतलब सर जो ये होगी और ये क्या होगी सर डायमीटर सो दिस इसे दी कॉर्ड विच हज दी मैक्सिमम लेंथ ये आपकी मैक्सिमम लेंथ वाली कॉर्ड होगी जो की क्या हो जाएगी जो की बेसिकली होगी क्या डायमीटर याद ए रहा है क्या आई होप आप मेरी बात समझ का रहे हो लेकिन ये कौन सा खास डायमीटर है विच पासेस थ्रू डी डायमीटर विच पासेस थ्रू व्हाट X1 कमा y1 कोई दिक्कत तो नहीं है तो ये वो कार्ड है जिसकी लेंथ मैक्सिमम है सर एक मिनट क्या होगी आपकी परपेंडिकुलर क्या बात आप समझ पाए सर अगर वो इस पॉइंट से पास हो रही है और सर्कल के सेंटर को आई होप आपस में समझ का रहे हो सर्कल के सेंटर को इस पॉइंट से कनेक्ट करने वाली लाइन अगर उसे कोड पर परपेंडिकुलर है तो ये कार्ड कौन सी होगी सर ये जो कॉर्ड होगी अब ये आपकी मिनिमम लेंथ वाली कॉर्ड होगी क्या ये बेसिक सा फंडा याद रखोगे ये के याद रखने का का रहे हो सर आप ये इसलिए याद रखने का का रहा हूं स्टूडेंट्स क्योंकि इस पे क्वेश्चंस बनेंगे वो मिनिमम लेंथ पर भी सवाल कर सकता है मैक्सिमम लेंथ पर भी सवाल कर सकते हैं ये जो मिनिमम लेंथ वाली कॉर्ड है इसकी इक्वेशन याद है ना सर इसकी इक्वेशन जो है ये जो आपकी अब की इक्वेशन है वो क्या है सर वो हम देते हैं टी = S1 से अब ये लिंगो से आप थोड़े एक्वेंनटेंस हुए या नहीं सर क्या मतलब मतलब पी क्या होता है भाई टी होता है x² = x² को X1 से रिप्लेस करिए Y2 ए ए वैन से रिप्लेस करिए एक्स को एक्स + x1/2 से और ए को y1 ए + y1 Y2 से रिप्लेस करिए और S1 क्या होता है सर उसी कर्व यानी सर्कल की फिलहाल इक्वेशन में एक्स और ए की जगह 10 और y1 रिप्लेस कर दीजिए जैसे ही आप करेंगे तो आपको इस अब कोड के क्वेश्चन मिल जाएगी क्या इसकी इक्वेशन निकलने में कोई दिक्कत सर नहीं है क्योंकि सेंटर के कोऑर्डिनेट्स पता ही होंगे और मिड पॉइंट के अकॉर्डिंग पता ही होंगे तो X1 y1 और -3 - एस से पास होने वाली स्ट्रेट लाइन की इक्वेशन नहीं लिख सकते क्या सर ये तो सबसे आसान कम है इसके लिए तो ज्यादा सोचना ही नहीं है तो वो भी निकल सकते हैं बट आई होप आपको क्लियर सी एप्रोच समझ ए रही है की चीज कैसे इंटरनल हो रही हैं उसकी मेकैनिज्म क्या है वो चीज कैसे बन रही हैं क्रिएट हो रहे हैं और सबसे जरूरी बात आज के लेक्चर की मत बोलना जिससे आपको नोट डाउन करके रख लेना है की टीका ये मतलब होता है क्योंकि ये बात जो आज आपको बताई जा रही है ये ना सिर्फ किसी सर्कल के लिए बल्कि हम एल्स पैराबोला हाइपर बोला या किसी भी रैंडम कर्व के लिए कनक्लूड करेंगे कोई करके मतलब यह जो टी वाली बात है ना यह टी बहुत कृष्ण टर्म है इस पर बहुत सारी बातें निकलेंगे यह जो एस वैन वाली बात है ये बहुत कृष्ण टर्म है इस पर बहुत सारी बातें होंगी आने वाले दिनों में तो इन दोनों बातों को प्लीज गांठ बंद कर रख लीजिए इन पर बहुत सारे डिस्कशन हम करने वाले हैं हिंदी अपकमिंग डेज़ विले डिस्कसिंग ऑल दी चैप्टर parabolis हाइपरबोला एवं सर्कल के ही दौरान अभी हम बहुत बातें करेंगे इस बारे में यह सर ऐसा क्यों कर रहे हो आप सीधे क्यों नहीं कर रहे हो जैसे आपने मेकैनिज्म सिखाई ये तरीका आसान है जनाब ये तरीका बहुत अच्छा है ये बहुत सिंपल है इससे बहुत जल्दी बातें हो जाती है अब देखोगे की बहुत सारे चैप्टर बहुत कॉम्प्लिकेटेड बहुत ही अजीब सी प्रॉब्लम बातें करेंगे क्या यह बातें याद रहेंगी अगर ये बातें याद रहेंगे और इसका प्रूफ आप मुझे दे सकते हो एक मजेदार से क्वेश्चंस को सॉल्व करके क्या दिया है क्वेश्चन सर वो पूछ रहा है suniyega ध्यान से फाइंड डी इक्वेशन ऑफ दी कार्ड तो सर एक कॉर्ड के क्वेश्चन चाहिए जो की इस सर्कल की है ठीक है सर जो की इस पॉइंट से पास होती है और उसकी डिस्टेंस उसकी लेंथ जो है वो शॉर्टेस्ट लेंथ है आपको रिलाइज हो रहा है अगर एक शॉर्टेस्ट लेंथ की कॉर्ड पूछी जा रही है मतलब सर ये वही कोड है ये वही कोड है जिसका मिड पॉइंट है 2 3 अब बात समझ का रहे हो बड़ी सिंपल सी बात आपसे कहना चाह रहा हूं मैन लो ये ऑफ कोर्स रहा आपका क्या ए एक्सिस मतलब ए-एक्सिस एक्स-एक्सिस को तो लाओ ही मत पिक्चर में मैन लो ये आपका सर्कल है ना दिस इस योर सर्कल हर और इस सर्कल के सेंटर के कोऑर्डिनेट्स क्या होंगे भाई सारे सर्कल का सेंटर यहां कहीं होगा जिसके सेंटर के कोऑर्डिनेट्स क्या होंगे सर वैन कमा तू आई होप आप ये निकलना सिख गए इनका हाफ करके नेगेटिव ले लेते हैं तो वो जाएगा वैन कमा तू तो इसके सेंटर के कोऑर्डिनेट्स हो जाते हैं वैन कमा तू आप ऐसे भी सोच सकते हो इनके अगर आप चाहो तो और सर इस सर्कल में अंदर कहीं पॉइंट है 2 3 इस सर्कल के अंदर कहीं ना कहीं पॉइंट है लेट से यहां कहीं हमें मैन लेता हूं ये तू है ना वो एक और यूनिट यहां पे एक्स कार्ड तू है तो यहां पे ही मैं मैन लेता हूं जैसे ये पॉइंट है 2 3 अब जब आपसे कोई कहे जब आपसे कोई कहे इसके कोऑर्डिनेट्स क्या मैन लेते हैं 2 3 अगर कोई कहे की ये 2 3 से पास होने वाली शॉर्टेस्ट लेंथ वाली क्वाड निकालो तो आई थिंक सर शॉर्टेस्ट लेंथ वाली कौर तो बड़ी मजेदार सी होगी कैसी मजेदार सी होगी सर क्योंकि उसे कार्ड में मुझे जो दिखाई देगी चीज वो क्या होगी सर उसको ऐड में जो मुझे चीज दिखाई देगी वो ये की उसे कार्ड में जो वो पॉइंट है 2 3 जिससे वो पास हो रही है उससे अगर मैं सर्कल के सेंटर से कनेक्ट करूं तो वो परपेंडिकुलर हो जाएगी स्क्वाड पर या टेक्निकल वो जो पॉइंट है तू कमा थ्री वो स्क्वाड का मिड पॉइंट होगा की आप ये बातें समझ का रहे हो स्टूडेंट्स बिना ज्यादा कंफ्यूज हुए बिना ज्यादा घबरा अटेंशन लिए आई थिंक सर हम समझ का रहे हैं है ना आई थिंक मेरे ख्याल से तो आप सभी ये बात समझ का रहे हो की ये क्या हो जाएगी बायसेक्स खैर इतनी सारी बातें करने की जरूरत नहीं थी आपको बस समझने के लिए विजुलाइज करवाने के लिए मैंने बताने की कोशिश की चाहो तो आप इसे लिख सकते हो क्या लिखना है ये आपका एक पॉइंट है और ये आपका पॉइंट है तो इससे इस लाइन के स्लोप निकल जाएगी उसका नेगेटिव रिसिप्रोकल से इस लाइन की स्लोप ए जाएगी वो 2 थ्री से पास होती है पर सर इतना नहीं सोचना है ये कर सकता हूं मैं पर इसके अलावा भी कोई तरीका है क्या है कैसे सर याद करो हमने तो सिखा है टी = S1 अगर किसी कोड के मिड पॉइंट के कोऑर्डिनेट्स दे दिए जाएं किसी सर्कल की किसी कॉर्ड के मिढ्वाइंट के कोऑर्डिनेट्स दे दिए जाएं तो सर्कल की इक्वेशन लिखना बहुत आसान है टी = S1 से टी क्या होता है आई होप अब मुझे सीखने की जरूरत नहीं है टीम मतलब x² को रिप्लेस करिए एक्स एक्स वैन से यानी की से 2X से ए स्क्वायर को रिप्लेस करिए ए ए वैन से यानी किस 3y से कोई दिक्कत तो नहीं है सर एक्स को रिप्लेस करिए आई होप यू - 2 का तो 1/2 हो जाएगा यानी -1 तो एक्स की जगह रख दोगे एक्स + x1/2 तो वो तू से माइंस वैन बन गया है अब एक्स की जगह रखोगे एक्स + एक्स वो तू हटा दिया मैंने ना तो क्या हो जाएगा एक्स + 2 है ना बाहर माइंस साइन तो है ही तो माइंस ऑफ एक्स + 2 फिर क्या सर ये तू कैंसिल ए की जगह क्या रखोगे ए + y1 / 2 यानी ए + भाई तू तो भाई तू से फोर कैंसिल तो बाहर क्या बचेगा सर बाहर बचेगा -2 और अंदर बचेगा ए + 3 और क्या बच जा रहा है सर और बच रहा है -4 और ये किसकी इक्वल है खबरदार जो आपने इसे जीरो के इक्वल रखा नहीं इक्वल है S1 के एस वैन क्या है सर इसी सर्कल की इक्वेशन में तू कमा थ्री आप पास कर दीजिए तो 2 3 पास किया तो तू का स्क्वायर फोर थ्री का स्क्वायर 9 2 2 4 3 4 12 और क्या -4 ये आपकी उसे कार्ड की इक्वेशन है बात खत्म बात सच में खत्म सर्विसेज सिंपलीफाई करके निकल सकते हैं क्या निकल लो भाई कौन सी बड़ी बात है देखो जरा ध्यान से देखना स्टूडेंट्स यह है 2X ये है माइंस एक्स तो कितना बचेगा एक्स यह है 3y और यहां पर है -2y तो ये कितना बचेगा ये बचेगा ए कोई दिक्कत अब सुनना ये है माइंस तू और ये है -3 और ये है -4 क्या आपको बात समझ ए रही है फिर से लिख देता हूं ये है -2 ये है -6 एम रियली सॉरी ये होगा -6 तो -2 -6 कितना -8 - 8 - 4 - 12 और -12 से -12 कैंसिल हो आपको दिख रहा है तो बच के ए जा रहा है सर फोर से फोर कैंसिल तो बच्चा 9 - 4 कितना 5 5 को इधर ले तो कितना -5 तो या फिर आप इसे ऐसे ही रख लो और ये जो फाइनली आपकी इक्वेशन बन कर आई है एक्स + 5 = 5 ये उसे कट की इक्वेशन है जो की शॉर्टेस्ट लेंथ कैरी करती है क्योंकि इस पॉइंट से पास होती है इस सर्कल की डू यू अंडरस्टैंड इजीली कहीं पर भी कोई डाउट या परेशानी आई होप सबसे जरूरी बात आज के लेक्चर के जो आपको नोट डाउन करके रखनी है जो आपको याद रखनी है की टी क्या होता है सर टी क्या होता है टी मतलब याद रखिएगा स्टूडेंट्स x² को एक्स एक्स वैन से रिप्लेस करना इसका मतलब y² को yy1 से रिप्लेस करना इसका मतलब एक्स को एक्स + x1/2 से रिप्लेस करना इसका मतलब ए को ए + y1 Y2 से रिप्लेस करना आई होप ये बात याद रखिए और S1 मतलब क्या आई होप S1 को लेके तो ज्यादा डिस्कशन करने की जरूरत नहीं है S1 मतलब उसे कर्व की इक्वेशन में जहां भी एक्स दिखे उसको X1 से रिप्लेस कर दो और जहां भी वे दिखे कांस्टेंट कोई वैल्यू होगी क्या यह बात याद रहेंगी और बस इसी का कंक्लुजन था जो आज हमने पढ़ा की सर क्या t=s1 ये किसकी इक्वेशन है जिसके मिड पॉइंट्स हो कौन X1 y1 और वही चॉर्ड्स शॉर्टेस्ट लेंथ की होगी और मैक्सिमम लेंथ की कौन सी कार्ड होगी जो की उससे पास होने वाला डायमीटर होगा कोई डाउट स्टूडेंट्स कोई परेशानी आई थिंक चीज स्मूथ हैं चीज आसान है चीज बड़ी ही क्लियर्स हैं अब आते हैं सीधे आज के असाइनमेंट क्वेश्चंस की तरफ ये हैं आपके असाइनमेंट क्वेश्चन जिन्हें आप का सकते हैं बिल्कुल होमवर्क क्वेश्चन भी तो यह रहा आज का पहला होमवर्क क्वेश्चन जिसका आपको ले लेना है स्क्रीनशॉट और यह क्वेश्चन करना है आपको खुद से अटेम्प्ट मैं समझा जरूर देता हूं वो का रहा है ट्रायंगल पिक इस इनस्क्राइब इन दिस सर्कल तो इस सर्कल के अंदर आपको ट्रायंगल पिक इनस्क्राइब करना है कैसे की के और आर के कोऑर्डिनेट्स आपको दिए हैं ये है के और ये है आर ठीक है क्या और वो का रहा है की आपको एंगल कपर ये बहुत आसान क्वेश्चन आप थोड़ी सी प्रॉपर्टीज उसे करना तो सारी अभी तक हमने बड़ी क्वेश्चन हो जाएगा सर कैसे करेंगे विजुलाइज करना ड्रा करना देखो हिंट मिल रही है जीरो कमा जीरो रेडियस 5 तो सबसे पहले तो ये की सर इस सर्कल का सेंटर हो जाएगा 0 और उसकी रेडियस हो जाएगी 5 कुछ तो समझ ए जाएगा उससे के और पी पॉइंट्स बन जाएंगे बस फिर चीज आपको दिख जाएंगे फिर चीज मेरा यकीन करिए देखते जाएंगे बहुत टू नहीं है क्वेश्चन आई होप आप ये बात याद रखोगे सर अगर ये बात हमें समझ आई होप पिक क्या है एक ट्रायंगल है जो आपने इनस्क्राइब किया है इसके बाद क्या सर इसके बाद अगर मैं आपसे बात करूं तो आता है आज का अगला असाइनमेंट क्वेश्चन ये आपका सेकंड होमवर्क क्वेश्चन है स्टूडेंट इससे भी आपको ट्राई करना है और इसे भी आपको आंसर से मैच करना है एक बार जरा इस क्वेश्चन का भी स्क्रीनशॉट ले लीजिए स्टूडेंट्स को देख लीजिए की इस क्वेश्चन को क्या पूछ रहा है अगर ये क्वेश्चन आपको समझना है तो क्या लिख रहा है वो लेंथ ऑफ डी टांगें लेंथ ऑफ डी टांगें ड्रॉन फ्रॉम फाइव तू दिस सर्कल लेंथ ऑफ डी tenjent मुझे बताने की जरूरत नहीं है हमने कुछ तो पढ़ा था कुछ तो पढ़ा था स्टूडेंट्स याद करो की हमने लेंथ ऑफ टैसेंट निकलना नहीं सिखा है बस यह क्वेश्चन खत्म हो गया मेरा यकीन करिए वेरिएबल अननोन है बट लेंथ दे दी है ना तो बस वो लेंथ निकलना उसके इक्वल रख देना तो शायद के की वैल्यू ए जाएगी और वो आपका आंसर ए जाएगा क्या ये एक आसान सा बेसिक सा मजेदार सा डायरेक्ट सा क्वेश्चन है जो आप कर लेंगे याद का आपका सेकंड असाइनमेंट क्वेश्चन हुआ क्वेश्चन थर्ड असाइनमेंट क्वेश्चन जो आपका निकल कर आता है देखिएगा ध्यान से वो बोल रहा है आई होप अपने थर्ड क्वेश्चन का स्क्रीनशॉट ले लिया है तो रेस्ट क्वेश्चन का स्क्रीनशॉट क्लिक कर लीजिए देखो क्या बोल रहा है वो समझेगा वो का रहा है फाइंड डी लेंथ ऑफ डी tenjent ड्रोन फ्रॉम अन्य पॉइंट है पॉइंट ऑन डी सर्कल तू डी सर्कल अब क्या फर्क इन दोनों सर्कल्स में पहली बात तो आप नोटिस करो यह दोनों सर्कस को सेंट्रिक है आप समझ का रहे हो दोनों ही सर्कल्स को सेंट्रिक है जिनके सेंटर्स के cardinate हैं -जी कमा माइंस है वह यह पूछ रहा है की सर इसका रेडियस में ही फर्क आएगा यहां पे C1 ए जाएगा यहां पे C2 ए जाएगा तो बस रेडियस में ही थोड़ा फर्क है वो ये का रहा है फाइंड डी लेंथ ऑफ डी टांगें ड्रॉन फ्रॉम अन्य पॉइंट ऑन दिस सर्कल तू दिस सर्कल याद करो एक ऐसा क्वेश्चन किया था हमने दो कंसंट्रिक्स सर्कल वाला क्वेश्चन किया था जहां पर हमने बात की थी सर एक सर्कल ये बाहर है और एक सर्कल आपका अंदर है है ना एक सर्कल्स द अब आपको पता है सर इस सर्कल पर मौजूद किसी भी पॉइंट से अगर मैं कोई यू नो टांगें ड्रॉ करता हूं तो कुछ-कुछ चीज या बहुत कुछ चीज कही जा सकती बस उसी दौरान आपको इस लेंथ को निकलना था अब यकीन करिए क्वेश्चन आप बहुत इजीली कर लेंगे आपको इस लेंथ पर बात करनी है आपको हर एक जो चीज जरूरी है पढ़ की है और आपको फाइनली ये प्रूफ करना है की वो ये होता है सर नहीं करें तो ये क्वेश्चंस ये रहने दे दो मत रहने दीजिएगा मतलब क्वेश्चंस कर लीजिएगा कुछ कम होगा कुछ कारण होगा जो आपको जरूरी होगा जिस वजह से डेलाइब्रेटली नोइंग ली विलिंग ली एक क्वेश्चन यहां पर रखे गए हैं इन पर बेस्ड सवाल आएंगे स्टूडेंट्स यह प्रॉपर्टी है जो की अगर आपको पता होगी और नहीं भी रात पाए नहीं भी याद रख पाए जो की ऑफकोर्स नहीं कहा जा रहा की हर चीज राखी है तो आप अगर ये एप्रोच सीखे की देखो इससे इसपे कोई टेंशन ड्रॉप किया जाए तो उसकी लेंथ इस तरीके से कुछ ऐसे निकल जाती है तो क्वेश्चन जब आएगा तो आप ऐसे ही मेथड वापस उसे करके कर लेंगे बना लेंगे और ये तभी होगा जब आपके दिमाग में चीज तोड़ दूंगी जब आपने ऐसी प्रैक्टिस की होंगी जब आपने ऐसे क्वेश्चंस सॉल्व किए होंगे आई होप ये बातें आप समझ का रहे हो जितनी ज्यादा प्रॉब्लम सॉल्व करोगे उतने दिमाग के तार खोलेंगे उतनी अलग अलग अप्रोचों देख पाओगे सिख पाओगे उतना ब्रेन आपका चीज है डेवलप कर पाएगा अलग-अलग प्रॉब्लम से डील करना है इस एग्जाम को क्रैक करने का तरीका आपको वैरायटी ऑफ क्वेश्चंस का एक्स्पोज़र रहना चाहिए वर्ण बात नहीं बनेगी इसमें देखो भाई क्या लिखा है वो का रहा है की एक सर्कल है जो पॉइंट ऑफ इंटरसेक्शन ऑफ दिस एंड दिस से पास होता है एक बात बताओ स्टूडेंट्स अगर ये जो स्ट्रेट लाइंस के पॉइंट ऑफ इंटरसेक्शन से वो पास होता है तो बहुत सारी बातें निकल कर आते हैं पॉइंट ऑफ इंटरसेक्शन ऑफ डी cardinate एक्सेस विद दिस लाइन एंड दिस लाइन यहां पर आपकी थ्योरम पिक्चर में ए जाएगी स्टूडेंट्स याद करो एक ऐसा क्वेश्चन किया था की अगर एक सर्कल है जो की इससे और इसके और ऑफ कोर्स आपके दो और लेने कौन-कौन सी एक तो इस लाइन के पॉइंट ऑफ इंटरसेक्शन इसके साथ और एक इस लाइन के पॉइंट ऑफ इंटरसेक्शन इसके साथ है ना जैसे मैन लो यही है इसके साथ तो मैं कहूंगा की वो जो पॉइंट ऑफ इंटरसेक्शन आएंगे कौन-कौन से वो जो पॉइंट ऑफ इंटरसेक्शन आएंगे उनसे काफी कुछ कहा जा सकता है स्टूडेंट अन सर्कल पासेस थ्रू डी पॉइंट ऑफ इंटरसेक्शन ऑफ डी cardinate एक्सिस विद डीज लाइंस अगर मैं इसे थोड़ा और क्लीयरली लिखो ना तो मैं थोड़ा इसको और सही से बनाना चाहूंगा स्टूडेंट्स बस मुझे दो सेकंड दो आप एक तो ऑफ कोर्स आपका कैसा कुछ होगा एक तो ऑफ कोर्स आपका कुछ ऐसा कुछ होगा ये जो पॉइंट है ये यहां से जारी होगी लाइन और एक लाइन आपकी यहां से कुछ ऐसी जा रही होगी वो बहुत अच्छे से मैं बना नहीं पाया हूं तो ये वो दो लाइंस हैं और आपके वही बात स्टूडेंट्स याद करो ये क्वेश्चन मैंने पढ़ाया था जब आपको आज ही कहा था लास्ट लेक्चर है ना की आपको इसके पिछले वाले लेक्चर में जब हमने बात की थी की वो दो लाइंस हैं ए वैन एक्स + b1 y+c1=0 और दूसरी लाइन है आपकी a2x + B2 = 0 वो cardinate एक्सिस पर इंटरसेक्ट करती हैं तो फिर conscikalik वाली आपकी सेकंड वाली थ्योरम ए गई थी ओए इन अब विल बी इक्वल तू ओक इन दी बस उसी प्रॉपर्टी को उसे करेगा आपका एक क्वेश्चन सॉल्व हो जाएगा और इसीलिए उसे वक्त आपको पढ़ते वक्त कहा गया था की देखो यहां पे हम इसे ऐसा थ्योरम आज अन डेरिवेशन इसे अन प्रॉपर्टी पढ़ रहे हैं पर उसी की जगह आपको कुछ क्वेश्चंस बना के दिए जाएंगे और कुछ वैल्यू पूछ ली जाएंगी और इस तरह के क्वेश्चंस आपको आईआईटी जी मैं आंसर एडवांस में निकल कर आएंगे तो ये आज का आपका अगला होमवर्क क्वेश्चन है जो आपको कर लेना है कहीं भी कोई परेशानी हो तो जरा इसका स्क्रीनशॉट ले लीजिए ताकि आप इसे भी सॉल्व कर लें स्क्रीनशॉट ले लिया क्या चलिए नेक्स्ट क्वेश्चन देखते हैं जो आपको ऐसा होमवर्क दे दिया जा रहा है सो ये तो बड़ा अजीब सा लग रहा है इसका आंसर भी बहुत अजीब सा लग रहा है पहले तो घबराओ मत वो का रहा है थ्री cosentric सर्कल्स कंसंट्रिक्स मतलब तीनों सर्कल्स का सेंटर से है ऑफ विच डी बिगेस्ट इस दिस ठीक है जो तीनों की रेड है ना प्लूरल हो जाएगा रेड आय रेडियस का वो अर्थमैटिक प्रोग्रेशन ए-आर ए और ए + आर और जो बिगेस्ट रेडियस है वो कौन है वैन तो ए + आर की वैल्यू कितनी हो जाएगी वैन क्या बात समझ का रहे हो फिर क्या बोल रहा है वो एक लाइन है ए = एक्स + 1 कट ऑल डी सर्कल्स आते रियल एंड डिस्टिंक्ट पॉइंट करती है बस इससे बहुत मजेदार बात निकल के आएगी आप सोचना दें फाइंड डी इंटरवल इन विच डी कॉमन डिफरेंस ऑफ आप विल लाइक देखो मैं क्वेश्चन आपको समझने की कोशिश करता हूं बहुत बेहतरीन क्वेश्चन है ये क्वेश्चन कुछ ऐसा है जैसे पहले तो आपके पास क्या है तीन सर्कल्स आपके पास 3 सर्कल्स हैं एक और सर्कल आपको बनाना होगा जो कलेक्शन यहां पर बना लेता हूं कुछ ऐसा क्योंकि अब वो ये कहना चाह रहा है आपसे जस्ट गिव मी अन कौन सी एक्स ए = एक्स + 1 है ना ए = एक्स + मैन लेता हूं कोई एक ऐसी स्ट्रेट लाइन रही होगी जो की है वही इस इक्वल्स तू एक्स प्लस वैन इसमें जो सबसे बाहर वाला सर्कल है सबसे जो बाहर वाला सर्कल है वो ऑफ कोर्स क्या है आपका x² + y² = 1 तो आपको ये तो पता चल गया सर की इन तीनों सर्कल्स की जो सेंटर के कोऑर्डिनेट्स हैं वो है ओरिजिन यानी जीरो कमा जीरो क्यों का रहे हो क्योंकि तीनों को centrif है तो जो इसकी सेंटर है वही इसके भी हैं आई होप ये बात आपको समझ ए रही है अब जब ये का रहा है की ये जो लाइन है ये जो लाइन है ये इन तीनों कोड इंस्टेंट पॉइंट्स पर इंटरसेक्ट करती मतलब अंदर वाले को भी दो पर बाहर वाले को दो पर सबसे बाहर वाले को भी ऊपर तो याद है ना दो डिस्टिंक्ट पॉइंट इंटरसेक्ट करना मतलब मतलब सर कहानी बड़ी सिंपल सी है की अगर इसकी जो परपेंडिकुलर लेंथ है वो उसकी रेडियस से कम होगा इस लाइन की सेंटर से जो परपेंडिकुलर लेंथ है वो भी उसके रेडियस से कम होगा और इस लाइन की इसी सेंटर से जो परपेंडिकुलर लेंथ है जो की तीनों के लिए से है वो इसकी रेडियस से भी कम होगा बस यह कंडीशन आप अप्लाई कर दीजिएगा आपका क्वेश्चन सॉल्व हो जाएगा ये क्वेश्चन तू बी वेरी ऑनेस्ट मैसेज का होना तो जी एडवांस्ड लेवल का क्वेश्चन है जो की अच्छे तरीके से अगर आप सॉल्व करें तो आपके दिमाग के बहुत सारे तार खुलेंगे अब बहुत एंजॉय करोगे इस क्वेश्चन को सॉल्व करने के जो भी कॉन्सेप्ट चाहिए हमें सब पढ़ा है तो ये होना चाहिए ये सच में होना चाहिए इसको लेकर हर एक चीज हम जानते हैं इसके बाद अगर आज के में अगले असाइनमेंट क्वेश्चन की तरफ बढ़ो ये आज का आपका अगला असाइनमेंट क्वेश्चन है जो की बेसिकली आपकी डीपी पीस की तरह है जिन्हें आप कहते द डेली प्रैक्टिस प्रॉब्लम शीट उसी तरीके से अब हम हर टॉपिक के ओवरऑल कंप्लीशन पर आपको ऐसे क्वेश्चन देते हैं ताकि पूरे टॉपिक आपका रिवीजन हो जाए वो का रहा है थ्रू अन फिक्स्ड पॉइंट ह कमा के सी केस आर ड्रोन तू दिस सर्कल आई होप इसको देख के आपको समझ ए रहा है जीरो कमा जीरो सेंटर और रेडियस आर शो डेट डी lokass ऑफ डी मिड पॉइंट्स ऑफ डी सी केड्स बाय डी सर्कल्स इस दिस वह क्या कहना चाह रहा है एक पॉइंट है ना ह कॉम के उससे आपने क्या किया उससे आपने सेकंट ड्रॉ की उसे आपने क्या ड्रॉ की ड्रॉ की अब वह यह कहना चाह रहा है की जो सीक्वेंस आपने बनाई है याने ऑफ कोर्स ये आप किसी केंट है कौन सी एक बी और एक सेकंड क्या होगी आपकी सीडी कोई दिक्कत तो नहीं है और ये किस पॉइंट से बने हैं पी से जैसे मैं का देता हूं क्या ह कॉम के वो आपसे पूछना चाह रहा है मैं पूरे क्वेश्चंस लगभग आपको एक्सप्लेन कर देना बस आपको ट्राई करना है ये सर्कल आपका बेसिक सा बड़ा सिंपल सा शॉर्ट ट्रिक्स और सर्कल था जिसके सेंटर के कोऑर्डिनेट्स 0 और इसकी रेडियस में जानता हूं कितनी है आर है ना ये यहां पर भी कनेक्ट करोगे तो और होगी यहां पर भी रोगी और यहां पर भी आर ही होगी तो वो आपसे ये कहना चाह रहा है भाई बड़ी सिंपल सी बात की सर जो मिड पॉइंट है किसका सर इसका मिड पॉइंट्स ऑफ डी सेकंड है ना जो मिड पॉइंट है सेकंड के एक तो मिड पॉइंट यहां अगर एक मिड पॉइंट क्या होगा इन मिड पॉइंट का लॉकर्स आप ये प्रूफ कर दो मेरा हकीम करिए बहुत आसान क्वेश्चन है बहुत इजीली आप प्रूफ कर दो की अगर आपने वही सारी बेसिक प्रॉपर्टीज अप्लाई की जो सब कुछ हमने सीखी है एक बहुत ही कृष्ण और एक बहुत जरूरी मैन लो मेरे पास एक कर्व है लेट्स दिस इसे अन बहुत ध्यान से सुनना बहुत कम की बात है स्टूडेंट्स और की कोई ना कोई क्वेश्चन है ना कुछ भी क्वेश्चन है लेट्स से भी थोड़ी देर के लिए का देता हूं एफ ऑफ एक्स मैं थोड़ा सा कैलकुलस यानी डर को उसे करना चाह रहा हूं घबराना मत कर साहब मैं आपसे अगर पूछूं की सर यहां पर ना एक पॉइंट है लेट्स से यहां पर एक रैंडम कोई पॉइंट है इसके cardinate हैं X1 कमा y1 suniyega बहुत कम की बात है स्टूडेंट है ना स्कार्फ को आप क्या कहते हैं आप कहते हैं ए = एक्स राइट आपसे जो बात मैं पूछ रहा हूं जो मैं का रहा हूं वो ये है स्टूडेंट्स की मैन लो अगर मैं इस पॉइंट पर एक टांगें ड्रॉ करता हूं इफ आई हैपेंड तू ड्रा अन टैसेंट तू दिस करीब आते दिस पॉइंट लेट से मैं इस पॉइंट पर एक टांगें ड्रॉ करता हूं कुछ इस तरीके से क्या ये बात आप समझ का रहे हो स्टूडेंट्स बात इतनी सी है बस suniyega ध्यान से क्या मैं किसी तरीके से कैलकुलस में एक एप्रोच आपको दी जाती है क्या मैं किसी तरीके से यह जो टैसेंट मैंने इस कर्व के एक्स वैन कमा y1 पॉइंट पर जो ड्रा की है इसकी स्लोप निकल सकता हूं कैलकुलस कहता है हान एक ऐसा तरीका है ऐसा क्यों होता है कैसे होता है इस बारे में बात करेंगे जब हम कैलकुलस पढ़ेंगे क्वाड्रेंट ज्यामिति के बाद हम कैलकुलेशन शुरू करेंगे फिर हम डर पढ़ रहे होंगे तब मैं आपको समझा रहा हूं की देखो चीज ऐसी होती हैं बट फिलहाल अभी हमारे लिए ये समझना बहुत जरूरी है की चीज क्या है ये बात क्या है suniyega ध्यान से बात बस इतनी सी है की अगर मुझसे कोई पूछे की ये जो आपने टैसेंट ड्रॉ की है इसकी अगर मुझे स्लोप निकालनी है तो कुछ नहीं कर रहे हैं आपको जो भी कर्व आपको दिया गया है उसकी एक अलजेब्राइक एक्सप्रेशन मतलब इक्वेशन दी जाएगी आपको कुछ नहीं करना है आपको उसका कर देना है डर आपको उसका कर देना है डर फर्स्ट डेरिवेटिव निकलना है और उसमें जो एक्स और ए वैरियेबल्स रखे होंगे ना उनकी जगह आपको क्या पास कर देना है वो पॉइंट जिस पॉइंट पर आप ने ये X1 y1 जो आपका पॉइंट है जिस पर आपने ये टेंशन ड्रॉ की है बहुत कन्ज्यूरिंग तो नहीं हो रही है जैसे अगर मैं सिंपल सा एग्जांपल लेकर आपको समझाना चाहूं जैसे लिखा है मेरे पास 2x² + 3X + 2 मैन लो यह कर है मुझे नहीं पता सर कौन सा कर वे बट ये एक कर्व है ना और लेट्स से एक पॉइंट है एक पॉइंट है -1 2 मैं पूछ रहा हूं आपसे की क्या आप मुझे इस कर्व के पॉइंट -1 2 पर ड्रा की गई tanagent की इक्वेशन बता सकते हो सर कैसी बातें कर रहे हो मतलब कैसे निकलेंगे मैथ्स इतनी मजेदार सी चीज है की इसने एक तरीका क्रैक करके आपको दिया है तरीका ही है की आप पहले तो scurv का डर निकल लीजिए तो पहले आप एफ ऑफ एक्स का डेरिवेटिव निकल लीजिए जैसे आप कहते हो एफ-एक्स कैलकुलस की बातें कर रहा हूं घबराना मत स्टूडियो अगर आप निकलोगे तो कितना ए जाएगा सर वो ए जाएगा 4X आई होप आप बात समझ का रहे हो प्लस थ्री और ये जीरो आई होप डर अब इस एफ-एक्स में अगर आपको एक्स और ए दिख रहे हैं तो उनकी जगह माइंस वैन कमा तू रख दीजिए ठीक है कर रख देते हैं तो एफ-1 कमा तू निकल रहा हूं तू तो है नहीं वैसे वहां पर तो बस माइंस वैन ही रखना होगा तो क्योंकि वही नहीं है ना तो मैं सीधे माइंस वैन ही रख देता हूं कोई घबराहट वाली बात तो नहीं है स्टूडेंट तो बिना घबरा बस ये लिख लो की एफ ऑफ एफ- ऑफ माइंस वैन है हमें निकलना है -1 यहां रखा तो माइंस फोर माइंस फोर प्लस थ्री कितना सर माइंस फोर प्लस थ्री मेरे अकॉर्डिंग होता है -1 तो एफ-1 आता है -1 दिस इस नथिंग दिस इस नथिंग बट डी स्लोप ऑफ डी स्ट्रेट लाइन ये जो है ना ये उसे स्ट्रेट लाइन के स्लोप है किसकी जो की tagent है तो दिस इसे दी स्लोप ऑफ डी टैसेंट सर कहां पर कौन सी ट्रेन तू डी गिवन कर्व कौन सा कर सर वह जो आपको दिया हुआ है मतलब ये देखकर थोड़ा बहुत अगर आपको आइडिया लग रहा है तो आपको ये समझ आएगा की सर ये बेसिकली एक पेरा बोला है ये एक पैराबोला है इसमें कहीं ना कहीं एक पॉइंट रहा होगा जो की कौन सा रहा होगा जो की रहा होगा -1 2 तो अगर मैंने सर इस माइंस तू पर एक टांगें ड्रॉ की होती अगर माइंस ऊपर मैंने टांगें ड्रॉ की होती तो उसे tyanjan की जो स्लोप आती है ना उसे स्ट्रेंजर की जो स्लोप आती वो स्लोप क्या होती है सर उसे स्ट्रांग की जो स्लोप आती वो होती - 1 क्या ये सारी बातें आपको समझ ए रही है और जो अगर उसने पूछा होता की इक्वेशन बता दो तो सर कहानी तो सुलझ गई ना क्यों क्योंकि टैसेंट के स्लोप पता है और वो किस पॉइंट से पास हो रही है पता है तो एक स्ट्रेट लाइन की स्लोप और वो किस पॉइंट से पास हो रही है पता है तो टैसेंट की इक्वेशन है ना स्ट्रेट लाइन की इक्वेशन हम नहीं लिख सकते हैं क्या देख सकते हैं तो क्या माइंस कितना माइंस वैन टाइम्स एक्स माइंस एक्स माइंस माइंस एक्स + 1 इस सिंपलीफाई कर लेना यही आपकी इस टैसेंट की जो आपको टेंशन दी गई थी इस टैसेंट की स्लोप होगी बहुत परेशानी तो नहीं हो रही है स्टूडेंट्स इस कॉन्सेप्ट को डाइजेस्ट करने में इस बात को समझ लेने में आई थिंक ये बड़ा आसान सा कॉन्सेप्ट है जो आपको क्लियर हो रहा है सर ये सब हम क्यों पढ़ रहे हैं ये सब पढ़ने का रीजन है स्टूडेंट्स की मैन लो अगर मुझसे कोई कहे की सर एक सर्कल है अब अभी हम रैंडम भी एक जनरल किसी कर्व के लिए जनरलाइज की बात कर रहे द और चैप्टर तो हमारा सर्कल चल रहा है ना तो मैन लो आपके पास एक जैसे हम कहते हैं सर्कल है ना तो सर्कल की इक्वेशन हम पढ़ते आए हैं सर x² + ए स्क्वायर एक कोऑर्डिनेट्स होगा क्या - जी - एफ मतलब सर्कल का बेसिकली सेंटर मैं आपसे यह पूछना चाह रहा हूं सर की इस सर्कल पर मैन लो कहीं ना कहीं कोई पॉइंट है कोई पॉइंट है ना जैसे वो पॉइंट है क्या वो पॉइंट के cardinate हैं X1 कमा y1 तो क्या आप मुझे इस पॉइंट पर ड्रा की गई tenjent की इक्वेशन बता सकते हो कैन यू हेल्प मी इन फायरिंग आउट विच टी टांगें ड्रोन तू दिस कर क्यूट इजीली क्योंकि हम कैलकुलस जानते हैं मैं हमेशा आपको raycommend करूंगा की आप कैलकुलस वाली ही तकनीक उसे करना घबराना मत कैलकुलेशन वाली तकनीक का मतलब क्या मतलब आप पहले तो ऐसे कर देना डिफरेंटशिएट बिल्कुल कर देंगे सर और उसमें ये X1 y1 रख देना तो उससे क्या ए जाएगी स्लोप और टेंशन की स्लोप ए जाएगी तो क्या करना है सर तेनजिंग की स्लोप अगर ए गई है तो टेंशन की इक्वेशन नहीं लिख पाएंगे क्या लिख देंगे आई होप आप मेरी बात समझ का रहे हो मैं आपसे कहना चाह रहा हूं की एक तकनीक जो आपको दिख रही होगी वो क्या की सर मैं अगर कर्व की बात करूं तो कर्व क्या है सर एफ ऑफ एक्स जो की क्या है x² + ए स्क्वायर मतलब ये अलजेब्राइकली लिखा है एफसी और ये सब तो अभी ये बहुत कन्ज्यूरिंग लग सकता है बट रियलिटी में या नंबर होगी ना x² + y² 3X + 4y - सी -3 ऐसा कुछ तो आई थिंक डिफरेंटशिएट करना आसान होगा तो अब आप क्या करो सबसे पहले सबसे पहले आप इसे डिफरेंटशिएट कर दोगे तो डिफरेंटशिएट करेंगे तो क्या हो जाएगा सर वो ए जाएगा एफ-एक्स एफ-एक्स क्या ए जाएगा सर 2X है ना y² का डर 2y दश + कोस कितना 2G और यहां पे क्या डिफरेंस यहां पे क्या ए जाएगा इसका डर डर 2f है ना और अगेन अभी यहां पे क्या करोगी इस ए का डर क्या हो जाएगा सर वो हो जाएगा wydasht कोई तकलीफ तो नहीं है और प्लस ही है ना सी का डेफिनेशन जीरो अब आपको क्या करना है suniyega अब आपको ये निकलना है सर की वायर दश की वैल्यू निकालनी है तो पूरा आप इधर कॉमन ले लोग पापा और यहां से इसमें आप एक्स और ए की वैल्यू जहां जहां एक्स और ए दिख रहा है उनकी जगह क्या रख दोगे आप रख दोगे 109 y1 उससे आपकी ए जाएगी स्लोप यहां से स्लोप ए जाने के बाद जब आप X1 और y1 वैल्यू पास कर दोगे तो इस tagent की आप क्या बना लोग इक्वेशन क्या ये बात आप समझ का रहे हो आई होप ये बात अच्छे से आपको डाइजेस्ट हो रही है जो मैं कहना समझाना आएगा बोल ना आपसे चाह रहा हूं लेकिन सर इन सारी बातों में एक बात क्या ये सब करना जरूरी है नहीं अभी तो आप कोऑर्डिनेट्स जानते हो अभी तो आप कैलकुलस नहीं जानते हो तो सर कैलकुलस पढ़ते वक्त ये सारी बातें करेंगे अभी आप सीधे पॉइंट पे आओ कम की बात बताओ अभी क्या करना होगा तो अभी जो हम करेंगे वो क्या होगा suniyega ध्यान से आई होप आपको ये बात समझ ए रही है की ये अभी फिलहाल जो मुझे दिख रहा है वो ये की एक सर्कल दिया हुआ है उसका सेंटर है माइंस जी कमा माइंस है और इस सर्कल पर मुझे एक पॉइंट दिया है X1 इस पर एक टांगें ड्रॉ करनी है अगर मुझे कैलकुलस नहीं भी आती है तो मैं कहूंगा क्या मैं कहूंगा सर की एक कम करो X1 y1 से X1 y1 से सेंटर को कनेक्ट करो ये पॉइंट था जिस पर मुझे टांगें ड्रॉ करनी है और इसका सेंटर है क्या इस इक्वेशन इस लाइन पीसी की मैं स्लोप बता सकता हूं सर मुझे अगर लाइन पीसी की कोई स्लोप पूछता है इफ संबदी आस्क मी डी स्लोप ऑफलाइन पीसी आई थिंक सर पीके आर्डिनेंस पता है सी के कोऑर्डिनेट्स बताएं तो क्या पीसी किस स्लोप नहीं निकल सकते अरे निकल सकते हैं नहीं भाई सर निकल सकते हैं और अगर पीसी की स्लोप निकल ली है तो यह जो टैसेंट है ये जो टैसेंट है ये टैसेंट जो होगी वो कहना है इसे की सर पीसी लाइन पर जो की बेसिकली रेडियस वाली लाइन है उसे पर परपेंडिकुलर होगी तो अगर मैं पीसी की स्लोप निकल लूंगा y1 - एफ / X1 - -जी यानी y1 + एफ / एक्स + 0 जो किसकी स्लोप होगी सो क्या इसकी स्लोप की हेल्प से मैं इस tagent की इक्वेशन नहीं निकल सकता आपको जानते हो ना सर की जो पॉइंट लाइन है वो परपेंडिकुलर है पीसी पर तो जो स्लोप होगी आपकी पॉइंट की वो क्या होगी होगी -1 / स्लोप ऑफ क्या पीसी जो की निकल लूंगा और सर इतने आसान से तरीके से अगर आप इसकी स्लोप निकल का रहे हो किस की टांगें की जो की इस पॉइंट से पास होती है तो क्या आप टेंशन की इक्वेशन नहीं लिख सकते बिल्कुल लिख सकते हो cardinate ज्यामिति वाला तरीका जो बिल्कुल आसान और सोराइट्स है और कैलकुलस वाला तरीका जो की मैं कमेंट करूंगा उसे करना बट ऑन अन लेटर स्टेज ऑफ दिस चैप्टर अभी फिलहाल नहीं अभी फिलहाल आप कोऑर्डिनेट्स वाला ही तरीका बट तू मेक यू अंडरस्टैंड की देखो टैसेंट से डील करने के लिए हम कैलकुलस का तरीका अक्सर उसे करेंगे और हर कर भी उसे करेंगे पैराबोला एलिप्स हाइपरबोला या कोई भी रैंडम सा ग्राफ है उसमें किसी पॉइंट पर उसे कर्व की टैसेंट निकालनी होगी तो मैं उसका डेरिवेटिव निकल के ही उसे करूंगा हम बात करेंगे इस बारे में और फिलहाल क्या इस बात से किसी भी स्टूडेंट को भी तकलीफ की सर ये तो बहुत आसान तरीका है आई होप ये तरीका तो आपको समझ ए ये तो हमारा पड़ा हुआ तरीका है ना तो सर इस तरीके पे बात करते हैं ठीक है इस तरीके पे बात करते हैं और इसका कंक्लुजन देखते हैं तो suniyega ध्यान से मैं आज फिलहाल बस बात इतनी करना चाह रहा हूं किस तरह की सर्कल गिवन है तो उसके किसी पॉइंट पर अगर मैं tenjent ड्रॉ करूं तो उसकी इक्वेशन क्या होती है suniyega बहुत कम की बात है पहले तो हम स्टैंडर्ड ऑपरेटिंग प्रोसीजर की चीज निकल कैसे जाती है उसकी इंटरनल मेकैनिज्म क्या है वो समझ लेते हैं तो सर कल क्या होगा आपका x² + y² यही वाला है ना सर्कल का सेंटर दिख रहा है क्या ड्रा करना चाह रहा हूं टैसेंट ठीक है सर सर्कल का सेंटर हम निकल चुके हैं एक बात बताओ सीधी सीधी सी बात इस सेंटर से इस पॉइंट पी को जो लाइन ज्वाइन कर रही है उसकी स्लोप क्या होगी सर हम ये निकल चुके हैं y1 - एफ / X1 - जी यानी ये तो सर क्लियर सी बात है जो टैसेंट पी से पास हो रही है इसकी स्लोप क्या होगी सर इस टेंशन की जो स्लोप होगी वो इस स्लोप का नेगेटिव रिसिप्रोकल होगी तो माइंस ऑफ एक्स + जी / y1 + एफ बहुत बढ़िया अब सर बेसिक सी बात है मैं आपसे बस यही कहना चाह रहा था की ये जो बात यहां पर लिखी गई है जो उसने एक हल्का सा आपको हिंट दिया है वो यही की किसी भी कफ की टांगें की स्लोप डर से भी निकल जा सकती है जिस बारे में हम कब बात करेंगे जब हम कौन सा चैप्टर पढ़ेंगे जब हम हमारी कैलकुलस वाला पार्ट्स सॉल्व कर रहे होंगे जहां पर हम लिमिट्स कंटिन्यूटी डिफरेंशियल एप्लीकेशन ऑफ डेरिवेटिव इंटीग्रेशन इंटीग्रेशन अंडर डी कर्व और उसके बाद डिफरेंशियल इक्वेशन यह सारी चीजों को सॉल्व करेंगे तब हम इस बारे में बात करेंगे अभी फिलहाल यू कीप योर सेल्फ रिस्ट्रिक्टेड तू दिस ऑल डी इससे भी कुछ क्वेश्चंस हम करना सिख लेंगे घबराओ मत और इससे भी उतनी आसानी जितना ये दोनों आसान है पर क्यों पढ़ रहे हैं दो तरीके सब बिल्कुल अगर ये कम नहीं कर रहा हो तो ये उसे कर लेंगे ये नहीं लग रहा होगा तो उसे कर लेंगे तरीके तो होना चाहिए ना इट्स लाइक अलग-अलग हथियार होना ये नहीं कम कर रहा है तो ये वाली गण चला रही है तो ये वाली गण चला दी हमारे पास चीज तो हो ताकि हम जो भी तरीका वहां फटाफट अप्लाई हो रहा हो उसके थ्रू आंसर ले आए तो बात करते हैं वापस इस पॉइंट की की सर tenjent की इक्वेशन निकलने के लिए सर मुझे टैसेंट की स्लोप पता चल गई टैसेंट किस पॉइंट से पास होती है ये पता चल गई तो तेनजिंग की इक्वेशन कौन सी बड़ी बात है अगर ये बात आपको समझ आई तो बस ये आपकी tagent की इक्वेशन निकलने का डर वाला तरीका इस पे बात करते हैं इस पे बात करते हैं ये वो डर वाला तरीका है वही बात स्टूडेंट्स आपने x² का डर किया आपने ए का किया और यहां से डीएक्स निकलोगे जो तरीका आपको यहां पर मैं बताना चाह रहा था इक्वेशन किसके इक्वल थी इक्वेशन जीरो के इक्वल थी तो जब आप जीरो को डिफरेंशिएट करोगे जब आप जीरो को डिफरेंशिएट करोगे तो क्या ए जाएगा जीरो का डर जीरो तो यहां से फाइनली क्या रख दोगे आप एक्स की जगह X1 ए की जगह y1 है ना ऐसा रख दोगे तो फाइनली आपको वाइड दश मिल जाएगा ये ए' ही क्या है ये ए दश ही आपका एफ-एक्स या फिर आपका क्या है दी / डीएक्स जिसे आप क्या कहते हो ए एफ तो यहां से जब आप X1 y1 रख दोगे तो आपका ए - ए जाएगा इस बारे में डिटेल डिस्कशन कैलकुलस के वर्क करेंगे और अभी फिलहाल भी क्वेश्चंस के थ्रू मैं samjhaunga अभी अगर मैं अलजेब्रा के samjhaunga ना आप कंफ्यूज हो जाओगे आप बोलोगे नहीं सर ये तो हम छोड़ रहे हैं ये वाला लेक्चर हमको नहीं समझ ए रहा है और ये ये तो हमको चीज अजीब सी लग रही हैं सर ऐसे मत करो तो मेरा कहना है अभी फिलहाल cardinate वाला तरीका समझो पर मैं इस तरीके पर भी बहुत डिटेल में आपसे बात करूंगा पर देखो भाई ऐसे ही बस बात नहीं है तो देख लो x² का डर तू एक्स ए स्क्वायर का डर 2y wydash यानी दी / डीएक्स फिर 2gx का डर 2G और तू फी का डर तू फी दश ए दश यानी दी / डीएक्स और सी का डर जीरो और जब यहां फिर से अपने डिवाइस वाली टर्म को कॉमन लेकर स्लोप की यू नो टांगें सॉरी टांगें की स्लोप स्लोप की टेंशन की जब आपने स्टॉप निकल X1 y1 पर तो यहां पे आपने रख दिया एक्शन यहां पे अपने रख दिया y1 तो जैसे ही आपने निकाला तो ऑफ कोर्स वापस बात वही आई तो या तो वैसे निकालो डिफरेंटशिएट करके या आप सीधे-सीधे अपनी सिंपल सी cardinate ज्यामिति में X2 - Y2 - y1 अपॉन X2 - X1 वाली तकनीक से निकल लो बात वही है बात वही है तो बात क्या है सर बात ये है सर की हम जानते हैं अगर ये लाइन अगर ये लाइन किस से X1 कमा y1 से पास हो रही है और इसकी स्लोप ये है तो फाइनली जब मैं इसको सिंपलीफाई कर रहा हूं फाइनली जब मैं इसको सिंपलीफाई कर रहा हूं तो इसकी इक्वेशन जो बन के निकल के ए रही है वो कुछ ऐसी ए रही है सारी कहानी आप पढ़ लीजिए इसकी इक्वेशन कुछ ऐसी निकलकर ए रही है एक्स एक्स वैन प्लस ए ए वैन प्लस जी एक्स + X1 + f5 प्लस फाइव वैन प्लस सी सर ये तो कुछ पड़ा हुआ सा लग रहा है ये तो कुछ पड़ा हुआ था लग रहा है और लगना भी चाहिए क्योंकि कुछ लेक्चरर्स पहले आपको ये कहानी सिखाई गई की अगर एक कर्व की इक्वेशन है जैसे की हमारा सर्कल जैसे की हमारा सर्कल तो कहानी तो अब इतनी आसान हो गई है जनाब की आपको कुछ नहीं सोचना है दोनों ही तरीके भूल जाइए आप दोनों ही तरीके भूल जाइए ना तो मुझे डिफरेंटशिएट करके स्लोप निकल लो फिर दुनिया भर की कहानी पढ़नी है नाम है टेक्निकल स्लोप निकालो फिर इक्वेशन में रखो और फिर एक क्वेश्चन बनाओ नहीं है आपको कुछ नहीं करना है आप मेरा हकीम कारी आपको कुछ नहीं करना है आपको एक सर्कल की इक्वेशन दे दी जाएगी आपको एक सर्कल की इक्वेशन दे दी जाएगी और पूछा जाएगा सर्कल पर एक पॉइंट है जो की है X1 y1 इस पॉइंट पर ड्रा के लिए इंजन के क्वेश्चन निकल लो तो ये रही वो तेनजिंग की क्वेश्चन खत्म सर ये तो कुछ समझ नहीं ए रही है मैं बात ये क्या होता है हमने suniyega फिर से ध्यान से बड़ी कम की बात है ये सर्कल की एक इक्वेशन है जिस पर एक पॉइंट लाइक करता है और उसे पर ड्रा की गई tenjent की इक्वेशन होती है t=0 क्या बात कर रहे हो सर इतना आसान यस इतना आसान तो टी = 0 क्या होता है ये तो हमने डिस्कस किया है t=0 एक्स स्क्वायर की जगह क्या रखेंगे एक्स एक्स वैन ए स्क्वायर बाय जी टाइम्स एक्स + X1 सिमिलरली डी से थिंग हैपेंस विद दिस थिंग इसे वेल वेरी रिप्लेस दिस वही वही वही + y1/2 डेट तू कैंसिल सर थैंक्स ए प्लस y1 फाइनली दिस सी और यह जो मिला वह क्या है सर यह जो जनाब आपको मिला है ना यह कुछ नहीं है यह आपकी टांगें की इक्वेशन है बस इतना आसान है इतनी सारी कुछ चीज पढ़ने की जरूरत नहीं है बस इससे और जल्दी इजीली इफेक्टिवली आपकी इक्वेशन ए जाएगी मेरा यकीन करिए इतना आसान है मेरा यकीन करिए इतना आसान है मतलब एक एग्जांपल देख के बताओ सर एग्जांपल बस इतना था जैसे मैन लो आपका एक सर्कल है कुछ भी एक सर्कल प्लस ए स्क्वायर करता हूं मिल गया जैसे कोई भी अपॉइंटमेंट मिल गया जैसे रैंडम भी उसमें का दिया जैसे ए बी है मैं बस का देना नहीं चाह रहा हूं अभी इसके तो वो जो ए कमा भी उसने बोला है ये पॉइंट इस सर्कल पर लाइक करता है तो इंजन की इक्वेशन क्या होगी तो मैं कहूंगा बड़ा सिंपल सा कैसा है एक्स स्क्वायर बी की जगह कोई वैल्यू होगी जैसे होगा -1 2 कुछ भी होगा तो उससे सीधे एक लीनियर इक्वेशन बन जाएगी और जो वो इक्वेशन बनेगी वो आपकी tagent की इक्वेशन होगी इतनी आसान से सिंपल से शॉर्ट ट्रिक से बात है क्या ये बात याद रहेगी अच्छा ये बात ना और भी तरीके से याद रखना जैसे की मैन लो अगर एक सर्कल दे दिया जाए कैसा असर अगर एक सर्कल दे दिया जाए x² प्लस ए स्क्वायर टेंशन के क्वेश्चन बताओ इस सर्कल पर 5 0 पर टांगें की इक्वेशन बताओ खैर वो आप देख के ही बता सकते हो बट फिर भी मेरा मैन हो रहा है मैं कैसे निकलूं सर मैं तो निकलूंगा टी = 0 तो टी = 0 क्या कहता है वो कहता है एक्स एक्स वैन यानी क्या फाइव एक्स प्लस ए ए वैन यानी क्या 0 ए = एन - 25 इधर रख लो ये उधर रख लो इस इक्वल्स तू 25 ये बात आपको समझ आई थिंक और कुछ नहीं करना है सर क्या अरे 2G एक्स 2 फी है ही नहीं ना उसमें तो सीधी सी बात है बस ये बचेगा तो 5x 25 के इक्वल आया है तो एक्स के इक्वल है सर एक्स आता है फाइव के इक्वल एक्स किसके इक्वल आता है फाइव के और जब X5 के इक्वल आता है तो मैं क्या कहूंगा भाई X5 के इक्वल आने का मतलब क्या हुआ एक्साइज के पैरेलल ये आपकी एक्स = 5 स्ट्रेट लाइन हुई ना क्या यह बात क्या यह स्टेटमेंट क्या है कंक्लुजन आपको समझ ए रहा है बिना किसी तकलीफ परेशानियों हेसिटेशन के कोई दिक्कत कोई डाउट कोई परेशानी स्टूडेंट्स आई थिंक चीज बड़ी आसान से सिंपल सी सर ऐसा क्यों हो रहा है अरे देखो ना ये ओरिजिन पर सर्कल ऐसा कुछ जा रहा होगा ये सर्कल ऐसा कुछ जा रहा होगा तो यहां पे जो tangedraw होगी वो ऐसी होगी कोई दिक्कत अच्छा सर इसके अलावा कोई और तरीका भी हो सकता है क्या देखो मैन लो कभी वह पैरामेट्रिक फॉर्म में दे दे सर पारंपरिक फॉर्म मतलब याद करो अगर आपके पास है x² + y² = r² तो इस सर्कल पर लाइक करने वाले किसी रैंडम पॉइंट के कोऑर्डिनेट्स पैरामीटर फॉर्म में हमने क्या पढ़े हैं सर हमने पढ़े हैं आर कोस थीटा आर सिन थीटा याद ए रहा है तो क्या इस पर ड्रा होने वाले tenjent के क्वेश्चन निकल सकते हैं फिर वही बात टी = 0 उसे करिए क्वेश्चन जीरो का मतलब मतलब आप जानते हैं क्यों एक्स स्क्वायर साइन थीटा ये हो जाएगा आर सिन थीटा क्या कुछ चीज अच्छे से या सारी चीज बहुत क्लेरिटी से आपको समझ ए रही हैं आई होप कोई कन्फ्यूजन या कोई डाउट का स्कोप अभी तक तो नहीं उसी बात को कहने को आपको घुमा firakar चीज का रहा है बस इतना याद रखो की जब आप से कोई भी एक्सप्रेशन की बात करें तो आप क्या करोगे हमेशा x² को X1 y² को ए वैन तू एक्स को है एक्स + X1 सही या एक्स को एक्स + x1/2 से ए को ए + y1 / 2 से या 2y को ए + 5 से रिप्लेस करेंगे बस ये बात याद रखना अगर मैन लो इन केस वो मुझे सर्कल के क्वेश्चन को अच्छे से फॉर्म मिलते द एक्स - x² + 5 - के स्क्वायर ही देते हैं हम लोग पूरा कन्वर्ट करके नहीं दिया सिंपलीफाइड करके जनरल इक्वेशन नहीं थी उसने ह कमा किया सेंटर और आर रेडियस की फॉर्म में हमें ऐसा लिख के दे दिया तो भी क्या X1 कमा y1 पे सर यू नो टेंशन की इक्वेशन लिख सकते हैं बिल्कुल लिख सकते हैं पहली बात या तो आप सिंपलीफाई करके निकल लो वही अपनी टी = 0 वाला केस या अगर आपको ऐसा नहीं जाम रहा है तो आप लिख सकते हो क्या एक्स - टाइम X1 - ह देखो क्या है इसका मतलब सुनना इसका मतलब ये है स्टूडेंट जैसे मैन लो आपका सर्कल है का सर्कल है एक्स - 3² + ए - 2² = 6 आई होप अब ये इंटरप्रिटेशन आप निकल लेते हो इक्वेशंस देखकर तो सर क्या कहना चाह रहे हो आप मैं कहना चाह रहा हूं जैसे मैन लो इसमें कोई पॉइंट दे दिया होता है वो वही कहता है लेट्स से ए कमा भी वो कहता है लेट्स से ए कॉम भी जैसे मैन लेता हूं कोई पॉइंट अगर मुझे ढूंढना है आपको समझने के लिए तो 36 सही है ना 6 चलो अभी ले लो है ना ए कमा भी ले लो तो आप मतलब यहां पे कोई पॉइंट हो सकता है जो सर्कल पर लाइक कर रहा है वही बात मैं ढूंढना नहीं चाह रहा हूं तो मैं इसे रैंडम सा एक पॉइंट मैन लेता हूं एक कॉम अभी तो सर क्या निकाला जा सकता है बिल्कुल निकाला जा सकता है सुनना कैसे मेरा आपसे कहना है देखो आप क्या करो पहले तो लिख दो एक्स - 3 फिर आप क्या लिख दो इन दोनों का डिफरेंस बट ऑफ कोर्स आप क्या लिखोगे X1 - h1 - ह मतलब क्या ए माइंस थ्री तो यहां पे आप क्या लिख दोगे ए माइंस थ्री सिमिलरली आप यहां पे क्या लिखोगे यहां पे आप पहले तो लिख लो वायर माइंस तू और क्या लिख लो स्टोरेज ये बी -2 सॉरी हान बी -2 = वही बात वापस क्या आपका r² जो की क्या है कोस यहां पर 36 ये जो इक्वेशन है स्टूडेंट्स ये भी आपकी क्या है इस सर्कल की ए बी पर ड्रा की गई tanagent की इक्वेशन क्या किया हुआ है एक्स - 3 ये वही बात है घुमा फिर के बात वही जा रही है तो आप इसे एक्सपेंड कर देते हैं जनरल फॉर्म में कन्वर्ट कर देते हैं टी = 0 निकल देते हैं तब भी यही आता पर मैं डायरेक्ट सीखना चाह रहा हूं की अगर ऐसा कुछ दिखे तो ऐसा भी लिख दोगे तो कोई बुराई नहीं है क्या ये बात याद रहेगी अच्छे से क्या कोई डाउट या परेशानी है स्टूडेंट्स यहां तक अगर नहीं है तो वापस आते हैं कम की बात पर ये तो हम डिस्कस कर चुके हैं की सर अगर ऐसा एक सर्कल दिखे तो आपके ए कोस थीटा और ए साइन थीटा ने वही जो रेडियस होती है आपकी ये उसे केस में ये क्वेश्चन होती है जिस बारे में हम डिटेल डिस्कशन अच्छे से कर चुके हैं बहुत अच्छे से कर चुके हैं कोई परेशानी की सर अगर किसी सर्कल पर कोई पॉइंट है X1 तो इस पॉइंट से पास होने वाली इस पॉइंट से पास होने वाली है तीन तरीके से एक तरीका क्या था सर एक तरीका था की अगर आपको सेंटर के कोऑर्डिनेट्स के थ्रू स्लोप निकल लेनी है इस लाइन की तो इसकी परपेंडिकुलर लाइन की स्लोप और इस पॉइंट के थ्रू निकल लोग एक तरीका था की आपका स्क्रब की इक्वेशन क्या होती है सर्कल के क्वेश्चन होती है x² + y² + 2X + 2fy प्लस सी आप इसे डिफरेंटशिएट कर दीजिए वाइड दश आएगा ए दश में X1 y1 रख दीजिएगा तो इसकी स्लोप ए जाएगी इक्वेशन ऑफलाइन ए जाएगी या सबसे आसान सबसे बेहतर सबसे अच्छा तरीका तो सर हमने ऐसे कहा की सारी बातें छोड़ो सर टी इस इक्वल्स तू जीरो अप्लाई करो और इसमें हमने क्या सिखा था की सर x² की जगह क्या रख देना सर x² की जगह रख देना एक्स एक्स वैन ए स्क्वायर की जगह क्या रख देना सर उसकी जगह रख देना वाई-वा1 2G एक्स की जगह आप एक्स की जगह रखना एक्स + x1/2 तो वो तू से तू कैंसिल हो जाएगा तो ये हो जाएगा जी टाइम्स एक्स + X1 सिमिलरली तू फी की जगह आप क्या रख देंगे वही ए + 51 / 2 तो ये हो जाएगा व्हाइट टाइम्स ए + y1 एंड ऑफ कोर्स यहां पे क्या बचेगा प्लस सी इसे इक्वल तू जीरो बस इतना सा ये कम है बस इतना सा ये करना है बाकी कुछ ज्यादा दिमाग लगाने की जरूरत नहीं है इसीलिए एक अच्छा तू लेके बस इसके थ्रू हो उसमें बार-बार इस बात पे एक्सरसाइज इसलिए कर रहा हूं स्टूडेंट्स क्योंकि ये एक काफी कृष्ण फॉर्मूला है जो ना सिर्फ अभी बल्कि इसी चैप्टर के दौरान कई बार किसी और फॉर्म में कहीं ना कहीं किसी तरीके से या फिर एलबीएस पैराबोला हाइपरबोला है कर्व मैं कहीं ना कहीं हम इसकी जरूरत पड़ेगी तो इसे अच्छे से गांठ के समझ लो क्योंकि इस पर सवाल बनेंगे इस कॉन्सेप्ट की हमें कई बार उसे होगी नीड लगेगी इस तरीके से उसे करने पुलिया का मत चलो अब आते हैं सीधे पॉइंट पे एक क्वेश्चन ट्राई करते हैं और समझते हैं की क्वेश्चंस कैसे बनते हैं टैसेंट पर बेस जो कॉन्सेप्ट अभी हमने पढ़ा उसे ध्यान में रखें वो का रहा है टैसेंट तू डी सर्कल तो सर ये एक सर्कल है इसे देखकर आपको कुछ समझ आए सिर्फ दो बातें क्या की सर इस सर्कल का सेंटर होगा जीरो कमा जीरो पर और उसकी रेडियस होगी क्या √5 कोई डाउट तो नहीं है इस सर्कल पर इस पॉइंट पर इस पॉइंट पर अपने टेंट ड्रॉ की और वह यह कहना चाह रहा है की जो आपने टांगें ड्रॉ किया है ना सर वो एक और सर्कल को टच करती है वो एक और सर्कल की टैसेंट है मतलब वो कमेंट है ठीक है सर तो वो पूछ रहा है की वो जो आपने इस सर्कल पर इस पॉइंट पर जो टैसेंट ड्रा किए विच इस बिहेविंग विच इस अलसो बिहेविंग इस ए tenjent तू दिस सर्कल इस सर्कल को वहां किस पॉइंट पर टच कर रहे होंगे सर क्या बात कर रहे हो कैसे निकलेंगे मैं का रहा हूं थोड़ा सा पॉज करो स्क्रीन को और एक बार जरा क्वेश्चन ट्राई कर लो मेरा यकीन करो यह क्वेश्चन टू नहीं है यह क्वेश्चन आज के सबसे आसान क्वेश्चंस में से एक क्वेश्चन है जो आप आसानी से क्रैक कर सकते हो कैसे करेंगे सर suniyega कम की बात सबसे पहला थॉट तो ये सारी बातें छोड़ो सर मुझे तो आप इसकी टेंशन बता सकते हो क्या सर हमारे पास सर्कल की इक्वेशन पर draculate इंजन के क्वेश्चन नहीं लिख सकते सर सबसे आसान कहां है आज की तारीख में क्योंकि आपको बस क्या लिखना है टी = 0 कर सकते हो की नहीं सर ये अगर मैंने किया तो क्या मैं तेनजेन के क्वेश्चन बना लूंगा बिल्कुल हान तो सर बनाया तो क्या रखा सर x² को एक्स एक्स वैन तो 1 एक्स यानी कितना एक्स y² को रिप्लेस किया y1 सो - 2 ए तो ये कितना हो जाएगा स्टूडेंट्स ये हो जाएगा -2y और ऑफकोर्स इसे इक्वल तू फाइव तो माइंस फाइव इस इक्वल्स तू जीरो मैं लिख लूं क्या तो ये निकल कर आती है आपके इस सर्कल की टैसेंट की इक्वेशन मतलब बात समझने की कोशिश करना स्टूडेंट्स बात बहुत ध्यान से सुनना आपका एक सर्कल था इस सर्कल पर आपने क्या किया इस सर्कल पे आपने तेनजिंग ड्रॉ की इस सर्कल के एक पॉइंट पर अपने टैसेंट ड्रॉ की विच फॉर्चूनेटली और ओबवियसली हैपेंस तू बी अन टैसेंट तू अंदर सर्कल एस वेल क्या आप मेरी बात समझ का रहे हो ये आपका सर्कल था कौन सा x² + y² इस सर्कल की भी क्या है भाई tenjet अब अब आपको यह थॉट यह आइडिया क्यों नहीं ए रहा है सर की अगर यह लाइन इस सर्कल की टेंशन है तो इस लाइन को और इस सर्कल को मैं साथ में सॉल्व करूं तो बनेगी यह क्वाड्रेटिक और उसे क्वाड्रेटिक के दो रूट्स नहीं होने चाहिए क्योंकि ये लाइन इस सर्कल के सिक्के नहीं है एलाइंस सर्कल को दो पॉइंट्स में इंटरसेक्ट नहीं कर रही है उसे क्वाड्रेटिक के इमेजिनरी मतलब जीरो रूट्स भी नहीं आने चाहिए क्योंकि यह लाइन ऐसा नहीं हो रहा है की सर्कल को टच ही भी ना कर रही हो ऐसा नहीं हो रहा है की लाइन से दूर से निकल रही है यह लाइन इस सर्कल को एक्जेक्टली एक पॉइंट पर इंटरसेक्ट कर रही है मतलब उसे क्वाड्रेटिक के इक्वल एंड रियल रूट्स होने चाहिए मतलब उसे क्वाड्रेटिक कार्ड डिस्क्रिमिनेंट जीरो होना चाहिए अगर आपको बात याद ए रही है या तो ऐसा कर लो या ये क्लिक नहीं हो रहा है तो बस इतना सा कम करके मुझे बता दो स्टूडेंट्स इस इस स्ट्रेट लाइन पर एक पॉइंट मैन लो अगर मैन लो ह कमा के है तो वह के इस सर्कल पर भी लाइक करेगा तो पहले तो इस सर्कल को सेटिस्फाई करेगा दूसरी बात दूसरी बात उसे ह ए की इस पॉइंट से जो डिस्टेंस होगी या रदर इस सर्कल के सेंटर से स्ट्रेट लाइन की जो परपेंडिकुलर डिस्टेंस होगी वो सर्कल की रेडियस जो की एक लंबा तरीका है तो आसान तरीका क्या है आस आय होप आपको तरीके समझ रहे हैं आसान तरीका बड़ा आसान सा है साहब क्यों की आपको बस ये कर देना है की आपकी जो टैसेंट की इक्वेशन बन रही है और जो सर्कल की इक्वेशन बनी इट्स लाइक सी आर सॉल्विंग बेसिकली तू कब्ज हम दो कर्व सॉल्व करें जैसे एक कर्व ये रहा एक कर्व ये रहा इनका पॉइंट ऑफ इंटरसेक्शन निकलना है तो क्या करते हो एक कर्व होता है एफ ऑफ एक्स एक कर्व होता है जिओ ऑफ एक्स्ट्रा पॉइंट ऑफ इंटरसेक्शन मतलब सर वो पॉइंट जहां पे एफ ऑफ ह और फिक्स इक्वल हो जाए क्योंकि वो कॉमन पॉइंट है जहां पे आपके दोनों को सेटिस्फाई कर रहा है वो तो हम क्या करेंगे अगर मुझे इस लाइन का और इस सर्कल का पॉइंट ऑफ इंटरसेक्शन निकलना है तो मैं इन दोनों को इक्वेट कर दूंगा इन दोनों को इक्वल रख दूंगा क्या मेरी बात समझ का रहे हो दो तरीके या तो यहां से ए की वैल्यू निकालो एक्स की वैल्यू निकल लो वो यहां रिप्लेस कर दो या दोनों को सॉल्व कर लो जो आपको ठीक लगे मेरे लिए जो आसान तरीका है वो ये होगा स्टूडेंट्स की मैं यहां से ए की वैल्यू निकल रहा हूं या एक्स की निकाल लो कोई फर्क नहीं पड़ता मैन लो मैं एक्स की वैल्यू निकलता हूं यहां से एक्स की वैल्यू क्या ए रही है सर यहां से एक्स की वैल्यू आती है 2y + 5 और ये जो एक्स की वैल्यू है ना तू y5 ये यहां पे रख दो रखते हैं suniyega ध्यान से आपकी सर्कल की इक्वेशन क्या है सुनना सर सर्कल की इक्वेशन जो है हमारे पास वो है x² + y² माइंस 8 प्लस 5 मैं यहां रखने जा रहा हूं है ना तो यहां पे रखा क्या 2y + 5 तो ये हो जाता है इसका क्या स्क्वायर फिर क्या लिखा हुआ है सर ये लिखा हुआ है y² यहां पे भी सॉरी ये है -8x तो यहां पे रखा -8 और ये रख दिया क्या 2y + 5 यहां पे क्या जा रहा है सर ये है 6y और ये है प्लस 20 आप देख रहे हो या तो एक्स में या ए में आपको एक क्वाड्रेटिक बनती है दिख रही है बिल्कुल दिख रही है सर थोड़ा ऐसे सिंपलीफाई करते हैं 25 + 5 का होल स्क्वायर तो ये हो जाता है 4y स्क्वायर 10 2 क्या आप मेरी बात समझ का रहे हो तो जाता है सर ये दिख रहा है -16 ए कितना दिख रहा है भी एरर मत कर देना स्टूडेंट्स कोई भी सीरियल है आप बस चीज ठीक से कैलकुलेट करते हुए चलना ताकि कहीं भी कोई गलती ना हो एक बार फिर से देख लो 4y² + 25 + 25 ए स्क्वायर -1 - 40 65 + 20 बिल्कुल सही सर अब क्या करना है अब अगर मैं इसमें सिंपलीफिकेशन का स्कोप देख रहा हूं तो देखो भाई क्या-क्या दिख रहा होगा सर 4y² + y² ये कितना हो जाएगा सर ये हो जाएगा पहले तो 5y स्क्वायर बहुत बढ़िया सर और क्या करना है और सर अगर मैं इसे ध्यान से देखो तो मुझे एक और बात निकल के आएगी क्या निकल कर आएगी suniyega कम की बातें ध्यान से देखना एक और बात जो निकल के आएगी वो ये की 20y है ना -1 कितना हो रहा है कितना ये हो जाता है आपका 10वाय कोई डाउट तो नहीं है आई थिंक नहीं है और क्या कर लेना चाहिए सर और एक और बात अगर आप ध्यान से देखो तो माइंस 40 प्लस 20 माइंस 20 25 - 20 सर 25 - 20 मेरे अकॉर्डिंग होना चाहिए 5 पता नहीं क्यों दिमाग कम नहीं कर रहा है 25 - 20 आप नहीं कर का रहे हो तो आई थिंक सर ये एक इक्वेशन बन रही है क्वाड्रेटिक सर इसे सिंपलीफाई कर सकते हैं क्या आप कोशिश करिए अगर मैंने इसे सिंपलीफाई करना चाहा तो क्या बनेगा देखो 5 पूरा कॉमन लिया तो ये बन जा रहा है y² + 2y + 1 क्या ऐसे में 1 स्क्वायर लिख डू आपको डिस्क्रिमिनेंट निकलने की जरूरत भी नहीं हुआ ने जीरो ही होगा क्योंकि इसके रूट्स कितने ए रहे हैं a² + b² + 2ab तो ये ए रहा है कनक्लूड करके y+1 का होल स्क्वायर और अगर इसका परफेक्ट स्क्वायर जो ये जीरो के इक्वल है मतलब क्या ए जो है सर वो माइंस वैन के वैल्यू और अगर सर आपका ए - 1 के इक्वल है तो यहां रखा -1 तू टाइम्स माइंस वैन माइंस तू फाइव माइंस तू कितना थ्री तो एक्स की वैल्यू कितनी ए रही है थ्री तो ये -1 अगर पास किया तो एक्स की वैल्यू आती है थ्री तो ये जो पॉइंट निकलकर आया है स्टूडेंट्स ये जो पॉइंट निकल कर आया है कौन सा ये जो पॉइंट निकलकर ए रहा है थ्री कमा माइंस वैन ये वो पॉइंट होगा जहां पर ये आपकी जो टैसेंट है वो इस सर्कल को टच करेगी जो की उसे सर्कल को कहां टच कर रही थी कहां टच कर रही थी भाई वैन कमा -2 ऊपर और ये कौन सा सर्कल है आपका जो गिवन था x² + y² - 8x + 65 + 20 आई थिंक आसान क्वेश्चन था बहुत बहुत डिफिकल्ट नहीं था की आपको ये पूरा आंसर समझ ए रहा है की थ्री कमा - 1 कैसे आया किसी भी स्टेप में कोई डाउट है तो पूछ लो स्टूडेंट्स आई थिंक चीज आसान है क्वेश्चन बड़ा मजेदार सा था बड़ा बेसिक सा सिंपल सा क्वेश्चन था दिस वाज डेलाइब्रेटली केप तू मेक यू अंडरस्टैंड डी कॉन्सेप्ट की टांगें की इक्वेशन से कैसे डील करना है और इन सारी बातों में जो मैं बार-बार जिस बात पर जोर दे रहा हूं जो आपको याद रखनी है वो क्या टी = 0 व्हाट इसे गोइंग तू हेल्प यू आउट हर इस टी = 0 यही आपका ट्रक है की आपको ग = 0 के हेल्प से तेनजिंग की इक्वेशन निकालनी है कब माइंड यू एवरीवन कब जब वो पॉइंट उसे सर्कल पर लाइक कर रहा हूं मतलब टेंशन और हाथों में बनती बनती है उसे बारे में बात करेंगे पर अगर वह पॉइंट तो सर्कल पर लाइक कर रहा है तो उसे पॉइंट से पास होने वाली टांगें की इक्वेशन टी = 0 से दी जाती है बस ये बात अच्छे से याद रखना स्टूडेंट्स उसे पे क्वेश्चंस बनते हैं और बहुत घुमा फिर के बहुत सारे अलग-अलग अजीब से तरीकों के बजाय की सर ये स्लोप निकालो या डिफरेंटशिएट करके सूप निकालो सबसे अच्छा तरीका है यही चीज इक्वल्स तू जीरो इसलिए मैं बार-बार सीखना चाह रहा हूं की ये हम करेंगे बार-बार याद रहेगा और ये अभी सिखा रहा हूं इसका मतलब की इसकी बहुत जरूरत पड़ने वाली और भी चैप्टर के दौरान इसलिए इसको अच्छे से समझ लो मैं बार-बार इस रेफरेंस को उसे करूंगा की देखो हमने सर्कल में पढ़ा था तो फिर बात वही ए जाएगी ना की हमने यहां पर अच्छे से पढ़ लिया तो आगे जाकर फटाफट कर लेंगे तो सीक्वेंस है एक ऑर्डर है और इसी में अंदर में आप चीज पढ़िए फिर क्या सर अगर हम यहां से आगे थोड़ा बड़े तो एक और क्वेश्चन ले लेते हैं और एक और कॉन्सेप्ट समझते हैं ये क्वेश्चन मैं आपको समझने की कोशिश करता हूं मैन लो सर आपके पास एक सर्कल है लेट्स नाज़ एंड डेट देयर इस अन सर्कल एक सर्कल हम मैन लेते हैं और क्वेश्चन ये है सर की इस सर्कल पर आप दो पैरेलल टैसेंट करते हो क्वेश्चन suniyega ध्यान से आप इस सर्कल पर दो पैरेलल tesence ड्रॉ करते हो और एक और थर्ड टेंशन ड्रॉ करते हो दो तो पैरेलल और एक थर्ड आप क्वेश्चंस समझ का रहे हो मैं एक हल्का सा आइडिया देने की कोशिश करता हूं इसमें एक छोटा सर्कल बना लेता हूं ताकि मैं आपको बात समझता हूं की एक सर्कल है आपके पास इस सर्कल के पास आपने दो टैसेंट ड्रॉ की एक किए और एक ये इन दोनों टांगों की खास बात ये थी की ये दोनों टैसेंट आपकी क्या थी पैरेलल सवाल ये है की मैन लो एक और इस सर्कल की क्या है tagent एक और इस सर्कल की है टैसेंट तो सवाल ये है की क्या आप मुझे बता सकते हो की इस थर्ड इंजन ने जो एंगल सब ये पॉइंट्स में इंटरसेक्ट जो किया है इस टांगें को यहां और इस इंजन को यहां तो इन दोनों पॉइंट से सर्कल के सेंटर पर कितना एंगल सब्सटेंड हो रहा है क्या आप वो निकल सकते हो सर बहुत वर्ड बहुत अजीब और बहुत ही बहुत ही अजीबोगरीब बात नहीं कर रहे आप कुछ तो भी का रहे हो ना सर्कल की इक्वेशन गिवन है ना सर्कल के किस पॉइंट पे आपने टैसेंट ड्रॉ की वो गिवन है ना थर्ड टेंशन की इक्वेशन गिवन है आपने बस मतलब उठा के ये पूछ लिया की ये थर्ड है जो इन दोनों paralleltanges को इंटरसेक्ट करती है और जिन दो पॉइंट्स में इंटरसेक्ट कर रही है उन दोनों पॉइंट से उसे सर्कल के सेंटर पे कितना एंगल सब्सटेंड हो रहा है निकल सकते हैं अब अगर क्वेश्चन पूछा गया है तो कर ही सकते होंगे तो क्वेश्चन आपके सामने है अगर आपको इसका रिटर्न फॉर्म चाहिए तो एक क्वेश्चन है जरा इस क्वेश्चन को जल्दी से ट्राई कर लो और मुझे बता दो इस क्वेश्चन का आंसर और अगर आपने बता दिया तो भाई मैन जाएंगे की आप कुछ चीज समझ ए रही है और मजा ए जाएगा की हान भाई चीज बहुत आसान है बहुत इजी हैं और आपको होने लगी हैं चीज क्लियर तो एक कम करते हैं स्टूडेंट्स क्वेश्चन को सिंपलीफाई करते हैं जनरल सा क्वेश्चन है तो मैं भी तो अपने लिए सबसे आसान तरीका चुगा ना सर मैं क्यों अपने जीवन की परेशानियों बधाई कम है क्या हमारे सामने आपकी जेनरेशन के साथ तो बहुत है साथ तो मैं कहूंगा की सर जीवन को थोड़ा आसान बना है तो अच्छा तो जीवन को आसान बनाने में इंटरेस्टेड हूं जो आपका बेसिक ली ओरिजिन percented तो हमने कौन सा सर्कल लिया सर हमने लिया x² + y² = a² किसी भी स्टूडेंट को भी तकलीफ अब सर जब आपको दो पैरेलल दो आपको जो पैरेलल वो टैसेंट माननी ही थी तो हर कुछ क्यों मानोगे आप कुछ अच्छी वाली टैसेंट मारना ना तो सर एक टैसेंट तो ये मैन लूंगा मैं और एक टैसेंट के मैन लूंगा की बात आप समझ का रहे हो जो मैं कहना चाह रहा हूं तो ये जो टैसेंट है हो सकता है शायद वो पैरेलल ना देखें क्योंकि मेरी ड्राइंग बहुत अच्छी नहीं है पर ये जो दोनों हैं ये दोनों मैंने पैरेलल टैसेंट बनाई और सर मैं सीधी सीधी सी बात जानता हूं अगर ये सर्कल जीरो कमा जीरो पर सिचुएशन है तो ये ए कमा जीरो हो गए -एक कमा जीरो होगा ये आपका जीरो कमा ए होगा और क्या होगा जीरो के पैरेलल है तो क्या आप मुझे नहीं सिखा सकते की इस लाइन के क्वेश्चन क्या होगी सर ये होगी ए = ए और इस लाइन की इक्वेशन होगी ए = -ए कोई डाउट कोई परेशान हान बिल्कुल तैसी बातें और हम बिल्कुल सिंपल सोचूंगा मैं भी तो अपने कन्वीनियंस के अकॉर्डिंग चीज देखूंगा मैं भी तो सबसे आसान तरीका ही दूंगा बिल्कुल ढूंढ लीजिए आप चलो सर आप की आपके संतुष्टि के लिए आपकी मर्जी के लिए एक कम आसान सा कर लेता हूं जो थर्ड दे तेनजेन मानता वो कोई रैंडम से मैन लेता हूं वो क्या रे रैंडम सिग्नल मैन लेता हूं मतलब लेट्स से वोट टैसेंट में ये मैन लेता हूं क्या मैं उसे टैसेंट कोई टेंशन मैन बिल्कुल मैन लीजिए तो टेक्निकल मैंने इस पॉइंट पर इस पॉइंट पर एक टांगें ड्रॉ की है पैरामेट्रिक फॉर्म के थ्रू क्या मैं इस पॉइंट के कोऑर्डिनेट्स सोच सकता हूं देखो x² + ए = a² बड़ा है याद करो स्टूडेंट सर पारंपरिक फॉर्म कहती है की इस पॉइंट के कोऑर्डिनेट्स आप मैन सकते हो ए कोस्थेटा कमा ए साइन थीटा क्या किसी भी स्टूडेंट को इस बात से कोई आपत्ति इस बात से कोई आपत्ति कोई परेशानी हान सर नहीं है मतलब एक क्या है डिस्टेंस और थीटा क्या है इस पॉइंट को ओरिजिन से ज्वाइन करने वाली लाइन का पॉजिटिव एक्स सेक्शन बनाया हुआ है एंगल याद है स्टूडेंट्स पैरामीटर फॉर्म तो एक रैंडम पॉइंट हमने इस कर्व पर यानी सर्कल पर मैन लिया और इस पॉइंट पे आप ड्रॉ कर रहे हो 10 मिनट 1 मिनट टेंशन टेंशन की इक्वेशन तो मैं बता सकता हूं सर देखो टेंशन की इक्वेशन क्या होगी ध्यान से याद करो स्टूडेंट सर सर्कल की इक्वेशन है x² + y² इस पर एक पॉइंट लाइक करता है कौन सा ए कोस थीटा कमा ए साइन थीटा आसान है कैसे क्या निकलेंगे भाई टी = 0 वो क्या कहता है सर वो कहता है आप एक्स स्क्वायर को करो एक्स एक्स वैन से रिप्लेस तो आप क्या लिखोगे भाई ए कोस्थेटा टाइम्स एक्स ए स्क्वायर को ए ए वैन से तो ये हो जाएगा क्या ए साइन थीटा साइंस ए = कोस आपका ये कांस्टेंट है जो की है a² सर एक छोटा सा कम आपको दिख नहीं रहा है पता नहीं क्यों ये ए और ये ए मिलकर इस स्क्वायर वाले एक ए को कैंसिल कर देंगे तो आपकी टैसेंट के जो इक्वेशन आएगी वो होगी कोस थीटा सिथेंटा टाइम्स ए है ना = टैसेंट यह आपकी इस टांगें की इक्वेशन है इस टांगें की इक्वेशन क्या ए रही है सर यह ए रही है क्या ए रही है यह ए रही है मैं ऐसा लिख डन क्या एक्स कोस्थेटा + ए सिन थीटा = ए क्या आप मेरी बस इतनी सी मदद करेंगे स्टूडेंट्स की आप मुझे ये पॉइंट और ये पॉइंट निकल के दे देंगे जो की क्वेश्चन के अकॉर्डिंग क्या है जो की क्वेश्चन के अकॉर्डिंग है आर और के तो मैं मैन लेता हूं ये पॉइंट है आर और ये पॉइंट है के सर ये कुछ नहीं है आर और के क्या है सर इस स्ट्रेट लाइन का इस स्ट्रेट लाइन के साथ पॉइंट ऑफ इंटरसेक्शन अरे दो लाइंस को सॉल्व करते आपको आता है ना तो ए = ए ए = ए इसमें रख दो तो सर अगर आर निकलना चाहा अगर आर निकलना चाहा तो ए = ए रखते हैं तो ए = ए अगर पास किया तो देखो ए की जगह क्या रख दिया ए तो क्या ए जाएगा सर ध्यान से देखना एक्स टाइम्स कोस थीटा ए की जगह ए रख दिया तो इसको वहां शिफ्ट किया तो कितना ए जाएगा सर ये ए जाएगा ए - ए सिन थीटा तो ए कॉमन ले लिया तो कितना हो जाएगा सर ये ए जाएगा 1 - सिन थीटा बहुत बढ़िया तो ऐसे एक्स की वैल्यू ए रही है कैसे देखो ना आप कोस थीटा डिनॉमिनेटर में पहुंचा दो तो कोस थीटा अगर आपने डिनॉमिनेटर में पहुंचा तो ये हो जाएगा ए टाइम्स 1 - साइन थीटा अपॉन कोस थीटा और ये किसकी वैल्यू है एक्स की मतलब आपका जो वो आर पॉइंट है उसके एक्स और ए koardinate क्या हो जाएंगे सर वो हो जाएंगे ए टैंक्स वैन माइंस साइन थीटा अपॉन कोस थीटा और ए मतलब ये एक्स स्क्वाड इन अवर कार्ड तो क्या मैं एक्स और ए का ऑर्डर है लिख का रहा हूं उसे लाइन के हान सर मैं लिख का रहा हूं ये एक्स कोऑर्डिनेट्स है और ए कार्ड कितना है ए ये किस पॉइंट के कोऑर्डिनेट्स की बात हो रही है सर ये आपका बेसिकली है पॉइंट आर कोई तकलीफ किसी भी स्टूडेंट को किसी भी स्टूडेंट को कोई परेशानी मेरे ख्याल से तो नहीं होनी चाहिए है ना मतलब ये क्लीयरली आपका excordinate है तो जैसे मैंने यहां पर ऐसे लिख दिया और क्लियर है तो इसे मैंने यहां पर ऐसे लिख दिया है आई होप ये बात अच्छे से आप समझ का रहे हो ठीक है सर मैन लिया आपकी बात यह बात समझ आई क्या ए गई सर अब सुना और अगर हमने निकल लिया तो क्या आप क्यों नहीं निकल सकते क्यों नहीं निकल सकते क्यों क्यों नहीं निकल सकते क्यों क्या है वही इसे इक्वल तू माइंस ए और इस स्ट्रेट लाइन का क्वेश्चन रख सकते हैं सर तो ए की जगह माइंस ए रहा देखो कुछ नहीं बदलेगा ध्यान से सुनना ए की जगह माइंस ए रखा तो इधर चला गया तो ये हो जाएगा ए + ए साइन थीटा तो बस इतना सा बदलाव है इतना सा बदलाव है क्या की सर यहां पर माइंस की जगह प्लस हो जाएगा वह होगा माइंस ए ए जो कोऑर्डिनेट्स होगा वो होगा माइंस ए तो बस आपके जो के के कोऑर्डिनेट्स आएंगे आई थिंक यू ही है तो जो के के कोऑर्डिनेट्स जो आप लिखोगे वो क्या लिखोगे स्टूडेंट्स आर के रहे कोऑर्डिनेट्स और अगर मैंने के के कोऑर्डिनेट्स लिखे तो वो क्या लिखूंगा सर के पॉइंट के जो कोऑर्डिनेट्स होंगे वो होंगे आई थिंक ए टाइम्स 1 + सिन थीटा बिल्कुल सही ए टाइम 1 + सिन थीटा / में ऑफ कोर्स किया कोस्थेटा कमा - ए माइंड दिस इस गोइंग तू बी माइंस ए ये सब क्यों निकल रहे हो सर सिर्फ फायदा क्या हो रहा है एक सेकंड थोड़ा पेशेंस रखिए अब आप एक बात का जवाब दो अब आप एक बात का जवाब दो सर अगर अगर मुझसे ये पूछा जा रहा था की सर्कल के सेंटर पर ये आर मतलब आर मतलब ये लाइन को सेंटर से कनेक्ट किया और ये के मतलब के को आपने सेंटर से कनेक्ट किया तो टेक्निकल वो ये एंगल जानने में इंटरेस्ट है सर इस एंगल को निकलने का शायद मैंने स्ट्रेट लाइंस के चैप्टर में पढ़ा है कैसे पड़ा है भाई अगर मैं आर ओ लाइन की स्लोप निकल लूं और अगर मैं के ऑनलाइन की स्लोप निकलूं तो दो लाइंस की स्लोप्स के हेल्प से उनके बीच का एंगल निकलती आता है या नहीं हान सर बात तो सही का रहे हो निकल सकते हैं तो निकले बिल्कुल निकल दें सुनेगा यहां से आर ओ लाइन के स्लोप कितनी आशाएं निकलना सर जो आर के कोऑर्डिनेट्स हैं ना मैन लो जैसे की मैन लो X1 y1 तो y1 - 0 अपॉन X1 - 0 तो y1 अपॉन X1 तो आर ओ लाइन के स्लोप क्या हो जाएगी सर आर ओ लाइन की जो स्लोप होगी वो बड़ी आसानी से होगी यहां से अगर मैं पूछूं आपसे आर ओ लाइन की स्लोप है ना तो स्लोप ऑफ डी लाइन आर ओ इस गोइंग तू बी इसका ए कोऑर्डिनेट्स अपॉन इसका एक्स कोऑर्डिनेट्स आप समझ का रहे हो तो क्या हो जाएगा ए / ए टाइम से वैन माइंस साइन थीटा जापान में कोस्थेटा कोई दिक्कत तो नहीं स्टूडेंट कोई परेशानी आई होप आपको बात समझ ए रही है हम क्या करेंगे अगर मैं इसे मैन लो ए / 1 तो कोस थीटा ए के साथ चला जाएगा और वैन के साथ ये चला जाएगा और ए से ए कैंसिल हो जाएगा तो फाइनली ये जो सिंपलीफाइड हो के आपको दिखेगा फाइनली ये जो सिंपलीफाइड हो के दिखेगा क्या दिखेगा सर ये दिखेगा कोस्थेटा / कोस्थेटा / 1 - सिन थीटा क्या ये स्टेटमेंट आप सभी को समझ ए रहा है ये आपकी स्लोप आई है आर ओ की सर इसी तरीके से क्या मैं के ओके स्लोप निकल सकता हूं आई थिंक यू की भी स्लोप क्या होगी वही अगर इसके कोऑर्डिनेट्स है X2 Y2 तो हो जाएगा Y2 - 0 / X2 - 0 नहीं 2 / X2 यानी इसके जो ए कोऑर्डिनेटर एक्स कॉर्डिनेट है उनका डिवीज़न है ना तो अगर मैं स्लोप निकलूं किसकी अगर मैं स्लोप निकलना चाहता हूं के ओके तो उसे लोग क्या होगी सर माइंस ए अपॉन में ये एक्सप्रेशन जो की क्या है ए टाइम्स वैन प्लस साइन थीटा कोस थीटा कोई दिक्कत तो नहीं है ए से ए कैंसिल कोस थीटा neumerator में मतलब जो ये स्लोप आएगी ये क्या होगी सर ये होगी माइंस कोस थीटा माइंस कोस थीटा / वनप्लस साइन थीटा गौर से देख लो यह माइंस कोस थीटा / 1 + सिन थीटा और यहां पर जो दूसरी स्लोप आपको दिख रही है वो है कोस्थेटा / 1 - सिन थीटा आपको अगर कुछ समझ नहीं ए रहा है तो ध्यान से सोच के देखो ध्यान से सोच के देखो फिर से ये है माइंस कोस थीटा / 1 + सिन थीटा हम यहां पे लिख देता हूं ताकि आपको देखें जो ये स्लोप हमने निकल है किसकी के ओके वो क्या है सर वो है माइंस कोस थीटा / 1 + सिन θ दो बातें दो बातें क्या सर एक तरीका तो ये की अगर मुझे दो लाइंस की स्लोप्स पता है अगर मुझे दो लाइंस के स्लोप्स बताए तो उनके बीच में एंगल निकल सकता हूं क्या बिल्कुल निकल सकते हैं दो लाइंस की अगर स्लो होता है अच्छा बाय डी वे यही थीटा और ये थीटा अलग है ना मैंने कुछ गलत तो नहीं दिया ये थीटा वो थीटा नहीं है ये थीटा तो इन दोनों लाइंस के बीच में एंगल है जैसे मैं कुछ नहीं कहता हूं अभी फिलहाल के वो थीटा क्या था वो थीटा बेसिकली आपका ये था इसको इससे ज्वाइन करने वाला एंगल है ना तो तो एक अलग बात है जैसे भूल जाओ पर बात जो अब निकल के आई है उसे सुनना ध्यान से सर अगर इन दोनों लाइंस के बीच का एंगल निकलना है तो मैं जो जानता हूं वो ये की दो लाइंस की स्लोप मुझे पता है दो लाइंस की स्लोप मुझे पता है तो उन दोनों इसके बीच का एंगल अगर मैं कहूं दोनों लांस के बीच कैंडल तो हम पढ़ते हैं सर क्या होता है m1 - M2 / 1 + m1 M2 क्या ऐसा कुछ हमने पढ़ा है असर पड़ा तो ये पुरी मेथड उसे करते हैं पर मेरा नॉलेज यह का रहा है बस मुझे एक चीज दिख रही है की सर चांसेस हैं की शायद ये लाइंस परपेंडिकुलर हो ऐसा बस मुझे पता है मुझे लग रहा है और ऐसा अगर मुझे फुल हो रहा है क्योंकि मुझे चेक करना है बस तू तू नॉट यूज्ड दिस होल प्रक्रिया कैन आई डू वैन थिंग सर अगर दो लाइंस परपेंडिकुलर होते हैं इफ है ना इफ तू लाइंस परपेंडिकुलर लाइन L1 जो है वो L2 पर परपेंडिकुलर है तो हम जानते हैं सर m1 और M2 का प्रोडक्ट क्या होता है -1 या उल्टा डायरेक्शन तो परपेंडिकुलर होते हैं आय होप ये बातें आपको पता है तो इस पुरी मेथड यह उसे करने की बजाय क्या मैं ये मेथड ट्री कर लूं क्योंकि सर से तो नहीं है से होती है तो मैं कहता हूं पैरेलल जो तो चांस ही नहीं बन रहा है पैरेलल होने का तो मैं का रहा हूं की सर इनकी स्लोप्स का प्रोडक्ट निकलती हैं इनकी स्लोप्स का प्रोडक्ट निकलती हैं मतलब मैं क्या निकलने साहब ने स्टूडेंट्स हम निकल लें जा रहे हैं ये बात की एमआरओ मतलब आर ओ लाइन के स्लोप और को के लाइन के स्लोप्स का प्रोडक्ट क्या है तो ये कितना लिखा है कोस थीटा / 1 - सिन थीटा यहां लिख लेता हूं मैं कोस थीटा अपॉन में वैन माइंस साइन थीटा है ना और सर यहां पर क्या लिखा है ये लिखा है माइंस कोस थीटा / 1 + सिन थीटा तो ये लिखा है दिखाना चाहिए आपको है ना ऐसा दिखाना चाहिए क्योंकि मुझे दिखा कुछ अगर मैंने न्यूमैरेटर को ठीक से ऑब्जर्व किया होता तो मुझे दिखेगी सर शायद बात बन जाएगी क्यों बन जा रही है सर आपको दिख रहा है क्या आपकी बात बन गई बात कैसे बन गई सर न्यूटन में कोस थीटा कोस थीटा तो कितना हो जाएगा सर ये हो जा रहा है माइंस कोस स्क्वायर थीटा और डिनॉमिनेटर में क्या हो रहा है सर डिनॉमिनेटर में हो रहा है ए - बी ए + बी तो a² - b² 1 - sin² थीटा 1 - sin² थीटा कितना होता है सर वो होता है कोस स्क्वायर थीटा तो ये कितना हो जाएगा सही हो जाएगा कोस स्क्वायर थीटा और क्या आपको मैं कुछ बताऊं आप मुझे बता रहे हो सर कोस स्क्वायर थीटा से कोस स्क्वायर θ कैंसिल बचा कितना -1 कहानी खत्म हो गई क्या कहानी खत्म हो गई क्या सर कहानी ऐसी खत्म हो गई की इन दोनों लाइंस की स्लोप्स का प्रोडक्ट ए रहा है माइंस वैन और अगर दो लाइंस की स्लोप्स का प्रोडक्ट -1 है तो मुझे ज्यादा सोचने की जरूरत नहीं है सर वो दोनों लाइंस कैसी होती हैं वो दोनों लाइंस परपेंडिकुलर होती हैं यानी उनके बीच एंगल होगा 90 डिग्री क्या यह सारी बातें आपको समझ आई यहां तक तो कोई डाउट या परेशानी नहीं है इससे अच्छे से चेक कर लो स्टूडेंट्स किसी भी स्टेप में कोई डाउट है तो अच्छे से देख लो ये क्वेश्चन क्लियर हुआ है स्टूडेंट्स एक-एक स्टेप एक-एक एप्रोच एक-एक थॉट एक-एक आइडिया आई थिंक यहां तक कोई परेशानी नहीं है मैं अलजेब्राइक बहुत ज्यादा डिटेल में इस चीज को डिस्कस नहीं करना चाहूंगा क्योंकि वो शायद आपको कंफ्यूज कर दें पर बस एक बात का जवाब दो की अगर मैं आपसे कहूं की ये एक सर्कल है एक सर्कल है इस सर्कल पर मैं आपसे कहता हूं की एक स्ट्रेट लाइन है जिसकी स्लोप है एम जिसकी स्लोप है जो की का दिया जाएगा क्या क्योंकि का दिया जाएगा x² + y² + 2G एक्स + 2xy + सी = 0 मतलब अगर आपसे कोई कहे की सर इस सर्कल पर एक टांगें बनाया जिसकी स्लोप है एन तो आप कैसे बनाओगे लाइन होगी तो मैं का सकता हूं की सर वो शायद ऐसे ही जा रही हो पर क्या जरूरी है सिर्फ ऐसे ही जा रही हो क्या कोई और तरीके से नहीं जा सकती सर स्लोप है ना तो यहां के बजाय इधर कहीं से भी तो जा सकती है सर यहां से भी तो जा सकती है अगर मैं इसे थोड़ा सा इसके पैरेलल बनाओ तो आप मेरी बात ध्यान से समझने की सुनने की और अच्छे से उसे यू नो इंपोर्ट करने की कोशिश करना बात समझो स्टूडेंट्स मेरा आपसे बस ये कहना है की सर जरा ध्यान से देखो चाहे यह लाइन हो या चाहे ये लाइन हो यह दोनों लाइंस क्या है पैरेलल ये दोनों ही लाइंस क्या है पैरेलल और इन दोनों की स्लोप क्या है इन दोनों की स्लोप है एन पर आप बात समझो अगर मुझसे कभी भी कोई कहे की सर्कल पर एक टांगें ड्रॉ करो जिसकी स्लोप है एम तो वो शायद यहां से जा सकती है या यहां से जा सकती है दोनों जगह से जा सकती है तो बेहतर होगा कहना की अगर कभी मुझे कोई स्लोप दे दी जाए तो उसे केस में सर्कल से दो टांगेंट्स ड्रॉ की जा सकती है उसे गिवन स्लो की एक सर्कल के इस तरफ और एक उसे सर्कल के इस तरफ और उसे तरफ और इस तरफ का मतलब है की ये जो स्लोप्स के जो टैसेंट जो अपने ड्रा की इनके जो एक्सट्रीम होंगे ना अगर आप इनसे मिले एक्सट्रीम को मिले इस पॉइंट को इस पॉइंट से तो ये कुछ नहीं होगा ये आपका डायमीटर होगा सीधे पॉइंट पर आते हैं स्टूडेंट्स की अगर मुझे इस टांगें की चाहे इसकी या इसकी इक्वेशन चाहिए तो वो मैं कैसे लूंगा सर सबसे पहली बात तो ये की हम जानते हैं सर इस इंजन की इक्वेशन क्या होगी मेरे अकॉर्डिंग जो इक्वेशन होगी वो होगी वही इसे इक्वल तू एमएक्स + सी + सी वहां ले लिया है तो मैं मैन लेता हूं एमएक्स + अल्फा अब वो फर्स्ट इसकी भी कुछ ना कुछ होगी बट फिलहाल मैं मैन लेता हूं की एक ही लाइन इसकी टेंशन है देख लेंगे जो होना होगा हो जाएगा अगर ये लाइन इस इस सर्कल की tenjent है तो मैं सर दो बातें जानता हूं की सर्कल का जो यू नो सेंटर होगा इस सर्कल का जो सेंटर होगा यहां कहीं होगा तो सेंटर के कोऑर्डिनेट्स में जानता हूं सर क्या होंगे माइंस जी कमा माइंस है और अगर उसे सेंटर से मैंने इस टैसेंट पर परपेंडिकुलर ड्रॉप किया तो किसके इक्वल होगा इसकी रेडियस के और रेडियस कितनी होती है सर रेडियस हम जानते हैं कितनी होती है स्टूडेंट्स अंडर रूट ओवर प्लस एक्स स्क्वायर जब परपेंडिकुलर ड्रॉप करूंगा तो क्या मैं किसी पॉइंट से किसी स्ट्रेट लाइन पर ड्रॉप किए गए परपेंडिकुलर के लिए निकल सकता हूं आई होप आप ये भूले नहीं हो हमारी स्ट्रेट लाइन की इक्वेशन क्या है हमारी स्ट्रेट लाइन के क्वेश्चन है ए = एमएक्स + अल्फा है ना तो मैं क्या लिख सकता हूं एमएक्स - ए + अल्फा ये आपकी स्ट्रेट लाइन है इसकी आप निकल रहे हो क्या परपेंडिकुलर डिस्टेंस किस से हम निकल रहे हैं किस माइंस जी कमा - एफ से मैं आपसे का रहा हूं ये जो परपेंडिकुलर डिस्टेंस आप निकल रहे हो सर ये इसकी रेडियस की इक्वल होगी ना इसकी रेडियस केक को लोग तो एक तरफ तो आप रख दोगे रेडियस जो की क्या होगी अंडर रूट ओवर क्या G2 + x² - सी अब अभी अलजेब्राइक बहुत अजीब लग रहा है ये पर रियलिटी में बेसिक नंबर होगा एक रेडियस आप समझ रहे हो और ये किसके इक्वल होगी सर ये परपेंडिकुलर डिस्टेंस के इक्वल होगी किसकी परपेंडिकुलर डिस्टेंस सर इस सर्कल के सेंटर से इस स्ट्रेट लाइन की और जो की मुझे आता है आई होप आपको याद ए रहा है ax1 यानी -जी एम तो ये कितना हो जाएगा - एमजी देखो आपको बातें समझ ए रही है फिर क्या सर फिर आप लिखोगे क्या ए मतलब बी ए वैन सो - एफ -1 से कितना हो जाएगा + एफ और ऑफ कोर्स फिर क्या आपका प्लस यानी अल्फा और अब डिनॉमिनेटर में क्या लिखते हो सर आपके एक्स और ए के kofician से स्क्वेयर्स के सबका अंडर रूट तो हो जाएगा m² + 1 सर ये तो बहुत अजीब लग रहा है ये अभी अजीब लग रहा है क्योंकि आपके पास ये एफसी बहुत सारे वैरियेबल्स हो गए मैं यही कहना चाह रहा हूं की अलजेब्राइक से मत करो अब देखो आपको सब पता होगा आपको क्या पता होगा जीएफ और सी पता होगा सर्कल की इक्वेशन में दी होंगी मतलब रेडियस है बेसिकली तो पता चली जाएगी आपको एम दिया होगा क्योंकि गिवन स्लोप है ना वो कोई स्लोप देगा आपको आपको जी और एफ पता है तो आपको सब पता है और इस इक्वेशन को सॉल्व करके आप करने वालों के अल्फा लेकिन जब ये मोड आप hataoge मतलब इसको अनमोड जब करोगे ऐसा कोई वर्ड पता नहीं होता है या नहीं पर जैसे आप उन्माद करेंगे तो ये यहां पे प्लस माइंस साइन दे देगा उससे क्या होगा सर उससे मुझे अल्फा की दो वैल्यूज मिलेंगे और वो जो अल्फा की दो वैल्यू मिलेगी जब मैं रखूंगा तो मुझे एक इसकी और एक इसकी इक्वेशन मिलेगी मैं बस आपको मेकैनिज्म समझाना चाह रहा हूं की जब आप ऐसे करोगे मतलब जब आपसे सर्कल की कोई गिवन स्लोप देकर एक टेंशन की इक्वेशन पूछेगा की सर्कल स्लोप की दो टैसेंट बन सकती हैं सर एक ऊपर एक नीचे एक लेफ्ट में राइट में ऐसा कुछ ना कुछ बन रहा होगा और उन इक्वेशंस को ढूंढने का डी बेस्ट वे डी बेस्ट वे अभी फिलहाल अत दिस स्टेज कूद बी दिस की सर क्या करूं सर्कल के सेंटर से परपेंडिकुलर ड्रॉप करो वो परपेंडिकुलर लेंथ निकल लो उसे स्ट्रेट लाइन की सर्कल के सेंटर से और वही परपेंडिकुलर लेंथ और सर्कल की रेडियस के इक्वल रख दो बस बात खत्म जो अल्फा आपको यू नो वही एक्सेस का इंटरसेक्ट निकलना है ए = एमएक्स + अल्फा वाली फॉर्म में क्योंकि एम पता है एक्स और ए तो वेरिएबल है अल्फा यहां से ए जाएगा और आपकी क्वेश्चंस ए जाएंगे दो इक्वेशंस आई होप आपको ये प्रक्रिया क्लियर है अब इसे अलजेब्राइकली मत समझिए ए अलजेब्राइकली आपको कंफ्यूज कर देगा देखो अलजेब्रा के लिए मैं शो जरूर करता हूं बट इसे मत समझना आपको पता है स्टूडेंट्स क्या हो रहा होगा सर ये आपका सर्कल है x² + y² + 2X + 2 इस सर्कल के सेंटर के क्वाड्रेंट होंगे - जी कमा माइंस है अब हम जानते हैं सर अगर इससे अगर मेरे पास एक स्ट्रेट लाइन है ए = एक्सप्रेस अल्फा जो की टांगें है तो मैं जानता हूं सर क्या की इस एमएक्स ए = एक्स + अल्फा लाइन की इससे जो परपेंडिकुलर डिस्टेंस होगी वो किसके इक्वल होगी वो इसकी रेडियस के इक्वल होगी फाइनली जब आप सिंपलीफाई करोगे तो आपकी अल्फा की ये वैल्यू आएगी क्या एमजी अब ये बहुत अजीब सा एक्सप्रेशन है अलजेब्रा के लिए कोई न्यूमेरिकल एक बहुत आसान सा क्वेश्चन होगा कैसे समझो मैं आपको न्यूमेरिकल समझने की कोशिश करता हूं बजाय इस तरीके से आपको अन्य से रेली कंफ्यूज करने के जैसे मैन लो आपको एक सर्कल दे दिया जाए जैसे दे दिया जाए x² + y² लेट्स से +4x है ना माइंस तू ए वाले से -3 = 0 क्या बात आप समझ का रहे हो और वो क्या कहे वो कहे की सर एक टैसेंट बनाओ एक टैसेंट बनाओ जिसकी स्लोप है जिसकी स्लोप है लेट्स से लेट्स से थ्री एक आप tenjent की इक्वेशन बताओ जिसकी स्लोप है थ्री क्या इक्वेशन लिख सकते हैं बहुत आसान क्वेश्चन है पहले तो बताओ अगर मुझे बोला है की एक ऐसी टैसेंट के क्वेश्चन बताओ जिसके स्लोप है थ्री तो मैं क्या कहूंगा मैं कहूंगा सर वो tagent होगी मेरे लिए वाइस रिक्वेस्ट तू एमएक्स + सी लेकिन सी अल्फा जो भी आपको ठीक लगे क्या बात आपको समझ ए रही है कोई कन्फ्यूजन तो नहीं तो ये हमारे लिए उसे टैसेंट की इक्वेशन है अब हम बस से ढूंढ ले बात खत्म हो जाएगी अब बात बस इतनी सी है की अगर इस इक्वेशन को मैं थोड़ा और सिंपलीफाई करूं मैं इसे लिख सकता हूं ना 3X - ए + सी = 0 सो टेक्निकल ये टैसेंट है इस सर्कल की सर्कल और उसकी टैसेंट हमको दी है तो मेरा कहना तो ये है सर इस सर्कल के सेंटर के cardinate क्या होंगे अरे सर्कल के सेंटर के cardinate होंगे -2 अच्छा सर सर्कल की रेडियस निकल सकते हो क्या बिल्कुल तू का स्क्वायर 4 + 1 5 - -3 यानी 5 + 3√2 तो इसकी रेडियस कितनी ए गई 2√2 अब बात समझ का रहे हो अब क्या करना है सर इस सर्कल के सेंटर से इस स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस इस रेडी इक्वल होगी बस इतनी सी बात है बस सच में इतनी सी बात है मतलब क्या करना है सर suniyega ध्यान से इस सर्कल के सेंटर से इस लाइन की परपेंडिकुलर डिस्टेंस कितनी होगी सोच के देखो - 2 3 ये कितना -6 मुझे नहीं पता ये क्या है पर डिस्टेंस है तो मोड लगाना पड़ेगा थ्री का स्क्वायर 9 और 1² 9 + 1 कितना 10 तो यहां क्या हो जाएगा डिनॉमिनेटर में √10 = व्हाट इसे इक्वल तू डी रेडियस ऑफ डी सर्कल विच इस 2√2 कोई डाउट अरे भाई सर्कल के सेंटर से उसकी टांगें पर जो परपेंडिकुलर ड्रॉप किया उसकी लेंथ क्या होगी रेडियस के इक्वल कोई डाउट तो नहीं है आई थिंक आप अब बातें समझ का रहे हो और जब ऐसा किया आपने तो देखो उसे दौरान क्या मिल जाएगा सर ऐसा करने के दौरान हमें अगर सिंपलीफिकेशन फाइनली मिल रहा है तो मैं क्या बोलूंगा बिकॉज -6 - 1 कितना -7 तो ये हो जाता है -7 + सी का मोड लेकिन इस तरफ अगर देखे तो अंडर रूट 10 2 √10 √2 डेट √20 √20 यानी 5 4 5 4 यानी 4 का अंडर रूट कितना तू और अंडर रूट में फाइव तो 2 4 तो कितना हो जाएगा 4√5 अब सीधी सी बात जो मुझे निकल कर ए रही है सर वो ये की अगर इसका मोड में हटाओ तो मुझे क्या मिलेगा मुझे मिलेगा -7 + सी मुझे लेफ्ट हैंड साइड पर तो बिल्कुल माइंस सेवन प्लस सी मिलेगा लेकिन राइट हैंड साइड पर जो है 4√5 इसके आगे क्या ले लेंगे प्लस माइंस फोर रूट फाइव बस यही बात मैं आपको कहने की कोशिश कर रहा हूं की ये जो वैल्यू आई है इस तरीके से ए रही है प्लस माइंस अब आप क्या करोगे अब आप क्या करोगे बेसिक कम ही करोगे की सर देखो -7 को उधर पहुंचा है तो सी क्या आता है सर सी आता है सेवन प्लस माइंस फोर रूट फाइव तो एक बार सी की पॉजिटिव वैल्यू ले लो एक बार नेगेटिव से आपकी स्ट्रेट लाइन की इक्वेशन ए जाएगी जो की क्या है ए = एमएक्स यानी क्या थ्री एक्स प्लस सी प्लस सी यानी प्लस 7 प्लस 4√5 और एक दूसरी इक्वेशन क्या होगी ए सी इक्वल्स तू एमएक्स अब ये वही प्लस सी लेकिन सी इस बार क्या निगेटिव साइन के साथ तो ये कितना हो जाएगा प्लस 7 - 4√5 ये आपकी दोनों जो बन गई है स्लोप है और ये क्लीयरली पैरेलल लाइंस है पर इनमें कुछ डिस्टेंस है क्या ये बात आपको समझ आई होप यू बात आप समझ में बस इतनी सी बात है यह जो बाहर निकल कर ए रही है बट यहां पे बहुत सारे वैरियेबल्स लिखे हैं तो आप उन नेसेसरी कंफ्यूज हो जाएंगे मैं का रहा हूं इतना मत करो इतना इतना कंफ्यूज मत हो पर आई होप आपको ये मेथड ये चार लाइन की स्टेप समझ आई बस इतना सा करना है टेक्निकल इतना ही करना है पर अलजेब्रा के लिए वो बहुत कॉम्प्लिकेटेड सी प्रक्रिया है इतना मत पढ़ो इतना मत पढ़ो मेरा बस आपसे ये कहना है की आपको क्या मिलेगी आपको दो टांगेंट्स मिलेंगे और वो दोनों कैसे होंगे सर पैरेलल होंगी और उनके जो एंड पॉइंट्स होंगे वो सर एक से डायमीटर के ही एंड पॉइंट्स होंगे आई होप यहां तक कोई डाउट नहीं अच्छा एक और तरीका हो सकता है की सर अल्फा निकलने का कोई और तरीका हो सकता है क्या आप सोचो ना आपको एक तरीका हमने ज्यामिति के लिए ये भी तो समझे सर सीधी सीधी सी बात है सर सीधी सी बात है अगर ये एक सर्कल है और ये स्ट्रेट लाइन इसकी क्या है तो एक सर्कल है तो इस सर्कल और इस स्ट्रेट लाइन को सॉल्व करूं जैसे मैं मैन लेता हूं ए की जगह क्या रख देता हूं 3x+3 यहां और यहां तो ये ए में क्वाड्रेटिक बन जाएगी अब जब इन दोनों को सॉल्व करूंगा तो क्वाड्रेटिक के कितने रूट्स आने चाहिए सर इस क्वाड्रेटिक का एक ही रियल रूट आना चाहिए क्योंकि ये इसे एक ही पॉइंट पे टच या इंटरसेक्ट कर रहा है तो बस में क्या करूंगा मैं सिंपल सा कम ये करूंगा देखो एक और प्रक्रिया में बता ही देता हूं हान तो ये आपका क्वाड्रेंट ये आपका सर्कल का इक्वेशन है क्या सर्कल का एक क्वेश्चन है x² प्लस ए स्क्वायर रख दीजिए तो क्या हो जाएगा सर ये ए जाएगा x² हो जाएगा 3X + सी का होल स्क्वायर प्लस क्या है सर 4X यहां पे फिर माइंस तू ए की वैल्यू क्या है सर वो है 3X + सी समझ में ए रहा है क्या और ये सब -3=0 तो आप इस स्ट्रेट लाइन और इस सर्कल को सॉल्व कर रहे हो और जैसे ही आप इसे सॉल्व करते हो तो आपको क्या मिलेगा गौर से देखिए स्टूडेंट्स जैसे ही आप इसे सॉल्व करते हो तो आपको मिलेगी दो चीज ये पहले तो बनेगा पूरा एक क्वाड्रेटिक बना दो देखो ये हो जाएगा x² ये हो जाएगा ये दिखेगा 3 2 कितना 6x + सी मतलब तू इन थ्री तो - 6 - 3 = 0 अब आप जरा कुछ चीजों को थोड़ा आइसोलेट करके रख लीजिए आइसोलेट मतलब पहले x² वाली टर्म से एक साथ रख लो तो नाइन प्लस वैन कितना सही हो जाता है 10x² ठीक है सर अब एक्स वाली टर्म से एक साथ देख लो तो - 6x - 4X - 2X है ना और ये है प्लस 6cx तो क्या लिख रहा हूं मैं -2x तो 2c लिख ले रहा हूं आई होप आपको बात समझ ए रही है बहुत अच्छे से हम रिलीज सॉरी - 2X तो ये यहां से मैं -2 ले रहा हूं और यहां से क्या ले ले रहा हूं 6c कोई दिक्कत तो नहीं स्टूडेंट्स अब आपको बात समझ ए रही है अगर मैं इसे थोड़ा और ठीक से लिखूं तो सुनेगा ध्यान से क्या मैं ऐसे ऐसे लिख सकता हूं यहां से देखना 6c - 2 टाइम्स एक्स देखो सेक्सी - 2 6 बहुत बढ़िया और जो कांस्टेंट है उसमें से अगर मैंने साइन कॉमन लिख के रख के लिखना चाहो तो ये हो जाता है 2c + 3 = 0 अब आपको बस इतना सा सोचना है की सर बड़ी बेसिक्स आदर एंड से सिंपल सी बात ये नज़र ए रही है की ये क्वाड्रेटिक ठीक है एक्स अब यह क्वाड्रेटिक एक्शन मतलब मुझे तो सॉल्व करके देगा वो इस एक्स koardinate की वैल्यू देगा और टेक्निकल सर्कल और ये टैसेंट एक ही पॉइंट पे इंटरसेक्ट कर रहे हैं क्योंकि टैसेंट है तो इस क्वाड्रेटिक के रूट्स कितने होंगे सर दो तो हो नहीं सकते क्योंकि 2.3% नहीं है जीरो भी नहीं हो सकते क्योंकि एक पॉइंट पे इंटरसेक्ट कर ही रही है तो कितने होंगे रियल अन्य इक्वल रूट्स मतलब एक ही रूट और अगर इस क्वाड्रेटिक का सर मेरे नॉलेज के अकॉर्डिंग एक ही रूट है तो हम जानते हैं सर क्वाड्रेटिक का इक्वल एंड रियल रूट तब होता है जब उसका डिस्क्रिमिनेंट जीरो हो डिस्क्रिमिनेंट मतलब b² यानी मैं क्या करूंगा मैं लिखूंगा 6c - 2 का होल स्क्वायर माइंस 4ac -4 सो -4 ए कितना 10 और सी कितना ये पूरा टर्म जो की क्या है माइंस ऑफ तू सी माइंस थ्री और ये क्या होना चाहिए जीरो तो डिस्क्रिमिनेट जब आप जीरो रख दोगे तो सी में क्वाड्रेटिक बन जाएगी इस सी को सॉल्व कर लेना तो सी की कितनी वैल्यू हो जाएगी दो दो वैल्यू क्यों आएंगे सर क्योंकि सी की दो वैल्यूज नहीं आएंगे क्योंकि आपको दो पैरेलल लाइंस से डील करना है क्योंकि वो टैसेंट सिर्फ एक ही tagent नहीं है एक्चुअली दो टांगेंट्स है आपको याद अगर ए रहा है तो एक ऐसी कुछ और एक ऐसी कुछ याद ए रहा है भाई भूले नहीं हुए बातें क्या यह एप्रोच क्लियर हो रही है आई थिंक दोनों मैथर्ड क्लियर है और इसको अगर मैं अलजेब्रा के लिए समझने जाऊंगा तो बहुत सारे वैरियेबल्स पिक्चर में लाने पड़ेंगे और बहुत लंबी चौड़ी कहानी आपको समझनी पड़ेगी तो वैरियेबल्स के थ्रू मत समझो आप इसकी मेकैनिज्म न्यूमेरिकल की जरूरत फटाफट क्लिक कर जाओगे की हान सर चीजे ऐसी ऐसी होनी है तो दो तरीके स्टूडेंट्स ये दोनों तरीके समझ आए क्या अच्छे से नोट डाउन करके चले गए सारी बातें बस वही जो दूसरा तरीका सिखा है वो इन तीन लाइन में उन्होंने डिस्क्राइब किया पर आप उसे तरीके को अच्छे से समझ के रखना इस पर क्वेश्चंस बनेंगे तो हम इन दोनों में से कोई भी तरीका उसे करेंगे अपनी रिटायरमेंट के अकॉर्डिंग है ना सर अगर यहां से मैं आगे बढ़ो तो यहां से फिर वही बात अगर मैं एक बात करूं सर x² + y² = a² मैन लो सर अगर आपका सर्कल है सर्कल कैसा है सर्कल है x² + y² = a² और मुझे इस सर्कल में बोला की इस सर्कल की आप tenjent बताओ जिसकी स्लोप है एन तो मैं कहूंगा सर मेरे लिए तो के क्वेश्चन ए = एमएक्स प्लेयर इस सर्कल का सेंटर का सेंटर है 0 इस सर्कल की रेडियस कैसा है रेडियस है फिर वही बात इस सर्कल के सेंटर से इस स्ट्रेट लाइन की जो परपेंडिकुलर डिस्टेंस होगी वो इसकी रेडियस के इक्वल होगी क्या बात समझ पाए फिर वही बात वापस सब चीज कर लेनी है पर मैं का रहा हूं ऐसा राठोड ऐसा रतो मत की अगर एक सर्कल है x² + y² = a² इसकी टांगें है जिसकी स्लोप है एम तो उसकी इक्वेशन जो होगी वो होगी ए = एमएक्स + - ए टाइम्स अंडर रूट ओवर m² + 2 मत यार ये तकलीफ वाली बात है हर चीज कितनी रात होगी कितना याद करोगे कितनी चीज घुटने पड़ेंगी इतना मत करो है ना rafcourse ये बातें आपको समझ के अच्छे से चलनी चाहिए मेरा तो बस ये मारना है मेरा बस ये मारना है की सर क्वेश्चन में क्वेश्चन कुछ दिया होगा ऑफ कोर्स वो कैसे देगा वो ऐसे दे देगा वो दे देगा x² + y² = 36 तो आप बोलोगे सर इसकी रेडियस जो है वो है सिक्स है ना और बोल देगा की उसे उसे टैसेंट की इक्वेशन बताओ इसके स्लोप है लेट से 4 तो अब बस यह का दोगे अब मैं कैसे सॉल्व करूंगा सर इसे मैं कैसे सॉल्व करूंगा मैं कहूंगा सर इक्वेशन है 4X - ए + अल्फा = 0 तो इस स्ट्रेट लाइन की इसके सेंटर से जो परपेंडिकुलर डिस्टेंस है वो क्या जाएगी सर जीरो जीरो प्लस अल्फा तो अल्फा के ऊपर क्या लगा दिया मोड डिवाइडेड बाय क्या फोर का स्क्वायर प्लस वैन का स्क्वायर यानी कितना 16 + 1 यानी कितना अंडर रूट 17 ये किसके इक्वल होगा सर इसकी रेडियस के यानी 6 तो यहां से अल्फा की वैल्यू क्या ए रही है सर अल्फा की वैल्यू ए रही है प्लस माइंस कितना सिक्स टाइम्स रूट 17 क्या बात आपको समझ आई और बस सर अल्फा की दो बार की तरफ दो आपका आंसर ए जाएगा मैं का रहा हूं कितना आसान है ये चार लाइन का भी क्वेश्चन नहीं है दो लाइन में खत्म हो जाएगा अगर आप चीज समझ का रहे हो तो इतनी सारी चीज मत रखो यार इतना सारा याद नहीं करने की जरूरत मतलब ये बहुत ही अजीबो गरीब बात है अलजेब्राइक भी बहुत अजीब अजीब चीज है की सर अगर मुझे ढूंढना है तो उसे तरीके से पर कितना कॉम्प्लिकेटेड हो जाएगा कितनी अजीब अजीब शब्द है इतना आपको जीवन में कभी नहीं रखना चाहिए मतलब स्टूडेंट्स को कभी भी इस तरीके से तो नहीं होना चाहिए की सर कितना याद रखना आईआईटी जी मैं आंसर एडवांस कभी भी कभी भी कोई रत्न वाला क्वेश्चंस नहीं पूछेगा वो कभी आपके रत्न की नॉलेज चेक नहीं करेगा मतलब रत्न की एबिलिटी चेक नहीं करेगा वो क्या आप कितना अच्छा रात लेते हैं कितना अच्छा याद रख पाते कितना अच्छा घोट के पी पाते नहीं वो हमेशा ऐसी बातें करेंगे की आपने कॉन्सेप्ट्स को कितने अच्छे से समझा आपको उनके अंदर की इंटरनल मेकैनिज्म पता है या नहीं की वो चीज ऐसी होती तो क्यों होती है और कैसे होती है और उन पर बेस्ड एप्लीकेशन बेस्ड क्वेश्चंस पूछेगा की चलो आपने पढ़ा है ना तो इसको अप्लाई करके यहां पे सोचो की यहां पे कैसे बनेगा और हर प्रॉब्लम अपने आप में मजेदार यूनिक से शानदार से चैलेंजिंग प्रॉब्लम होगी जिससे सॉल्व करके आपके दिमाग के बहुत तर्क खुले क्या आप मेरी बातें समझ का रहे हो तो रत्न नहीं है है ना चीज समझो ना अच्छे से आप बात खत्म ये बहुत सारे फॉर्मूले हो जाएंगे इतना कब तक याद रखोगे और अभी तो सर्कल है साहब फिर पैराबोला एलिप्स हाइपर बोला हर जगह से थोड़ी तो मत रखिए याद रखिए समझते हैं जो घुमा फिर करें बहुत सारे वैरियेबल्स में का रहा हूं सर ये क्वेश्चन ऐसे हो गया ना बस बात खत्म अब इक्वेशन निकल लेंगे हम क्यों ज्यादा सोचना बिल्कुल नहीं सोचना है बस ऐसे ही करना है हान उसने कंडीशन दिए कंडीशन क्या दी है वो सुनता है तू फाइंड डी पॉइंट ऑफ कॉन्टैक्ट सी सॉल्व डी लाइन इन डी सर्कल सिर्फ वेयर वही बात सर अगर मुझे पॉइंट ऑफ कॉन्टैक्ट पता करना होता की सर वो किस पॉइंट पे कॉन्टैक्ट करें किस पॉइंट पे टच कर रही है तो मैं क्या करूंगा सर मेरे पास लाइन है लाइन क्या है ए = 4X + अल्फा तो उसकी वैल्यू यहां रख देता हूं तो जैसे ही यहां रखा तो 4X + अल्फा सर्कल की इक्वेशन को मैं सॉल्व कर रहा हूं जैसे ही सॉल्व कर तो क्या मिलेगा देखो x² तो x² ही है ए स्क्वायर तो ए की जगह क्या रखा 4X + अल्फा तो ये हो जाएगा 4X + अल्फा का होल स्क्वायर अब बात समझ का रहे हो 36 को इधर ले आए तो -36 = 0 इसे आप सॉल्व करिए इससे आप सॉल्व करिए और फिर वही बात जैसे ही आप इसे सॉल्व करोगे तो क्या मिलेगा तो जैसे ही आप इसे सॉल्व करेंगे तो आपको एक क्वाड्रेटिक मिलेंगे क्वाड्रेटिक से क्या होगा सर क्वाड्रेटिक से आपको यह इंचार्ज करना है की आपकी क्वाड्रेटिक का डिस्क्रिमिनेंट जीरो क्योंकि सर क्योंकि ये सर्कल और ये टैसेंट एक ही जगह टच कर रहा है और जब डिस्क्रिमिनेंट जीरो रखोगे फिर वही बात जो भी जस्ट हमने फॉलो की थी उससे आपकी अल्फा की वैल्यू हो जाएगी और अल्फा की वैल्यू ए जाने से वो वैल्यू रख देना इस इक्वेशन में तो एक्स की वैल्यू हो जाएगी एक्स की वैल्यू ए गई तो इसकी वैल्यू यहां या यहां रख दें तो ए की वैल्यू हो जाएगी तो आपके दोनों पॉइंट ऑफ कॉन्टैक्ट ए जाएंगे बहुत कंफ्यूज तो नहीं कर रहा हूं स्टूडेंट क्या आप बात समझ का रहे हो कंडीशन यही है की जब पॉइंट ऑफ कॉन्टैक्ट पूछे तो सर्कल और स्ट्रेट लाइन को सॉल्व कर लो स्ट्रेट लाइन में क्या गिवन होगा उसकी स्लोप गिवन होगी ये ए इंटरसेक्ट नहीं दिया होगा तो हम क्या करेंगे सर इन दोनों को जब सॉल्व करेंगे तो हमें पता है की ये सर्कल और स्ट्रेट लाइन एक ही पॉइंट पे इंटरसेक्ट कर रहा है तो जो क्वाड्रेटिक बनेगी एक्स में ए में किसी में भी बना लेना उसे क्वाड्रेटिक का एक ही रियल रूट होना चाहिए यानी इक्वल रियल रूट होना चाहिए मतलब क्वाड्रेटिक का जो भी ये बनेगी इसका डिस्क्रिमिनेंट जीरो कर देना यहां या यहां फिर से आपके यहीं रख देना अल्फा तो एसईसी ए जाएगा यहां रख देना तो उससे ए ए जाएगा क्या यहां तक कोई परेशानी है स्टूडेंट्स आई थिंक चीजें आसान है और बहुत टू या बहुत डिफिकल्ट नहीं है यही याद रखना है ना की कैसे सॉल्व करना है मैं फिर से बता देता हूं क्या करना है हम दो चीज करेंगे हम सर्कल और स्ट्रेट लाइन की इक्वेशन को क्या करेंगे सॉल्व है ना जब उन दोनों को सॉल्व करेंगे तो सॉल्व करने पे क्या बनेगी ये क्वाड्रेटिक फिर उसको क्वाड्रेटिक इक्वेशन का क्या करेंगे सर उसको ऑर्डर एक क्वेश्चन का डिस्क्रिमिनेंट हम जीरो के इक्वल रख देंगे जीरो रखने से क्या होगा सर उससे आपको क्या मिल जाएगा अल्फा की वैल्यू मिल जाएगी जब अल्फा की वैल्यू मिल जाएगी तो उसको क्वाड्रेटिक इक्वेशन पास कर देना तो उससे क्या मिल जाएगा इधर एक्स और ए है जिसकी भी वैल्यू मिली है उसे या तो यहां रख देना या यहां रख देना उससे आपको क्या मिल जाएगा उपाय और जो एक्स और ए मिलेंगे वही क्या होंगे जो एक्स और ए आपको मिलेंगे वही क्या होंगे आपके पॉइंट ऑफ कॉन्टैक्ट क्या आप मेरी बातें समझ का रहे हो कॉन्टैक्ट का क्या मतलब होता है जिस पॉइंट पे वो टच कर रहा है जिस पॉइंट पर वो टच कर रहा है वो निकलने की प्रक्रिया या मेथड मैं आपको समझने की कोशिश कर रहा हूं बस यही हम बार-बार करेंगे अगर ऐसा कुछ आता है तो क्या बोल रहा हूं मैं इस इक्वेशन को सॉल्व करेंगे उससे आपकी क्वाड्रेटिक बन जाएगी क्वाड्रेटिक में सिर्फ एक ही पॉइंट ऑफ इंटरसेक्शन आना चाहिए तो डिस्क्रिमिनेंट जीरो रख देंगे उससे आपकी वो अल्फा की वैल्यू हो जाएगी क्योंकि स्लोप गिवन होगी इस लाइन की क्या गिवन होगी स्लोप गिवन होगी कोऑर्डिनेटर ए जाएगा उसको आप स्ट्रेट आएंगे सर्कल के क्वेश्चन में रख देना उससे आपका जो दूसरा वाला cardinate है वो भी ए जाएगा तो वो जो एक्स और ए कार्ड आएंगे वही आपके पॉइंट ऑफ कॉन्टैक्ट रहे होंगे कोई और फॉर्मूला रत्न की जरूरत नहीं है मेरा यकीन करो वहां भी यही सब कर रहे होंगे तो इससे अच्छा तो है सर ये कर लें और मेरा यकीन करो ये आसान है याद रहेगा अच्छे से चलो अगर याद रहेगा तो थोड़ा आगे बढ़ते हैं और समझते हैं वो यही बातें सारी लिख के दे रहे हैं की वो पॉइंट ऑफ कॉन्टैक्ट है मत बैठो आप तो बस ये याद रखो ये मेथड समझ लो ना अगर हमें समझ लिया तो बस हर बार कर लेंगे सर ये वही बात है ना गिव अन मैन फैशन यू कैन फीड इन फॉर अन दे बट फिश यू कैन फीड हिम फॉर डी लाइफ टाइम अगर आप एक इंसान को कुछ करना ही सिखा दें तो फिर तो खुद से कर ही लेगा ना तो क्यों किसी को कोई फॉर्मूला और डिपेंड करना की अभी तो इस सिनेरियो में इन हालत में ये हो रहा है कहीं और कुछ बदल दिया तो उससे अच्छा सिख लो इंटरसेक्शन निकलना है तो उन दोनों को सॉल्व करो बात खत्म पर यहां कंडीशन बड़ी मजेदार सी है की उनका पॉइंट ऑफ इंटरसेक्शन ऐसा है की ये कर दो जगह इंटरसेक्ट नहीं करते एक ही जगह इंटरसेक्ट करते हैं और हम जानते हैं वो क्वाड्रेटिक बना के देगा तो डिस्क्रिप्शन में जीरो रख दो बात खत्म इतना सा यह मेथड है क्या याद रहेगा और अगर ये याद रहेगा तो थोड़ा आगे बढ़ते हैं और कम की बातों पर आते हैं ये बहुत सारी घुमा firakar बातें कहीं गई है मैं नहीं कहूंगा की आप ये सब बातें याद रखिए बिल्कुल भी मत रखिए है ना ये वही बात है बेसिकली उसने पूरा उसको अलजेब्राइक सॉल्व किया और पूरा सॉल्व करके निकल के दिया जो की आप कभी भी लाइफ में याद नहीं रखेंगे आप सीधे सीधे इस तरीके से क्वेश्चंस को सोचेंगे कैसे suniyega ध्यान से अगर मैं इस क्वेश्चन को करना चाहूं तो देखो मैं कैसे करूंगा पढ़ो फाइंड डी इक्वेशन ऑफ टांगें तू डी सर्कल विच आर पैरेलल तू डी स्ट्रेट लाइन कैसे सोचेंगे स्टूडेंट बन रही है अगर यह लेने जिसके पैरेलल वो टैसेंट बन रही है तो मैं तो ये कहूंगा सर इस लाइन की स्लोप क्या है इस लाइन के स्लोप निकालो अच्छा इसके अलावा कोई और तरीका सुना हमसे क्या पूछ रहा है फाइंड डी इक्वेशन ऑफ दी 10 तो दो तरीके हैना सुनना मैं पहले तो इससे स्लोप ही निकल लेता हूं इससे अगर स्लोप निकल ले तो क्या ए जाएगा देखो ए = एक्स + सी है ना तो क्या हो जाएगा 3y = क्या -4x + 5 है ना तो ए = ए = -4 / 3 एक्स + 5 लेकिन हमारी जो टैसेंट है उसमें फाइव तो नहीं मिलेगा उसमें तो कुछ और इंटरसेप्ट होगा तो हमारी जो टैसेंट होगी उसका जो इंटरसेप्ट होगा उससे मैं थोड़ी देर के लिए सी अल्फा कुछ भी का लेता हूं हम ऐसे करेंगे ऐसे कई बात आपको समझ आई ठीक है सर तो आपके अकॉर्डिंग ये आपकी टेंशन है बिल्कुल सही बात हमारे अकॉर्डिंग ये हमारे टेंडर तो हमने क्या सिखा था हमने बड़ी बेसिक सी बात सीखी है सर की अगर ये जो लाइन है ये इस सर्कल की पेंडेंट है तो दो तरीके एक तरीका की सर्कल के सेंटर से इस लाइन की परपेंडिकुलर डिस्टेंस निकालिए और उसे उसकी रेडियस के इक्वल रख दीजिए दूसरा तरीका इस लाइन और इस सर्कल को सॉल्व करिए discretement जीरो रख दीजिए आपके आंसर ए जाएंगे दोनों तरीके से सुनना मैं सिंपल सा तरीका उसे करता हूं जो बड़ा बेसिक सा है की सर इस स्ट्रेट लाइन की इक्वेशन को थोड़ा और बेहतर तरीके से लिख सकता हूं क्या तो मैं लिखूंगा ध्यान से मैं इसी लाइन को थोड़ा और बेहतर तरीके से रखना चाह रहा हूं सुनना यही लाइन क्या सुनना 4X + 3y बस 5 की जगह क्या होगा कोई ना कोई अल्फा तो हमारे लिए स्ट्रेट लाइन हो जाएगी बस 5 की जगह कुछ और होगा ना इसी के पैरेलल वो लाइन है तो हमारे लिए वो लाइन होगी अच्छा ये सर्कल के सेंटर के कोऑर्डिनेट्स क्या होंगे सर ये सेल का सर्कल का सेंटर के क्वाड्रेंट - 3 कमा तू तो इसके सेंटर के अकॉर्डिंग हो जाएंगे -3 कॉमन अच्छा इसकी रेडियस क्या होगी सर ठीक है 25 तो ये 25 का अंडररूट 25√5 इसकी रेडियस कितनी है 5 तो 5 तो है इसकी रेडियस भाई है ना और थ्री कमा -3 क्या है इसका सेंटर आई होप आप बातें समझ का रहे हो अब सर अगर सेंटर से साइन की परपेंडिकुलर डिस्टेंस निकल लें तो किसके इक्वल होगी रेडियस के बस ये सिंपल सी तकनीक है कुछ नहीं सोचना है क्यों कोई फॉर्मूला रत्न है सर तो अगर ये तकनीक का अप्लाई की जाए तो क्या -3 से इसकी परपेंडिकुलर डिस्टेंस तो -3 4 कितना -12 + 3 कितना सिक्स प्लस बनेगी उसे स्लोप की पैरेलल आई होप बात समझ का रहे हो है ना डिवाइडेड बाय फोर का स्क्वायर + 3² वही बात 435 है ना तो यहां पर क्या ए जाएगा फाइव इसे इक्वल तू इसकी रेडियस जो की कितनी है 5 आई थिंक सर ये फाइव को हटा के यहां से सामने ले आओ तो 5 5 कितना हो जाएगा सर 5 5 हो जाएगा 25 सो 25 में भी एक और बात आप देख रहे हो क्या जैसे की क्या अगर ये मोड और ये मोड यहां से हटाया तो ये कितना हो जाएगा सर ये हो जाएगा प्लस माइंस 25 सर माइंस 12 प्लस 6 कितना होता है सर माइंस 12 प्लस 6 अगर मैं देखूं तो -12 + 6 - 6 -6 को उधर भेजा तो कितना शुरू हो जाएगा प्लस सिक्स आपको दिख रहा है तो अल्फा की वैल्यू 31 और अगर -25 लिया तो -25 + 6 कितना -29 हो जाएगा -29 कुछ गलत कर रहा हूं मैं बिल्कुल गलत कर रहा हूं माइंस 19 होगा ना आपको कैलकुलेट करते नहीं आता है सर माइंस 25 प्लस एक्स माइंस 19 होगा -29 पता नहीं क्यों बोला मैंने है ना कुछ गलती रही होगी दिमाग में अब क्या क्वेश्चन खत्म हो गया जी हान ये दो लाइन में क्वेश्चन खत्म हो गया है कैसे सर आप से अल्फा की अगर वैल्यूज निकल गई हैं तो अल्फा की वैल्यू रख दो आपका आंसर है कौन सा सर एक तो आपकी इक्वेशन होगी 4X + 3y प्लस 31 इस इक्वल तू जीरो और दूसरी इक्वेशन होगी 4X + होते हैं -19 = 0 ये जो दो इक्वेशंस आई है ना भाई ये जो दो इक्वेशंस आई हैं सर यही आपकी वो स्लोप्स एम गिवन जब दी जाए तो उससे उसे स्लोप को कैरी करने वाली tenjins की इक्वेशन है सर्कल के आई थिंक ये तो बहुत आसान सा तरीका है क्यों इतने सारे फॉर्मूले रखने हैं सर क्यों इतनी अजीबो गरीब बातें करनी है अच्छा सर अगर पॉइंट ऑफ इंटरसेक्शन निकलना होता तो क्या करते सुनना मैं अब इस लाइन को इस सर्कल के साथ सॉल्व कर लूंगा या फिर इस लाइन को सॉल्व कर लूंगा ए जाएंगे पॉइंट ऑफ कॉन्टैक्ट इसके अलावा कोई और तरीका एक और तरीका याद करो स्टूडेंट्स अगर मैं सीधे-सीधे आपसे बात करूं कौन सा तरीका अब बस मैं उसको बता देना चाह रहा हूं सॉल्व आप खुद करिएगा हम क्या करेंगे यह जो आपकी सर्कल ये जो स्ट्रेट लाइन की इक्वेशन है इसमें एक्स या ए की टर्म्स में से रिप्रेजेंट कर देंगे है ना यहां से मैं इसे ए की टर्म्स में लिख दूंगा कुछ ना कुछ जो की आप सब लिख सकते हो जो वैल्यू आएगी उसे मैं यहां और यहां रिप्लेस कर दूंगा तो ये जो पुरी इक्वेशन है ये पुरी इक्वेशन बन जाएगी है और क्योंकि एक ही पॉइंट ऑफ कॉन्टैक्ट है तो क्या बोलेंगे सर उसका डिस्क्रिमिनेंट जीरो डिस्क्रिमिनेंट जीरो रखने से अल्फा की वैल्यू हो जाएगी यार आदर वैल्यूज हो जाएंगी क्योंकि क्वाड्रेटिक होगी वो भी तो वही आएंगे आप करके देख लीजिएगा फिर पॉइंट ऑफ कॉन्टैक्ट निकलना है तो कुछ मत करना वो जो इक्वेशन है ये फिर वही बात इसको इसके साथ सॉल्व कर लेना ए जाएंगे आई होप यह प्रक्रिया यह मेथड भी क्लियर है मेरा यकीन है उससे उतना घुमा फिर के करने की बजाय ये और जल्दी हो जाती है ऐसा कर लीजिए ये भी सही है पर मेरे ख्याल से और आसान है जब जरूरत लगी तो बेशक वैसा भी करेंगे और आई थिंक ये एक आसान और अच्छा तरीका है हमने क्या सिखा अभी हमने सिखा की सर अगर किसी सर्कल में कोई tenjent बनानी हो जिसकी स्लोप हमें दे दी जाए तो हम कैसे बनाते हैं एक आसान सा बेसिक सा सैंपल सा तरीका था बहुत डिफिकल्ट हमने बहुत घुमा फिर कर बातें नहीं की और बेशक हमने यह बात भी सीखी सर की इसमें दो टेंट बनती है जो की पैरेलल होती है वो डायमीटर की एक्सट्रीम पर होती है उन पैरेलल टैसेंट के बीच की डिस्टेंस डायमीटर दे देगी आपको सन रहे हो आप अच्छा सर अब सीधे ऐसा ही क्वेश्चंस आएगा क्या बिल्कुल नहीं आने वाला है ये तो बस आपको सीखने के लिए रखा गया है हम तो जी मैंस और एडवांस लेवल की तैयारी कर रहे हैं तो अच्छे क्वेश्चंस आएंगे तो इसी तरीके से सारे क्वेश्चंस को अगले लेक्चर में कंटिन्यू करेंगे इस इन सारे वीडियो का जॉब्स इंगेज मैथमेटिक्स और आईआईटी जी मेंस एंड एडवांस पढ़ा रहे हैं इसका एक प्लेलिस्ट बना के दे दीजिए बिल्कुल इसका प्लेलिस्ट ऑलरेडी बना हुआ है इसी वीडियो के डिस्क्रिप्शन में जब आप जाएंगे तो आपको वो प्ले लिस्ट का लिंक दिखाई देगा इंगेज मैथमेटिक्स और आईआईटी जी मेंस एंड एडवांस लिखा हुआ है उसे जरूर से करके रखें अब बात करते हैं इस पॉइंट की की सर जहां पर मुझे किसी सर्कल पर देखो फंडा वही है फंडा मेरे यकीन करो बार-बार वही है क्या सर की आपके पास एक सर्कल है इस सर्कल पर किसी एक्सटर्नल पॉइंट बना दिया आपने आई होप आप समझ का रहे हो और एक टैसेंट आपने कुछ ऐसी बना दी है लेट्स चाहिए अब मेरा यह कहना है की सर मैन लेते हैं ये वो पॉइंट है एक्सटर्नल पॉइंट पी जहां से आपने ये दो एक्सटर्नल पॉइंट मतलब एक एक्सटर्नल पॉइंट पी से इस सर्कल पर ये दो टांगेंट्स आपने जो ड्रॉ किए अब वो यही बात वापस का रहा है क्या इस टेंशन की इक्वेशन लिखी जा सकती है देखो सिंपल सी बात बहुत सिंपल सी बात इस इंजन की क्या खास बात है आप सोच के देखो सर इससे पहले जो हमने लेक्चर पढ़ा था वहां पे देखा था की अगर मुझे टांगें की स्लोप दे दी जाए तो मुझे बस वो वही इंटरसेक्ट निकलना था वायरस = एमएक्स + अल्फा और अल्फा की हम निकल लेते द वैल्यू कैसे सर्कल के सेंटर से परपेंडिकुलर डिस्टेंस रेडियस के इक्वल रख देते द तो इस बार उसने आपको स्लोप नहीं दी है पर इस बार उसने पॉइंट दे दिया है क्या पॉइंट दे दिया है की सर ये जो tagent है ये किसी एक फिक्स्ड पॉइंट लेट से X1 y1 से पास होती है फिर वही बहुत स्टूडेंट्स आपको इनडायरेक्टली डायरेक्टली करना वही है सर्कल की इक्वेशन होगी क्या x² + y² + 2G एक्स + 2fy प्लस सी सर इसे देख के क्या समझ आता है इसे देख के सबसे अच्छा कम जीवन का आप जो कर सकते हैं वो क्या सर मुझे तो ये समझ ए जाता है की सर्कल का सेंटर जो होगा वो होगा -जी कमा -एफ बहुत बढ़िया अब अगर मैं इस इस टांगें की मैन लो अभी फिलहाल ये जो एक्सटर्नल पॉइंट्स आपने ड्रॉ किए हैं वो पी आर और लेट से पॉइंट है जो ऑफ कोर्स में ड्रा कर पाऊंगा ना क्योंकि ये यहां जा रही हो की यहां जा रही होगी मैं थोड़ी देर के लिए मैन लेता हूं की सर हमारी जो तेनजिंग की इक्वेशन है वो है पर तो मैं जानता हूं सर इसकी कोई ना कोई स्लोप रही होगी तो आप इक्वेशन क्या लिखोगे ए - y1 = एमएक्स - एक्स अब क्या आपको पता है की नहीं पता है X1 y1 पता है एन नहीं पता है आप समझ रहे हो बहुत टू नहीं है क्यों नहीं है सर अब बात समझो स्टूडेंट्स क्योंकि सर फिर वही बात सर्कल के सेंटर के कोऑर्डिनेट्स हैं और अगर पुरी सर्कल की इक्वेशन है तो सर्कल की रेडियस नहीं निकल सकते कैसे रेडियस का फॉर्मूला तो मैं जानता हूं क्या होता है सर यहीं से निकल लेंगे जी स्क्वायर प्लस एक्स स्क्वायर माइंस सी अब क्या करूं सर अब आप मुझे बताओगे क्या क्या कर रहे हैं सर मेरे पास पर लाइन की इक्वेशन है जो की है सर्कल की टेंशन और अगर मैं सर्कल के सेंटर से ये बड़ी मजेदार से प्रॉपर्टी है की सर्कल के सेंटर्स अगर मैंने परपेंडिकुलर ड्रॉप किया इस टांगें पर तो क्या मैं वो परपेंडिकुलर लेंथ निकल सकता हूं अननोन वैरियेबल्स में सिर्फ एम मतलब टेक्निकल -3 से इस स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस जो आप निकलेंगे वो किसके करोगे सर इसकी रेडियस के फिर वही बात सर उससे तो एक ही क्वेश्चन आएगी नहीं उससे दो श्लोक मिलेगी क्योंकि जब आप परपेंडिकुलर डिस्टेंस निकल लोग तो न्यूमैरेटर में वहां मोड लगाओगे तो वह जो इक्वेशन बनेगी जिसमें एम आपका वेरिएबल होगा उसमें मोड लगा होगा मोड hataoge तो प्लस माइंस लग जाएगा और उससे आपको दो एम मिल जाएंगे तो एक तो आपके पर वाली लाइन का स्लोप होगा और एक पॉइंट वाली लाइन का स्लोप होगा इतना आसान सा सिंपल सा कर सकते हैं अगर मैं पूरा अलजेब्राइकली आपको डेरिव करके नहीं देना चाह रहा हूं मैं बस आपको मेकैनिज्म समझाना चाह रहा हूं की देखो चीज आसान है बस अच्छे से अप्लाई करना क्वेश्चंस हो जाएंगे क्या आपको यह थॉट यह प्रक्रिया यह स्टैंडर्ड ऑपरेटिंग प्रोसीजर समझ ए रहा है आई होप अगर यहां तक क्लियर है तो जल्दी से देख लेते हैं वो आपको ये कहना चाह रहा है अब उसमें पूरा अलजेब्राइकली सॉल्व कर दिया है नेवर वुड वांट यू तू लर्न क्योंकि algebraicly आप कंफ्यूज हो जाओगे बेहतर है न्यूमेरिकल करो किसी क्वेश्चन के चर्चा किया पर अच्छे से समझ पाओ बट इट इस इट्स अन गुड आइडिया ऑलवेज की आप इन्हें अलजेब्रा के लिए डेरिवेशन की टर्म्स में थ्योरी वाइजर थ्योरम विजय चीजे जानते हो वो क्या का रहा है ऑफ कोर्स हम जानते हैं दिस इस बार उसने एक एग्जांपल ले लिया है ना जैसे की सर क्या एक सर्कल प्लस ए स्क्वायर इस इक्वल्स तू लेट से वो 2 की जगह मैन लो a² लिखा हुआ है अब सर क्लियर सी बात है मैं मैन लेता हूं की इसकी जो स्लोप है tagent है उसकी ये स्लोप होगी एंड तो उसे केस में हम ये अटेंडेंट के क्वेश्चन लिखना चाहते हैं मैं कहता हूं हमने तो नहीं रहता है सर हमने तो यार नहीं है मैं तो बना लूंगा मैं तो क्वेश्चन देख के बना लूंगा बिल्कुल परफेक्ट कोर्स विद टाइम अगर आप प्रैक्टिस कर करके इसे याद कर लें तो जरूर कर लीजिएगा ये थोड़ा जल्दी कम कर देगा आपका बस ये स्टैंडर्ड केस तो एटलिस्ट याद रखना की जब भी आपको ये वाला सर्कल दिखे कौन सा स्टूडेंट्स x²+y² = a² इसकी टांगें की इक्वेशन अगर स्लोप फॉर्म में पूछे जाए की स्लोप आपको एम दे दी जाए तो आप स्लोप फॉर्म में सीधे-सीधे इसकी इक्वेशन लिख दोगे ए = एमएक्स + अल्फा टाइम्स 1 + m² अल्फा क्या है वो इंटरसेप्ट सर अगर मुझे स्लोप पता ना हो मतलब एक क्वेश्चन ना मैंने याद की हो तो बिल्कुल चिंता की बात नहीं हमारे पास तो अपने मैथर्ड हैं ए = एमएक्स + अल्फा और चीज वही बात वापस सर्कल का सेंटर से ड्रॉप की केस लाइन पर परपेंडिकुलर लेंथ रेडियस के ऊपर लोग और वहां से आप वापस यही चीज बना लेंगे अब कर लेंगे मेरा यकीन करिए हमने हर एक कॉन्सेप्ट को हर एक पहलू को बहुत बारीकी में एक्सप्लोर किया है मेरा यकीन करिए आपको आती है चीजें तो क्या सर तो या तो याद कर लीजिए और नहीं भी हो तो इससे समझने की कोशिश करिए वो इन सारी बातों पर ही आपसे बात करना चाह रहे हैं और सीधे घुमा फिर के वो कनक्लूड करना चाह रहा है की आप क्वेश्चंस को मेकैनिज्म मतलब वो मॉडल अप्रेंटिस की तरह समझ के करिए की चीजें कैसे होती हैं जैसे इस क्वेश्चन के थ्रू मैं आपको थोड़ी क्लेरिटी देता हूं देखो वो का रहा है फाइंड डी इक्वेशन ऑफ टांगें तू डी सर्कल तो एक सर्कल है कहां से सर ड्रोन फ्रॉम पॉइंट पी विच तू कमा 3 क्वेश्चन आपको याद है ना हमें कोई फॉर्मूला रत्न की जरूरत नहीं हमें बेसिक कॉन्सेप्ट पता हो की चीज कैसे रियलिटी में बिहेव करती हैं कैसे सर मतलब मेरा आपसे कहना है सर्कल की इक्वेशन बनाओ सर्कल बना दो सर्कल है मतलब ये आपको बस समझ अच्छा ऐसे ए जाए इसलिए मैं थोड़ा प्रॉपर्ली ड्रा करूं वर्ण इसमें ज्यादा कोई लोड लेने वाली बात है नहीं ये क्या हो गई सर आपकी दोनों टांगेंट्स कोई दिक्कत तो नहीं भाई मेरे ख्याल से तो नहीं होनी चाहिए है ना तो ये हो गई आपकी टैसेंट ठीक है सर यह एक्सटर्नल पॉइंट पी जिसके कोऑर्डिनेट्स आप का रहे हो सर तू कमा 3 बिल्कुल सही बात है ये आपका सर्कल है जिसकी इक्वेशन है ये क्या सर्कल को देखकर आप सर्कल के सेंटर के कार्ड बता सकते हो सर क्या पूछ रहे हो आप तो यूं ही बता देंगे हम बोलो भाई ये कितना हो जाएगा वैन और ये कितना हो जाएगा -2 दिख रहा है क्या तो सर सर्कल का सेंटर के कोऑर्डिनेट्स हो जाएंगे वैन कमा -2 क्या यहीं से आप मेरी सर्कल की यू नो रेडियस निकलने में हेल्प करेंगे बिल्कुल कर देंगे सर कौन सी बड़ी बात है अब तो हम सब जानते हैं वैन का स्क्वायर 1 - 2 का स्क्वायर तू मतलब तू का स्क्वायर तू तो स्क्वायर है तो नेगेटिव पॉजिटिव कोई फर्क नहीं पड़ता तो 4 + 1 कितना फाइव फाइव माइंस माइंस फोर फाइव माइंस माइंस फोर तो वो जाता है 99√3 तो इसकी रेडियस जो हमें निकल का रहा हूं सर वो कितनी आती है थ्री कोई तकलीफ आपको यहां तक क्या यह क्वेश्चन खत्म होता है दिख रहा है सर ये क्वेश्चन खत्म हो गया बस मतलब यह भी आपको समझने के लिए मैंने लिखा है ऐसा मैन में सोचना ये बनाना नहीं है आप सीधे-सीधे जानते हो सर मेरे लिए क्या होगी इक्वेशन ए = मतलब मैं क्या बोलूंगा ए - y1 तो ए - 3 = एम एक्स - एक्स ऍम तुम्हें क्या लिखूंगा एक्स - 2 आप इसे थोड़ा सिंपलीफाई कर लीजिए ये क्या हो जाएगी एमएक्स ए इधर आएगा तो -ए ये ऑलरेडी - 2m है है ना और थ्री इधर आएगा तो +3 = 0 मेरा बस आपसे ये पूछना है की सर अगर मैं वैन कमा - 2 से इस स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस निकलूं जो की हमारे लिए अभी ये है तो क्या मैं कहूंगा वो थ्री के इक्वल होगा बस इतना सा करना है क्या ये बहुत टू बहुत मुश्किल बात है सर नहीं है अब चीज सीधे निकालो आंसर है आएगा निकलते है भाई ये कितना है -2m प्लस वैन तो अंडर रूट ओवर एम स्क्वायर प्लस वैन होगा किसके रेडियस के चौकी कितनी है ना तो ये हो जाएगा 5 - मोड लगा हुआ है आप चाहें तो लगा लीजिए और उसे तरफ जब गया तो ये हो जाता है थ्री टाइम्स अंडर रूट ओवर 1 + m² अब आप मेरी एक छोटी सी बात पे एक और करिए स्टूडेंट्स अगर मैं दोनों तरफ स्क्वायर कर डन बाय डूइंग डेट एल टेक्निकल गेट रीड ऑफ बोथ मतलब क्या मुझे मोड से भी मुक्ति मिल जाएगी और मुझे अंडर रूट से भी मुक्ति मिल जाएगी और जैसे ही आप ऐसा करेंगे तो यहां लग जाएगा स्क्वायर और यहां लग जाएगा स्क्वायर क्योंकि मोड का स्क्वायर मेक अन्य डिफरेंस कोई तकलीफ आय होप आपको रिलाइज हो रहा है अब बनेगी एक क्वाड्रेटिक और यहां से मिलेंगे आपको एम की दो वैल्यू बस इस तरीके से आप चीज सॉल्व कर लेंगे बिल्कुल सर अगर मैंने यहां से सॉल्व किया तो क्या बन जा रहा है सर ये बन जा रहा है 25 प्लस डाइजेस्ट कर लो तो उधर लेकर जाते हैं जो 9 m² - m² कितना हो जाएगा 10 लाइन 25 - 9 मेरे लिए कितना होता है आई थिंक 16 तो ये हो जाएगा -16 क्या आपको दिख रहा है की सर आप तू कॉमन ले सकते हो आई थिंक मैं तू कॉमन ले सकता हूं जैसे ही तू कॉमन लिया तो देखो क्या दिखेगा सर ये दिखेगा 4m² + 5m - 8 = 0 आई थिंक सही है क्या सिर्फ दो बातें 32 अब मुझे दो ऐसी टर्म्स चाहिए जिनका प्रोडक्ट कितना हो - 32 और जिनका सैम ऑफ फाइव ढूंढ सकते हैं क्या कोशिश करते हैं 32 के लिए मुझे क्या देखता है 16 और 2 भी तो देखता है पर बात नहीं बनती है ना जैसे मैं इसका फैक्टराइजेशन करता हूं देखो 32 मतलब 2 की पावर फाइव मैं ऐसे सोच लेता हूं मैन में मैं ऐसे करता हूं 32 मतलब मेरे लिए क्या तू की पावर फाइव अब तू को ब्रेक करने की कोशिश करते हैं तू को ब्रेक किया तो एक तो हो जाता है 8 और 4 बट आते और फोर का डिफरेंस नहीं बनाएगा इसके अलावा एक और क्या बनेगा 16 और 2 तो 16 और 2 से भी बात नहीं बनेगी फाइव तो नहीं ए रहा होगा कहीं से कहीं तक नहीं ए रहा है इसके अलावा और क्या बना सकता हूं सर इसके अलावा आई थिंक 32 और 1 से तो बात बन्नी नहीं है तो मुश्किल लग रहा है लग रहा है क्या एक कम करते हैं सर अगर बात नहीं बन रही है इन केस बन भी रहा हो पर आपके दिमाग में उसे वक्त आइडिया नहीं ए रहा है तो आपके पास सबसे अच्छा तरीका है श्रीधर आचार्य मेथड श्रीधराचार्य मेथड क्या कहे की सर वो कहेगा की सर आप सीधे सीधे निकालो एम की वैल्यू कितना एम की वैल्यू आप कहोगे माइंस बी यानी कितना माइंस फाइव प्लस माइंस अंडर रूट ओवर b² b² यानी 5 का स्क्वायर 25 - 4ac - 4ac मतलब मतलब माइंस माइंस अब देखो सुनना 16 का 10 टाइम्स कितना होता है 160 16 का डबल कितना होता है 32 16 में से 32 सूत्र करना है 160 में से 30 सब्सट्रैक्ट है तो 130 2 और सब्सट्रैक्ट किया तो कितना 128 कोई तकलीफ तो नहीं तो ये कितना ए जाएगा 25 प्लस ऑफ कोर्स 128 / 2a सो था 4 2 8 तो चीज कुछ ऐसे निकल कर ए रही है जो भी है वो आपके सामने है है ना तो सर यहां से आप एम की दो वैल्यू निकलोगे कौन-कौन सी फिर ये हो जाएगा -5 + -√ होगा सर 128 में 25 ऐड करना है तो 128 में अगर 28 किया तो 138 14 ट्रिकी आया है ऑल डी एग्जामिनेशन हॉल में इतने ट्रिकी नंबर्स आपको नहीं देंगे ताकि ऐसी कैलकुलेशन मैंने बट अभी फिलहाल बन गई है तो हम इसे वर्कआउट करते हैं ये आपका एम ए चुका है और एम की जब आप वैल्यूज रखेंगे कहां यहां रदर यहां जहां भी आपको ठीक लगे उससे आपकी दो इक्वेशन मिल जाएगी क्योंकि आप एक बार प्लस आएंगे उसे करेंगे माइंस साइन और वही दोनों इक्वेशंस इन दोनों टांगेंट्स की होंगी आई थिंक एक बड़ा ही बेसिक सा सिंपल सा ये एप्लीकेशन है इससे पहले वाले लेक्चर में जब हमने बात की थी की अगर स्लोप दे दी जाए तो इक्वेशन कैसे निकलती हैं इस बार हमने बात की अगर वो पॉइंट दे दिया जाए तो स्लोप निकल कर कैसे हम वो इक्वेशंस लिखते हैं कोई परेशानी कोई बेहतरीन सब बेसिक सा सिंपल मेकैनिज्म समझिए अब वो स्टैंडर्ड ऑपरेटिंग प्रोसीजर समझे की चीज है कैसे फ्लो हो रही है और आप आसानी से चीज वर्कआउट लेंगे कोई दिक्कत कोई परेशानी तो नहीं आई थिंक चीजें आसानी से समझ ए रही है कौन-कौन से हैं ज्यामिति कैलकुलस वेक्टर एंड 3D अलजेब्रा एंटरप्राइजेज ये सारी चीज हम डिटेल में करेंगे और बहुत अच्छे से करेंगे यहां पर जब बात की जाए तो बात बस इतनी सी अगर मैं आपको समझाऊं की देखो हम क्या करने वाले हैं हम क्या पढ़ने वाले हैं तो मैन लो अगेन ये आपका एक सर्कल है इस सर्कल में आपने क्या किया इस सर्कल में आपने ये किया की सर ये एक टैसेंट है एक्सटर्नल पॉइंट्स है और किसी एक एक्सटर्नल पॉइंट से ये extendent है ठीक है सर मैन लिया आपकी बात अब ये जो दो टांगेंट्स आप बना रहे हो सर ये जो दो आप बना रहे हो इन दोनों टांगें की पैर ऑफ टांगेंट्स तो या पैर ऑफ स्ट्रेट लाइंस वाले कॉन्सेप्ट कैसे लिखेंगे इस बार मैं चाहूंगा की आप एक आसान सा तरीका याद कर लें तरीका क्या होगा सुनेगा ध्यान से आपके पास सर्कल होगा होगा सर आपके पास ये पॉइंट होगा जहां से अपने वो टांगें ड्रॉ किए हैं बिल्कुल होगा बस इतना ही काफी है सर्कल के क्वेश्चन दे दी जाएगी आपको वही बात वापस जो की मेरे ख्याल से अब आप अच्छे से जानते हैं की आपकी सर्कल की इक्वेशन होगी इसे देखकर आप इसके सेंटर और इसकी रेडियस को निकलने में कोई कसर बाकी नहीं रखते अब अगर पूछा जाए की ये पैर ऑफ टैसेंट लिखनी है मुझे एक ही क्वेश्चन से याद है ना दो स्ट्रेट लाइंस के कंबाइंड क्लब इक्वेशन हम लिखा करते द कैसे सर याद करो हम लिखते द L1 और L2 दोनों का प्रोडक्ट ले लिया करते द याद ए रहा है स्टूडेंट्स पैर ऑफ स्ट्रेट लाइंस बारे में बात की थी तो उसी बात को आज हम इस तरीके से समझेंगे की डायरेक्ट कोई तरीका है क्या बिल्कुल है suniyega ध्यान से एक नई चीज आप sikhoge और पुरानी चीजों को रिकॉल करते हुए सीखो की देखो यह बहुत कम की चीज है suniyega आपको याद आएगी वो बोल रहा है मैन लो आपके पास एक सर्कल है वो अभिनेश अब एक स्टैंडर्ड सर्कल ले रहे हैं जैसे की x² + y² = a² तू मेक यू अंडरस्टैंड बाय जो की X1 y1 है यहां से अपने दो टांगेंट्स ड्रॉ की अब सीधा सीधा फैन है मैं मैन लेता हूं की अब एक लोकस में लेना चाह रहा हूं भाई ये ह को हम ए के अल्फा कमा बेटा अब इस पॉइंट ने कौन सा पथ ट्रेवल किया है इस पॉइंट ने यह पथ ट्रेवल किया है की वो या तो इस लाइन पर रह रहा है या इस लाइन पर रह रहा है suniyega ध्यान से बहुत कम की बात है मैं आपसे यह कहना चाह रहा हूं की अगर मैं एक इस लाइन पर पॉइंट लेता हूं किस बात पर या तो इस लाइन पर यह लाइन पर कहीं पर भी एक पॉइंट लेता हूं लेट्स से ह कॉम के तो क्या मैं यह का सकता हूं सर क्या मैं यह का सकता हूं की जो मैं पॉइंट ले रहा हूं इस पॉइंट की और स्पीक की जैसे मैं इस पॉइंट को का देता हूं लेट से थोड़ी देर के लिए कोई भी पॉइंट लेट्स से तो ये जो पॉइंट पथ ट्रेवल कर रहा है कौन सा वाला पार्ट ये पॉइंट इस लाइन पर और इस लाइन पर रह रहा है इन पर घूम रहा है इस पॉइंट की खास बात ये है की यह पॉइंट और यह पॉइंट से बनने वाली स्ट्रेट लाइन या आई शुड से स्ट्रेट लाइंस क्योंकि या तो ये स्ट्रेट लाइन या ये स्ट्रेट लाइन यह जो स्ट्रेट लाइंस हैं इनकी इस सर्कल के सेंटर से परपेंडिकुलर डिस्टेंस रेडियस के इक्वल होगी जो बात में कई दफा आप से का चुका हूं अगर मैं यह लोकस की कंडीशन सेटिस्फाई करवाना तो क्या मैं ये का सकता हूं की सर इस पॉइंट का जो पथ होगा इस पॉइंट का जो ट्रेवल्स पथ होगा इस पॉइंट का जो लोकस होगा वो आपको पैर ऑफ टैसेंट दे देगा यानी इन दोनों स्ट्रेट लाइंस की क्लब या कंबाइंड इक्वेशन क्या आप यह बात थोड़ी बहुत डाइजेस्ट कर का रहे हो कंडीशन समझ ए रही है हम लोकस निकलने के लिए कौन सी कंडीशन अप्लाई करेंगे कंडीशन होगी की इससे और इससे गुजरने वाली स्ट्रेट लाइन जो की या तो सर ये होगी या आज कॉम के यहां भी जा सकता है तो या तो ये होगी और इन दोनों स्ट्रेट लाइंस की इस सर्कल के सेंटर से परपेंडिकुलर डिस्टेंस इस सर्कल की रेडियस के इक्वल होगी मैं इस कंडीशन को अप्लाई करके वो लोग कस या वो इक्वेशन निकल लूंगा किसकी इस और इस लाइन की कंबाइन या अब डी क्वेश्चन आई होप ऑपरेशन क्लियर किया घुमा firakar जब बातें निकल कर आएंगे तो मैं नहीं करना चाह रहा हूं आपसे पुरी कहानी डिस्कस मैं सीधे कंक्लुजन पर आना चाह रहा हूं वापस जो सब ऑपरेशंस आप करके कनक्लूड करोगे वो ये होगा की आप सीधे सीधे लिखोगे सर ss1=t² सर ये क्या होता है सर यह क्या होता है ss1 = t² suniyega ध्यान से जब कोई आपसे का रहा है ss1 = t² तो याद करो स्टूडेंट्स ss1 = t² का मतलब क्या होगा ये ये बाई डी वे क्या है ये वो पैर ऑफ टांगेंट्स की क्लब डी इक्वेशन है ऐसे कैसे हैं सर इसका प्रूफ देने से ज्यादा जरूरी है मैं समझता हूं की ये क्या है पहले मैं आपको वो समझाऊं फिर हम प्रूफ पे वापस जाएंगे पहले तो एस क्या होगा एस आप हमेशा जानिए एस आपके सर्कल की इक्वेशन मतलब x² + ए स्क्वायर प्लस तू फी प्लस सी दिस इसे गोइंग तू बी दी क्वेश्चन याद रहेगा क्या अच्छा सर ठीक है बात आपकी मैन ली फिर एस वैन क्या रहेगा सर S1 कुछ नहीं होगा जिस एक्सटर्नल पॉइंट जिस एक्सटर्नल पॉइंट X1 y1 यानी जैसे आप का लो थोड़ी देर के लिए पी से आप वो पैर ऑफ टांगेंट्स ड्रॉ कर रहे हो उसे एक्सटर्नल पॉइंट को एक्स और ए की जगह पर प्लेस कर दो इसकी इक्वेशन में तो वो बना के दे देगा आपको S1 x1² + y1² + 2gx + 2X 51 + सी अब देखने को ही बहुत बड़ा एक्सप्रेशन दिख रहा है पर ये नंबर होगा क्योंकि यहां पे कोई नंबर आएगा ये भी नंबर है ये भी नंबर है ये भी नंबर है ये आइडिया सब्सट्रैक्ट एक सिंगल नंबर दे देंगे और वो नंबर इससे मल्टीप्लाई हो जाएगा क्योंकि आप लिख रहे हो एस एस वैन और क्या मुझे आपको याद दिलाने की जरूरत है की t² क्या होता है आप मुझे याद दिलाई है स्टूडेंट्स से t² तो हम जानते हैं t² तो हमने पढ़ा है t² से पहले मैं बात करूं अगर आपसे टी की तो ठीक क्या होता है सर सर x² की जगह एक्स एक्स वैन ए स्क्वायर की जगह ए ए वैन तू जी एक्स तो एक्स की जगह एक्स + x1/2 ये हो जाएगा जी टाइम्स एक्स + X1 2fy में आप क्या लिख दोगे एफ ए + y1 और सी से आप कोई छेड़खानी नहीं करेंगे तो ये जो टर्म बनेगी आप इसका कर देंगे स्क्वायर और वो इसके और इसके प्रोडक्ट के इक्वल रख देना सर वो आपकी पैर ऑफ टांगें की इक्वेशन हो गया उसने क्या किया पता है उसने आपको समझने के लिए एक सिंपल सा एग्जांपल लिया सिंपल सा एग्जांपल दो पूरा ये वाला सर्कल नहीं लिया उसने आपको समझने के लिए प्रॉब्लम को आसान बनाने के लिए उसने एक ऐसा सर्कल ले लिया विच इसे सेंटर डेट ओरिजिन और जिसकी रेडियस कितनी है ए और कंडीशन वही अप्लाई कर रहा है वो जो लोकस का इक्वेशन बनाने के लिए हम कंडीशन लगा रहे हैं की इसको मार्क की ओर इससे पास होने वाली स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस सेंटर से उसकी रेडियस के इक्वल है जब उसने यही कंडीशन अप्लाई की यही कंडीशन जब उसने अप्लाई की और इसमें कुछ एडजेस्टमेंट किए कुछ चीज है ऐड और सब्सट्रैक्ट की तो फाइनली कंक्लुजन पर पहुंचा की सर मैं कुछ इस तरीके से जनरलाइज कर का रहा हूं मैं कुछ इस तरीके से जनरलाइज कर का रहा हूं की सर्कल की इक्वेशन मैं सर्कल की इक्वेशन X1 और y1 पास कर रहे हो वैल्यूज का प्रोडक्ट नहीं है क्या आपका सर्कल अगर उसे बड़े सर्कल पूरे क्लब के बजाय ओरिजिनल तो ये क्वेश्चन होती है तो S1 क्या हो जाएगी सर X1 और y1 पास कर दो और टी क्या हो जाएगा ये आई होप इन बातों को तो आप अच्छे से अब जानते हो फाइनली आपको ये बात समझाना चाह रहा हूं की ये बहुत टू या बहुत कॉम्प्लिकेटेड कॉन्सेप्ट नहीं है पर आपको क्या याद रखना है आपको याद रखना है की अगर पैर ऑफ अगर मुझसे कभी भी कोई पैर ऑफ टांगें पूछता है मुझसे कभी भी कोई क्या पूछता है पैर ऑफ टांगेंट्स मतलब उन दोनों टांगें की कंबाइंड इक्वेशन प्लग की प्लग यू नो पीरियड जो स्ट्रेट लाइंस का है ना जो की tenjent से बेसिकली उनको अलग-अलग निकल के फिर उनका प्रोडक्ट लिखने की जरूरत नहीं है आप और आसान तरीके से का सकते हो ज्यादा खुद सोचने की जरूरत नहीं आपको तो इस पॉइंट से वो पैर ऑफ tesence निकल है ना की इस पॉइंट से ड्रॉ किए गए वो दोनों टांगेंट्स की इक्वेशन क्या है आप कुछ मत करो आप सर्कल की इक्वेशन लो सर्कल की इक्वेशन में वो X1 y1 पास करो उन दोनों का प्रोडक्ट लो और उसको t² के इक्वल रख दो क्वेश्चन खत्म हो जाएगा यस क्वेश्चन खत्म हो जाएगा तो अभी आपको इस लेक्चर में नोट्स में क्या बनाना है ये जो स्लाइड मैंने आपको यहां पर प्रेजेंट की है इससे पूरा अच्छे से नोट डाउन कर लीजिए बस यही आज के लेक्चर का रक्षा जो आपको याद रखना है और ये फैक्ट याद रखना है इस फैक्ट को याद rakhvane के लिए सबसे अच्छा तरीका मेरे पास जो रहेगा वो क्या होगा आय होप ये उसने किसके लिए कनक्लूड कर दिया आपके पूरे कंप्लीट वाले के लिए वो क्या का रहा है अगर सर कल आपका पूरा वाला है है ना तो S1 क्या हो जाएगा सर उसमें एक्स और ए की जगह आप क्या रिप्लेस कर दो X1 y1 ये सब और टी तो हम जानते ही हैं और फिर वही बात सर एस और S1 का प्रोडक्ट t² के इक्वल रख दो आपकी पैर ऑफ टांगें की इक्वेशन ए जाएगी जो आपने इस सर्कल पर ड्रोन की है ड्रा की है एक sunvayu से अपनी बातें समझ का रहे हैं और थोड़ी क्लेरिटी देता हूं मैं आपको इस क्वेश्चन के थ्रू suniyega बहुत मजेदार क्वेश्चन है मजेदार क्यों है सर क्योंकि ये समझना वो का रहा है आपसे फाइंड डी इक्वेशन ऑफ पैर ऑफ टांगें डॉन तू डी सर्कल फ्रॉम दिस पॉइंट पहले इतना करते हैं फिर आगे का क्वेश्चन करेंगे बाकी आगे का क्वेश्चन क्या है ये बहुत मजेदार बात है ये ये चीज आसान हो जाएंगे अगर पैर ऑफ टांगें लिखा तो करेंगे इस बारे में बात पहले तो सर हम हमारा जीवन आसान बनाते हैं तो सर्कल की इक्वेशन इन्होंने एस क्या होता है सर एस होता है सर्कल के क्वेश्चन जो की है x² + y² आपको एग्जाम में ऐसे नहीं लिखना है बस आपको समझ ए जाए ये चीज आप इन लिंगो से vakyap हो जाएं इसलिए मैं ये चीज अभी लिख रहा हूं ताकि आपको ये बातें समझ आने लगे बाकी आपको ऐसे नहीं लिखना है आपको तो फटाफट करना है अभी जैसे सर जैसे हमने फटाफट एस निकाला अब मैं क्या निकल लूंगा एस वैन तो S1 क्या हो जाएगा सर कुछ नहीं - 2 3 इन में रख दो तो -2 का स्क्वायर कितना फोर थ्री का स्क्वायर कितना 9 - 2 - 2 4 - 3 4 - 12 और ये -4 अगर मैं इसे सॉल्व करूं तो मुझे मिलेगा suniyega ध्यान से सर सीधी-सीधी बातें 4 से 4 कैंसिल 9 और 413 और 13 में से 12 सब्सट्रैक्ट किया तो वैन तो S1 कितना आया सर वैन कोई दिक्कत अब मैं अगर बात करूंगा तो टी होगा इस इक्वेशन में चीज एडजस्ट करनी है x² को करेंगे एक्स एक्स वैन सो - 2X कोई दिक्कत तो नहीं है ए स्क्वायर को रिप्लेस करेंगे ए ए वैन सो ये कितना हो जाएगा 3y -2x तो एक्स रिप्लेस होगा किस एक्स + X1 तो एक्स - 2 है ना तो यहां पर क्या यार प्लेस करूंगा मैं एक्स - 2 / 2 और ए की जगह क्या ए जाएगा ए + 3y + y1 तो ये क्या हो जाएगा y+3 और क्या बचाव देख रहा है सर यहां पर एक टर्म और बची दिख रही है माइंस फोर ये आपका क्या हो गया भाई सर ये हो जाता है टी बट इसमें और थोड़ा सिंपलीफिकेशन का स्कोप है क्या हान सर थोड़ी गुंजाइश तो है तो वही अप्लाई कर देते हैं देखिएगा ध्यान से सर सबसे पहले तो जो हमें दिख रहा है वो ये माइंस तू एक्स और - एक्स तो कितना हो जाएगा सही हो जाएगा -3x ये क्या निकल रहे हैं सर हम टी का सिंपलीफाइड वर्जन 3y और 2y तो ये कितना हो जाएगा सही हो जाने वाला है 5y ठीक है सर इसे भी रख दीजिए 5/ और क्या सर और अगर आप आगे देखें तो आपको दिखेगा माइंस माइंस प्लस तू प्लस तू प्लस सिक्स कितना हो जाएगा आपको चीज बहुत आसान दिख रही हैं मेरा आय होप आप यकीन कर का रहे होंगे किसी से बहुत सिंपल सी और सॉर्टेड से हैं अब हम कैसे लिखते हैं सर वो पैर ऑफ टांगें की इक्वेशन ये बेसिकली पता है किसकी इक्वेशन आई होप आप समझ का रहे हो सर्कल था आपका एक पॉइंट था बाहर से उसे पर उसे पॉइंट से इस सर्कल पर जो दो अपने tangej बनाई है ना इन दोनों टांगें की इन दोनों स्ट्रेट लाइंस की कंबाइंड और क्लब डेकोरेशन जो की हमने डिस्कस किया है पैर ऑफ स्ट्रेट लाइंस में ये दो स्ट्रेट लाइंस की एक सिंगल इक्वेशन भी लिखी जा सकती है बस वही आपको सीखने कहने बोलने या बताने की कोशिश कर रहा हूं क्या बात है समझ ए रही है आपको तो क्या करेंगे सर ss1 इसे इक्वल तू अब देखो एस एस वैन में ज्यादा सोचने की जरूरत नहीं है क्यों सर क्योंकि एस एस वैन तो S1 कितना है वैन तो वो पैर ऑफ इंजन की इक्वेशन क्या लिखेंगे में लिख दे रहा हूं अभी डिटेल में आपको एग्जाम में ऐसा नहीं करना है आपको फटाफट आंसर निकलना है बट अभी आपको समझ ए जाए अब कन्फ्यूज्ड ना रहे इस कॉन्सेप्ट में इसकी तो यहां पर आप क्या लिखेंगे इक्वेशन ऑफ टांगेंट्स है ना यार अगर इक्वेशन ऑफ पैर ऑफ टांगें है ना वो जो दोनों टांगेंट्स का पैर है उसे आप कैसे लिखेंगे आप लिखते हो उसे S1 = t² से तो एस क्या है सर एस है आपका x² + y² - 2X + 4y -4 यही है क्या आई थिंक सर यही है एक बार बस वेरीफाई कर लेता हूं - 2X + 4y -4 और इससे आप मल्टीप्लाई करेंगे S1 से इसे आप मल्टीप्लाई करेंगे S1 से जो की कितना है 1 = t² अब t² क्या है -3x+5y+4 वो क्या है सर वो है ध्यान से suniyega -3x प्लस फाइव प्लस फोर स्क्वायर वही है x² + y² - 2X + 4y - 4 = है अब ये ए + बी + सी का होल स्क्वायर सर a² + b² + c² + 2ab + 2bc+2c है - साइन लगा है चिंता मत करो अब वैसे ही ले लेना वो एडजस्ट हो जाएगा a² यानी -3x का स्क्वायर कितना सही हो जाएगा 9x² है ना 5y कितना हो जाएगा 25y² और 4 कितना हो जाएगा 16 अब डबल करके दो दो लेंगे है ना तो देखना 3 5 कितना माइंस 15 क्योंकि -3 है -152 कितना - 15 2 मैंने कुछ गलत तो नहीं लिखा है 16 हान बिल्कुल सही है ना फिर से देखते हैं -325 - 15 - 15xy क्या बात आपको समझ ए रही है कोई दिक्कत तो नहीं और क्या स्टूडेंट्स ध्यान से देखना अब 3X 4 - 12 - 12 का डबल कितना माइंस 24 तो ये हो जाएगा - 24x बहुत बढ़िया सर और क्या और अगर आगे मैं करता हूं तो देखना 5 4 20 20 का डबल 40 तो ये कितना हो जाएगा सही यहां पे एक और टर्म दिखेगी जो की होगी आपकी 40 बाय आई थिंक सर अब वक्त है की आप इसे फटाफट चीज आप आसानी से लिख रहे हैं तो मैं कम करता हूं इस पूरे को उठा के उधर रखता हूं तो नाइन नाइन एक्स स्क्वायर माइंस एक्स स्क्वायर बहुत बढ़िया फिर क्या सर फिर अगर आप देखें तो 25y² - y² तो कितना ए जाएगा सर ये होगा 24 y² किसी भी स्टूडेंट को कोई तकलीफ फिर सर 30xy का जो आप कुछ नहीं कर सकते तो इसे आगे इट इसे रख दीजिए - 30xy क्या आपको कुछ याद ए रहा है कुछ देजा वो टाइप हो रहा है की सर ऐसा तो कुछ हमें पढ़ा है पैर ऑफ स्ट्रेट लाइंस आपका सेकंड चैप्टर याद करिए आई थिंक थर्ड चैप्टर स्ट्रेट लाइंस था है ना फिर क्या सर फिर अगर आप देखें तो देखो आपको क्या दिख रहा होगा -24 और यहां से -2 उधर गया तो वो ऐड हुआ ओह सॉरी सब्सट्रैक्ट हुआ तो -24 + 2 यानी कितना - 22x तो यहां पर आप लिखेंगे -22 एक्स और क्या सर वहां लिखा है 45 होप आपके बाते समझ ए रही है ये टर्म्स कंसीडर कर लिया है ना वहां लिखा है 45 और यहां से गया है 4y तो फोर ए से स्टार्ट होगा तो कितना बचेगा सर ये बचेगा - 36 ए और क्या सर और सीधी सी बात है तो और अच्छा रहता कॉमन फोर ए रहा होता सर अगर कुछ चीज कॉमन ए रहा है तो जैसे ही तू कॉमन निकाला तो शायद मुझे एक क्वेश्चन सॉर्टेड सिंपलीफाइड सी दिखेगी क्या दिखेगी देखना 4x² + 12y² तो ये क्या देखेगी सर ये दिखेगी 4 x² + 12y² और क्या सर माइंस 15 एक्स ए अब मुझे जो आप सेल्फ चाहिए तो थोड़ा सा मैं हेल्प जरूर कर देता हूं हमने पढ़ा है सर ए x² प्लस बी ए स्क्वायर होती थी इसमें बेसिकली दो लाइंस मौजूद हुआ करती थी और आप मेरी बस इतनी सी हेल्प कर दो की यहां पर वो जो दो लाइंस मौजूद हैं उनके बीच के एंगल का फॉर्मूला क्या होता था बहुत आसान सा फॉर्मूला था जो हमने पढ़ा था बस अगर आप मुझे वो फॉर्मूला बता दें तो एक क्वेश्चन खत्म हो जाएगा अगर आप मुझे वो फॉर्मूला बता दें तो ये क्वेश्चन फ्रेक्शंस ऑफ सेकंड में खत्म हो जाएगा और आपका आंसर ए जाएगा बिना ज्यादा मेहनत किए बिना ज्यादा परेशान हुए बिना ज्यादा तकलीफ में आए हुए क्या मुझे जल्दी से अब वो फॉर्मूला बता सकते हैं स्टूडेंट सोच के देखो फॉर्मूला क्या पढ़ा है हमने आई होप आपको याद ए रहा है वो फॉर्मूला हमने पढ़ा है जो तन थीटा जो होता है वो कितना होता है क्यूट एंगल होता है दोनों लाइंस के बीच में वो होता है तू टाइम्स अंडर रूट ओवर स्क्वायर माइंस है यह फॉर्मूला आप अप्लाई कर लीजिए और आपका आंसर आपके सामने होगा बिना ज्यादा मेहनत किए बिना ज्यादा परेशान हुए बस एक बार चीज क्रॉस चेक कर लें कहीं कोई गलती तो हमने नहीं की क्वेश्चन में कहीं कोई मिस्टेक तो हमने नहीं किए आपकी सर्कल की इक्वेशन जो आपको दी गई है वो क्या है सर x² + y² - 2X है ना + 4y - 4 = 0 और ये है आपका -2 3 यहां से आपने जब tagent ड्रॉ करनी चाहिए तो क्लीयरली सर आप देख का रहे द की आपका ये जो वैल्यू ए रही है t² ये तो हम सर सीधे सीधे निकल पाए क्या t² तो आई थिंक किसी भी स्टूडेंट को भी तकलीफ नहीं आई होगी जब हमने टी सिंपलीफाई किया जो की फाइनली हमारा क्या ए रहा था - 2X है ना टी हमारा जो ए रहा है वह -3x + 5 + 4 ठीक है सर है ना और इसी तरीके से हमने एस और S1 भी निकले आई होप आपको एसआरएस में कोई तकलीफ नहीं बस अगर मैं एक बार क्रॉस चेक कर लूं S1 तो ये तू रखा तो कितना ए जा रहा है फोर ये ए जा रहा है 9 या ए जा रहा है 4 ये ए जा रहा है आई थिंक ये क्वेश्चन मैंने सही लिखिए बिल्कुल सही लिखी ये आज ए रहा है 12 मैंने -12 क्यों लिखा आई होप आपको ये गलती समझ ए रही है पता नहीं क्यों बस एक छोटा सा एरर मैंने किया है स्टूडेंट्स आई होप अब देख का रहे हो 3 4 ये होगा प्लस 12 ये कितना होगा प्लस 12 बस इसी छोटे से करेक्शन के साथ इस क्वेश्चन को ठीक करेंगे और आगे बढ़ेंगे बस थोड़ा सा पेशेंस रखिए ज्यादा फर्क नहीं आएगा इसमें यह कितना हो जा रहा है 9 और 413 और 12 कितने होते हैं 25 तो यहां पे आएगा बेसिकली 25 तो छोटा सा करेक्शन जरूर करेंगे स्टूडेंट्स यहां होगा 25 ऐसे छोटे से सीनियर मत करना स्टूडेंट्स देखो एक माइंस प्लस का फेर था पता नहीं क्यों ये थ्री इन फोर मैंने कैसे -2 लिख दिया बस एक माइंस और उससे पूरा आंसर गलत हो जाता है है ना और अगर मैं इससे थोड़ा आगे बढूं तो टी आई थिंक हमने सही नहीं निकाला था बस एक बार क्रॉस चेक कर लेते हैं तो टी कितना था सर - 2X बिल्कुल सही + 3y - एक्स - 2 और - + 2y + 3 - 4 बिल्कुल सही है ना इसे हमने सिंपलीफाई किया तो - 2X - एक्स - 3 एक्स 3y+2y5y और ये हो जाता है +2 और वो हो जाता है प्लस सिक्स तो तू प्लस सिक्स एट और एट माइंस फोर फोर बहुत से बस एक गलती की सर जो S1 आपका जो है S1 कितना होगा सर वो होगा 25 और उसे वजह से जो बदलाव आएंगे जरा उन बारे में उनके बारे में बात कर लें तो आई थिंक मैं इसे पूरा ही हटा दो अनलेस से रेली सारे चेंज में करूंगा इससे हमें इसे थोड़ा एक बार ये राइट कर लूं तो आपको भी परेशानी तूने 2 सेकंड लगेंगे स्टूडेंट्स और हम चीज है फिर से ठीक कर लेंगे ताकि हमारी कैलकुलेशन में कोई प्रॉब्लम नहीं सिर्फ एक छोटा सा प्लस माइंस वाली गड़बड़ की उससे पूरा आंसर गलत ए जाएगा है ना वो एंगल होगा लेकिन वो एंगल आपके ऑप्शन में मौजूद नहीं होगा सब कुछ सही है बस वो एक माइंस वाली गलती के कारण और ऐसी गलतियां होती हैं ऐसी गलतियां होती हैं है ना तो इनका थोड़ा सा ध्यान रखो स्टूडेंट्स जैसे की क्या बोलूंगा मैं आई थिंक सर 25x² तो ये सारी टर्म्स में क्या ए जाएगा सर यहां पर दिखेगा 25x² है ना फिर क्या दिखेगा सर ये दिखेगा 25y स्क्वायर और क्या दिखेगा सर माइंस 25 कितना - 50x और ये कितना हो जाएगा प्लस 100 ए और ये कितना हो जाएगा -100 कोई दिक्कत तो नहीं फिर यहां पे तो आई थिंक ये वाली टर्म आप लिख ही रहना चाहते तो अच्छा होता है क्योंकि ये तो वही है इसमें कोई फर्क नहीं है तो ये कितना था सर ये था 9x² है ना फिर ये कितना था और ये कितना था सर प्लस 16 फिर कितना ए रहा था 5 3 15 और 15 का डबल कितना 30 - 30xy और क्या सर - 3 4 हो जाता है -12 -12 का डबल - 24 तो माइंस 24x और यहां पे आप कहते हैं तो 5 20 का डबल 40 तो ये हो जाता है यहां तक कोई तकलीफ नहीं है सिंपलीफाई कर लीजिए आई थिंक अब आपकी लाइफ थोड़ी सॉर्ट आउट हो जाएगी अब सारी टर्म्स उठा के यहां लेकर आते हैं सुने जाते हैं यहां से मैं कहां से शुरू कर रहा हूं एक सेकंड के लिए ध्यान से देखना यहां से शुरू कर रहा हूं माइंस नाइन एक्स स्क्वायर गया तो कितना 16 तो पहली जो टर्म दिखेगी वो कितनी होगी सर वो होगी 16x² सही है क्या सर एक बहुत मजेदार सी बात ये हो जा रही है देखो 25 y² यहां भी है और वहां भी 25y² ही है तो सर ये y² और ये y² वाली टर्म पुरी कैंसिल तो ये बचेगा आपका जीरो y² ड्यूल गेट डेट यहां भी 25y² है और यहां भी आपका 25y² ही है क्या ये सारे स्टूडेंट्स को दिख रहा है की सर y² वाली टर्म कैंसिल हो जा रही है और थोड़ा हमें आगे बढ़ो तो सर एक्स ए वाली टर्म से कुछ लेना देना है नहीं तो वो ऐसे कैसे उठ कर ए जाएगी तो ये हो जाएगा प्लस 30 एक्सी कोई डाउट स्टूडेंट्स आई थिंक यहां तक तो कोई परेशानी नहीं है तो मैं इससे भी कंसीडर कर चुका हूं अब अगर मैं सर ध्यान से देखो तो ये है माइंस 50 एक्स और ये है प्लस 24x ये उधर आया तो ये हो जाएगा प्लस 24x मतलब माइंस से इधर ए जाएगा तो प्लस सो 50 में से 24 सूत्र क्या है तो 26 लेकिन नेगेटिव साइन का तो ये हो जाएगा माइंस 26 एक्स सही लिख रहा हूं क्या आई थिंक सही लिख रहा हूं फिर क्या करना होगा सर एक छोटा सा कम और करते हैं क्या सर ये है 100 वो है 41 तो 100 में से 40 अगर सब्सट्रैक्ट किया तो क्या आप मुझे बताएंगे कितना ए रहा है सर 100 में से अगर 40 सब्सक्राइब किया तो ये ए जाएगा 60 और ऑफकोर्स पॉजिटिव यानी क्योंकि पॉजिटिव -100 और कितना प्लस 16 कहीं और कोई और टर्म तो नहीं बच रही है लेकिन सर एक मिनट 16 प्लस नहीं है 16 - हो जाएगा तो माइंस 100 -16 मतलब ऐड होना है तो कितना ए जाएगा -116 ये कितना ए जाएगा -116 = 0 अब आप एक स्कोप जरूर देख का रहे हो की हान सर अभी भी सारी टर्म्स आपकी क्या है एवं और अगर सारी टर्म्स ए वैन है तो क्या कुछ कॉमन लिया जा सकता है हान सर कॉमन आने का स्कोप तो दिख रहा है और वही कॉमन लेकर ऐसे थोड़ा सिंपलीफाइड वर्जन में लिखूंगा तो सिंपलीफाइड वर्जन क्या होगा देखना सबका हाफ कर रहा हूं तो 16x² का हाफ कितना सर वो हो जाएगा आपका 8x² है ना ये टर्म है नहीं छोड़ दे रहा हूं बेसिकली वो है जीरो ताकि आपको बी कंपेयर करने में आसानी रहे फिर आप लिख लो 2h5 तो 30 का हाफ कितना 15 ये हो जाएगा 15 एक्स ए कोई डाउट्स आई थिंक नहीं है फिर क्या सर माइंस 20 एक्स मतलब यह हो जाएगा 13 और ये हो जाएगा तो 13x और 30y है ना तो माइंस 13 देखो 100 का आधा 50 16 का आधा 8 तो 58 लेकिन - 58 तो कितना हो जाएगा सही हो जाएगा माइंस 158 आई थिंक अभी हमने सही किया बस कस छोटे से सिली एरर का ध्यान देखो एक माइंस बस एक माइंस की वजह से पूरा सॉल्यूशन चेंज हो गया पूरा आंसर बदल जाएगा आई होप आपको ये बात समझ का रहे हो आई थिंक सर ये चीज तो हमने कहा पढ़ राखी है पैर ऑफ स्ट्रेट लाइंस में तो इस पे ज्यादा घबराने की या टेंशन लेने की जरूरत नहीं है आई थिंक यहां से चीज निकले तो ये क्या निकल कर आएगी ध्यान से देखना सर सीधा सीधा 10 थीटा निकलना है तन थीटा क्या होता है सर तू टाइम्स स्क्वायर माइंस अब सुनना ध्यान से कम की बात x² - अब अच्छा बी कितना है बताओ मैं सीधे-सीधे बोलना चाह रहा हूं ए यहां से निकलोगे तो ए कितना आएगा देखो भाई सर ए जाएगा 8 बी कितना आएगा सर बी दिख रहा है हम सभी को कितना जीरो ये मैं आपको समझने के लिए लिख रहा हूं ऐसा आपको नहीं करना है ना अच्छा सर ह कितना ए रहा है सर ह की अगर मैं बात करूं तो ह जो ए रहा है यहां से देखो 2hxy तो 2h कितना है 15 तो ह कितना ए रहा है सर ह ए रहा है 15/2 कोई डाउट अब suniyega ध्यान से क्वेश्चन कितना आसान बन गया सर जो आप एंगल निकलना चाह रहे हो उसके लिए आप जो कहोगे वो क्या कहोगे स्टूडेंट्स आप जो कहोगे तू टाइम्स है ना अब देखना tantheta जो ए रहा है वो ए रहा है तू टाइम्स अंडर रूट ओवर देखो भाई ए और बी का प्रोडक्ट तो जीरो 8 0 0 S2 कितना 15 / 2 तो 15 / 2 के स्क्वायर का अंडर रूट तो कितना बचा के देगा वो देगा 15 / 2 आई होप आप मेरी बात समझ में तो न्यूमैरेटर में आपको क्या मिलेगा 15 / 2 कोई डाउट तो नहीं है आई थिंक नहीं है अच्छा डिनॉमिनेटर में कितना ए जाएगा ए + बी यानी 8 + 0 तो आई थिंक आपका आंसर अब ए चुका है बस इसे आप सॉल्व कर लीजिए शायद आप तन थीटा निकल चुके हो आय होप यहां तक कोई डाउट नहीं है क्या ए रहा है भाई जल्दी बता दो मुझे तन थीटा कितना ए रहा है सर देखो आप सीधी-सीधी बात है तू से तू कैंसिल हो जा रहा है 15 / 8 तो 10 थीटा अगर आया 15 / 8 तो थीटा कितना होगा सर थीटा होगा तन इन्वर्स 15 / 8 तो ये होगा आपका उन दोनों टांगें के बीच का एंगल आलो जो दोनों टांगों से हमने ड्रॉ की थी ना उनके बीच का थीटा इस गिवन की दिस क्या ये बहुत कॉम्प्लिकेटेड सी बात है आई थिंक सर ये सब तो हमने पढ़ा है और इस तरीके के क्वेश्चंस पूछे जाएंगे मतलब वो शायद आपसे डायरेक्टली ना पूछे की आप पैर ऑफ टांगेंट्स बताओ पर उन दोनों के बीच का क्या बताओ एंगल मैं कहूंगा एक अच्छा तरीका है सर क्लब डिक्टेशन निकालो और ये फॉर्मूला लगा जोर बना दो सर इसके अलावा आसान तरीका नहीं है क्या है कैसे सर आप दोनों की अलग-अलग क्वेश्चन निकालो दोनों का एम निकल लो वही फॉर्मूला वापस की स्लोप फॉर्म में आप इक्वेशन निकल लो दो स्लोप ए जाएंगे तो दो टांगेंट्स की एक क्वेश्चन हो जाएंगे दोनों का प्रोडक्ट कर दो तो उससे आपकी वो कंबाइंड इक्वेशन ए जाएगी पैर ऑफ 10 दोनों की श्लोक मुझे पता होगी तो दोनों की अगर मुझे स्लोप पता होगी तो दो लाइंस की स्लोप पता होने से क्या उनके बीच का एंगल में निकल सकते हो बिल्कुल सर वैसे निकलेंगे तो पैर ऑफ टांगेंट्स या पैर ऑफ स्ट्रेट लाइंस तो एक एक्सटेंडेड कॉन्सेप्ट है कहां से वहीं से जो पैर ऑफ जो आप स्ट्रेट लाइंस पढ़ा करते द भाई तो इस क्वेश्चन को ऐसे नहीं करना है तो ऐसे तो कर ही सकते हो ना जैसे आपने सिखा है तो एक तरीका ये जो एक अच्छा तरीका है मेरे ख्याल से पर ये ना जा मैं तो जो तरीका हम पढ़ते आए हैं वो कौन से तरीके की बात कर रहे हो मैं आपसे का रहा हूं बड़ी सिंपल सी बात की ये आपका सर्कल है इस सर्कल पर किसी एक एक्सटर्नल पॉइंट्स आपने टैसेंट ड्रॉ की है उसे एक्सटर्नल पॉइंट के cardinate दे दिया उसने क्या X1 और y1 जो भी आपको दिए तो आप क्या बोलोगे सर की इस पॉइंट से ड्रॉ की गई कोई टेंशन की कोई ना कोई स्लोप होगी तो मैं क्या कहूंगा उसे ए - y1 = एम टाइम से एक्स - X1 और जो मुझे सर्कल दिया है उसके सेंटर से इस पर जो परपेंडिकुलर ड्रॉप किया है तो उसे परपेंडिकुलर लेंथ को निकल लूंगा वो जो -3 का वो माइंस एफ है उससे इस स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस निकल लूंगा तो वो एक फॉर्मूला बन जाएगा वो एक फॉर्मूला बन जाएगा कुछ ना कुछ फाइनली सॉल्व करोगे तो एम में एक क्वाड्रेटिक बन जाएगी और एम की आपको दो वैल्यू मिलेगी एम की दो वैल्यू क्यों मिलेगी सर क्योंकि एक आपकी इस लाइन की स्लोप होगी और एक आपकी साइन किस स्लोप होगी वो जो दो स्ट्रेट लाइंस बनेगी उनका प्रोडक्ट कर देना आय होप आप बात समझ का रहे हो वो जो जो स्ट्रेट लाइंस बनेगी उनका प्रोडक्ट कर देना तो वही आपकी पैर ऑफ स्ट्रेट लाइंस बनेगी और ये वही बना के देगी क्या आपको फाइनली ये जो इक्वेशन हमने यहां निकल थी ये भी आपको यही बना कर दे देगी फाइनली आपको ये यहां पहुंचा देंगे पर वो थोड़ा कॉम्प्लिकेटेड तरीका है मेरा ऐसा मानना है और अगर मुझसे पूछा जाता जो पूछा गया था की इन दोनों के बीच का एंगल कितना है तो सर एक लाइन जिसकी स्लोप है m1 और दूसरी लाइन जिसकी स्लोप है M2 तो मैं दो लाइंस के बीच में एंगल निकलना जानता हूं वो कितना होता है तन थीटा कितना स्टूडेंट्स याद करो m1 - M2 / 1 + 1 M2 तो आप देख सके ये तरीका उसे कर सकते हो इसमें कोई बुराई नहीं है ये तरीका आपके पास हमेशा है तो मैं का रहा हूं ये तो हमने पढ़ा हुआ है ना ये तो हमने सिखा हुआ है तो ये अप्लाई करने में कोई बुराई नहीं है पर मैं का रहा हूं की इससे भी तो एक आसान तरीका है सर कैसे आसान हो गया आप क्या निकल लो एस पता है S1 निकल लो t² निकल लो कितनी दो तीन लाइन में तो आंसर ए गया पुरी मेहनत से बच गया परपेंडिकुलर डिस्टेंस क्वाड्रेटिक उसके सॉल्व करना रूट्स निकलना कुछ नहीं करना पड़ा बस तो यहां पर नंबर थोड़े टिपिकल नंबर नहीं देखा कैलकुलेशन बहुत टफ हो जाए तीन या चार स्टेप में आपकी इक्वेशन ए जाएगी बहुत जल्दी और बस ये फॉर्मूला लगा की आप उनके बीच का एंगल दे सकते हो तो क्या मैं यह का सकता हूं की सर शायद यह थोड़ा बेहतर तरीके का है तरीके दोनों हैं आपके पास आपकी मर्जी आप जो उसे करना है जिसके लिए आप कंफर्टेबल पर मैं प्रिफरेबली आपको सजेस्ट करूंगा हाईली रिकमेंड करूंगा यार आदर इंसिस्ट करूंगा की आप प्लीज ये मेथड सीखे ये मेथड क्या कहती है सर की अगर किसी एक्सटर्नल पॉइंट से अगर किसी एक्सटर्नल पॉइंट से आप पैर ऑफ टांगें बनाते हो तो उन पैर ऑफ टांगेंट्स की क्लब कंबाइंड टुगेदर जो एक सिंगल इक्वेशन दी जाती है वो इस ss1= t² से दी जाती है आई होप आगे याद रखेंगे भूलेंगे नहीं क्योंकि यही आज के इस लेक्चर की क्या है लर्निंग पंच पंच में डिवाइडेड है cardinate ज्यामिति कैलकुलस एंड 3D अलजेब्रा और ट्रिगोनोमेट्री इनमें आपका पूरा 11 12 आईआईटी जी मांस और एडवांस का सिलेबस कंप्लीट है इट्स के हर चैप्टर के हर कॉन्सेप्ट को हम डिटेल में पढ़ने वाले हैं और हर चीज को बहुत अच्छे से सीखने वाले हैं अब अगर मैं सीधे आपसे मसाले की बात करूं तो मसाला यह है की मैन लो आपके पास एक सर्कल है बिल्कुल है सर इस सर्कल में लेट से देयर इसे अन कार्ड ठीक है सर देयर इसे अन कार्ड नजर हमारी भी तो एक स्ट्रेट लाइन है ना तो मैंने उसे थोड़ा एक्सटेंड किया है कोई परेशानी तो नहीं है इस बात से सर अब एक कम करते हैं यहां से एक टैसेंट ड्रा करते हैं तो सर एक टांगें तो रही है जो की यहां से जा रही होगी जो की इस सर्कल kafkarj यहां टच कर रही होगी इस पॉइंट पे सिमिलरली इस पॉइंट पे जब इसने टच किया ना तो मैंने यहां से भी लेट्स चेक टैसेंट ड्रॉप किया मैंने यहां से भी सर एक टैसेंट लेट से ड्रॉ की क्या बात है आपको समझ ए रही है हमने क्या किया हम एक सर्कल दिया गया और सर्कल में एक और थी उसे कार्ड के पॉइंट ऑफ कॉन्टैक्ट द कौन-कौन एक यहां और एक यहां और उन दोनों पॉइंट्स पर हमने क्या किया दोनों पॉइंट से पास होने वाली टांगें ड्रॉ की जो ऑफ कोर्स कहीं ना कहीं इंटरसेक्ट कर रही होंगी सर ये सारी बातें तो कुछ-कुछ हद तक हमने पढ़ी है की सर किसी पॉइंट से अगर आप दो टांगें ड्रॉ कर रहे हो तो टांगें की पैर एक्शन लास्ट लेक्चर स्टूडेंट्स और इसी तरीके से बहुत सारी और भी चीज एक्सप्लोर किए और उन्हें कुड़ियां को आज हम आगे बढ़ाएंगे इस लेक्चर में कोई दिक्कत कोई डाउट तो नहीं अभी तक बड़ी बेसिक सी बात है मैं आपसे का रहा हूं समझ ए ही रही होंगी कानन तौर पर अब अगर इस कॉन्सेप्ट को हम समझना चाहे तो आपसे जो मेरा कहना है वो बस इतना सा स्टूडेंट्स suniyega ध्यान से अगर मैं tenjings ड्रा करता हूं कहां से वही एक पॉइंट पी X1 से एक एक्सटर्नल पॉइंट ऑफ कर सकती जैसे आप मैन लेते हो ये यही बातें वापस जिसके cardinates हैं X1 y1 ये आपका पी पॉइंट है इससे फ्रॉड सर आप दो टांगें ड्रॉ कर पाते हो किसी भी सर्कल पर किसी भी एक्सटर्नल पॉइंट से यू कैन ड्रॉ मैक्सिमम तू टांगें आते आल विल यू ऑल रिमेंबर डेट अब क्या सर अब अगर मैं इसमें आगे बात करूं तो ये जो दो आपने यू नो टेंसेज ड्रा की है वो सर्कल को लेट से के और आर 2 पॉइंट्स लेट्स से के और आर पर टच करते हैं क्योंकि इंजन से है ना तो के और आर के कार्ड लेता हूं लेट्स से एक्स तू Y2 और लेट्स से X3 y3 अब मैं पूरे डेरिवेशन को आपको डेमोंस्ट्रेट नहीं करूंगा पूरे यू नो एप्रोच को डेमोंस्ट्रेट नहीं करूंगा अगेन आस्क यू तू रिमेंबर डी डायरेक्ट कंक्लुजन जो रिजल्ट आपके कम पे जैसे लास्ट लेक्चर अगर यहां से कोई टेंशन ड्रॉ की जाए तो उन टेंसेज के पेड एक्शन क्या होती है ss1=t^2 और यही कंक्लुजन यही क्रैकर्स हमारे कम आने वाला है इसी तरीके से आज भी कुछ ऐसा ही डिस्कस करेंगे अब suniyega ध्यान से अब अगर मैं कुछ अपनी बेसिक से ज्यामिति अप्लाई करो जो सब कुछ हम पढ़ते हैं की सराफ कोर्स आपके सर्कल की एक इक्वेशन होगी बिल्कुल सर होगी है ना अब ऑफ कोर्स सर टैसेंट जो पीके की इक्वेशन होगी वो मैं लिख लूंगा टी = 0 से आप समझ का रहे हो तो टेंशन की इक्वेशन कैसे लिखते हैं सर मैं अगर एक दूसरा गणित लगाऊं दूसरा गणित क्या कहता है सर एक गणितीय कहता है की सर आप तो बड़ी बेसिक सी बात बताओ ये के पॉइंट है तो के पॉइंट पर सर्कल पर अगर टांगें ड्रॉ की है तो कैसे आएगी याद ए रहा है ना स्टूडेंट सर्कल की इक्वेशन अगर x² + y² + 2gx + 2fy प्लस सी इसे इक्वल तू जीरो तो आप कैसे टेंशन ड्रॉ करते हो मैं कहूंगा सर के से पास होने वाली इस सर्कल पर जो टैसेंट होगी वो होगी तीस इक्वल्स तू जीरो बट स्टीम में कौन पास होंगे X2 Y2 यानी एक्स एक्स तू + ए / 2 + जी एक्स + x1/x1 / 2 तो 2 से 2 का आंसर प्लस एफ तन ए प्लस y1 और यहां पर सी तो वो किसकी इक्वेशन हो जाएगी इस टैसेंट पी के की सिमिलरली क्या मैं पी आर की भी इक्वेशन निकल सकते हो कोई टू तास में ही वही टी = 0 अप्लाई करेंगे लेकिन किस तरीके से सर यहां पे X3 ए सी ए जाएंगे तो एक्स एक्स थ्री ए बाय थ्री जी टाइम से एक्स + X3 एफ टाइम्स ए + y3 + सी = 0 तो ये आपकी दोनों टांगेंट्स पी आर और पीके की इक्वेशन ए जाएंगी मेरा बस ये कहना है आपसे की अगर मैं पी आर और पीके को सॉल्व करूं अगर मैं इन दोनों टांगें पी आर और पीके को सॉल्व करूं तो क्या मैं ऐसा कहूं सर की उन दोनों को सॉल्व करने से इन दोनों का पॉइंट ऑफ इंटरसेक्शन पी एंड X1 y1 मिल जाएगा क्या एक और तरीके से अगर मैं आपसे बात कहूं तो कैसा का सकता हूं की सर पीके टांगें क्या पर टांगें दोनों पर एक पॉइंट पी X1 y1 लाइक करता है सो इट मस्ट सेटिस्फाइड डी इक्वेशंस पीके एंड पी आर बोथ वुडन इट आई होप ये स्टेटमेंट आप समझ का रहे हो की ऑफ कोर्स सीधी सीधी सी बात है जो अपने पीके इंजन की क्वेश्चन की है जो पी आर टेन के क्वेश्चन लिखी है उसे X1 कमा ए वैन सेट है एस ए करेगा तो बस इसी बात तक मैं आना चाह रहा हूं इसी की अगर X1 वायरल सेटिस्फाई करेंगे तो आपकी पहली बात तो ये दोनों टांगें की इक्वेशन ए गई है और ये दोनों ही टेंशन जो आपकी हैं ये जो दोनों टांगें से पी आर और पीके वो X1 y1 से पास होती है तो वो X1 y1 को क्या करेंगे सेटिस्फाई हमने क्या किया इस इक्वेशन में जहां-जहां एक्स ओर ए द वहां 10 और ए वैन रखा इस इक्वेशन में भी जहां-जहां एक्स ओर ए द वहां भी 10 और ए बन रहा कोई तकलीफ आई थिंक कोई परेशानी नहीं है इन सारी बातों को मध्य नजर रखते हुए आप एक फाइनल कंक्लुजन पर पहुंचेंगे सुनेगा ध्यान से की आप एक्चुअली के और आर पॉइंट के लिए एक कॉमन इक्वेशन कनक्लूड कर पाएंगे ये सर जो के और आर जो है ना वो इस लाइन पर लाइक करते हैं जो के और ए रहे हैं जब इन दोनों को आप सॉल्व करेंगे और आप चीज निकलेंगे जब आप X1 X2 और X3 या y1 Y2 और y3 के लिए जो दोनों इक्वेशन सॉल्व करेंगे तब आप ये कंक्लुजन पर आएंगे किस कंक्लुजन पर आएंगे suniyega ध्यान से ये जो आपने निकाला है ना ये जो वनप्लस 5 वनप्लस ग्रंथि 6 प्लस 61 प्लस 5 प्लस 5 प्लस सी ये ऑफ कॉस्ट = 0 पर किसके लिए X1 y1 के लिए तो ये जो इक्वेशन आई है ना दिस इसे दी इक्वेशन ऑफ कोड ऑफ कॉन्टैक्ट विद रिस्पेक्ट अब मैं इसकी बात आपसे करना चाह रहा हूं आज के लेक्चर का बस यही रख है कोड ऑफ कॉन्टैक्ट विद रिस्पेक्ट तू पॉइंट पी suniyega बहुत कम की बात है बस इसी पे बात हुई suniyega ध्यान से बहुत कम की बात है जैसे मैन लो आपके पास कोई क्वेश्चन आया कोई क्वेश्चन आया सुनना है ना कोई क्वेश्चन क्या है जैसे की आपके पास कोई क्वेश्चन नहीं आया की सर एक सर्कल है उसे सर्कल की उसने एक क्वेश्चन दे दिया आपको कुछ भी लिख कर एक्स स्क्वायर प्लस ए स्क्वायर वह यह कहता है मैन लो की सर इस सर्कल पर मैंने किसी एक्सटर्नल पॉइंट लेट से किसी एक्सटर्नल पॉइंट X1 कमा y1 जैसे हम दिनो कर देते हैं पी से इससे मैंने टांगें ड्रॉ की ठीक है सर आप tangenced ड्रा कर लो हमें कोई परेशानी नहीं तो जैसे ही मैंने इससे tangenced ड्रा किया तो एक tagent ऑफ कोर्स किए जा रही होगी है ना और एक टांगें ऑफ कोर्स आपकी क्या जा रही होगी ये वाली मुझे सर इन tesence में कोई इंटरेस्ट नहीं है आई हैव नो इंटरेस्ट इट ऑल इन डीज तू टेंडनंस व्हाट आई एम आई इंटरेस्टेड इन इस की इन दोनों टांगें ने इस सर्कल को जिन पॉइंट्स पे टच किया है उन पॉइंट्स में भी हमें कोई इंटरेस्ट नहीं है जिन पॉइंट्स पर टच किया है अगर उन पॉइंट्स को मिलाकर एक लाइन बनाई जाए जो की ऑफ कोर्स जो की ऑफ कोर्स इस सर्कल की क्वाड होगी क्या मैं बस इस पॉइंट और इस सर्कल की इक्वेशन की हेल्प से उसे क्वाड की इक्वेशन लिख सकता हूं बिल्कुल लिख सकते हो हम इस बात को करना चाह रहे हैं क्या आप कोई बात समझ ए रही है और सच में मेरा यकीन करिए उसकी इक्वेशन बस इतनी सी है suniyega ध्यान से ये खास कार्ड है जिसे हम कहते हैं कॉर्ड ऑफ कॉन्टैक्ट ऑफ दिस सर्कल विद रिस्पेक्ट तू पॉइंट पी ऑफ कोर्स विद रिस्पेक्ट तू पॉइंट पी क्या आप मेरी बात समझ का रहे हैं मतलब सर ग = 0 मतलब हमने कुछ पढ़ा है कुछ तो याद ए रहा है क्या कुछ याद ए रहा है आपको जो मैं बात करना चाह रहा हूं जो आपको याद आना चाहिए वो ये की X1 y1 यहां रिप्लेस कर दीजिए तो ये वापस क्या हो जाएगा एक्स एक्स वैन प्लस ए ए वैन प्लस X1 आई होप आप मेरी बात समझ का रहे हो प्लस ऑफ कोर्स एफ टाइम्स ए + y1 + सी = 0 ये जो इक्वेशन है और इस कॉर्ड को मैथमेटिकली हम ज्यामितीय में सर्कल के कॉन्टेक्स्ट में कैसे दिनो करते हैं vikolet कोड ऑफ कॉन्टैक्ट कोड ऑफ कॉन्टैक्ट विद रिस्पेक्ट मिन व्हाट दस इट इमले व्हाट दस इट इंडिकेट की सर यह एक्सटर्नल पॉइंट पी से जब आपने दो टांगेंट्स drawking सो ने सर्कल को जिन पॉइंट्स पर टच किया उन पॉइंट्स को कनेक्ट करने से जो कॉर्ड बनती है उसे कोड के क्वेश्चन इतनी आसानी से लिखी जा सकती यू नीड नॉट एवं फाइंड डी स्लोप ऑफ दिस स्ट्रेट लाइन यू नीड नॉट एवं फाइंड डी पॉइंट ऑफ पॉइंट्स थ्रू विच डी लाइन पासेस यू बेसिकली व्हाट यू नीड तू हैव दिस कौन सा सर्कल है उसे पर कौन से एक्सटर्नल पॉइंट से आपने ड्रॉ किए बस वो पॉइंट और सर्कल के क्वेश्चन दे दो हम बना देंगे गोद ऑफ कॉन्टैक्ट इसका कंक्लुजन क्या था इसका कंक्लुजन आता था की हमने इस तरीके से चीज निकल की सर हमने मैन लिया X2 Y2 और X3 y3 पर हो जाती है तो X2 Y2 से एक वो हमारी तेनजिंग बन जाएगी हम नई कंडीशन अप्लाई कर दी की हमारा पी पॉइंट उन दोनों को सेटिस्फाई करेगा और उसकी हेल्प से मैंने X2 व्हाइट और X2 y3 को एल्यूमीनियम करने की कोशिश की जिस radhoj है हमें मैं अपनी कोड ऑफ कॉन्टैक्ट की इक्वेशन बना लाया है बना पाए बस यही हमारी एप्रोच यही हमारी मेथड है की हम उसे इक्वेशन को बना पाए जिस पे के और आर लाइक करते हैं और के और आर अगर उसे इक्वेशन पर लाइक करते हैं मतलब वो हमारी क्या है वो हमारी कार्ड ऑफ कॉन्टैक्ट है हमने ये उसे किया बस इतनी सी बात है कोई तकलीफ और सर ये तो हम जानते हैं की अगर पॉइंट पी कर्व पर ही ए गया तो फिर वो क्या बन जाएगी आपकी आई होप मुझे कुछ कहने बताने की जरूरत नहीं मैन लो सर अगर ये पॉइंट पी जो था अगर ये पॉइंट पीस कर्व पर ही ए जाता तो फिर तो ये टांगें बन जाती है ना फिर कोर्ट ऑफ कॉन्टैक्ट कैसे बनती है आप बात सीधी-सीधी सी है सर ये तो बात हमने शुरुआत से पढ़िए आई थिंक मेरा जो पहला लेक्चर रहा होगा टांगेंट्स के ऊपर उसे दौरान तुमने शायद यही बात की थी की सर्कल है अगर पी पॉइंट बाहर नहीं है मैन लो वो पी पॉइंट इस कर्व पर ही लाइक कर रहा है जिसके cardinate क्या है सर X1 y1 है ना तो अब कोड ऑफ कॉन्टैक्ट क्या होगी सर पी से पहले आप टेंशन ड्रॉ कर लो लेट्स से मैंने पी से tenjent ड्रॉ की तो पी से जब एक टैसेंट जा रही है तो ये रही वो टैसेंट तो अब अगर मैं कोड ऑफ कॉन्टैक्ट बनाऊंगा तो क्या होगी सर अब अगर मैं टी = 0 दूंगा ना आप अगर मैं टी = 0 दूंगा तो वो ऑफ कोर्स ये रही ये रहेगा ये xx1 ए ए वैन प्लस ऑफ कोर्स जी टाइम्स एक्स + एक्स ए + y1 + सी = 0 लेकिन ये जो अब इक्वेशन बन रही है यह जो इक्वेशन बन रही है आपको कॉन्टैक्ट बनने का स्कोप ही नहीं है ना क्योंकि इस टैसेंट ने इस टैसेंट ने जो पी से निकली है जो पी से गुजर रही है वो टेक्निकल पी पर ही इसकी टैसेंट है वो कर पर ही लाइक कर रहा है तो सर ये अब उसे टैसेंट की इक्वेशन बन गई सर पी बाहर होता तो वह दो पॉइंट पर टच करती फिर उनके इंटरसेक्शन जो वह कर सेंटर से कर दी उन पॉइंट से पास होने वाले कोड ऑफ कॉन्टैक्ट बना था मैं विद रिस्पेक्ट तू पी वो भी यही होती पर यह तो यह सिनेरियो बन गया की अगर पी उसी पर लाइक कर गया तो फिर वो कोड ऑफ कॉन्टैक्ट नहीं बनेगी वो वो टैसेंट बन जाएगी और स्ट्रांग के क्वेश्चन ये देना जानते हैं की अगर किसी कर पर कोई पॉइंट लाइक करता है तो scurv में टी = 0 लिख दो वही उसे कर्व की टांगें के क्वेश्चन होती है उसे पॉइंट पर जो पॉइंट उसे कर पर लाइक करता है यह बात मैं पैराबोला एलिप्स हाइपर बोला या किसी भी कर के रिस्पेक्ट में भी बोलूंगा तो अभी तुम सर्कल के रिस्पेक्ट में पढ़ रहे हैं बट इसे याद रखिएगा बहुत कम की बात याद है ना मैंने डिस्कस किया है की सर डर से भी स्लोप निकल सकते हो आप अपनी बेसिक अंडरस्टैंडिंग से वो माइंस 00 - एक्स से भी निकल सकते हो पर इतना सब कुछ करने से अच्छा तरीका हमने क्या पढ़ा है टी = 0 बस इसे मत भूलना आय होप ये बातें आप याद रखोगे तो बस वही कहना चाह रहे हैं की अगर वो कोड ऑफ कॉन्टैक्ट बनाते कभी ये हो जाए की वो जो एक्सटर्नल पॉइंट पी था वो scurv पर ही लाइक करने लग गया तो कोड ऑफ कॉन्टैक्ट बनेगी ही नहीं उसके इसमें क्या बन रही होगी उसे केस में इस सर्कल की टांगें बन रही होगी जो पी से पास हो रही होगी किसके थ्रू टच = 0 के थ्रू कोई तकलीफ कोई परेशानी कोई डाउट स्टूडेंट्स क्या यह बातें अच्छे से आपको समझ ए रही हैं बिना किसी फ्लकचुएशन हेसिटेशन या कन्फ्यूजन के और अगर ये बातें समझ ए रही हैं तो मेरा तो हमेशा से माना यही रहा है की बातें अगर समझ ए जाएं आपको तो आप जस्टिफाई कर देंगे इन क्वेश्चंस को सॉल्व करके और क्वेश्चंस सॉल्व करने का हमारा प्रोटोकॉल तो बस इतना सा है सर की आप क्वेश्चंस खुद से एक बार ट्राई करिए मतलब आप मतलब आप आप ट्राई करिए स्क्रीन पास करके क्वेश्चन ट्राई करिए और देखिए अगर आप कर पाते हैं अगर नहीं भी कर का रहे हैं तो घबरा मत जब मैं क्वेश्चन आपको एक्सप्लेन करूं तो एक बार फिर से स्क्रीन पीओएस करिए और फिर से ट्राई करिए फिर भी ना हो तो थोड़ा बहुत सॉल्यूशन देखिए और फिर ट्राई करिए पर हमेशा क्वेश्चंस खुद से ट्राई करते रहिए थोड़ी हेल्प लेकर कारी है पर करिए पूरा क्वेश्चन देखकर समझ ए जाए की सर नहीं पूरा हमसे तो नहीं हुआ आपका सॉल्यूशन देख कर आया तो फिर से एक बार आप ट्राई करिए पर जब तक आप चीजें नहीं करेंगे बात नहीं बनेंगे मुझे करते हुए देख कर आपको समझ जरूर ए जाएगा पर आपकी योग्यता आपकी दक्षता आपकी काबिलियत आपकी एबिलिटी डिवेलप करने में एक ही चीज हेल्प करेगी वो है आपका खुद से क्वेश्चंस को सॉल्व करना इस क्वेश्चन को ट्राई करते हैं स्टूडेंट्स क्या लिख रहा है वह का रहा है tenjent आर ड्रोन तू दिस तो देखो उसमें एक अच्छा सा कर्व दे दिया है हमको बहुत शॉर्ट ट्रिक्स जिसका टी = 0 ढूंढना बहुत मुश्किल कम नहीं है फ्रॉम अन्य आर्बिट्रेरी पॉइंट पी तो एक आर्बिट्रेरी पॉइंट पी है जो स्ट्रेट लाइन पर लाइक करता है आपसे ये का रहा है प्रूफ डेट कॉरस्पॉडिंग ऑफ कॉन्टैक्ट पस्त थ्रू अन फिक्स्ड तू पॉइंट एंड फाइंड डेट पॉइंट बहुत ही अच्छा बहुत ही बेसिक बहुत ही सिंपल सा डायरेक्ट सा क्वेश्चन है मेरा यकीन करिए इसमें बहुत बेसिक सी और सिंपल सी चीज दी गई है कैसी है सर ध्यान से सुनना पहले तो एक बात बताओ सर क्या मैं सर्कल ड्रा कर सकता हूं सीरियस सर्कल जो है ये बेसिक सेक्स सर्कल है विच इस सेंटर डेट ओरिजिन है ना विच इस सेंटर डेट ओर 0 इसकी रेडियस 7 सो दिस सर्कल इस सेंटर डेट ओरिजिन और इस सर्कल की अगर मैं आपसे रेडियस पूछूं तो वो कितनी है वैन यहां तक तो कोई डाउट नहीं है तो हम देख कर इट्स क्यूट अप्रेंटिस वेरी एवीडेंट के सर हान यही दिख रहा है वह का रहा है एक स्ट्रेट लाइन एक स्ट्रेट लाइन है कौन सी 2X + ए - 4 है ना सो स्ट्रेट लाइन है लेट्स से 2X + 5 - 4 मैं बना लेता हूं एक्स स्ट्रेट लाइन है ना ये ले वो स्ट्रेट लाइन है जो की है क्या जो की है 2X + ए - 4 अब वो आपसे ये का रहा है suniyega ध्यान से कम की बात की सर इस लाइन पर एक रैंडम आर्बिट्रेरी पॉइंट है मतलब ये पॉइंट मूव करता है इस लाइन इस पॉइंट से जब मैंने कोई कार्ड ड्रॉ की इस पॉइंट से मैंने कोई कार्ड ड्रॉ की तो क्या आप मुझे बता सकते हो उसे पॉइंट से जब आप कोई क्वाड मतलब सॉरी इस पॉइंट से जब कोई टेंशन ड्रॉ की तो इसके रिस्पेक्ट में वही कोड ऑफ कॉन्टैक्ट की वो बात कर रहे हैं तो कोड ऑफ कॉन्टैक्ट का मतलब क्या है सर इस पॉइंट पी से हम रिलीज सॉरी इस पॉइंट पी से आपने क्या किया इस पॉइंट पी से पहले तो आपने क्या ड्रॉ की सर आपने टेंशन ड्रॉ एक टेंसेज ये बनेगी और दूसरी टांगें आपकी क्या बनेगी हम रिलीज हो रही है फॉर दिस इसे नहीं रखेंगे और दूसरी टेंशन जो आपकी बनेगी सर वह आपकी बनेगी यह वाली कुछ ऐसी बन जाएंगे इनसे क्या बनेगी कोट्स ऑफ कोर्स इस पॉइंट और इस पॉइंट का जवाब मिलेंगे तो ये बनेगी तो सर अगर आपने इस पिक को थोड़ा चेंज भी किया है स्पीक को थोड़ा नीचे भी ले तो कुछ बन जाएंगे दो वापस उनसे भी एक और बनेगी ये जितनी भी कोट्स बनेंगे ना ये सारी कोर्स एक फिक्स्ड पॉइंट से पास हो रही होंगी वह का रहा है यह जो सारी कोट्स आप बनाएंगे सारी कोट्स एक फिक्स्ड पॉइंट से पास हो रही होंगी तो आपको पहले तो ये प्रूफ करना है की वो सारी कोट्स एक फिक्स्ड पॉइंट से पास होती है और प्रूफ करते-करते आपको ये भी निकल लेना है की वो पॉइंट के अकॉर्डिंग क्या है क्वेश्चन समझ ए रहा है मेरा कहना है सारी बातें छोड़ो क्या मैं इस पी पॉइंट के कोऑर्डिनेट्स लिख सकता हूं ह ए लिखना इस नॉट अन गुड आइडिया मैं कहूंगा सर मुझे पता है की इस लाइन पर लाइक करता है पी जो है वो इस लाइन पर लाइक करता है क्या मैं ऐसा लिख सकता हूं सोच के देखना स्टूडेंट्स इसी लाइन की इक्वेशन को ध्यान से देखना क्या मैं इसे ऐसे लिख सकता हूं की मैं 2X - 4 को उसे तरफ भेजूं तो ए को मैं लिख सकता हूं सुनेगा ध्यान से क्या 4 - 2X ऐसा क्यों लिख रहे हो सर अब सुनना ध्यान से अगर मैं इसका एक्स कोऑर्डिनेट्स मैन लूं टी अगर मैं इसका एक्स कोऑर्डिनेट्स मैन लूं टी इस पॉइंट टी का तो मैं टी यहां रखता हूं 4 - 2t तो ये कितना हो जाता है तो इसका एक्सपोर्टर नेट एक्स मैन लेते हो जाता 4 - 2X पर फिर आप नेसेसरी कन्फ्यूज्ड हो जाते हर चीज है अननेसेसरी बहुत बहुत सी हो जाती है तो मैंने कहा की सर कुछ भी के आरटीआर लामबीडीए कुछ भी ले लिया तो मैंने मैन लिया ये जो पॉइंट पी है जो की इस स्ट्रेट लाइन पर लाइक करता है उसका एक्सपोर्ट्स ए कोऑर्डिनेट्स क्या होगा 4 - 2t क्योंकि वो स्ट्रेट लाइन पर लाइक करता था तो कंडीशन को तो undoli वो सेटिस्फाई करेगा कोई दिक्कत अब क्या आप यह नहीं बता सकते मुझे की सर एक बड़ी सिंपल सी बात है अगर इस पॉइंट पी के रिस्पेक्ट में मैं स्क्वाड ऑफ कॉन्टैक्ट की quashion निकलना चाहूं तो क्या वो मैं निकल सकता हूं बस मेरी रिक्वेस्ट आपसे इतनी सी है जो जस्ट अभी हमने पढ़ा है वो थॉट वो इनफॉरमेशन अप्लाई करो और मुझे बस इतना सा बता दो की अगर मैं आपसे पूछूं कोड ऑफ कॉन्टैक्ट की इक्वेशन तो उसे आप क्या कहोगे क्या आप सब जानते हो सर कोड ऑफ कॉन्टैक्ट हमारे लिए हमेशा होता है ग = 0 और अगर आपने जीरो रखा तो क्या आप आंसर निकल का रहे हो स्टूडेंट्स आई थिंक बहुत टू बहुत डिफिकल्ट क्वेश्चन नहीं है कैसे सर सुनेगा ध्यान से मैं ग = 0 निकालो पहले किसी भी बात को ज्यादा कमाए फिर आए बिल्कुल निकल लो सर तो जैसे ही आपने ग = 0 निकलना चाहा तो सर्कल की इक्वेशन क्या है x² + y² - 1 = 0 तो सर्कल है आपका x² + y² - 1 और आपका पॉइंट क्या है जहां से आप ड्रॉ कर रहे हो सर पॉइंट है आपका टी कमा 4 - 2t आई थिंक 4 - 2t अब अगर मैं वही अपनी इक्वेशन मैं लिख रहा हूं कोड ऑफ कॉन्टैक्ट है ना विद रिस्पेक्ट तू ऑफ कोर्स पी है ना अब जो मैं उसकी इक्वेशन लिखना चाह रहा हूं वो कैसे लिखूंगा भाई हम सीधे-सीधे लिखेंगे सर कितना X1 तो ये हो जाएगा कितना ये हो जाएगा सर टी एक्स प्लस ए ए वैन तो ये कितना हो जाएगा सही हो जाएगा 4 - 2t टाइम्स ऑफ कोस क्या ए - 1 = 0 किसी भी स्टूडेंट को इस बात से कोई आपत्ति परेशानी तकलीफ दिक्कत आई थिंक नहीं होनी चाहिए सर इसे थोड़ा रिवाइट करें क्या कोशिश करते हैं भाई ध्यान से देखना जैसे की मैं फिर से थोड़ा बेहतर तरीके से लिखूं तो मैं एक छोटा सा कम ये करना चाह रहा हूं देखो ये टक्स आगे इट इसे रखा ए का अंदर मल्टीप्लाई किया तो ये हो जाता है 4y ये हो जाता है -2 व्हाइट टाइम लिख लो अगर आपको तकलीफ है तो है ना और ये कितना है माइंस वैन एक छोटा सा कम जो मैं करना चाह रहा हूं वो यहां पर ये हो गया प्लीज गौर से सुना ये क्रैक से इस क्वेश्चन का यहां पर खत्म हो जाएगा ये क्वेश्चन देखो मैं टी वाली टर्म से एक साथ लिखता हूं तो टी एक्स और माइंस तू टी ए एफ कोस टी कॉमन ले ले रहा हूं तो ये क्या हो जा रहा है सर ये हो जाता है एक्स - क्या 2y अब समझ का रहे हो और इसमें से क्या कॉमन ले लिया है टी और बच क्या जा रहा है सर बच रही हैं दोनों टर्म्स जो की है 4y - 1 तो ऐसे यहां पर लिख लेता हूं फोर ए माइंस वैन और ऑफ कोर्स ये क्या है प्लस टी एक्स टाइम्स एक्स - एक्स पर मैं एक ब्रैकेट लगा ले रहा हूं सर ऐसा क्यों कर रहे हो बस आपको अभी नजर ए जाएगा की ऐसा हमने क्यों किया अब सारी बातें छोड़ो आप बस इसे देखो आप सारी बातें छोड़ो अब बस इससे देखो सर क्यों देखें अगर आपको क्लिक नहीं किया तो मैं फिर से बात करता हूं आपसे आपको अगर याद ए रहा है एक लेक्चर हमने बात की थी फैमिली ऑफ स्ट्रेट लाइन करंट लाइंस तो याद करो हमने कुछ ऐसा पढ़ा था की सर L1 प्लस लामबीडीए l2=0 मुझसे कभी भी कोई कहे जहां पर L1 और L2 दो स्ट्रेट लाइंस है तो L1 + लामबीडीए L2 ऑन इंफिनिटी स्ट्रेट लाइंस की इक्वेशंस को जाहिर करता है उन infyanite स्ट्रेट लाइंस की इक्वेशन बताता है जो सारी इस L1 और L2 के पॉइंट ऑफ इंटरसेक्शन से पास होती हैं मैन लो ये आपकी लाइन है L1 और ये L2 तो L1 और L2 जहां इंटरसेक्ट कर रही है ना उनसे पास होने वाली सारी infainight स्ट्रेट लाइंस की क्लब डी इक्वेशन इसे गिवन बाय एल वैन प्लस जो अलग-अलग वैल्यूज ए जाएगा और वो आपको अलग अलग स्ट्रेट लाइन की वैल्यूज देता जाएगा आपको याद ए रहा है क्या तो कहना है कनक्लूड सर की मुझे बहुत मजेदार एक क्वेश्चन मिली की डी वे क्वेश्चन क्या है सर ये क्वेश्चन कोड ऑफ कॉन्टैक्ट की कोड ऑफ कॉन्टैक्ट की मतलब गोद ऑफ कॉन्टैक्ट मतलब मतलब थोड़ी सी और क्लेरिटी देना चाहूं तो थोड़ा और अच्छे से समझता हूं देखो मैन लो जैसे ये पॉइंट मूव करके यहां के यहां जाता था मैन लो यहां कहीं से तो क्या यहां से भी डेंजरस बन रही होती सर यहां से भी टेंशन बन रही होती एक टैसेंट यहां कहीं ए रही होती और सिमिलरली एक टैसेंट यहां कहीं ए रही होती आप मेरी बात समझ का रहे हो और अगर इन दोनों टांगें का हमें पॉइंट ऑफ इंटरसेक्शन निकलता तो वो ये होता तो वो ये कहना चाह रहा है क्वेश्चन में की जितने भी ये पी से आप tenjans बनाओगे ना उनका पॉइंट ऑफ कॉन्टैक्ट होगा और उसका जो कोर्ट बनेगी वो सारी कार्ड एक फिक्स्ड पॉइंट से पास हो रही होंगी तो आई थिंक सर जब मैंने पी के थ्रू वह इस कंडीशन को अप्लाई करके कोड ऑफ कॉन्टैक्ट निकल तो वो मुझे infainight स्ट्रेट लाइंस की इक्वेशन दे रहा है जो किसी फिक्स्ड तू पॉइंट से पास होती है जो की L1 और L2 के इंटरसेक्शन से पास होती है प्रूफ तो कर का रहा हूं की हान सर दो लाइंस के पॉइंट ऑफ इंटरसेक्शन से वो सारी कोट्स पास होंगे मेरी बात समझ का रहे हो या नहीं और वो किन दो पॉइंट से मतलब किस कोऑर्डिनेट्स एक ही पॉइंट्स ऑफ कोर्स होगी क्या वो फिक्स्ड पॉइंट मैं बता कर सकता हूं सर वो इन दोनों लाइंस का पॉइंट ऑफ इंटरसेक्शन होगा कौन सी दो लाइंस फॉर ए - 1 कौन सी दो लाइन सर एक लाइन तो है कौन सी फोर ए माइंस वैन जीरो और दूसरी लाइन कौन सी है सर एक्स - 2y = 0 दूसरी लाइन कौन सी है एक्स - 2y = 0 सर इन दोनों को अगर मैं सॉल्व करना चाहूं तो मुझे बड़ा मजेदार सा कन्फ्यूजन दिखता है की ए की वैल्यू यहां से आती है कितनी 1/4 और यहां से अगर मैं देखूं तो ए की वैल्यू कितनी आती है सॉरी ए की वैल्यू नहीं ए की वैल्यू वैन बाय फोर अगर यहां राखी तो हो जाता है 2 / 4 यानी 1 / 2 1 / 2 वहां गया तो एक्स की वैल्यू आती है 1 / 2 तो सर ये जो सारी इंफिनिटी कोड्स हैं किस से पास हो रही होंगी सर ये वैन बाय तू कमा 1 / 4 से पास हो रही होगी ये जो सारी इंफिनिटी कोट्स हैं ये किस पॉइंट से पास हो रही है क्योंकि इसमें मैं का डन क्यों थोड़ी देर के लिए तो 1 / 2 तो ये इसको फिक्स्ड पॉइंट है कोड ऑफ कॉन्टैक्ट बनेगी ऐसी चॉर्ड्स बनती चली जाएंगी अलग-अलग बहुत सारी इंफिनिटी ऑफ कॉस्ट पर आपको जो खास बात मिलेगी वो क्या मिलेगी सर की वो सारी की सारी कोट्स इस फिक्स्ड पॉइंट से पास होगी जिसके कार्ड होंगे 1 / 2 आपको समझ ए रहा है कितना मैजिकल कितना कितना मजेदार सहायक कॉन्सेप्ट है कैसे चीजें अलग-अलग तरीके से सोचिए जा सकती है और मैथ्स कितनी आसानी से चीज विजुलाइज करने में आपकी हेल्प करता है आपने तो बस रेट रखा आपने अपने कॉन्सेप्ट अप्लाई किया और मैथ्स ने खुद आपको जवाब दे दिया इसलिए चीजे बहुत मजेदार है इसलिए चीज आसान है और हम चीज ड्रा करें बनाए देखें इससे अच्छा है कोऑर्डिनेट्स ज्यामिति आप तो पॉइंट रखो सर इक्वेशन से मैं डालो वो खुद बना के आपको रिजल्ट्स देगा इसलिए ये कॉन्सेप्ट हम पढ़ रहे हैं की हमें ज्यादा दिमाग लगाने की जरूरत नहीं है हमने मैथर्ड बना ली है फीड करिए कंप्यूटर में प्रोग्रामिंग करिए वो रिजल्ट आपको खुद देगा आई होप आप बातें समझ का रहे हैं प्रोग्राम लाजिक्स वो बना कर दे देगा आपको आउटपुट तो इस तरीके से हम चीज करेंगे और क्रिएट करेंगे और नए-नए रिजल्ट्स आएंगे अगर बातें समझ का रहे हो और अगर इस क्वेश्चन में कोई डाउट है तो लिख दीजिए कमेंट में हम कभी ना कभी इसे रिजॉल्व जरूर करेंगे कोई दिक्कत होगी तो वर्ण हम बढ़ते हैं आगे इस क्वेश्चन को जरा अच्छे से देखने के बाद आई होप ये क्वेश्चन आप सभी को समझ आया अगला क्वेश्चन ट्राई करते हैं स्टूडेंट्स एक और क्वेश्चन के थ्रू चीजे समझते हैं आप जरा ये तीन इक्वेशंस को ध्यान से देखो मैं आपको क्वेश्चन बताता हूं क्वेश्चन क्या है ये जो तीन इक्वेशंस है इसमें ऐसा ही मैन लो की ये जो पहली इक्वेशन लिखी है आपकी x² + y² = a² मैन लो यहां से आपने इस सर्कल पर लाइक करने वाले किसी पॉइंट से आपने कोड ऑफ कॉन्टैक्ट बने किस पर बनाएं कोड ऑफ कॉन्टैक्ट आपने बनाई है दूसरे वाले सर्कल पर जो की ये है और फिर आपने क्या किया एक और बात है की सर वो जो कोड ऑफ कॉन्टैक्ट बनाई है ना सर आपने वो कोड इस तीसरे वाले सर्कल को टच करती है अब बातें समझ का रहे हो क्या जैसे समझना बात को मैं एक बेसिक सा सिनेरियो देता हूं पहले तो क्या तीन चीज आप देख का रहे हो की सिर्फ कोर्स ये जो तीन चीज हैं इसके इसके और इसके इन तीनों को देखकर आप ये प्रिडिक जरूर कर का रहे हो की इन सर्कल का जो सेंटर है वो ऑफ कोर्स को सेंट्रिक है जो की तीनों ओरिजिन पर सेंटर है क्या ये बात आप समझ का रहे हो क्वेश्चन ये है suniyega ध्यान से मैं आपको पहले सिनेरियो समझा देता हूं फिर हम क्वेश्चन पे बात करते हैं आपके पास पहले तो तीन kosentax सर्कल्स हैं यू हैव थ्री को सेंट्रिक सर्कल्स है ना तो तीन सर्कल्स हैं बेसिकली सर आपके पास मैं तीन को centrif सर्कल बनाने की जड़ों ज़हर में लगा हुआ हूं और अगर मैं इसे बनाऊं तो क्या को सेंटर दिख रहा है थोड़ा सा और इसे शिफ्ट कर दीजिए सर तो लगेगा है ना बिल्कुल सही और एक और सर्कल हम ड्रॉ कर लेते हैं जो की इसके अंदर होगा है ना तो ये आपका एक और सर्कल हो जा रहा है मेरे ख्याल से थोड़ा सा शिफ्ट कर लेंगे है ना अब suniyega ध्यान से ये तीनों सेंटर्स हैं सर्कल्स हैं आपके एक पॉइंट है एक पॉइंट है ऐसे ये आपका ये वाला सर्कल है कौन सा है ये वाला सर्कल एक्स स्क्वायर जो की कौन सा है आपका ये वाला सर्कल क्या आपको बात समझ ए रही है तो जब आपने कॉन्टैक्ट नहीं तो कैसे बन रही होंगी सर आई होप आपको इस बात में तो कोई परेशानी नहीं है की सर से जो कोड ऑफ कॉन्टैक्ट आपने कॉन्टैक्ट बना है वो कुछ इस तरीके से बनाई लेट्स से कुछ यहां पे है ना वो मेरे लिए थोड़ा मुश्किल हो जाएगा आपको बताना फिर भी मैं कोशिश करता हूं अब जाने की अब जो ये कोड ऑफ कॉन्टैक्ट आपकी बनी है सर ये जो आपकी कोड ऑफ कॉन्टैक्ट बनी हैं इसमें आपने एक बात ये नोटिस की की सर जब आपने ये कोड ऑफ कॉन्टैक्ट कैसे बनेगी भाई ये कोड ऑफ कॉन्टैक्ट कुछ ऐसी बन रही होगी ना मतलब मेरे लिए वो बनाना थोड़ा मुश्किल हो जा रहा है पर मैं कोशिश करता हूं की वो ये का रहा है की ये जो आपने कोड ऑफ कॉन्टैक्ट बनाई है suniyega ये जैसे वो कोड ऑफ कॉन्टैक्ट कुछ ऐसी ही होगी वो ये का रहा है की ये जो कोड ऑफ कॉन्टैक्ट है ये इस अंदर वाले थर्ड सर्कल को टच करती है तो मैं थर्ड सर्कल थोड़ा बड़ा बना देता हूं ताकि आपको सिनेरियो अच्छे से मैं समझा पाऊं तो थर्ड सर्कल में एक बार और बनाता हूं suniyega ध्यान से ताकि हमारी चीज है बहुत क्लियर सी और शॉर्टकट सी हो तो मैं थर्ड सर्कल आपके सामने ड्रॉप करने जा रहा हूं जो की ये है क्या अब आपको समझ ए रहा है बहुत सिंपल सा बेसिक सा ये वो आपका थर्ड सर्कल है क्या का रहा है यह क्वेश्चन क्वेश्चन का रहा है पहले तो क्वेश्चन जो की सबसे बाहर है आपका इस सर्कल पर किसी भी लाइक करने वाले पॉइंट से आपने कोड ऑफ कॉन्टैक्ट बनेगा कॉन्टैक्ट मतलब क्या इससे इस सर्कल पर टैसेंट के पॉइंट ऑफ कॉन्टैक्ट को मिला है तो कोर्ट बन गई वो कोड ऑफ कॉन्टैक्ट जो आपने इस सर्कल पर बनाई जो की था ये वाला सर्कल वो कार्ड ऑफ कॉन्टैक्ट इस सर्कल के लिए टैसेंट की तरह बिहेव कर रही है तो वो आपसे ये पूछ रहा है की आपको ये प्रूफ करना है क्वेश्चंस suniyega स्टूडेंट्स क्वेश्चन अब ए रहा है आपको ये प्रूफ करना है की ए बी सी किस्म है मतलब पता करना है की अर्थमैटिक प्रोग्रेशन ज्यामिति प्रोविजन हार्मोनिक प्रोग्रेशन एंड नॉन ऑफ दिस चार ऑप्शन दिए जाएंगे तो ए बी सी किस्म है मतलब aphpgp किस्म होंगे अगर ऐसी कोई कंडीशन फुलफिल हो रही होगी तो क्या यह क्वेश्चन क्लेरिटी से आप सभी को समझ आया आई थिंक क्वेश्चन तो बड़ा मजेदार है जिसमें कोई कन्फ्यूजन डाउट या परेशानी यहां तक नहीं होगी आप सभी को अब करना क्या है सर अब निकलेंगे कैसे चीजों पर कम कैसे होगा सबसे पहली बात तो क्या आपको ये दिख रही है सर की हम जानते हैं सर की ये जो सर्कल है सर्कल पर लाइक करता है तो सर अगर ह के जिस पॉइंट से आपने जिस पॉइंट के रिस्पेक्ट में कोड ऑफ कॉन्टैक्ट सर्कल की इक्वेशन को सेटिस्फाई करेगा बिल्कुल करेगा सर तो पहली तो सबसे जरूरी बात जो मैं निकल पाए वो क्या की सर h² + K2 विल बी इक्वल तू ए स्क्वायर बिल्कुल सर सही बात है h2 + K2 विल बी इक्वल तू ए स्क्वायर अब दूसरा स्टेटमेंट ये निकल कर आता है क्या आप कोड ऑफ कॉन्टैक्ट की इक्वेशन निकलना जानते हो रिपीट माय स्टेटमेंट निकाला जाते हो सर बहुत बड़ी बात नहीं है बेसिक सी बात है ये आपका सर्कल है जिस पर वह कोड ऑफ कॉन्टैक्ट बन रही है तो हमने पढ़ा है तो मेरा अगला सवाल गोद ऑफ कॉन्टैक्ट बनाई है तो उसे कोड ऑफ कॉन्टैक्ट यही बातें हम करते हुए आए हैं तो सर इसकी इक्वेशन तो मैं लिख सकता हूं कैसे लिखूंगा सर देखो एचसी 1 + के ए ये हो जाता है y1 और ये आपका -b² = 0 या फिर आप इसे इक्वल तू b² लिख लें जैसे आपको ठीक लगे तो सर एक कोई रिलेशन नहीं ए रहा है एक कोई रिलेशन नहीं ए रहा है ठीक है हम सारी बातें करेंगे उसने एक और मजेदार और जरूरी बात आपसे की आपने ध्यान नहीं दिया सर ये जो सबसे अंदर वाला सर्कल है ये जो सबसे अंदर वाला सर्कल है इसकी रेडियस कितनी है बताओ सर इसकी रेडियस है सी इसकी रेडियस ऐसी और यह जो तीनों सर्कल्स द इनका सेंटर क्या था सर इन तीनों सर्कस का सेंटर था 0 अगर यह जो स्ट्रेट लाइन है जिसकी इक्वेशन क्या है सर इसकी इक्वेशन निकल कर आई है एचसी + दी स सॉरी ह एक्स + के ए = b² अगर मैं मेरी नॉलेज से कहूं तो मेरे पास जो इक्वेशन निकल कर ए रही है वो इस कोड ऑफ कॉन्टैक्ट की क्वेश्चन ए रही है एचसी प्लस के ए = b² तो सर ये जो आपकी टांगें की इक्वेशन है अगर इस टैसेंट पर इससे मैं परपेंडिकुलर ड्रॉप करूं तो shailise की सर वो इसकी रेडियस के इक्वल होगी और उसकी रेडियस कितनी है सर सर इसकी रेडियस है करवा कर देखें ट्राई करते हैं तो मैं जीरो कमा जीरो से इस लाइन की परपेंडिकुलर डिस्टेंस निकल रहा हूं निकलती हैं सर जीरो कमा जीरो से मैंने निकल रहा हूं परपेंडिकुलर डिस्टेंस किस स्टेट लाइन की मैं परपेंडिकुलर डिस्टेंस निकलने में इंटरेस्टेड हूं सर कौन सी हमारी यह वाली स्ट्रेट लाइन कौन सी ह प्लस है ना माइंस बी स्क्वायर निकल सकते हैं देखो ध्यान से फटाफट कैसे X1 है ना तो ये हो जाएगा जीरो इन ह फिर 0 के उनको ऐड करेंगे सब जीरो क्योंकि ए जाएगा -20 और -बी स्क्वायर है तो मुझे समझ ए रहा है की मोड लगा के क्या करना है b² अन्य वे पॉजिटिव वैल्यू होगी तो माइंस है जाएगा मोड में से क्या बात समझ ए रही है डिवीज़न में अंडर रूट ओवर डिवीज़न में अंडर रूट ओवर क्या एक्स और ए के कॉएफिशिएंट्स के स्क्वायर सैम तो वो हो जाएगा h² + K2 सर ये ए रही है आपके अकॉर्डिंग परपेंडिकुलर डिस्टेंस जो की किसकी इक्वल है जो की इक्वल है इस सर्कल की रेडियस के जो की कितनी है सी जो की कितनी है भाई सी तो मैं क्या कर सकता हूं इसको सी के में लिख सकता हूं क्या लाइफ में समझ ए गया बहुत कुछ लाइफ में सॉर्ट आउट हो गया पहली बात तो ये की माइंस बी स्क्वायर तो बी जो भी हो पॉजिटिव नेगेटिव बी स्क्वायर पॉजिटिव अब एक बात जो डिनॉमिनेटर में लिखा है प्लस के स्क्वायर प्लस के स्क्वायर ए जाएगा सर तो ए और सी का प्रोडक्ट ए जाएगा क्या अब मैं कुछ बोलूं या आप मुझे बता रहे हैं जो मुझे का देना चाहिए जो आपको शायद पता चल गया सर क्या घुमा फिर के बातें कर रहे हो अरे भाई क्या आपको नहीं पता है की अगर तीन नंबर्स मेरे पास हैं आई हैव थ्री नंबर्स ए बी और सी और अगर मुझे पता चल जाए की b² एक के इक्वल है तो मैं कहूंगा सर बी जो है वो ज्यामितीय मिन है ए और सी का मतलब आप कनक्लूड करोगे की ए बी सी किस्म है सर जीपी में क्या यह बात आप सभी को डाइजेस्ट हुई क्या यह बात आप सभी को डाइजेस्ट हुई तो मैं क्या कहूंगा सर जब मुझसे पूछा गया था की अगर इस तरीके से कोई कंडीशन फुलफिल हो की एक सर्कल जिसका कोई भी एक पॉइंट से अगर आप कोड ऑफ कॉन्टैक्ट बनाएं तो कोड ऑफ कॉन्टैक्ट उसे थर्ड सर्कल पर मतलब जो सेकंड सर्कल पे आपने कोड ऑफ कॉन्टैक्ट बनाएंगे वह थर्ड सर्कल के लिए टांगें की तरह बिहेव करें तो इन तीनों की रेडियस जो की कोस एंड ट्रिक सर्कल द इन तीनों की रेडियस ए बी और सी ये अर्थ सॉरी ज्यामितीय प्रोग्रेशन में होगी तो ए बी सी क्या होंगे सर ए बी और सी एक ज्यामिति प्रोग्रेशन में रहेंगी क्या यह क्वेश्चन और यह एप्रोच और यह सॉल्यूशन आप सभी को समझ आया और इसी तरीके के क्वेश्चंस आपको आईआईटी जी मैंस और एडवांस पूछेगा वहां पर कुछ और दे देगा कुछ और पूछने का बट चीज इसी तरीके के प्रॉपर्टीज पर बेस्ड या डेरिवेशंस पर बेस्ड या इन formulon पर भेज दिया इन कॉन्सेप्ट्स पर बेस्ड होगी यह बातें आप सभी को समझ ए रही हैं आई होप इस क्वेश्चन में कोई डाउट या परेशानी नहीं है कुछ भी हो कोई भी दिक्कत हो आज पार्ट्स में डिवाइड है मतलब पंच अलग-अलग बुक्स हैं कार्डिनल ज्यामिति कैलकुलस एक्ट्रेस इन 3D अलजेब्रा और त्रिकोणमिति हम हर एक बुक को अच्छे से करेंगे और आपको हर एक कॉन्सेप्ट अच्छे से कराएंगे आईडी के मेंस और एडवांस लेवल तक मैथमेटिकल सब्जेक्ट की अगर मैं पार्टिकुलरली बात करूं अब डायरेक्टर सर्कल का कॉन्सेप्ट समझने की कोशिश करना सुनेगा यहां से डेफिनेशन में ही सब कुछ है और मैं क्लेरिटी के साथ जरूर samjhaunga इस सर्कल विच इस कॉल्ड डायरेक्टर सर्कल नहीं समझ आया सर यह तो हमको समझ आता ही नहीं है परिभाषाएं जो होती है सर यह तो बहुत ही अजीब सी होती है आप ऐसे समझो सुनना ध्यान से मैन लो आपके पास एक सर्कल सो दिस इस डी सर्कल अब आप ऐसा मैन लो की सर यहां पर कहीं ना कहीं ना सो यहां कहीं लेट से कोई एक पॉइंट है मैं इस एक्सटर्नल पॉइंट से सर इस पर टांगेंट्स ड्रॉ करता हूं तो लेट्स से मैंने एक टैसेंट ड्रॉप की जो की है कोई तकलीफ तो नहीं है स्टूडेंट्स और सर हमने एक और तेनजिंग जो ड्रा किया है जो की लेट से यह है ना यह बहुत अच्छी ट्रांस नहीं है क्योंकि इस बात से मैं अपनी बात प्रूफ नहीं कर पाऊंगा तो मैं थोड़ा इस पॉइंट को और दूर लेता हूं है ना तो उसको दूर ले जाने के लिए पहले मैं सर्कल को थोड़ा छोटा बना लेता हूं है ना तो एक अच्छी एप्रोच यह होती की मैं थोड़ा छोटा सर्कल बना था और अब अगर मैं एक दूर एक कोई पॉइंट लेता हूं जैसे मैंने कहीं दूर पॉइंट लिया आई थिंक इतना काफी होगा कोशिश करके देख लेते हैं अब जब सर आप यहां से यहां से एक टैसेंट बनाओगे तो एक टैसेंट ऑफ कोर्स आप किए जा रही होगी के बाद आप समझ का रहे हो हान सर मैन लिया आपकी बात और एक टेंट मेरी ख्याल से मैं जितना दूर जाऊंगा उतना ही एंगल छोटा होता चला जाएगा सो इट्स एडवाइस की मुझे तो एंगल बड़ा चाहिए तो मैं पास लता हूं तो मैन लो मैंने पॉइंट्स यहां बनाया है ना अब सुनना ध्यान से सर मैन लो यह तो आपकी एक टिनेंट रही यहां क्या बात समझ का रहे हो की आप इसको पास लेट गए और पास लेट गए जैसे यहां से थोड़ा और पास ले जैसे की इस पॉइंट को थोड़ा पास में लाया कैसे थोड़ा और पास लेकर आए तो जैसे की हमने उसे पॉइंट को और क्लोज बनाए लेट्स से यहां कहीं बनाया बहुत कम की बातें suniyega ध्यान से अब जब मैं इस पॉइंट को यहां इतना पास ले आए तो देखना क्या होगा एक टैसेंट ये बन रही है और एक टैसेंट जब बनाओगे सर आप एक टैसेंट ऑफ कोर्स ये है और दूसरी टांगें जब आप बनाओगे आप तो ये दूसरी टैसेंट है जो मैं कहना चाह रहा हूं सब थोड़ा और अगर पास ला तो देखो अब थोड़ा और पास ला रहा हूं तो suniyega कम की बात अब कम की बात आपके सामने आने वाली है की सर अगर ये पॉइंट थोड़ा और पास होता जो ये पॉइंट थोड़ा और पास होता है तो सर एक टैसेंट तो आपकी ये होती और ऑफ कोर्स दूसरी टांगें आपकी होती क्या आपको कुछ नजर ए रहा है जो मैं आपसे कहना चाह रहा हूं क्या आपको कुछ नजर ए रहा है जो मैं कहना चाह रहा हूं सर इतना ऑप्टिकल डिस्टेंस पर मिलेगा ऐसे की सर उसे पॉइंट से जब मैंने टांगें ड्रॉ की उसे पॉइंट से जब मैंने tangenced ड्रा किया तो वो टेंसेज जो थी सर वो परपेंडिकुलर थी अब क्या सर सिर्फ यही एक पॉइंट होगा जहां से tesence परपेंडिकुलर होगी नहीं सर यहां पर और भी कई सारे पॉइंट्स होंगे जहां से जब आप tenjins ड्रॉ करोगे तो पैर वाइस वो टेंसेज परपेंडिकुलर होंगी तो इस पॉइंट का जो लो कस होता है ना ये पॉइंट जहां जहां हो सकता है इस सर्कल के रिस्पेक्ट में वह बेसिकली जब आप निकलोगे मैं मैथमेटिक्स प्रूफ करेंगे जब वो बेसिकली आप निकलोगे तो वो भी एक सर्कल ही बनेगा मतलब एक्चुअली एक्चुअली जो आपका ये निकल कर आएगा वो भी एक सर्कल ही बनेगा मतलब टेक्निकल वो आपका एक सर्कल ही बनेगा की आप मेरी बात समझ का रहे हो स्टूडेंट्स ही बनेगा मतलब आप जब देखोगे ना ध्यान से तो आपको ये नजर आएगा की सर्कल पर लाइक करने वाले किसी भी पॉइंट से जब आप दूसरे अंदर वाले सर्कल पर टांगें ड्रॉ करोगे तो वो saritanges 90 डिग्री पर इंक्लाइंड होगी मतलब सर यह जो पॉइंट है यह जो पॉइंट है इसका जो लोकस है यह जो इस सर्कल पर ट्रेवल कर रहा है यहां पर से इस सर्कल पर ड्रा की गई सारी टेंशंस परपेंडिकुलर है आपस में पैर वाइस ऑफ कोर्स आई होप आप समझ पाएं इसका ये मतलब है की इस पॉइंट का जो ये लोकस है जो की बेसिकली एक सर्कल है वो सर्कल ना सिर्फ ऑर्डिनरी सर्कल है बल्कि वो सर्कल इस सर्कल का डायरेक्टर सर्कल है क्या आपको अब थोड़ी बहुत परिभाषा समझ आई डायरेक्टर सर्कल मतलब क्या मतलब एक ऐसे पॉइंट का लॉस जहां से ड्रा की गई सर्कल क्या हो भाई परपेंडिकुलर आपस में ऑफ कोर्स फेर वाइस क्या यह स्टेटमेंट समझ आया अगर यह स्टेटमेंट डाइजेस्ट हुआ तो उसे पर क्वेश्चंस बनेंगे तो याद रखिएगा वो इतनी सारी बातें घुमा फिर के नहीं कहेगा की सर यहां से ड्रॉप किया तो यहां से टैसेंट बनाई तो टांगें पर वाइस आपस में परपेंडिकुलर इतनी बातें नहीं करेगा वो आपसे वो तो आपका नॉलेज के अंडरस्टैंडिंग चेक करेगा ये बोल करके एक डायरेक्टर सर्कल है सर्कल का है तो अब आपको यह स्ट्राइक करना चाहिए की डायरेक्टर सर्कल मतलब हान सर हान मतलब इंटरनल का और होगा की ये जो सर्कल दिया है इससे बाहर एक सर्कल हो रहा होगा जहां से अगर आप पॉइंट्स यू नो जहां से अगर आप किसी भी पॉइंट से उसे सर्कल पर लाइक करने वाले किसी भी पॉइंट्स अगर आप टांगें ड्रॉ करोगे तो वो tesence कैसी होंगे आपस में पैर विजय परपेंडिकुलर हप में कंफ्यूज नहीं कर रहा हूं आपको आप बातें समझ का रहे हैं तो डायरेक्टर सर्कल सुनकर यह बात आपके ब्रेन में आणि चाहिए ये बात आपको क्लियर होनी चाहिए तो अब क्या मतलब है suniyega यहां से मैन लो सर सीधी-सीधी सी बात है अगर मेरे पास एक सर्कल है मैं एक बेसिक सा एग्जांपल ले के आपको समझाना चाह रहा हूं हम कोशिश करेंगे की हम सिंपल फॉर्म से ले वापस हम उसको जर्नलाइज कर देंगे तो अगर मैं एक सिंपल लेता हूं सर की मैन लो ये हमारा एक स्टैंडर्ड सर्कल है ये हमारा जो था ये बेसिक सा स्टैंडर्ड जिसे मैन लेता हूं x² + y² = a² मतलब मैं ऐसा इसलिए मैन ले रहा हूं ताकि मुझे इसके सेंटर के कोऑर्डिनेट्स का मिल जाए ओरिजिनल पर और ऑफकोर्स की रेडियस क्या मिल जाए मुझे सर इसकी रेडियस मुझे मिल जाए ए क्या आप मेरी बात समझ पाए कोई दिक्कत तो नहीं अब एक बात सोच के बताओ स्टूडेंट्स अगर ये ए है रेडियस तो क्या ये भी ए होगी रेडियस तो अगर ये परपेंडिकुलर है तो ये भी क्या होगा परपेंडिकुलर क्या मेरी बातें समझ का रहे हो तो सर अगर ये ये 90 डिग्री है क्वॉड्रिलैटरल में तो कहना है इसे टोटल क्वॉड्रिलैटरल की तीनों इंटीरियर चारों इंटीरियर्स को हम 360 होता है तो ये भी कितना होगा 90 डिग्री है और सर किसी क्वॉड्रिलैटरल में अगर चारों साइट्स के बीच में जो एडजेसेंट साइड्स के बीच में एंगल है अगर वो 90° है और आप यह बात जानते हो की अगर यह आपकी टैसेंट है तो ये भी क्या होगी टैसेंट अमरेली सॉरी अगर यह आपकी रेडियस है तो ये भी क्या होगी रेडियस क्योंकि आपने टैसेंट पे क्या ड्रॉ किया परपेंडिकुलर तो दो एडजस्ट है साइंस इक्वल है और उनके बीच में एंगल है 90° तो सर पहले तो मैं का रहा था 90° देख कर के रेक्टेंगल होगा पर अब तो मैं कहूंगा दोनों एडजेसेंट साइड मतलब आपका लेंथ भी और विथ भी दोनों से लेंथ के हैं तो actingal नहीं सर ये स्क्वायर है सर ये एक्टिंग नहीं एक स्क्वायर है जिसकी साइड की लेंथ है ए अब अगर बस मैं आपसे इतना पूछना चाहूं बेसिक सा कॉमन सेंस लगा के की सर की सर यहां से यह जो डिस्टेंस है कितनी होगी तो आप क्या कहोगे सर इस स्क्वायर के डायग्नल की लेंथ हाफ से पूछ रहा हूं और अगर आप पाइथागोरस थ्योरम लगा रहे हो तो देखो बेस परपेंडिकुलर और हाइपोटेन्यूज तो ए स्क्वायर प्लस ए स्क्वायर और 2a² तो इस वाले डायरेक्टर सर्कल की रेडियस कितनी ए जाएगी 2a की डी वे अभी मैंने ये प्रूफ नहीं किया है की इस पॉइंट का लॉस एक सर्कल ही होगा अभी तो मैंने वो डायरेक्टर सर्कल की आपको रेडियस दी है दूसरे तरीके से निकल कर पर आप कैसे का का रहे हो सर की यह जो पॉइंट है जिससे आपने यह tenjans ड्रॉ की हैं जो की 90° एंगल पर इंक्लाइंड है यह सारी टांगेंट्स जो है यह सारी टांगें जो हैं जहां जब इस तरीके से ड्रॉ हो रही है क्योंकि बीच में 90° एंगल है तो जिस पॉइंट से आप ड्रॉ कर रहे हो उसे पॉइंट का जो लोकस है वो एक सर्कल ही होगा ये आप कैसे इतनी सुरेती से का रहे हैं पहली बात तो थोड़ा तो एविडेंस है सर थोड़ा अपेरेंट है दिख ही रहा है की अगर यहां से बन रहा है तो अगला कोई पॉइंट मिलेगा तो वो एक सर्कल वाली चीज पर ही लाइक करना होगा क्योंकि वो सब कुछ सिमिट्रिकल होगा ना सर्कल के ही केस में बात बन पाए पर फिर भी नहीं सर आप मैथमेटिकली प्रूव्ड करके बताओ तो मेरा कहना है की क्या मैं क्या मैं बड़ी बेसिक सी बात है सुनेगा ध्यान से क्या मैं किसी एक्सटर्नल पॉइंट से ड्रॉ किए गए टांगे की इक्वेशन निकल सकता हूं सर किसी एक्सटर्नल पॉइंट जैसे मैन लो अगर आपको नहीं समझ ए रहा है तो क्या मैं ऐसा का सकता हूं की सर इस एक्सटर्नल पॉइंट थोड़ी देर के लिए मैं कहता हूं ह कॉम आके इस एक्सटर्नल पॉइंट को अगर मैं थोड़ी देर के लिए कहता हूं ह कमा के जिसमें सर्कल की इक्वेशन क्या है सर x² + y² = a² जिसमें सेंटर है जीरो कमा जीरो रेडियस ए सर्कल की इक्वेशन क्या है x² + y² = a² मुझे बस ये बताओ अगर मुझे टैसेंट की इक्वेशन चाहिए तो मैं टांगें की इक्वेशन क्या कहूंगा सर तेनजिंग की इक्वेशन में बोलूंगा ए - y1 तो ए - कितना के = एम जो की स्लोप है जो की मुझे नहीं पता है अभी फिलहाल बात करेंगे इस बारे में टाइम्स क्या एक्स - एक्स - ह इससे मैं थोड़ा सिंपलीफाई करना चाहता हूं तो मैं क्या लिखूंगा ऐसे में सारी चीज उसे तरह शिफ्ट करता हूं तो ये हो जाएगा एमएक्स है ना माइंस वही है ना प्लस के माइंस एम यह आपकी स्ट्रेट लाइन की इक्वेशन होगी जो की बेसिकली क्या है ये ह कमा कैसे पास होने वाली स्ट्रेट लाइन जिसके स्लोप है एम जिसे फिलहाल मैं ट्रीट कर रहा हूं आगे अन टेंट ठीक है सर अच्छी बात है आपकी सारी बात मैन ली हमने अब एक बात बताओ मैं तो बड़ा बेसिक सा कॉन्सेप्ट सर हमेशा सही है अप्लाई करता हूं जब तक मुझे 10 दिन से डील करने को कहा जाता है की सर सर्कल के सेंटर से टैसेंट पर ड्रॉप की गई परपेंडिकुलर लेंथ जो वो होती है उसकी रेडियस के इक्वल होती है तो जीरो कमा जीरो से क्या इसकी लेंथ निकल सकते हैं बिल्कुल निकल सकते हैं तो जीरो को अच्छा पूरे वो स्टैंडर्ड सर्कल में भी रख सकते द आप x² + y² + 2G + 2y + सी = 0 अब मैं का रहा हूं वहां से चले जाएंगे पर मैं का रहा हूं एक आसान से सिंपल से केस में देख लो यार वहां भी कर सकते हो कोई बड़ी बात नहीं है पर मैं का रहा हूं सर यहां देख लेंगे यहां समझ लें उतने कॉम्प्लिकेटेड तरीके पे क्यों करना है हम तो चीजों को आसान सिंपल सुलझा के रखना चाह रहे हैं ना तो हम सर्कल को ओरिजिन पर ले आए सेंटर को और रेडियस ए मैन ली तो उसमें हम आसानी से देख पाएंगे होगा वहां भी वही कैलकुलेशन सब कुछ होगी बट थोड़ी सी कॉम्प्लिकेटेड होगी और आप कर सकते हो आसानी से कर सकते हो मैथमेटिकली आप प्रूफ कर सकते हो तो मैं आपको इस से नारी को समझाना चाह रहा हूं तो मैं क्या करना चाह रहा हूं सर मैंने माना ये जो तेनजिंग की इक्वेशन है वो है स्लोप फॉर्म में मतलब एम और ह के फॉर्म में ह के वो पॉइंट है जिसका लोकस हमें निकलना है बस मेरा कंसर्न ये है सर की अगर ये इसकी टैसेंट है तो जीरो कमा जीरो से इस लाइन की परपेंडिकुलर डिस्टेंस इसके लिए एक के इक्वल होगी जीरो कमा जीरो मतलब क्या जीरो जीरो तो बचा क्या सर बचा के-एम कोई दिक्कत तो नहीं है सर इस पे आप क्या लगाओगे मोड डिवाइडेड बाय अंडर रूट ओवर एक्स और ए के कॉएफिशिएंट के स्क्वायर सैम का अंडररूट तो एम और 1 मतलब एम और माइंस वैन तो m² + 1 और ये किस्म आएगा सर ये आएगा अंडर रूट में आई होप बातें समझ ए रही है अब क्या सर यह जो परपेंडिकुलर डिस्टेंस है सर वह इसकी रेडियस यानी किसके इक्वल होगी सर वो इक्वल होगी ए के क्या इस बात से कोई परेशानी सर अभी तक तो नहीं है अब एक बात बताओ एक बात का जवाब दो अगर मैं अब लोकस निकलना चाहता हूं अगर अब मैं rockage निकलना था तो मैं क्या करूंगा सर मैं चीज थोड़ी सी ना शॉट आउट करूंगा मैं चीज थोड़ी सी ऑर्गेनाइज्ड करूंगा देखो मैं क्या करूंगा इनको उधर भेजूंगा तो मेरे पास क्या ए जाएगा सर मेरे पास आएगा के माइंस एम दिख रहा है क्या आप सभी को और इसका ऑफ कोर्स क्या लिया है आपने मोड = ए टाइम्स अंडर रूट ओवर 1+m2 सर एक सिंपल सा छोटा सा आप कम कर लो यही बात बेसिकली उसने यहां पर लिखिए अगर आप ध्यान से देखो तो ये वही लिखा है वॉइस रिक्वेस्ट तू एमएक्स + ए टाइम्स m² + 1 ये वही बात है जो आपको यहां पर वो लिख कर दे रहा है ये वही बात है बस उसने ह और के की जगह ए और एक्स रिप्लेस कर दिया तो सिर्फ वही बन गया अब आप एक छोटा सा कम ये करो सर की अगर मैं दोनों तरफ स्क्वायर कर देता हूं स्क्वायर बोथ दी साइड्स तो क्या मिलेगा देखो इधर मिलेगा के माइंस एम का होल स्क्वायर टाइम्स सी है कोई परेशानी नहीं होनी चाहिए सर अब अगर मैं चीज निकलना चाहूं तो कम की बातें सुनेगा यही से पूरा क्वेश्चन आपका सॉल्व होगा सर इसका परफेक्ट स्क्वायर क्या होता है निकलती हैं स्टूडेंट्स के - एमएस का होल स्क्वायर है ना ये कितना हो गया सर ये होगा के स्क्वायर प्लस m² - माइंस ऑफ कोर्स 2K mh2k और छोटी सी बात स्टूडेंट्स क्या एक और छोटी सी बात ये है सर की यहां पर देखोगे तो हो जाएगा a² + a² यहां पर क्या हो जाएगा सर ए स्क्वायर प्लस ए स्क्वायर एम स्क्वायर अब अगर मैं चीज सिंपलीफाई करना चाहूं अगर मैं चीज सॉल्व करके सॉर्ट आउट करना चाहूं तो मुझे जो दिख रहा है वो ये की सर ये एक m² में क्वाड्रेटिक बन रही है अब देख का रहे हो स्टूडेंट्स ये एक m² में मतलब एम वेरिएबल में क्वाड्रेटिक बन रही है जहां पे के और h2 के ऐप के वो एक्सटर्नल पॉइंट और ए क्या है नॉन हमारे सर्कल की रेडियस तो हमारे अननोन क्या है अभी फिलहाल क्या बात है आपको समझ ए रहे हैं तो सर मैं एक बहुत जरूरी सी बात जानता हूं की अगर ये दोनों लाइंस की स्लोप्स को रिप्रेजेंट कर रहे हैं वो दोनों पैर ऑफ लाइंस की तो क्या मैं जानता हूं इसके रूट्स होंगे दो m1 और M2 अगर मैं मैन लूं और वो दो लाइंस वो दोनों लाइंस अगर एक दूसरे पर परपेंडिकुलर हैं अगर वो दोनों लाइन एक दूसरे पर परपेंडिकुलर हैं तो क्या मैं सीधी सी बात कहूं दोनों स्लोप्स का प्रोडक्ट ऑप्शन होगा m1 M2 -1 होगा ध्यान से इस कंक्लुजन को लाने की कोशिश कर रहा हूं इस वाले से नारी में पहले तो मैं इसे क्वाड्रेटिक में कन्वर्ट कर लेता हूं सुनना ध्यान से क्वाड्रेटिक में जब कन्वर्ट करने की कोशिश की तो पहले तो टर्म आती है आपकी कौन सी है तो a² m² और m²h² तो मैं क्या m² को कॉमन ले सकता हूं तो अंदर क्या बचेगा सर अंदर बचेगा a² - h² तो ये दिखेगा a² - x² कोई दिक्कत तो नहीं है भाई अब क्या सर अब suniyega ध्यान से ये है -2 ख एम तो मैं इसे लिख देता हूं -2 खान राइट बस यहां पे क्या लगा लो भाई प्लस क्योंकि ये ए जा रहा है प्लस कोई डाउट तो नहीं है अब क्या करेंगे सर अब मेरे ख्याल से सर देखो हमने क्या लिख लिया हमने लिख लिया a² ये वाला टर्म है ना a² - x² टाइम्स m² बहुत बढ़िया सर a² - के स्क्वायर तो ये जो आपकी टर्म बच रही है बज रही है a² - K2 अब मेरा जो आपसे कहना है वो ये की सर क्या आप इसे एक क्वाड्रेटिक की तरह देख का रहे हो अगर मैं कहूं आपसे तो बिल्कुल सर क्वाड्रेटिक कर रहा देख का रहा हूं देखो ना आप ये एम में क्या होगा सर क्वाड्रेटिक आप समझ का रहे हो ना ये मा क्वाड्रेटिक है अब वापस वही बात निकल कर ए रही है सर की हम जानते हैं ये एम एक तो इसकी स्लोप है और एक इसकी स्लोप है और अगर इन दोनों की स्लोप एम है तो वो ऑफ कोर्स m1 और M2 होगी जो की क्वाड्रेटिक के दो रूट्स होंगे और एक छोटी सी बात स्टूडेंट्स क्या मैं जानता हूं की अगर यह दोनों लाइंस परपेंडिकुलर हैं तो इन दोनों लाइंस की स्लोप्स का प्रोडक्ट -1 होगा मतलब सर अगर मैं आपसे यहां पर एक बड़ी जरूरी सी बात पूछूं तो क्या कहोगे आप कहोगे सर सैम ऑफ डी रूट्स जानते हैं प्रोडक्ट ऑफ डी रूट्स जानते हैं मैं बात करूंगा प्राउड ऑफ जरूर इसके दो रूट्स होंगे सर m1 और M2 बिल्कुल होंगे सर इस क्वाड्रेटिक के जो की एम में है इसके दो रूट्स होंगे म और M2 और अगर वो दोनों लाइंस एक दूसरे पर परपेंडिकुलर हैं तो जो m1 और M2 स्लोप्स हैं उनके प्रोडक्ट -1 होगा लेकिन हमारे अकॉर्डिंग क्वाड्रेटिक में प्रोडक्ट ऑफ डी रूट्स क्या होता है सर वो होता है सी अपॉन ए क्या देना कांस्टेंट टर्म अपॉन x² का कोई फिश तो ये कितना हो जाएगा सर ये हो जाने वाला है a² - के स्क्वायर अपॉन ऑफ कोस कितना a² - x² कोई दिक्कत तो नहीं है अब हम इस पर बात करेंगे वो ये ध्यान से देखना मैं इसे यहां लता हूं तो ये हो जाता है क्या स्क्वायर प्लस ऑफ कोर्स के स्क्वायर प्लस ए स्क्वायर यानी तू ए स्क्वायर और लोकस में तो सर हमने ये जानना हमेशा सिखा है की ह और के को अंत में आप एक्स कमा ए से रिप्लेस करते हो तो आपके पास फाइनली जो इस डायरेक्टर सर्कल की इक्वेशन ए रही है वो ए रही है x² + K2 यानी आप क्या लिखोगे x² + y² = 2a² तो 2a को मतलब तू को क्या मैं अंडर रूट 2 टाइम्स ए का होल स्क्वायर लिख सकता हूं क्या कोई बात आपको समझ ए रही है जो मैं कहना चाह रहा हूं क्या आप मेरी एक सिंपल सी बात समझ का रहे हो जो मैं यहां पर आपको कहना चाह रहा हूं जरा इसे ध्यान से देखो और प्लीज अच्छे से देखो और मुझे बताओ क्या आप समझ का रहे हो जो मैं आपसे कहना चाह रहा हूं हमने सर क्या किया हमने इस डायरेक्टर सर्कल का लोकस निकाला तो हम प्रूफ कर का रहे हैं ना की हान सर ये जो पॉइंट है जिससे जब आप tenjans ड्रा कर रहे हो इन सच अन मैनर डेट डोज तू टेंशन परपेंडिकुलर ऑन इ आदर यह पॉइंट एक ऐसे पथ को ट्रेवल कर रहा है जो सर्कल है और उसे सर्कल की जो रेडियस है उसे सर्कल की जो रेडियस है वह अंडर रूट तू टाइम्स मतलब उसे पुराने सर्कल की रेडियस यही की यही बात आप उसे अपने स्टैंडर्ड सर्कल पर भी प्रूफ कर सकते हैं 5 + सी पर मेरा कहना है वहां भी अब आती है तो डायरेक्टर सर्कल के बारे में जो जरूरी कंक्लुजन हम दे का रहे हैं वो क्या की ऑफ कोर्स सर आपके पास एक सर्कल होगा जिसके सेंटर के कुछ cardinate होंगे जिसकी अवकाश आपके पास रेडियस होगी और जो इस तरीके से आप फाइनली प्रूफ करके निकल कर लेकर आओगे है ना जैसे आप क्या का रहे हो डायरेक्टर सर्कल मेरा बस आपको ये साला है सर की आप जो फाइनल कंक्लुजन दोगे वो ये फाइनल कंक्लुजन दोगे वो ये की जो भी आपका सर्कल हो हमें फर्क नहीं पड़ता जैसे मैन लो आप अगर इस पर बात करना चाह रहे हो x² + y² + 2gx + 2fy प्लस सी ये आपका एक सर्कल है हान सर मैन लिया इस सर्कल का जो डायरेक्टर सर्कल बनेगा इस सर्कल का जो डायरेक्टर सर्कल बनेगा उसके भी सेंटर के कार्ड - जी फ रहेंगे मतलब इस सर्कल के और उसके डायरेक्टर सर्कल दोनों को सेंट्रिक होते हैं उन दोनों के सेंटर के अकॉर्डिंग ही रहेंगे लेकिन अगर इस सर्कल की रेडियस कितनी है सर इस सर्कल की रेडियस अगर मैं कहता हूं है g² + x² - सी तो उसे सर्कल की रेडियस इसका अंडर रूट तू टाइम्स होगी जो डायरेक्टर सर्कल होगा उसे सर्कल की रेडियस इसका अंडर रूट तू टाइम्स होगी क्या आप बातें समझ का रहे हैं तो टेक्निकल क्या किसी भी सर्कल को देखकर किसी भी सर्कल को देखकर उसके डायरेक्टर सर्कल की क्वेश्चन किए जा सकती है डायरेक्टर सर्कल भूल तो नहीं रहे हो ना सर डायरेक्ट के सर्कल मतलब वो सर्कल जिस पर लाइक करने वाले सारे पॉइंट से ड्रॉ की गई tenjins आपस में वो जो पैर विजय म्यूचुअल परपेंडिकुलर होती हैं इस सर्कल पर आई होप आप बात समझ का रहे हो मतलब मैं सर किसी भी सर्कल को देख के उसका डायरेक्टर सर्कल बना सकता हूं कैसे सेंटर के कोऑर्डिनेटर जैसे होंगे मतलब वह centrif होंगे और रेडियस जो इस सर्कल की होगी इसका अंडर रूट तू टाइम्स कर दूंगा बात खत्म हो जाएगी यानी इस सर्कल के डायरेक्टर सर्कल की इक्वेशन क्या होगी सर सेंटर पता है रेडियस पता है तो आप नहीं लिख सकते क्या बिल्कुल लिख सकते हैं अभी तो क्या लिखेंगे भाई हम लिखेंगे अगर हमारा स्टैंडर्ड सर्कल है कौन सा x² + y² + 2j + 2 / 3 = 0 तो डायरेक्टर सर्कल के भी सेंटर के cardinate तो -डी कॉमन -एफ ही रहेंगे तो मैं क्या लिखूंगा एक्स - जी का होल स्क्वायर प्लस ए माइंस माइंस एफ का होल स्क्वायर इस इक्वल्स तू रेडियस का स्क्वायर और रेडियस क्या है सर रेडियस है √2 √g² + x² - सी और उसे पूरे का स्क्वायर किया तो अंडर रूट कैंसिल तो तू टाइम्स g² + x² - सी क्या यह पूरा कॉन्सेप्ट अब आपको अच्छे से डाइजेस्ट हुआ आज के लेक्चर का बड़ा सिंपल सा क्रक्स और कनक्लूड याद रखेगा अगर मैं किसी भी सर्कल के डायरेक्टर सर्कल से उसको अगर कंपेयर करूं अगर किसी भी सर्कल में और उसके डायरेक्टर सर्कल में रिलेशन निकालो तो पहली बात तो क्या की सर दोनों के सेंटर क्यों होंगे भाई जल्दी बताओ तो दोनों के सेंटर से होंगे दूसरी बात क्या सर इस सर्कल की जो रेडियस होगी इस सर्कल की जो रेडियस होगी इस सर्कल की जो रेडियस होगी उसका क्या उसका अंडर रूट तू टाइम्स उसका अंडर रूट तू टाइम्स आपके डायरेक्टर सर्कल की रेडियस होगी तो यह सर्कल ऑफ कोर्स इसका ये मतलब होता है क्या यह बात आपको याद रहेगी मेरा कहना है सर बातें याद रखने का मेरे लिए तो सबसे अच्छा तरीका होता है सवालों को सॉल्व करना तो ये क्वेश्चंस करते हैं और और कंसोलिडेटेड करते हैं तो पहले तो आप स्क्रीन पीओएस करके ट्राई कर लीजिए तब तक मैं आपको बताने की कोशिश करता हूं वो क्या पूछा है suniyega ध्यान से फाइंड डी लेंथ ऑफ डी कॉर्ड ऑफ कॉन्टैक्ट विद रिस्पेक्ट तू डी पॉइंट ऑन डी डायरेक्टर सर्कल ऑफ डी सर्कल सर जीवन में समस्या यही है समस्या क्या है सर की क्वेश्चन तो समझ आए तब तो हम क्वेश्चन सॉल्व करें तो क्वेश्चन कैसे समझ आता है सर क्वेश्चन समझ आने का तरीका है क्वेश्चंस को पार्ट्स में ब्रेक करो अब सुनना दो-तीन टर्म्स है जो आपको ध्यान से देखनी चाहिए उसने बोला लेंथ ठीक है सर लेंथ निकलेंगे वह पॉइंट डायरेक्टर सर्कल पर है किसके इस सर्कल थोड़ी बातें कुछ तो है सर पर कुछ समझ नहीं ए रहा है मैं कम करता हूं मैं चीज विजुलाइज करता हूं मैं चीज इमेजिन करता हूं क्वेश्चंस को प्लॉट करके ये एक सर्कल है मैं इसको बनाता हूं तो एक सर्कल है सर्कल बनाना कौन सी बड़ी बात है सर ये सर इसका डायरेक्टर सर्कल बना लो ठीक है सर इसका डायरेक्टर सर्कल बना लिया तो मैंने इसका क्या किया इसका मैंने डायरेक्टर सर्कल भी बना लिया किसी भी स्टूडेंट को इससे कोई आपत्ति तो नहीं है सर नहीं है अब क्या करना चाह रहे हो अब मुझे एक बात का जवाब दो स्टूडेंट्स अगर इस सर्कल का डायरेक्टर सर्कल है तो मैं क्या कहूंगा की सर इस सर्कल के डायरेक्टर सर्कल पर अगर कोई पॉइंट है अगर कोई भी पॉइंट है लेट्स से एक रैंडम पॉइंट है अगर मैंने इस पर क्या ऐसी दिखेंगे और सर इन दोनों सर्कल्स का सेंटर से होगा क्योंकि ये कोर्स एंट्री है ना मैं सर्कल के सेंटर को थोड़ा शिफ्ट कर लूंगा अपनी सहेलियों के अकॉर्डिंग और यहां से अगर मैंने इसे ऐसे कनेक्ट किया अपनी बात समझ का रहे हो मैं जहां आपको ले जाना चाह रहा हूं की वो क्या बात है अगर मैंने इसको थोड़ा सा ऐसा रखा है ना थोड़ा सा मैं चीज ज्यामिति के लिए करेक्ट बनाना चाह रहा हूं ताकि आप कन्फ्यूजन से बचाए तो suniyega सर बहुत कम की बात है ये ऑफ कोर्स आपके सर्कल का सेंटर है जो की दोनों ही सर्कल्स के सेंटर हैं और ये बाहर वाला सर्कल आपका डायरेक्टर सर्कल है अब suniyega वो आपसे क्या पूछ रहा है वो आपसे पूछ रहा है कॉल्ड ऑफ कॉन्टैक्ट की लेंथ ये एक पॉइंट पी था जो की इस गिवन सर्कल का डायरेक्टर सर्कल इससे आपने कोड ऑफ कॉन्टैक्ट बने कोड ऑफ कॉन्टैक्ट क्या होती है जब आप टेंज्ड रॉक करोगे ना तो जब टांगेंट्स का पॉइंट ऑफ कॉन्टैक्ट होगा उन कॉन्टैक्ट को अगर आप मिला दोगे तो ये बनेगी आपकी क्या कोड ऑफ कॉन्टैक्ट के बात आपको समझ ए रही है ये जो लाइन बनेगी आपकी ये क्या होगी सर ये हो गया आपकी गोद ऑफ कॉन्टैक्ट अब मेरा आपसे बस ये पूछना है मेरा बस आपसे ये पूछना है मैं इसे थोड़ा सा और ठीक से बना लूं ताकि बस आप कंफ्यूज ना हो जाए यह कोड ऑफ कॉन्टैक्ट जो बनी है सर यह जो कोड ऑफ कॉन्टैक्ट बनी है इसके बारे में आपके क्या विचार है सारी बातें छोड़ो मुझे तो बस कुछ बातों का जवाब दो पहली बात सुना ध्यान से ये कोड ऑफ कॉन्टैक्ट पर हम बाद में आएंगे सर क्या यह एंगल 90 डिग्री है हान है सर क्यों सर्कल के सेंटर से सर्कल की टांगें पर अगर परपेंडिकुलर ड्रॉप करोगे तो रेडियस है तो बिल्कुल ही 90° है क्या यह 90 डिग्री एंगल है बिल्कुल है सर ये 90 डिग्री एंगल है क्या हान ये तो हो गई ना क्योंकि ये डायरेक्टर सर्कल है तो डायरेक्टर सर्कल्स अगर मैंने टांगें ड्रॉ की तो वो परपेंडिकुलरली तो होती है ठीक है सर यह यह 90 हो गया तो सर यह तो तय है की यह क्या होगा 90 क्या अब आप मुझे आंसर बता रहे हो क्या अब आप मुझे आंसर बता रहे हो अगर अभी भी नहीं ए रहा तो मैं फिर से बोलता हूं बड़ा सिंपल सा आंसर है जनाब suniyega ध्यान से की सर आपसे कोड ऑफ कॉन्टैक्ट की लेंथ पूछी जा रही है मतलब आपसे कौन सी लें पूछी जा रही है आपसे ये वाली लेंथ पूछी जा रही है सर ये जैसे भी निकालो जो भी निकालो मुझे नहीं पता पर क्या मैं ये का सकता हूं की सर अगर ये स्क्वायर है तो इसके दोनों daigonals इक्वल लेंथ के होंगे स्क्वायर है ना दोनों daigonals इक्वल लेंथ के होते हैं तो अगर आपसे कोड ऑफ कॉन्टैक्ट की लेंथ पूछी जा रही है जैसे मैं इसे का डन थोड़ा देर के लिए आर और इसे का दो के है ना और इसे मैं का डन थोड़ी देर के लिए सी तो जो लेंथ आपसे पूछी जा रही है वो क्या पूछी जा रही है वो पूछी जा रही है r³ जो लेंथ आपसे पूछे जा रही है वो क्या पूछी जा रही है r³ लेकिन आप उसे लेंथ को क्या कहोगे आप उसे लेंथ को कहोगे पीसी आप उसे लाइन को क्या कहोगे आप उसे कहोगे पीसी स्टेटमेंट जो लेंथ आपसे एक क्वेश्चन में पूछे जा रही है वो क्या है r³ और उसे लेंथ को आप क्या कहोगे पीसी समझ में ए रहा है ना भाई सर वो दोनों ही तो diagnals हैं आपसे जो पूछा है वो अर्क है पर मैं आरके नहीं निकलना चाह रहा हूं निकल लूंगा पर मैं उसको इस तरीके से निकल लेता हूं की सर r³ जो है वो पीसी ही तो है क्योंकि ये स्क्वायर है c³ पी आर तो टेक्निकल टच निकल लो तो आर क्यों निकल जाएगी पर पीसी कैसे निकलेंगे सर बता रहे हो क्या पीसी कैसे निकलेंगे सर पीसी तो निकाला रखा है पीसी क्या है सर पीसी है डायरेक्टर सर्कल की रेडियस दिख रही है क्या डायरेक्टर सर्कल की रेडियस तो मैंने सीखी है क्या अंडर रूट तू टाइम्स होती है होती है की नहीं तो क्या मैं गिवन सर्कल की रेडियस निकल सकता हूं निकल लो उसका क्या कर दूंगा वो क्या होगा आंसर समझ में ए रहा है क्या कुछ तो सर आप जो का रहे हो वो करते हैं पहले तो मैं क्या करता हूं ये जो गिवन सर्कल है गिवन सर्कल मतलब भाई ये ये आपका गिवन सर्कल है ना इसकी रेडियस क्या होगी सर इसकी रेडियस क्या होगी अच्छा देखो यहां से अगर सेंटर के क्वाड्रेंट निकले तो क्या हो जाएगा -ए तो इसके सेंटर के कोऑर्डिनेट्स क्या हो गया माइंस ए बी अन्य डेट वाज नॉट नीडेड अन्य वेट वैसे नॉट नीडेड बट फिर भी हम निकल लेते हैं ना तो अगर मैं निकलूं G2 + x² तो ये हो जाएंगे आपके a² + b² तो कितना हो जाएगा a² + b² समझ ए रहा है क्या जी स्क्वायर प्लस एक्स स्क्वायर माइंस सी माइंस सी मतलब माइंस ऑफ दिस यू कितना हो जाएगा - a² और माइंस माइंस प्लस कितना बी स्क्वायर ये क्या है आपके गिवन सर्कल की रेडियस गिवन सर्कल की रेडियस मतलब जैसे अगर आप कहना चाहो तो आप क्या का सकते हो अगर आपको भूलना नहीं तो आप इसे का सकते हो या तो कर और उसमें अंडर रूट में लगाऊंगा है ना इसको आप का सकते हो सी आर या ऑफ कोर्स आप इसे का सकते हो सी के ये बात समझ आई की नहीं भाई जल्दी जवाब दो ये बात समझ ए गई क्या सर आगे a² कैंसिल तो बच्चा बी स्क्वायर बी स्क्वायर तो ये हो जाएगा तू बी स्क्वायर तो तू भी स्क्वायर का अंडर रूट लिया तो कितना हो जाएगा सही हो जाएगा √2 बी क्योंकि b2b तो ये क्या हुई सर आपके गिवन सर्कल की रेडियस ये हुई आपकी गिवन सर्कल की रेडियस सर मुझे गिवन सर्कल की रेडियस में इंटरेस्ट नहीं है है पर कब मुझे तो डायरेक्टर सर्कल की रेडियस है ये तो वो गिवन सर्कल की रेडियस का अंडररूट तू टाइम्स होगी तो मैं क्या बोलूंगा सर मैं कहूंगा की जो आपको चाहिए वो क्या है सर आपको जो चाहिए वो है r³ आपको जो चाहिए वो है आरके ऊपर मैं कहता हूं आरके से अच्छा शब्द होगा आप उसे का लीजिए पीसी और पीसी को मैं कैसे निकलूंगा पीसी को मैं निकल लूंगा इस गिवन सर्कल की रेडियस का अंडर रूट तू टाइम्स इस गिवन सर्कल की रेडियस का अंडर रूट तू टाइम्स जिसे मैं क्या कहूंगा √2 √2 ऑफ बी तो 2 2 मतलब तो कितना हो जाएगा तू बी मेरा बस आपसे ये कहना है एक समझदार स्टूडेंट ये सब नहीं लिखेगा तो आपने क्यों लिखा आपको समझने के लिए मैंने लिखा था की आपको समझ ए जाए आसानी से पर आप एग्जाम में क्या करेंगे आप बस ये सीनरी करेंगे और ये देखते ही आपको क्लिक हो जाएगा की सर इसकी रेडियस निकालो उसका रूट टाइम्स कर दो इसकी आप यहां दिमाग में निकलोगे फटाफट सर ए स्क्वायर प्लस बी स्क्वायर है ना मतलब a² + b² - a² + b² मतलब इसका नेगेटिव करेंगे तो वो हो जाएगा तो हो जाएगा √2 बी और अंडर रूट तू भी का रूट तू टाइम्स तू भी आप तू भी ऑप्शन मार्क करके आगे बढ़ जाओ आपको रिलाइज बाइक अच्छा मजेदार इंटरेस्टिंग और एक बेहतरीन सा एक क्वेश्चन था जिसके थ्रू आपने काफी सारी इनसाइड सीखी मेरा क्रश आज के लेक्चर में आज के आज के लेक्चर के दौरान सिर्फ यही है की आपको मैं समझा पाऊंगी डायरेक्टर सर्कल होता क्या है और उसके और गिवन सर्कल में रेडियस में क्या रेडियस में उसके सेंटर में क्या रिलेशन होता है बस यह बात याद रखना क्योंकि जब भी किसी भी क्वेश्चन में कोई एक टर्म उसे कर दे है ना एग्जाम में वो बोलेगा नहीं वो आपको परिभाषाएं थोड़ी लिख के देगी क्योंकि डायरेक्टर सर्कल क्या होता है तो ये आपके दिमाग भी आना चाहिए की डायरेक्टर सर्कल का मतलब क्या होता है और आप ऐसे चीज सोच पाए बना पाए देख पाए ये आपकी काबिलियत आपकी क्षमता होगी क्या पैसा कर लेंगे स्टूडेंट्स उसे पर क्या चीज निकल कर आती हैं कॉन्सेप्ट बड़ा आसान और सिंपल सा है सुनेगा ध्यान से मैन लो आपके पास एक सर्कल है यह कॉन्सेप्ट एक्चुअली और आसान है क्योंकि समझना बात को की आप ये बात समझ का रहे हो की सर अगर किसी सर्कल पर अगर किसी सर्कल पर आपने कोई tagent ड्रॉ की लत से ये tagent है तो नॉर्मल क्या होगा सर नॉर्मल कुछ नहीं होगा उसी पॉइंट पर अगर मुझे नॉर्मल चाहिए तो जो ये टांगें आपने ड्रॉ की है उसे पर क्या बेसिकली सर उसे टांगें पर परपेंडिकुलर उसे टैसेंट पर क्या सर उसे tyanjan पर परपेंडिकुलर क्या मैं आपको अपनी बात समझा का रहा हूं क्या आप मेरी बात समझ का रहे यह जो निकल कर आया है ना यह जो निकल कर आया है दिस इस कॉल्ड डी नॉर्मल और ये क्या थी आपको क्या नॉर्मल की परिभाषा समझ ए रही है की नॉर्मल एक्जेक्टली होता क्या है ठीक है सर आपकी बात मैन ली लेकिन इसमें क्या चीज होंगी क्या निकल कर आएंगे बातें उनके बारे में बात करेंगे क्या आप बिल्कुल करिए पहली बात तो सर मैं हमेशा tenjent को निकल सकता हूं बहुत सारे तरीके से जिनके बारे में उन्होंने बात की जैसे की मेरी सीधी सीधी सी एक एप्रोच या थॉट हमेशा रहेगा की सर tenjent क्या होगी सर्कल की टांगें होती है जो पॉइंट हमें दे दिया जाता है जनरली एक केस में के सर X1 कमा y1 आपकी क्या हो की इस पॉइंट से पास होने वाली टैसेंट एक बहुत ही मजेदार सी बात है सर्कल्स को लेकर स्पेसिफिकली की सर्कल के सारे ही नॉर्मल सर्कल के सेंटर से पास होते हैं सर्कल के सारे ही नॉर्मल सर्कल के सेंटर से पास होते हैं का क्या मतलब है सर की अगर सर्कल की इक्वेशन है x² + y² + 2gx + 2fy + 6 = 0 अब दो बातें दो क्या बात है सर या तो मुझे क्या पता करना होगा या तो मैं यह पता कर लूं की उसे नॉर्मल की स्लोप क्या है क्योंकि मुझे पता होगा वो नॉर्मल डेफिनेटली इस पॉइंट से तो पास हो ही रहा है तो स्लोप और उसे पॉइंट से मैं सर्कल के नॉर्मल की इस पॉइंट पर इस स्लोप वाले नॉर्मल की इक्वेशन बना दूंगा क्योंकि मुझे पता है ना क्या की सर सर्कल किस पॉइंट से पास हो रहे हैं और सर्कल के स्लोप क्या है सर अगर स्लोप नदी जाए तो अगर स्लोप ना दी जाए तो मुझे कहीं ना कहीं से ये आइडिया दिया रहा होगा की ना किस तरीके से की वो नॉर्मल किस और एक्स्ट्रा पॉइंट से पास हो रहा है अपार्ट फ्रॉम सर्कल के सेंटर से पास होने के उससे क्या होगा सर मुझे एक स्ट्रेट लाइन की क्वेश्चन चाहिए टेक्निकल जिसके मुझे दोनों पॉइंट से पीओएस होने के कोऑर्डिनेट्स दिए गए हैं की वो किन दो पॉइंट से पास होती है तो क्या इन दोनों की हेल्प से मैं उसे नॉर्मल की इक्वेशन नहीं लिख सकता दोनों बातें आपके पास है क्या बात है आप समझ का रहे हो या तो क्या स्लोप दे दी जाए और वो किस पॉइंट से पास होता है हमें पता है क्वेश्चन बना देंगे या फिर वो किन दो पॉइंट से पास होती है ये पता हो और हम उसे नॉर्मल की यानी स्ट्रेट लाइन की इक्वेशन बना देंगे का दोनों बातें आपको समझ आई थिंक सर आसान है तो ऐसा करना है बस इतना सा ही कॉन्सेप्ट है बस यह सच में एक्चुअली नॉर्मल्स की बात पता है कहां से आएगी हर बार टांगें से आएगी तो वो टैसेंट से घुमा फिर के नॉर्मल कॉप को रिलेट करेगा तो आपको डील करना होगा टैसेंट आगे अन कॉन्सेप्ट बहुत फास्ट है जहां पे हमने देखा सर सर्कल पर अगर कोई पॉइंट लाइक कर रहा है और उससे अगर टैसेंट पास हो रही तो टी = 0 और सर्कल के बाहर किसी एक्सटर्नल पॉइंट से आपने टेंट्स का पैर बनाया तो कैसे या फिर अगर आप यू नो कोड ऑफ कॉन्टैक्ट के थ्रू वहां पर बात कर रहे हो तो कैसे ही ऐसे अलग-अलग तरीके से हमने बात किया और हर एक तरीका हमने एक्सप्लोर किया उन्हें सारे त्रिकोण के एक्सटेंशंस निकल कर आते हैं नॉर्मल्स में पर मैं का रहा हूं नॉर्मल के लिए बस आप ये बात याद रखें की नॉर्मल इस नथिंग बट अन स्ट्रेट लाइन स्पेशल कैटिगरी और काइंड ऑफ स्ट्रेट लाइन विच डेफिनेटली पासेस थ्रू डी सेंटर ऑफ डी सर्कल अगर मैं सर्कल के कॉन्टेक्स्ट में नॉर्मल की बात करूं और क्या सर दूसरी बात ये निकल कर आती है की नॉर्मल अगर सर्कल के सेंटर से पास होता है तो नॉर्मल यानी स्ट्रेट लाइन की इक्वेशन लिखने के लिए मुझे दो में से कोई एक चीज की और आवश्यकता होगी या तो नॉर्मल यानी उसे स्ट्रेट लाइन की या तो स्लोप या फिर वो एक और किस पॉइंट से पास होता है वो बहुत कन्ज्यूरिंग तो नहीं हो रहा है ये मतलब जनरली आप जब नॉर्मल ऐसा कॉन्सेप्ट पढ़ोगे ना तो आप क्या पढ़ोगे जैसे मैन लो ये कर्व है ये कर्व है तो आपसे मैन लो स्कार्फ पे कोई कहे की पॉइंट है और इस पॉइंट पर आप मुझे नॉर्मल ड्रॉ करके दीजिए तो मैथमेटिकली नॉर्मल क्या मतलब होता है सर की आप क्या करो पहले तो उसे करके उसे पॉइंट पर आप बना दो एक टैसेंट उसे कर्व के उसे पॉइंट पर आप क्या बना दोगे सर एक टैसेंट के उसे पॉइंट पर अब जो आपने क्या बना दीजिए एक परपेंडिकुलर आप एक परपेंडिकुलर ड्रॉप कर दीजिए जैसे ही उसे टैसेंट पर आप एक परपेंडिकुलर ड्रॉप करेंगे वह जो परपेंडिकुलर होगा वही आपका कुछ नहीं आपका नॉर्मल होगा ये होता है नॉर्मल ऐसा कॉन्सेप्ट जो हम डर में नॉर्मल की बात करूं तो जो मेरी अंडरस्टैंडिंग कहती है वो क्या की सर्कल के कॉन्टेक्स्ट में नॉर्मल निकलने के दो तरीके किए हैं क्या बजे बात करें बिल्कुल सर कर लीजिए तो अगर नॉर्मल पर बात की जाए तो नॉर्मल कॉन्सेप्ट आई होप आपको क्लियर हो रहे हैं सुनेगा ध्यान से सर्च डी लाइन परपेंडिकुलर तू डी टैसेंट तू डी सर्कल आते अन्य पॉइंट ऑफ कॉन्टैक्ट पासेस थ्रू डी सेंटर ऑफ डी सर्कल बिल्कुल सर ये तो सही बात है अभी जस्ट हमने डिस्कस किया नॉर्मल तू डी सर्कल इसे लाइन पासिंग थ्रू डी सेंटर ऑफ डी सर्कल इस इनफेक्ट ऑल डी डायमीटर्स ऑफ डी सर्कल आर डी नॉर्मल तू डी सर्कल क्योंकि हर एक और डायमीटर जो सर्कल का सेंटर से पास होता है इफ इट इस एक्सटेंडेड फॉर डी वुडन इट बी आर नॉर्मल वुडन इट बिहेव वुडन इट बी बिहेविंग लाइक अन नॉर्मल यस इट विल बी अन नॉर्मल आप मेरी बात समझ का रहे हो एंडिंग स्टूडेंट्स अगर मैं यहां से आगे बात करूं तो मैन लिया वही बात क्योंकि मेरे पास एक सर्कल है x² + y² + 2gx + 2xy + सी = 0 तो सर नॉर्मल की जो इसकी इक्वेशन आएगी मैन लो अगर सर्कल पर एक पॉइंट है X1 कमा y1 सर्कल के सेंटर है माइंस जी कमा एफ तो मुझे नॉर्मल किन दो पॉइंट से पास होता है पता चल ही चुका है तो क्या मैं नॉर्मल की इक्वेशन नहीं लिख सकता सर इससे आसान तो कुछ हो ही नहीं है ना लाइफ में वही बात सर ए - y1 = ए - एफ / एक्स - जी टाइम्स एक्स - X1 और इसे आप सिंपलीफाई करके लिख सकते हो या मैन लो मुझे ये ना कहे वो की सर नॉर्मल जो है आपका वो X1 कमा ए वैन सर्कल पर एक पॉइंट है उससे पास होता है बल्कि वो ये का दे की सर आपका जो नॉर्मल है उसकी एक स्लोप है तो बिग बॉस में एक बात तो जानते ही हूं या तो X1 y1 से पास हो रहा है या फिर -0 - एक्स या दोनों से पास हो रहा है तो माइंस जी एक्स वैन ए वैन उसे कर लेंगे और मुझे जब स्लोप दे दी जाएगी तो मैं स्लोप उसे कर लूंगा का दोनों बात आप समझ का रहे हो डाइजेस्ट कर का रहे हो इसको समझने का एक आसान तरीका मेरे पास है जो की हमेशा क्या होगा एक क्वेश्चन तो ये कुछ असाइनमेंट क्वेश्चंस हैं जो आप देखते जाइए और इन्हें सॉल्व कर लीजिए क्योंकि इन्हीं पर बेस्ड हम बातें करने वाले हैं तो पहला आज का आपका असाइनमेंट क्वेश्चन ये रहा आई होप ये आपका असाइनमेंट क्वेश्चन दिख रहा है जो आज का आपका पहला असाइनमेंट क्वेश्चन और मेरा यकीन है आपने स्क्रीनशॉट लिया क्या करना है सर इस क्वेश्चन में फाइंड डी इक्वेशन ऑफ टांगें तू डी सर्कल विच आर परपेंडिकुलर तू दिस लाइन आई थिंक ये तो क्वेश्चन आसान है क्यों आसान है सर क्योंकि इस सर्कल की tenjent की जब आप इक्वेशन nikalwaoge और वो इस लाइन के परपेंडिकुलर होगी तो क्या मैं सबसे पहले तो इस लाइन की स्लोप निकल लूंगा सर जो भी इस लाइन की स्लोप आएगी जो आप देख कर ही का सकते हो तो इसकी परपेंडिकुलर लाइन की स्लोप निकल सकता हूं क्या बिल्कुल निकल सकता हूं सर वही बात जो भी इस लाइन के स्लोप आएगी उसका नेगेटिव मुझे पता है की इस पर जो आपने टेंशन ड्रॉ की है यार अगर tesence जो ड्रा किया उनकी क्या स्लोप है तो क्या बार-बार से बात क्या या तो स्लो फॉर्मेट इंजन के लिख लो और उसे टैसेंट की सर्कल के सेंटर के कोऑर्डिनेट्स से परपेंडिकुलर डिस्टेंस सर्कल की रेडियस के इक्वल रख दो बात खत्म या फिर उसे टैसेंट की इक्वेशन को क्योंकि स्लोप फॉर्म में आपने निकल और इस सर्कल की इक्वेशन को सॉल्व कर लो और बात खत्म हो जाएगी कैसे क्योंकि सर आपको ये पता होगा की ये क्वाड्रेटिक बनेगी ए एक्स पे और उसे क्वाड्रेटिक के दो डिस्टिंक्ट रूट्स होने चाहिए सॉरी क्वाड्रेटिक का डिस्क्रिमिनेंट जीरो होना चाहिए दो नहीं होना चाहिए एक ही रूट होना चाहिए क्या आप ये बात समझ में क्योंकि वो पॉइंट को टच कर रही है वो सर्कल को टच कर रही है वो इंटरसेक्ट नहीं कर रही है दो पॉइंट्स पे तो दो डिस्टेंस रूट्स नहीं होंगे एक ही रूट होगा आई होप आपको ये बातें याद है हमने एक बहुत अच्छे लेक्चर में बहुत डिटेल में ये दोनों अप्रोचों डिस्कस किए हैं आई विल प्रेफर गोइंग विद डी फर्स्ट एप्रोच यहां पे सीधा-सीधा तरीका है ए = एमएक्स + सी फॉर्म में एक क्वेश्चन लिखूंगा एम यहां से निकल लूंगा इसका नेगेटिव रिसिप्रोकल इसकी स्लोप का और उसे लाइन की इसके सेंटर से बहुत आसान है परपेंडिकुलर डिस्टेंस निकल लूंगा और वो इसी सर्कल के रेडियस के इक्वल रख दूंगा आसानी से क्वेश्चन हो जाएगा दो वैल्यूज आएंगे सी की वो जब आप रखेंगे तो ये आपके दो इक्वेशंस ए जाएंगे आई थिंक ऐसे क्वेश्चंस हमने किए हैं तो ये आज का आपका पहला असाइनमेंट क्वेश्चन जैसे आप का सकते हो होमवर्क क्वेश्चन तो क्या आप सभी ने इसका स्क्रीनशॉट ले लिया है मैं चाह रहा हूं की मैं क्वेश्चन आपको करवाना ताकि आपको चीज मैं समझा पाऊं की आज जो हम नॉर्मल वाला कॉन्सेप्ट पढ़ रहे हैं उसे हम कैसे करने वाले हैं बात बस इतनी सी जैसे मैन लो एक सर्कल है x² + y² - 2X = 0 और सर इस सर्कल पर आप एक नॉर्मल ड्रॉ करना चाह रहे हो जो की इस लाइन के पैरेलल है इस लाइन पर इसे सर्कल पर आप एक नॉर्मल ड्रॉ करना चाह रहे हो जो की इस लाइन के पैरेलल है इस लाइन के पैरेलल होने का क्या मतलब है इसका मतलब है जैसे अगर मैं ये लिखूं तो मैं इस लाइन को क्या लिख सकता हूं मैं इस लाइन को लिख सकता हूं सर एक्स + 2y = 3 तो एक्स को उधर लेकर गए तो 2y कितना ए जाएगा सर 2y ए जाएगा -एक्स + 3 तो यहां से अगर मैं ए निकलूं तो ए कितना ए जाएगा सर ए ए जाएगा -1 / 2 टाइम्स एक्स + 3 आपसे बस मेरा ये कहना है की अगर इस लाइन के आप स्लोप देखोगे तो ये कितनी आती है सर माइंस 1 / 2 तो जो इसका नॉर्मल है उसकी भी स्लोप कितनी होगी -1 / 2 क्योंकि वो नॉर्मल इस लाइन के पैरेलल है बाय डी वे सर क्या मैं जानता हूं की हर सर्कल का कोई भी नॉर्मल उसके सेंटर से पास होता है है तो मुझे नॉर्मल की स्लोप पता है और नॉर्मल यह पता चल जाए किस पॉइंट से पास हो रहा है तो क्या normaltask होगा नहीं सर तो मैं थोड़ी अगर कोशिश करूं तो देखो इस सर्कल के सेंटर के कोऑर्डिनेट्स के होंगे सर 2gf 2gx और 2fy से निकल लेते हैं हम तो एक्स के ऑफिशल का नेगेटिव हाफ -2 का हाफ वैन माइंस मतलब आप समझ रहे हो ना -2 का हाफ -1 और उसका नेगेटिव वैन तो सेंटर के क्वाड्रेंट क्या ए जाएंगे वैन सर ए वाली टर्म तो है ही नहीं मतलब जीरो तो ये क्या हो जाएगा जीरो तो टेक्निकल नॉर्मल वैन कमा जीरो से पास हो रहा है और वो इतनी स्लोप कैरी कर रहा है तो क्या उसकी इक्वेशन नहीं लिख सकते सर ये आज का सबसे आसान क्वेश्चन होगा क्योंकि नॉर्मल की इक्वेशन होगी ए - y1 = एम जो की कितना है -1 / 2 टाइम्स एक्स - X1 जो की होगा एक्स - 1 और अगर मैं क्वेश्चन को लिखना चाहूं तो मैं फाइनली क्या लिखूंगा सर तू को इधर ले आते हैं तो ये हो जाता है 2y यहां पर हो जाएगा माइंस और माइंस माइंस है सर क्या निकल कर ए रहा है एक्स + 2y = ऑफकोर्स क्या वैन ये किसकी इक्वेशन ए रही है सर ये ए रही है आपके नॉर्मल की इक्वेशन क्या इसमें कहीं भी किसी भी स्टेप में आपको कोई डाउट आई थिंक सर ये तो एक आसान क्वेश्चन था बस ये समझने के लिए रखा गया था की नॉर्मल बहुत टू कॉन्सेप्ट नहीं है स्पेशली तब जब आपने टेंसेज के अलग-अलग सिनेरियो से डील करना सिख लिया है वापस मूव करते हैं आज के असाइनमेंट क्वेश्चंस की तरफ देखिएगा आज का आपका अगला असाइनमेंट क्वेश्चन ये भी आपका आज का अगला या सेकंड होमवर्क क्वेश्चन है जो आपको करना है देख लेते हैं इस क्वेश्चन में वो क्या बोल रहा है वो बोल रहा है फाइंड डी इक्वेशन ऑफ पैर ऑफ टांगेंट्स पैर ऑफ टांगेंट्स आई होप आपको कुछ याद ए रहा है ड्रोन फ्रॉम ओरिजिन तू दिस सर्कल सबसे आसान बात है जीवन की है क्यों सर क्योंकि एक्सटर्नल पॉइंट से अगर पैर ऑफ निकालनी है तो वो निकलती हैं ss1 = t² आप समझ रहे हो ना एक सर्कल है इसको ओरिजिन से आपने पैर ऑफ टांगें बना कर दे दी तो वो ऑफ कोर्स कैसे होंगे सर ss1 = t² और एस क्या होता है सर एस आपका सर्कल S1 क्या सर एक्स और ए की जगह जिस पॉइंट्स है यानी क्या ओरिजिन यानी 0 पास कर दो तो आपकी एक वैल्यू हो जाएगी 20 है ना तो 20 से इस इक्वेशन को मल्टीप्लाई करना है और वो किसके इक्वल रखना है t² के और टी तो आपको याद है सर 0 से आपको टी बनाना है तो X1 यानी 0 + ए ए वैन यानी जीरो फिर यहां पे क्या रखोगे एक्स + X1 यानी क्या हो जाएगा एक्स + 0 / 2 तो ये 10 ए + y1 यानी ए + 0 Y2 यानी फिर ये 10 + 20 आई थिंक बहुत आसान से एक क्वेश्चन बनेगी इस क्वेश्चन का आंसर होता है वर्बल बस इसे देखकर ही दे सकते हो एक या दो लाइन लिखोगे क्योंकि जीरो कमा जीरो है ना ss1 और टी निकलने में बहुत आसानी से हेल्प कर देगा आपकी आई थिंक ये चीज हमने पढ़ी है और ये डिटेल में आपने पढ़ा है बस ये इधर रिवीजन एक होमवर्क क्वेश्चन रखा गया है ताकि आप ये कॉन्सेप्ट ही कॉल कर लें की किसी एक्सटर्नल पॉइंट से अगर पैर ऑफ टांगें ड्रॉ किए जाते हैं तो कितनी आसानी से लिखी जा सकती है तो यह मत bhuliyega की ss1=t^2 हमारी हेल्प करता है बिना ज्यादा सोचते समझे आई होप आपको ये याद हो चुका है अगर इसके बाद में मूव करूं नेक्स्ट हमारे असाइनमेंट क्वेश्चन की तरफ तो ये आपका थर्ड असाइनमेंट क्वेश्चन है स्टूडेंट्स इसका भी स्क्रीनशॉट ले लीजिए मेरा यकीन है आपने इस क्वेश्चन का स्क्रीनशॉट ले लिया है और अब मैं आपसे चाह रहा हूं इस क्वेश्चन का स्क्रीनशॉट लेने को क्या बोल रहा है वो सुनेगा ध्यान से इफ डी टैसेंट आते थ्री कमा -4 तू दिस सर्कल कट दिस सर्कल है आपका इस सर्कल पर आपने टांगें ड्रॉ की है तो सर किसी सर्कल पर अगर किसी पॉइंट है मतलब ये सर्कल पर एक पॉइंट है कौन सा थ्री कमा -4 तो यह जो टैसेंट ड्रॉप की गई इसकी इक्वेशन क्या होती है सर इसकी इक्वेशन दी जाती है टी = 0 मैंने आप क्या करोगे 3X - 4y - 4 एक्स + 3 / 2 + 2 4 / 2 - 5 वो आपकी टेंशन हो जाएगी फिर वो का रहा है वो जो टैसेंट है वो किसी और सर्कल को किसी और सर्कल मतलब इस सर्कल को 2 पॉइंट्स पर टच कर रहे हैं तो आप क्या करना पड़ता है वह जो टैसेंट की इक्वेशन आएगी ना और इसको सॉल्व करना अलग टेंशन होगी मतलब एक्स + बी / सी फॉर्म में कोई एक लीनियर इक्वेशन तो वहां से एक्स को ए की टर्म्स में या फिर ए को एक्स की टर्म सिर्फ यू नो एक्सप्रेस कर देना तो उससे क्या होगा सर वो वैल्यूज यहां रख देना मतलब या तो ये पुरी इक्वेशन एक्स में ए जाएगी या पुरी ए में ए जाएगी और ये क्या होगी क्वाड्रेटिक अब मैं आपसे कहूंगा की सर अगर ये क्वाड्रेटिक है तो उसके दो रूट्स आने चाहिए क्योंकि इन दोनों कार्स को यानी सर्कल और स्ट्रेट लाइन को सॉल्व किया एक तो स्ट्रेट लाइन की कुछ नहीं की टेंशन की यहां से और सर्कल की इन दोनों को अगर मैंने सॉल्व किया तो मुझे दो रूट्स निकलने चाहिए क्योंकि दो जगह इंटरसेक्ट करेगी ना और दो रूट्स मतलब डिस्क्रिमिनेंट ग्रेटर दैन जीरो होना चाहिए इक्वल तू जीरो भी नहीं ग्रेटर दें जीरो और बस वो कंडीशन अगर हम अप्लाई कर देंगे तो सर बात खत्म हो जाएगी वो कंडीशन अगर हम अप्लाई कर देंगे तो बात इसलिए खत्म हो जाएगी सर क्योंकि फिर आपके वो दोनों पॉइंट ऑफ इंटरसेक्शन निकल जाएंगे है और अगर आपके वह दोनों पॉइंट ऑफ इंटरसेक्शन निकल जाएंगे तो वह क्या होगा इसका मिड पॉइंट पर मैं एक बात कहूं इसको एक और आसान तरीके से निकल सकते हो कैसे सुनना कितना आसान है सॉल्व करने की जरूरत ही नहीं है एक तरीका और सुनो यहां पर इसके सेंटर की कोऑर्डिनेट्स निकल लोग क्या सर ये सर्कल के सेंटर के कोऑर्डिनेटर क्या होंगे -8 -1 तो सर्कल के सेंटर के cardinate होंगे -8 कमा करना है अगर इस कॉल्ड के मिड पॉइंट निकलना है अगर इस कॉल्ड के मिड पॉइंट निकलना है तो सर कॉर्ड का मिडिल पॉइंट इस नथिंग बट सर्कल के सेंटर से ड्रॉप किए गए परपेंडिकुलर के फीट और स्ट्रेट लाइन चैप्टर में हमने पढ़ा है की सर किसी पॉइंट से किसी स्ट्रेट लाइन पर अगर परपेंडिकुलर ड्रॉप किया जाए तो उसके फीट के कोऑर्डिनेट्स क्या होते हैं आपको इक्वेशन या फॉर्म मिला याद होगा एक्स - एक्स 1 / ए = ए - y1 अपॉन बी = ax1 आई होप A1 एक्स + b1y + सी मतलब ये जो आपके होते हैं एक्स एक्स में जो भी आप लिखते हो वो दोनों मतलब कॉएफिशिएंट और इस इस इन वैल्यूज का प्रोडक्ट अपॉन में a² + b² अंडर रूट नहीं लगाते हैं याद ए रहा है स्टूडेंट्स बस आगे आप माइंस साइन लगाते हो और अगर इमेज लेनी होती तो माइंस तू लगाते हो याद ए रहा है क्या ये फॉर्मूले और ये सब कुछ हमने पढ़ा है स्टूडेंट तो या तो पॉइंट ऑफ इंटरसेक्शन निकालो फिर मिड पॉइंट निकालो जो की एक लंबी प्रक्रिया है ये आसान तरीका होगा की सेंटर से इस लाइन पर ड्रॉप किए गए परपेंडिकुलर के फीड जो की और आसान तरीका है और मेरे ख्याल से आप ये जानते हो और आप ही कर लोग तो क्या ये क्वेश्चन हो जाएगा पहले तो तेनजेन निकालनी है ग = 0 से और उससे फिर स्ट्रांग पे ड्रॉप किया गया फीट के कोऑर्डिनेट्स स्टैंडर्ड से बेसिक से फॉर्मूले से आई होप ये क्वेश्चन हो जाएगा और एक अच्छा क्वेश्चन है होना चाहिए आप सभी से और आंसर मैच करेगा स्टूडेंट्स आना चाहिए अगर नहीं फॉर्मूला यह जो बात अभी एक्सप्लेन कीजिए सर फिर से suniyega कर लेंगे इस क्वेश्चन को फिर क्या सर आज का अगला असाइनमेंट क्वेश्चन तो ये हो जाएगा आपका अगला साइन इन क्वेश्चन है पहले से इस क्वेश्चन का क्या ले लेंगे आप लोग इस क्वेश्चन का ले लीजिए एक स्क्रीनशॉट फिर क्या क्वेश्चन क्या है वो पढ़ते हैं सर सर क्वेश्चन में जो बातें लिखी गई है उन्हें ही समझने की कोशिश की जाए तो ए बी डी सेंटर ऑफ डी सर्कल तो सर इस सर्कल का सेंटर है जो की ऑफ कोर्स क्या है जो की ऑफ कोर्स होगा वैन कमा 2 तो ये किसके कोऑर्डिनेट्स हो जाएंगे सर एक एक सपोस डेट डी टेंसेज आते पॉइंट बी एंड डी ऑन डी सर्कल मीत आते डी पॉइंट सी तो आपको क्वॉड्रिलैटरल एबीसीडी का एरिया निकलना है आपको ये पता है बी पता है दी पता है वो क्या पूछ रहा है वो सुनना आपको बी पे टांगें ड्रॉ करनी है तो जब बी पर टांगें ड्रॉ करनी है तो इसी का t=0 निकलना है बी के लिए वैन कमा 7 रख के सर्कल -2 रखें बन रहा है उसका कोई मुझसे एरिया पूछेगा तो क्या आपको आपकी स्टेप मेथड आती है और इसके मोड के आगे आप क्या लगाते हो भाई तू आई थिंक बस इतना सा आसान सा क्वेश्चन है इसके अलावा भी एक और तरीका है तो उसके लिए मैं यह क्वेश्चन आपके ऊपर छोड़ रहा हूं की इसका एक और अल्टरनेटिव मेथड हो सकती है क्या जो हम सोच सकते हैं बिल्कुल देयर वेट कैसे सोचना है यह आपका बस ज्यामिति के लिए इसको ड्रॉ करना है आप कुछ चीज दिख जाएंगे और देर आर वैन और तू मोर वेज ऑफ सॉल्विंग डी से क्वेश्चन मैंने एक स्टैंडर्ड ओबवियस एप्रोच बताइए की सर ऐसा तो कर ही सकते हैं पर इसके अलावा भी यहां पर तरीके हैं जैसे आप सोच सकते हो फिर क्या सर फिर आते हैं एक और क्वेश्चन पर और एक और आज के लिए आपका लेते हैं असाइनमेंट क्वेश्चन तो ये आज का आपका अगला असाइनमेंट क्वेश्चन पहले तो सबसे पहले क्या कर लीजिए स्क्रीनशॉट ले लीजिए स्क्रीनशॉट लिया क्या आप समझते हैं इस क्वेश्चन में उसने क्या बोला है वो का रहा है यह जो है आपका ये ऑफ कोर्स फेर ऑफ टांगें है हमने अगर कहीं बाहर से किसी एक्सटर्नल पॉइंट से किसी सर्कल पर पैर ऑफ टांगेंट्स बनाई है अगर किसी सर्कल पर किसी एक्सटर्नल पॉइंट से पैर ऑफ टांगें बनाई हैं तो यह जो भी पॉइंट है हम तो सर उसको बनाने का तरीका जानते हैं ss1 = t² तो यह पैर ऑफ टांगेंट्स का और जो आपने यह एक्सटर्नल पॉइंट से पैर ऑफ़ टेंशन बनाई है ना वह है आपका ओरिजिन जीरो कमा जीरो सर्कल पर बनाएंगे जिस सर्कल की रेडियस कितनी है थ्री ठीक है क्या बनाया क्योंकि ओरिजिनल तो फिर यहां होना चाहिए था तो उसे तरीके से आप बना लेना की ओरिजिन आपकी इस तरफ मेरी बात समझ का रहे हो हेलो ड्रॉप ऐसे कर लेना फिर वो क्या का रहा है इफ ए इस वैन ऑफ डी पॉइंट ऑफ कॉन्टैक्ट तो एक पॉइंट ऑफ कॉन्टैक्ट ऑफ कौन है सर इसका ए तो वह पूछ रहा है फाइंड डी लेंथ ऑफ ओम मेरा यकीन करिए इससे आसान क्वेश्चन आपने कभी नहीं देखा होगा क्या आपको कुछ क्लिक हुआ एक आर्ट देता हूं क्या अभी कुछ क्लिक हुआ आई थिंक अब इससे ज्यादा बताऊंगा तो ये क्वेश्चंस सॉल्व हो जाएगा तो अब आप खुद से सोचेगा अब आप सच में खुद से सोच सकते हो आपके पास पैर ऑफ टांगें की इक्वेशन है इससे क्या हो जाएगा सर इससे अगर आप चाहे इससे अगर आप चाहे तो उन्हें दो अलग-अलग फॉर्म्स में एक्सप्रेस करना है उससे आपको दो अलग-अलग स्ट्रीक लाइंस हो जाएगी जो की ये और ये मतलब इसकी और इसकी इक्वेशन होगी फिर क्या करना है फिर आप sochiyega अगर मैं पूरा बता दूंगा तो क्वेश्चन सॉल्व हो जाएगा और क्वेश्चन खत्म हो जाएगा तो इसके आगे का थोड़ा पार्ट में आपके लिए छोड़ता हूं जो आप करेंगे आई होप यह असाइनमेंट क्वेश्चन है होमवर्क क्वेश्चन है जो आप ट्राई करेंगे फिर क्या सर फिर आगे देखते हैं एक और असाइनमेंट क्वेश्चन जैसे आप चाहें तो का सकते हैं क्या कहेंगे भाई आप इसे होमवर्क क्वेश्चन याद का आपका अगला असाइनमेंट क्वेश्चन है ये बेसिकली डीपी पीस की तरह है जब आप कहते हो की सर हम आईआईटी जी मांस और एडवांस क्या-क्या तैयारी कर रहे हैं मैथमेटिक सब्जेक्ट की तो हम ज्यादा से ज्यादा क्वेश्चंस प्रैक्टिस करना चाहेंगे उसी के थ्रू कॉन्सेप्ट्स कंसोलिडेटेड होंगे बिल्कुल सही बात है तो ये क्वेश्चंस आपके प्रैक्टिस क्वेश्चंस असाइनमेंट क्वेश्चंस या डीपी भी होमवर्क क्वेश्चंस की तरह ट्रीट किए जा सकते हैं ये आपके लिए दिल्ली रखे जाते हैं ताकि आप inconsps को रिकॉल कर ले रे विजिट कर लें और रिवाइज करले ताकि आपकी हर एक चीज अच्छे से बनी रहे सर अगर इस क्वेश्चन को देखा जाए तो क्या लिखा है फाइंड डी डिस्टेंस बिटवीन डी कोड्स ऑफ कॉन्टैक्ट ऑफ डी टैसेंट तू डी सर्कल ऑफ डी टांगें तू डी सर्कल फ्रॉम डी ओरिजिन एंड डी पॉइंट जी कमा एफ ठीक है सर तो क्वेश्चंस समझ ए रहा है क्या क्वेश्चन क्या है डिस्टेंस बिटवीन डी कोड्स ऑफ कॉन्टैक्ट ऑफ टांगें तू डी सर्कल आई होप भी आप निकल सकते हो कार्ड ऑफ कॉन्टैक्ट के बाद हमने की थी सर कोड ऑफ कॉन्टैक्ट भी टी = 0 से ही दी जाती फ्रॉम दी ओरिजिन तो उसकी जो कोड ऑफ कॉन्टैक्ट ऑफ टांगें है उसकी जो आपकी बेसिकली जो डिस्टेंस है किस-किस से निकालनी है ओरिजिन है और ऑफ कोर्स जी कमा एक्स है एक अच्छा आसान और बेहतरीन सा क्वेश्चन जो आप जरूर ट्री करेंगे ऐसा मेरा मानना है और इसे भी आप करके देखिएगा की क्या फाइनली आंसर निकल कर आता है अगर यह क्वेश्चन आपने देखा तो एक और क्वेश्चन लीजिए अच्छे से लीजिए और सारे क्वेश्चंस करिएगा यह आज का अगला असाइनमेंट क्वेश्चन या फिर मैं कहूं होमवर्क क्वेश्चन जो की आपको ट्राई करना है पढ़ लीजिए क्या लिखा हुआ है स्क्रीनशॉट हवा में वर्बल इंटूटिव सॉल्व कर सकते की यह दो लाइंस के रिस्पेक्ट में बात की थी एंगल बाईसेक्टर की ऐसा ही कुछ एक आपको पर्सपेक्टिव देता है रेडिकल एक्सरसाइज क्या होता है थोड़ा बात करते हैं जैसे मैन लो आपके पास है दो सर्कल्स जैसे मैन लो आपके पास क्या है सर दो सर्कल्स एक सर्कल ये रहा और एक सर्कल ये रहा अब मैन लो इनके बीच में कहीं से कोई एक स्ट्रेट लाइन जा रही है मैन लो इनके बीच में कहीं से कोई एक स्ट्रेट लाइन जा रही है अब इस स्ट्रेट लाइन की खास बात ये है suniyega ध्यान से मैथमेटिक्स भी प्रूफ करेंगे पर इस स्ट्रेट लाइन की खास बात ये है की जब इस स्ट्रेट लाइन पर मौजूद किसी भी पॉइंट से आप tagent ड्रॉ करेंगे ना जब इस स्ट्रेट लाइन पर मौजूद किसी भी पॉइंट से आप tagent ड्रॉ करेंगे ना एक सेकंड दीजिए बस मुझे ये आपकी हो गई टैसेंट आई थिंक नहीं ये होगी है ना तो जब इस स्ट्रेट लाइन पर मौजूद आप किसी भी पॉइंट से टैसेंट ड्रॉ करेंगे तो उन टेंसेज की लेंथ इक्वल होगी कुछ समझ नहीं ए रहा है समझने वाली बात बस इतनी सी है क्या आपके पास दो सर्कल्स होंगे और उन दो सर्कल्स पर इस स्ट्रेट लाइन पर मौजूद किसी भी पॉइंट से जब आप tenjes ड्रा करेंगे तो वह टेंसेज होगी इक्वल लेंथ की तो यह जो स्ट्रेट लाइन है ना जिस पर मौजूद हर पॉइंट से ड्रॉ की टांगेंट्स के लिए इक्विवेलेंट इस पॉइंट से जब आप करें तो इक्वल है किसी और पॉइंट से अत्यंत ड्रा करते हो जैसे मैन लो यहां से ड्रॉ की तो ये टैसेंट ऑफ कोर्स और कौन सी ये टैसेंट ऑफ कोर्स कौन सी ये वाली है ना यह वाली अब मैं ऑफ कोर्स बहुत अच्छा बना नहीं रहा हूं बहुत अच्छा ड्राइंग कर रहा हूं बट आय होप आप मेरी बात समझ का रहे हो मेरी बात बड़ी आसान सी है की ये लेंथ है ना और यह लेंथ क्या होगी दो सर्कल्स का रेडिकल एक्सिस उसे पॉइंट का लोकस होता है उसे पॉइंट का पात होता है जहां से ड्रा की गई tengence की लेंथ इक्वल होती है दोनों सर्कल्स पर कुछ समझा का रहा हूं क्या मैं क्या आप यह बात समझ का रहे हो आई होप थोड़ा बहुत क्लेरिटी डिवेलप हो रहा है बस इसी बात को थोड़ा डाइजेस्ट करते समय रेडिकल एक्सेस एक्जेक्टली होता क्या है तो रेडिकल एक्सेस ऑफ तू सर्कल्स इस डी लोकल्स ऑफ अन पॉइंट वही बात किसी एक पॉइंट का बात जो वो ट्रेवल करता है इस तरीके से की सर लेंथ ऑफ डी टेनिस ड्रोन फ्रॉम इट इस पॉइंट से इस पॉइंट से ड्रॉ करो उन दोनों सर्कल्स पर तो उन टेनिस की लेंथ क्या हो भाई बात समझ आई या नहीं आई थिंक चीज क्लियर है तो यह जिस पथ को ट्रेवल कर रहा है जिस जिस रास्ते पर यह मूव कर रहा है रेडिकल एक्सेस क्योंकि ये तकनीक है बेसिकली एक स्ट्रेट लाइन कैसे होती है बात कर लेते हैं बड़ी सिंपल सी बात है स्टूडेंट्स मैन लो आपके पास दो सर्कल्स हैं पहला सर्कल वही x² + y² + 2g1x + 2f1y + C1 और दूसरा सर्कल है x² + y² + 2g2 2fy और सी तू है ना अब मेरा आपसे ये कहना है की सर अगर कोई पॉइंट है जिससे ड्रा की गई इस सर्कल पर टैसेंट और इस सरकार पर तेनजेन की लेंथ इक्वल है तो क्या आपको याद है की आपको याद है हमने फॉर्मूला पढ़ा था जिसमें कोई बहुत बड़ी विशेष बात हम नहीं की थी हमने डिस्कस किया था याद करो स्टूडेंट्स की सर किसी सर्कल पर किसी एक्सटर्नल पॉइंट्स अगर कोई टेंशन ड्रॉ की जाए किसी एक्सटर्नल पॉइंट से कोई टेंशन ड्रॉ किया जाए जैसे की यार एक्सटर्नल पॉइंट है लेट से X1 y1 है ना तो क्या आप इस टांगें की लेंथ बता सकते हो याद करो स्टूडेंट बड़ा ही बेसिक सा फॉर्मूला पढ़ा था और हम बार-बार इन सब पॉइंट्स को बात करते हैं ये होती है अंडर रूट S1 याद आया क्या सर कुछ नहीं होता है ये सर्कल जो होता है ना जैसे अगर ये लिखा है x² + y² + 2G एक्स + 2fy प्लस सी आपके सर्कल की इक्वेशन है तो आप कुछ मत करो जहां भी एक्स और ए दिख रहा है ना उसकी जगह x1² + 2G X1 ये सब कुछ वैल्यूज होंगी राइट कुछ ना कुछ न्यूमेरिकल वैल्यूज होंगी और इसका क्या ले लेंगे सर इसका ले लेंगे हम अंडर रूट भेजो की इस पर एक पॉइंट लाइक करता है जिसका cardinate क्या है ह के तो ह ए की इस सर्कल से जो टैसेंट की लेंथ है और ह कमा कैसे इस सर्कल की जो टैसेंट की लेंथ है इसकी लेंथ की ए जाएगी सर एक्स और ए की जगह क्या रख दो ह कमा के तो ये हो जाएगा x² + ए के स्क्वायर प्लस 2g1 के ह प्लस तू ऍफ़ वैन के प्लस सी वैन अंडर रूट में ऐसे ये भी अंडर रूट में अंडर रूट वाला पार्ट है जाएगा ये पार्ट इस पार्ट से कैंसिल हो जाएगा क्योंकि ये स्क्वायर प्लस के स्क्वायर प्लस के स्क्वायर है तो अल्टीमेटली आप ध्यान से देखो स्टूडेंट्स क्या मिलेगा आपको अल्टीमेटली अगर आप ध्यान से देखोगे तो आपको जो मिलेगा वो कुछ ऐसा मिलेगा मैं बात करता हूं इस बारे में जो उन्होंने डायरेक्टली आपको लिख के दे दिया तो आपको जो फाइनली कंक्लुजन क्या मिलेगा की सर देखो आप ये और ये ये और ये कैंसिल हो जाएगा तो फाइनली आपको 2g1 - G2 टाइम्स ह प्लस तू F1 - f2 टाइम्स के प्लस C1 - C2 मिलेगा जो की टेक्निकल कुछ नहीं है जो की टेक्निकल आपका S1 - S2 है सर S1 - S2 का क्या मतलब S1 - S2 का मतलब अगर एक सर्कल है x² + y² + 2 वैन ए प्लस सी वैन इसे इक्वल तू जीरो suniyega ध्यान से दूसरे सर्कल की इक्वेशन है वही x² + y² पर चूंकि एक दूसरा सर्कल है तो इस केस सेंटर और रेडियस कुछ और होंगे ऑफ कोर्स है ना तो ये जो आपका दूसरा सर्कल है बेसिकली आपके क्या हो गए ये बेसिकली आपके हो गए दो सर्कल्स या आपका पहला सर्कल है इसमें और ये आपका दूसरा सर्कल S2 अगर कोई पूछे की इनके रेडिकल एक्सेस की इक्वेशन क्या है तो रेडिकल एक्सिस मतलब एक ऐसी स्ट्रेट लाइन जहां से ड्रा किए गए टेंसेज की लेंथ इन दोनों सर्कल्स पर इक्वल होती है तो रेडिकल एक्सिस बड़ा ही साधारण सिंपल और सुलझा हुआ सा कॉन्सेप्ट है जो की होता है S1 - S2 = 0 यानी क्या ये पार्ट और ये पार्ट कैंसिल अब आप इन दोनों का डिफरेंस लेंगे तो क्या हो जाएगा सर 2X कॉमन ले लीजिए तो अंदर क्या दिखेगा जीवन माइंस G2 यहां पर तू ए कॉमन ले लीजिए तो यहां पर क्या दिखेगा सर F1 - f2 और यहां पर ऑफ कोर्स क्या दिखेगा सर C1 - C2 और यही आपका क्या होगा रेडिकल्स मतलब बेसिकली ये एक स्ट्रेट लाइन है जहां से जहां से ड्रा की गई टेंशन इन दोनों सर्कल पर इक्वल लेंथ कैरी करेंगी कोई परेशानी कोई डाउट कोई दिक्कत याद रखना स्टूडेंट्स या रेडिकल एक्सेस होता है भूल जाते हैं बच्चे मतलब वो गलतियां करते हैं ये की वो भूल जाते हैं की ये कुछ नहीं है ये बेसिकली क्या है आपका दिस इसे नथिंग बट व्हाट योर रेडिकल एक्सेस अब या तो क्वेश्चन में रेडिकल एक्सिस लिख देगा क्या वो बात लिख देगा तो बात लिख देगा तो यहां से पहुंचने आपको आना चाहिए ऐसे रेडिकल एक्सेसिव लिख दे तो ये क्लिक आपको होना चाहिए की कुछ नहीं है ये बेसिकली आपका क्या है S1 - S2 यही आज के लेक्चर का करेक्ट सही आज के लेक्चर का जस्ट है सर है एक लाइन जहां से किसी भी पॉइंट से अगर आप tenjent ड्रॉ करो एक लाइन जिस पर मौजूद किसी भी पॉइंट से इन दोनों सर्कस पर टैसेंट ड्रॉ करो तो उन दोनों टेंसेज की लेंथ इक्वल होगी थॉट्स इट थॉट्स इट अगर इस बारे में बात आगे बढ़ाएं तो क्या सर ध्यान से सुनना रेडिकल एक्सेस के बारे में क्या मैं एक बड़ा सिंपल सा कैलकुलेशन दे सकता हूं सर की देखो आप रेडिकल एक्सिस की स्लोप की आएगी रेडिकल एक्सिस की स्लोप मतलब क्या इसको उठा के उधर रखा तो वही ए = एमएक्स + सी फॉर्म में कन्वर्ट किया और जैसे ए = 2X + सी फॉर्म में कन्वर्ट किया तो ये आपके रेडिकल एक्सेस के क्या ए जाएगी स्लोप जो की आई थिंक बहुत डिफिकल्ट बात नहीं है अच्छा एक और बात आपको सोचने लगे एक और बात पे आप गौर करना चाहिए वो बात क्या है स्टूडेंट्स की जब आप बहुत गौर से देखोगे ना तो बहुत अच्छा एक ऑब्जर्वेशन आप करोगे तो आपको ये बात दिखाई देगी सर की जो रेडिकल एक्सेस होता है ना और सर्कल के सेंटर्स को कनेक्ट करने वाली जो लाइन होती है उनमें एक रिलेशन होता है क्या रिलेशन होता है वो सुनेगा आई रिपीट माय स्टेटमेंट जैसे मैन लो आपके दोनों क्या है सर यह दोनों आपके सर्कल है इस सर्कल का सेंटर यहां है इस सर्कल का सेंटर यहां और मैं इन दोनों को ज्वाइन करने वाली एक स्ट्रेट लाइन बना दो इन दोनों को ज्वाइन करने वाली एक्सट्रेटोरेंट बनाता हूं के बाद आप समझ का रहे हो स्टूडेंट्स तो यह जो स्ट्रेट लाइन है इन दोनों के सेंटर्स को ज्वाइन कर रही है अब अगर आप इसका रेडिकल एक्सेस बनाओगे ना तो यहां कहीं से गुजरेगा पहली बात तो और दूसरी बात यह होगी की वो जो रेडिकल एक्सिस होगा वो परपेंडिकुलर होगा आप मेरी बात समझ का रहे हो मतलब सर्कल का सेंटर था C1 इस सर्कल का सेंटर था मैन लो C2 तो जो C1 और C2 को ज्वाइन करने वाली लाइन है ना वो रेडिकल एक्सेस के परपेंडिकुलर होगी ये बात हमेशा ट्रू होगी अच्छा दूसरी बात क्या दूसरी बात यह स्टूडेंट्स की आई होप आपको यह बात समझ ए चुकी है की अगर मुझे पता है अगर मुझे पता है क्या की सर आपके मतलब इधर यू कैन प्रूफ इट इधर यू कैन प्रूफ इट मतलब देखो ऐसे प्रूफ कर सकते हो आप यह बताओ इन दोनों के सेंटर को ज्वाइन करने वाली लाइंस की स्लोप क्या होगी सर ये ऑफकोर्स क्या होगा आपका -1 कमा -f1 इसकी स्लोप स्किप को नेट क्या होंगे -2 कमा - f2 तो क्या आप C1 C2 की स्लोप निकल लोग बिल्कुल निकल लेंगे अब जब आप C1 C2 की स्लोप निकलोगे तो क्या आपको रेडिकल एक्सिस की स्लोप पता है हान सर C1 C2 की जो स्लोप निकलेंगे हम वो ये ए जाएगी और रेडिकल एक्सेस के स्लोप जब आप निकलोगे कहां से इस इक्वेशन से तो वो ये ए जाएगी जो भी हमने निकल और मेरा बस आपसे ये कहना है की अगर आप इन दोनों का प्रोडक्ट देखोगे तो देखो ये से कैंसिल बचेगा क्या -1 क्योंकि आपको realaiz हो रहा है यहां से मैथमेटिकली प्रूफ कर सकते हैं की C1 C2 है ना दोनों सर्कल्स के सेंटर को ज्वाइन करने वाली जो लाइन है वो रेडिकल एक्सिस पर क्या होगी सर वो लाइन रेडिकल एक्सरसाइज पर परपेंडिकुलर होगी इस बात को नोट डाउन करना है स्टूडेंट्स उसे पर क्वेश्चंस बनेंगे इस बात को अच्छे से नोट डाउन कर लेना इस पर क्वेश्चन बनते हैं दूसरी बात रेडिकल एक्सेस आपकी जो कमेंट इंजन है ना उनको बायसेक्स करती है रिपीट माय स्टेटमेंट कौन सी लेंथ को बायसेक्स करती है सुनना ध्यान से इस पॉइंट को मैं आपको समझने की कोशिश करता हूं जैसे अगेन मैन लो आपके पास क्या है लेट्स से दो सर्कल है थोड़ा ऐसे बना रहा हूं ताकि मैं चीज आपको अच्छे से दिखा पाऊं ठीक है अब मैन लो आपके पास क्या है सर रेडिकल एक्सेस होगा और रेडिकल एक्सरसाइज से भी पहले मैं आपसे कहूंगा की लिस्ट से यह जो लाइन है यह लेट्स रन की क्या है commentagent है सन रहे हो क्या आप लोग और इसका रेडिकल एक्सरसाइज यहां कैसे गुजर रहा होगा तो रेडिकल एक्सेस की क्या खास बात होती है पहले मैं कॉमन टांगें आपको समझा डन स्टूडेंट्स कॉमन टांगें का मतलब है सर्कल्स क्या बात कर रहे हो सर ऐसा होता है बिल्कुल दोनों सर्कल्स के लिए टेंशन की तरह बिहेव कर सकती है मतलब इसके लिए भी टैसेंट और उसके लिए भी टेंशन है सर ऐसी एक ही हो सकती है क्या नहीं आप थोड़ा ध्यान से देखिए से ऐसा कैसा एक टेंशन मुझे यहां भी तो देख सकता था सर क्या सिर्फ ये दो ही दिखेंगे और नहीं दिख सकती क्या नहीं आप थोड़ा और ध्यान से देखिए आपको एक टैसेंट शायद कुछ ऐसी भी तो दिख सकती है सर क्या सिर्फ इतना ही है नहीं थोड़ा और ध्यान से देखिए आपको शायद एक टैसेंट कुछ ऐसी भी तो देख सकती है पर आप मेरी बात समझ का रहे हो ना आप मेरी बात डाइजेस्ट कर का रहे हो बात बस इतनी सी है की समझना बात को सुनना अब बहुत कम की बात मैं आपसे करने का जैसे मैन लो इस इंजन का पॉइंट ऑफ कॉन्टैक्ट है ना ये पी के को तो इक्वल पार्ट्स में डिवाइड करता है सन रहे हो क्या अच्छा सर ये जो आपकी टैसेंट है जिसका ये पॉइंट है ए और ये पॉइंट है लेट से बी तो जो रेडिकल एक्सरसाइज में डिवाइड करता है और क्या सर मैन लो यह वाली जो टेंट है इसके पॉइंट्स है कॉन्टैक्ट लिस्ट से कुछ भी ले लो यार है ना क्या बोलते हैं एसएससी और दी बोल देते हैं तो सीडी को भी ये रेडिकल एक्सेस दो पॉइंट्स में दो-दो इक्वल पार्ट्स में बायसेक्स कर रहा है कौन रेडिकल एक्सिस ये पॉइंट ये पॉइंट यहां ये पॉइंट यहां और ये पॉइंट यहां और सिमिलरली अगर मैं इस इस कॉमन टांगें की भी बात करूं जिससे अगर मैं अभी समझने के लिए आपको कैसे कुछ भी है ना इससे एन और एन तो यह एन एन भी यहां पर वही वही से बात से क्या हो रही है बायसेक्स क्या यह सारी बातें आपको समझ ए रही हैं बात बस इतनी सी है की रेडिकल एक्सेस कॉमन टांगेंट्स को इन दोनों सर्कल्स के ऑफ कोर्स को बायसेक्स करता है सर क्यों इतना ध्यान से पढ़ा रहे हो इस बात को पढ़ जा रही है बात अगर ध्यान से मतलब उसके पीछे कोई कारण होगा कारण बस इतना सा है की ये जो दो बड़ी इंपॉर्टेंट सी प्रॉपर्टीज है ना इन पर क्वेश्चंस बनते हैं स्टूडेंट्स तो क्वेश्चंस तब ही सॉल्व कर पाओगे आप जब आपको ये प्रॉपर्टीज याद रहेंगे तो दोनों प्रॉपर्टी क्या एक तो दोनों सर्कल्स के सेंटर को ज्वाइन करने वाली लाइन आपके रेडिकल एक्सिस पर परपेंडिकुलर होती है दोनों सर्कल्स की जितनी भी कॉमन टांगें है वो सारी committents रेडिकल एक्सिस से बायसेक्स होते हैं दो इक्वल पार्ट्स में डिवाइड होती है याद रहेगा अच्छे से याद रहेगा क्या और अगर अच्छे से याद रहेगा तो उससे कंसोलिडेटेड करने के लिए हमारे पास जो सबसे अच्छा तरीका होता है वो है सर हम एक क्वेश्चन सॉल्व एक सर्कल कर दो तो रेडिकल एक्सेस की इक्वेशन ए जाएगी पर जीवन सच में इतना आसान है क्या जीवन इतना आसान होता तो आप यह क्वेश्चन करके मुझे आंसर बता देते और उसे स्ट्रेट लाइन की इक्वेशन आणि चाहिए सर अगर मैंने इसमें से इसको सब्सट्रैक्ट किया ना तो क्या बचेगा सर x² + y² + 3y फिर क्या बचेगा आई थिंक फाइव है ना पर ये तो सर एक सर्कल की इक्वेशन बन रही है ये तो एक स्ट्रेट लाइन के क्वेश्चन नहीं बन रही है स्ट्रेट लाइन की इक्वेशन तो सर एक लीनियर इक्वेशन होती है तो सर रेडिकल एक्सरसाइज आपने जो पढ़ाया है एक स्ट्रेट लाइन होती है सारी बातें एक तरफ रखो भाई मेरा आपसे ये पूछना है की सर अभी फिलहाल ये सर्कल की क्वेश्चन है लेकिन क्या यह स्टैंडर्ड सर्कल की इक्वेशन है नहीं क्यों सर क्योंकि इस इक्वेशन में एक प्रॉब्लम है स्क्वायर बना लो या फिर यहां इससे भी तू बना लो जो आपको ठीक लगे और फिर इक्वेशन एक्सेस की इक्वेशन निकालो क्या आप मेरी बात समझ पाए तो मेरा कहना है सर इसको हाफ करूंगा 13/2 हो जाएगा बहुत कैलकुलेशन हो जाएगी जैसे इसको डबल कर लेने हैं कोई ज्यादा फर्क नहीं पड़ेगा ऑफ कोर्स आप इसको हाफ करके भी सब्सट्रैक्ट कर सकते हो बिल्कुल सही बात है वो x²y² फिर इनसे माइंस हो जाएंगे कोई दिक्कत नहीं है और आप चाहो तो उसको डबल करके भी कर सकते हो जैसा आपको ठीक लगे डबल करके अगर आपको नहीं बात हो रही है डाइजेस्टिव ऑफ करके कर लो अगर आपको चीज अच्छी नहीं लग रही है तो इसकी इक्वेशन क्या है सर ये है x² + y² - 3X + 5 - 7 = 0 ये आपका फर्स्ट सर्कल है दूसरा सर्कल को अगर मैं सिंपलीफाइड वर्जन में लिखूं तो मैं तू से पूरा डिवाइड कर रहा हूं तो मैं इसे लिखूंगा एक्स स्क्वायर प्लस ए स्क्वायर करवा रहा हूं याद करो स्टूडेंट दोनों सर्कल के सेंटर को ज्वाइन करने वाली लाइन पर वो परपेंडिकुलर होती है तीसरी बात रेडिकल एक्सेस दोनों सर्कल की कॉमन टांगेंट्स को बायसेक्स करती है चौथी बात रेडिकल एक्सेस की इक्वेशन क्या होती है सर वो होती है S1 - S2 = 0 पर ये कब अप्लाई करना है जब आपके सर्कल्स कैसे फॉर्म में हो स्टैंडर्ड फॉर्म में होप यू आर गेटिंग माय पॉइंट तो यहां से क्या करेंगे सर दिस दिस जीरो माइंस 3X - 2X तो यहां पे क्या मिलता है - एक्स वही आई होप की बातें समझ ए रही है तो -7 + 13 / 2 तो कितना हो जाएगा सही हो जाएगा -7 + 13 / 2 आई थिंक आप चाहो तो पुरी इक्वेशन को तू से मल्टीप्लाई कर दो और इक्वल तू जीरो रख दो ऑफ कॉस्ट तो ये क्या हो जाएगी आपकी उसे कॉमन रेडिकल एक्सेस की इक्वेशन अगर आपको पूरा फाइनल कंप्लीट सॉल्यूशन चाहिए तो एक छोटा सा कम करो या तो यहीं पर अब तू से मल्टीप्लाई कर देते हैं इसको भी और फिर सब्सट्रैक्ट कर देते हैं बिना इसको सिंपलीफाई किए तो भी आंसर वही ए जाता जो अभी आने वाला है क्या सर देखो मैं पुरी इक्वेशन को दूसरे मल्टीप्लाई कर रहा हूं तो ये जाता है कितना -2x नेगेटिव साइन कॉमन ले रहना चाह रहा हूं तो ये हो जाएगा 2X - 2y ये कितना होगा -1 और - कॉमन लिया तो कितना +1=0 तो दिस इसे गोइंग तू बी और इक्वेशन आई सी डी सी डी सॉल्यूशन था कोई टू कोई डिफिकल्ट बात थी क्या बस ट्रिकी पार्ट ये था की कुछ स्टूडेंट्स इसे देखिए बिना सीधे सॉल्व कर देते हैं बस डायरेक्टली s1-s2 = 0 लगाकर जो की गलत होता क्योंकि सर्कल का स्टैंडर्ड फॉर्म में होना बहुत जरूरी है ये ऑफ कोर्स आपके 11th 12th का कंप्लीट सिलेबस लेते हुए आपको आईआईटी जी मांस और एडवांस लेवल की तैयारी करवाता है बेसिकली ये जो पंच बुक्स हैं ये है cardinate ज्यामिति जो की अभी हम कर रहे हैं इसके बाद हम कैलकुलस करेंगे फिर हम करेंगे अलजेब्रा और अंत में हम देखेंगे ट्रिगोनोमेट्री इस तरीके से आपका पूरा सिलेबस टाइमली हम कंप्लीट करवाएंगे जैसा हम कर रहे हैं की आपके लिए हम दो लेक्चरर्स रिलीज कर रहे हैं पहले समझते हैं की रेडिकल एक्सेस के कॉन्सेप्ट को जो की हमने लास्ट लेक्चर में पढ़ा था याद ए रहा होगा की रेडिकल एक्सेस एक्जेक्टली होता क्या है याद ए रहा है स्टूडेंट्स बेसिकली एक स्ट्रेट लाइन होती है जिस पर लाइक करने वाले किसी भी पॉइंट से जब उन दोनों सर्कल्स पर आप tenjens ड्रा करते हैं तो उन टेंसेज की लेंथ इक्वल होती है याद ए रहा है क्या यही बात आपको इस बार वापस में समझाना चाह रहा हूं जैसे मैन लो आपके पास तीन सर्कल्स है ना एक सर्कल से ये रहा एक सर्कल है और एक सर्कल लेट्स से ये रहा ऐसे लेट से तीन सर्कल्स हैं ऐसे ही बस रैंडम से मैं बना ले रहा हूं है ना ज्यादा मैंने सोचा नहीं अब आपसे अगर मैं कहूं मैन लो बस ऐसे ही ऐसे ही मैन लो है ना की इन दोनों सर्कल्स का लेक्चर रेडिकल एक्सिस था ये आप समझ का रहे हो इन दोनों सर्कल का रेडिकल एक्सरसाइज चाहिए और इन दोनों सर्कल का लेक्चर रेडिकल एक्सरसाइज बना लिया इन दोनों को कंसीडर किया तो यह रेडिकल एक्सरसाइज इन दोनों सर्कल्स को कंसीडर किया तो ये इनका रेड एलेक्सिस बना और इन दोनों को कंसीडर किया तो ये inkaratic कलेक्शन बना आई होप आप ये termrilology समझ का रहे हो की रेडिकल एक्सेस का मतलब क्या होता है रेड कलर एक्सिस मतलब एक बेसिक सा एक्सेस या स्ट्रेट लाइन जिस पर लाइक करने वाले किसी भी पॉइंट से जब आप इन दोनों सर के उसे पर टांगेंट्स ड्रॉ करेंगे तो उन टेंसेज की लेंथ इक्वल होगी याद ए रहा है रेडिकल एक्सेस के ही कॉन्सेप्ट को अगर मैं थोड़ा एक्सटेंड करूं सुनेगा ध्यान से जैसे मैन लो अगर मैं इसको थोड़ा एक्सटेंड करूं मैं इससे भी थोड़ा एक्सटेंड करूं तो मैं यह पाऊंगा की यह तीनों जो आपके रेडिकल एक्सेस हैं यह किसी एक पॉइंट से पास होते हैं यानी यह तीनों रेडिकल एक्सेस कौन करंट होते हैं मतलब यह किसी एक से पॉइंट से पास होते हैं यह जिस पॉइंट से पास होते हैं उसे पॉइंट को हम कहते हैं रेडिकल सेंटर क्या कहते हैं भाई इस पॉइंट को भी कॉल डी रेडिकल सेंटर अब सर इस रेडिकल सेंटर की क्या खास बात है इस रेडिकल सेंटर की ये खास बात है की रेडिकल सेंटर से जब आप तीनों सर्कल्स पर टैसेंट ड्रा करेंगे तो उन तीनों ही tegens की लेंथ इक्वल होगी की आपको स्टेटमेंट समझ ए रहा है रिपीट माय स्टेटमेंट बहुत ध्यान से suniyega स्टूडेंट्स रेडिकल सेंटर की खास बात यह होती है की जब इस रेडिकल सेंटर से आप तीनों सर्कल्स पर जो टेंसेज ड्रा करते हैं जो टैसेंट आप ड्रॉ करते हैं उन तीनों ही मतलब ऑफ कोर्स यहां पर वो दो होंगी यहां पर दूसरी दो होंगी आई होप आप ये बात समझ का रहे हो और उसी तरीके से इस सर्कल के लिए जो तीसरा सर्कल आपका है उसके लिए अलग आपकी दो होंगी क्या ये बात आप मेरी समझ का रहे हो मेरा बस आपसे ये कहना है की ये जो रेडिकल एक्सेस है आपने टेंसेज ड्रा की हैं आई थिंक मैं बना का रहा हूं इससे अच्छा तो मैं ऐसे अंदू ही कर डन और मैं एक दूसरा ही एक तीसरी ही बना देता हूं जो की मेरे ख्याल से कुछ ऐसी जा रही होगी और इसी तरीके से एक ये जो मैं बनाऊंगा ये वाली जो मैं बनाऊंगा यह जो 6 मिनट बनाई जो की पैर वाइस दो-दो के महत्व है तो टेक्निकल मैं कहूं तीन जो टेनसन आपने ड्रा किया जो की टेक्निकल है यह सारी की सारी टांगें की जो लेंथ होगी ये सारी की सारी tenjins की जो लेंथ होगी वो स्टूडेंट्स क्या होती है सर वो होती है इक्वल क्या आपको ये बात समझ आई ने रेडिकल सेंटर कॉन्सेप्ट आपको क्लियर हो रहा है अब हम बात करने वाले रेडिकल सेंटर तक पहुंचा कैसे जाता है मतलब अगर मुझे तीन सर्कल दे दिए जाएं तो उनका रेडिकल सेंटर कैसे ढूंढ जाता है रेडिकल सेंटर में तो बेसिकली वो पॉइंट जहां से तीनों सर्कल्स पर जब आप टांगें ड्रॉ कर रहे हैं तो उनकी लेंथ इक्वल है या और अच्छी भाषा में टेक्निकल क्या की तीनों रेडिकल एक्सेस का जो पॉइंट ऑफ इंटरसेक्शन होता है यानी ये तीनों रेडिकल एक्सिस कौन करंट होते हैं तो यह जहां एक सिंगल जिस पॉइंट से पास होते हैं उसे पॉइंट को हम कहते हैं रेडिकल सेंटर कोई तकलीफ अब बात समझने की कोशिश करते हैं की चीज एक्चुअली वर्कआउट कैसे हो रही है जैसे बात समझना स्टूडेंट्स हमारे पास दो सर्कल हैं इन दोनों सर्कल का रेडिकल एक्सिस क्या हो जाएगा S1 - h2o = 0 याद ए रहा है अगर मैं पहले सर्कल की इक्वेशन को कहूं S1 दूसरे सर्कल की इक्वेशन को कहूं S2 तो ये आपके पहले सर्कल के रेडिकल पहले दोनों सर्कल जैसे दो ये सर्कल द आपके और इन दोनों सर्कल का जो रेडिकल एक्सेस है उसकी इक्वेशन क्या हो जाएगी S1 - h2s = 0 अब मैन लो आपके पास एक और सर्कल है आपके पास मैन लो भाई एक और सर्कल है उसे सर्कल की इक्वेशन है S3 और उससे भी एक रेडिकल एक्सरसाइज है तो इस S3 की मतलब इस वाले रेडिकल एक्सिस की इक्वेशन क्या होगी सर इस वाले एक्सेस की इक्वेशन होगी S2 - S3 इसे इक्वल तू जीरो आई होप आपने फैमिली ऑफ लाइंस चैप्टर मैं बहुत डिटेल में डिस्कशन पढ़ा है की सर मैं ये बहुत खास बात जानता हूं की मैन लो ये है लाइन L1 सर और लेट्स से ये है आपकी L2 तो मैं उन सारी इंफिनिटी स्ट्रेट लाइंस जो की L1 और L2 के पॉइंट ऑफ इंटरसेक्शन से पास होती हैं वो सारी infainight स्ट्रेट लाइंस जो की L1 और L2 के इंटरसेक्शन से पास होती है इन सब की कंबाइंड इक्वेशन देता हूं L1 + लामबीडीए टाइम सेल तू इस इक्वल्स तू जीरो से आपको याद ए रहा है क्या ये बात आपको याद ए रही है यानी मेरा आपसे ये कहना है की सर अगर यहां पर मैं सीधी-सीधी आपसे ये बात कहना चाहूंगा की सर इन दोनों रेडिकल एक्सेस से के पॉइंट ऑफ इंटरसेक्शन से पास होने वाली वह सारी इंफिनिटी स्ट्रेट लाइंस स्ट्रेट लाइंस की इक्वेशन को मैं कैसे लिखूंगा एस वैन माइंस S2 प्लस लामबीडीए टाइम्स S2 - S3 = 0 क्या बात आप समझ पाए एक छोटी सी गुजारिश है मेरी आपसे प्लीज ध्यान से suniyega इस बात को मैन लो थोड़ी देर के लिए अब कोर्स हम जानते हैं सर लामबीडीए की आप अलग-अलग वैल्यूज ट्री करते जाएगा आपको अलग-अलग स्ट्रेट लाइन के क्वेश्चन मिलती चली जाएगी क्या बात आपको याद है पर उन सारी स्ट्रेट लाइंस की खास बात ये होगी की वो इस पॉइंट से डेफिनेटली पास हो रही होंगी मैं मैन लेता हूं की सर मैं लौंडा की वैल्यू वैन ट्री करता हूं तो वैन ट्राई करती क्या होगा देखो S2 से S2 कैंसिल तो क्या बचेगा तो बचेगा S1 - S3 S1 - X3 का क्या मतलब है अब बता सकते हो सर S1 - S3 का मतलब है एस वैन और S3 का रेडिकल एक्सरसाइज एस वैन और S3 का रेडिकल एक्सिस और सर हम ये देख का रहे हैं करते हैं की असफल और स्त्री का जो रेडिकल एक्सरसाइज है वह भी इसी पॉइंट से पास हो रहा है क्योंकि ऑफ कोर्स फॉर्म ने किस से बनाया है इससे और ये किस खास लाइंस की इक्वेशन है ये उन फैमिली ऑफलाइन की एक क्वेश्चन है जो इस पॉइंट से पास होती है और उसी में लामबीडीए की एक खास वैल्यू हमने ट्राई की जिससे हमें आपके इन दोनों की भी रेडिकल एक्सेस की इक्वेशन मिली तो क्या आप ये प्रूफ समझ का रहे हो की इन दोनों का रेडिकल एक्सिस इन दोनों का रेडिकल एक्सिस और इन दोनों का रेड कलर एक्सिस कौन करंट होते हैं कौन करंट का मतलब ये तीनों किसी एक ही पॉइंट से पास होते हैं अगर यह कॉन्करेंट है तो हम यह बात जान चुके हैं सर इस पॉइंट को जिससे ये तीनों रेडिकल एक्सरसाइज होते हैं उसे हम कहते हैं रेडिकल सेंटर और इस सेंटर से तीनों ही सर्कल्स की डिस्टेंस इक्वल होगी किस महीना में टांगें के बयानों में यानी की उसे सेंटर से यहां पर ड्रॉप के गेट इंजन के लिए यहां पर यहां पर ड्रेंस के लेंथ वो सारी की सारी टेंड्स विल बी ऑफ इक्वल लेंथ क्या यह कॉन्सेप्ट बड़े आसान से मैटर में आप सभी को समझ आया अगर आया तो इसी बात को दोहरा लेते हैं यहां पर जो चीज लिखी गई हैं वही बात है की मैन लो ये आपके तीन सर्कल्स हैं S1 S2 और S3 ये रेडिकल एक्सिस हो जाएगा एक्सिस हो जाएगा मतलब सन - H3 और सन-s2 तो दो से अगर मैंने मिलके उनके पॉइंट ऑफ इंटरसेक्शंस हैं पास होने वाली वो सारी इनफिनिटी स्ट्रेट लाइंस की इक्वेशंस बनाई तो मैं क्या कहूंगा सर वो होगी आपकी L1 + क्लाइमेट टाइम से L2 जो की ये हो जाएगी अब अगर मैं लामबीडीए की वैल्यू वैन ट्री करता हूं तो उससे मुझे क्या मिल जाती है उससे मुझे थर्ड रेडिकल एक्सेस की इक्वेशन मिल जाती है और मैं ये प्रूफ कर पता हूं की सर तीनों के तीनों रेडिकल एक्सरसाइज आपके कौन करंट है अब बात यह निकल कर आती है की अगर तीनों के तीनों रेडिकल एक्सेस कौन करें जहां से तीनों पास हो रहे हैं उसे पॉइंट से तीनों ही सर्कल्स की टांगें की लेंथ क्या होगी भाई इक्वल तीनों सर्कल्स पर रेडिकल सेंटर से ड्रा किए tesence के लिए इक्वल होगी ये फैक्ट ये स्टेटमेंट याद रखिएगा यही आज के लेक्चर कक्ष है क्या कहना चाह रहा हूं डी पॉइंट ऑफ कॉन्करंसी ऑफ थ्री रेडिकल एक्सिस इस कॉल्ड डी रेडिकल सेंटर पहली बात दूसरी बात रेडिकल सेंटर इस डी ओनली पॉइंट इन डेट प्लेन ऑफ ठोस थ्री सर्कल्स यहां से तीनों सर्कल्स की tenjins की लेंथ क्या होगी भाई इक्वल सेंटर का है तो पहले सर्कल पर अगर टेंशन रॉक की तो का बनती है दूसरे सर्कल पर टांगें ड्रॉ की गई तो उसकी लेंथ बनती है बीवी और तीसरे टांगेंट्स पर अगर दूसरे सर्कल पर अगर उसकी लाइन बनती है पीसी तो पापी और पी सी विल बी ऑफ इक्वल लेंथ याद रहेगा तो इन सारी बातों को कंसोलिडेटेड करेगा सबसे अच्छा सबसे आसान और मेरे ख्याल से सबसे बेहतर तरीका है क्वेश्चन सॉल्व करना है और आप ट्राई मुझे बताइए की इसका क्या आंसर आप लिखेंगे वो का रहा है डी इक्वेशन ऑफ थ्री सर्कल्स तो एक सर्कल तो है आपका x² + y² = 1 बड़ा ही सॉर्ट था 0% और इसकी रेडियस 1 दूसरा सर कल x² + y² - 8x + 15 तो क्लीयरली सर दिख रहा है फोर कमा जीरो परसेंट है और ऑफ कोर्स फोर का स्क्वायर 16 16 - 15 कितना 1 तो इसकी रेडियस भी कितनी वैन क्योंकि वैन का रूट वैन इसी तरीके से अगर इस सर्कल को देखें तो x² + y² + 10y + 24 तो +10y मतलब इसके सेंटर के क्वाड्रेंट जीरो कमा -5 आई होप आप देख का रहे हो और इसकी रेडियस क्या होगी फिर से वही बात है 25 - 24 एन है वैन वैन का अंडर रूट वैन तो इसकी रेडियस भी वैन तो तीनों सर्कस रेडियस 5 से लेंथ का है एंड ऑफ कोर्स एक बात और की इसका सेंटर जीरो कमा जीरो इसका फोर कमा जीरो और इसका 0 - 5 आप चाहे तो यहां से कर सकते हैं फिर वो पूछ क्या रहा है अब आगे सुनना डिटरमिन डी कोऑर्डिनेट्स ऑफ डी पॉइंट पी सच डेट डी टांगेंट्स ड्रॉन फ्रॉम इट तू डी सर्कल्स आर इक्वल इन लेंथ अब देखो यहां कहीं भी उसने जिक्र तक नहीं किया है आपसे की ये एक रेडिकल सेंटर या रेडिकल एक्सेस के कॉन्सेप्ट पर बेस्ड क्वेश्चन है उसने तो आपको तीन सर्कल दे दिए और उसने कहा एक ऐसा पॉइंट ढूंढ के बताइए जहां से ड्रा की गई टेंशन की लेंथ इन तीनों सर्कल पर इक्वल हो अब या तो मैं अपनी कन्वेंशनल तरीका उसे करूं मैं मैन लूं ह के फिर उससे इसकी डिस्टेंस इसकी डिस्टेंस और उसकी टांगें की डिस्टेंस निकालो फिर इक्वल रखूं एम निकालो ह कमा क्या निकलूं और फिर करके कुछ ना कुछ करके वो पॉइंट मैं निकल लूं या फिर मैं इतनी मेहनत करने के बजाय बस इतना सा कॉन्सेप्ट जानता हूं सर बस बड़ी बेसिक सी बात ये जानता हूं सर की आपके जो तीनों रेडिकल एक्सेस होते हैं वो कौन करंट होते हैं और तीनों छोड़ो सर दो ढूंढ लो आपने कोई दो भी रेडिकल एक्सेस ढूंढ लिया जैसे मैन लो मैंने इससे और इससे एक बना लूंगा और इससे और इससे भी बना लूंगा तो वो जो दोनों रेडिकल एक्सरसाइज बनेंगे उन रेडिकल एक्सेस का जो पॉइंट ऑफ इंटरसेक्शन होगा उन दोनों रेडिकल एक्सेस का जो पॉइंट ऑफ इंटरसेक्शन होगा क्या वो रेडिकल सेंटर होगा सोचिए थोड़ा ध्यान से और अगर वह रेडिकल एक सेंटर मुझे मिल गया तो वही तो वो पॉइंट होगा जहां से तीनों सर्कल्स की टांगें की लेंथ इक्वल होंगी मैंने आपको लगभग हर एक जो जरूरी चीज इस क्वेश्चन को सॉल्व करने के लिए बता दी है बस अब आपसे मेरी है की जरा इस क्वेश्चन को देखिए ट्राई करिए और मुझे बताइए की आपके अकॉर्डिंग इस क्वेश्चन का आंसर क्या होगा या बेसिकली उसे पॉइंट पी के कोऑर्डिनेट्स के आएंगे सर बड़ी आसान सी बात है आप पता नहीं क्यों इतना कॉम्प्लिकेट कर देते हो चीजों को जीवन तो बड़ा आसान है सर देखो आप पता नहीं क्यों इतना दिक्कत में रखते हो हमें सर पहले क्या कर लेंगे पहले से अगर मैं इस सर्कल को का लूं S1 और इस सर्कल को का लो S2 तो पहला रेडिकल एक्सिस क्या मिल जाएगा सर वो मिल जाएगा S1 - S2 जो की ऑफकोर्स क्या हो जाएगा देखो x² x²y²y² कैंसिल दिख रहा है क्या सबको अब क्या बचेगा सर देखो सब्सट्रैक्ट किया तो -8x क्या हो जाएगा प्लस 8x अब सुनना ध्यान से बहुत कम की बात है स्टूडेंट्स ये वैन को मैं इधर लाया तो ये क्या हो जाएगा -1 -1 - 15 तो -1 - 15 कितना होता है सर वो हो जाता है -16 तो ये हो जाएगा 8x - 16 = 0 यहां से तो सर इक्वेशन छोड़ो पॉइंट ही मिल गया क्या मिल गया सर जो एक्स हो जाएगा वो क्या होगा 16 / 8 यानी कितना तू सर ऑफ कोर्स आपका जो रेडिकल एक्सेस है उसका एक्स कोऑर्डिनेट्स तो मुझे मिलेगी क्वेश्चन समझ ए रहा है यहां तक भाई अब क्या सर अब एक कम करते हैं S2 S3 भी निकल लेते हैं तो S2 S3 निकलने के लिए क्या करेंगे देखो भाई S2 और S3 का डिफरेंस हम ऑफ कोर्स को किसके इक्वल रखते हैं जीरो के हम इसे किसके इक्वल rakhten हैं जीरो के तो जब आप लिखते हो s1-s3 जो से यहां से जब इसे सब्सट्रैक्ट किया x²y² कैंसिल आई होप आप बात समझ का रहे हो तो यहां क्या बचेगा -10y सर यहां बचेगा -10y एंड ऑफ कोर्स क्या ध्यान से देखिएगा सर ये है माइंस वैन और ये है -24 क्योंकि ये सब्सट्रैक्ट हो रहा है -1 - 24 - 25 तो ये हो जाएगा कितना -25 = 0 आई थिंक सर जीवन बड़ा आसान है आप बेवजह घबराते हो क्या करेंगे देखो भाई 25 को उधर भेज रहा है तो प्लस 25 10 से डिवाइड किया तो ए कितना ए जाएगा सर ए ए जाएगा 25 / 10 विद एन नेगेटिव साइन तो फाइव का फाइव टाइम्स फाइव का तू टाइम्स तो ये हो जाएगा - 5 / 2 ए कोऑर्डिनेट्स ऑफ डी रेडिकल सेंटर और मेरा यकीन करिए आप दो लाइन में इसका आंसर निकल चुके हैं जो की क्या होगा जो की होगा 2 - 5 / 2 यही आपके पी पॉइंट के कोऑर्डिनेटर इसी बात तक इसी दिन को दिखाने के लिए हमने इतनी मेहनत की है किसे देखो लाइफ आसान है लाइफ तब आसानी जब आपको फंडे पता हो जब आपको कॉन्सेप्ट्स बताओ जब आपने बारीकियां में चीज पड़ी हैं और जब आपको हर एक चीज आती है तो जो क्वेश्चन एक कन्वेंशनल एप्रोच या ट्रेडिशनल एप्रोच से होने में एक नान तौर पर 10 से 15 मिनट लेगा वो दो लाइन में खत्म हो जाएगा और इसे ही पता नहीं क्यों लोग शॉर्ट ट्रिक्स का देते हैं मैं नहीं कहूंगा कभी भी की एक शॉर्ट ट्रिक है यह तो एक तकनीक है ना जिसकी बारीकी आपको पता है तो आपने इस प्रश्न को ऐसे किया पर यह क्यों हुआ यह कैसे आया इसके पीछे की कहानी क्या है अब आपको यह पता है और मेरा यकीन करिए आईआईटी जी मांस और एडवांस ऐसी ही एग्जाम है जो के पर बात करती है जो लॉजिक पर बात करती है जो इंटेलिजेंस चेक करती है हर बार आपसे कहता हूं ये आपकी रिटर्निंग की एबिलिटी काबिलियत चेक नहीं करती है ये आपको स्टेप्स के मार्क्स नहीं दे रहे हैं की आपने कितने अच्छे से सुंदर एंड्रॉयड में चीज लिखी है और कितने अच्छे से गिवन सॉल्यूशन और सबको और हंस प्रूव्ड और ऐसा एलएस रहा लाइन यह नहीं ऐसा खेलती है यह एग्जाम तो देखती है की क्या आपने हर चीज अच्छे से पढ़ी और हर चीज अच्छे से पढ़िए है और फिर उसके नॉलेज को बड़ी मजेदार तरीके से आपसे क्रॉस चेक करते हैं पर कोई एक समझदार स्टूडेंट कम वक्त में जल्दी तब कर लेगा जब उसको ये बात पता हो की सर रेडिकल एक्सेस का पॉइंट ऑफ इंटरसेक्शन होता है मेडिकल सेंटर और वहां से तीनों सर्कल की टांगें की ड्राइंग लेंथ की इक्वल होती है और बस इतना सा कॉन्सेप्ट है पर ये कॉन्सेप्ट अगर पता है तो चीज है ऐसी हो जाए और यही बात मैं कहता हूं बड़ा ही आसान सा मजेदार सा ये कॉन्सेप्ट है डिसजॉइंट सर्कल का सीधा सा सुलझा हुआ सा मतलब ये है की अगर आपके पास लेट से दो सर्कस हैं और वह दो सर्कल से इस तरीके से हैं की वो पूरे तरीके से आइसोलेटेड है वो पूरे तरीके से सेपरेटेड है मतलब इसका कोई भी रीजन इसके साथ कॉमन नहीं है इसका कोई भी रीजन इसके साथ कॉमन नहीं है एक ही बात है बेसिकली और यह एवं एक दूसरे को टच भी नहीं कर रहे हैं पूरे तरीके से डिसजॉइंट है पूरे तरीके से डिसजॉइंट है इस तरीके के सर्कस को जो की कुछ भी पार्ट अपना शेयर नहीं कर रही एक दूसरे से इनमें कोई इंटरसेक्शन नहीं है ना तो एवं ये एक दूसरे को टच कर रहे हैं ये टच तक नहीं कर रहे हैं अब बात समझ रहे हो इस तरीके के सर्कल्स को हम कहेंगे डिस्ज्वाइंट्स सर्कस और आप हमेशा एक बड़ा ही खास सा ऑब्जर्वेशन देखेंगे की सर मैन लो अगर इस सर्कल का सेंटर इससे यहां और इस सर्कल का सेंटर से यहां और अगर इन दोनों सर्कल्स को मैंने कनेक्ट किया अगर इन दोनों सर्कल्स को मैंने करेक्ट किया तो क्या बड़ी मामूली सी डायरेक्ट सी सीधी सी बात देख का रहे हो की सर देखो कितना आसान होगा मेरे लिए कहना की सर अगर इस सेट करके सेंटर के कोऑर्डिनेट्स हैं C1 और इस सर्कल के सेंटर के cardinate को मैं सीटों से नोट कर रहा हूं तो अगर C1 और C2 के बीच की डिस्टेंस निकालो है अगर मैं C1 और C2 के बीच के डिस्टेंस निकालो जैसे मैं दिनो कर देता हूं C1 C2 से तो shailise फॉर अन डिसजॉइंट सर्कल C1 C2 जो होगा सर क्लीयरली आप देख पाओगे इस सर्कल की रेडियस को आप क्या कहोगे सर इस सर्कल की रेडियस को लेते हैं आप कहते हो R1 और इस सर्कल की रेडियस को क्या कहोगे लेट से आप उसे कहते हो r2 तो क्लीयरली सी वैन सी तू जो होगा सर वो आपका R1 + r2 से ज्यादा होगा आय होप आप बात समझ का रहे हो जो सी वैन और सी तू ने सर्कल्स के सेंटर्स के बीच की जो डिस्टेंस है वो आपके ऑफ कोर्स दोनों सरफेस की रेडियस के हिसाब से ज्यादा होगा बस ये बेसिक फेनोमेना है ये बेसिक प्रॉपर्टी हम अप्लाई करेंगे जब भी किस तरीके सर्कल्स की बात हो रही होगी जब भी सर आपको किस तरीके के सर्कस की बात हो रही होगी दिस इसे ज्वाइन सर्कल्स की यहां तक कोई परेशानी नहीं एक बेसिक सा कॉन्सेप्ट जो आपको समझने की कोशिश की जा रही इसे ही थोड़ा कंसोलिडेटेड करते हैं देखिएगा ध्यान से वो का रहा है तू डिसजॉइंट सर्कल डेफिनेशन वाइस निडर हैव कॉमन रीजन नो टच शो इन डी फॉलोइंग फिगर इस फिगर में अब देखो ये सर्कल पूरा अलग है पूरा सर्कल अलग है काफी सारी चीज हम बात करेंगे S1 और S2 S1 है आपका वही 2G x1251 और C1 वाला पॉइंट लेकर और ये है G2 f2 और C2 वाले आपके वेरिएबल का रहे हैं कांस्टेंट लेकर अब बड़ी आसान सी बात है सर इसका सेंटर है -1 - F1 और इसका सेंटर है माइंस जी तू कमा - f2 अब suniyega ध्यान से इसकी रेडियस निकलने में क्या तकलीफ है जीतू स्क्वायर प्लस तू स्क्वायर माइंस सी तू अब suniyega कम की बात बहुत जरूरी बात सर हमें बड़ी सिंपल सी बात दिख रही है की अगर ये दोनों सर्कल्स जॉइंट है दोनों सर्कस डिस्क जॉइंट है तो इन दोनों सर्कल्स के बीच की डिस्टेंस मतलब इसका अगर सेंटर में मैन लूं क्या -1 - F1 और इसके सेंटर के cardinate मैन लूं - G2 हम सॉरी मैं इसके सेंटर के कोऑर्डिनेट्स मालूम - G2 f2 तो मैं जानता हूं सर की इसके और इसके बीच की डिस्टेंस क्या हम डिस्टेंस फॉर्मूला उसे कर सकते हैं आई थिंक कर सकते हैं सर तो मैं क्या बोलूंगा माइंस जीवन माइंस माइंस G2 तो उसे समय लिख सकता हूं क्या उससे मैं लिख सकता हूं जी तू माइंस जीवन का होल स्क्वायर प्लस इसी तरीके से माइंस माइंस लगाऊंगा जो की होगी इन दोनों के बीच की डिस्टेंस लगाऊंगा अभी है ना और ये जो डिस्टेंस होगी सर ये हमेशा ज्यादा होगी किस से ये जो डिस्टेंस है ये इन दोनों सेंटर्स के बीच की जो डिस्टेंस है ये हमेशा ज्यादा होगी इनके रेडियस के हिसाब से और रेडियस का संप्रदाय होप आप कर सकते हो सर दोनों सर्कस के रेडियस सी जो वैल्यू आएगी वो होगी जीवन ² + F1 स्क्वायर माइंस सी प्लस दूसरे सर्कल की रेडियस क्या रही होगी सर वो रही होगी g2² + f2² - C2 और -c1 कंक्लुजन जो अलजेब्राइकली शायद थोड़ा डिफिकल्ट आपको समझ ए रहा है की चीज कैसे होने वाली है अगर हम बात करें किसकी अगर सर हम बेसिकली बात करें दो डिश जॉइंट दो डिस्ज्वाइंट्स के क्या कोई दिक्कत यहां तक आई थिंक बड़ा बेसिक स्ट्रेट फारवर्ड आर डायरेक्शन जिसमें कुछ ज्यादा हर फेयर नहीं है अगर नहीं है तो कम की बात पर आते हैं suniyega ध्यान से एक कंक्लुजन आप हमेशा निकल सकते हो एक जरूरी कन्फ्यूजन निकल सकते हो मैं पूरे प्रूफ पर नहीं जा रहा हूं मैं बस डायरेक्टली आपको यहां पर ये दिखाना चाह रहा हूं की जब आप चीज से देखोगे तो आप इस कंक्लुजन पर आओगे की ध्यान से सुनना ये आपके दो सर्कल्स हैं suniyega मैं थोड़ा सा इसको सॉर्टेड मैनर में बना रहा हूं बहुत कम की बात है और इस पर बेस्ड क्वेश्चंस बनते हैं और उसे पर क्वेश्चंस आते हैं है ना अनलॉक का एक सर्कल तो ये है ना और लेट से दूसरा सर्कल ये है सन रहे हो क्या और सर यह जो सर्कल्स हैं आपके इनके सेंटर्स के कोऑर्डिनेट्स से मैन लो एक सर्कल का सेंटर यहां है और इस सर्कल का सेंटर ऐसे यहां अब suniyega जो कम की बात है वो ये की मैन लो सर इन दोनों को मैंने मिलाया ऑफ़ कोर्स इट इस जॉइंट है इन दोनों के सेंटर्स के कोऑर्डिनेट्स को आपने क्या किया मिला दिया है ना तो इन दोनों के सेंटर्स के cardinors को अपने इस तरीके से मिला दिया कोई दिक्कत तो नहीं है आई थिंक नहीं होनी चाहिए अब suniyega कम की बात कम की बात ये है सर की मैन लो मैं क्या बनाता हूं मैं इनकी कुछ टैसेंट बनाता हूं मैं क्या करता हूं सर इनकी कुछ टांगेंट्स बनाता हूं मैं पहले तो एक ऐसी टैसेंट बनाता हूं जो की इससे इस तरह से और इसे इस तरफ से टच कर रही हो यह ट्रांसफर कुछ इस तरीके से इन दोनों सर्कल्स को क्या कर रही होगी भाई टच के बात आप समझ का रहे हो आई थिंक बहुत ज्यादा प्रॉब्लम मैटिक नहीं हो गए सर इसी तरीके से अगर आप देखो इसी तरीके से अगर आप देखो तो आप एक और टैसेंट बना पाओगे जो की इस तरीके से आपकी जा रही होगी आई थिंक मैं सही बना रहा हूं क्या थोड़ा सा टिल्ट कर दें या थोड़ा सा ऐसे बिल्कुल तो मैं थोड़ा सा ऐसे यहां लाना चाह रहा हूं जो की मुझे यहां पर चीज देगा तो इस तरीके से आप ये देखोगे ऑफकोर्स मैंने बहुत ज्यादा अच्छे से नहीं बनाई है चीज पर आप मेरी बात समझने की कोशिश करना की यह जो आपकी टैसेंट है जो ट्रांसफर सेल कमेंट्री जो की इससे यहां और उसे दूसरी साइट पर इसे यहां और इसे दूसरी साइड पर कनेक्ट करती है तो यह टच करने में जहां पर ये दोनों टांगें आपस में एक दूसरे को इंटरसेक्ट करती है वह पॉइंट कॉन्करेंट हो जाता है इन दोनों के सेंटर्स को ज्वाइन करने वाली लाइन के साथ आप सारी बातें छोड़िए आप तो यह समझिए जो दोनों आपकी ट्रांसपोर्टेशन कमेंट टैसेंट है और जो इन दोनों के सेंटर्स को ज्वाइन करने वाली लाइन है आई होप आप ये समझ का रहे हो इन तीनों में एक बात खास होगी की इन तीनों लाइंस कौन करंट होंगी यह तीनों लाइंस कौन करंट होंगे यानी एक किसी एक खास पॉइंट से पास होंगे इसके साथ-साथ आप एक बात और याद रखिएगा यही बात मैं आपसे करना चाह रहा हूं सुनेगा ध्यान से की यह जो आपका पॉइंट टी है ना यह जो आपका पॉइंट ये कॉमन टांगें के लिए जो लिया है ये जो पॉइंट आपका टी है ये जो पॉइंट टी है भाई इस कंक्लुजन पर मैं ए रहा हूं सीधे ये पॉइंट टी है ध्यान से देखना कहां लिखा है ये है ना तो पॉइंट टी जो होगा आपका वो C1 C2 को इंटरनल उन दोनों की रेडियस के रेश्यो में डिवाइड करेगा क्या कहना चाह रहा हूं यह जो टी पॉइंट है यह C1 C2 को C1 C2 को किस रेशों में डिवाइड करेंगे आई होप आप याद रखोगे मैं पूरे अलजेब्राइक प्रूफ पर नहीं जा रहा हूं जो हम प्रूफ करेंगे सर ये ट्रायंगल हो रही है traangle सिमिलर है और सिमिलरिटी लगा और फिर मैं प्रूफ करूं मैं आपको डायरेक्ट कंक्लुजन दे रहा हूं इसे याद रखिएगा इस पर क्वेश्चंस बनते हैं इस पर क्वेश्चन आते हैं क्या कंक्लुजन है की ये जो आपकी ट्रांसफर कॉमन टेंट है और यह जो इसके सर्कल के सेंटर्स को ज्वाइन करने वाली स्ट्रेट लाइन है तीनों कॉन्करेंट होते हैं और वो किसी पॉइंट पर इंटरसेक्ट होते हैं और जहां यह इंटरसेक्ट करते हैं या टेक्निकल ये दोनों टांगें इस पॉइंट को इस इस लाइन को जो है इंटरसेक्ट करती है जहां ये कौन करंट होते हैं वो पॉइंट इन दोनों को ज्वाइन करने वाली लाइन को इनकी रेडियस के रेश्यो में इंटरनल डिवाइड करता है आई होप ये बात आपको याद रहेगी बिना कुछ ज्यादा तकलीफ में आए हुए और क्या कहना चाह रहे हो सर मैं एक और आपसे बात कहना चाह रहा हूं जिसे ध्यान से suniyega ये बात और कृष्ण है जैसे मैन लो मैं थोड़ा सा ऐसे अलग तरीके से ही बनाता हूं सुनते जाइए ध्यान से जैसे मैन लो हम एक और सिनेरियो बनाते हैं कैसे ध्यान से सुनना बहुत कम की बात है मैन लो ये आपका एक क्या हुआ सर्कल है ना और मैन लो सर यहां पर एक और आपका यहोवा सर्कल सन रहे हो क्या आप लोग क्या करते हो सर आप एक डायरेक्ट कॉमन टांगें बनाते हो डायरेक्ट कमेंट इंजन कैसी होती है मुझे 1 सेकंड दीजिए बस डायरेक्ट कॉमन टांगें जो होती है वो कुछ इस तरीके से होती है वो कुछ इस तरीके से होती है की वो एक ही साइड से गुजर रही होती है वो क्या हो रही होती है भाई वो एक ही साइड से गुजर रही होती है ये आपकी एक डायरेक्ट कॉमन अटेंडेंट होगी क्या आप मेरी बात समझ पाए क्या सर एक ही डायरेक्टर होगी नहीं अगर डिसजॉइंट सर्कल है तो जैसे दो ट्रांसवर्सल कमेंट टैसेंट थी ऐसे ही डायरेक्ट कमेंट इंजन भी दो होंगी तो अगर यहां पर एक डायरेक्ट कॉमन टेंट है तो आप एक और डायरेक्ट कॉमन टांगें बना सकते हो सर इस तरीके से क्या किसी भी स्टूडेंट को कोई आपत्ति जो अभी हमने किया उससे आई थिंक बड़ी मजेदार सुलझी और साधारण सी डायरेक्ट सी बातें मैं आपसे कर रहा हूं जो आपको अच्छे से समझ ए रही होंगी जिम आपको कोई भी आपत्तियों परेशानी नहीं हो रही होगी आई थिंक कोई दिक्कत तो नहीं है भाई यहां तक ये आपकी क्या होगी ये होगी आपकी कमेंट है अब क्या करना है सर अब suniyega ध्यान से अब कम की बात क्या है अगर मैं अगर मैं अगर क्या करूं इसके और इसके सेंटर को ज्वाइन कर कर एक स्ट्रेट लाइन बनाओ suniyega बहुत कम की बात है पर क्वेश्चंस बनते हैं इसलिए इसे बहुत ध्यान से सुनना अगर मैं इसके और इसके सेंटर को ज्वाइन करके एक स्ट्रेट लाइन बनाओ मैं क्या कर रहा हूं इसके सेंटर को इसके सेंटर से मिलते हुए अगर बनाओ तो आपको एक बहुत ही मजेदार sirochak सी बहुत ही इंटरेस्टिंग सी बात ये मिलेगी की इस बार भी इस बार भी आपकी ये तीनों लाइंस कौन सी तीन लाइंस आपकी C1 और C2 दोनों के सेंटर्स को ज्वाइन करने वाली लाइन जो की आगे मैं फर्ज और एक्सटेंड कर रहा हूं और आपकी ये डायरेक्ट कमेंट्स फिर से वही कहानी आप सिमिलर प्रूफ कर दोगे बहुत ज्यादा डिफिकल्ट बातें नहीं है आप आसानी से अर्थमैटिक के लिए ओके मैं डायरेक्ट ये तीनों जो डायरेक्ट कॉमन टैसेंट हैं जहां पर कॉन्करेंट है जिस पॉइंट पर ये इंटरसेक्ट हो रही है या फिर ये दोनों डायरेक्ट कॉमन टैसेंट के C1 सीटों को ज्वाइन करने वाले लाइन को जहां इंटरसेक्ट कर रही हैं उसे पॉइंट को एफ आय कॉल दी फॉर ए तो बिल्कुल सर एक और बात आप देखोगे एक और बात ये होगी की क्या मैं ऐसे देखता हूं यहां पर ये इस बार जैसे पिछली बार इंटरनल हो रहा था ना इस बार क्या होगा सुनना ये वही बात वापस आप सारी कहानी प्रूफ करोगे और आप ये कंक्लुजन पर आओगे सर की वो जो दी है ना वो C1 C2 को externalli R1 r2 के रेश्यो में डिवाइड करता है वह C1 C2 को एक्सटर्नल R1 r2 के रेश्यो में डिवाइड करता है इसका क्या मतलब होगा suniyega ध्यान से आपका जो ये C1 C2 है इसको externalli R1 r2 के रेश्यो में कौन डिवाइड करता है दी यानी की C1 से जो दी तक डिस्टेंस होगी उसका रेशों और ऑफ कोर्स C2 से जो d2 तक की डिस्टेंस होगी उसका रेशों क्या रे क्या यह स्टेटमेंट समझ से कोई परेशानी क्या यह बात अच्छे से डाइजेस्ट हो रही है यह दोनों बातें प्लीज आप नोट डाउन करिए स्टूडेंट्स आज के लेक्चर पहली बात तो यही हमने जो पढ़ी है की अगर सर दो सर्कल्स दिस जॉइंट है तो एक कंडीशन फुलफिल होना जरूरी है की उनके बीच की उनके सेंटर्स के बीच के डिस्टेंस उनके रेडियस के सबसे ज्यादा होगी ऑफ कोर्स यह तो अंडरस्टूड सी बात है यह बड़ी साधारण ऑर्डिनरी सिंपल सी बात लगती है पर इससे आपको याद रखना जरूरी है क्योंकि इस पर ही सवाल बनती है और ये स्टूडेंट्स अप्लाई करना भूल जाते हैं तो एक कंडीशन तो हम ये लगा रहे होंगे जब भी जो भी क्वेश्चन आएगा उसे पर है ना जब हमें डिसजॉइंट सर्कल की बात हो रही होगी दूसरी कंडीशन जो हमें सोचनी होगी वो ये की अगर मुझे कभी भी कभी भी ट्रांसवर्सल कमेंट इंजन से डील करना हुआ तो मैं कहूंगा की C1 और C2 को ज्वाइन करने वाले लाइन को वो टांगेंट्स R1 इस तू आर तू के रेश्यो में डिवाइड करते हैं और डायरेक्ट कॉमन टांगें की बात हुई अगर डायरेक्ट कॉमन टेंट की बात हुई है डायरेक्ट कमेंट है इसकी और इसकी दोनों की टैसेंट है ये इसकी और इसकी दोनों की टैसेंट है तो वो C1 और C2 को ज्वाइन करने वाली लाइन को R1 इस तू आर तू के रेश्यो में एक्सटर्नल डिवाइड करती है बहुत तकलीफ बहुत परेशानी कोई डाउट कोई फ्लकचुएशन कोई भी दिक्कत तो नहीं है स्टूडेंट्स ऐसे बड़े बेसिक से मजेदार से कॉन्सेप्ट से अच्छा एक बात का जवाब दोगे की सर आपने सारी बातें कही हमको अच्छी लगी सारी बातें हमने समझी भी लेकिन क्या आप मुझे सवाल का जवाब दे सकते हो क्या आप मुझे सवाल का जवाब दे सकते हो की मैन लो सर यही सर कल से सर्कल मतलब बिल्कुल से सर्कल कुछ फर्क नहीं अगर से सर्कल मतलब बिल्कुल से मतलब वही लेंथ वही हाइट बिल्कुल वही सर्कल हो मतलब टेक्निकल वही लेंथ वही हाइट कम मतलब वही रेडियस का सर्कल मतलब दोनों सर्कल से रेडियस के हो तो इनकी डायरेक्ट कमेंट को लेकर आपका क्या विचार है तो इनकी डायरेक्ट कमेंट इंजन कैसे बनेगी सर इनकी जो डायरेक्ट इंजन बनेगी वह टेक्निकल कैसी होंगी क्या मैं ऐसा नहीं का सकता की सर जो अब डायरेक्ट कमेंट बनेगी वह बेसिकली दो पैरेलल लाइंस होंगी अरे रिपीट माय स्टेटमेंट अब जो डायरेक्ट कॉमन इंजन बनेगी वह बेसिकली दो पैरेलल लाइंस होंगी और चूंकि यह दोनों लाइंस पैरेलल होंगी तो क्या ये कभी भी इंटरसेक्ट करेंगी नहीं करेंगे तो उसे केस में डिवीज़न नहीं होगा ये कब की बात है स्टूडेंट्स ये तब की बात है जब आपके दोनों ही सर्कल की रेडियस क्या हो सी मतलब वो दोनों रेड आय अगर से है तो फिर आपकी जो डायरेक्ट कमेंट टैसेंट है दे विल नेवर इंटरसेक्ट दे विल ऑलवेज दे पैरेलल तू एच एडम आई होप बड़ी बेसिक्स सी बात क्या करेगी तीन बातें फिर से सन लीजिए दिस इस जॉइंट सर्कल्स मतलब अलग-अलग कोई भी रीजन कॉमन या इंटर सेट करने से भी ज्यादा बड़ी बात सी वांट टच भी नहीं करेंगे दूसरे को तीसरी बात दोनों के सेंटर्स के बीच के डिस्टेंस उनकी रेडियस के सबसे ज्यादा होगी चौथी बात उनका जो ट्रांसफर सेल आपकी जो टैसेंट होगी वह उन दोनों के सेंटर को ज्वाइन करने वाली लाइन को इंटरनली R1 और r2 के रेश्यो में डिवाइड करेगी नोट बना लीजिए गैस का और सबसे जरूरी बात की जो डायरेक्ट कॉमन टांगेंट्स का पॉइंट ऑफ इंटरसेक्शन होगा वह सी वैन और C2 को R1 इस तू आर तू के रेश्यो में एक्सटर्नल डिवाइड करेगी यह सारी बातें याद रहेंगी दिस इस ऑल अबाउट दिस जॉइंट सर्कस आई थिंक की अच्छी अंडरस्टैंडिंग अपने बेड की एक अगर अच्छी अंडरस्टैंडिंग बिल्ड की है अगर ये सारी बातें आपको समझ आई है तो क्या हम आगे बड़े क्या हम आगे बड़े आई होप ये वही सारी बातें लिखी गई हैं जो आपको समझने की कोशिश की जा रही है वो आपसे ये जॉइंट सर्कस पर बात कर रहा है वो एक और नया कॉन्सेप्ट आपको सीखना चाह रहा है की क्या मैं कॉमन टेंसेंट की लेंथ निकल सकता हूं क्या मैं वो कमेंट निकल सकता हूं मैं कहूंगा निकल तो सकता हूं सर बट कैसे जरा इस शुरुआत करते हैं देखो जरा बड़ा बेसिक सा फंडा है बड़ा मजेदार सा फंडा है मैन लो सर एक सर्कल ये रहा है ना और एक सर्कल की बात करूं देखो भाई डायरेक्ट कमेंट इंजन की अगर मैं आपसे बात करूं तो क्या होगा ध्यान से देखना बहुत मजेदार बात है सर अगर ये आपकी एक डायरेक्ट कमेंट टैसेंट है क्या आप सब एग्री करते हुए डायरेक्ट कमेंट्री से यहां पर भी बनेगी और डायरेक्ट कॉमन टांगें की लेंथ से क्या मतलब है मतलब ये जो पॉइंट ऑफ कॉन्टैक्ट है और ये जो पॉइंट ऑफ कॉन्टैक्ट है इन दोनों के बीच के डिस्टेंस मतलब वो कौन सी लेंथ निकलवाना चाह रहा है आपसे वो आपसे ये लेंथ निकलवाना चाह रहा है कैसे निकलेंगे सर एक मिनट थोड़ा पेशेंस रखिए आप बहुत मजेदार तरीके से इसका जवाब दोगे जब आप से मैं पूछूंगा की देखो भाई इसका सेंटर क्या होगा सर इसका सेंटरलेस से C1 है इसका सेंटर क्या होगा C2 कोई तकलीफ तो नहीं है अभी तक तो नहीं होनी चाहिए अब मैंने अगर आपसे कहा की चलो एक कम करो C1 से stanjan पर परपेंडिकुलर करो तो ये हो जाएगा सर C2 से भी stanzon पर परपेंडिकुलर ड्रॉप करो तो ये हो जाएगा आई थिंक आप बातें समझ का रहे हो आपको क्या समझने की कोशिश करूं की हमने एक तो ये परपेंडिकुलर ड्रॉप कर दिया एक ये परपेंडिकुलर ड्रॉप कर दिया टेक्निकल ये क्या हो गई सर टेक्निकल ये रेडियस होगी तो ये इस सर्कल की रेडियस होगी r2 और ये सर्कल की रेडियस हो गई क्या हम रियली सॉरी ये वो R1 है और ये सर्कल की रेडियस क्या होगी r2 अब सर एक छोटा सा कम और कर लीजिए आप क्या कर लीजिए आप एक कम करो अब C1 से C2 को मिला सकते हो क्या मिला देता हूं कोई दिक्कत तो नहीं मैंने C1 से C2 को मिला है अब एक छोटा सा कम और कर लो सर आप बस क्वेश्चन खत्म हो जाएगा क्या C1 से C2 को जो मिलने वाली लाइन है इसके पैरेलल एक लाइन यहां ड्रॉ कर सकते हो क्या इसके पैरेलल एक लाइन यहां ड्रा कर सकते हो क्या आई थिंक कर सकता हूं बहुत बड़ा सा फिगर में नहीं बना पाया अनफॉर्चूनेटली बट आई थिंक मैं कुछ तो बना का रहा हूं जो आप मेरी बात समझ का रहे हो क्या आप समझ का रहे हो जो मैं आपसे कहना चाह रहा हूं मैं जो कहना चाह रहा हूं मैं उसे बात पर आता हूं मैं आपसे कहना चाह रहा हूं की यह लाइन और यह लाइन पैरेलल है और यह लाइन और यह लाइन पैरेलल है आपको दिख रहा है क्या सर आपको क्या निकलना है इस पे बात करते हैं मैं कुछ पॉइंट्स को नाम दे देता हूं जैसे मैं इसे का देता हूं थोड़ी देर के लिए लेट से t1 और इस समय का देता हूं टी तू और इस पॉइंट को अलग से मैं का देता हूं मैं पिक कुछ भी दे डन अपना लेट्स से कहते हु पी इस पॉइंट को मैं कहता हूं पी इस पॉइंट को मैं कहता हूं क्यों और इस पॉइंट को यह जो यहां देख रहा है कोई दिक्कत तो नहीं भाई अब आप क्या निकलना चाह रहे हो सर अगर मैं ट्रायंगल पिक को देखूं तो शैल आईक्लाउड की सर जो ये आपका ट्रायंगल पिक है वो तो सर एक राइट एंगल ट्रायंगल है क्या आप ये बात देख का रहे हो क्या आप सब यह बात नोटिस कर का रहे हो हान सर बढ़िया आसान सी बात है ट्रायंगल पिक एंगल पी पर वर्टेक्स पी पर राइट एंगल दे बिल्कुल सही बात है जब मुझसे कोई पूछे की आप मुझे पीके बताओ तो आप मुझसे बेस पूछ रहे हो हाइपोटेन्यूज और लेंथ ये रही तो अगर मैं पाइथागोरस थ्योरम अप्लाई करूं तो मैं क्या कहूंगा सर मैं कहूंगा की पर स्क्वायर प्लस पी के स्क्वायर विल बी इक्वल तू आर के स्क्वायर सर ये सब क्यों पढ़ा रहे हो मैं आपको यह सिखा रहा हूं की आप कोई फॉर्मूला ना रखें और आप यह मेकैनिज्म समझ ले और आप हर बार इसी तरीके से क्वेश्चंस को सॉल्व करें फॉर्मूला रात लेना आपका मैन हो तो आपको अगर रत्न अच्छा लगता है अगर आप याद रख पाते हो बहुत सारी चीज तो आप याद कर लेना मुझे कोई आपत्ति नहीं है पहले समझ तो लो की वो चीज हो क्यों रही हैं क्योंकि ये बात पे तो हम बार-बार जोर दे ही रहना की ये एग्जाम रत्न की एग्जाम नहीं है एग्जाम समझने की एग्जाम में जहां पर आपकी सब्जी हुई चीजों को चेक किया जाएगा आप यकीन करिए जिसने जितना अच्छा समझा होगा ना उसका उतना अच्छा स्कोर आएगा अच्छे से समझो बात आखिर में आती जिसने जितना अच्छे से समझा है जिसने जितना अच्छे से कॉन्सेप्ट्स को समझे रिवाइज किया है पढ़ा है और क्वेश्चंस में इंप्लीमेंट करना सिखाए उसका उतना अच्छा पेपर ये जाने वाला है उसकी उतनी अच्छी अंडरस्टैंडिंग होने वाली और कहीं कुछ नहीं आने वाला है सब धारा का धारा रह जाएगा ये रता हुआ कहीं भी कम नहीं आएगा रात के होना क्या मतलब मुझे सेंसी समझ नहीं आता है क्या आप रेड के पता नहीं क्या करते हो मतलब आप एक्चुअली अपनी लाइफ अपना टाइम अपने बहुत सारे इयर्स वेस्ट कर रहे हो अगर आप चीज से रात रहे हो तो प्लीज मत अटो रत्न पद रहा है मतलब कोई मजबूरी है कोई प्रॉब्लम नहीं हो रही है अवॉइड करिए आप चीज अच्छे से समझिए लाइफ एंजॉय करिए अच्छे से सब्जेक्ट को इंजॉय करिए कितना मजेदार सब्जेक्ट है ठीक से समझ कर पढ़ा जाए तो समझ गया रे टिए मत है ना पता नहीं क्यों लोग फॉर्मूले रखते रहते हैं तो देखो भाई क्या लिखा है वो लिख रहा है की सर आपसे मैं जानना चाह रहा हूं आरके के डिस्टेंस मैंने बोला सर इस राइट एंगल ट्रायंगल पिक में देखो आप बोलोगे की आर जो ये पीके है हम क्या निकाला चाह रहे हैं पीके के डिस्टेंस तो हम कहना चाह रहे हैं की r³ के स्क्वायर में से अगर आप आरपी का स्क्वायर करोगे तो पीके का स्क्वायर ए जाएगा बिल्कुल सही वही बात हमने करनी चाहिए अब मेरा कहना है सर पिक का स्क्वायर आप लोग पर क्या आपको ये बातें पता है थोड़ा ध्यान से देखो मुझे पहले तो चाहिए r³ मुझे पहले तो चाहिए आर यू क्या आपको ये नहीं दिख रहा की सर आर सी वैन सी तू क्यों ये एक पैरेललोग्राम है पैरेलल पैरेलल पैरेलल पैरेलल तो एक पैरेललोग्राम hailogram कैन आई कनक्लूड की सर अगर आपसे कोई r³ पूछे तो आप तो का देना सी वैन सी तू अगर मुझसे कोई पूछ रहा है आरके का स्क्वायर तो मैं कहूंगा सर C1 और C2 का स्क्वायर और C1 और C2 मतलब क्या दोनों के सेंटर्स के बीच के डिस्टेंस दोनों सर्कल दिए रहे होंगे भाई तो दोनों के सेंटर्स के कार्ड भी पता चल जाएंगे तो दोनों के सेंटर्स के बीच की डिस्टेंस का स्क्वायर ए जाएगा बिल्कुल ए जाएगा सर अब क्या अब अगर मैं निकल लूं किसी भी तरीके से पी आर का स्क्वायर अगर मैं निकलूं किसी भी तरीके से प्यार का स्क्वायर कोई हेल्प करेगा क्या पर का स्क्वायर निकलना है अब बात समझो स्टूडेंट्स पर का जो स्क्वायर है यह जो पर है क्या मैं पी आर को tc1 - आरसी वैन लिख सकता हूं क्या इसको मैं पीसी वैन माइंस आर सी वैन लिख सकता हूं क्या मैंने लिखा pc1 - rc1 अच्छा बताओ पीसी वैन आर वैन है और rc1 क्या है तो क्या आपको बात समझ ए रहा है दिस इस नथिंग बट व्हाट दिस इस R1 - r2 दिस इस नथिंग बट व्हाट R1 - r2 का होल स्क्वायर एंड डेट इस क्लीयरली डेट इस क्लीयरली = व्हाट पी के का स्क्वायर और अगर मैं यहां से स्क्वायर हटाओ तो क्या मैं √ यहां पोस्ट कर डन सर कर दीजिए और आई थिंक यह दोनों बातें आपको पता होंगी सर दोनों के सेंटर के कोऑर्डिनेट्स आपको पता होंगे दोनों की रेडियस पता होंगे तो दोनों के सेंटर्स के बीच का डिस्टेंस का स्क्वायर में से दोनों के रेडियस के डिफरेंस के स्क्वायर को सब्सट्रैक्ट करके उनका अंडर रूट ले लीजिएगा आपको उसकी डायरेक्ट कॉमन टांगें की लेंथ मिल जाएगी क्या यह बात आपको समझ आई मैं का रहा हूं की यह फॉर्मूला रत्न की जरूरत क्यों है यह फॉर्मूला तो दो लाइन ही तो था अब अगर यह तरीका आपको समझ ए गया है तो फॉर्मूला आपके पास हमेशा रहेगा क्योंकि आप कुछ चीज बनाते आती हैं फॉर्मूले पर बेस्ड क्वेश्चन नहीं पूछेगा वो इस तरीके पर क्वेश्चन पूछेगा की आपको यह चीज सोचते आता है या नहीं आप ऐसे चीज देख पाते हो या नहीं आपको इस एप्लीकेशन पर भी क्वेश्चन आएंगे इस फॉर्मूले में वैल्यू C2 की वैल्यू पूछेगा वहां कर लेना शांति से भाई आई होप ये बात समझ का रहे हो सर ठीक है आपकी बात मैन लिया हमने हम डायरेक्ट कमेंट इंजन की तो लेंथ निकलेंगे इस पॉइंट को नोट डाउन किया या नहीं भाई इस पॉइंट को अच्छे से नोट डाउन किया है डायरेक्ट कमेंट इंजन की लेंथ का फॉर्मूला हमें समझ आया इसका फॉर्मूला भी बना सकते हो कोशिश कर सकते हैं कोशिश ही तो कर सकते हैं हमारे हाथ में है ही क्या कोशिश तो ट्रांसफर कॉमन तेनजेन में क्या हो रहा होगा एक ऑफ कोर्स आपका ये सर्कल रह रहा होगा और एक ऑफ कोर्स यहां पर एक और सर्कल रह रहा होगा जो की ये रहा होगा है ना फिर से वही बात अगर मैं आपसे करूं तो मेरा आपसे कहना है सर वही सारी बातें दोहरे जाए की पहले तो इनकी ट्रांसवर्सल कॉमन एक टैसेंट बनाते हैं और ऑफ कोर्स जो इस वाली की होगी वही इस वाली की लेंथ होगी ना इसमें कोई दो राय नहीं है अब अगर फिर से अगर मैं आपसे वही सब बातें करूं तो क्या आप देख सोच या समझ का रहे हो की आपको तो चीज कुछ इस तरीके से देखनी चाहिए सर आपको तो चीज कुछ इस तरीके से देखनी चाहिए सर मैं थोड़ा इसको ट्रीट कर रहा हूं ताकि चीज हम ऐसी दिखा पाएं बना पाए है ना तो हमें दिख रही है को इमेजिन करके आप मुझे बना के बता दो की अब आप यह जो ट्रांसवर्सल कमेंट है इसकी लेंथ आप कैसे निकलेंगे जल्दी से बताइए स्टूडेंट्स मेरा कहना है की सर मैं कैसे निकलूंगा मैं निकलूंगा एक बड़ी बेसिक से एप्रोच उसे करके देखो भाई यह आपकी टांगें है किसी के अलग nazariyat है तो क्या करेंगे यह इस सर्कल का सेंटर्स के सेंटर को ज्वाइन करने वाली लाइन होगी और ऑफ कोर्स सर यकीनन तौर पर आप देख का रहे हो की यहां से अमरेली सॉरी अमरेली रियली सॉरी पता नहीं की उठ के चला जाता है सॉफ्टवेयर थोड़ा सा परेशान करते हैं हमको है ना अब अगर मैं इस सेंटर से यहां पर एक परपेंडिकुलर ड्रॉप करो तो आई होप कुछ कहने की जरूरत नहीं ये इसकी क्या होगी रेडियस इस सेंटर से यहां पर एक परपेंडिकुलर ड्रॉप करू तो ये क्या हो गई भाई इसकी रेडियस आई थिंक कोई दिक्कत नहीं कोई दिक्कत तो नहीं अच्छा इनको हम लेवल कर लेते हैं कुछ बातों को जैसे की सर इस सर्कल का सेंटर हो गया C1 इस सर्कल का सेंटर हो गया C2 इस सर्कल की रेडियस होगी R1 और इस सर्कल की रेडियस हो गई यार तू अब आप निकलना क्या चाहते हो मैं निकलना चाह रहा हूं की सर इसमें अगर इसे यहां टच किया था ए पर और उसने अगर यहां टच किए से बी पर तो ये अब डिस्टेंस हम निकलना था इंटरेस्टेड इन दिस डिस्टेंस अब सर कैसे निकलेंगे क्या करेंगे क्या सोचेंगे कैसे थॉट आएगा बात करते हैं मेरा कहना है सर अगर आप अब डिस्टेंस निकलना ही चाह रहे हो इफ यू आर रियली इन दिस डिस्टेंस अब नौ व्हाट आर डी डिफरेंट वेज क्या एक तरीका यह हो सकता है की मैं aoc2 से कनेक्ट करूं और बी को सी वैन से कनेक्ट करू तो शायद कोई क्वार्टर लैटरल बने जिसमें कुछ ऐसा सिनेरियो या ऐसी पॉसिबिलिटी देखता है या फिर कोई ऐसा तरीका की सर मैं यहां पर कोई राइटिंग बनाओ यहां कोई ऐसा तरीका की सर जिससे मैं ये प्रूफ कर पाऊं जिससे मैं ये प्रूफ कर पाऊं की देखो सर अगर मैं इसके कॉमन पॉइंट को जहां पर इंटरसेक्ट करेंगे लेट से मैसेज थोड़ी देर के लिए कहूं टी तो अगर मैं यहां पर कोई रेडियस वाला कॉन्सेप्ट अप्लाई कर पाऊं तो वहां से अगर मैं सारी बातें भूल जाऊं तो एक बात तो मुझे पता है एक बात मुझे क्या पता है suniyega इससे क्वेश्चन खत्म हो सकता है अगर आप दिमाग से सोचते हो सर देखो बड़ी बेसिक सी बात है सर क्या आप ये देख का रहे हो ये एक 90° एंगल है हाथ से 90 डिग्री एंगल है क्या आप यह देख का रहे हो एक 90 डिग्री एंगल हान सर ये 90° का एंगल क्या आप ये देख का रहे हो की ये वर्टिकल अपोजिट एंगल्स हैं हान सर ये वर्टिकल अपोजिट साइकिल से जो की इक्वल होंगे तो यह एंगल और ये एंगल इक्वल ये एंगल और ये एंगल इक्वल तो क्या मैं ऐसा नहीं का सकता की सर ये एंगल और ये एंगल भी इक्वल होंगे और यही बात मैं आपको समझाना चाह रहा था की इस तरीके से आप इन्हें सिमिट्रिकल प्रूफ कर दोगे किस-किस को ए सी वैन टी को और bc2t को कोई दिक्कत तो नहीं है काफी हद तक एक अच्छी लंबी चौड़ी हिंट मैंने आपको दे दी अब बस आपको अब के डिस्टेंस निकालनी है कैसे करेंगे जल्दी बताओ भाई बोलो भाई क्या निकल कर ए रहा है कैसे करोगे एक सिंपल सा आइडिया सिंपल सा सजेशन मैं आपको देता हूं देखो ऐसे सोचो सर एक कम करते हैं C1 ए को मैं थोड़ा एक्सटेंड करता हूं मैं इसे इतना ही एक्सटेंड करूंगा की वो bc2 के पैरेलल और इक्वल हो ऑफ कोर्स पैरेलल तो आना तय बट मैं इसको इस तरीके से एक्सटेंड करता हूं देखो अब जैसे ही एक्सटेंड किया तो मैं यहां पर कुछ मजेदार सा होता हुआ देख का रहा हूं क्या आप देख का रहे हो लेट्स कॉल इट पॉइंट पी मैं इसे कहता हूं लेट से पॉइंट पी अब आप मेरी बात बहुत ध्यान से सुनिए अगर मैं आपसे कहूं की सर आप ट्रायंगल सी वैन पीस तू में देखो आप ट्रायंगल सी वैन पी सी तू में देखो क्यों देखो सर देखो ध्यान से C1 पी सी तू बहुत ध्यान से सुनना बहुत कम की बात है क्या मैं C1 pc2 में कहूं की सर मैन लो थोड़ी देर के लिए बस ऐसे ये तो 90° होना आता है क्यों सर ये 90 डिग्री है इसी को एक्सटेंड किया था यहां पर भी 90 डिग्री तो यह जो आप है यह इस तरीके से हमने बनाया है की वो bc2 के इक्वल जो bc2 के इक्वल है मतलब ये और ये अगर इक्वल है तो ऑफ कोर्स आप की वैल्यू कितनी ए जाएगी सर आप की वैल्यू हो जाएगी आर तू अब सन रहे हो क्या मैं अब नहीं निकल लूंगा आप समझो ना यह एबीसीडी निकलूंगा तो मैं अब निकल लूंगा तो जो मैं ए बी निकलना चाह रहा हूं जो की अभी फिलहाल है पी सी तू के पी सी तू के जिसमें अगर मैं आपसे बात करूं pc2 के टेक्निकल पाइथागोरस थ्योरम आई होप आप देख का रहे हो बेस परपेंडिकुलर और हाइपोटेन्यूज है ना तो pc2 कितना हो जाएगा सोच के देखना सर C1 C2 का स्क्वायर में से ये हाइपोटेन्यूज का स्क्वायर में से अगर मैं परपेंडिकुलर यानी कौन सी वैन ए और ap1 यानी क्या c1p और c1p क्या है सर C1 ए + आप और C1 ए क्या है R1 और आप क्या है r2 तो ये क्या हो जाएगा सर ये हो जाएगा R1 प्लस r2 suniyega ये क्या हो जाएगा R1 + r2 का होल स्क्वायर और इसका आप क्या लेते हैं सर आई होप आप समझ का रहे हो अंडर रूट इसलिए क्योंकि इसका स्क्वायर अगर यह वैल्यू चाहिए तो उसका स्क्वायर इस तरफ जाके अंडर रूट बन जाएगा और अगर आप ध्यान से देखो तो फिर वही बात दोनों के सेंटर्स के बीच के डिस्टेंस के स्क्वायर में से इस बार दोनों की रेडियस का सैम का स्क्वायर करके अंडर रूट लिया जब बात डायरेक्ट कॉमन टैसेंट की थी जब बट डायरेक्ट कॉमन टैसेंट की थी तो हमने डिफरेंस का स्क्वायर किया था जब बात transversial कॉमन टेंट की लेंथ की है तो हमने दोनों रेडियस का स्क्वायर का मतलब दोनों रेडियस के सैम का स्क्वायर किया है आई थिंक बहुत मुश्किल बहुत टू बातें नहीं है बट इन्हें अच्छे से नोट डाउन कर लेना फॉर्मूले इसको इन इन कॉन्सेप्ट्स को अच्छे से नोट डाउन कर देना एक तो ये हो जाएगा आपका जिस पर हम बात करेंगे एक ये हो जाएगा आपका हम बात करेंगे अच्छे से नोट डाउन करने का मतलब ये बात याद रखना की चीज कैसे आई हैं अगर आप लगा पाए तो सीधे फॉर्मूला लगा के फटाफट आंसर ले आइएगा या फिर आप ये एटलिस्ट एप्रोच से ये कॉन्सेप्ट्स याद रखना क्योंकि इन पर बेस सवाल आपके बनेंगे वही परेशानी कोई तकलीफ यही सारी बातें उन्होंने भी कहेंगे घूम जाकर मैं कहूंगा इस तरीके से आप चीज निकलोगे अब एक और कॉन्सेप्ट निकल कर आता है जहां पर बात आती है सर अगर वो कॉमन टांगें की इक्वेशन चाहिए तो इक्वेशन कैसे बनाएंगे मैन लो सर मुझे वो कॉमन टांगें की इक्वेशन चाहिए तो वो इक्वेशंस हम कैसे निकलेंगे तो उनके अकॉर्डिंग एक सिंपल सी एप्रोच बता रहा हूं ध्यान से सुनना वैसे तो हम कंडीशन अप्लाई करेंगे अब कंडीशंस कैसे होंगे सुनना कंडीशन ऐसी होंगी जैसे फिर से वही बात आपके पास क्या होगा एक ये सर्कल होगा ऑफ कोर्स सर्कल होगा मैं एक किसी भी commentanger पर बात कर रहा हूं suniyega ध्यान से मैं किसी भी कॉमन टेंसेंट पर बात कर रहा हूं आपसे सुनना ध्यान से कॉमन तेनजिंग पर बात करने से मेरा आपसे जो तात्पर्य है वो ये suniyega बहुत ध्यान से इस कम की बात को की सर ये आपकी मैन लो वो कमेंट टेंट है अब इस कॉमन टेंट में किसी भी स्ट्रेट लाइन की इक्वेशन की जरूरत है पॉसिबल ही अगर मुझे इसकी स्लोप या यह किस पॉइंट से पास होती है वो मिल जाए तो बात बन जाएगी या फिर मुझे ये पता चल जाए या फिर मुझे ये पता चल जाए की एक किन दो पॉइंट से पास होती है तो हम इसकी इक्वेशन बना लूंगा क्या ये एप्रोच इसके बातें आपको समझ ए रही है अब इन सारी बातों के दौरान बहुत सारी बातें आपको बताऊंगा मैं ये हिंट यह आइडिया आपके ऊपर छोड़ता हूं की वो कंडीशन और सोच लो जहां पर हम बात क्या करते द की अगर कोई लाइन कोई लाइन किसी की टांगें है तो उसकी परपेंडिकुलर डिस्टेंस उसे सर्कल के सेंटर से उसकी रेडियस के इक्वल होती है बस ये दो बातें जल्दी से अप्लाई करो और मुझे ये बता दो की इस केस में आप यहां पर इस कॉमन टांगें की इक्वेशन कैसे निकलोगे हिंट आपके सामने है कंडीशन आपको बता दिया और इसी पर बेस नंबर अभी तक क्वेश्चंस करते हुए आए हैं स्टूडेंट्स और यह चीज अब आपको क्लीयरली नजर आणि चाहिए की आप कैसे सोचोगे आप चीज कैसे निकलेंगे और आई थिंक बहुत डिफिकल्ट बहुत टू टास्क नहीं है अच्छा सोचो क्यों नहीं है टू एक बात बताओ जैसे मैन सर यह पता है की टैसेंट कहीं ना कहीं जाके एक बात सोचो यह tagent कहीं ना कहीं जाके एक्सटेंड होगी और एक्सटेंड होते होते एक बड़ा मजेदार सा वाक्य मुझे मिलेगा की सर इन दोनों के सेंटर से ये जाकर इंटरसेप्ट करेगी इस दोनों सर्कल से सेंटर से जाकर कहीं ना कहीं इंटर सेट करेगी आप मेरी बात समझ का रहे हो क्या ऐसे थोड़ा और ठीक से बनाऊं तो देखो ध्यान से मैं इसी लाइन को थोड़ा एक्सीडेंट कर रहा हूं घबरा मत मैं इसी लाइन को थोड़ा एक्सीडेंट करके आपको ये कहना चाह रहा हूं की ये कहीं ना कहीं जाके क्या करेंगे भाई एक्सीडेंट कहीं ना कहीं करेंगे ऑफ कोर्स तो वो जहां एक्सटेंड करेगी वो जो भी योनो जहां इंटरसेक्ट करेंगी वो एक पॉइंट ऑफ इंटरसेक्शन होगा मैं इसे थोड़ा प्रॉपर्ली बना देता हूं ताकि आपकी समस्या का निवारण हो जाए और आप अच्छे से समझ पाओ चीज suniyega ध्यान से जैसे की मैन लो मैं इस तरीके से बना रहा हूं मैं आपको ये दूसरा एग्जांपल लेता हूं आपको दो सेकंड के अंदर क्लिक हो जाएगा किसी से कैसे सोचनी है बाय डी वे ये अब आपको क्लिक होना चाहिए था फिर भी खैर अगर नहीं समझ ए रहा है तो सुनना क्या मैं ये जानता हूं सर इसकी जो ये कॉमन टांगें है इसकी जो सर ये कॉमन टांगें है इसकी खास बात यह होगी सर की यह जो कॉमन टेंट होगी सर यह बड़ी बेसिक सी टैसेंट है इसकी खास बात यह होगी की इन दोनों के सर्कल्स के सेंटर्स को जब आप ज्वाइन करोगे तो वो किसी ना किसी पॉइंट पर इंटरसेक्ट करती है लेट्स क्वालिटी फॉर अवर लेट्स कॉल इट्स सी वैन फॉर अन लेट्स क्वालिटी अगर दोनों सर्कल्स दिए रहेंगे तो क्या मुझे सर्कल की रेडियस और इस सर्कल की रेडियस नहीं दी गई होगी सर मुझे बेसिकली ये पता है की सर्कल के सेंटर के कोऑर्डिनेट्स कितने हैं इस सर्कल के सेंटर कितने हैं मुझे इसकी रेडियस बताओ की इसकी रेडियस बताओगी तो सिंपल सी बात है सर अगर C1 और C2 के कोऑर्डिनेट्स पता है अगर C1 और C2 के कोऑर्डिनेट्स पता है तो क्या मैं तक कोऑर्डिनेट्स पता कर सकता हूं क्योंकि मैं जानता हूं जब ये ट्रांसवर्सल ट्रांसवर्सल कमेंट टेंट आती है पिक्चर में तो ये जो टी पॉइंट होता है जहां पर वो C1 और C2 को ज्वाइन करने वाली लाइन से इंटरसेक्ट होती है ये टी पॉइंट तो मैं बिल्कुल सर इस टी के कोऑर्डिनेट्स निकल सकता हूं क्या आप मेरी बात समझ का रहे हो जैसे की ऑफ कोर्स ये एक सर्कल है इस सर्कल को मैं कहता हूं थोड़ी देर के लिए लेट्स सेल्स वैन और इस सर्कल को कहता हूं इस तू पुरी मेकैनिज्म समझा रहा हूं ध्यान से सोचना S1 क्या है सर आपका सर्कल जिसको आप कहते हो x² + y² + 2g1x + 2f1y + C1 और सेकंड सर्कल को आप क्या का रहे हो सर उसे आप का रहे हो x² + y² + 2X + 2y + C2 आई होप आई नॉट टेल यू बड़ी सिंपल सी बात आप देख का रहे हो की सर्कल का सेंटर C1 है अच्छा ये C1 C2 और ये C1 C2 अलग है ना अगर आपको कन्फ्यूजन हो रहा है तो मैं एसएससी वैन और सी तू नहीं कहता हूं इसे कुछ और कर लेता हूं इसे मैं का देता हूं जैसे इस सर्कल का सेंटर है ना तो मैं कहता हूं इसे ओ वैन और इसे मैं का देता हूं o2 अब अगर मैं इस सर्कल के सेंटर की बात करूं ओवन की तो क्या हो जाएगा सर वो हो जाएगा - जीवन कमा कोस - F1 और उसे सर्कल के सेंटर की बात करूं तो हो जाएगा o2 जो की हो जाएगा माइंस के तू कमा - f2 आप क्यों ये नहीं सोच का रहे हो की तक जो cardinate होंगे वो क्या होंगे सर सर R1 और r2 के रेश्यो में डिवाइडेड है कौन ओवन और o2 को ज्वाइन करने वाली लाइन तो मैं क्या कहूंगा मैं कहूंगा R1 टैंक - G2 मैं क्या कहूंगा R1 टैंक - G2 प्लस प्लस नहीं माइंस ही आने वाला है क्योंकि वहां भी माइंस ही है तो R1 टाइम्स जी तू मतलब माइंस G2 - r2 टाइम्स जीवन अपॉन में एफसी क्या अपॉन में R1 + r2 सिमिलरली ऐसे ही आप इसका वे कोऑर्डिनेट्स भी निकल लेंगे वो क्या हो जाएगा सर हो जाएगा - R1 टाइम्स f2 - r2 टाइम्स एफ वैन अपॉन में क्लीयरली वापस क्या होगा R1 + r2 ये आपके क्या ए जाएंगे सर ये आपके ए जाएंगे बड़ी सिंपल सी बात है ये आपके ए जाएंगे क्या इस टी के cardinate से मेरा क्या मतलब अगर मैं इससे कहूं थोड़ी देर के लिए और इसे कहूं बी तो ये अब लाइन किस पॉइंट से पास होती है मुझे पता चल गया अब सन रहे हो क्या अगर मुझे ये पता चल गया की ये ए बी लाइन के इस पॉइंट से पास होती है तो क्या मैं थोड़ी देर के मल हूं अच्छा ये अलजेब्राइक बहुत अजीब लग रहा होगा पर ये नंबर्स होंगे यहां लिखा होगा कुछ ना कुछ जैसे 2 3 या -5 कमा 7 ऐसा कोई एक नंबर होगा ये इतना अजीब तो अभी अलजेब्राइकली लग रहा है तो फिर मैं फाइनली जब मैं जानता हूं जैसे मैं मैन लेता हूं की ये जो टी पॉइंट के कोऑर्डिनेट्स हैं जिसे मैं थोड़ी देर के लिए आपको समझने के लिए लिख देता हूं लेट्स से की सर ये पॉइंट्स हैं लेट्स से X1 y1 आपको समझने के लिए जो की रियलिटी में कुछ ऐसे बनकर आएंगे तो क्या टी पॉइंट से पास होने वाली से स्ट्रेट लाइन अब की इक्वेशन लिख सकता हूं सर अगर स्लोप को मैं थोड़ी देर के लिए मैन लूं एम अगर मैं इसकी स्लोप को क्या मैन लूं अगर मैं इसकी स्लोप को थोड़ी देर के लिए मैन लूं एम तो मैं इसकी इक्वेशन को क्या कहूंगा मैं इक्वेशन ऑफ अब क्या कहूंगा सर मैं इक्वेशन ऑफ ए बी जो कहूंगा वो होगी ए - 5 जो की ये होगा है ना की मैं y1 लिख ले रहा हूं इस इक्वल तू एम टाइम्स एक्स - X1 जो की ये होगा है अब यह ए जाएगी आपकी इक्वेशन अब यह क्वेश्चन इसलिए खत्म हो जाने वाला है स्टूडेंट्स क्योंकि सोच के देखो यह जो अब स्ट्रेट लाइन है यह टैसेंट है तो या तो अब की इस सेंटर से परपेंडिकुलर डिस्टेंस निकल लेना या अब की इस सेंटर से परपेंडिकुलर डिस्टेंस निकल लेना और वो किसके इक्वल होगी रेडियस के रिपीट में स्टेटमेंट अगर मैं निकलूं क्या परपेंडिकुलर डिस्टेंस कहां से सर किसकी अब की कहां से लेट से कहां से फ्रॉम ओवन तो वो क्या होगी सर इसकी रेडियस के इक्वल जो की क्या होगी R1 या फिर मैं क्या उसे कर सकता हूं या तो मैं ये कर लूंगा या फिर क्या कंडीशन अप्लाई कर दूंगा या फिर ये कंडीशन अप्लाई कर दूंगा की सर वही परपेंडिकुलर डिस्टेंस किसकी लाइन आपकी अब की बट कहां से सर कहां से o2 से वो कितनी होगी r2 के इक्वल अगर आपने ये कंडीशन अप्लाई कर दी तो अभी आप का आंसर ए जाएगा क्योंकि जब आप परपेंडिकुलर डिस्टेंस का फॉर्मूला लगाओगे तो बस अननोन वेरिएबल कौन बचेगा मोड में कुछ फॉर्मूला बनता है स्क्वायर स्क्वायर ए जाता है क्वाड्रेटिक बन जाती है तो आप ऐसा नहीं सोच रहे हो क्या की आपके एम की दो वैल्यूज आएंगे रानी भी चाहिए स्टूडेंट्स एम की दो वैल्यू इसलिए आएंगे क्या कोई स्टूडेंट्स कमेंट में लिख रहा है की दो वैल्यूज के ए रही होंगी सो दो वैल्यूज आना तय है क्योंकि इस पॉइंट से सिर्फ यही एक commentanging नहीं है सर इस पॉइंट से एक और कॉमन टैसेंट जाएगी पता नहीं की आप सोच नहीं का रहे हो सर ध्यान से क्या आपको मेरी बात रिलाइज हो रही है स्टूडेंट्स आई थिंक बड़ी आसान सी बातें मैं आपसे कहना करना चाह रहा हूं जो आप देख का रहे हो की बिल्कुल सर इसी पॉइंट की से एक और कॉमन टांगें जा रही होगी तो उसकी भी तो स्लोप दे रहा होगा तो टेक्निकल एक को सॉल्व करते गए द आपको दूसरे रूप मिल जाएगी कोई डाउट सर यह तो हुई ट्रांसवर्सल कमेंट इंजेक्ट क्या इसी तरीके से हम डायरेक्ट कमेंट में निकल सकते हैं क्यों नहीं निकल सकते फिर वही बात इसे एक्सटेंड करेंगे इससे एक्सटेंड करेंगे वो कहीं जाके इंटरसेक्ट कर रही होंगी वो जहां जाके इंटरसेक्ट कर रहे होंगे वहां पे एक्सटर्नल डिवीज़न वाला फॉर्मूला लगा देना तो उससे भी आपकी टांगें की इक्वेशन ए जाएगी स्लोप पता नहीं होगी तो स्लोप हम एम मैन लेंगे और वही परपेंडिकुलर डिस्टेंस को रेडियस के इक्वल रख देंगे सेंटर से तो हमको क्या ए जाएगी वो एम की वैल्यू वापस दो वैल्यूज आएंगे तो यह वाली डायरेक्ट कमेंट होगी और एक ये वाली डायरेक्ट कॉमन टैसेंट होगी आई थिंक चीज आसान है अगर आप अच्छे से मेकैनिज्म समझ ले आप स्टैंडर्ड ऑपरेटिंग प्रोसीजर समझते आप समझ पाएं की चीज कैसे हो रहे हैं आई थिंक ये बात आपने अच्छे से नोट डाउन की यहां तक तो कोई डाउट नहीं है इससे अच्छे से समझ लेना स्टूडेंट्स ये आप अच्छे से इसका स्क्रीनशॉट ले लीजिएगा जो भी आपको लगे इससे अच्छे से देख लीजिएगा है ना इसे अच्छे से प्लीज नोट डाउन करना ये पुरी पुरी मॉडल ऑफ इंडिया याद रहनी चाहिए की चीज कैसे निकल जाती है अगर मुझे कॉमन टांगें की इक्वेशन बनानी है तो मैं कितनी कहानी से गुजरूंगा ये मेकैनिज्म समझ के रखना क्योंकि इस पर बेस्ड क्वेश्चंस बनेंगे तो हमारे क्या क्या ऑब्जर्वेशंस रहेंगे बात समझो स्टूडेंट्स मैन लो ये एक सर्कल है और मैन लो यहां पर एक और सर्कल है और ये दोनों सर्कल्स एक दूसरे को लेट्स से यहां कहीं एक्सटर्नल टच करते हैं ये एक्सटर्नल टच किया तो क्या बात समझ का रहे हो सर जहां अभी तक मैं बात यह करता आया हूं की सर देखो कॉमन टांगेंट्स का जो मैं पहलू आपको दिखाना चाह रहा हूं वह समझो सर की देखो आप एक टैसेंट तो यहां कहीं कुछ ऐसी बना पाओगे थोड़ा सा इसको मैं शिफ्ट करूंगा तो आई थिंक ये कमेंट इंजन बन जाएगी बिल्कुल सही कहा सर आपने एक कॉमन टांगें शायद आप यहां कहीं कुछ ऐसी बना पाओगे जैसे मैं थोड़ा सा घुमा डन तो चीज आसान हो जानी चाहिए हॉपफुली है ना ये आपके कमेंट टेंट हो जाएगी आप सर यहां पर जो एक ट्रांसवर्सल कॉम बना सकती है देख का रहे हो ना ये दोनों डायरेक्ट कॉमन टैसेंट हैं बट यहां पे दोनों सर्कस जो की externalli एक दूसरे को टच कर रहे हैं तो वो जो आप ट्रांसफर कमेंट द वो तो सर इस बार दो तो बनने से रही दो क्यों बनने से नहीं जनाब क्योंकि सर दो अगर बन जाती तो फिर तो कहानी अलग ही हो जाती है ना दो कैसे बन जाएंगे इस बार जहां दोनों सर्कल से एक दूसरे को टच कर रहे हैं उसी पॉइंट पर यह ट्रांसवर्सल आपकी जो टैसेंट है ये भी तो वही उसी पॉइंट पर टच करेगी क्या आप यह पॉइंट समझ का रहे हैं यानी की सर पहला ऑब्जर्वेशन तो जो निकल कर आता है वो ये की अगर दो सर्कल से एक दूसरे को externalli टच करें तो jahanind सर्कल्स में आप चार कॉमन टांगेंट्स बना पाते द दो डायरेक्ट दो ट्रांसफर सेल यहां तीन ही बना पाएंगे दो डायरेक्ट और सिर्फ एक ट्रांसवर्सल और बिल्कुल सर डायरेक्ट पर ट्रांसवर्सल कमेंट उसे पॉइंट से पास होगी जहां पर दोनों सर्कल्स एक दूसरे को एक्सटर्नल टच कर रहे होंगे क्या ये बात आप सभी को समझ आई थिंक इस बात से तो कोई आपत्ति नहीं है अब बात करते हैं अगर सर दो सर्कल्स एक दूसरे को externalli टच करें तो क्या हो रहा होगा मेरा कहना है की अगर ये इसका सेंटर है और इसका सेंटर है इन दोनों के अगर ये सेंटर्स हैं तो सर सीधी सीधी सी बात है ज्यादा सोचने का कहने की जरूरत है नहीं की अगर दोनों सर्कस ने जो एक दूसरे को टच किया लेट्स से ये सर्कल का सेंटर सी वैन और जो सर्कल का सेंटर सी तू और इन दोनों सर्कस की अपनी अपनी रेडियस अलर्ट से R1 और r2 तो एक्सटर्नल अगर ये दोनों एक दूसरे को टच करें तो क्या आपको ये सीधा सा ऑब्जर्वेशन नहीं दिख रहा है की दोनों सर्कस के सेंटर्स के बीच की डिस्टेंस उन दोनों सर्कल्स की रेडियस के सैम के इक्वल होती है मैं फिर से रिपीट करता हूं दोनों सर्कस के सेंटर्स के बीच के डिस्टेंस दोनों सर्कल्स के क्या भाई सेंटर्स के बीच की डिस्टेंस उन दोनों सर्कल्स की रेडियस के सैम के इक्वल होती है आई थिंक ये तो बिल्कुल सही बात है सर क्या इस बात को समझने में किसी भी स्टूडेंट को कोई परेशानी आई थिंक ये तो बड़ी सिंपल सी शॉर्टकट सी डायरेक्ट सी बात है जो आप सभी को समझ आई होंगी अब इन्हीं बातों को थोड़ा कंटिन्यू करते हैं थोड़ा आगे बढ़ते हैं और थोड़ा और अच्छे से समझते हैं और देखते हैं चीज कैसे निकल कर आपकी आएंगे है ना तो इस बारे में बात करते हैं स्टूडेंट्स कैसे ध्यान से देखना पहला फैक्ट हमारे पास दो सर्कल्स हैं कहां से दो सर्कल सर एक सर्कल S1 जो की कुछ ऐसा है और एक सर्कल आपका S2 जो की कुछ ऐसा है हम जानते हैं सर इस सर्कल के सेंटर के कोऑर्डिनेट्स क्या हो जाएंगे -जी वैन कमा - F1 और इस दूसरे वाले सर्कल के सेंटर के कोऑर्डिनेट्स क्या हो जाएंगे सर वो हो जाएंगे -जी2 कोई दिक्कत तो नहीं है दोनों की रेडियस भी निकल इस सर्कल की रेडियस क्या हो जाएगी वही बात जीवन स्क्वायर प्लस f1² - C1 इनसाइड एंड अंडर रूट और इस सर्कल की रेडियस ऑफ निकलना चाहोगे तो क्या लिखोगे g2² + f2² - C2 इनसाइड अंडर रूट सर ये तीन बातें या चार बातें क्यों कर रहे हो क्योंकि बड़ी बेसिक सी बात जो मैं आपको समझाना चाह रहा हूं वो ये किन दोनों के सेंटर्स के बीच की जो डिस्टेंस होगी वो क्या होगी भाई याद करो -1 - F1 से -2 - f2 की जो डिस्टेंस होती है सर वो होती है आपकी वही बात G2 - G1 का होल स्क्वायर प्लस ऑफ कोर्स क्या f2 - F1 का होल स्क्वायर प्लस f1² - सी और दूसरे की रेडियस क्या है सर वो है g2² मतलब -c1 है ना d2² + f2² - C2 आई थिंक ये कंडीशन जो की देखने में शायद बहुत लेंडी है बहुत अजीब सी लग सकती है पर रियलिटी में ये कैसे कम कर रही है आई होप आपको समझ ए रही है बड़ी सिंपल सी वैल्यूज होंगी और कहीं ना कहीं कोई ना कोई वेरिएबल या अननोन क्वांटिटी होगी जो आपको निकल दी होगी बस इस कंडीशन को अप्लाई करके तो पहले तो ये फंडा याद रहेगा क्या एफ तू सर्कल्स एफ तू सर्कल्स टच एच आदर externalli अगर वो एक दूसरे को externalli टच करें तो यू नो मसाला या मसरा कुछ ऐसा रहेगा आई होप ये बात आपको याद रहेगी अगर आप ये बात याद रखते हैं तो अब आते हैं आगे की बात पर suniyega ध्यान से बहुत ध्यान से सुनना ये कंक्लुजन तो हम दे ही चुके हैं है की दोनों सर्कल्स के सेंटर्स के बीच की डिस्टेंस उन दोनों की रेडियस के सबके इक्वल होगी और क्या कहना चाह रहे हो भाई उन दोनों के रेड आय कम के इक्वल हो गया सर ये बात भी हम कर चुके हैं दो तो डायरेक्ट कॉमिटेंट और एक आपकी transvar कमेंट्री कोई दिक्कत तो नहीं है अच्छा एक और बात वापस इस बार तो शायद मुझे प्रूफ करने की भी जरूरत नहीं है सर की आप खुद देख का रहे हो की ये जो आपकी ट्रांसवर्सल कमेंट टैसेंट है ये जिस भी पॉइंट पर इंटरसेक्ट कारी या जो उनका पॉइंट ऑफ कॉन्टैक्ट है जिस पॉइंट पर ये दोनों सर्कल्स एक्सटर्नल एक दूसरे को टच कर रहे हैं क्या उसे पॉइंट को निकलना बहुत डिफिकल्ट आस्क है नहीं क्योंकि इसके कोऑर्डिनेट्स बताएं और इन दोनों को यह डिवाइड कर रहा है बात खत्म और अगर यह पॉइंट इन दोनों को R1 के रेश्यो में डिवाइड कर रहा है तो क्या इसके निकलना बहुत डिफिकल्ट है आई थिंक मेरे ख्याल से तो नहीं है तो पी के अकॉर्डिंग इनके इस मुझसे किसी ने पूछे होते तो मैं क्या कहता है सर पीके जो कोऑर्डिनेट्स होते हैं मैं पीके cardinates क्या कहता गोद farmaiega C1 को हमने क्या माना है हमने माना है जीवन जीवन है ना जीवन कमा - F1 और C2 को आपने क्या माना है आपने माना है - G2 - f2 ये -1 क्लियर सी बात है सर बहुत दूर नहीं जाना है आपको सीधे-सीधे सोचना है वही बात r2 टाइम्स -1 तो हो जाएगा - r2 जीवन - R1 G2 / क्लीयरली दोनों के उसे रेश्यो का स्थान और इसी तरीके से क्या आप yordinate भी निकलेंगे जो की क्या हो जाएगा आपका कोर्स - r2 टाइम्स F1 - R1 टाइम्स f2 अपॉन में क्या R1 + r2 तो ये क्या होंगे भाई ये क्लीयरली algebratali बहुत अजीब लग रहा होगा बट दिख जाएगा क्लियर बहुत आसान सा दिख जाएगा तो इनके सेंटर्स को R1 इस तू आर तू के रेश्यो में आपका ये पी पॉइंट जो की ट्रांसफर कॉमन टेंसेंट है जो की टेक्निकल उसे पॉइंट पी से पास होगी जहां पर ये दोनों externalli एक दूसरे को टच कर रहे हैं तो वो एक्सटर्नल कहां टच कर रहे हैं वो पॉइंट निकलने के लिए मेरे पास ये सीधा सीधा तरीका है नोट डाउन करते रहिएगा स्टूडेंट्स ये बहुत कम की बहुत कृष्ण सी हम बातें करते चले जा रहे हैं जिन्हें आपको अच्छे से बनाना है ये मैंने थोड़ा शायद कर दिया बट आप थोड़े से प्रॉपर नोट्स banaaiyega आप इसे सिस्टमैटिक रखिएगा यहां लिखते जाएगा की पी क्या है भाई वो पॉइंट ऑफ कॉन्टैक्ट है ना ऐसे चीज सिस्टम बाद में रेफर करेंगे तो आपको पढ़ के समझ जाना चाहिए की आपने क्या लिखा था और वही रिवीजन के लिए आपका बेस्ट सोर्स होना चाहिए क्योंकि वहां पुरी चीज बहुत थेरली और कंप्रिहेंसिवली डिटेल मैनेजमेंट में आपने लिखी होंगी इन वीडियो को रेफर करके अब क्या अब अगर मैं कम की बात पर आओ तो कम की बात क्या निकल कर आती है सर सुनेगा ध्यान से हम देख चुके हैं सर आपकी दो डायरेक्ट कॉमन टेंट और इस तरीके से आपका जो पॉइंट दी है जो अभी हमने कहा पी आपके सिमन सी तू को इंटरनल हम रिलीज सॉरी दी तो वो पॉइंट है जो डायरेक्ट कमेंट इंजन को डिवाइड कर रहा है एक्सटर्नल और अगर मैं एक और पॉइंट मैन लूं कौन सा टी जो की आपका ये पॉइंट ऑफ कॉन्टैक्ट भी है उसको ऑफ कोर्स आपका इस तरीके से डिवाइड करता है अच्छा एक और बात मैं आपसे कहूं एक बहुत मजेदार सी बात आपसे कहूं suniyega ध्यान से बहुत ध्यान से सुनना है ये बहुत मजेदार सी बात है बहुत कम की बात है और बहुत कृष्ण बात है suniyega बस ये बहुत इंपॉर्टेंट बात है प्लीज इसे ध्यान से सुनना आपके जो सर्कल्स हैं ना स्टूडेंट्स जैसे मैन लो ये एक पहला सर्कल है और यहां पर एक दूसरा सर्कल है ना ये जो दो सर्कल्स हैं इन दोनों सर्कल्स ने आपस में एक दूसरे को जहां टच किया है उसे पॉइंट पर अगर मैंने टांगें ड्रॉ की थी मैन लो उसे पॉइंट पर अगर मैंने टांगें ड्रॉ की थी तो यह जो ट्रांसफर कॉमन टेंट है ना आई रिपीट माय स्टेटमेंट यह जो लाइन है जो की आप का रहे हो ट्रांसफर आप का रहे हो इस केस में इस केस में जब ये दोनों सर्कस क्या कर रहे हैं जब दोनों सर्कल्स एक दूसरे को एक्सटर्नल टच कर रहे हैं ना इस केस में जब दोनों सर्कल दूसरे को टच करें एक्स्ट्रा कॉमन टांगें ही नहीं है बल्कि ये आपके लिए एक रेडिकल एक्सेस की तरह भी कम करती है और सर रेडिकल एक्सेस क्या होता है ये आप हमें सिखा सकते हो क्या अब तो मैं बिल्कुल नहीं सिखाऊंगा याद ए रहा है क्या रेडिकल एक्सेस क्या होता है स्टूडेंट्स सर रेडिकल एक्सरसाइज यानी एक ऐसे पॉइंट का लुक जहां से ड्रा की गई इन दोनों सर्कल्स पर टांगें की लेंथ इक्वल होती है यानी की सर इस स्ट्रेट लाइन पर अब कोई भी पॉइंट चुन लीजिए आपकी मर्जी से कहीं पर भी चुन लीजिए उसे पॉइंट से उसे पॉइंट से जब आप टांगेंट्स ड्रॉ करेंगे उसे पॉइंट से जब आप टेंसेज ड्रा करेंगे तो उन टेंसेज की लेंथ इक्वल होगी वो 10g = लेंथ की होगी आई होप की बात तो हमने पढ़ी है तो ये जो आपका ट्रांसवर्सल कमेंट टैसेंट है ये ना सिर्फ ये कॉमन टांगें है डायरेक्ट कॉमन टांगेंट्स की बात नहीं कर रहा हूं मैं ये इसके साथ-साथ एक रेडिकल एक्सेस भी है और अगर मुझे ज्यादा मेहनत नहीं करनी है जीवन में ज्यादा परेशान नहीं होना है तो दो तरीके पहला तरीका अगर मुझसे इस कॉमन इंजन की कोई इक्वेशन पूछता तो मैं क्या कहूंगा सर मैं कहूंगा की C1 और C2 जो इसके सेंटर्स के cardinate हैं उनको ज्वाइन करने वाली जो स्ट्रेट लाइन है उसे पर ही परपेंडिकुलर होगी क्योंकि टैसेंट है ना तो इनकी जो रेडियस है इनकी जो रेडियस उसे पर ये परपेंडिकुलर होगी तो ऑफ कोर्स यह जो लाइन है उसे पर भी परपेंडिकुलर होगी तो इसकी स्लोप निकल लूंगा और फिर इसकी स्लोप तो परपेंडिकुलर स्लोप हमें निकलती हैं आता ही है m1 M2 -1 होता है दूसरी बात आई थिंक C1 और C2 को R1 और r2 के रेश्यो में डिवाइड करने वाला पॉइंट होगा तो पॉइंट से पास होने वाली स्ट्रेट लाइन लिखूंगा पर एक बेवकूफ शख्स ये करेगा इतना करने की जरूरत ही नहीं है साहब आपको ये पॉइंट ऑफ कॉन्टैक्ट निकलना उसके बाद इस साइन की स्लोप निकलना है इतनी जीवन में परेशानियों लेने की जरूरत ही नहीं है क्योंकि सर आप क्यों नहीं देख का रहे हो की रेडिकल एक्सेस की इक्वेशन को निकलना हमने सिखा है सिखा है की नहीं भाई सर अगर यह सर्कल है S1 यह सर्कल है S2 तो रेडिकल एक्सिस की इक्वेशन क्या होती है जल्दी से बताओ भाई वह हो जाती है s1-s2 = 0 और यह क्वेश्चन खत्म हो जाएंगे इतना सा करते हुए कुछ सोचना नहीं है ना पॉइंट ऑफ कॉन्टैक्ट निकलना है ना इनकी स्लोप निकालनी है फिर उसका नेगेटिव रासी पालन इसकी स्लोप निकल लिए फिर लाइन की इक्वेशन लिखनी है नहीं साहब इतनी मेहनत नहीं करनी है जीवन में इसलिए तो हमने चीज पड़ी हैं इसलिए तो हमने अपना वक्त इन्वेस्ट किया है इन अलग-अलग चीजों को सीखने पर ताकि जब मजेदार सी चीज करनी हो तो हम इतनी आसानी से उनको वर्कआउट कर पाए सिर्फ इतना सा पता चल जाने से की ये एक रेडिकल एक्सेस की तरह बिहेव कर रही है ट्रांसफर सेल कमेंट्री कई सारी प्रॉब्लम्स रिजॉल्व हो जाएंगे पर ये सारे हालत आप कहां अप्लाई करेंगे ये सारे कंसर्न्स आपके तब होंगे जब आपके सर्कल्स क्या कर रहे हो जब आपके सर्कस टच कर रहे हो एक्सटर्नल आई थिंक चीजें आसान है चीज मुश्किल नहीं है चीज डिफिकल्ट नहीं है आई होप ये बात आपको अच्छे से समझ आई यहां तक कोई परेशानी तो नहीं है स्टूडेंट्स आई थिंक अगर यहां से चीज हम आगे बढ़ाएं तो आगे बढ़ते हैं बड़ी सिंपल सी बेसिक सी बात है भाई हम ये बात तो देख ही चुके हैं सर की जो टी होगा पॉइंट यानी जो पॉइंट ऑफ कॉन्टैक्ट होगा वो C1 C2 को इंटरनल R1 इस तू आर तू के रेश्यो में डिवाइड करेंगे की हम पढ़ते चले आए हैं सर वो अगर टच ना भी कर रहे हो तो यही होता है टच ना भी कर रहे हो तो अगर ये आपका एक एक्सटर्नल लाइन है तो उनके सेंटर्स को डिवाइड करने वाली लाइन ये टी जो है वो R1 इस तू आरटीओ में डिवाइड करती है तो हम पढ़ते चले आए हैं ये तो engineeral वैसे भी ट्रू होता है ये तो हम देखते हुए आए हैं बोलो भाई आए हैं की नहीं चलो सर आगे बढ़े इसके बाद क्या करना चाह रहे हो लेंथ ऑफ डायरेक्ट कॉमन टांगें इस गिवन बाय दिस फॉर्मूला तो ये फॉर्मूला कैसे बना ये फॉर्मूला कैसे बना सर मुझे फॉर्मूले रेट नहीं आता ऐसे कितने फॉर्मूले पढ़ूंगा मैं कौन कौन से फॉर्मूले के पढ़ते रहूंगा मतलब क्या क्या रेट में हो गया सारा तो मैं नहीं का रहा हूं आप रखिए आप समझ लीजिए क्या आपको लास्ट लेक्चर में याद है हमने डायरेक्ट कमेंट्स जिनकी लेंथ निकलने का एक एप्रोच सिखा था इस केस में वह एप्रोच क्या हो जाएगी ज़रा बात करते हैं देखो यह आपका एक सर्कल रहा होगा और ऑफकोर्स दूसरा भी आपके पास एक सर्कल रहा होगा और यह दोनों सर्कल्स में एक दूसरे को क्या किया सर इन दोनों सर्कल्स में एक दूसरे को टच किया अब जब इन दोनों सर्कस ने एक दूसरे को टच किया तो हम बात कर रहे हैं डायरेक्ट कमेंट इंजन की जो की कुछ ऐसी जा रही होगी सर बिल्कुल सर ऐसी जा रही होगी इसको थोड़ा सा आप ठीक कर लो आई थिंक चीज है कुछ ऐसी मुझे दिखेंगे थोड़ा सा इसको और नीचे कर लेते हैं है ना तो ये आपकी डायरेक्ट कमेंट टैसेंट है इसकी लेंथ से आपका क्या तात्पर्य है आप एक बात बताओ स्टूडेंट्स आप बस एक बात बताओ मैन लो इसका ये सेंटर इन दोनों के अपने-अपने सेंटर हैं इन दोनों के सेंटर्स को मैंने आपस में मिलाया मैंने क्या किया दोनों के सेंटर्स को आपस में ले हैं और क्या किया हमने इसके सेंटर से इस पर एक परपेंडिकुलर ड्रॉप किया और से इसके सेंटर से इस पर एक परपेंडिकुलर ड्रॉ किया की आपको नजर ए रहा है आई थिंक सर अगर मैं यहां से एक लाइन ऐसी ड्रॉ करूं तो सर ये तो कुछ मेरा पड़ा हुआ था मुझे दिख रहा है ये तो कुछ मैंने पढ़ा है या सिखा है सर ये बात तो मुझे कुछ ना कुछ कहीं सिखाई गई मुझे सर आती है मुझे पता है मुझे कैसे पता है सर suniyega ध्यान से एक क्वेश्चन ऐसे आप करोगे आप प्लीज इस बात का ये आपका सर्कल था फर्स्ट ये आपका सेकंड सर्कल था है ना इसके इस पॉइंट को लेट से मैं का लेता हूं ए इसे मैं का लेता हूं बी और समय लेता हूं सी ऑफ फाइनली निकलने में इंटरेस्टेड हो बी सी आप फाइनली निकलने में ये जो भी लिखा है मैं इसमें नहीं जा रहा हूं ना फाइनली निकलने में शैडो भी इसी ई ही तो आपकी डायरेक्ट कमेंट इंजन की लेंथ है और ऐसी की ऐसी बनेगी वो भी वही लेंथ कैरी करेगी ठीक है बिल्कुल सिमिट्रिकल होगी तो ई की लेंथ को लेकर आपका क्या ओपिनियन होगा सर ई की लेंथ को मेरा ओपिनियन बड़ा सिंपल सा है साहब कैसा आप एक बात बताओ भाई आप एक बात का जवाब दे दो मुझे क्या ये 90 डिग्री है अरे है की नहीं सर ये 90 डिग्री है तो एक राइट एंगल ट्रायंगल हो गया जो की वर्टेक्स सी पर राइट एंगल है तो अगर मुझे ई चाहिए तो फिर वही बात हाइपोटेन्यूज अब के स्क्वायर ने हाइपोटेन्यूज के स्क्वायर में से किसका एक के स्क्वायर का आप ध्यान से देखो ना अब क्या है सर ये जो ए बी सी तू C1 है आपको नहीं दिख रहा ये एक पैरेललोग्राम है आपको नहीं दिख रहा क्या है पैरेललोग्राम में और अगर ये पैरेललोग्राम है तो क्लीयरली सर हम जानते हैं अगर आपसे कोई अब पूछ लेता तो आप उसे का दोगे C1 C2 और C1 C2 क्या है सर C1 C2 इस नथिंग बट उनकी दोनों की रेडियस का सैम उन दोनों की रेडियस का सैम तो अगर C1 C2 मुझसे पूछा जा रहा है तो मैं C1 C2 यानी क्या अब क्या हो जाएगा सर दो-दो की रेडियस का सैम तो मैं लिखूंगा R1 + r2 का स्क्वायर किसी भी स्टूडेंट को कोई तकलीफ तो नहीं है यहां तक अच्छा सर इसी के साथ-साथ अगर मैं आपसे पूछूं एक क्या है तो एक क्या है जनाब सर एक ये रहा है की नहीं ऐसी क्या होगा यह माइंस दिस यह हो जाएगा आप समझ रहे हो ना मैं अगर इसमें से इसको सब्सट्रैक्ट कर डन तो मुझे ये मिल जाएगा बोलो भाई कोई तकलीफ तो नहीं इस बात में तो जब एक जानना हुआ तो सर वो होगा R1 - r2 क्योंकि पैरेललोग्राम है ना तो ये जो ए सी वैन है वही ई वैन है तो R1 - r2 तो कितना हो जाएगा सर ये हो जाएगा R1 - r2 का होल स्क्वायर और इस पे क्या लगाना बाकी है अंडर रूट ये आप क्या निकल रहे हो डायरेक्ट कॉमन टांगें के लिए मैं का रहा हूं इस तरीके से सोचो आप आपको क्यों ज्यादा चीज करनी है और अगर यहां से मैं फॉर्मूला बनाने की तरफ आगे बढूं तो देखना बहुत ध्यान से सुनना r1² + r2² + 2r1r2 फोर आर वैन आर तू और फोर आर वैन आर तू को लेके आप क्या कहोगे भाई 4 R1 r2 का जब अंडर रूट में लूंगा तो फोर बाहर ए जाएगा और अंडर रूट में क्या ए जाएगा दोनों की रेडियस का प्रोडक्ट अंडर रूट में क्या ए जाएगा अभी दोनों की रेडियस का प्रोडक्ट और इस तरीके से आप ये निकलोगे जो की वो वहां भी यही कनक्लूड करना चाह रहा है तो अगर कोई सर्कल्स कनेक्ट करें टच करें externalli अगर दो सर्कल्स एक दूसरे को externalli टच करें तो मैं कहूंगा उनकी जो डायरेक्ट कॉमन टांगें होती है उनकी जो डायरेक्ट कमेंट होती है उसकी लेंथ इस नथिंग बट दोनों की रेडियस के प्रोडक्ट का अंडर रूट का डबल इंजन की लेंथ होती है इसके लिए नहीं निकलोगे क्या अब इसकी लेंथ कैसे निकलोगे आप मुझे बताओ तो इंफिनिटी जा रही है ना उसको कोई रोक नहीं रहा है कहां रुक रही है बस यही तो टच हो रही है कोई दो पॉइंट होते तो मैं समझता तो यहां पर तो इसकी लेंथ निकलना मुश्किल है तो बिल्कुल इसकी लेंथ नहीं निकलेगी बस आप क्या निकलेंगे डायरेक्ट कॉमन टेंट की लेंथ जब कुछ ऐसा सिनेरियो बने इस फॉर्मूले से क्या इस बात से किसी भी स्टूडेंट को कोई परेशानी क्या इस बात से किसी भी स्टूडेंट को कोई तकलीफ और अगर आपने अभी तक पढ़ की हर बात को बहुत डिटेल में अच्छे से क्लियर से मिनर में समझो तो क्या हम ये क्वेश्चन ट्राई करें देखिएगा भाई क्या लिखा हुआ है वो लिख रहा है शो डेट डी सर्कल्स स्क्वायर प्लस ए स्क्वायर प्लस 14 प्लस 22 0 टच एच आदर कौन सी बड़ी बात है सर यह तो हम कर सकते हैं क्या यह बात पर कोई परेशानी नहीं होनी चाहिए फिर क्या पूछ रहे हो मुझे आता है जहां तक मुझे याद है एंड डी इक्वेशन ऑफ डी कॉमन टांगें आते डी पॉइंट ऑफ कॉन्टैक्ट सर तीनों बातें मैं निकल सकता हूं ये तीनों बातें वही बातें हैं जिनका अभी तक आपको चीज पढ़ गई है सो टेक्निकल क्वेश्चन आपको जो सब अभी तक पढ़ाया गया है उसका रिवीजन है और मेरा मानना है आप यह क्वेश्चन बहुत आसानी से कर सकते हो सर पहले तो समझते हैं यह जो सर्कल है इसका सेंटर क्या होगा सर इसका सेंटर होगा 5 - 2 होगा की नहीं सर इस सर्कल का सेंटर क्या होगा सर्कल का सेंटर होगा माइंस सेवन प्लस 3 तो इसका सेंटर होगा -7 कमा प्लस 3 किसी भी स्टूडेंट को इस बात से कोई तकलीफ तो यह हुआ आपका कौन सा ये आपके पहले सर्कल का सेंटर हुआ जिसे मैं C1 का लेता हूं यह आपके दूसरे सर्कल का सेंटर हुआ जिसमें C2 का लेता हूं अब एक बात का जवाब दीजिए भाई क्या आप पहले सर्कल की रेडियस निकल सकते हो जिससे मैं का लेता हूं आर1 तो बिल्कुल निकल सकते हैं 25 + 429 - 20 यानी 29 + 29 यानी 49 और 49 का अंडर रूट 7 कोई तकलीफ भाई याद है ना तो फिर क्या सर आर तू निकल लेते हैं 49 प्लस 949 कितना होता है सर क्या कर रहे हो 58 - 22 8 मैसेज जब आपने 22 सब्सट्रैक्ट किया तो आपको दिख रहा है क्या दिख रहा है क्या 58 में से 22 तो 8 में से तू गए तो 6 5 में से तू गया है तो थ्री यानी 36 और 36 का अंडर रूट कितना सर 36 का अंडर रूट सिक्स आर वैन और r2 R1 और r2 कितने ए रहे हैं 7 और 6 अब अगर दो सर्कस externalli टच करें तो मैंने तो जीवन में बस इतना सा सिखा है सर की आपके दोनों सर्कल्स की रेडियस का सैम उनके बीच के सेंटर्स के डिस्टेंस के इक्वल होना चाहिए सेंटर्स के डिस्टेंस के इक्वल होना चाहिए हान या ना तो निकलती हैं तो मुझे कितना लाना है इसे 13 13 किसे लाना है 13 लाना है इनके सेंटर्स के बीच के डिस्टेंस 5 - 7 और 5 12 का स्क्वायर कितना वैन 44 किसी भी स्टूडेंट को कोई तकल्लुफ है तो बताएं जरा और क्या जनाब और अगर आप आगे बड़े -2 का स्क्वायर 25 और क्या आपको दिख रहा है सर क्यों ज्यादा लिखने का आपको शौक ए रहा है यह आता है 169 और उसका अंडर रूट हो रहा है 13 तो मैं कुछ आगे लिखूं या समझाऊं आप समझ गए तो सर एक बात तो तय हो गई की ये दोनों सरफेस एक दूसरे को externalli टच करते हैं अब क्या करना है सर अगर यह सर्कल्स एक दूसरे को एक्सटर्नल टच करें या ना करें जो भी बात रही होगी वह जो ट्रांसवर्सल tenjent होगी जो ट्रांसफर कॉमन टांगें होगी वो एक ही बनेगी अगर ये एक दूसरे को टच कर रहे हैं externalli और वह टेंट इनके पॉइंट ऑफ कॉन्टैक्ट से पास हो रही होगी और वो टैसेंट इनके सेंटर को ज्वाइन करने वाली लाइन को R1 और r2 के रेश्यो में डिवाइड कर रही होगी मतलब इन सर्कल्स का जो पॉइंट ऑफ कॉन्टैक्ट होगा वो इनके सेंटर्स को ज्वाइन करने वाली लाइन का R1 रेश्यो में डिवाइड करने वाला पॉइंट ऑफ कॉन्टैक्ट ठीक होगा मेरी बातें डाइजेस्ट हो रही है या नहीं भाई तो वह निकल लेते हैं आई थिंक वो भी आप निकल सकते हो तो जरा ये पार्ट आप मुझे सॉल्व करके बता दो और मुझे आगे का आंसर बता दो कैसे कर भाई इसके आगे का आंसर निकाला या नहीं देखो सर बढ़िया आसान सी बात है जब कोई पूछ रहा है फाइंड डी आर्डिनेंस ऑफ डी पॉइंट ऑफ कॉन्टैक्ट ही आपका पॉइंट होगा जो कॉन्टैक्ट का पॉइंट होगा दोनों सर्कल्स के एक्सटर्नल जब टच होने पे वो पॉइंट बनेगा तो सर निकल सकते हैं निकल लो भाई कैसे सुनेगा कहां से 5 6 प्लस 6 जो की कितना हो जाएगा - 2 6 कितना -12 1 और फिर वही बात 7 प्लस 6 कितना 13 यह हो जाएगा आपका पॉइंट ऑफ कॉन्टैक्ट यह होगा आपका क्या पॉइंट ऑफ कॉन्टैक्ट जैसे जनरली हम टी से दिनो कर देते हैं अगर सिंपलीफाई करना चाहो तो क्या थोड़ा सिंपलीफाई करके लिख सकता हूं सर कोशिश करते हैं नाइन तो ये हो जाएगा -9/13 तो कितना ए जाएगा सर ये ए जाएगा -9/13 एंड दिस इसे गोइंग तू बी डी पॉइंट ऑफ कॉन्टैक्ट अब क्या अब मेरा बस आपसे ये कहना है अगर वह ट्रांसफर कॉमन टेन जिनकी इक्वेशन निकालनी है तो क्या हम जानते हैं रेडिकल एक्सेस के बारे में कुछ कुछ तो अगर मैं आपको रेडिकल एक्सेस की हिंट दे चुका हूं तो क्या बस इक्वेशन को देख कर इन दोनों इक्वेशंस को देख कर ही आप मुझे आंसर नहीं बता का रहे हो क्या सर आई थिंक चीज तो बड़ी आसानी है आप बस चीजें सोच नहीं का रहे हो आप बस चीजें देख नहीं का रहे हो बस एक चीज मैंने गलत लिख दी है 21 - 12 अब रिलीज सॉरी फॉर दिस तो यहां पर हम क्या लिखेंगे प्लस नाइन बाय 30 है ना अब आते हैं कम की बात पर अब अगर आप पूछना चाह रहे हो सर क्या आप पूछना चाह रहे हो कॉमन टेंसेंट तो कॉमन टांगें क्या होती है सर हम बड़ी बेसिक सी बात पढ़ चुके हैं की कॉमन टांगें नासिर कॉमन टांगें होती है बल्कि वो रेडिकल एक्सिस भी होती है याद आया क्या और अगर वो रेडिकल एक्सेस होती है तो सर देखो x² तो गए तो बचा क्या -10x - 14x यानी - 24x ये कितना हो जाएगा सर यहां से मिलेगा आपको पहले तो माइंस 24x और क्या सुनेगा 4y - - 6y ये माइंस 6y है माइंड सो 4 - 6 यानी 4 + 6 तो एक्स कितना है 10थ दिस इस गोइंग तू बी प्लस 10y और आगे सुनते हैं स्टूडेंट्स कम की बात -20 - 22 तो ये हो जाएगा कितना -42 सो अगली टर्म जो मिलेगी वो क्या होगी सर वो हो जाएगी - 42 = 0 क्या एक छोटा सा कम आपके दिमाग में ए रहा है की सर हाफ कर दो ना तो 1/2 किलो कितना ये हो जाएगा मतलब नेगेटिव हाफ ले रहा हूं है ना तो ये हो जाएगा कितना 12x ये हो जाएगा -5y और ये कितना हो जाएगा प्लस प्लस कितना 21 और सर ये जो इक्वेशन निकल कर आई है ये जो इक्वेशन बन कर आई है दिस इस डेट कॉमन टेंस इट वो जो दोनों के टच होने पर वो ट्रांसफर सेल कमेंट्री दोनों के पॉइंट ऑफ कॉन्टैक्ट से पास हो रही है ये वो कॉमन टेंशन है तीनों बातें क्लियर है तीनों बातें क्लियर मतलब मैं आपको चीज visilise करवाना चाहूं तो मैं आपको ये समझाना चाह रहा हूं रियली सॉरी फॉर दिस अगर मैं आपको चीज visilise करवाना चाहूं तो आपके दो सर्कल्स द बेसिकली इन दोनों सर्कल्स ने एक दूसरे को टच किया हमने प्रूफ किया कैसे इनके सेंटर्स के बीच के डिस्टेंस को उनके रेडियस के इक्वल रख दिया अब आपसे दो बातें और पूछी जा रही थी की ये कहां पर टच कर रही हैं तो वो पॉइंट हमने निकल लिया कौन सा भाई ये और पूछा जा रहा था की जब वो कॉमन टांगें है उसकी इक्वेशन क्या होगी तो वो भी हम ने निकल बहुत डिफिकल्ट बहुत टू के क्वेश्चन तो नहीं था आई होप ये बात याद रखिए बस एक आइडिया देना चाह रहा हूं एक और चीज जो आप याद रखिएगा प्लीज हमेशा याद रखना स्टूडेंट्स की आपके जो सेंटर्स को कनेक्ट करने वाली जो लाइन है ना C1 C2 को उसे पर परपेंडिकुलर होती है आपकी ट्रांसलेशन अभी फिलहाल इस सिनेरियो में का लो या रेडिकल एक्सिस तू बी मोर प्रेसिडेस हैं वुड डेफिनेटली लाइक तू तेल यू विच हैव ऑलरेडी मेंशंड एलियर अन लॉट में टाइम्स की सर्कल्स के सेंटर्स को कनेक्ट करने वाली लाइंस पर परपेंडिकुलर होता है रेडिकल एक्सरसाइज कहना क्या चाह रहे हो सर मैं आपसे यह कहना चाह रहा हूं स्टूडेंट्स की अगर मुझे C1 और C2 के कोऑर्डिनेट्स बताते जो की पता द मुझे है ना तो यहां से C1 और C2 को ज्वाइन करने वाली लाइन के स्लोप ए जाती और अभी जो रेडिकल एक्सेस है वही आपकी कॉमन रिसोर्सेस टेंसेंट है क्योंकि परपेंडिकुलर है तो C1 C2 की स्लोप ए जाती है तो इसकी स्लोप भी ए जाती है क्योंकि m1 M2 - 1 होता है अगर दो लाइंस परपेंडिकुलर हो तो और यह ट्रांसफर जो आपकी टांगें है ये एक खास पॉइंट से पास होती है जो की होता है पॉइंट ऑफ कॉन्टैक्ट जोगी में इस तरीके से निकल चुका हूं तो उसे तरीके से भी आप इसकी इक्वेशन निकल सकते द याद रखिएगा ऐसा क्यों बता रहे हो सर क्योंकि हो सकता है कभी शायद ये कम ना करें कम तो ये करेगा पर शायद इससे बात ना बन पाए तो मैं फिर ये आइडिया भी उसे करूंगा बातें समझ ए रही है क्या बातें डाइजेस्ट हो रही हैं क्या अगर बातें डाइजेस्ट हो रही है तो एक और क्वेश्चन कर लेते हैं इस क्वेश्चन को अगर हम ध्यान से देखें तो आप अगर चाहे तो मेरी दरख्वास्त तो यही है की आप इसे कर लीजिए ट्रीट आज अन होमवर्क क्वेश्चन हेस एन असाइनमेंट क्वेश्चन एक बार जरा ट्राई जरूर करेगा बट एनीव्हेयर लाइक तू गिव डी हिट की इस क्वेश्चन को सोचना कैसे तो बात करते हैं इस बारे में भाई वो का रहा है दो सर्कल्स हैं है ना इफ तू सर्कल्स ax² + y² = + c² = 2 और x² + y² + c² - 2b वे इसे इक्वल तू जीरो एग्जाम में क्या आएगा ऐसा नहीं लिखेगा दें प्रूफ डेट वो पूछेगा इफ दे टच इ आदर externalli दें और क्वेश्चन मार्क लगा देगा फिर ऑप्शन में ऐसी अलग-अलग बातें लिखी होंगी तो आपको पता करना होगा की आंसर क्या होगा तो एग्जाम में इस तरीके से आपको चीज बनाकर पूछ ली जाएंगी तो अब करना क्या है इस क्वेश्चन में वो समझते हैं इस क्वेश्चन को लेकर मेरा आइडिया तो बस इतना सा है सर देखो क्या आप मुझे सर्कल की इक्वेशन में सेंटर के अकॉर्डिंग बता सकते हो सर देखो 2ax को इधर ले तो -2ax है तो एक्स का कॉएफिशिएंट क्या है -2a उसका नेगेटिव हाफ क्या होगा ए और ए का तो कोई कॉएफिशिएंट है ही नहीं तो इसका सेंटर क्या हो जाएगा सर वो हो जाएगा ए कमा जीरो इसका सेंटर क्या हो जाएगा सर इसका सेंटर हो जाएगा देखो एक्स वाली टर्म नहीं है तो जीरो कमा भी ये क्या हो गया इसका सेंटर कोई तकलीफ अच्छा सर इन दोनों की रेडियस अगर मैं निकलूं आपसे तो रेडियस क्या आएगी देखो भाई ध्यान से a² + 0² - c² तो इसकी रेडियस क्या है की सर इसकी जो रेडियस आएगी वो होगी a² - c² कोई तकलीफ तो नहीं है भाई obbviesli इनसाइड एंड अंडर रूट वैन रूट नहीं होगा कोई तकलीफ तो नहीं सिमिलरली इस सर्कल की सर्कल की रेडियस होगी क्या जीरो स्क्वायर प्लस बी स्क्वायर माइंस बी स्क्वायर माइंस सी स्क्वायर करना क्या चाह रहे हो सर घुमा फिर के बात इतनी बोल रहे हो कम क्या करना है मेरा कहना है कम खत्म हो गया जनाब वो लोग कैसे बोला तो था उसने दे टच एच आदर टच एच आदर एक्सटर्नल और क्या इसको लेकर मुझे कोई कंडीशन पता है वो लोग दोनों सर्कल्स अगर एक दूसरे को टच करें externalli तो दोनों के सेंटर्स के बीच की डिस्टेंस उन दोनों की रेडियस के सब के इक्वल होती है रेड आय के इक्वल होती है बिल्कुल होती है सर तो क्या तो देखो भाई दोनों के सेंटर्स के बीच की डिफरेंस ए - 0 a² पर क्या लगाया अंडर रूट ये दोनों के सेंटर्स के बीच के डिस्टेंस अंडर रूट ओवर क्या बी स्क्वायर - c² के सैम के इक्वल होगी होगी या नहीं जवाब दीजिए बिल्कुल होगी सर अब क्या अब कम की बात सुना अब अगर मैं सीधे पॉइंट पर आऊं सर यहां तक पहुंचना है ना तो एक कम करते हैं इसका स्क्वायर करते हैं जैसे इसका स्क्वायर करते हैं इसका स्क्वायर करते हैं क्या हो जाएगा सर इसका स्क्वायर करते ही लेफ्ट हैंड साइड पर बचेगा a² + b² है ना राइट हैंड साइड पर अगर मैं इस पूरे का स्क्वायर करता हूं तो इसका स्क्वायर प्लस इसका स्क्वायर तो वो क्या बचेगा सर वो बचेगा a² - c² + b² - c² + 2 टाइम्स है ना प्लस तू टाइम्स अंडर रूट ओवर अंडर रूट ओवर व्हाट ये है ना a² - c² और इसी में ये मल्टीप्लाई हो जाएगा क्या b² - c² तो ये आपको एक एक्सप्रेशन मिलेगा कुछ सिंपलीफाई कैंसिल होता हुआ दिख रहा है क्या सर कुछ तो दिख रहा है देखो ध्यान से यह a² से a² b² से b² गया यह बचेगा -2c स्क्वायर स्क्वायर है जाएगा तो बिल्कुल सर पहले तो ये तू हटा देते हैं तो जो c² है आपका वो इसके इक्वल होगा इसके स्किपर मैं इसका भी स्क्वायर कर दूंगा तो इसका स्क्वायर कर दूंगा तो c² का स्क्वायर क्या हो जाएगा सर c² का स्क्वायर हो जाएगा सी ^ 4 और जब इसका स्क्वायर करेंगे तो √ कैंसिल हो जाएगा तो बचेगा क्या उसे ध्यान से देखना जो बचेगा वो होगा a² - c² टाइम्स क्या बी स्क्वायर कोई तकलीफ कोई परेशानी सर इसे थोड़ा सा एमप्लीफायर करते हैं बिल्कुल करिए जब इसे हमने सिंपलीफाई करना चाहा तो क्या मिलेगा बहुत ध्यान से देखना ये पार्ट थोड़ा सोशल है c²b² तो सी तू डी पावर फोर से कैंसिल अब जो बचेगा वो क्या है सोना a² और b² तो पहले तो क्या दिख रहा है सर पहले तो दिख रहा है आपको एक जो टर्म वो है a² b² कोई दिक्कत कोई परेशानी इसी तरीके से सर आप ध्यान से देखो आपको एक और टर्म मिलेगी क्या मिलेगी ध्यान से देखा माइंस सी स्क्वायर बी स्क्वायर तो एक टर्म क्या मिलेगी सर वो मिलेगी माइंस बी स्क्वायर सी स्क्वायर और क्या मिलेगा सर ध्यान से देखो और टर्म जो मिलेगी वो मिलेगी a² - c² - a² c² तो ये क्या हो गया - a² c² और ऑफ कोर्स इस इक्वल तू जीरो इस इक्वल तू जीरो क्यों सर क्योंकि जो बचेगा सी तू डी पावर फोर यहां वो इससे कैंसिल हो जाएगा अब मुझे कुछ ऐसा पाना है कुछ ऐसे तक पहुंचना है कोशिश करते हैं भाई एक कम करिए क्या ये जो एक्सप्रेशन इसको उठा के उधर रख दे बिल्कुल रख दो मैंने क्या किया इस एक्सप्रेशन को उठाकर उधर रख दिया या फिर आई की बात ऐसा किया है ना तो जैसे ही ऐसा किया तो यहां पर क्या जाएगा ये इक्वल साइन कोई दिक्कत तो नहीं कोई दिक्कत तो नहीं अब सर मैं एक छोटा सा कम ये करना चाह रहा हूं की मैं तीनों तरफ a² b² से डिवाइड करता हूं मैं तीनों को ए स्क्वायर बी स्क्वायर से डिवाइड करता हूं तो ये लिखा है a² b² अपॉन में क्या आता है a² b² c² यहां क्या लिखा है b² c² तो यहां पर भी क्या ए जाता है a² b² और यहां पर भी ये लिखा हुआ है क्या a2c² और इसके अपॉन में भी क्या ले आता हूं a² b² c² तो दोनों साइड एलएस और रहा को ए स्क्वायर बी स्क्वायर सी स्क्वायर से डिवाइड किया तो ये यहां कैंसिल यहां कैंसिल और यह कैंसिल तो कनक्लूड मैं कर का रहा हूं की सर वैन अपॉन सी स्क्वायर जो यहां बच रहा है वो यहां से 1 / a² + यहां से 1 / b² के इक्वल होगा एंड दिस वुड बी डी फाइनल करेक्ट ऑथेंटिक आंसर क्या इसमें कहीं भी कोई डाउट आई थिंक सर एक जेनुइन से आसान सा क्वेश्चन था यहां पे आपको सर्कल के बेसिक्स से डील करते हुए इस कॉन्सेप्ट पर बात करनी थी इस इस कॉन्सेप्ट पर बात करनी थी है ना इस कॉन्सेप्ट पर बात करनी थी आसान सा क्वेश्चन था स्टूडेंट दो सर्कल्स अगर मेरे पास हैं यह एक सर्कल है मैन लो और एक और सर्कल लेट्स से इस देयर एक और सर्कल से इस देर और उसे सर्कल ने क्या किया उसे सर्कल ने सर इस सर्कल को इंटरनली टच किया था सर्कल ने इसको यहां कहीं इंटरनली टच किया अच्छी बात है सर हो क्या रहा है यहां पर हो ही रहा है की ये आपका बिगड़ सर्कल लेट्स कॉल इट एस वैन ये आपका इंटरनल सर्कल लेट्स कॉल इट एस तू मुझे कुछ बहुत कृष्ण सी बातें नज़र ए रही हैं सर क्या अगर मैं इस सर्कल के सेंटर की बात करूं लेट्स कॉल इट यहां पर कहीं होगा और इसे हम मैन लेते हैं क्या C1 इस सर्कल के सेंटर की बात करू तो लिसन यहां कहीं होगा सम में मैन लेता हूं सी तू एक बड़ी जरूरी सी बात जो आप सब नोटिस कर रहे होंगे की सर अगर इन C1 और C2 से मैं यहां पर कोई एक लाइन देखूं जो की मैं इनको पॉइंट ऑफ कॉन्टैक्ट तक ड्रॉ करूं तो मुझे दोनों सर्कल की रेडियस मिलती हैं दोनों सर्कल की रेडियस मिलने का मतलब है ऑफ कोर्स बड़े सर्कल के रेडियस जो की क्या होगी R1 और इस छोटे वाले सर्कल की रेडियस जो की होगी r2 आई होप आप ये फर्क समझ का रहे हैं मेरा आपसे यह कहना है की अगर इन दोनों की रेडियस को अगर स्तानजिन से मिलाया जाए जो की इनकी कॉमन टांगें है जो की इनकी क्या है कॉमन टांगें तो क्लीयरली वो क्या होगी परपेंडिकुलर एक जरूरी सा ऑब्जर्वेशन जो आप सबको देखना चाहिए की शायद मैं यहां पर डायरेक्ट कॉमन टैसेंट ड्रा नहीं कर का रहा और शायद से अच्छा शब्द होगा सर ये भी आप डायरेक्ट कमेंट हैं जिनको ड्रॉ नहीं कर का रहे हो क्योंकि दोनों के लिए कोई डायरेक्ट कमेंट टैसेंट है ही नहीं मतलब एक ही कॉमन ट्रांजैक्शन डायरेक्ट transffersel जो का रहे हैं का लीजिए जहां पर ये दोनों एक दूसरे को इंटरनल टच कर रहे हैं बस वहीं से एक कॉमन टांगें निकले क्या ये बात आपको नजर आई कोई परेशानी है डाउट तो नहीं है तो आज हम बात करेंगे की अगर सर दो सर्कल से एक दूसरे को इंटरनल टच करें तो क्या-क्या बातें होती हैं क्या-क्या चीज सोच ही जाते हैं और उनसे टांगें से कैसे रिलेशन होते हैं और उन पर कैसे चीज सोचनी होती है टेंशन की जो भी सब चीज सोचनी है आज उसे बारे में डिटेल डिस्कशन करेंगे अच्छा एक बात क्या आपको नजर आई क्या आपने ऑब्जर्व की सर अगर दो सर्कस ने एक दूसरे को इंटरनल टच किया इंटरनल तो एक बार मजेदार था ऑब्जर्वेशन आपको देखना चाहिए जैसे मैं हल्का सा फिर से आपके अलग तरीके से दिखाने की कोशिश करता हूं जैसे मैन लो ये आपका एक सर्कल है ना और उसी तरीके से आपका एक और सर्कल और इन दोनों ने एक दूसरे को टच किया इंटरनल इस तरीके से अब ऑफ कोर्स आपके पास ये जो बड़ा सर्कल है इसका ये सेंटर है यहां पर और ये जो छोटा सर्कल है इसका सेंटर है बहुत कम की बात है स्टूडेंट्स suniyega एक बात बताओ अगर मैं आपसे पूछूं प्लीज उसे बात को गौर से सुनार बताना की इस वाले सर्कल की रेडियस कैसे इस वाले सर्कल की रेडियस है ये जैसे मैं का लेता हूं r2 कोई दिक्कत तो नहीं है अच्छा सर बड़े वाले सर्कल की रेडियस के सर बड़े वाले सर्कल की रेडियस की पुरी है जिसे आप का सकते हो की R1 इसका सेंटर था जिससे C1 और उसका सेंटर था C2 आप खुद जरा सोच के बताओ इस बात को प्लीज थोड़ा ध्यान से देखो सोच के बताओ की सर इस सर्कल के सेंटर से इस की रेडियस की जो लेंथ है ऑफ कोर्स क्या है अभी जस्ट हमने वेलवेट की क्या R1 इस सर्कल के सेंटर से इसकी सरकम्फ्रेंसेस तक की जो डिस्टेंस है जो की इसकी रेडियस है r2 अब मेरी बात समझ का रहे हो अगर मैं R1 में से r2 सब्सट्रैक्ट करता हूं अगर मैं R1 में से आर तू सब्सट्रैक्ट करता हूं तो shailise मुझे सर ये डिस्टेंस मिल रही होगी मुझे यह डिस्टेंस मिल रही होगी जो की टेक्निकल कुछ नहीं है जो दोनों सर्कल्स के सेंटर्स के बीच की डिस्टेंस है आपको समझ का रहे हो जो मैं कहना चाह रहा हूं मैं महज इतनी सी बात आपसे कहना चाह रहा हूं की अगर दो सर्कस एक दूसरे को इंटरनल टच करें तो उनके बीच के सेंटर्स के डिस्टेंस उनकी रेडियस के डिफरेंस के इक्वल होगी क्या इस बात से कोई आपत्ति या कोई परेशानी आई रिपीट दो सर्कल्स अगर एक दूसरे को इंटरनल टच करें तो उनके सेंटर्स के बीच की डिफरेंस नहीं उनके सेंटर्स के बीच की डिस्टेंस उन दोनों की रेडियस के डिफरेंस के इक्वल होगी आई होप और मास्टरिंग तू कन्वे सो दिस इस गोइंग तू बी डी पैरामीटर थ्रू विच विल बी एसेसिंग वेदर तू सर्कल आर टचिंग एच आदर इंटरनल और नॉट डेट इस हो विल बी वर्किंग आते अपॉन है ना अब हम क्या करेंगे हम इसके ऊपर कम करना चाह रहे हैं तो मैं कहूंगा की सर इसके बारे में भी बात कर लेंगे सारी बातें कर लेंगे तो एक-एक करके हर एक चीज समझते हैं अगर यहां तक आपने चीज समझेगा तो suniyega ध्यान से एंड इमेजिंग डेट इस डी सर्कल जैसे हम कहते हैं S1 और S1 आपका है सर्कल जैसे हम पढ़ते हुए आए हैं x² + y² 2G और इस बारे में कुछ कहने की जरूरत नहीं है इसका जो सेंटर होगा वो हो जाएगा -1 - F1 और इसकी रेडियस क्या हो जाएगी सर जीवन स्क्वायर प्लस f1² - सी इनसाइड एंड अंडर रूट सिमिलरली एक और चैनल है सर S2 और उसमें भी चीज ऐसी हो जा रही है से वही बातें क्या यह वही तो यह भी हो जाएगा -d2 - f2 और वहां पर क्या ए जाएगा जी तू स्क्वायर प्लस f2² - C2 ये भी रेडियस हो जाएगी इनसाइड एंड अंडर रूट अब बात कम की जो निकल कर ए रही है वो क्या सर मैं बिल्कुल आसानी से कनक्लूड करूंगा की दोनों के सेंटर्स के बीच के डिस्टेंस उनके दोनों के रेडियस के डिफरेंस के इक्वल होगी मतलब कंक्लुजन वही है स्टूडेंट्स अगर आप ध्यान से देखें मैं यहां पर क्या कहूंगा दोनों के सेंटर्स के बीच की डिफरेंस है ना डिस्टेंस तो वही बात ये है -1 -1 - F1 ये क्या है भाई -1 - F1 और ये cardinate कैसे ये cardinate है आपका -g2 कमा -f2 आई होप आपके पास समझ का रहे हो तो सर इन दोनों के बीच के डिस्टेंस निकालो तो फिर वही बात आप भूले तो नहीं हो G2 - जीवन का होल स्क्वायर प्लस f2 - F1 का होल स्क्वायर लेते हो अंडर रूट में सर ये किसकी इक्वल होगा ये इन दोनों की रेडियस के डिफरेंस तो अगर मैं इस सर्कल की रेडियस निकालो R1 की है ना तो वैल्यू कितनी होगी भाई वो भी अंडर रूट ओवर जीवन स्क्वायर प्लस f1² - 1 और इसमें से प्राप्त करेंगे r2 जो की क्या होगी भाई जो की होगी gtu² + f2² - C2 आई थिंक जब भी ऐसा कुछ कंडीशन फुलफिल होती हुई दिखेगी हम कहेंगे ये क्लीयरली दोनों सर्कल्स एक दूसरे को इंटरनल टच कर रहे हैं याद रहेगा अच्छा सर ये बात तो हमने आपकी मैन ली और क्या और सुनते हैं ध्यान से बड़ी सिंपल सी बात है यहां पर वो मोड इसीलिए लगा रहा है की उससे नहीं पता अभी की R1 और r2 में कौन बड़ा है तो आप इस तरीके से लीजिएगा की आप बड़ी रेडियस में से छोटी रेडियस सब्सट्रैक्ट करें इस तरीके से ताकि आपको सिर्फ क्या मिले उन दोनों का डिफरेंस याद रहेगा चलो सर आगे बढ़ते हैं यहां पर अगर मैं एक जरूरी बात अगर ऑब्जर्व करो तो यहां पर एक जरूरी बात जो दिखेगी वो क्या सर यहां पर एक ही कॉमन टेंट मिल गया आपको एक ही कॉमन इंजन मिलेगी पॉइंट और सीधा सीधा तरीका सर वो कॉमन टैसेंट अभी फिलहाल इससे नारियों में वापस क्या बन जाएगी रेडिकल एक्सेस और रेडिकल एक्सिस तो जहां भी वर्ड दिखे मैं क्या आऊंगा एस वैन माइंस S2 काश तू जीरो आई रिपीट माय स्टेटमेंट फिर से सुनेगा ध्यान से जो यहां पर कमेंट टैसेंट बन रही है जो यहां पर कॉमन तेनजेन कर रही है वो कौन सी है सर वो आपकी ये कॉमन टांगें है ना आई एम रियली सॉरी इसको मैं थोड़ा सा इंक्लिन करूंगा और चीज यहां पर लाने की कोशिश करूंगा ऐसे थोड़ा और हम एक लाइन करेंगे और आई थिंक चीज है हम यहां पर इस तरीके से पुश करेंगे तो ये आपका कमेंट टेंट है दिस लाइन हर इस दी कॉमन टेंसेज दिस इस अगेन नॉट जस्ट बिहेविंग एस अन कॉमन टैसेंट बट अलसो एस दी रेडिकल एक्सरसाइज और अगर ये रेडिकल एक्सेस की तरह बिहेव करें दिस इस नॉट जस्ट दी कॉमन टांगें बट ये क्या बन गया ये बन गया आपका रेडिकल एक्सेस और अगर सर ये रेडिकल एक्सेस की तरह बिहेव कर रहा है तो मैंने तो जीवन में इतना जाना है की इस पर कोई भी पॉइंट जो होगा वो वहां से जब आप इनका इंजन ड्रा करोगे तो इक्वल लेंथ के होंगे और रेडिकल एक्सिस यानी इस कॉमन टांगें की इक्वेशन क्या मैं लिख सकता हूं सर सर्कल वैन की इक्वेशन में से सर्कल तू की इक्वेशन सब्सट्रैक्ट कर दो तो आपको एलेक्सिस की इक्वेशन यानी कॉमन टेंसेंट की इक्वेशन मिल जाएगी आई थिंक चीजें आसान है बिल्कुल आसान है तो इसकी इक्वेशन क्या हो जाएगी भाई फिर वही बात x²y² है जाएगा तो आपके वो जी एफ और सी वाले टर्म्स बचेंगे उनमें एक्स और ए में आपको एक लीनियर इक्वेशन यानी एक स्ट्रेट लाइन की इक्वेशन मिल जाएगी बस यही बातें यहां पर का कर लिखकर आपको समझने की कोशिश की गई है एक और मजेदार और जरूरी बात निकल कर आती है जो की आपको याद रखनी है और इन्हें अच्छे से नोट डाउन करना स्टूडेंट्स ये दो तीन बातें ऐसी हैं जिन पर क्वेश्चंस बनते हैं तो प्लीज इन्हें अच्छे से नोट डाउन करना ये बात ये है सुनिए ध्यान से यह जरूरी बात है की ये जो आपकी जो पॉइंट है ना ये जो पॉइंट है ये जो पॉइंट है इसमें थोड़ा और अच्छे से बना कर आपको ड्रॉ करके दिखाता हूं ताकि बहुत मजेद अपना हो चीजे बहुत हॉटस्पॉट ना हो और आप चीज अच्छे से देख पाएं है ना जैसे की क्या ये आपका एक सर्कल है और एक सर्कल से इंटरनली ऐसे टच करता है कोई तकलीफ तो नहीं है अब जब उसने टच किया तो suniyega ध्यान से जिस पॉइंट पर इन दोनों ने एक दूसरे को टच किया है इसे मैं का लेता हूं क्या टी याद ए रहा है कुछ है अगर इसने यहां पर टच किया तो मैन लो सर इसका सेंटर यह रहा और इसका सेंटर यह रहा है ना और इन सारे सेंटर्स को जब आपने मिलाया है ना C1 C2 और इस टी को पॉइंट ऑफ कॉन्टैक्ट को टू कोर्स आपका C1 था और ये आपका था C2 तो हम बिल्कुल मतलब इसमें तो मुझे कुछ कहने की जरूरत नहीं है ना C1 क्या है C1 से टी तक के डिस्टेंस क्या है R1 कुछ बोलूं या आप खुद मुझे बता देंगे और सर C2 से टी तू टी तक के डिस्टेंस क्या है सर ये है r2 तो आप खुद कहोगे सर आप खुद कहोगे क्या की टी ने एक्सटर्नल ec1 C2 को R1 इस तू आर तू के रेश्यो में डिवाइड किया है तो क्या मैं का सकता हूं टी एक्सटर्नल डिवाइड करता है ना डिवाइड करता है मतलब मुझे अगर कभी भी कोई पूछेगा की दो सर्कल से एक दूसरे को इंटरनल टच कर रहे हैं तो मैं क्या कहूंगा सर अगर वो पॉइंट ऑफ कॉन्टैक्ट पता करना है ना तो एक्सटर्नल डिवीज़न वाला फॉर्मूला अप्लाई करो बात खत्म एक्सटर्नल डिवीज़न वाला फॉर्मूला अप्लाई करो बात खत्म आपका आंसर ए जाएगा याद रहेगा क्या आई थिंक चीजें आसान है चीज सली हुई है और चीज बेसिक हैं और चीज सिंपल हैं और इस बेसिस पे हम सोच पाएंगे क्या यहां तक कोई डाउट ये दोनों बातें प्लीज याद रखिएगा और अगर ये तीनों बातें बस तीन ही बातें हैं इंटरनल डिवीज़न दोनों के सेंटर्स के बीच के डिस्टेंस उनकी रेडियस के डिफरेंस के इक्वल दूसरी बात उनकी जो कमेंट होगी वही उनका रेडिकल एक्सेस होगा वह होगा S1 - h2 = 0 और क्लीयरली वो पॉइंट ऑफ कॉन्टैक्ट पता करने का तरीका होगा की वो जो C1 और C2 को ज्वाइन करने वाले लाइन है उसे ये पॉइंट ऑफ कॉन्टैक्ट R1 इस तू आर तू के रेश्यो में externalli माय डियर वैन एक्सटर्नल डिवाइड कर दो एक्सटर्नल डिवीज़न वाला फॉर्मूला आप अप्लाई करेंगे यहां तक कोई दिक्कत कोई परेशानी कोई डाउट स्टूडेंट्स अगर नहीं है तो मेरा तरीका आसान क्या हम तो करेंगे क्वेश्चंस सॉल्व और देखेंगे की चीज कैसे हो रहे हैं इस क्वेश्चन को पढ़िए समझिए अच्छा क्वेश्चन है और एक्चुअली आपके दिमाग के कई सारे तारिक क्वेश्चन खोलेगा इंटरनल जब सर्कल एक दूसरे को टच करें इंटरनल उसे कॉन्सेप्ट को लेकर कैसे देखोगे फाइंड डी इक्वेशन इक्वेशन निकल है सर ऑफ डी स्मॉलर सर्कल्स ऑफ डी स्मॉलर सर्कल्स दिस एंड पासेस थ्रू डी पॉइंट फोर थ्री बताओ इंटरनल इंटरनल टच करें ऐसा तो कुछ समझ ही नहीं ए रहा है ऐसा कैसे लग रहा है देखेंगे ना अभी हम क्वेश्चन ट्राई करेंगे की कंडीशन जो दी है उसे अप्लाई करेंगे और सारी बातें छोड़ो आपको तो एग्जाम में नहीं पता रहेगा ना की इंटरनल इंटरनल ही टच करने पर बेस्ड है एक्सटर्नल टेस्ट करने वाले क्वेश्चन लिख के थोड़ी देगा वो की ये सर्कल का इस टॉपिक का इस चैप्टर का ही ऐसा ऐसा क्वेश्चन है ऐसा तो बिल्कुल नहीं देगा फिर तो वो खुद ही कर ले भाई है ना तो वो आपकी इंटेलिजेंस चेक करना चाह रहा है वो ये देखना चाह रहा है की आप क्वेश्चन पढ़ के समझ पाते हो या नहीं की आपको exaktali करना क्या है मतलब क्वेश्चन की लैंग्वेज को समझना उसे इंटरप्रेट करना और उसे इंटरप्रेट करने के बाद उसे प्रक्रिया करना है की यहां पर करना क्या है यही लॉजिक यही इंटेलिजेंस आपकी कंप्रीहेंशन और इंटरप्रिटेशन एबिलिटी के साथ चेक किया जा रहा है और इसी एग्जाम को हम कहते हैं आईआईटी जी है ना यहीं से रेडबस दोनों इसी तरीके से चेक करती है लेवल अपग्रेड करके इस क्वेश्चन को जब मैं पढ़ूंगा ना तो मैं कहूंगा सर की इस क्वेश्चन में मुझे एक जरूरी बात जो नज़र आती है वो ये की ये जो बात है ध्यान से देखो x² + y² = 1 सर्कल जिसका सेंटर है ओरिजिन पर और जिसकी रेडियस है वैन ये बात तो मुझे नजर आई सर बहुत बढ़िया ये बात बहुत कम की और इस बात को आप गौर से देख लो इसे ऐसे सारी बातें होंगी कम की बात थी अब अगर मैं चीज विजुलाइज करना चाहूं तो सुनना ध्यान से आपके पास ए एक्सिस होगा आपके पास एक्स एक्सेस होगा बिल्कुल होगा सर और एक सर्कल होगा जो की सेंटर होगा कहां जो की सेंटर होगा ओरिजिन पर कोई तकलीफ तो नहीं है उसकी रेडियस है वैन तो वैन कमा जीरो जीरो कमा 1 - 1 0 0 - 1 ऐसे पॉइंट से पास हो रहा होगा अब मुझे एक ऐसा सर्कल चाहिए जो पहली बात तो फोर कमा थ्री से पास होता है और जो की इस सर्कल को टच करते हैं तो एक कम करते हैं सर मैं कुछ सर्कल्स ऐसे इमेजिन करने की कोशिश करता हूं जो कहां से पास हो रहे हैं फोर कमा थ्री फोर थ्री यहां कहीं मिल रहा होगा तो लेट्स से ये जो पॉइंट है यहां कहीं दूर जाकर यहां कहीं मुझे मिलता है क्या 4 3 ये जो पॉइंट है दिस इस फोर कमा थ्री कैन यू तेल मी और कैन यू गिव मी एक ऐसा सर्कल जो फोर कमा थ्री से पास हो रहा है और इस सर्कल को टच कर रहा है सर मैं इसे दो सर्कल बना सकता हूं मैसेज दो सर्कल बना सकता हूं कौन से दो सर्कल जनाब सर एक सर्कल तो ऐसा एक सर्कल तो ऐसा जो मुझे बस थोड़ा सा वक्त दीजिए आप एक सर्कल तो ऐसा जिससे पास हो रहा होगा और ऑफ कोर्स इसे टच कर रहा होगा ऑफ कोर्स इसे टच कर रहा होगा कुछ इस तरीके से आई थिंक अभी हम काफी क्लोज हैं इसके आप मेरी बात समझ पाए क्या ये बिल्कुल इस सर्कल को इंटरनल टेस्ट कर रहा है और फोर कमा थ्री से पास है सर कोई और भी संभावना है क्या देख लेते हैं भाई एक और पॉसिबिलिटी मुझे नजर आती है सर जो अगर आपने एक ऐसा सर्कल बनाया होता अगर आपने ऐसा सर्कल बनाया होता तो इस सर्कल की भी तो खास बात यही थी सर मैंने थोड़ा इसे प्रॉपर्ली नहीं बनाया है मुझे थोड़ा सा वक्त दीजिए मैं इसका पॉइंट ऑफ कॉन्टैक्ट अगर यहां निकालो तो सर इस सर्कल की भी तो खास बात यह है यह जो सर्कल बना है इसकी भी तो खास बात यही है की यह भी सर्कल को टच कर रहा है बट externalli काफी कुछ आपके सामने हैं साहब काफी कुछ आपको बता दिया गया है अब आप मुझे इस क्वेश्चन को सॉल्व करके आंसर बताएंगे आपको अगर जानना है तो ये सर्कल के बारे में हम जानते हैं सेंटर डेट ओरिजिन जीरो कमा जीरो सर्कल एंड दिस सर्कल बोथ हैव वैन थिंग इन कॉमन डेट दे टच दिस सर्कल रिस्पेक्टिवली कैसे externalli और इंटरनल और इनकी खास बात ये है की फोर थ्री से पास होता है आई थिंक मेरे ख्याल से इतना विजुलाइजेशन इस क्वेश्चन का मेरे ख्याल से काफी है अब आप करेंगे इस क्वेश्चन को ट्राई और मुझे बताएंगे की सर इस क्वेश्चन में होना क्या है क्या यहां तक चीज आपको क्लियर है यहां तक तो कोई कन्फ्यूजन नहीं है अब थॉट क्या ए सकता है सर सर स्मालेस्ट सर्कल जो है जो ढूंढने को वो का रहा है जब वो लिख रहा है की आपको स्मॉलर सर्कल ढूंढना है तो स्मॉलर सर्कल के लिए क्या एप्रोच रहेगी आई थिंक मेरे ख्याल से सर बड़ा छोटा देखने से अच्छा आइडिया होगा की आप खुद देख लो की कौन सा सर्कल है एक्सटर्नल या इंटरनल कौन सा सर्कल जो इसे टच कर रहा है externalli इंटरनल इनमें छोटा कौन होगा बड़ी बेसिक सी बात है आप खुद ऑब्जर्व करके का दोगे की छोटा सर्कल अब जो से इसकी इक्वेशन मुझे निकालनी है तो मेरा यह कहना है मेरा बस आपसे कहना है की अगर आप से मैं कहूं बड़ी सिंपल सी बात सर इसकी इक्वेशन आपको निकालनी है ना आई होप आपने इस क्वेश्चन को ट्राई किया मैंने कुछ हिंट और देने की कोशिश की उसमें आपसे क्या पूछा उसने आपसे बोला स्मॉलर सर्कल निकल है उसने आपसे बोला स्मॉलर सर्कल निकल तो सर सीधी सी बात है 4 3 से पास होगा जैसे यहां टच करेगा तो ये बड़ा होगा दिख भी रहा है क्यूट अपेरेंट और एवीडेंट फोर कमा 3 से पास होगा और इसे एक्सटर्नल यहां टच करेगा आई थिंक ये सर वो छोटा सर्कल होगा डेफिनेटली वो छोटा सा करो सो सी आर लुकिंग फॉर दिस सर्कल सी आर नॉट लुकिंग फॉर दिस सर्कल डिड यू ऑल गेट दिस आइडिया कॉल की हमें क्या करना है मैन गए सर ये बात मैन ली अब क्या मैं एक बात जानता हूं ये आपका ओरिजिन है क्या आप ये बात समझ का रहे हो की ओरिजिन यानी सर्कल का सेंटर है और इस सर्कल के सेंटर से अगर मैंने परपेंडिकुलर ड्रॉप किया सॉरी नॉर्मल ड्रॉप किया किस तरीके से इस तरीके से क्यों इनके पॉइंट ऑफ कॉन्टैक्ट पर मिलता है तो सर ये बात तो होता है ये बात बिल्कुल होता है की सर इस सर्कल का जो सेंटर होगा ना इस सर्कल का जो सेंटर होगा सर्कल का जो सेंटर होगा और इस सर्कल का जो सेंटर होगा उनके लिए वो एक कॉमन नॉर्मल की तरह बिहेव करेगा आप सन का रहे हो जो मैं कहना चाहता हूं मैं फिर से रिपीट करता हूं एक बात सुनेगा ध्यान से ये बहुत कृष्ण और कम की हंटर शायद इसी से एक क्वेश्चन हो जाएगा अगर आप चीज सोच का रहे हैं तो मुझे बस ये बताओ इन दोनों सर्कल्स का पहले तो हमने इस बाहर वाले सर्कल को हटा देना है हमने उससे कोई मतलब नहीं हम इस अंदर वाले सर्कल पर बात करें अंदर वाले सर्कल के अंदर मुझे दिख रहा है सर की अगर इस सर्कल का एक और इस सर्कल का कॉमन नॉर्मल बनाओ तो कॉमन नॉर्मल क्या होगा सर कॉमन नॉर्मल वो नॉर्मल होगा suniyega ध्यान से मैं आपको और अच्छे से समझने की कोशिश करता हूं जो इनकी कॉमन टांगें है जो इनकी कमेंट है उसे पर परपेंडिकुलर होगा तो जो कॉमन नॉर्मल होगा ना सर वह बेसिकली इस कॉमन टांगें पर क्या होगा परपेंडिकुलर मैंने चीज अच्छे से ली नहीं या बनाई नहीं इसलिए मुझे चीज अच्छे से दिख नहीं रही हैं बट मैं थोड़ी सी चीज बेहतर तरीके से बनाने की कोशिश जरूर करूंगा ताकि आपको अच्छे से चीज समझा ही जा सके है ना आई थिंक सर चीज कुछ ऐसी हो तो ज्यादा अच्छा ये मेरे ख्याल से बहुत अच्छे तरीके से दिख रहा है सर ये इन दोनों सर्कस के सेंटर से क्यों पास हो रहा है बेशक ये होगा ये होगा ये क्यों नहीं होगा सर नॉर्मल तो सारे सर्कल के सेंटर से पास होते हैं ना तो अगर यह ऐसा नॉर्मल है जो इसका भी नॉर्मल है और इसका भी नॉर्मल है तो ऑफकोर्स वो कैसा नॉर्मल है ये कॉमन नॉर्मल है और कॉमन नॉर्मल कहां बनेगा बन पाओगे चीज की चीज कहां और कैसे जा रहे हैं बस इतना सा आपको कर लेना है देखो इस नॉर्मल को लेकर मैं आपसे बार-बार इसलिए का रहा हूं क्योंकि आप इस नॉर्मल को देखो पहले तो यह पॉइंट समझना थोड़ा सा हमने प्रोबेबली ड्रॉ नहीं किया तो ये यही पॉइंट है जो आपको दिया हुआ है लेट्स कॉल इट ए फॉरवर्ड है ना इसे मैं का लेता हूं क्या बात आपको समझ ए रही है अब एक बात बताओ सर सिमेट्री से ये तो दिख ही रहा है सर की जहां ये टच कर रहे हैं वहीं ये टच हो रहा है वही है टच हो रहा है बस वो properally मैंने एक्सेस ठीक से ड्रॉ नहीं किए वो थोड़े से टिल्टर है इसलिए चीज ऐसी आपको नजर ए रही होंगी अब सर कम की बात क्या है एक बात बताओ ये जो कॉमन नॉर्मल है ये जो कॉमन नॉर्मल है जो की कॉमन टांगें के कारण वो कॉमन नॉर्मल है तो यहां पे भी वो कॉमन नॉर्मल होगा क्योंकि यहां भी वो कॉमन टांगें है यू गेटिंग माय पॉइंट और यहां पर भी वो कॉमन नॉर्मल होगा क्योंकि यहां भी वो कमेंट टैसेंट है यानी वो बेसिकली इन पॉइंट ऑफ कॉन्टैक्ट से ही पास होगा यानी ये जो पॉइंट ऑफ कॉन्टैक्ट था जो आपका 43 यहां से पास होगा अब आपका क्या कहना है सर मेरा आपसे ये कहना है क्या आप मुझे इस स्ट्रेट लाइन ओ एक ही क्वेश्चन बता सकते हो ऑफ कोर्स ये आपके 0 है जैसे मैं कहता हूं ओ तो सर ओ इक्वेशन क्या होगी सर सीधी सी बात है क्या सोचना है 430 है ना तो मैं क्या लिखूंगा ए - 0 यानी ए = एन एम कितना होगा 3 - 0 / 4 - 0 तो कितना हो जाएगा 3/4 कोई दिक्कत तो नहीं टाइम्स एक्स - 0 तो ये कुछ ऐसा हो जाएगा कोई तकलीफ किसी भी स्टूडेंट को आई थिंक नहीं होनी चाहिए अब सर जो ये स्ट्रेट लाइन है जो ये स्ट्रेट लाइन है ये इस सर्कल को ये इस सर्कल को ये सर्कल कौन सा है ये जो सर्कल है ये x² + ए = 1 इससे इंटरसेक्ट करती है इससे कहां इंटरसेक्ट करती है सर इससे इंटरसेक्ट करती है वो इन दो पॉइंट्स पर किस-किस पॉइंट पर यहां पर और यहां पर तो क्या मैं ऐसा कहूं क्या मैं ऐसा कहूं की सर मैंने अगर इस स्ट्रेट लाइन को और इस सर्कल को सॉल्व किया तो शायद में ये दोनों पॉइंट्स ऑफ इंटरसेक्शन निकलूंगा बिल्कुल निकल लूंगा तो ए की वैल्यू है 3x/4 यहां रिप्लेस करते हैं तो ये है x² + यहां पर रिप्लेस करेंगे 3x/4 तो 3/4 3 का स्क्वायर 9 4 का स्क्वायर 16 और ये हो जाएगा x² = 1 कोई तकलीफ स्टूडेंट्स आई थिंक यह बात आप समझ का रहे हो अब सीधी-सीधी सी बात ध्यान से देखना स्टूडेंट्स पुरी इक्वेशन को 16 से मल्टीप्लाई किया जैसे ही 16 से मल्टीप्लाई किया तो यहां से 16 गया दिख रहा है क्या यहां पर 16 आया और यहां पर 16 आया कोई दिक्कत तो नहीं है 16 + 9 कितना हो जाएगा सही हो जाएगा 25 ये कितना हो जाएगा भाई ये हो जाएगा 25 तो यहां पर आपको दिखेगा 25x² आई थिंक 25 यहां से सर अच्छा होता आप यहीं लिख देते तो आपने कहां लिखा 25 यहां तो अगर x² है 16/25 एक्स की जो वैल्यू आएगी वो क्या होगी प्लस माइंस फोर बाय फाइव ऑप्शन का रहे हो समझ का रहे हो प्लस माइंस फोर बाय फाइव अगर एक्स की वैल्यू प्लस माइंस फोर बाय फाइव यहां रखता हूं तो ध्यान से देखना एक्स की वैल्यू अगर प्लस माइंस फोर बाय फाइव यहां रखता हूं तो ए की वैल्यू क्या आएगी देखो प्लस माइंस फोर बाय फाइव रख देता हूं फोर से फोर कैंसिल तो क्या बचेगा 3/5 तो ए की वैल्यू क्या ए जाएगी सर वो ए जाएगी 3/5 आप सुनेगा ध्यान से जब आपने प्लस फोर बाय फाइव रखा तो ये जो कार्ड है ये क्या मिला सर आपको ये मिला 4/5 सुनिए ध्यान से बहुत कम की बात -3/5 अगर मैं इस पॉइंट को थोड़ी देर के लिए का लेता हूं लेट्स से बी सो कैन आई कनक्लूड की सर आप एक ऐसे सर्कल के क्वेश्चन ढूंढ रहे हो जिसके डायमीटर के कोऑर्डिनेट्स आपको पता है हान या ना तो सर ये ऐसा तो हमने पढ़ा है अगर मैं इसे मुझे एक सर्कल के इक्वेशन पूछी जाए जब डायमीटर के अकॉर्डिंग बताओ तो कुछ ऐसा पढ़ की सर एक्स - X1 एक्स - X1 यानी एक्स - 4 अरे हान या ना टाइम एक्स - x2 एक्स - x2 यानी एक्स - 4 / 5 यस और नो ऑनलाइन स्टूडेंट्स जवाब दीजिए और यहां पर प्लस ए माइंस y1 ए - y1 यानी कितना स्टूडेंट्स थ्री कोई दिक्कत तो नहीं टाइम्स ए - Y2 जो की कितना होगा भाई थ्री बाय फाइव तो इस तरीके से आप इसे लिखेंगे जो की होगा इक्वल तू जीरो के क्या कोई बात समझ आई अब चाहे तो इसे सिंपलीफाई कर दीजिए मेरा कहना है यहां पर भी रोक दो सिंपलीफाई करके भी लिख सकते हो ये कौन से सर्कल की इक्वेशन होती है सर ये इस सर्कल के क्वेश्चन अब जो की ऑफ कोर्स आपका कौन सा है जो की आपका स्मॉलर सर्कल है मैन लो इसकी जगह उसने लार्जेस्ट सर्कल पूछा हो तो क्या करते हो आप लार्जेस्ट सर्कल कौन सी बड़ी बात है लार्जर सर्कल का डायमीटर क्या है सर वो है ए और इस समय का देता हूं ऐसे थोड़ी सी तो वो हो जाता है एक और एक इसे डी डायमीटर के भी तो cardinate मुझे पता है तो क्या वहां से क्वेश्चन नहीं निकल सकते मैन लो उसने पूछा था लार्जर सर्कल तो वो भी हम निकल लेते हैं वही बात होती एक्स - 4 टाइम्स एक्स - 4 / 4X + 4 / 5 + ए - 3 टाइम्स ए + 3 / 5 = 0 आई होप आपको बातें समझ ए रही है की आप दोनों सर्कल की इक्वेशंस इसकी और इसकी निकल सकते हो क्या ये एक आसान सा क्वेश्चन था बेसिक सिमेट्री और बेसिक अंडरस्टैंडिंग अप्लाई करनी थी और इस क्वेश्चन को सोचना था आई थिंक चीजें बहुत सिली हुई सी है अगर आपको कॉन्सेप्ट पता है अगर आपको चीज पता है अगर आपने बारीकियां ठीक से पद कोई डिफिकल्ट क्वेश्चन तो नहीं था स्टूडेंट से अगला क्वेश्चन ट्राई करेंगे ठीक है सर अगर यह क्वेश्चन समझ आए तो लेट्स मूव टुवर्ड्स डी नेक्स्ट क्वेश्चन आई होप आप ये निकल लोग ऐसे सिंपलीफाई कर लोग बहुत बड़ा टास्क नहीं है अगर इन केस आपको लगता है नहीं सर कर दो हम मैच करना चाह रहे हैं हमारा आंसर तो फटाफट कर लो भाई देखो एक्स और एक्स ये कितना हो जाएगा सर ये हो जाएगा x² है ना यहां से क्या हो जाएगा सर ये ए जाएगा -4x ये कितना हो जाएगा सर ये हो जाएगा -4x/5 कोई दिक्कत तो नहीं और ये हो जाएगा कितना 4 4 16 / 5 बहुत बढ़िया प्लस ए ए हो जाएगा y² ये कितना हो जाएगा -3y ये हो जाएगा -3y/5 और क्या सर और ध्यान से देखो ये हो जाएगा प्लस गड़बड़ किया गया नहीं प्लस नाइन बाय फाइव इसे इक्वल तू जीरो अगर आपको थोड़ी सी चीज है सुलझी हुई चाहिए अगर आपको चीज थोड़ी सी सिंपलीफाइड चाहिए तो मेरी आपको हिदायत ये होगी की आप पुरी इक्वेशन को 5 से मल्टीप्लाई कर दो अगर आपको चीज आसान चाहिए तो जैसे ही आप फाइव से मल्टीप्लाई करोगे तो देखो क्या हो जाएगा समझना बात को फाइव से मल्टीप्लाई करते ही हो जाएगा 5x² हान या यह फोर की जगह 20x यहां पर से यह फाइव है जाएगा यहां से ये फाइव है जाएगा सन का रहे हो क्या आप लोग यहां पर क्या ए जाएगा सर यहां पर ए जाएगा 5y स्क्वायर यहां पे 5 3 कितना हो जाएगा सभी हो जाएगा 15 ये आई थिंक अब आप समझ का रहे हो मैं जो करना है कहना चाह रहा हूं यहां पे फाइव मल्टीप्लाई करके हटा देगा और यहां पर भी फाइव हो गया हटा देगा तो अब अगर आप सिंपलीफाइड वर्जन पूछे तो ये है 5x² और ये है 5y² तो ये हो जाएगा 5x² + 5y स्क्वायर और क्या सर देखो भाई -20 - 4 सो था - 24x फिर क्या सर माइंस 15 माइंस थ्री सो था - 18y और क्या सर 16 प्लस 9 आई थिंक डेट क्लीयरली नथिंग बट व्हाट था cleali 25 इस इक्वल्स तू जीरो दिस इस गोइंग तू बी योर आंसर आप चाहे तो वही x²y² पूरा 5 हटा भी सकते द बट मैंने कहा की ऐसा सिंपलीफाइड रखते हैं वहां फ्रेक्शंस रखने के बजाय तो आप ऐसा आंसर भी प्रेफर कर सकते हैं नेक्स्ट क्वेश्चन की तरफ आई होप काफी सारे स्टूडेंट्स का आंसर मैच हुआ होगा यहां दिख रहा है suniyega तो निकलना है तू आर मेकिंग एंड एंगल ऑफ 60 डिग्री विद इ आदर नहीं समझ ए रहा है मतलब कुछ समझ नहीं ए रहा है सर तो मेरा कहना है की विजुलाइज करो ड्रा करो चीज समझ में आएंगे अगर आप चीज ड्रा करोगे सुनना ध्यान से मैं सारी बातें खत्म करता हूं सर इस सर्कल को ड्रॉ करने का सबसे अच्छा आइडिया होगा पहले सर्कल का सेंटर ढूंढता हूं इस सर्कल के सेंटर होंगे वैन कमा तू दिख रहा है क्या तो सर इस सर्कल का सेंटर है वैन कमा तू तो वैन कमा तू पर एक सेंटर सर्कल है जिसकी रेडियस अगर मैं निकालो तो रेडियस कितनी होगी सोच के बताओ 1 स्क्वायर 123 तो सर टेक्निकल आप एक थ्री यूनिट रेडियस वाला एक सर्कल ड्रा कर रहे हो जो इसका सेंटर है वैन कमा तू ऊपर इसका सेंटर है वैन कमा 2% दिख रहा है क्या दिख रहा होगा है ना और इसकी रेडियस कितनी दी है सर आपने इसकी रेडियस आपने ली है कितनी भी इसकी रेडियस सर आपने ली है 3 यूनिट जैसे मैं थोड़ा यहां पर लिख दे रहा हूं आई होप आपको इस बात से कोई आपत्ति नहीं होनी चाहिए इसकी ये जो है ये रेडियस है और ये कितनी यूनिट्स है ये थ्री यूनिट्स हैं वह यह कहना चाह रहा है आपसे फाइंड डी लोकस ऑफ डी सेंटर ऑफ डी सर्कल सेंटर ऑफ डी सर्कल सब निकल लेंगे सर्कल की खास बात क्या है उसे सर्कल की खास बात ये है की वो जो सर्कल है वो इस सर्कल को टच करता है इंटरनल इस सर्कल को वो जो सर्कल का सेंटर का लोकस पूछा जा रहा है वो उसको इंटरनल टेस्ट करता है तो इसके अंदर कहीं बनेगा दूसरी बात उसे सर्कल पर जो आपको इसके अंदर बनाना है उसे सर्कल पर जब आप tangenced ड्रा करोगे ना वैन कमा तू से है उसे सर्कल पर जब आप टेंशन ड्रॉ करोगे वैन कमा तू से मतलब मतलब इसी सर्कल के सेंटर से जब वो सर्कल पर जब आप tangenced ड्रा करोगे तो उन टेंसेज के बीच में एंगल कितना होना चाहिए 60° करेंगे सर बात सारी बातें कर लेंगे एक कम करो एक कम करो भाई ये बताओ हमारे पास एक सर्कल चाहिए जो इसको इंटरनली टेस्ट कर रहा हूं मैं एक रैंडम सा कहीं भी बना लेता हूं कोई परेशानी आपत्ति तो नहीं है आपको इस बात से नहीं है सर तो ये हमने रैंडम सा सर्कल बना लिया थोड़ा छोटा कर लेता हूं तो आपको 60 डिग्री टाइप दिख जाएगा है ना तो ये एक ऑफ कोर्स आपका ये एक सिनेरियो बन रहा है है ना अब सर अगर आपने दो टांगेंट्स बनाई होती हैं यहां से अगर आपने दो टांगेंट्स बनाई होती तो कैसी दिखती भाई ये टेंशन जा रही होती है एक्सटेंड कर डन तो चलेगा क्या परेशान तो नहीं हो जाएंगे भाई और सिर्फ कोर्स आपकी एक और टेंशन जा रही होती की कुछ ऐसी जा रही होती आई होप इस बातों से किसी को कोई आपत्ति नहीं और ऑफ कोर्स एक और बात आपको ए रही होगी की सर अगर आप इन दोनों के सेंटर से कुछ पास करवा लाइन तो वो इनकी कॉमन जो टैसेंट है उसे पर परपेंडिकुलर नॉर्मल की तरफ जा रही होगी क्योंकि इसका भी सेंटर होगा और इसका भी कहीं ना कहीं सेंटर होगा अब बात समझ का रहे हो अब सुना मुझे इसके सेंटर का लोकल चाहिए तो इसका सेंटर यहां कहीं लाइक कर रहा होगा जिससे मैं थोड़ी देर के लिए का लेता हूं ह कॉम के कोई तकलीफ कोई परेशानी ये है वैन कमा तू अब क्लीयरली सर क्लीयरली चीजें बिल्कुल सिमिट्रिकल है ये आपकी टांगें ये टैसेंट इन दोनों के बीच में 60° एंगल तो कहना इससे की सर ये जो एंगल होगा ये होगा एक 30 डिग्री एंगल क्योंकि एंगल बाईसेक्टर की तरह बिहेव करेगा हो मुझे इसको मैथमेटिकली प्रूफ करने की जरूरत नहीं है आपकी बातें अच्छे से एवीडेंटली यू नो अपेरेंटली देख का रहे हो समझ का रहे हो बातें जच रही है दिमाग में है ना अब बात सुनना अब मैं जो अगली हिंट आपको देने वाला हूं इस क्वेश्चन को सॉल्व करने के लिए सर ये 30 डिग्री मिल चुका है ये ह कॉम के मिल चुका है ये दो tesence है अब आपके पास मेरे ख्याल से इस क्वेश्चन को सॉल्व करने के लिए सब कुछ है ऐसा कैसे है सब कुछ एक मिनट सब कुछ इसलिए है क्योंकि अगर मैं आपसे पूछूं ये वैन कमा तू से वैन कमा तू से अगर यहां तक की इस पॉइंट की डिस्टेंस अगर मैं पूछूं तो क्या आप बता सकते हो कितनी क्या बता सकते हो कितनी आई थिंक आपके पास कुछ है मेरी बस गुजारिश है की अगर आप यहां से परपेंडिकुलर ड्रॉप कर दें अगर आपको अच्छा नहीं लग रहा है यहां से ड्रॉ कर दे और इसको ये परपेंडिकुलर कहते हैं अगर आप इनको लेवल करना चाहे तो लेवल कर लीजिए जैसे मैं इनको नाम देना चाहूं तुम्हें कुछ नाम दे देता हूं जैसे मैं इसे का लेता हूं जैसे ए जैसे का लेता हूं बी इसे का लेता हूं सी से का लेता हूं दी इसे का लेता हूं आई है ना और कुछ तो करना बाकी नहीं है लेवल आई थिंक आई के अलावा और कुछ तो बाकी नहीं है ना तो ये हो जाता है इसके बाद हमारे पास चीज है तो अब जरा आप इस पूरे फिगर को ध्यान से देखो और मुझे बस इतना बताओ की चीज कैसे शुरू करनी होगी चीज कहां से हम निकलेंगे और चीज किस तरीके से आप मुझे निकल कर बताओगे बहुत बेसिक सा ऑब्जर्वेशन है जो ट्रायंगल आपको मैं बना कर दे रहा हूं ना ये जो अंदर वाला बन रहा है इसमें अगर आप एक और नाम लिखना चाहो तो मैं यहां पे इस पॉइंट को भी नाम दे देता हूं लेट्स क्वालिटी एफ जिसका हमें लुकास निकलना है तो हिंट बस में इतनी सी ही देना चाह रहा हूं स्टूडेंट्स आपसे की आप अगर ट्रायंगल ए बी एफ में देखोगे तो आपका क्वेश्चन खत्म हो जाएगा अगर आप रिपीट माय स्टेटमेंट अगर आप ए बी एफ में देखोगे तो आपका क्वेश्चन खत्म हो जाएगा अगर आप चीज ध्यान से देख का रहे हो तो कैसे खत्म हो जाएगा सर आप तो कुछ तो भी बडबड रहे हो नहीं मैं कुछ तो भी नहीं बडबड रहा हूं स्टूडेंट्स मैं आपको एक बहुत अच्छी हिंट दे रहा था जैसे शायद आपने अमल में ले होते तो ये क्वेश्चन खत्म हो चुका था देखो ये आपका ट्रायंगल है ए बी एफ ऑफ कोर्स सर इस ट्रायंगल एबीसी में ये जो एंगल है ये भी कितना होगा सर ये एंगल भी कितना होगा सर ये भी 30 डिग्री होगा कोई तकलीफ इस बात से अब मैं इस ट्रायंगल अब का चेरी आपको समझाना चाह रहा हूं सर ये एंगल है 30° क्या में इस ट्रायंगल अब में sin30 निकल सकता हूं निकल तो सकते हो सर sin30 क्या होगा सर साइन 30 क्या होता है एंगल के सामने वाली परपेंडिकुलर साइड जैसे आप क्या कहोगे जैसे आप कहोगे बीएफ अरे हान या ना अपॉन ऑफ कोर्स का हाइपोटेन्यूज अब स्ट्रगल का हाइपोटेन्यूज क्या है सर वो है एफ / एफ होगा ना तो इसका हाइपोटेन्यूज क्या होगा सर वो होगा ए एफ क्या यहां तक किसी भी स्टूडेंट को कोई आपत्ति किसी भी स्टूडेंट को कोई परेशानी अब क्या करने वाले मेरे ख्याल से अब आप सोच लो अब आपका क्वेश्चन खत्म हो जाएगा अब आपको सब बता दिया अब इससे ज्यादा अगर मैंने बताया होता ना तो इस क्वेश्चन में आप कर दोगे गड़बड़ बीएफ और एफ में अब आपको चीज सोचनी है और आपको निकलना है इसका फाइनल आंसर मेरा यकीन करिए इस क्वेश्चन को इससे ज्यादा एक्सटेंड मत करिए अब यहां से आप चीज बिल्कुल सीधी सीधी सोच पाओगे बहुत आसान है चीज सोच पाना आप देखो भाई मैं बस आपसे इतना कहना चाह रहा हूं की एफ के बारे में मुझे ज्यादा जानकारी नहीं है आपका हाइपोटेन्यूज सर सोच लेंगे अब के बारे में पर क्या मैं एक बड़ी सिंपल सी बात आपसे ये का सकता हूं सर suniyega बहुत कम की बात है इसे ध्यान से suniyega प्लीज बीएफ क्या है सर बीएफ है इस सर्कल पर जो टैसेंट अपने ड्रा की है उसमें इस सर्कल के सेंटर से ड्रॉप के परपेंडिकुलर इस नथिंग बट दिस स्मॉलर सर्कल्स रेडियस और अगर वह बीएफ है अगर वो एफडीबी तो होगी अरे स्मॉलर सर्कल की रेडियस या तो आप का लो बीएफ है या फिर क्या है एचडी तो कहना है एचडी फर्स्ट ऑफ ऑल यूएफसी ऑल काइंड सो सिन 30 गेट्स रिप्लेसिड की 1 / 2 विच सी ऑलवेज वैन बाय तू एंड बा फॉर्म replaceing बाय एफ दी विच शुड नॉट बदर अन्योन अत ऑल आई थिंक एवरीवन विथ मी सो फार सर एचडी तो ठीक है आपका समझ आया अब आप एफ का क्या करने वाले हो मैं का रहा हूं एचडी के बारे में ही सोचते हैं एफ को वहीं रखिए क्यों अब realaiz करोगे की मैंने एफ को qintak रखा एफ को क्यों नहीं छेड़ा आप बात करो एचडी के क्या मैं एचडी को ऐसा नहीं का सकता सुनेगा ध्यान से एचडी को क्या मैं एड-एफ का दो फिर से सुनिए एड-एफडी नहीं होगा क्या ऐसा क्यों कर रहे हो सर मुझे कुछ दिख रहा है जो आपको भी दिखेगा जैसे ही मैं ऐसा लिखूंगा मैं एचडी को लिख रहा हूं एड -एफ है ना तो ये तो है 1 / 2 जैसे मैं ऐसा ही रखता हूं एचडी को मैं लिखता हूं क्या एचडी को मैं लिखता हूं एड-अस इसे मैं लिखता हूं ए दी - ए एफ डिवाइडेड बाय ए एफ पता नहीं सर क्या-क्या कर रहे हो आप हमें तो कुछ समझ नहीं ए रहा है थोड़ा सा पेशंस और रखें क्वेश्चन खत्म हो जाएगा ऐसा कैसे खत्म हो जाएगा सर थोड़ा सा पेशेंस रखिए भाई बोलो भाई यहां देखो इस ऐप को यहां मल्टीप्लाई करता हूं तू को उधर multiplaye करता हूं एफ को यहां वैन से मल्टीप्लाई किया तो ये हो जाता है एफ और तू को जब वहां मल्टीप्लाई करने में गया ना जब तुको मैं मल्टीप्लाई करने गया तो मुझे एड दिखा मुझे ए दी देख कर को चमक क्या आपको चमक क्या सर एड देख के मुझे ये स्ट्राइक हुआ की एड इस नथिंग बट इस सर्कल की रेडियस बड़े वाले सर्कल की और अगर ये बड़े वाले सर्कल की रेडियस है तो सर वो तो मैं निकल चुका हूं कितनी है भाई थ्री सो कैन आई डू नथिंग बट रिप्लेस सी कॉल कैंसिल एड गेट्स रिप्लेसिड की 3 हर व्हाट आई लेफ्ट नौ सी आर लेफ्ट तो इसको रख देते हैं तो ये मिलता है एफ सर अब क्या करने वाले हो आप सॉल्व करेंगे देखेंगे कुछ चीज तो बंद कर आएंगे तो suniyega ध्यान से suniyega ध्यान से 2 3 = 6 और 2 एफ - 2f -2f इधर आए तो कितना ए रहा है 3 टाइम्स ए एफ और वहां क्या मिलता है सर वहां मिलता है सही का रहा हूं आपको चीज समझ ए रही है -3 - 2af इधर आए एफ की वैल्यू कितनी आती है सर ये एफ की वैल्यू आती है तू जब आपसे कोई का रहा है की आपके एफ की वैल्यू ए रही है तू तो ये वाली लेंथ कितनी है तू यूनिट्स वैसे क्वेश्चन खत्म हो गया आप चाहे तो किसी और तरीके से कर लीजिए या फिर ऐसे कर लीजिए आप चाहे तो ऐसे कर सकते हो की सर एफ की वैल्यू आई है तू तो ये वैल्यू वैन होगी और उसकी रेडियस 1 है और उसके सेंटर के अकॉर्डिंग पता है सर्कल के क्वेश्चन लिख सकता हूं ऐसे कर लो या ऐसे नहीं करना है तो एक और तरीका बताता हूं अगर af2 ए रहा है तो बस ऐसे कर लो ए एफ अगर तू ए रहा है मतलब ह कमा के की वैन कमा तू से डिस्टेंस कितनी है रिपीट माय स्टेटमेंट ह कॉम के यानी एफ की वैन कमा तू डिस्टेंस है ए एफ ओ कितनी है तू निकल सकते हैं क्या तो बिल्कुल सर ह कमा के की वैन कमा तू से डिस्टेंस तो ह कमा के की वैन कमा तू से डिस्टेंस कोई तकलीफ कोई तकलीफ सर्कल दिख रहा है आखिर में हम क्या करते हैं आखिर में हम यह कर देते हैं तो ये बन जाएगा एक्स - 1 का होल स्क्वायर प्लस ए माइंस सर्कल है विच इस सेंटर अगेन आते वैन कमा तू जिसकी रेडियस है तू यानी की ये जो आपका एफ है ना ये पॉइंट ये एक ऐसा लोकस फॉलो कर रहा है ये एफ एक ऐसा लॉकर्स फॉलो कर रहा है जिसका सेंटर है वैन कमेंट्री जिसकी रेडियस तू इस लॉक कस पर कहीं पर भी आप एक सर्कल बनाना जैसे इंटरनल ही टच कर रहा होगा वो बेशक 60° की ही टांगेंट्स इंक्लाइंड बना रहा होगा ये तय बात है कहीं भी किसी भी स्टेप में कोई परेशानी सर ये पार्ट कैसे सोचा बस यही ट्रिकी पार्ट है जिसमें दिमाग लगाना होगा इसमें मेहनत करनी होगी बाकी मेरे ख्याल से टू चीज नहीं है एक छोटा सा कम और करिए स्टूडेंट्स इस प्लेलिस्ट को से करके रख लीजिए आपको क्या करना है इस वीडियो के डिस्क्रिप्शन में जाना है वहां फुल प्लेलिस्ट लिंक इंगेज मैथमेटिक्स फॉर आईआईटी जी मैं इनसे एडवांस दिखाई देगा वहां क्लिक करके प्लस बटन आएगा उसे से करके रखिए ताकि यह प्लेलिस्ट हमेशा आपके पास रहे मैन लो ये दो सर्कल्स थोड़ा बड़ा सर्कल शायद मैंने बना दिया बट ठीक है ना एक और सर्कल है सर जो की टेक्निकल इससे कहीं ना कहीं क्या कर रहा है सर इंटरसेक्ट सही है क्या बिल्कुल डिस्क जॉइंट रहता है या इंटरनल टेस्ट कर रहा होता है एक्स्ट्राऑर्डिनरी टच कर रहा होता है या फिर इसके अंदर होगा बिना टच किए जैसे वो बाहर डिस्ट जॉइंट था उसे तरीके से जहां टच नहीं कर रहे द या फिर अगर वो टच करने के अलावा कोई और सिनेरियो या पॉसिबिलिटी मुझे शो करेगा तो वो ये होगी की भाई ये दोनों सर्कस एक दूसरे को दो पॉइंट्स पर इंटरसेक्ट करें अब जब ऐसा होता है तो क्या-क्या होता है इस बारे में आज हम बात करेंगे बहुत ध्यान से सुनना कई सारे कंक्लुजंस हैं पहले तो क्या आप ये देख का रहे हो की बेशक सर अगर ये दो जगह इंटरसेक्ट कर रहे हैं तो कुछ कुछ तो दिख रहा है जैसे की क्या सुनना ध्यान से सर इस सर्कल का सेंटर का मैं मालूम है यहां C1 और सर्कल का सेंटर मैं मैन लूं यहां से C2 तो क्या एक और कंक्लुजन आपको दिख रहा है की सर C1 और C2 को मिलाया तो एक्स स्ट्रेट लाइन ऐसी मुझे दिखाई देगी बिल्कुल दिखाई देगी सर इसमें तो कोई दिक्कत नहीं है भाई अब इसके अलावा सर अगर आप चीज है थोड़ी ध्यान से देखें तो आप ये नहीं का सकते क्या suniyega ध्यान से की सर C1 से अगर मैंने इसकी ये लेट से मतलब अभी तो मेरे से नहीं ड्रॉ करता हूं अभी तो मैं चीज बस इतनी दिखाने की कोशिश करता हूं पहला कंक्लुजन जब भी कभी मुझे वेरीफाई करना होगा क्या की सर ये दोनों सर्कल्स एक दूसरे को इंटरसेक्ट कर रहे हैं दो पॉइंट्स पर तो मैं हमेशा ये कहूंगा मेरा बेसिक चेक जो होगा जिसके बेसिस पर मैं कनक्लूड करूंगा वो ये होगा की सर बड़ी साधारण सी बात है आप इस सर्कल की रेडियस देखो ये इतनी है जो की ऑफकोर्स कितनी है जो की है ऑफ कोर्स R1 आप सर इस सर्कल की रेडियस देखो जो की इतनी है जो की ऑफ कोर्स कितनी है सर ये है r2 और अगर मैं आपसे पूछूं की आप R1 और r2 का सैम करिए अगर मैं आपसे पूछूं आप R1 और r2 का सैम करिए तो क्लीयरली आप देख का रहे हो सर इन दोनों सर्कल्स जो रेडियस का सैम ज्यादा है इन दोनों के बीच के सेंटर्स के डिस्टेंस है क्या यह बात आप समझ का रहे हो इंदौर सर्कस के सेंटर्स के बीच की डिस्टेंस है ज्यादा है इन दोनों सर्कल्स की इंडिविजुअल रेडियस या राजा सब बड़ी बेसिक्स सी बात है आई थिंक बहुत टू बहुत मुश्किल बातें हैं नहीं हैं इससे क्या कंक्लुजन निकलता है इससे कम की बात जो निकल कर आती है वो ये suniyega बहुत ध्यान से की सर जब भी मुझे कहा जाएगा की दो सर्कल्स इंटरसेक्ट कर रहे हैं एक्जेक्टली दो पॉइंट्स पर ऑफ कोर्स तो उसे केस में मैं कंडीशन अप्लाई करूंगा ऐसी हमने हर केस के लिए निकल है जब सर्कस डिज़ाइन द जब सर्कल से एक्सटर्नल टच कर रहे द जब सर्कस इंटरनली टेस्ट कर रहे द और बेशक जब सर्कस दो पॉइंट्स पर इंटरसेक्ट कर रहे हैं याद रहेगा आगे आते हैं अब नेक्स्ट पॉइंट को हम ऑब्जर्व या एलिमिनेट या फिर आउट करने की कोशिश करते हैं अच्छा एक बात बताओ सर हमने हर बार टांगें की बात की है बिल्कुल की तो इस बार मुझे सोच के बता सकते हो कैसे अटेंडेंट बन रहे होंगे सर इस बार जब आप टेंशन बनाओगे ना तो कुछ ऐसी बन रही होगी इस बार जब आप बनाओगे ना तो यह टैसेंट कुछ ऐसी थी बन रही होंगी ऑफ कोर्स बिल्कुल सही बात है सर ये आपकी एक टैसेंट मिल जाएगी सिमिलरली अगर यहां से टैसेंट आप ड्रॉ करेंगे बिल्कुल तो एक टैसेंट कुछ ऐसी सी भी बन रही होगी क्या इस बात से कोई आपत्ति है भाई बिल्कुल सर ये बात भी आपकी मैन ली ये बात भी अच्छे से आपकी हमने समझ ली की एक टैसेंट ऐसी भी आपकी बन रही होगी बिल्कुल सही बात है पर क्या ये दोनों टांगें तो कैसी है सर ये हमारी डायरेक्ट कॉमन टेंट है पर इस बार क्या मुझे वो ट्रांसवर्सल कमेंट जैसा कुछ मिल पाएगा सर शायद वो मुश्किल है वैसा तो कुछ नहीं मिलेगा पर अगर मैं वैसा कुछ बनाने की कोशिश करो तो मुझे कुछ एक बहुत ही मजेदार सी चीज मिलेगी जिससे मैं अभी फिलहाल कहूंगा की सर इससे हम कहते हैं कॉमन कोड आपको मेरी बात समझ ए रही है मैं फिर से रिपीट करता हूं अपना से पॉइंट के साथ बड़े ध्यान से देखो बड़ा सिंपल सा ऑब्जर्वेशन है अगर आप इस कार्ड को देखें तो ये वो कॉर्ड है जो की टेक्निकल दोनों ही सर्कल्स के अकॉर्डिंग क्योंकि आप देखना सर्कल के लिए भी ये लाइन कट की तरफ कर रही है और इस सर्कल के लिए भी ये लाइन क्वाड की तरह भी है सो दिस लाइन इस बेसिकली बिहेविंग लाइक अन कॉमन कोड और अगर ये इसकी कॉमन चोर्ड है तो मैं एक और कंक्लुजन पर आऊंगा जो की आप में से कुछ स्टूडेंट्स गेस्ट कर चुके होंगे की सर क्या मैं इसे रेडिकल एक्सेस भी का सकता हूं आप फिर से suniyega क्या मैं इसे रेडिकल एक्सेस ठीक का सकता हूं बिल्कुल का सकते हो रेडिकल एक्सरसाइज जानते हो आप लोग रेडिकल एक्सेस मतलब बिल्कुल सिंपल सा पॉइंट एक ऐसे पॉइंट का लॉक जहां से ड्रा की गई टांगें की लेंथ दोनों सर्कस पर हमेशा इक्वल होती है यानी की यह आपके ना सिर्फ एक कॉमन कोड है बल्कि रेडिकल एसेसरीज भी है तो इस लाइन को आप क्या कहोगे इस लाइन को आप ऑफ कोर्स एक स्टेटमेंट तो यही कहोगे की सर ये आपकी क्या है कॉमन कोड बट अलांग विथ डेट अगेन दिस इस गोइंग तू बी डी रेडिकल एक्सेस और सर रेडिकल एक्सेस के थ्रू क्या मैं कॉमन कोड की इक्वेशन निकल सकता हूं अगर यह सर्कल है जिसकी इक्वेशन आप कहते हो S1 इस इक्वल्स तू जीरो इस सर्कल के क्वेश्चन कहते हो S2 = 0 तो कॉमन चोर्ड जो की टेक्निकल यहां पर है आपका रेड कलर एक्सिस विल बी गिव इन की S1 - S2 = 0 नहीं है भाई ये बात आई होप बड़ी आसान सी डायरेक्ट सी सुलझी हुई इसे सिंपल सी बात है और यकीनन तौर पर ये जो कॉमन कॉर्ड होगी इन दोनों सेंटर्स को ज्वाइन करने वाली लाइन पर क्या होगी परपेंडिकुलर क्यों होगी इस बारे में तो कुछ कहने की जरूरत नहीं है एक और बात आप जान रहे हो क्या एक और जरूरी बात आप जान रहे हो क्या की सर ये जो डायरेक्ट कॉमन टांगेंट्स है यह जो डायरेक्ट कॉमन टांगें से जहां भी आगे जाके मिल रही होंगे कन्वर्ज हो रही होगी लेट्स कॉल डेट पॉइंट दी कन्वेंशनली आते अन एडीशनली अभी तक दी कहते आए हैं उसे तो जहां वो कनेक्ट हो रही है जहां हम इंटरसेक्ट कर रही है वो इस C1 C2 को ज्वाइन करने वाली लाइन पर ही इंटरसेक्ट करेंगे दोनों डायरेक्ट committents और C1 और C2 को ज्वाइन करने वाली लाइन कौन करंट होती है और वो हम जानते हैं C1 C2 को इंटरसेक्ट करती है सर वो लाइन वो पॉइंट जो डायरेक्ट C1 और C2 को R1 में इंटरसेक्ट करता है यह बात हम कई बार कई दफा डिस्कस कर चुके हैं आई होप यह स्टेटमेंट आपको याद है बहुत सारी कृष्ण बातें इस दौरान मैंने कहीं आई होप आपने इन सारी बातों को नोट डाउन किया और इनको अच्छे से ब्रेन में रजिस्टर किया सर समझ में तो सब ए गया पर समझ में आने के साथ जरूरी है की सारी चीज आप नोट डाउन करें नोट डाउन करें मतलब नोट्स बनाएं इन सारी चीजों को अच्छे से नोट डाउन करेगा स्टूडेंट्स यही काफी कृष्ण कम की बात है जिन पर आज हम बात करेंगे आई थिंक चीज आसान है चीज मुश्किल नहीं है अब क्या-क्या कंक्लुजंस निकलेंगे कैसे क्वेश्चन सॉल्व करेंगे क्या-क्या मेकैनिज्म या हमारी यू नो स्टैंडर्ड ऑपरेटिंग प्रोसीजर होगी की हम कैसे चीजें सोचेंगे क्वेश्चंस कैसे सॉल्व करेंगे इस बारे में बात करते हैं चलिए एक-एक करके चीजे देखते हैं अभी तक जो हमने डिस्कस किया मेजर्ली किया पहला पार्ट तो यही सर क्या करिए इंटर सेट करेंगे तो हम बिल्कुल यह देख पाते हैं सर की दोनों सर्कस के सेंटर्स के बीच के डिस्टेंस उनकी रेडियस के हिसाब से कम होगी ये डेफिनेटली ट्रू है आई होप यहां तक कोई तकलीफ नहीं है एक और बात ऑब्जर्व करोगे आप की सर दोनों सर्कस के सेंटर्स के बीच के डिस्टेंस उन दोनों रेडियस के डिफरेंस से ज्यादा होगी है ना इस केस में अगर ऐसा नहीं हो ऐसा नहीं हो तो फिर सर ये भी हो सकता है क्यों सर्कल पूरा ही अंदर ए जाए अब बात समझो अगर ये नहीं होता अगर ये नहीं होता आप समझना बात को की सर अगर मैं R1 में से suniyega ध्यान से अगर मैं R1 में से r2 को सब्सट्रैक्ट करूं अगर मैं R1 में से r2 को सब्सट्रैक्ट करू तो देखो भाई क्या मिलेगा मुझे सीधा-सीधा सा पॉइंट मिलेगा ध्यान से देखना 11 और ये r2 है तो R1 में से जब r2 सब्सट्रैक्ट किया तो shailise है ये देखो कितनी लेंथ चली जाएगी वो समझना जब R1 में से r2 से अट्रैक्ट कर रहे हो तो एक तो आप इतनी लेंथ कवर बैठ रहे हो आप समझ रहे हो और इतनी लेंथ और गवा बैठे हो क्या आप पॉइंट ऑब्जर्व कर का रहे हो और जब इस पॉइंट को आप ऑब्जर्व कर रहे हो तो आप इस कंक्लुजन पर ए रहे हो सर बिल्कुल उन दोनों के बीच के सेंटर्स की डिस्टेंस R1 और r2 से ज्यादा होनी चाहिए अगर ऐसा नहीं होगा तो उसे केस में उसे केस में पता क्या होगा उसे केस में ये हो रहा होगा की ये सर्कल पूरा इसके अंदर ए जाएगा हम बात करेंगे इस पॉइंट पर और आई होप ये दोनों बातें आपने समझी यानी की जब दो सर्कल्स इंटरसेक्ट करें जब दो सर्कल्स इंटरसेक्ट करें तो दो कंडीशंस फुलफिल होना जरूरी ना सिर्फ ये की दोनों के सेंटर्स के बीच के डिस्टेंस R1 + r2 क्योंकि आप बात समझो आप बात समझो मतलब दोनों के सेंटर्स की डिस्टेंस R1 + r2 से कम क्योंकि बात समझो अगर मैं आपको एक और सिनेरियो समझता हूं ध्यान से देखो आप खुद इस बात से बहुत स्ट्रांग्ली एग्री करेंगे अगर आप इस चीज को समझेंगे तो देखो ये है मैन लो एक आपका सर्कल और इसके अंदर लेट्स से देयर इस वैन मोर सर्कल है ना अब आप खुद सोचो यह सर्कल का सेंटर है जिसे हम कहते हैं C1 और यह इस सर्कल का सेंटर जैसे हम कहते हैं C2 अब बात समझना स्टूडेंट्स बहुत ध्यान से सुनना है स्टेटमेंट को सर में अगर आपसे पूछूं तो इस सर्कल की बड़े वाले सर्कल की रेडियस ऑफ कोर्स कितनी होगी R1 जो की कौन सी डिस्टेंस होगी जो की ये वाली डिस्टेंस होगी समझ का रहे हो क्या और सिर्फ कोर्स जो ये छोटा वाला सर्कल है इसकी रेडियस कितनी है सर इसकी रेडियस ये है जो की ऑफ कोर्स क्या r2 कोई तकलीफ तो नहीं है और क्लीयरली आप इस सर्कल में भी कंडीशन सेटिस्फाई होती हुई देख पाओगे कौन सी कंडीशन की सर सीधी-सीधी सी बात है ध्यान से देखो इस कंडीशन को की सर सीधी-सीधी सी बात है जो दोनों के सेंटर्स के बीच के डिस्टेंस है वो दोनों की रेडियस से कम है आई होप आप देख का रहे हो बड़ी बड़ी ऑफिस बड़ी appearansi बात है की दोनों के सेंटर्स के बीच की जो डिस्टेंस है वह दोनों की रेडियस के हिसाब से डेफिनेटली कम है अब मेरी बात डाइजेस्ट कर का रहे हो तो सिर्फ यह सफिशिएंट और मैंडेटरी कंडीशन नहीं है की सर दो सर्कल्स दो पॉइंट्स पर इंटरसेक्ट करें ये बेशक वैन ऑफ डी कंडीशन से बट इसके साथ एक और कंडीशन लगेगी की दोनों का डिफरेंस से ज्यादा होनी चाहिए वर्ण फिर बाद बिगड़ जाएगी वर्ण फिर बाद बिगड़ जाएगी और यहां पर यही हो रहा है की दोनों के डिफरेंस है वो ज्यादा नहीं रिपीट मैं स्टेटमेंट जो दोनों के बीच के सेंटर्स के डिस्टेंस है वो इन दोनों के डिफरेंसेस ज्यादा नहीं है इस केस में क्या ये बात आपको याद रहेगी अगर यहां तक चीज याद रहेंगे तो थोड़ा आगे बढ़ते हैं अब अगले पॉइंट पर तो मेजर्ली जो कंबाइंड या क्लब डिक्टेशन है या कंडीशन है वो आप ये अप्लाई करेंगे जब-जब आपसे इस पर सवाल पूछे जाएंगे की दो सर्कल्स इंटरसेक्ट कर रहे हैं याद रहेगा क्या चलो आगे बढ़ो इन दिस केस ओनली डायरेक्ट कॉमन टांगेंट्स कैन सी ड्रा हम बात कर चुके हैं की सर एक मतलब दो आप डायरेक्ट कमेंट्स से ड्रा कर पाते हो और एक और बात जो हम शुरुआत से डिस्कस करते हैं की जहां यह डायरेक्ट कमेंट एंड रिसीव करेंगे अगर लेट से उसे हम पॉइंट को दी कहें तो दी पॉइंट सी वैन और C2 को ज्वाइन करने वाली लाइन को R1 इस तू r2 के रेश्यो में एक्सटर्नल डिवाइड करेगा ये बात हम शुरुआत से करते हैं इसमें कोई तकलीफ कोई परेशानी कोई इशू नहीं होना चाहिए अब और क्या एक बात जो हमें निकल कर आती है जो हमें समझ आणि चाहिए वो बड़ी सिंपल जी की जब भी मुझसे पूछा जाए जब भी मुझसे पूछा जाए की सर ये दोनों सर्कल्स की कुछ ना कुछ रेडियस होगी और अगर दोनों सर्कल्स की रेडियस से हुई अगर दोनों सर्कस की रेडियस से है तो ये बात याद है ना की वो उनकी जो डायरेक्ट कमेंट टैसेंट है वो पैरेलल चलती चली जाएगी और अगर वो डायरेक्ट कमेंट टैसेंट पैरेलल चलती चली गई तो उसे केस में क्या वो दोनों इंटरसेक्ट करेंगे बिल्कुल नहीं तो उसे केस में फिर दी एक्जिस्ट ही नहीं करेगा ये तो बड़ी आसान सी बात है और क्या सर और बड़ी इंपॉर्टेंट सी क्रोशिया से बात आय होप ये हम देख का रहे हैं की जब कोई सा भी सिनेरियो हो इस इस सिनेरियो को जब हम अप्लाई करेंगे जब जब मुझे ऐसी कंडीशंस मिलेगी तो मैं हमेशा ही ऑब्जर्व करूंगा की दोनों सर्कल्स दो पॉइंट्स पर इंटरसेक्ट कर रहे हैं और उन दोनों सर्कस का अगर मैं पॉइंट ऑफ इंटरसेक्शंस मैन लूं ए और बी तो उसे पॉइंट इंटरसेक्शन जो ए और बी बनता है तो वो जो अब बनती है हम उसे देख का रहे हैं वो हमारे लिए कॉमन कोड है है और हमने यह भी प्रूफ किया है की वह ना सिर्फ हमारे लिए कॉमन कोड है बल्कि वह हमारे लिए एक रेडिकल एक्सेस की तरह भी बिहेव कर रही है और रेडिकल एक्सेस की इक्वेशन मेरे ख्याल से सर मैं तो निकलना जानता हूं एस वैन माइंस S2 = 0 कोई दिक्कत कोई परेशानी तो नहीं है क्या यह स्टेटमेंट आप सभी को समझ आया आई थिंक रेडिकल एक्सेस का एक डिटेल डिस्कशन हमने इस लेक्चर में किया है बट इस बात को याद रखना क्योंकि इस पर काफी सारी चीज आएंगे कैसे आएंगे बात करेंगे पहले से याद रखना और क्या सर अब अगर मैं बात करूं डायरेक्ट कॉमन टेंट की लेंथ तो सर इसको तो मत ही पढ़ो ये हमने पढ़ा है याद ए रहा है क्या सर डायरेक्ट कॉमन टांगें की लेंथ जो होती है वो आपके दोनों सेंटर्स के बीच के डिस्टेंस का स्क्वायर - दोनों सर्कल्स की रेडियस के डिफरेंस का स्क्वायर का अंडर रूट होती है और जो वो ट्रांसफर कमेंट आया करती थी जो इस सिनेरियो में कॉस्ट नहीं बन रही है जब वो दोनों जॉइंट सर्कल हुआ करते द तो उसे केस में बस ये फर्क ए जया करता था की यहां पर R1 प्लस r2 होता था याद है कर रही हूं यह बहुत कृष्ण स्टेटमेंट से स्टूडेंट जो प्लीज आप याद रखिए इन्हें अच्छे से नोट डाउन करेगा ये बातें हम कई बार कर चुके हैं और इसे अब अच्छे से आप याद कर लेंगे ऐसा मेरा मानना है है ना पर चूंकि हम हर अलग-अलग सिनेरियस में दो सर्कल्स को कंसीडर कर रहे हैं तो हर बार उन पर बनने वाले केसेस को एक्सप्लोर करना जरूरी है याद रहेगा चलिए आगे बढ़ते हैं स्टूडेंट्स कम की बात पर आते हैं अब मैन लो मुझसे कभी भी कोई पूछे की पॉइंट ऑफ इंटरसेक्शन कैसे निकलना है suniyega बहुत कम की बात है सर आपके पास लेट से ही एक सर्कल है और लेट्स हैव वहां पर एक और दूसरा सर्कल भी है है ना और इन दोनों सर्कल्स का सर आपसे पूछा जाए पॉइंट ऑफ इंटरसेक्शन निकलना तो मैं कहूंगा सर इस सर्कल का इससे सेंटर यह था और इस सर्कल का सेंटर यह है अब इनका पॉइंट ऑफ इंटरसेक्शन क्या है सर यह जो पॉइंट है और यह जो पॉइंट है इन्हें आप कहेंगे इनका पॉइंट ऑफ इंटरसेक्शन जैसे-जैसे मैं कहूं ए और इसे हमें कहूं बी क्या तरीका हो सकता है क्या तरीका हो सकता है क्या सर इन दोनों सर्कल्स की इक्वेशन को सॉल्व कर ले बेशक आप कर सकते हैं पर मैं कहना चाह रहा हूं जैसे मैन लो अब जानते हो की सर ये जो सर्कल है इसकी इक्वेशन क्या है S1 ये जो सर्कल है इसकी इक्वेशन क्या है S2 suniyega बहुत कम की बात कहने वाला हूं मैं क्या ऐसा का सकता हूं सर जो इन दोनों सर्कल्स का पॉइंट ऑफ इंटरसेक्शन है भूल जाओ सारी बातें मुझे तो आप यह बताओ की सर क्या मैं यह जानता हूं की इनकी कॉमन कोर्ट की ये कोई इक्वेशन होगी ये वो कॉमन कॉर्ड होगी और इस कॉमन कोड की इक्वेशन हमने लिखना सिखा है सर होती है S1 - S2 = 0 सर खा रहे हो अब बात समझो स्टूडेंट अलजेब्राइकली आपको समझने की कोशिश करता हूं बहुत कम का ये कंक्लुजन है इससे बहुत मजेदार सा फेनोमेना आप देखोगे सर मैंने आपका सर्कल है जैसे आप कहते हो x² + y² + 2 वैन बाय प्लस सी वैन ऑफ कोर्स बता सकते हो सर कौन सी बड़ी बात है कॉमन कोड क्या हो जाएगी सर वो हम निकल लेंगे s1-s2 से तो यहां से जब आप निकलोगे तो वो कॉमन कॉर्ड यानी लीनियर इक्वेशन ए जाएगी अब एक बात ऑब्जर्व करो आप एक बात को ऑब्जर्व करो आपने इस स्ट्रेट लाइन की इक्वेशन निकल ली बिल्कुल निकल ली सर और आप इसमें से कोई भी एक सर्कल को उठा लो चाहे इससे और चाहे ऐसे मैं आपसे ये कहना चाह रहा हूं की अगर मैं इस स्ट्रेट लाइन और इस सर्कल को सॉल्व करूंगा तो भी तो मुझे दो पॉइंट्स मिलेंगे क्यों मिलेंगे सर क्योंकि ये स्ट्रेट लाइन भी कोई सा भी एक सर्कल उठा लो उसे दो पॉइंट्स पर इंटरसेक्ट करती है डाइजेस्ट हुई क्या जो बात मैंने कही भाई मैं कहना चाह रहा हूं यह जो स्ट्रेट लाइन है वैसे दोनों बट कोई भी एक ले लो उससे भी तो दो पॉइंट्स पर इंटरसेक्ट करती है और अगर ऐसा हो रहा है तो मैं बड़ी आसान सी बात कहूंगा सर बड़ी ही फिल्म सिंपल सी शॉर्टकट सी डायरेक्ट सी बात कहूंगा की सर देख क्यों नहीं रहे हो आप समझ क्यों नहीं रहे हो की सर सर देखो भाई ए बी से पास होने वाली है स्ट्रेट लाइन सर्कल को जब दो पॉइंट पर इंटरसेप्ट करेगी तो दो सर्कल्स की इक्वेशंस को सॉल्व करने से अच्छा होगा की उसे स्ट्रैटेजें और सर्कल की इक्वेशन को सॉल्व कर लें यानी एक तो आपकी इक्वेशन बन जाएगी S1 - S2 = 0 यहां से क्या बन जाएगा सर S1 - S2 = 0 और इनमें से कोई सा भी एक सर्कल ले लेना यहां पर ऑफ कोर्स एक्स और ए में एक अलजेब्राइक लीनियर एक्सप्रेशन होगी तो वहां से एक्स को या ए को किसी दूसरे की टर्म्स में एक्सप्रेस कर देना और उसे या तो यहां रख देना तो ये पुरी इक्वेशन एक्स में या ए में क्वाड्रेटिक बन जाएगी जैसे आप सॉल्व करके दोनों पॉइंट ऑफ इंटरसेक्शन निकल सकते हो व्हाट कन्फ्यूज्ड तो नहीं कर रहा हूं भाई बहुत आसान सी बात का रहा हूं यह बात न्यूमेरिकल मैं और अच्छे से समझा दूंगा आपको एक क्वेश्चन करवा कर चिंता मत करिए हम क्वेश्चन करेंगे पहला तो मैं आपको यह स्टेटमेंट समझाना चाह रहा हूं की अगर मुझसे कभी भी कोई पूछे की दो सर्कल्स का पॉइंट ऑफ इंटरसेक्शन निकलना है तो इंसटिड ऑफ सॉल्विंग उन दोनों सर्कल्स की इक्वेशन को टुगेदर जो की डायरेक्टली एंड डायरेक्टली मैं वही कर रहा हूं सर्टेनली वही हो रहा है लेकिन उसको करने का तरीका हमने ये आजाद किया की सर इन दोनों की कॉमन टांगें सॉरी कॉमन कोड जो की रेडिकल एक्सिस भी है जो की हो जाएगा S1 - S2 = 0 और उससे और इस सर्कल की इक्वेशन को अगर मैं साथ में सॉल्व कर लूं तो उससे भी तो पॉइंट ऑफ कॉन्टैक्ट या पॉइंट ऑफ इंटरसेक्शन टेक्निकल निकले जा सकते हैं बहुत कन्ज्यूरिंग सी लग रही हो बट मेरा यकीन करिए जब हम क्वेश्चंस करेंगे दें आई विल मेक यू रिलाइज की बहुत आसान कॉन्सेप्ट है जैसा बेसिकली हम दो कप स्कोर इंटरसेक्ट कर रहे हैं तो पॉइंट ऑफ इंटर सेक्शन कैसे निकाला जाता है सर एफ ऑफ एक्स को जिओ ऑफ एक्स की इक्वल रख रखा कर यही बात यहां पर सूची जारी है की पहले तो आपने क्या किया S1 - h2 किया तो आपकी यह जो कॉमन कोड है या फिर जो रेडिकल एक्सिस की इक्वेशन ए गई और चूंकि ये इनमें से किसी भी एक सर्कल के साथ ले लीजिए दो पॉइंट्स में इंटर सेट कर रहा है तो उसे रेडिकल एक्सिस और सर्कल के क्वेश्चन को सॉल्व करेंगे तो उससे आपके दोनों पॉइंट ऑफ इंटरसेक्शन ए जाएंगे यकीन करिए क्या ये बात आपको समझ आई सर ये बात तो समझ ए गई लेकिन एक कम और करके दे सकते हो क्या बिल्कुल करेंगे सर क्या आप इस अब के लेंथ निकल सकते हो सर क्या आप इस अब की लेंथ निकल सकते हो कैसे निकलेंगे अभी की लेंथ कैसे निकलेंगे सुनना कितना आसान है निकलना सोच के देखो भाई अगर मैं इस ए को इस सेंटर से कनेक्ट कर डन चाहो तो आप इससे भी कर सकते द मुझे कोई आपत्ति नहीं है और अगर इस सेंटर से मैं इस अब पर हम रिलीज सॉरी अगर मैं इस सेंटर से अब पर एक परपेंडिकुलर ड्रॉप कर दो suniyega ध्यान से इस सेंटर से अगर मैं अब पर एक परपेंडिकुलर ड्रॉप कर दो आप सन रहे हो क्या सुनेगा ध्यान से हमने क्या किया है गौर फरमाइए स्टूडेंट्स यह जो सर्कल है इसका सेंटर है C2 और इसमें जो परपेंडिकुलर ड्रॉप किया और सर्कल के सेंटर से हमने इस पर ड्रॉप किया है परपेंडिकुलर तो रिकॉर्ड हो जाएगी अब कम की बात suniyega जनाब एक तू क्या है सर A2 इस नथिंग बट सर्कल सर निकलना है अगर निकल गया तो एम का डबल कर देंगे अब ए जाएगा अच्छी बात है सर लेकिन mc2 कैसे निकलेंगे क्या कोई स्टूडेंट मुझे बता सकता है की सर mc2 कैसे निकलेंगे आइडिया सर आप निहायती बेवकूफी भारी बात पूछ रहे हो यह आज का शायद सबसे आसान सवाल होगा क्योंकि अब हम चीज बहुत अच्छे से पढ़ चुके हैं मुझे पता है C2 के कोऑर्डिनेट्स क्या हैं C2 सेंटर है जिसके कोऑर्डिनेट्स हैं माइंस जी तू कमा - f2 और इस स्ट्रेट लाइन की इक्वेशन मुझे पता है s1-s2=0 तो क्या आपको ये नहीं पता की सर किसी पॉइंट से किसी स्ट्रेट लाइन पर परपेंडिकुलर ड्रॉप करें तो उसकी परपेंडिकुलर लेंथ क्या होती है पता है की नहीं तो सर इस पॉइंट से इस स्ट्रेट लाइन की परपेंडिकुलर लेंथ निकल लेंगे जो की होगा आपका c2m आप सन रहे हो बहुत ध्यान से सुनना प्लीज यह बहुत कम की बात है तो जैसे ही हमारा c2m ए जाएगा हमें r2 पता है तो क्या मैं पाइथागोरस थ्योरम अप्लाई करके हाइपोटेन्यूज का स्क्वायर माइंस परपेंडिकुलर का स्क्वायर कर कर क्या मैं बेस का स्क्वायर नहीं निकल सकता और उसका अंडर रूट लेकर उसे डबल कर देंगे तो क्या अब नहीं ए जाएगा बहुत हैवी तो नहीं हो रही है बातें आई होप मेरी बातें मैं आपको समझा का रहा हूं मैंने क्या कहा मैंने कहा आपसे की ये जो एक तू है एक तू है यह तो क्या हो जाएगी भाई ये हो जाएगी सर रेडियस निकल लोग क्या इसके अलावा जब सी तू एम निकलोगे तो c2m क्या हो जाएगा सर c2m इस नथिंग बट व्हाट सर्कल के सेंटर से इस लाइन पर ड्रॉप की गई परपेंडिकुलर लेंथ आई होप ये भी निकल जाएगी यहां पर मैं क्या करूंगा इस पर मैं पाइथागोरस थ्योरम अप्लाई करूंगा तो उससे एम का स्क्वायर ए जाएगा लेकिन एम के स्क्वायर में से मुझे एम से मतलब है तो उसका माइंड रूट ले लूंगा पाइथागोरस थ्योरम और उन रूट लेने के बाद मुझे मिल जाएगी एम की वैल्यू और उसका अगर मैं डबल कर दूंगा तो मुझे अब की वैल्यू मिल जाएगी क्या आपको अब कार्ड की लेंथ निकलने की मॉडेस्ट ऑपरेटिंग ये जो प्रक्रिया है ये समझ ए रही है बहुत डिफिकल्ट तो नहीं है भाई आई होप ये बात याद रखोगे ये स्टेप बाय स्टेप नोट डाउन कर लेना मैंने आपको अभी वर्बल समझाया है क्वेश्चन के थ्रू मैं ऑर्गन्स ऑल एडिट करवाऊंगी peshup नहीं सैम कोच एंड है डेटिंग के लिए हमेशा अच्छे से समझेंगे पर फिर भी अभी गुजारिश यही है आपसे की आप इस स्टेप को इन सारी प्रक्रिया इसको अच्छे से नोट डाउन करके चीजें क्लियर है और चीज अच्छे से समझ ए रही हैं और यही सारी बातें यहां लिखी हैं देखो क्या लिख रहा है तू फाइंड डी पॉइंट्स ऑफ इंटरसेक्शन ऑफ डी सर्कल सी सॉल्व वैन ऑफ डी सर्कल एंड डी कॉमन कोड डेट इस रेडिकल एक्सिस यही बातें हमने की है एंड ऑफ कोर्स यूजिंग डिस्टेंस फॉर्मूला सी कैन फाइंड डी लेंथ ऑफ कॉमन कोड इस वेल कैसे सर सोच के देखना हमें पता है सर की आपके ट्रायंगल की रेडियस कितनी है सर्कल के सेंटर से उसे कोड पर ड्रॉप किए गए परपेंडिकुलर की लेंथ कितनी है तो उसे कार्ड की लेंथ का हाफ मुझे मिल जाएगा और टेक्निकल उसे लेंथ का अगर मैं डबल कर दूंगा तो क्या मुझे फाइनली आपको वो चीज देने को मिल जाएगी जो आप पूछ रहे हो या नहीं की इस कॉल्ड की लेंथ इस कॉमन कोड की लेंथ आई होप बहुत डिफिकल्ट टास्क के बहुत मुश्किल प्रक्रिया नहीं है और इसी बात को अगर मैं कनक्लूड करूं इसी बात को मैं कनक्लूड करूं तो हम इन कुछ रिजल्ट्स फॉर पहुंचने हैं अब मैं फाइनल बात करने वाला हूं की दो सर्कल्स के बीच एंगल कैसे निकाला जाता है पर अभी तक जो बातें पढ़ी और मैं तो कोई दिक्कत नहीं है यकीन करिए इन सारे कॉन्सेप्ट्स पर एक साथ लगातार मेरे करके मंगल करके बेहतरीन से अच्छे लेवल ऑफ क्वेश्चंस हम करेंगे पर पहले हम वो सारे cinerios एक्सप्लोर कर ले रहे हैं जब दो सर्कल्स इंटरसेक्ट करते हैं तो क्या-क्या होता है जब दो सर्कल्स दो पॉइंट्स पर इंटरसेक्ट करते हैं तो क्या-क्या चीज निकल कर आती हैं जैसे अब एक और कॉन्सेप्ट निकल कर आता है जिसे आप कहते हो एंगल बिटवीन तू सर्कल्स अब यह क्या होता है सर एंगल बिटवीन तू सर्कल से बहुत ही मजेदार सा कॉन्सेप्ट है suniyega ध्यान से कम की बात है एंगल बिटवीन तू सर्कल्स का मतलब है प्लीज गौर से सुनना ये बहुत क्रशर स्टेटमेंट है इसमें स्टूडेंट्स गलती करते हैं एक सर्कल ये रहा आपका बिल्कुल सर अच्छी बात है और एक और सर्कल लेते हैं यहां के ही रहा ये क्या इस बात से कोई तकलीफ एंगल बिटवीन तू सर्कल्स का मतलब है सर आप एक कम करिए सिमिलरली इस सर्कल के जो पॉइंट ऑफ इंटरसेक्शन है उसे पर भी आप एक टैसेंट ड्रा कर दीजिए कर दिया क्या आई थिंक कर दिया अब ये जो tesence आपने ड्रा किया इनके बीच जो एंगल है इनके बीच में जो एंगल डेट एंगल बिटवीन तू आई डिड नॉट कन्फ्यूज्ड ऑल आई रिपीट माय स्टेटमेंट एंगल बिटवीन तू सर्कल आप क्या करते हैं आप उनके पॉइंट ऑफ इंटरसेक्शन पर आप उनके पॉइंट ऑफ इंटरसेक्शन पर दोनों सर्कल की टांगें ड्रॉ करते हैं और और उन टेंसेज के बीच का एंगल को हम कहते हैं एंगल बिटवीन तू सर्कल जैसे मैन लो ये सर्कल है इसका सेंटर है C1 और ये सर्कल के क्वेश्चन है एस वैन ये सर्कल है जिसका सेंटर है C2 और इसकी इक्वेशन जानते हैं हम दो बहुत जरूरी सी बात ये जानते हैं सर की इस सर्कल से अगर मैंने इसके यहां पर कनेक्ट किया तो ये इस सर्कल की रेडियस हो जाएगी और ऑफ कोर्स इस सर्कल से इसकी पैरामीटर या सरकम्फ्रेंसेस को कनेक्ट किया तो ये इस सर्कल की रेडियस हो जाएगी कोई दिक्कत तो नहीं है नहीं है सर अब एक बात बताओ स्टूडेंट्स बड़ी जरूरी सी बात बताओ जब इस सर्कल से इसकी टांगें पर परपेंडिकुलर ड्रॉप किया तो आई थिंक वो रेडियस होगी जब इस सर्कल के सेंटर से इसकी टेंशन पर ड्रॉप किया तो वह इसकी रेडियस होगी और हम क्या निकलने चले हैं सर हम निकलने चले हैं यह वाला एंगल जो की है आपका क्या θ आई होप यहां तक किसी भी स्टूडेंट को कोई आपत्ति नहीं है किसी भी बात से कोई परेशानी नहीं है एक छोटी सी बात का जवाब दो स्टूडेंट्स प्लीज एक छोटी सी बात का जवाब दो क्या थोड़ी देर के लिए मैं इस एंगल को अल्फा और इस एंगल को बिता कहूं तो आपको कोई परेशानी तो नहीं है मैं क्या कहना चाह रहा हूं मैं इस एंगल को अल्फा कहना चाह रहा हूं और इस एंगल को यह जो यहां पर एक छोटा सा एंगल आपको जो दिख रहा होगा इस एंगल को मैं बिता कहना चाह रहा हूं क्यों का रहे हो तो थोड़ा सा पेशेंस रखिए एक बात का जवाब देना एक बात का सोच समझकर जवाब देना स्टूडेंट्स मैं अगर इस थीटा तक पहुंचना चाह रहा हूं तो मैं एक ट्रायंगल उसे करना चाह रहा हूं जिसे मैं कहता हूं की सर C1 और C2 को लेट्स हैव मिला देते हैं हम C1 और C2 को मिला देते हैं अब C1 और C2 को मिलाया और इनके पॉइंट ऑफ इंटरसेक्शन को नाम दे दिया ए तो अब आप जो भी सारे मेकैनिज्म होते हुए देखोगे वो है ए सी वैन और सी तू में क्या होगा इस बारे में बात करते हैं पहले इतना अच्छे से समझ लीजिए वो देख लीजिए अगर कोई डाउट है तो पूछ लीजिए देखो अब आप एक बहुत जरूरी ऑब्जर्वेशन देखिए इस जरूरी ऑब्जर्वेशन को कैसे देखेगा बात समझो सुनना मैं मैन लेता हूं ये जो टैसेंट है थोड़ी देर के लिए आपको समझने के लिए इसको का देता हूं और इस टांगें को का देता हूं टी - अब सुनना ध्यान से आप एक बात बताओ आप एक बात बताओ की सर ये जो C1 और ए है इसने टी के साथ जो एंगल बनाया अपने से कहा है अल्फा बिल्कुल मैंने इसे कहा है अल्फा अब कम की बात सुना अच्छा ये जो टी दश है ये जो टी दश है इस पे ये जो पूरा एंगल है ये जो टी दश है इस पे जो पूरा एंगल है ऑफ कोर्स 18 डिग्री है के बात आप 18 में से 90 तो यहां चला गया तो कितना बचा होगा वो होगा 90 - θ कोई तकलीफ तो नहीं है इसी आधार पर इसी बेसिस पर एक और बात ध्यान से सुनना अगर पिछली बार क्लियर नहीं हुआ तो आप समझ ए जाएगा सर देखो अगर मैं स्टील लाइन की बात करूं तो ये आपकी लाइन है टी जिसका पूरा एंगल होगा 180° है उसमें से यह 90 तो यहां चला गया तो अब उसे 180 में से 90 गए तो बचा कितना 90 और वह 90 कौन-कौन है सर थीटा और बिता का समय की अगर मैं आपसे फिर से वही बात dohraun तो मैं क्या ऐसा नहीं कहूंगा की सर जो बिता होगा वो होगा 90 - θ क्या आप मेरी बातें समझ और सिख का रहे हैं ये क्यों का रहे हो सर अब एक बात सुना अगर मैं अब एक ट्रायंगल लेता हूं कौन सा c1ac2 बहुत कृष्ण बात है ये बात बहुत कम आने वाली है इस चैप्टर के दौरान जब आपको दो सर्कल्स के बीच का एंगल निकलना होगा suniyega ध्यान से अगर मिल जाता तो ट्रायंगल c1ac2 इफ आई टॉक अबाउट डी ट्रायंगल कौन सा c1ac2 क्या आप देख का रहे हो C1 ए C2 में अगर सर आप ध्यान से देखो तो क्या आप हमें बता सकते हो अच्छा बाय डी वे सारी बातें छोड़ो ट्रायंगल पर मैं आऊंगा उससे पहले बस मेरा ऐसा ही मानो हो रहा है मेरा ऐसे ही बस मैन हो रहा है की मैं निकलूं अल्फा प्लस थीटा प्लस बिता हम बात करेंगे ऐसा क्यों निकाला आपको अभी दिख जाएगा की ऐसा क्यों निकाला बट मैं पहले से ही निकल के रख लेना चाह रहा हूं मैं निकलूंगा अल्फा प्लस थीटा प्लस बिता मतलब मैं c1ac2 एंगल निकलना चाह रहा हूं आप समझ का रहे हो मैं C1 एक तू एंगल निकलना चाह रहा हूं बाय डी वे दिस हज नथिंग तू डू दिस हज नथिंग तू डू विथ दी एंगल बिटवीन डी तू सर्कल्स वो तो थीटा है पर उसे तक ही हम पहुंचेंगे पहले से सुनिए अल्फा कितना था सर अल्फा था 90 - θ तो ये हो जाएगा 90 - θ ये है थीटा और थीटा कितना था सर बिता था 90 - θ सो टेक्निकल अल्फा प्लस हिट आप प्लस बिता विच इस क्या विच इस एंगल c1ac2 विच इस एंगल C1 ac2 इसे गोइंग तू बी 90 + 9180 ये थीटा से थीटा कैंसिल सो दिस इस गोइंग तू बी 180 - θ कोई तकलीफ अच्छा इन सारी बातों को छोड़ो आप तो वापस आओ एक ट्रायंगल के बारे में थोड़ी बातें करते हैं मुझे आपसे एक ट्रायंगल को लेकर बातें करनी है अगर आप थोड़ा सा सॉल्यूशन ऑफ ट्रायंगल को जानते हैं प्रॉपर्टीज ऑफ ट्रायंगल चैप्टर को जानते हैं तो अगर मेरे पास सर एक ट्रायंगल है लेट्स से ए बी सी तो आप इस बात को ध्यान से सुनना जैसे अगर मैं इस एंगल का कैसे लेता हूं अगर मैं इस ए एंगल का कोस लेता हूं तो आई होप आप पहले तो यह जानते हो की सर कन्वेंशंस से हमें जाते हैं एंगल ए के अपोजिट की साइड को स्मॉल ए एंगल बी के अपोजिट के साइड की लेंथ को स्मॉल भी और वर्टेक्स सी के अपोजिट की साइड की लेंथ को आप स्मॉल सी से दिनो करते हो तो ये जो कोस ए आप लेना चाह रहे हो ना आप इसे लिखोगे आप इसे लिखोगे बी स्क्वायर प्लस सी स्क्वायर माइंस ए स्क्वायर अपॉन तू बी सी अपॉन तू ई यह होता है आपका कोसिन इस बारे में डिटेल डिस्कशन करेंगे जब हम सॉल्यूशन ऑफ ट्रायंगल या फिर प्रॉपर्टीज ऑफ ट्राएंगल्स पढ़ेंगे बट अभी प्लीज इस बात को याद रखना आई होप आप कोसिन अगर मैं आपसे वापस बात करूं थोड़ी सी यहां पर तो अगर मैं यहां कोसिन रूल अप्लाई करना चाह रहा हूं तो मैं कोसिन रूल किस में अप्लाई करना चाह रहा हूं c1ac2 में एंगल c1ac2 पर तो मैं क्या अप्लाई करना चाह रहा हूं कोर्स क्या c1ac2 अब C1 ac2 क्या हो जाएगा सर C1 ac2 आपका ये एंगल है जो की कितना है जो की है 180 - θ बिल्कुल सही बात है सर ये किसकी इक्वल होगा सीधी-सीधी सी बात मैं कोस रूल कैसे याद रखता हूं शेयर करता हूं स्टूडेंट्स मेरा तरीका है की सर कोस ए निकलना है ना तो कोस ए यानी एंगल को बनाने के लिए कौन सी दो साइड्स लगी बी और सी तो उनके स्क्वायर का डिफरेंस डिवाइडेड बाय इन दोनों के डबल का कोई दिक्कत तो नहीं इन दोनों के प्रोडक्ट का डबल यानी की सर अगर यहां पर आपको साइन रूल अप्लाई करना चाह रहे हो तो आप कहोगे सर आप इस एंगल का आपके इस एंगल का निकल रहे हो आप इस एंगल का यू नो कॉस्ट ले रहे हो इस ट्रायंगल में तो मैं कैसे सोचूंगा स्टूडेंट्स पहले तो इस एंगल के क्रिएट करने के लिए जो दो साइड्स लगे वो कौन-कौन थी थी C1 ए और c2a और C1 ए और सी तू ए टेक्निकल है क्या सर C1 ए इस दी रेडियस ऑफ दिस सर्कल c2a इस दी रेडियस ऑफ दिस सर्कल तो C1 ए मतलब क्या मतलब r1² c2a मतलब क्या मतलब r2² - जो अपोजिट साइड है जो इस एंगल के अपोजिट साइड है वो कौन है सर वो है C1 C2 यानी दोनों सर्कल्स के बीच की सेंटर्स की डिस्टेंस है ना तो वो क्या हो जाएगी सर वो हो जाएगी C1 C2 का क्या भाई होल स्क्वायर डिवाइडेड बाय डिवाइडेड बाय तू टाइम्स आर वैन आर तू बस एक छोटा सा करेक्शन कर लेते हैं हम ढूंढ रहे द आपने हमें कोस 180 - देता दे दिया अब सारी बातें छोड़ो वही बातें वापस सोचने की कोशिश करते हैं क्या आपको थोड़ी बहुत क्वाड्रेंट थ्योरी याद है जहां पर हम बात करते हैं 180 मतलब क्या π में से कुछ सब्सट्रैक्ट क्या मतलब कौन सा क्वाड्रेंट सेकंड क्वाड्रेंट और सेकंड में कौन-कौन पॉजिटिव होते हैं सिन और इसका रिसिप्रोकल कोसेक तो अगर मैं आपसे पूछूं की कोसेक 180 - θ कितना होगा अगर मैं आपसे पूछूं कोसेक नहीं कोस 180 - थीटा कितना होगा तो पहले तो आप कहोगे सर पे के कारण कोस कोस ही रहेगा लेकिन ऑफ कोर्स π - θ मतलब सेकंड क्वाड्रेंट यहां पर कोस पॉजिटिव नहीं होता है तो नेगेटिव हो जाएगा तो ये हो जाएगा - कोस थीटा के हो जाएगा माइंस कोस थीटा यानी सर यहां पर अगर आप देखो आप तो कोस 180 - θ कितना हो जाएगा - कोस थीटा और वो माइंस अगर मैं यहां ट्रांसफर करता हूं तो फाइनली आप कहोगे कोस्थेटा आप फाइनली कहोगे कोस्थेटा जो की कितना होगा अगर नेगेटिव यहां ट्रांसफर किया था आप देख का रही हो क्या हो जाएगा सही हो जाएगा दोनों सर्कल्स के बीच के सेंटर्स का स्क्वायर माइंस दोनों सर्कल्स की रेडियस का स्क्वायर का डिफरेंस मतलब इन दोनों को माइंस किया इसमें से डिवाइडेड बाय तू टाइम्स क्या आपकी इन दोनों का प्रोडक्ट जो की कौन है दोनों सर्कल्स की रेडियस माइंस डिवाइडेड बाय 2 टाइम्स R1 r2 यानी की ये जो फॉर्मूला बन कर आया है ये जो फॉर्मूला बन कर आया है दिस गपशप दी एंगल बिटवीन तू सर्कल्स आपको क्या दो चीज चाहिए आपको बस ये पता होना चाहिए अगर कभी भी मुझे दो सर्कल्स के बीच का एंगल निकलना है तो मुझे क्या पता होना चाहिए मुझे बस ये पता हो सर की दोनों के सेंटर्स के cardinates क्या है ताकि मैं उन दोनों के सेंटर्स के बीच के डिस्टेंस निकल यह टेक्निकल उन दोनों सर्कल्स के बीच की सेंटर की डिस्टेंस मुझे पता हो और दूसरी बात मुझे उन दोनों सर्कल्स की रेडियस या रडाई पता हो अगर वो पता है तो मैं nihsan दे बिना ज्यादा घबरा हुए परेशान हुए इन दोनों सर्कस के बीच का एंगल निकलूंगा और सर्कल्स के बीच का एंगल मतलब क्या सर्कस के बीच का एंगल मतलब उनके पॉइंट ऑफ इंटरसेक्शन पर ड्रा की गई टांगें के बीच का है जो की क्या है ये थीटा और उसे हमने इस पुरी एक कंप्लीट स्टैंडर्ड मेकैनिज्म के थ्रू निकाला जिसको समझने में मेरी ख्याल से किसी स्टूडेंट को कोई परेशानी नहीं और इसके साथ जो सबसे जरूरी चीज आपको याद रखनी है वो है ये फॉर्मूला तो कोशिश कीजिएगा ऐसे याद रखने की सर फॉर्मूले याद नहीं होते मत घबरा हम क्वेश्चंस करेंगे अभी तो और भी बहुत सारी मजेदार चीज डिस्कस करना बाकी है जब हम वो कर लेंगे और क्वेश्चंस कर लेंगे तो ये बाय डिफॉल्ट आपके सब कॉन्शियस ब्रेन में स्टोर हो जाएगा आप अच्छे से प्रैक्टिस में ला पाएंगे और आप आसानी से इसे याद रख पाएंगे हमेशा तरीका यही है प्रैक्टिस करिए क्वेश्चन एक बेहतरीन सी एक बेसिक सी एक अच्छी सी अंडरस्टैंडिंग आप डिवेलप या बिल्ड कर का रहे हो इस कॉन्सेप्ट को लेकर की सर दो बातें जब कोई दो सर्कल्स इंटरसेक्ट करें दो पॉइंट्स पर तो क्या होता है पहला कंक्लुजन की सर उनकी एक कॉमन कोड बनती है उसे कॉमन कोड को ही आप रेडिकल एक्सेस कहते हो उसकी क्वेश्चन तो सर S1 माइंस तीसरा कंक्लुजन की सर अगर कभी दो सर्कल सेंटर सेट करें उनकी रेडियस के डिफरेंस से हमेशा ज्यादा होनी चाहिए और उनकी रेडियस के हिसाब से हमेशा कम होनी चाहिए याद रखिएगा ये मत bhuliyega चौथी बार उनका पॉइंट ऑफ इंटरसेक्शन निकलने के लिए उन सर्कल्स का पॉइंट ऑफ इंटरसेक्शन निकलने के लिए मैं उनकी कॉमन कोड यानी रेडिकल एक्सेस से किसी एक सर्कल को सॉल्व कर लूंगा और वैसे निकल लूंगा पांचवी बात अगर मुझे उसे कॉमन कोड की लेंथ चाहिए वो अब के लेंथ चाहिए तो मैं बड़ा आसान सा तरीका है उसे करूंगा मैं पाइथागोरस थ्योरम उसे करूंगा एक सर्कल की रेडियस pakdunga उसी के सेंटर से उसे कॉमन कोड पर परपेंडिकुलर लेंथ निकल लूंगा और फिर मैं पाइथागोरस थ्योरम से उसे कॉमन कोर्ट की हाफ डिस्टेंस निकल लूंगा और उसका डबल कर दूंगा तो बेशक मुझे उसे कॉमन कोर्ट की लेंथ मिल जाएगी आज का छठ पॉइंट आज के सिक्स्थ लर्निंग किन्हीं दो सर्कल्स के बीच में एंगल भी निकाला जा सकता है और उसका फॉर्मूला आपके सामने ये रहा कहानी यहां खत्म नहीं हो रही है इससे आगे और कहानी जाने वाली है थोड़ी और चीज सीखेंगे फिर कंबाइंड मंदर में क्वेश्चंस करेंगे ताकि ये सारे कॉन्सेप्ट अच्छे से ब्रेन में स्टोर हो जाए जल्दी से क्वेश्चंस कर दो ताकि सारी चीज अच्छे से समझ ए जाए पर हम थोड़े अच्छे लेवल के अच्छे स्टैंडर्ड थोड़े टू और चैलेंजिंग क्वेश्चंस करेंगे ताकि आपको मजा आए आप काफी सारी चीज सीखो तो उसके लिए मैं थोड़ी सी और चीज एक साथ सिखा डन और फिर हम साथ में सब चीजों को अप्लाई करते हुए क्वेश्चंस करें आई होप आप पेशेंटली वेट कर रहे हैं और हम शुरुआत करेंगे उसी पार्ट की पर आज के इस लेक्चर में तो कोई दिक्कत नहीं है यहां तक तो कोई परेशानी नहीं है मैं इसी पॉइंट पर आने वाला हूं की अगर दो सर्कल्स अगर दो सर्कल्स इंटरसेक्ट कर रहे द तो उनके बीच का एंगल आप कैसे निकलती हो और इसी को शायद हमने यहां पर इस तरीके से कनक्लूड किया अब मूव करते हैं आज की सेवंथ लर्निंग पर आज की सेवंथ लर्निंग क्या होगी आज की सेवंथ लर्निंग होगी की सर अगर दो सर्कल्स ऑर्थोंगोनल हैं ऑर्थोंगोनल मतलब ऑर्थोंगोनल होने का मतलब है की उन दोनों में जो एंगल बन रहा है जो एंगल ऑफ इंटरसेक्शन बन रहा है अगर वो 90 डिग्री है कैसी बात करेगा प्लीज इस बात को गौर से suniyega ये बात ये है की मैन लो आपके दो सर्कल हैं पहला सर्कल ये रहा यहां पर और इसी के साथ में एक और सर्कल बनाने की कोशिश करता हूं जो की लेट्स से मैं कुछ इस तरीके से बनाता हूं बस मुझे थोड़ा सा वक्त दीजिएगा मैं सही बना पाया तो हो जाएगी शायद मैं बना पाऊंगा आई थिंक अभी एक टैसेंट यह बन रही है थोड़ा सा जा रहा है यहां पर hopefulli बन्ना तो चाहिए मेरे कोशिश करता हूं वर्ण आप मैन के chaliyega की यहां पर कुछ जो बनेगा देखो आप ऐसे समझो सर बिल्कुल यह जो टैसेंट है आप सभी को ये वाली टैसेंट दिख रही है बड़ी सिंपल सी शॉर्ट ट्रिक्स सी हान सर ये वाली टैसेंट तो दिख रही है इसी तरीके से आपको एक और टैसेंट दिखाई देगी कौन सी है भाई ये वाली tenjent क्या आप सभी को ये वाली tenjent भी दिख रही है हान सर ये वाली इंजन भी दिख रही है तो ये इन दोनों सर्कल्स का पॉइंट ऑफ इंटरसेक्शन था जहां पर मैंने दोनों सर्कस की क्या ड्रा की टैसेंट और मुझे रिलाइज हुआ की हान सर दोनों टांगें के बीच जो एंगल है दोनों टांगें के बीच जो एंगल है डेट इस क्लीयरली उन 90° एंगल तो बातें आसान है बातें आसान कैसे हैं सर सारी बातें छोड़ो आप तो यह बताओ दो टांगेंट्स के बीच का एंगल कैसे निकलती हैं मतलब दो सर्कल्स के बीच का एंगल कैसे निकलती हैं सर हमने ये सिखा है दो सर्कल्स के बीच का एंगल का फॉर्मूला होता है कोस्थेटा जो की याद करो अभी पढ़ा था सर वो क्या होता है सर वो होता है दोनों सर्कल्स के सेंटर्स के बीच की डिस्टेंस जिसे आप कहते हो C1 C2 का स्क्वायर माइंस दोनों की रेडियस के स्क्वायर के सैम को आप सब्सट्रैक्ट कर देते हो और सर इसे आप डिवाइड कर देते हो तू टाइम्स दोनों की रेडियस के प्रोडक्ट्स हैं अच्छी बात है सर आपकी बात हमने मैन ली आपकी बात समझ ली पर यहां क्या अरे भाई यहां क्या यहां तो सर थीटा कितना है 90° है तो कोस 90 कितना होता है सर कोस 90 होता है जीरो तो आप कितना ले रहे हो θ सर यहां पर जो थीटा है डेट इस 90 तो ये कितना है सर ये है कोस्थेटा नहीं कोस 90 और कोस 90 यानी कितना जीरो और जीरो किसके इक्वल है इसके जीरो किसके इक्वल है सो जीरो है C1 C2 के स्क्वायर माइंस r1² - r2 का स्क्वायर डिवाइडेड बाय व्हाट डिवाइडेड बाय तू टाइम्स r1r2 और सर अगर मैं इस पार्ट को सॉल्व करूं तो ये इधर गया अब मुझे जो फाइनल कंक्लुजन मिला वो क्या मिला गौर से देखिएगा स्टूडेंट्स मुझे मिला C1 C2 का स्क्वायर - R1 का स्क्वायर माइंस आर तू का स्क्वायर जीरो के इक्वल एंड अब यहां से इंक्लूड करोगे की C1 C2 का जो स्क्वायर है वो किसके इक्वल है सर वो इक्वल है r1² + r2² के हान या ना अरे भाई हान या ना अच्छा सा ठीक है बातें हम समझ पाए लेकिन अगर मैं सर्कल्स की बात करूं तो इसका सेंटर क्या होगा सर अगर ये है सर्कल S1 तो इसका सेंटर हम जानते हैं क्या होता है C1 जिसके कार्ड क्या कहते हो आपने जनरली इक्वेशन से सीखे हैं - F1 सर इस सर्कल का सेंटर क्या था माइंस f2 अच्छी बात है सर अच्छा एक और छोटी सी बात बता दो क्या सर ध्यान से देखो अगर मैं सर्कल की आपसे इडियट्स पूछूं अगर मैं सर्कल की रेडियस जो की आप क्या कहोगे R1 तो वो क्या होगी सर वो भी अंडर रूट ओवर जीवन स्क्वायर प्लस f1² - 7 और सर अगर मैं आपसे इस सर्कल की रेडियस पूछो तो वो क्या होगी सर इस सर के लिए रेडियस होगी r2 जो की क्या होगी √1 जी तू स्क्वायर प्लस तू स्क्वायर माइंस सी तू ठीक है सर अगर इन सारी वैल्यू को मैं यहां रखता हूं तो पहले तो इन दोनों के सेंटर्स के बीच की डिस्टेंस C1 C2 है ना इन दोनों के सेंटर्स के बीच की डिस्टेंस तो होल स्क्वायर प्लस f2 - f1² से है जाएगा अब r1² + r2² लिखना है तो सर पहले सर्कल की रेडियस क्या हो जाएगी वो हो जाएगी जीवन स्क्वायर प्लस f1² - C1 और जो दूसरे सर्कल आई होप स्क्वायर है तो फिर से वो अंडर रूट से कैंसिल दूसरे सर्कल की रेडियस जब निकल लूंगा तो होगी g2² + f2² - C2 कोई दिक्कत यहां तक कोई भी परेशानी है तो पूछो सर अब क्या थोड़ा इससे सिंपलीफाई करते हैं देखो भाई यहां से क्या मिलेगा सर यहां से मिलेगा G2 प्लस जीवन स्क्वायर और क्या भाई माइंस तू जीवन जी यहां से क्या मिलेगा सर यहां से मिलेगा f2² + F1 स्क्वायर -2f1f2 और यहां पे जो चीज हैं जो की क्या है सर जो की है जीवन स्क्वायर पहले लिख लेता हूं फिर क्या दिख रहा है मुझे F1 स्क्वायर प्लस ऑफ कोर्स इस बात को ध्यान से देखो क्यों देखें सर ये और ये देखो ये और ये देखो यहां पर देखो आप क्या आपको एक आंसर होता दिख रहा है बहुत सारी तो बढ़िया अच्छी बात हो गई अब अगर मैं फाइनल कंक्लुजन की तरफ बडू तो क्या आपको दिख रहा है की सर ये नेगेटिव ये नेगेटिव कॉमन ले लो ये नेगेटिव ये नेगेटिव कॉमन ले लो तो जो आपको फाइनल चीज दिखी हो गया यहां से तू भी कॉमन ले लेना यहां से तू भी कॉमन ले रहा हूं तो आपको दिखेगा तू टाइम्स जीवन जी तू प्लस आई होप आपको दिख रहा है F1 f2 किसके इक्वल है सर C1 प्लस C2 के इसका क्या तात्पर्य है इसका क्या मतलब है सर ये के आप हमें पढ़ा रहे हो ये पढ़ने का मतलब है स्टूडेंट्स की अगर आपको दो सर्कल्स दिए जाए दो सर्कल्स दिए जाएं कौन-कौन से दो सर्कस है सर्कल S1 एंड सर्कल S2 और कहा जाए ये दोनों सर्कस क्या है ऑर्थो कोई मुझसे क्लीन करें की दोनों सर्कल्स के हैं औरतों को नल ऑर्थोंगोनल का क्या मतलब इन दोनों सर्कस के पॉइंट ऑफ इंटरसेक्शन पर ड्रा की टांगें के बीच एंगल है 90° और तो वर्ड जब भी आएगा ना तो आपका पिक्चर में आएगा 90 डिग्री आपका पिक्चर में आएगा परपेंडिकुलर बात याद रहेगी यानी की इन दोनों सर्कस के पॉइंट ऑफ इंटरसेक्शन पर जो टांगेंट्स ड्रॉ की गई है वो है 90 डिग्री तो आप हमेशा याद रखिएगा ये बात हमेशा याद रखेगा की बिल्कुल सर जब भी ऐसा होगा तो आपकी यह कंडीशन फुलफिल होगी क्या की दोनों सर्कल्स के वह G1 और G2 का प्रोडक्ट दोनों सर्कल्स के वह F1 और f2 का प्रोडक्ट का डबल उनके C1 और C2 के सैम के इक्वल होगा अगर कभी भी ऐसा हो अगर कभी भी ऐसा हो तो मैं क्या कहूंगा दोनों सर्कस ऑर्थोंगोनल है या दोनों सर्कल्स ऑर्थोंगोनल हैं तो कैसा होगा ऐसा क्या कोई बाई डायरेक्शनल कंडीशन समझ ए रही है इफ एन ओनली इफ दें क्या ये बात आपको समझ आई अगर ऐसा है तो ऐसा और ऐसा है तो ऐसा कोई कन्फ्यूजन आपको क्लियर हो रही जो यहां पर आपको का कर समझने की कोशिश कर रहा था की अगर दो सर्कल सेंटर सेट करें तो जो कोस्थेटा होगा आपका वो कितना हो जाएगा 90 और वहां से आप यह कंक्लुजन पर पहुंचेंगे जो की यहां पर लिखकर उसमें समझने की कोशिश की है जो की हमने भी अभी डिस्कवर किया कोई दिक्कत कोई परेशानी आई थिंक सर बड़ी आसानी बातें इंटरसेक्शन ऑफ तू सर्कल्स में तो कोई डाउट नहीं है आई थिंक सारे ही कॉन्सेप्ट्स आसान है पर ये आज की सेवंथ लर्निंग याद रहेगी इफ तू सर्कल आर इंटरसेप्टिंग ऑर्थोगोनली इसका ज्योमैट्रिकल मतलब क्या है मतलब सर उनके पॉइंट ऑफ इंटरसेक्शन पर ड्रा की टांगें परपेंडिकुलर है उनके बीच 90° एंगल है इसका हमारे लिए फॉर्मूला वाइस क्या मतलब है इसका मतलब है डबल ऑफ जीवन जी तू प्लस F1 f2 C1 + C2 के इक्वल होगा भूलोगे तो नहीं आई थिंक ये लर्निंग भी आप बहुत डिटेल में अच्छे से याद रखोगे बहुत अच्छे समझाना चाह रहा हूं की अगर मैन लो आपके पास क्या है अगर आपके पास बेसिकली ये दो सर्कल्स हैं और अगर यह किसी तीसरे सर्कल को ऑर्थोगोनली इंटरसेक्ट कर रहे हैं क्या आप यह सिनेरियो देख का रहे हो यहां पर ऑफ कोर्स यह दोनों सर्कस आपस में औरतों के लिए इंटरसेक्ट कर रहे हैं और यह दोनों सर्कल से आपस में orginalli इंटरसेक्ट कर रहे हैं तो इससे क्या मतलब है इसका यह मतलब है की इस सर्कल और इस सर्कल का जो रेडिकल एक्सेस होगा ना इस सर्कल और इस सर्कल का जो रेडिकल एक्सेस होगा जो की दोनों सर्कस किसी तीसरे सर्कल को ऑर्थोगोनली इंटरसेक्ट कर रहे हैं इन दोनों का रेडिकल एक्सेस इस तीसरे सर्कल के सेंटर से पास होगा अब कोई लाइन समझ आई क्या मैं फिर से दोहराता हूं suniyega थर्ड सर्कल ऑर्थोगोनली अगर एक तीसरा सर्कल है जिसे दो सर्कल्स आर्गेनाई इंटरसेक्ट कर रहे हैं तो आप याद रखिएगा जो रेडिकल एक्सेस होता है इन दोनों गिवन सर्कल का इसका और इसका जिसे आप कहते हो S1 - S2 = 0 ये रेडिकल एक्सिस आपके तीसरे सर्कल जिसको ऑर्थोगोनली इंटरसेक्ट कर रहे द उसके सेंटर से पास होगा याद रहेगी ये बात इसी बात को कहने का एक और तरीका क्या सर ध्यान से देखोगे इसी बात को कहने का एक और तरीका क्या 10 सेंटर 10 सेंटर ऑफ डी वेरिएबल सर्कल कटिंग डी तू गिव इन सर्कल्स ऑर्थोगोनली लिस ऑन डी रेडिकल एक्सेस ऑफ डी गिवन तू सर्कल्स इसका क्या मतलब है प्लीज इस बात को ध्यान से सुनना यह बात ज्यादा जरूरी है जो सेकंड स्टेटमेंट आए क्योंकि उसे पर बेस्ड क्वेश्चंस आते हैं पहली बात स्टेटमेंट था जिसका कंक्लुजन सेकंड बात या दोनों एक ही बात है पर क्या suniyega मैं तो आपको क्वेश्चन के थ्रू ऐसे samjhaunga की मैन लो कभी आपको किसी क्वेश्चन में कोई कहे की आपके पास दो सर्कल्स हैं एक सर्कल ये रहा और एक सर्कल ये रहा आपको मैन लो वो दो सर्कल दे दे रहा है ठीक है सर अच्छी बात है दोनों सर्कल ले लीजिए ये सर्कल के लिए क्या रहेगा आपका S1 = 0 और ये सर्कल क्या रहेगा S2 = 0 क्या आप मेरी बात समझ का रहे हो अब आपसे मेरा ये कहना रहा आपसे मेरा ये कहना रहा की अगर मैं अगर मैं ढूंढ रहा हूं सेंटर को किसके एक ऐसे सर्कल के या एक ऐसे वेरिएबल सर्कल के जो की इन दोनों पर परपेंडिकुलर है तो उसे सर्कल का सेंटर का वो सर्कल जो की इन दोनों पर ऑर्गेनाइज्ड वह हमेशा इन दोनों के रेडिकल एक्सेस पर ही होगा तो समझ नहीं ए रहा है सर मेरी बात प्लीज ध्यान से सुनिए रिप्लाई प्लीज बहुत ध्यान से सुनिए जिस बात को बहुत अच्छे से याद रखना क्योंकि इस पर ना सिर्फ क्वेश्चंस बल्कि स्टेटमेंट बनेंगे स्टेटमेंट का मतलब है मैन लो थोड़ी देर के लिए की ये जो लाइन है ये इन दोनों सर्कल्स का रेडिकल एक्सरसाइज है किन-किन का S1 और S2 का तो क्या मैं इस रेडिकल एक्सिस की इक्वेशन को S1 - S2 = 0 का सकता हूं बिल्कुल का सकते हो अब यहां पर मेरा आपसे ये कहना है की इस सर्कल पर इस इस रेडिकल एक्सिस पर किसी भी सेंटर का किसी पर किसी भी पॉइंट पर सेंट्रल लेकर आप सर्कल ड्रॉ कर लो बस ऐसे ड्रा करना बस ऐसे ड्रा करना की वह इन दोनों सर्कल्स को इंटरसेक्ट करें बस ऐसे ड्रा करना की वह दोनों सर्कस को इंटरसेक्ट करें जैसे मैंने मैन लिया यहां तो आप निश्चिंत होकर बैठे रहना सर ये सर्कल ये जो सर्कल है जिसका सेंटर है रेडिकल एक्सरसाइज करेगा ऑर्थोंगोनल इंटरसेक्ट करने का मतलब ortholi इंटरसेक्ट करने का मतलब है सर की यहां पर इससे और इससे जो आप इनके पॉइंट ऑफ इंटरसेक्शन पर टांगें ड्रॉ करोगे वह 90° होगी और इसके और इसके पॉइंट ऑफ इंटरसेक्शन पर जो आप टेंशन ड्रॉ करोगे वो भी 90° होगी कोई परेशानी आई होप ये बात याद रहेगी प्लीज इस बात को नोट डाउन कर लो और इसे अच्छे से समझ के रख लो क्योंकि इस पर क्वेश्चंस बनते हैं तो ये स्टेटमेंट जो सेकंड वाला है इसे प्लीज याद रखना मैं फिर से बोल देता हूं अगर कन्फ्यूजन हो रही है तो 10 सेंटर ऑफ डी वेरिएबल सर्कल वो वेरिएबल सर्कल जो की कहां लाइक कर रहा था और डायरेक्टली एक्सिस पर उसका सेंटर रेडिकल एक्सरसाइज पे अप्लाई करेगा किसका सर उसे खास सर्कल का किस सर्कल का वो वेरिएबल सर्कल का जो की ये जो दो गिवन सर्कल्स हैं जो की एक गिवन सरफेस है उनको ऑर्थोगोनली इंटरसेक्ट कर रहा है कहां लाइक करेगा वो सेंटर उन दोनों गिवन सर्कल्स के रेडिकल एक्सिस पर भूलोगे तो नहीं आई होप आप नहीं बोलोगे बस ये बहुत कृष्ण स्टेटमेंट है जैसे प्लीज आप याद रखिएगा चलो अगर ये सारी बातें याद रहेंगी तो आगे बढ़ते हैं और इस स्टेटमेंट को समझते हैं यहां पर जो बातें लिखी है वो देख लेते हैं हम क्या जान समझ गया देख का रहे हैं सर ऑफ कोर्स अगर आप उसे फिगर में देखें तो आप देख का रहे हो S1 और S2 आपके दो सर्कल्स हैं और उनका ऑफ कोर्स S3 वो सर्कल है जो की उन दोनों के रेडिकल एक्सरसाइज हम सब बड़ी बेसिक सी बात जान का रहे हैं की c1poc परपेंडिकुलर है हम क्या जान का रहे हैं सर हम जान का रहे हैं की आपके कौन सी 1 poc3p और सिमिलरली सी थ्री के और सी तू परपेंडिकुलर है बिल्कुल सर ये बात हम बहुत अच्छे से देख समझ सोच और सन का रहे हैं बिल्कुल सर समझ का रहे हैं अब क्या अब एक जरूरी बात ये की सर जो C3 है जो C3 है वो कहां लाइक करता है सी थ्री ऑफ कोर्स आपके रेडिकल एक्सेस पर लाइक करता है सर्कल S1 और S2 के और क्या देख का रहे सर c3p और C3 के क्या है सर ये लेंथ है किसकी टांगें की कहां-कहां से C3 से गिवन सर्कल्स पर मैं फिर से रिपीट करता हूं जो c3p और C3 के है फिर से देखो c3p और C3 के क्या है सर ये टांगे की लेंथ है और क्या मैं ये कनक्लूड करूंगा की सर C3 इसके रेडिकल एक्सेस पर लाइक कर रहा है तो सी थ्री पी और सी थ्री के की लेंथ इक्वल होगी आई रिपीट क्योंकि रेडिकल एक्सेस क्या होता है सर रेडिकल एक्सेस वो लाइन होती है अगर आप tenjent ड्रॉ करो गिवन सर्कल्स पर तो stenjan की लेंथ इक्वल होती है क्या यह बात आप समझ का रहे हो तो C3 अगर आपके इसके रेडिकल एक्सरसाइज पर लाइक कर रहे हैं तो c3p आपकी इस सर्कल की टैसेंट है और C3 की ओर ये सर्कल की टैसेंट है और इनकी लेंथ c3po और C3 = क्योंकि C3 आपका रेडिकल एक्सरसाइज पर लाइक करने वाला पॉइंट और रेड कलर से यही कंडीशन हमारी फुलफिल करता है याद है क्या अब सर अगर मैं आगे बढूं अब मैं आगे बढूं अगली लाइन क्या लिखिए आई होप ये बात आपको समझ ए रही है फ्रॉम दिस डिस्कशन सी अंडरस्टैंड डेट रेडिकल सेंटर इस डी सेंटर ऑफ डी सर्कल विच इंटरसेक्ट डी गिवन रेडिकल सेंटर की बात कर रहा हूं विच इंटरसेप्ट्स डी गिवन थ्री सर्कस ऑर्थोगोनली ये बात और कृष्ण है इसको suniyega फिर से ध्यान से रेडिकल सेंटर जो होता है वो सेंटर होता है उन सर्कस का जो की इंटरसेक्ट करते हैं गिवन थ्री सर्कल्स को क्या arthognary याद रहेगा क्या डी रेडियस ऑफ सर्कल इस डी लेंथ ऑफ डी टांगें फ्रॉम मेडिकल सेंटर तो यही जो लेंथ आप निकल थी कौन सी जो भी आपके आना चाह रहे द C3 की वाया सी थ्री पी ये जो लेंथ है ये ऑफ कोर्स आपकी क्या होगी ये लेंथ होगी आपकी रेडिकल सेंटर की उसकी सर्कल की रेडियस आई होप आप ये बातें याद रखेंगे इन सारी बातों से भी कृष्ण प्लीज सारी आज के लेक्चर के जो सबसे बड़ी बात है वो यही की इस बात को मत भूलना ये बहुत खुश है की सर क्या दो सेंटर्स का अगर कोई थर्ड सर्कल ऑर्थोगोनली इंटरसेक्ट कर रहा है और वो वेरिएबल सर्कल ढूंढ लूं तो उन सारे सर्कल्स का जो सेंटर होगा ना उन सारे सर्कस का जो सेंटर होगा ना वो इन दोनों सर्कल के रेडिकल एक्सिस पर ही लाइक करेगा आपको याद ए रहा है रेडिकल्स एक बेसिक सा कॉन्सेप्ट था पर कितना ज्यादा वाइडऐली उसे हो रहा है कॉमन कार्ड उसे हो जा रहा है वो टैसेंट उसे हो जा रहा था जब दो सर्कल इंटरनल ही टच कर रहे द और कितनी जगह यहां पर उसे हो जा रहा है तो रेडिकल एक्सिस से कृष्ण टर्म है और कितनी आसान सितम है S1 - X2 = 0 कितनी आसानी से हम उसे निकल सकते हैं बिना ज्यादा परेशान हूं तो काफी सिग्निफिकेंट है और काफी हेल्पफुल है टेक्निकल मतलब बहुत मदद करेगी आपको ये क्वेश्चंस को सॉल्व करना है दो सर्कल्स का पॉइंट ऑफ इंटरसेक्शन दोनों सर्कल्स को सॉल्व करने से अच्छा सर एक सर्कल और उनके रेडिकल एक्सरसाइज को सॉल्व कर लो हमको बातें कर्मचारी हैं तो बहुत मदद करेगा यह स्टेटमेंट लिख रहा है फाइंड दी एंगल अब हम इन क्वेश्चंस को सॉल्व करेंगे जो की आपके इन कॉन्सेप्ट्स पर बेस्ड है suniyega ध्यान से फाइंड डी एंगल आते विच एंड दिस इंटरसेक्ट सर एंगल निकलना है ना तो एंगल निकलने के लिए तो कुछ हमने पढ़ा है मेरे ख्याल से सर आप तो फॉर्मूला बता दो हम निकल देंगे तो मैं फॉर्मूला आपको बता देता हूं फॉर्मूला बता दे रहा हूं मैंने हिंट दे दे रहा हूं और आप मुझे आंसर बता देंगे की क्या इसका आंसर होगा जल्दी से देखो सर फॉर्मूला हमने पढ़ा है की सर दो सर्कल्स के बीच का एंगल जो होता है वो होता है कोस थीटा जो की क्या होता है सर उन दोनों सर्कल्स के सेंटर्स के बीच के डिस्टेंस का स्क्वायर मैं इसे उन दोनों की रेडियस का स्क्वायर आप डिफरेंस नो माइंस कर देते हो डिवाइडेड बाय डबल ऑफ दोनों की रेडियस का प्रोडक्ट याद ए रहा है फॉर्मूला हिंट आपके सामने हैं डाटा आपके सामने है आप क्वेश्चन सॉल्व कर दीजिए मुझे आंसर लाकर बता दीजिए मैं बहुत खुश हो जाऊंगा और मुझे बहुत खुशी होगी अगर आप मुझे आंसर बता दें तो सर यकीनन तौर पर यह क्वेश्चन बहुत टू बहुत डिफिकल्ट नहीं है आप चीज ध्यान से देखो आपको बहुत मजेदार तरीके से नजर आएंगे की हर चीज तो आपके पास है कैसे है सर देखो भाई सर सबसे पहली बात तो ये अगर ये सेंटर इस सर्कल के सेंटर ढूंढू तो ये क्या हो जाएंगे सर ये हो जाएंगे वैन कमा जीरो इस सेंटर के इस सर्कल की रेडियस ढूंढो तो हो जाएगी √10 इस सर्कल के सेंटर ढूंढ डन तो हो जाएंगे जीरो कमा तू और इसकी रेडियस क्या हो जाएगी सर इसकी रेडियस हो जाएगी √5 कोई तकलीफ तो नहीं है बिल्कुल नहीं है सर अब अगर मैं इन दोनों इन दोनों सर्कस के बीच का पॉइंट ऑफ इंटरसेक्शन जो भी रहा उसे पर टेंशन जो ड्रा किया उनके बीच का जो एंगल निकलना चाहो वही दो सर्कल्स के बीच में एंगल होता है वो कैसे निकलेंगे ध्यान से देखो सर दोनों के सेंटर्स के बीच के डिस्टेंस तो पहले मैं C1 C2 निकल लेता हूं C1 C2 क्या होगा देखो 1 - 0 1 1 का स्क्वायर वैन तू तू का स्क्वायर कितना फोर प्लस वैन फाइव फाइव यानी unroof 5√5 डिस्टेंस से अंडर रूट और स्क्वायर कैंसिल तो क्या बचा 5 तो कोस थीटा पहले तो क्या ए रहा है 5 - r1² r1² मतलब कितना √10 का स्क्वायर यानी कितना 10 - r2² यानी √5 का स्क्वायर यानी कितना 5 / 2 टाइम्स डिवाइडेड बाय तू टाइम्स R1 r2 R1 r2 कितना √10 √5 डेट्स अंडर रूट 50 और √ 50 को बस मेरा ऐसे मैन हो रहा है अंडर रूट 50 को मैं लिखना चाह रहा हूं 25 किसी भी स्टूडेंट को इस बात से 25 को आप क्या लिखोगे फाइव स्क्वायर फाइव स्क्वायर मतलब क्या ए जाएगा बाहर फाइव फाइव इन तू 10 10 से 10 कैंसिल तो ये बचेगा -1/√2 ये कितना बचेगा सर ये बचेगा -1/√2 अब सुनना कम की बात इस क्वेश्चन को सॉल्व करने का तरीका सर कोस्थेटा जो आपने निकाला वो कितना आया - 1 / √2 सर माइंस 1/√2 अच्छा नहीं लग रहा है तो 1 / √2 कितना होता है सर 1 / √2 होता है कोस 45 जैसे आप लिख सकते हो कोस π/4 लिख सकता हूं क्या कोई तकलीफ तो नहीं है सो लिख सकते हो अब जब मैं एंगल कंपेयर करना चाह रहा हूं तो मैं रखूंगा इसे कैसा कोस थीटा जब हम त्रिकोण पढ़ेंगे तब आप कोई और क्लेरिटी से समझ ए जाएगा की ऐसा क्यों कर रहा हूं अभी थोड़ा सा बस पेशेंस पेशेंट के लिए माइंस कोस थीटा मुझे मिल रहा है सर ये मिल जाता है मुझे कोस 180 - θ से कोस 180 - θ - कोस थीटा ही होता है तो क्या मैं इसे लिख सकता हूं कोस π - π/4 प्लीज इस बात को फिर से सुनिए पाई के कारण कोस कैसे रहेगा लेकिन π - के कारण ये सेकंड क्वाड्रेंट में जाएगा जहां पर आपका कॉस्ट पाई माइंस थीटा हो जाता है - कोस्थेटा कोई तकलीफ तो नहीं है अब जब आप इसे थोड़ा ध्यान से देखेंगे तो आप ये पाएंगे देखो भाई यहां पे तो ये लिखा हुआ है क्या कोस्थेटा लेकिन यहां पर ये लिखा है कोस कितना भी ये हो जाएगा 3π/4 और अब जब आप चीज कंपेयर करेंगे अब जब आप चीजें कंपेयर करेंगे तो थीटा कितना होगा π/4 क्या सर ऐसे हर बार कंपेयर किया जा सकता है नहीं अभी मुझे कुछ चीज पता है इसलिए मैं कंपेयर कर रहा हूं आपको थोड़े दिनों में रिलाइज हो जाएगी जब हम ट्रिगो पढ़ लेंगे तब आप भी इस ऐसे सोच पाएंगे बट अभी फिलहाल अब बस इसी से देखिए और समझी क्या यहां तक कोई दिक्कत ये जो थीटा की वैल्यू आई है 3π/4 डेट इस नथिंग बट दी एंगल बिटवीन डीज तू सर्कस इस क्वेश्चन का आंसर क्या होगा भाई 3π/4 आई थिंक बहुत डिफिकल्ट बहुत टू बहुत मुश्किल क्वेश्चन नहीं था सीधा सीधा डाटा पर बेस्ड क्वेश्चन था फॉर्मूला बेस्ड क्वेश्चन इट वास देयर डिलीवरी नो लिंक किया तो मेक योर अंडरस्टैंड डी कॉन्सेप्ट तू लर्न तू मेक यू लर्न डी फॉर्मूला याद रहेगा अगर ये याद रहेगा तो क्या मैं अब आपको अगला क्वेश्चन डन जो आप ट्राई करेंगे ये रहा आज का आपका अगला यानी सेकंड क्वेश्चन इसे पढ़ लेते हैं पहले फिर साथ में ट्री करेंगे क्वेश्चन में क्या लिखा है ध्यान से देखो भाई वो लिख रहा है तू सर्कल्स तो एक तो सर्कल ये है सर और एक ये सर्कल ये है है ना कट एच आदर ऑर्थोंगनली डीज तू सर्कल्स कट एच आदर arthognali कुछ याद ए रहा है कुछ याद ए रहा है तो वो के की वैल्यू निकलवा रहा है ये क्वेश्चन में नहीं बताऊंगा अगर मैं बता दूंगा तो सॉल्व हो जाएगा मतलब हिंट बताते मैं बस कुछ यहां से लूंगा कुछ यहां से लूंगा अगर नहीं समझ ए रहा है तो लास्ट लेक्चर फिर से देखो क्योंकि आना चाहिए फिर बहुत मेहनत से हमने पढ़िए चीज आपको क्यों नहीं याद ए रहा है अगर दो सर्कल्स और जनरली कट कर रहे हैं तो क्या होता है तो जल्दी से इस कंडीशन को अप्लाई कर लो और मुझे बता दो की यहां पर आप क्या सोचोगे और कैसे सोचोगे बस थोड़ा सा ध्यान से पढ़ना क्वेश्चन को कुछ गलती मत कर देना कुछ कुछ हड़बड़ाहट में कोई गलती मत कर देना क्योंकि ये क्वेश्चन होना चाहिए सर क्वेश्चन कैसे सॉल्व हो जाएगा आप बता दो हमें मेरा कहना है भाई अगर आप यहां से चीजे देखोगे तो यहां से मैं सिंपलीफाई कर सकता हूं क्या सर गड़बड़ है इस क्वेश्चन में x² और y² के कॉएफिशिएंट आपके वैन होने चाहिए तो क्या पुरी इक्वेशन को तू से डिवाइड कर दें कर दीजिए सर तो आपके पहले सर्कल की जो इक्वेशन आएगी वो होगी x² + y² - 3/2 टाइम्स क्या तकलीफ है किसी को कोई परेशानी है क्या सर दूसरा सर्कल के घर में बात करूं तो वह क्या दिया हुआ है सर वो दिया हुआ है आपको x² + ए स्क्वायर के सामने राखी हुई कोई डाउट या परेशानी तो नहीं अब क्या करना होगा सर अब जब मैं आगे बडूंगा अब जब मैं चीज सोचूंगा तो मेरा ये कहना है प्लीज ध्यान से सुनिए दो सर्कल्स अगर एक दूसरे को और जनरली टच करें तो वह क्या कर रहे होंगे सर अगर दो सर्कल से एक दूसरे को orthobanely टच करते हैं तो हमने कंडीशन पढ़ी थी आप कुछ मत सोचिए आप बस ये फॉर्मूला याद रखो सर ये हो जाएगा cos90 यानी जीरो तो ये उधर चला जाएगा तो दोनों के सेंटर्स के बीच के डिस्टेंस उनकी रेडियस के स्क्वायर्स के संकेत इक्वल होती है या तो यहां से ऐसे निकल दो या वो कंडीशन अप्लाई कर दो वो कंडीशन याद ए रही है क्या सर थोड़ी सी याद ए रही है क्या थी भाई याद करो सर अगर दो सर्कल से एक दूसरे को औरतों के लिए इंटरसेक्ट करें तो आपकी कंडीशन थी तू टाइम्स जीवन जीतू प्लस तू क्या फटाफट हो जाएगा ट्राई करते हैं स्टूडेंट्स देखो क्या लिखा हुआ है सर तू टाइम्स तो डबल लेना है किसका जीवन जो होगा वो होगा यहां से तो कितना हो जाएगा 3/2 का हाफ ऑफ कोस 3 / 2 का हाफ 3 / 2 का हाफ मतलब 3 / 4 लेकिन ये माइंस है यहां पर क्या ए जाएगा -3/4 कितना हो जाएगा -4 का हाफ डेट इस कितना -2 कोई दिक्कत तो नहीं है प्लस F1 f2 प्लस F1 f2 मतलब क्या यहां से 3 का हाफ यानी कितना 3 / 2 और यहां से 10 का हाफ तन का हाफ कितना भाई 5/2 सॉरी कोई दिक्कत तो नहीं है किसी को कोई परेशानी है तकलीफ हो तो बताओ अब क्या करने वाले हो सर आप अब मैं का रहा हूं सर ये जो आपने अभी तो बाय डी वे यह जीवन जीतू हो गया F1 f2 हो गया इससे किसके इक्वल रखना है C1 + C2 तो C1 + C2 मतलब क्या मतलब के / 2 + कोस 16 मैं सर सारा कम आसान कर लूंगा अगर मुझे चीज थोड़ी सी ऐसी दिख जाए ऐसी मैं कम करूं क्या दोनों तरफ फोर से अगर मल्टीप्लाई कर डन 4 के बाद तू से भी करेंगे तो चलेगा है ना दोनों तरफ तू से मल्टीप्लाई कर डन देखो तू से मल्टीप्लाई करने से फायदा क्या होगा क्यों कर रहे हो सर बस ऐसे ही थोड़ा कम आसान हो जाएगा जैसे ही यहां पर तू से मल्टीप्लाई किया तो ये हो जाएगा 4 जैसे ही यहां पर तू से मल्टीप्लाई करते हैं अभी कुछ दिखेगा क्या दिखेगा उससे तो कैलकुलेशन थोड़ी आसान हो जाएगी कैसे सर सुनना और ध्यान से यह फोर यहां गया तो यह फोर यह फोर कैंसिल अब यह फोर यहां गया तो सर एक तू हटा लेकिन उसे दो ज़हर में आपने एक और तू यहां पर ले आए कोई दिक्कत तो नहीं है नहीं है जब ये तू अंदर आया जब ये तू अंदर आया तो ये तू तो है गया सर लेकिन 16 का क्या हो गया डबल बिल्कुल सर डबल कर दो तो चीज कुछ ऐसी हो गई हैं चीज मेरे ख्याल से कुछ ऐसी हो गई हैं आई हो बहुत कॉम्प्लिकेटेड आसान बना देता हूं -3 -2 कितना भाई सर वो हो जाता है माइंस सॉरी 3 6 530 तो यहां पर मुझे क्या दिखेगा 30 उसे तरफ क्या देख रहे हैं सर उसे तरफ भी देख लेते हैं इस तरफ अगर ध्यान से देखें तो ये है के और ये है 32 तो कितना हो जाएगा सारी हो जाएगा के + 32 आई थिंक आप देख का रहे हो 36 - 32 कितना फोर क्योंकि किसकी वैल्यू है के की और आई थिंक यही हमसे पूछा गया था क्या मुझे सोच के बताओ ये एक टू क्वेश्चन था क्या ये एक बहुत मुश्किल क्वेश्चन था नहीं सर ये तो एक सीधा सीधा फॉर्मूला बेस्ड क्वेश्चन था ये दोनों क्वेश्चंस जानबूझ के रखे गए हैं ताकि वो बहुत सारी चीज जो हमने लास्ट लेक्चर में सीखी थी आपको सिखाई जा सके और रिकॉल करवाई जा सके की देखो ये फॉर्मूले मत बोलना हो सकता है कहीं ना कहीं किसी क्वेश्चन में डायरेक्टली एंड डायरेक्टली फॉर्मूला लग रहा हूं तो आप फटाफट आंसर निकल दो और अप्लाई कर दो चीज ले आओ तो बेशक ऐसा हो सकता है तो चीज प्लीज याद रखना जल्दी से हो जाएंगी क्वेश्चंस के साथ आंसर ए जाएंगे फटाफट आप चीजों को वर्कआउट कर लोग तो ये कंडीशन याद रहेगी की अगर दो सर्कल्स एक दूसरे को इंटरसेक्ट करें तो ज्यादा नहीं सोचना है क्वेश्चन खत्म हो जाएगा याद रखोगे क्या बिल्कुल याद रखेंगे और अगर आप ये याद रखोगे तो क्या मैं अब अगले क्वेश्चन की तरफ बढूं यह रहा आज का आपका तीसरा क्वेश्चन यहां क्या लिखा है सर पहले क्वेश्चन समझ लेते हैं तू सर्कल पासिंग थ्रू ए बी दो सर्कल्स हैं सर जो ए और बी से पास होते हैं समझ ए रहा है क्या मतलब एक सर्कल मेरी ख्याल से ए से बात हो पास होता होगा दूसरा बी से पास होता होगा ऐसा कुछ रह रहा होगा जो भी रह रहा होगा हम समझेंगे तो जब दो सर्कल्स ए और बी से पास हो रहे हैं तो दोनों सर्कल्स के बारे में मैं ये टिप्पणी तो नहीं करूंगा सर की एक सर्कल ऐसे और एक भी से दोनों का कुछ नहीं बोला ना तो मैं कहूंगा दोनों ही सर्कल्स ए और बी दोनों से पास हो रहे हैं अब अगर दो सर्कल्स दो पॉइंट से पास हो रहे हैं तो कहना चाहिए वो सिनेरियो बन सकता है जहां पर दो सर्कल्स दो पॉइंट्स पर इंटरसेक्ट कर रहे हैं कुछ क्लिक हुआ ठीक है सर यह जो है यह एक लाइन को टच करते हैं यह एक लाइन को टच करते हैं सी और दी पर इस तरीके से की सर ए सी ए दी एक पैरेललोग्राम है और इसमें ए क्या है वही तो पता करना है आपके पास दो सर्कल्स हैं आपके पास क्या है दो सर्कल्स यह रहे आपके दो सर्कल्स और उसने आपसे ये कहा भाई उसने आपसे ये कहा की ये जो दोनों सर्कल्स हैं आपके ये दोनों सर्कल्स आपके दो पॉइंट्स ए और बी से गुजरते हैं तो मैं ए को क्या का लेता हूं मैं ए को का लेता हूं क्या वैन कमा तू और मैं बी को का लेता हूं क्या मैं बी को का लेता हूं तो एक कमेंट नहीं होनी चाहिए तो ये वो कॉमन टैसेंट है जिसकी इक्वेशन क्या हो जाएगी सर ये हो जाएगी आपकी फॉरेक्स इन दोनों सर्कल्स को सी और दी पर टच किया अब उसने आपसे ये कहा की यह जो पैरेललोग्राम में कौन सा अकाद अब मेरा जो गणित या नॉलेज कहता है वो ये की सर आप पहले तो करो इकोसी से कनेक्ट फिर सर आप करो ए को दी से कनेक्ट अब आप कोशिश करो बनाने की आप कोशिश करो एक palogram बनाने की ठीक है सर मैं बनाता हूं तो पहले तो इसके पैरेलल एक लाइन खींचना हूं तो उसके पैरेलल लाइन कुछ ऐसी खींच डन क्या आई थिंक वो कुछ ऐसी हो जाएगी आई थिंक चीज आपको नजर ए रही है कोई तकलीफ तो नहीं है अब क्या करना होगा अब किस बात पर बात निकल कर आएगी उसे बारे में बात करते हैं क्या आपको ये नज़र ए रहा है की सर ये जो सीडी है दिस इस वैन ऑफ डी डायमीटर्स सॉरी daigonals ऑफ दिस पैरेललोग्राम बिल्कुल सर और जो दूसरा डायग्नल होगा वो क्या होगा सर दूसरा जो डायगोनल होगा वो आपका ये होगा जो दूसरा डायग्राम होगा वो आपका ये वाला होगा मैंने काफी कुछ आपसे कहा है काफी सारी बातें आपको समझने की कोशिश की है अब अगर आप चाहें तो आप मुझे आंसर बता सकते हैं अब आप चाहे तो मुझे ए के ऑर्डर्स ये जो आपकी आखिरी वर्टेक्स से चौथी यही आई है यह जो है यही आई है अगर आपको सहूलियत के लिए मैं इस क्वेश्चन को सॉल्व करना चाहूं तो मुझे ह कमा क कोऑर्डिनेट्स निकलने हैं एक हिंट देता हूं प्यार से सुनना सारी बातें भूल जाओ आप याद करो एक पैरेललोग्राम को आप याद करो एक पैरेललोग्राम को क्यों याद करें प्लीज याद करिए एक पैरेललोग्राम को लेकर मैं एक बहुत जरूरी बातें जानता हूं की सर पैरेललोग्राम के जो daigonals होते हैं ना सर पैरेललोग्राम्स होते हैं वह बायसेक्स करते हैं क्या आपको ये बात याद है बस यही कंडीशन आपको इस क्वेश्चन को सॉल्व करने में हेल्प करेगी पैरेललोग्राम के जो diagnals होते हैं वो बायसेक्स करते हैं मतलब मैं अगर आपसे सीधे-सीधे सिंपल सी बात कहूं तो एक बात बताओ ये जो अब का मिड पॉइंट होगा रिपीट माय सॉरी यह जो मिड पॉइंट होगा जो ए का मिड पॉइंट होगा जो की होगा यहां सर वह मिड पॉइंट क्लीयरली इस लाइन पर लाइक करेगा या आपकी होगी पहली कंडीशन यह हुई आपकी पहली हिंट दूसरी हिंट इस क्वेश्चन को सॉल्व करने की सर सीधी-सीधी सी बात है आप ध्यान से देखो जो आपकी स्ट्रेट लाइन है अब जो आपकी स्ट्रेट लाइन है अब उसे स्ट्रेट लाइन पर आपका ये पॉइंट लाइक करेगा कौन सा ए का मिड पॉइंट दोनों कंडीशन अप्लाई करो आप यह क्वेश्चंस आसानी से सॉल्व कर पाओगे मैंने काफी हद तक चीज दे दी है मैंने आपको हिंट बता दिए अब आपको महज ही एक क्वेश्चन सॉल्व करना है जो की बेशक आप कर सकते हो थोड़ा सा दिमाग लगाओ स्टूडेंट्स थोड़ा सा सोचो थोड़ी सी चीज देखो और यकीनन तौर पर आप निकल पाओगे सबसे पहला कम तो बस यही करो की मिढ्वाइंट निकल लो और मिड पॉइंट जो की क्लीयरली हमको दिख रहा है जो की दूसरा डायग्नल आपका वो कॉमन टांगें हो जाएगी वहां लाइक करेगा जल्दी से इसे अप्लाई कर लो भाई है और दूसरी बात में देख का रहा हूं सर यह जो लाइन अब है यह जो लाइन अब है यही आपका वो दूसरा वाला डायगोनल ए बना रही है क्या आप ये सोच का रहे हो आई होप आप ये बात सोच का रहे हो की वही वो कॉमन चोर्ड है और वहीं से वो चीज निकल कर आएंगे तो अगर वह आपकी दूसरी लाइन अब जो है वही आपकी दूसरा डायग्नल ए बना रही है तो जो आई cardinate होगा जो ए के कोऑर्डिनेट्स होंगे वो क्लीयरली लाइन अब पर भी लाइक करेंगे तो यकीनन तौर पर अगर आप अब की इक्वेशन निकल लें और ए के क्वाड्रन करवा दें तो ह कॉम के में एक एक क्वेश्चन यहां से इसको हम के में एक ही क्वेश्चन यहां से आप निकल सकते हो और आप कहां से ए जाएगा आई होप इन दोनों हैंड्स की बदौलत आपने इस क्वेश्चन को सोचा और ट्राई करने की कोशिश की मेरे अजिन बहुत टू बहुत मुश्किल बातें नहीं है स्टूडेंट्स एक बार जल्दी से इसे देखो सबसे पहला तरीका तो मेरा यही है सर की ए और आई का मिड पॉइंट निकलती हैं तो ए और ए का मिढ्वाइंट क्या होगा लेट्स कॉल इट कुछ भी का लो ए बी सी दी आई ले लिया ना तो मेरी सहूलियत के लिए का लेता हूं ऐप तो मेरे लिए जो होंगे वो होंगे सर ध्यान से देखना h+1/2 आप समझ रहे हो क्या और ऑफ कोर्स क्या है के + 2 / 2 और मैं ये एक बात जानता हूं की सर इन दोनों का इस इस डायगोनल का जो मिड पॉइंट है ऑफ कोर्स आपके इस डायगोनल पे लाइक करेगा तो वो 4X + 8y -7 को सेटिस्फाई करेगा फॉरेक्स जो की मैं रखने वाला प्लस 8 ए जो की मैं रखने वाला हूं ए की वैल्यू -7 को सेटिस्फाई करेगा एक्स और ए क्या है सर एक्स और ए है h+1/2 और के + 2 / 2 तो यहां हम रखेंगे h+1 ह प्लस वैन बाय तू और यहां पर मिल जाएगी ना तो ये हो जाएगा 2h+2 और ये कितना हो जाएगा 4K प्लस सिंपलीफाई कर लो फटाफट ये हो जाएगा 2K + 4 सॉरी 2h + 4K है ना और क्या आठ और दो 10 10 में से सात से प्राप्त किया तो कितना इस बात से कोई आप रखती नहीं है क्या मैं स्ट्रेट लाइन अब की इक्वेशन निकल सकता हूं आई होप आपको ये बात तो नजर ए रही है की ये जो एक और एड गए हैं ये क्लीयरली सर आपको नजर ए रहे होंगे आय होप आप ये बात समझ का रहे हो की ए से जो आपने पॉइंट्स ड्रॉ किए द कौन-कौन से पॉइंट ऑफ कॉन्टैक्ट उससे जो ये पैरेललोग्राम बन रहा है और ये जो डायगोनल आपका ए बन रहा है वो बी से पास हो गई एक सिमेट्री से आप देख का रहे होंगे तो मैं कहूंगा की सर बड़ी आसान सी बात है आप अब की इक्वेशन बताओ सर दो पॉइंट हैं तो अब लाइन की इक्वेशन क्या हो जाएगी मैं कहूंगा इक्वेशन जो अब की है वो क्या हो जाएगी सर वो हो जाएगी ए - y1 सो मैं लिख लेता हूं ए - 1 है ना इसे इक्वल तू Y2 -5 एंड सो 1 - 2 था कितना -1 है ना और 2 - 1 थॉट्स वैन टाइम्स एक्स माइंस कितना हो जाएगा एक्स - 2 और इससे आप थोड़ा सिंपलीफाई कर लीजिए सर आपको चीज दिखेगा माइंस वैन अंदर मल्टीप्लाई होगा तो - एक्स - एक्स इधर आए तो प्लस एक्स ये हो जाएगा एक्स + ए है ना माइंस वैन अंदर multiplayega तो + 2 + 2 इधर आया तो -2 और ये कितना हो जाएगा -30 = सो दिस हैपेंस तू बी डी स्ट्रेट लाइन कौन सी भाई ये वाली यह आपकी डायगोनल की इक्वेशन है और क्लीयरली सर अगर यह अब यानी ए डायगोनल की इक्वेशन है सो आई मस्ट सेटिस्फाई दिस इक्वेशन जीरो यहां पर ए जाएगा तो दूसरी इक्वेशन क्या आएगी ह प्लस के -3 = 0 ये आपकी दूसरी इक्वेशन ए गई अब अगर मुझे ऐसे सॉल्व करना है तो मैं कहूंगा सर एक कम करते हैं डबल कर देते हैं पुरी इक्वेशन को डबल किया तो 2h+2k और थ्री की जगह क्या ए जाएगा -6 सर सब्सट्रैक्ट कर देते हैं तो सब्सट्रैक्ट अगर किया तो प्लस है यहां और यहां है माइंस यहां प्लस है यहां -3 - और ये प्लस और ये क्या माइंस कोई दिक्कत तो नहीं है सर अब आप कर लीजिए सब्सट्रैक्ट तो 2X से 2h गया दिख रहा है क्या 9 = 0 तो सर यहां से आप जो के की वैल्यू निकलोगे वो होगा -9/2 के की वैल्यू आती है -9/2 अगर मैं माइंस नाइन बाय तू यहां रख डन क्योंकि ऑफ कोर्स इस इक्वेशन को सेटिस्फाई कर रहा था तो -9/2 यहां रखा तो एक्स की वैल्यू कितनी ए जाएगी सर एक्स की वैल्यू ए जाएगी -9/2 जो की उधर जाके प्लस हो जाएगा जो की उधर जाके प्लस हो जाएगा तो 9/2 + ऑफकोर्स कितना 3 और अगर में से ऐड करके लिखूं तो तीन और दो 3 2 = 6 9 + 6 15 तो ये टेक्निकल कितना हो जाएगा 15 / 2 ये टेक्निकल कितना हो जाएगा भाई 15 / 2 तो सर जो ह कमा क आपके कोऑर्डिनेट्स ए रहे हैं वो है 15 / 2 हम रिलीज सॉरी वो है 15 / 2 कॉमन 9/2 तो जो आपके ए के कोऑर्डिनेट्स होंगे जो आप कनक्लूड करोगे वो क्या होंगे भाई 15 / 2 9 / 2 आई थिंक आसान क्वेश्चन था ट्रिकी पार्ट जो रहा होगा जो शायद आपको क्रैक करने में मुश्किलात का सामना करना रहा होगा वो क्या होगा सर ये जो ए पॉइंट द जो रहा है आपका जो डायग्नल है ए बी से पास होता है पैरेललोग्राम का बी से पास होना तय है बस इस तरीके से आप दो इक्वेशंस बना के होमो के निकल लोग बहुत टॉप बोर्ड डिफिकल्ट क्वेश्चन तो नहीं है जहां पर आपके दो सर्कल्स इंटरसेक्ट हो रहे हैं याद रहेगा ये बातों के कंक्लुजंस इंटरप्रिटेशन उनके मीनिंग उनके ज्योमैट्रिकल इंटरप्रिटेशन कैसे निकले जाते हैं और उन्हें डाइजेस्ट करने की कोशिश करते हैं जहां हमसे कोई कहता है की सर एक सर्कल है और एक और सर्कल है और इस सर्कल के अंदर वो दूसरा सर्कल है तो होता क्या है करना क्या होता है बात क्या होती है वो समझते हैं पहले तो मैं सर्कल को समझने की कोशिश करता हूं की सर जब ये सर्कल में थोड़ा ऐसे ऐसा बना लेता हूं कोई आपत्ति किसी भी स्टूडेंट को अब आप खुद सोच के ये कहो बड़ी सिंपल सी साधन सी बात की सर देखो इस सर्कल का सेंटर यहां होगा इस सर्कल का सेंटर यहां होगा आप बताओ सर यह जो डिस्टेंस है यहां से सर्कल की सरकम्फ्रेंसेस से सर्कल की जो डिस्टेंस होगी इसे मैं का लेता हूं R1 और इसके सेंटर को क्या लेता हूं C1 इस सर्कल के सेंटर से इसकी जो डिस्टेंस होगी इसकी सरकम्फ्रेंसेस की ये जो डिस्टेंस होगी इसे आप क्या कहोगे r2 और इसके सेंटर की कोऑर्डिनेट्स को आप क्या कहोगे C2 कोई आपत्ति किसी भी स्टूडेंट को अब आप खुद गौर से देखोगे भाई आप खुद सर ध्यान से नोटिस करो आप देख पाओगे की सर जो C1 और C2 के बीच के डिस्टेंस है जो C1 और C2 के बीच के डिस्टेंस है और जो R1 और r2 का डिफरेंस है उसमें कुछ तो रिलेशन है आप खुद देखो ये पूरा R1 है मतलब अगर मैं आपको दिखाने का समझने की कोशिश करूं थोड़ा विजुलाइज करना घबराना मत बस ध्यान से सुनना ये पूरा आपका R1 है सन रहे हो क्या और ये इतना सा आपका r2 है और अगर उसे R1 में से r2 आपने सब्सट्रैक्ट भी कर दिया तो भी आपके पास काफी स्पेस बच जा रहा है कैसा वाला देखो ध्यान से ये R1 है आप समझ रहे हो और ये आपका r2 है समझ रहे हो क्या मतलब मैंने थोड़ा सा नीचे बनाया घबराना मत अब जब आप R1 में से आर तू अट्रैक्ट भी करते रहे हो तो सर आपको ऑफ कोर्स ये वाला स्पेस और ये वाला स्पेस बच जा रहा है आप मेरी बात समझ सर आप यह बातें क्यों कर रहे हो मैं आपसे ये बात इसलिए कर रहा हूं क्योंकि यही वो यही वो नेसेसरी और सफिशिएंट कंडीशन है जो हमारी हेल्प करेगी ये जानने में की कोई दो सर्कल्स इस तरीके से plesed है या नहीं सबसे पहले तो मेरा कैंसर यही होगा ना की उसे सर्कल को इस तरीके से जानना प्लीज करना या सिचुएशन करना उसे पैरामीटर को इंश्योर करके की वो इस तरीके के इन हालातो में है कहीं पर वो डिसजॉइंट हुआ करते द कहीं पर वो externalli टच करते द कहीं पर वो इंटरनल टच करते द कहीं पर वो दोनों दो पॉइंट्स पर इंटरसेक्ट किया करते द लास्ट लेक्चर तक आज उन दोनों ने एक दूसरे के अंदर ए चुके हैं और इंटरनल टच भी नहीं कर रहा है पूरा मतलब दूर है आइसोलेटेड है पर अंदर है तो कंटेन किया हुआ है तो यह सिनेरियो तो इसको इंश्योर करने के लिए मेरी पहली कंडीशन मेरा पहला पैरामीटर मेरा पहला चेक तो यही होगा की उन दोनों के सेंटर्स के बीच के डिस्टेंस हमेशा कम होगी उन दोनों के रेडियस के डिफरेंस से याद रहेगा बहुत मुश्किल बहुत टू बातें नहीं है इसे अच्छे से विजुलाइज करके रखना अच्छे से सोच कर रखना और इस बार इस बार अगर मैं आपसे पूछूं की सर इस बार टांगें को लेके क्या खास बात है आपकी कॉमन टांगेंट्स बनेगी क्या मुझे जरा सोच के बताओ आप लोग कॉमन टांगेंट्स कहां बनाओगे मतलब जो अगर मैं इसकी कोई टेंशन बनाने गया तो उसकी नहीं बनेगी क्योंकि सिर्फ इंटरसेक्ट करेगी और इसकी कोई टेंशन बनाई तो इसे टच नहीं करें क्या मेरी बात समझ का रहे हो तो सबसे खास सबसे जरूरी बात की यह वो स्पेशल केस है जहां पे कोई भी क्या नहीं होगी जहां पर कोई भी कॉमन टैसेंट नहीं होगी आई होप आप ये बात बहुत अच्छे से ध्यान रखेंगे इस बात का आपको हमेशा इल्म रहेगा और क्या सर और एक बात अभी भी तय है एक बात अभी भी तय है क्या सर अगर आप इन दोनों के सर्कल्स के सेंटर्स को कनेक्ट करते हुए जो लाइन जा रही है इसे देखोगे तो इसी पर परपेंडिकुलर आपको एक स्ट्रेट लाइन जाती हुई दिखेगी ऑफ कोर्स बहुत सारी दिख सकती हैं बट मैं एक खास स्ट्रेट लाइन की बात कर रहा हूं और उसे स्ट्रेट लाइन की खासियत ये होगी उसे स्ट्रेट लाइन की खासियत यह होगी सर की वहां से ड्रॉप किए गए वहां से ड्रॉप किए गए तेनजेन की लेंथ इस पर और इस पर इक्वल होगी है और वह क्या होती है आप मुझे सिखाओगे क्या सर लोकस ऑफ ए पॉइंट व्हिच मूव इन सच ए वे फ्रॉम डीज तू सर्कल्स आर ऑफ इक्वल लेंथ दें डेट लोकस इस नथिंग बट डी रेड कलर एक्सिस आई होप यू ऑल रिमेंबर डेट याद ए रहा है क्या कुछ याद ए रहा है सर तो दिस इसे नथिंग बट डी रेडिकल एक्सरसाइज और रेडिकल एक्सरसाइज का कॉन्सेप्ट तो बड़ा सुलझा हुआ है सर अगर ये आपका सर्कल S1 है और अगर आपके सर्कल S2 है सो दिस रेडिकल एक्सरसाइज गिवन बाय नथिंग बट F1 - S2 इसे इक्वल तू जीरो विच सी ऑल रिमेंबर वेरी प्रेसीजली कोई डाउट तो नहीं आई थिंक ये बात तो आप सभी को याद है सर की बिल्कुल सर दो सर्कल्स दो सर्कल्स का रेडिकल एक्सेस क्या होता है ये बात तो बिल्कुल मत बोलना क्योंकि इस पर बेस्ड कई सारी प्रॉब्लम्स को हमने बहुत अच्छे से सॉल्व कर लिया है इसकी हेल्प से और ये बहुत सिग्निफिकेंट है ये क्वेश्चंस को सॉल्व करने के लिए पर्सपेक्टिव नजरिया और कई बार चीजों को बहुत आसान बना देता है जो आपने हर तरीके में देखा होगा याद ए रहा है क्या तो आई थिंक ये दो तीन बड़ी मेजर सी बात है जो आपको समझनी चाहिए क्या पहला तो वही की दोनों के सेंटर्स की बीच के डिस्टेंस उसके रेडियस के डिफरेंस से हमेशा कम होगी अगर दोनों इस तरीके से प्लेस ने की एक सर्कल को दूसरे सर्कल ने कंटेन किया हुआ है विदाउट टचिंग इंटरनल दूसरी बात दूसरी बात सर रेडिकल एक्सरसाइज यहां पे मिलेगा तीसरी बात डेट विल बी नो कॉमन सेंस देयर विल बी जीरो कॉमन टेंसेज कोई कमेंट टैसेंट होने का चांस ही नहीं स्कोप ही नहीं है अब बना के बता दो मैं आपको रिवॉर्ड दूंगा आई विल गिव यू डी नोबल प्राइज एंड मैथ्स का होता नहीं है मैथ्स का ए बिल प्राइस होता है बट फिर भी आप बना के बता दिया मुझे बहुत खुशी होगी आप ढूंढ लीजिए आप आराम से वक्त लीजिए अपना और बनाया आप मत करिएगा वैसे तो पर अगर आपको मैन हो रहा है की नहीं है सर आप तो गलत हो मैं बना दूंगा तो बनाओ जब इस तरीके से प्लेसमेंट होगा तो कोई टेंशन नहीं होगी आई होप ये बातें याद रहेगी नॉट मीत डी सर्कल्स इन सर्कल्स हैव नो कॉमन टांगेंट्स प्लीज ये बातें याद रखिए यह बात प्लीज याद रखिएगा की सर्कस की कोई भी कॉमन टैसेंट नहीं होती है आगे बढूं क्या चलो आगे बढ़ते हैं कम की बात पर आते हैं अब यहां पर कम की बात यही है की हम तो सीधे कुछ क्वेश्चंस ट्राई करेंगे ऐसे कैसे सर कुछ पढ़ा ही नहीं आपने क्वेश्चन पे ए गए करो भाई कुछ पढ़ाया है मैंने वही अप्लाई करना है इस क्वेश्चन सोच के देखो ये क्वेश्चन टू नहीं है कैसे टू नहीं है सर ध्यान से देखते हैं पहले क्वेश्चन पढ़ते हैं वो का रहा है थीटा दिन पूर्व डेट डी लेंथ ऑफ कॉमन कोड ऑफ तू सर्कल बहुत मुश्किल हो बहुत बड़ी हो बहुत टू हो और उन्हें कंप्लेंट करना मुश्किल हो तो छोटे हसन में तोड़े और फिर सॉल्व करिए ब्रेक डी प्रॉब्लम इन पार्ट्स उन छोटे-छोटे पार्ट्स के छोटे-छोटे सॉल्यूशंस निकालिए और उन छोटे-छोटे सॉल्यूशंस को क्लब कर दीजिए आपकी पुरी प्रॉब्लम का सॉल्यूशन ए जाएगा और टेक्निकल इंटरनल में इस क्वेश्चन में यही करूंगा मैं थोड़े-थोड़े हिस्से उठाऊंगा थोड़े-थोड़े तंग उठाऊंगा उन्हें तोडूंगा क्योंकि बड़ी प्रॉब्लम को ऐसा होल सॉल्व करना शायद हमारे लिए मुश्किल हो और उसके छोटे-छोटे हिस्से आप यकीनन तौर पर सॉल्व कर सकते हैं क्योंकि इस क्वेश्चन में जो भी बातें चाहिए ना वो सब आपको पढ़ और सिखाई गई हैं आप कर सकते हैं मेरा यकीन करिए जैसे-जैसे मैं सॉल्व करते जाऊंगा ना आपको हर एक स्टेप पर एस होते चला जाएगा की सर ये तो आता है ये तो पता था ये तो पड़ा है ये तो पता था सर फिर आपने क्यों ऑफलाइन नहीं किया क्योंकि आप घबरा गए इस क्वेश्चन को देख कर इतना अजीबो गरीब एक्सप्रेशन लेकर और ये लिंगो की टर्मिनोलॉजी है भाषा पढ़ के और ये एग्जाम यही करेगी एग्जाम साइकोलॉजी के लिए आपके साथ खेलेगी एग्जाम साइकोलॉजी के लिए आप पर प्रेशर डालेंगे और उन्हें बैरियर्स को आपको ब्रेक करना है ब्रेक डी प्रॉब्लम इन स्मॉलर पार्ट्स छोटी-छोटी सीढ़ियां भी तो चढ़ सकते हो सर छोटी छोटी सीढ़ियां चढ़ने से क्या फायदा होगा छोटी-छोटी चिड़िया सन्डे से पहले तो ये फायदा होगा जनाब की आपको ऐसे लीव्स नहीं लेने पड़ेंगे इतने बड़े-बड़े जंप्स नहीं मरने पड़ेंगे दूसरा क्या फायदा होगा दूसरा फायदा देखो ना आप स्पीड बढ़ा लोग आपकी फटाफट निकल जाओगे तो कछुआ बनने में कोई बुराई नहीं है लोगों को लगता है की सर शायद कछुआ धीरे चलता है लेकिन मेरा यकीन करिए कछुआ धीरे और लंबा चलता है स्लो इन स्टडी विंस द रेस आप लंबी-लंबी छलांगा भर के शायद एक-दो बाल लगेगा की देखो कितना सारा कवर हो गया एक छलांगा में सर इतना सारा कवर कर लिया पर मेरा यकीन करिए आप लंबी रेस के घोड़े नहीं है फिर आप बहुत जल्दी थक जाएंगे तो धीरे लेकिन कंटीन्यूअसली कंसिस्टेंटली रेगुलरली चलते रहिए चलो ज्ञान बहुत हुआ कम की बात पर आते हैं मेरे ख्याल से बस ये इसलिए की आप इस क्वेश्चन को देखकर गई थी की घबराओ मत लाइफ में प्रॉब्लम बहुत डिफिकल्ट नहीं है ना आसान है चलो इस क्वेश्चन को ट्राई करते हैं देखो क्या लिखा है समझना मैं क्वेश्चंस समझता हूं फिर आप ट्राई करना आईएफ देता इस डी एंगल बिटवीन तू रहता है तो सर मेरे पास दो रेड आए हैं और रेड आय के बीच का एंगल है जो की वो का रहा है क्या है भाई थीटा क्या ये बात आपको समझ आई ठीक है सर तो दो सर्कल्स आपको दिए गए हैं दो सर्कल्स आपको दिख रहे हैं क्या एक तो है आपका x² + y² = a² और एक सर्कल है आपका एक्स - c² + y² = b² और इन दोनों की जो reaye है इन दोनों की जोड़ा है उनके बीच का एंगल है थीटा थीटा कहां से सर थीटा कहां से इस बात को पढ़ना ड्रोन फ्रॉम वैन ऑफ डी पॉइंट ऑफ इंटरसेक्शन ऑफ तू सर्कस मेरा यकीन करिए इस क्वेश्चन की सबसे जरूरी लाइन ये है बात समझने की कोशिश करना बात बहुत ध्यान से सुनना सर पहली बात तो आपके पास दो सर्कल्स हैं बना लेते हैं एक सर्कल रहा ये और दूसरा सर्कल रखा है इस बात को θ एक सर बात करेंगे इस बारे में वह का रहा है दो रेडियस के बीच में एंगल ठीक है सर कौन सी दो रेडियस ड्रॉन फ्रॉम वैन ऑफ डी पॉइंट ऑफ इंटरसेक्शन सर दो सर्कल मैक्सिमम दो पॉइंट्स पर इंटरसेक्ट होते हैं वर्ना ओवरलैप ही कर लेंगे तो दो सर्कल्स में जब दो पॉइंट्स में इंटर सेट किया तो उनमें से एक पॉइंट से एक पॉइंट से उसे पॉइंट से मैंने उनकी रेडियस बनाई दोनों सर्कल्स की रेड आय बनाएं सर दोनों सर्कल्स के बारे में एक मिनट बात करता हूं फिर ट्विटर पे आऊंगा पहला सर्कल है x² + y² = a² सर इससे अगर मैं देखता हूं ना तो इस सर्कल के बारे में मेरा कहना होता है किसका सेंटर होगा जीरो कमा जीरो पर और उसकी रेडियस होगी इससे अगर मैं देखता हूं तो इसका सेंटर होगा सी कमा जीरो पर और उसकी रेडियस ऑफ कॉस्ट कितनी होगी बी क्या मेरी बातें बहुत प्रॉब्लम सर्कस को थोड़ा बना देता हूं सर एक सर्कल तो सेंटर है जीरो कमा जीरो पर और उसकी रेडियस है ना बना लेंगे एक सर्कल आपका सेंटर है कहां जीरो सी कमा जीरो पर और उसकी रेडियस है बी ठीक है सर ये दोनों बातें हम कर लेंगे वो ये का रहा है की इनका जो पॉइंट ऑफ इंटरसेक्शन है ना उसमें से किसी एक पॉइंट से आपने दोनों सर्कल की रेडियस बनाएं दोनों सर्कल की रिटायर बने हैं एक तो ये रेडियस हो जाएगी भाई क्या आपको बात समझ ए रही है और इसी पॉइंट ऑफ इंटरसेक्शन से इसकी भी एक रेडियस मैंने आई थिंक ये हो जाएगी और इन दोनों के बीच का जो एंगल है ना सर इन दोनों के बीच का जो एंगल है वो है थीटा इन दोनों के बीच का जो एंगल है थीटा सर एक छोटा सा कम और कर लें क्या इन दोनों के सेंटर्स को ज्वाइन कर दें कर लो भाई मुझे कोई दिक्कत नहीं सर एक बात मैं जानता हूं फॉर सर कौन सी बात भाई एक बात नहीं यकीनन तौर पर जानता हूं की इस सर्कल की जो रेडियस थी वो कितनी थी ए तो ये जो लेंथ होगी ये कितनी होगी ए और ये लेंथ कितनी होगी बी आई थिंक अगर मैं आपको समझाना चाहूं तो इसे मैं का लेता हूं C1 और इसे हम का लेता हूं c2o इस पॉइंट को लेते में थोड़ी देर के लिए क्या लेता हूं ए कोई परेशानी तो नहीं है और थीटा उसने किसे आपको बता दी बताया है थीटा आपने उसने इस एंगल को कहा है θ आई होप बहुत चीज टू डिफिकल्ट और टू नहीं है अच्छा बस ऐसे ही शायद अगर कभी मुझे भविष्य में जरूरत पड़े तो क्या मैं C1 और C2 के बीच के डिस्टेंस निकल सकता हूं 0 से सी कमा जीरो की डिस्टेंस सी होगी तो ये जो C1 और C2 के बीच की डिस्टेंस है वो है सी ये पॉइंट का cardinate नहीं है सी ये डिस्टेंस है C1 और C2 के बीच मेरे ख्याल से मैंने आपकी काफी हद तक अब मदद कर दी है मैंने यकीन करिए मैंने आपकी ऑलमोस्ट मदद कर दी है बस अब आपको ये प्रूफ करना है की जो कॉमन कॉर्ड की लेंथ है इन दोनों सर्कल्स के बीच में उसकी लेंथ ये होगी जो आपको दी गई है मेरा सच में यकीन करिए अब ये क्वेश्चन बहुत टू नहीं है सर ऐसे कैसे टू नहीं है अच्छा आप सारी बातें छोड़ो आप तो मुझे ये बताओ क्या आपको याद है हमने कोसिन रूल पढ़ा था कोसिन रूल अगर आपको याद है तो क्या मैं इस ट्रायंगल A7 इस ट्रायंगल ए C1 C2 में कोसिन रूल अप्लाई कर सकता हूं क्या कहता है सर अगर आप इस थीटा का क्रॉस रेश्यो लेना चाहते हो वांटिंग तू टेक कोस्थेटा अगर आप कोस्थेटा लिखना चाहते हो तो इस थीटा को बनाने में किन दो साइड्स ने हेल्प की ac1 और ए सी टन है उनकी लेंथ क्या है ए और बी तो उनके स्क्वायर का सैम करिए पहले तो आप पहले लिख दीजिए a² + b² उसमें से सर आप सब्सट्रैक्ट करिए उसमें से किसे सब्सट्रैक्ट करिए जो थीटा के अपोजिट साइड है उसकी लेंथ तो वो क्या है सी तो उसके स्क्वायर को सब्सट्रैक्ट कर दीजिए और इससे डिवाइड कर दीजिए तू टाइम्स इन दोनों साइज के प्रोडक्ट से बिल्कुल अच्छी बात है इससे क्या इससे क्या मैं कुछ ना कुछ करके यहां तक पहुंच पाऊंगा आप ऐसे तो नहीं पहुंच पाओगे लेकिन एक बात पे बात करते हैं एक एक बात पर बात करते हैं कौन सी बात पर बात करते हैं मुझसे क्या पूछा जा रहा है इस कार्ड की लेंथ तो ये कार्ड क्या है सर ये आपकी कॉर्ड है आपसे इस कोड की लेंथ थी तो पूछी जा रही है आप समझ का रहे हो बात को आप से स्कॉट की लेंथ ही तो पूछी जा रही है इस कॉल्ड की जब लेंथ पूछी जा रही है तो मैं ये देख का रहा हूं बड़े अच्छे से सर बड़े साधारण बड़े सुलझे हुए तरीके से की सर जब सी वैन और सी तू को आप कनेक्ट करते हैं तो मैं जानता हूं की जो कॉमन गोद है ये आपके रेडिकल एक्सेस भी है और रेडिकल एक्सेस दोनों सेंटर्स को ज्वाइन करने वाली लाइन पर परपेंडिकुलर होती हैं तो मैं ये बिल्कुल अच्छे से समझ का रहा हूं की ये आपकी परपेंडिकुलर होगी क्या आप मेरी इस बात से एग्री करते हो की सर ये परपेंडिकुलर होना आता है आई रिपीट माय स्टेटमेंट जो कॉमन कार्ड होती है वो ऑफ कोर्स रेडिकल एक्सिस भी होती है इस केस में और वो सेंटर्स को कनेक्ट करने वाली लाइन पर परपेंडिकुलर होती है मतलब आप एक बात सुनना ध्यान से इस बात को सुनना अगर मैं इस पॉइंट को थोड़ी देर के लिए कहूं अगर मैं इस पॉइंट को थोड़ी देर के लिए कहूं अगर मैं किसी भी तरीके से एम निकल लो ना अगर किसी भी तरीके से एम निकल लूं तो जो एम का डबल होगा ना वही आपकी कमेंट की लेंथ होगी ऑफ कोर्स हम जानते हैं सर सर कल के सेंटर से अगर कोर्ट पर परपेंडिकुलर ड्रॉप किया जाए कट बायसेक्स हो जाती है तो जो कॉमन कोड तो एम का डबल ही स्क्वाड की लेंथ होगी आई होप में बहुत कंफ्यूज नहीं कर रहा हूं अब वापस आते हैं हम हमारे ही इसी नॉलेज पर की हम तो बात कर रहे द सर ए सी वैन सी तू पर हम तो बात कर रहे द अगर आप बात कर रहे द सर ए सी वैन सी तू पर तो मुझे दो चीज नजर ए रही हैं सर वो क्या क्या आप मुझे हेल्प कर सकते हैं इस ए C1 C2 ट्रायंगल का एरिया निकलने में मेरा क्वेश्चन है यार अगर मेरी हिंट जो आपको दी गई है वो ये की क्या आप मेरी मदद कर सकते हैं इस ट्रायंगल एक वैन सी तू का एरिया निकलने में सिर्फ बिल्कुल मदद कर सकते हैं कौन सी बड़ी बात है कौन सी बहुत मुश्किल वाली बात है आप दो तरीकों से इस ट्रायंगल का एरिया निकल सकते हो सर कौन से दो तरीके एक तो आप सिन थीटा उसे करके निकलती हो और एक हाफ बेस हाइट करके निकलती हो 2 मिनट सर आप कुछ-कुछ का रहे हो बट समझ नहीं ए रहा है एक मिनट ये सारी बात को हटाओ आप याद करो हमने ट्रायंगल के बारे में बात की अगर मैं ट्रायंगल का एरिया निकलना चाहता हूं तो सुनना ध्यान से स्टूडेंट्स एक तरीका से परपेंडिकुलर ड्रॉप करें और अगर आप कोई हाइट पता चल जाए और अगर आपको यह बेस पता हो अगर आपको हाइट और बेस पता है तो आप इस ट्रायंगल का एरिया कहते हैं हाफ बेस हाइट आई होप ये फॉर्मूले से आप सभी लोग परिचित हैं बिल्कुल परिचय का एक और तरीका होता है स्टूडेंट्स की अगर मैन लो मुझे यहां पर ये एंगल पता है मैन लो अगर मुझे एंगल पता है जैसे मैं थोड़ी देर के लिए का देता हूं लेट्स से कुछ भी कुछ भी है ना अभी लेट से वहां देता है तो यहां कुछ और ले लेते हैं लेट्स से ये एक साइड है ना ये है साइड आपकी बी [संगीत] लेंथ हो जाएगी और इस साइकिल लेंथ हो जाएगी तो अगर सर मुझे ये एंगल मैन लो पता है और मुझे इस एंगल को क्रिएट करने वाली दोनों साइड्स की लेंथ पता है तो भी मैं इस ट्रायंगल कैरियर लिख सकता हूं जो की हमारे अकॉर्डिंग होता है हाफ उन दोनों साइड्स का प्रोडक्ट जो की है ए और सी और उनके बीच में कंटेंट एंगल का साइन रेश्यो यानी साइन भी सर ऐसा कैसे होता है ये हम पढ़ेंगे जब हम प्रॉपर्टीज या सॉल्यूशंस ऑफ ट्रायंगल पढ़ेंगे तब हम बहुत डिटेल में इस बारे में बात करेंगे बहुत आसान सा प्रूफ है बहुत डिफिकल्ट बात नहीं है मैंने त्रिकोण उसे की है बेसिक राइट एंगल ट्रायंगल वाली और उसको इसी फॉर्मूले को उसमें कन्वर्ट किया बहुत मुश्किल लगी है तो यह दोनों ही फॉर्मूले आपके ट्रायंगल के एरिया के फॉर्मूले हैं सर ये क्यों पढ़ा रहे हो ये इसलिए पढ़ा रहा हूं स्टूडेंट्स क्योंकि प्लीज इस बात को ध्यान से सुनो अगर मैं यहां पर ट्रायंगल का एरिया निकलता हूं तो मैं हाफ बेस हाइट ही हेल्प से ये एम तक पहुंच सकता हूं सुनना ध्यान से अगर ट्रायंगल के लिए फॉर्मूला उसे करता हूं कौन सा फॉर्मूला इसी ट्रायंगल के लिए कौन सा एक वैन सी तू ए सी वैन सी तू का एरिया निकाला जा रहा हूं तो एक फॉर्मूला क्या कहता है एक फॉर्मूला कहता है हाफ 20 20 क्या है C1 C2 बेस है C1 C2 हाइट क्या है सर कोई दिक्कत तो नहीं है सर और एक फॉर्मूला और हमने पढ़ा है कौन सा सर एक फॉर्मूला और हमने पढ़ा है इसी ट्रायंगल के एरिया का जो की होता है 1/2 किन्ही दो साइड्स का प्रोडक्ट किन्हीं दो साइड्स का प्रोडक्ट तो वो मैं ले लेता हूं अब और उनके बीच कंटेंट एंगल का साइन रेश्यो तो वो मैं ले लेता हूं सिन थीटा क्या इन सारी बातों से किसी को भी कोई तकलीफ मैं ऐसा इसलिए कर रहा हूं क्योंकि शायद मुझे C1 C2 ए बी साइन थीटा पता है और अब मैं शायद ए एम तक पहुंच जाऊं और सोच का रहे हैं क्या शायद ध्यान से अगर आप देखो तो मैं वापस आता हूं इस बात पर लेकिन उससे पहले इससे थोड़ा देखते हैं क्या मैं 2ab को इधर ला सकता हूं सुनना ध्यान से ये क्या लिखा हो गया कोस थीटा सन रहे हो क्या आप लोग और उसके इक्वल क्या लिखा हुआ इसके इक्वल लिखा हुआ है a2+b2 कर सकता हूं कर सकते हैं सर तो ध्यान से देखना c² को मैं क्या लिखूंगा मैं c² को लिखूंगा ए स्क्वायर प्लस बी स्क्वायर प्लस बी स्क्वायर माइंस 2ab कोस्थेटा आई थिंक ऐसा मैं कर सकता हूं ऐसा क्यों कर रहे हो बस थोड़ा पेशेंस रखिए एंड आई थिंक यहां से मैं व निकल सकता हूं क्या कर दूंगा सी के लिए मैं यहां से स्क्वायर हटा दूंगा और इस तरफ मैं क्या पेस्ट कर दूंगा एक अंडर रूट आय होप ये अंडर रूट आपने पूरा इस पूरे एक्सप्रेशन पर पोस्ट किया है क्यों कर रहे हो सर बस थोड़ा सा पेशेंस रखिए अब एक बात का जवाब दीजिए क्या मैं 1 / 2 से 1 / 2 कैंसिल कर का रहा हूं बिल्कुल कर का रहा हूं अब सुनना इस एम को निकलने के लिए मैं क्या करूंगा C1 C2 को वहां पहुंचा दूंगा तो सुनना ध्यान से एम क्या आएगा आप ध्यान से देखना जो आपका एम आएगा सर वो होगा जो आपका एम आएगा सर वो होगा अब साइन थीटा तो न्यूमैरेटर में आई विल हैव ए बी साइन थीटा और डिनॉमिनेटर में क्या होगा सर डिनॉमिनेटर में कुछ चला जाएगा बस इस पार्ट को देखो क्वेश्चन खत्म हो जाएगा तो क्या आप सोच का रहे हैं स्टूडेंट्स की यहां से क्या करेंगे हम मेरे ख्याल से मेरा कहना है वही बात जो हमने निकालनी चाहिए की हम जब एम जानना चाह रहे हैं तो absintheta न्यूमैरेटर में और डिनॉमिनेटर में जाएगा C1 C2 क्योंकि ये हाफ से हाफ कैंसिल C1 C2 मतलब C1 से C2 के बीच के डिस्टेंस सी और सी हमारे लिए क्या है सी हमारे लिए है अंडर रूट ओवर a² + b² - 2ab टाइम्स कोस थीटा तो शैल आय से अब ये ए एम को जब हमने लिखा तो कितना ए रहा है सर ये ए रहा है a² + b² - 2ab कोस्थेटा एक छोटा सा कम सर क्योंकि आप एम में इंटरेस्ट रेट नहीं द आप तो एक कोड ऑफ कॉन्टैक्ट तो अलग बात हो जाएगी ये नहीं है कोड ऑफ कॉन्टैक्ट वो किसी और केस में बनती है ये तो है आपकी कॉमन कोड तो ये जो कॉमन कोड है आप इसकी लेंथ जानना चाह रहे द तो इसकी लेंथ जो होगी वो होगी एम का डबल तो ए एम का डबल करने के लिए मुझे क्या करना पड़ेगा मुझे बड़ा बेसिक सा डायरेक्ट साधारण सा कम करना है यह डबल सो दिस इस व्हाट दिस इस योर आंसर ओर दिस इस बेसिकली लेंथ ऑफ गोद आई थिंक एक आसान सा क्वेश्चन था ये बहुत टू बातें नहीं और ऐसे क्वेश्चंस आपको के डबली एडवांस्ड पूछेगा ऑफ कोर्स वहां पे हो सकता है क्या दिया हो वहां पर ऐसे वैरियेबल्स ना दिया हो मतलब हो सकता है ये कुछ वैल्यूज दे दी जाए यहां पर कुछ न्यूमेरिकल वैल्यू दीजिए और फाइनली ये एक्सप्रेशन पूछा जाए तो हमने अभी एक डेरिवेशन सिखा algebraicly पर इस तरीके से चीज निकल सोची और समझिए जाएंगे तो मैं कभी भी नहीं कहूंगा की आप फॉर्मूले रखती है आपने चीज अच्छे से समझिए ना तो आप आसानी से आंसर निकल लेंगे पर आंसर्स निकलने के लिए एक अच्छी अंडरस्टैंडिंग ऑफ ज्यामिति होनी चाहिए तब जाकर आप चीज सोच पाएंगे आई होप चीज है डिफिकल्ट नहीं लग रही और ग्रैजुअली आप नई नई चीज एक्सप्लोर करते हो सीखते चले जा रहे हैं और आप बहुत एंजॉय कर रहे हैं इन चीजों को भी देखो इस तरीके से एक्सप्लोर कर लेते हैं आई होप यह बात समझ आई अगर यह बात समझ आई तो क्या एक और क्वेश्चन ट्राई करें स्टूडेंट्स एक और क्वेश्चन देख लेते हैं यहां पर क्या लिखा है देखो भाई वो का रहा है इफ डी टांगें आर ड्रोन तू दिस सर्कल आते डी पॉइंट वेयर इट मीट्स डी सर्कल दें फाइंड डी पॉइंट ऑफ इंटरसेक्शन ऑफ दी स्ट्रेंज ये तो कुछ पड़ा हुआ ऐसा लग रहा है सर ऐसा तो हम जानते हैं या ऐसा निकाला जा सकता है कुछ बारीकियां अगर हमें याद है तो और क्या का रहे हो अलसो फाइंड डी लेंथ ऑफ कॉमन कोड लेंथ ऑफ कॉमन कोड यह शायद पिछले ही क्वेश्चन का एक्सटेंशन हो जाएगा अगर पूछा जाए तो आई थिंक अब चीज आप समझ पाई की कैसे सोचनी है बिल्कुल सर ट्राई करते हैं हमसे पहले तो कहा जा रहा है टांगें रॉन तू दिस सर्कल तो इस सर्कल पर टांगें ड्रॉ की जा रही क्वेश्चन सुनना क्वेश्चन हम करेंगे पहले अच्छे से विजुलाइज है ना तो सर एक सर्कल तो ये रहा आपका क्या ax² + y² = 12 और ये सर्कल इस सर्कल को कर रहा है इंटरसेक्ट तो लेट से यह सर्कल को इस तरीके से कर रहा है इंटरसेक्ट है ना मैं इसे बना लेता हूं थोड़ा छोटा क्योंकि मैं आपको कोई चीज बताना चाह रहा हूं जो मैं समझा पाऊंगा थोड़ा आसानी से वो जैसा भी हो बस में तो बना ले रहा हूं है ना मुझे नहीं पता इसकी रेडियस इससे ज्यादा कम है बस बना ले रहा हूं कहीं ना कहीं जाके चीजें कन्वर्ट या मंगल हो रही होंगी आप suniyega कम के बाद ध्यान से आप सर इन दोनों की क्या है इन दोनों पर मैंने एक चीज इस तरीके से ड्रा की जहां पर आपका ये पॉइंट ऑफ इंटरसेक्शन था जहां पर ये पॉइंट ऑफ इंटरसेक्शन था किसका इन दोनों सर्कल्स का उसमें मैंने किस पर इस सर्कल पर ने तो निकल लेते इसकी रेडियस कितनी होगी √12 इसकी रेडियस कितनी होगी √12 बस ऐसे ही निकल लेते हैं ताकि सहूलियत रहे वो विजुलाइज करने में सहूलियत है अंडर रूट 12 मोटा-मोटा तीन से चार के बीच होगा क्योंकि अंडर रूट 93 होता है रूट 164 होता है 3 से 4 क्वेश्चन मैन लेते हैं 3.5 बस अभी समझने के लिए है ना तो उसकी रेडियस में मैन लेता हूं 3.5 इसकी रेडियस क्या होगी सर 5/2 का स्क्वायर यानी कितना 25 / 4 बहुत लंबी हो जाएगा पर कर लेते हैं ये हो जाएगा वैसे इसकी जरूरत नहीं है जो मैं कर रहा हूं पर बस ये पता चल जाए इसलिए सुनना ध्यान से ये हो जाएगा 25/4 ये हो जाएगा 3/2 का स्क्वायर 3/2 का स्क्वायर कितना 9 / 4 आप समझ का रहे हो तो 25/4 + 9 / 4 25 + 9 कितना हो जाएगा 25 + 9 आई थिंक 34/4/4 में से तू हमें ऐड करना है तो 34/4 को क्या मैं 17 / 2 लिख सकता हूं तो हमें क्या करना है ध्यान से देखना 17 / 2 और उसमें क्या करना है तू ऐड तो क्या मैं 2 को 4/2 लिख सकता हूं बस ऐसे ही है ना और ये क्या हो जाएगी इसकी रेडियस तो आप देखो 17 और चार कितना होता है सही हो जाता है 21 21 / 2 21 / 2 होगा कितना 10 के आसपास और 10√ कितना होगा आई होप समझ का रहे हो दिस इस समथिंग अराउंड 10 और ये है अंडर रूट 12 ये है अंडर रूट 10 तो ये वाले सर्कल की रेडियस थोड़ी सी छोटी होगी है ना बस इतनी सी बात जानने के लिए मैंने ऐसा मैन में ही कैलकुलेट किया कोई बहुत खास बहुत बड़ा फर्क नहीं पद रहा है है ना तो ये आपका ये वाला सर्कल x² + y² = 12 अब सुनना नाम की बात वो ये का रहा है दो टांगेंट्स ड्रॉ की गई किस पर इस सर्कल पर मतलब इस पर कहां द्रव की गई उसे पॉइंट पर उसे पॉइंट पर जहां पर ये इससे इंटरसेक्ट करता है ठीक है सर ड्रा करते हैं suniyega ध्यान से मैंने जब ये टेंसेज ड्रा की तो मुझे क्या दिखेगा क्या समझ आएगा जरा उसे पे बात करते हैं देखो मैं यहां तेनजेन स्ट्रॉ कर रहा हूं थोड़ा सा रोते कर लिया जाए तो चीज आसान हो जाएंगी तो एक टांगें ऑफ कोर्स आपकी ये हो जाएगी कोई दिक्कत तो नहीं है सिमिलरली क्या मैं ऐसा कहूं की सर एक टांगें आपकी ये हो जाएगी क्या आप इस बात से एग्री करते हो की एक टैसेंट जो होगी आपकी कोई हो जाएगी और चीज काफी हद तक सिमिट्रिकल ही होंगी कोई तकलीफ तो नहीं है भाई कोई परेशानी नहीं है अच्छा सर एक और बात अगर मैं आपसे पूछूं तो क्या आपको यह नहीं दिख रही है की ये आपकी वो कॉमन कोड है यह आपकी वह कॉमन कोड है आप देख का रहे हो सर यह आपकी वह कॉमन कॉर्ड हो जाएगी आई होप आप समझ का रहे हो की यह है आपकी वह कॉमन कौन कोई दिक्कत तो नहीं है सो दिस इस गोइंग तू बी डेट कॉमन कट है ना ऑफ कोर्स यह डिस्टेंस अब अगर मैं आपको चीज समझाऊं तो सुनना ध्यान से सर आपको चाहिए क्या आप निकलना मैं निकलना चाह रहा हूं इंटेलिजेंस का पॉइंट ऑफ इंटरसेक्शन मैंने निकलना चाह रहा हूं इंटेलिजेंस का पॉइंट ऑफ इंटरसेक्शन अब आप एक बात बताओ अब आप एक बात बताओ इस पॉइंट ऑफ इंटरसेक्शन को अगर मैं थोड़ी देर के लिए मैन लूं थोड़ी देर के लिए लेट से ह कॉम के अगर इस पॉइंट ऑफ इंटरसेक्शन को मैं थोड़ी देर के लिए मैन लेता हूं ह ए सुनना है क्वेश्चन कितना आसान है क्या आप मुझे ह के से इस सर्कल पर ड्रा की गई कार्ड ऑफ कॉन्टैक्ट की इक्वेशन बता सकते हो याद करो स्टूडेंट्स यह बहुत डिटेल में बहुत अच्छे से हमने पढ़ा हुआ एक कॉन्सेप्ट है की सर अगर आपके पास ये सर्कल है अगर आपके पास एक सर्कल है ना और इस सर्कल पर आपने टैसेंट ड्रॉ की एक टैसेंट ये है सर मैन लो और ऑफ कोर्स एक टांगें आपकी लत से ये है है ना की आपको याद है स्टूडेंट्स और इस पॉइंट की अगर cardinate कुछ भी है लेट्स X1 y1 है ना और ये आपका एक सर्कल है जिसकी एक एक्शन है आपको याद है ना तो सर इन दोनों के जहां पॉइंट ऑफ इंटरसेक्शन निकल रहे हैं उसे पॉइंट पर जो बन रही है कोड ऑफ कॉन्टैक्ट उसकी इक्वेशन हम देते हैं हम देते हैं भूले तो नहीं हो क्या आपको यह बात याद है ऐसा क्यों पढ़ा रहे हो सर ऐसा मैं सिर्फ और सिर्फ इसलिए पढ़ा रहा हूं आपको क्योंकि अगर यहां पर भी मैं बात करूं तो ह कमा कैसे आपने इस सर्कल पर कोड ऑफ कॉन्टैक्ट ड्रा की है तो टी = 0 से अब टी = 0 में ह कमा के से मुझे इस पर ग = 0 बनाना है कोई तकलीफ कोई परेशानी नहीं है तो सर बिल्कुल बना लेते हैं कोई दिक्कत तो नहीं है भाई है ना तो इधर टांगें आर ड्रोन तू दिस सर्कल है ना एम रियली सॉरी मैंने थोड़ा सा उल्टा बना दिया है मैंने उल्टा इसलिए बना दिया क्योंकि मैंने tenjans इस वाले सर्कल पर बना दी मुझे टैसेंट इस वाले सर्कल के द्वारा बनानी चाहिए थी कोई बहुत बड़ा फर्क नहीं आने वाला है इस बात से कोई बहुत बड़ी विशेष खास बात नहीं हो जा रही है बस अगर मैं इसे ठीक से लिखना चाहूं तो मैं बस बड़ा मेजर सा फर्क ही ले आऊंगा की मैं इस वाली टांगें को कर देता हूं डिलीट और इस वाली टैसेंट को भी कर देता हूं डिलीट और मैं क्या कर लेता हूं यहां पर दो अलग-अलग टैसेंट ये बना लेता हूं तो एक टैसेंट तो आपको सर आपकी ये बनेगी है ना एक टांगें तो सरकार साहब की बनेगी यह आई होप यह बहुत ज्यादा कन्ज्यूरिंग नहीं है आपके लिए और सिमिलरली आपकी एक टैसेंट बनेगी ये जो की कुछ ऐसी जा रही होगी कोई तकलीफ किसी भी स्टूडेंट को और ये वो होगा पॉइंट ह कॉम के ना की है आई एम रियली सॉरी फॉर दिस वो मैंने इस वाले सर्कल को मैन लिया और उसे सर्कल्स का ही कन्फ्यूजन चल रहा था इसलिए और यह आपका सर्कल है जिस पर आपने टेंसेज ड्रा किया और यहां पर आपकी ये कोड ऑफ कॉन्टैक्ट है ये आपकी कोड ऑफ कॉन्टैक्ट कोई दिक्कत तो नहीं और ये सर्कल कौन सा है ये आपका ये सर्कल है कोई दिक्कत तो नहीं आई होप अभी चीजे डाइजेस्ट स्टोरी अब मेरा बस ये कहना है सर यहां पे टी = 0 बनाते हैं तो ह कमा के से इसके लिए t=0 क्या हो जाएगी सर X1 यानी क्या हो जाएगा एचसी + ए ए वैन यानी कितना हो जाएगा के ए और 12 को इधर ले आता हूं तो ये हो जाएगा -12 = 0 तो हमारे अकॉर्डिंग ये जो कोड ऑफ कॉन्टैक्ट है मतलब मैं इस पॉइंट को इससे का देता हूं ए और इस पॉइंट को का देता हूं अभी तो हमारे अकॉर्डिंग जो इक्वेशन है अब की वो क्या हो जाएगी वो ये हो जाएगी कोई दिक्कत तो नहीं है लेकिन एक मिनट आप ऐसे क्यों कर रहे हो इस क्वेश्चन को आप क्यों नहीं सोच रहे हो एक आसान सा तरीका जहां पर हमने ये सिखा है जहां पर हमने ये सिखा है सर की आप तो इस कोड ऑफ कॉन्टैक्ट को और आसान तरीके से निकल सकते हो आप तो इस कोड ऑफ कॉन्टैक्ट करो आसान तरीके से निकल सकते हो क्योंकि सर ये जो कोड ऑफ कॉन्टैक्ट है या रेडिकल एक्सिस भी तो है इसका और इसका और अगर इन दोनों का रेडिकल एक्सिस है अगर ये दोनों का रेडिकल एक्सरसाइज है तो सर आप क्या बोलोगे तो मैं कहूंगा की रेडिकल एक्सिस वाले नॉलेज से तो इसकी इक्वेशन में लिखूंगा s1-s2=0 लिखूंगा की नहीं तो लिख लो अब जब इसे सब्सट्रैक्ट किया तो ये क्या देगा सर ये आपको देगा 5x है ना -3 और ये है -12 मतलब माइंस ऑफ माइंस तू प्लस तू और ये इधर है माइंस 12 सो -12 +2 कितना हो जाएगा -10 तो सर आपकी कार्ड ऑफ कॉन्टैक्ट जो अपने यहां से निकल उसी को अगर आप इस तरीके से देखें तो यह तो रेडिकल एक्सेस है और उसे तरीके से अब की इक्वेशन ए जाती है मतलब आपके अकॉर्डिंग ये है और हमारे अकॉर्डिंग है सर दो लाइंस अगर से है आपको याद है क्या की अगर दो लाइन से है जैसे अगर मैं आपसे कहूं या दोनों लाइन से है a1x + b1y + C1 = 0 और एक और स्ट्रेट लाइन है जो की है a2x+b2y + C2 = 0 तो आप कहते हो अगर ये दोनों लाइन से है मतलब एक दूसरे पे एग्जैक्ट ओवरलैप कर रही हैं तो आप कहते हो इनके cofficients के रेश्यो से होते हैं मतलब A1 A2 b1 B2 और C1 C2 का जो रेश्यो होगा ना वो बिल्कुल से होगा आप ये कहते हो आप ये कनक्लूड करते हो अगर ये से लाइन का रिप्रेजेंट कर रहा है तो आप यही का रहे हो तो मैं क्या कहूंगा मैं कहूंगा सर ह / 5 मैं कहूंगा ह/5 विल बी इक्वल तू के अपॉन -3 और अगर सर आप इसे सॉल्व करें तो क्या आपको ह और क कोऑर्डिनेट्स मिल रहे हैं सर ह को देखो अगर ह/5 को 6/5 के इक्वल रखा तो फाइव से फाइव कैंसिल तो ह की वैल्यू कितनी आती है सिक्स सिमिलरली माइंस के / 3 को 6/5 के इक्वल रखा तो ए जाएगा -18 / 5 तो के की वैल्यू कितनी ए जाएगी सर के की वैल्यू होगी - 18 / 5 और जैसे ही मुझे ह कमा के मिले क्या मैं का सकता हूं मुझे उन टेंसेज का पॉइंट ऑफ इंटरसेक्शन मिला तो पहला पार्ट है इस क्वेश्चन कही खत्म हो गया ऑफ कोर्स एडवांस पूछ सकता है क्वेश्चन पर कितने आसानी से हम कर सकते हैं अगर हमें क्या पता हो अगर हमें बारीक है अगर हमें कॉन्सेप्ट्स अगर हमें हर एक चीज अच्छे से पढ़ राखी है तो इस तरीके से इतनी आसानी से आप चीज सोच पाओगे और ऐसे ही क्वेश्चंस आने वाले हैं ये आसान लगेंगे पर कब हमने हर चीज पढ़ राखी हो ठीक है सर आपकी ये बात मैन ली और क्या और अगर मैं देखूं तो क्या लेंथ ऑफ कॉमन कोड निकल जा सकती है सर लेंथ ऑफ कॉमन कोड कैसे निकल लोग आप लेंथ ऑफ कॉमन कोड का तरीका क्या है मैं कहूंगा लेंथ ऑफ कॉमन कोट्स और आसान है आप ऐसे सोचो स्टूडेंट्स बड़ी सिंपल सी बात है देखो भाई मैं क्या करूंगा मैं बोलूंगा सर की किसी भी सर्कल का सेंटर उठा लो अब जैसे मैन लो मैं सर्कल का सेंटर उठा ले रहा हूं जिसके अकॉर्डिंग क्या है सर ये आपका सर्कल ax² + y² = 12 तो इसके सेंटर के कोऑर्डिनेट्स के जीरो कमा जीरो सर एक बात बताओ इस सर्कल के सेंटर से अगर इस पॉइंट ए को कनेक्ट करें तो ये अगर मैं इसे का डन के लिए ओ तो ये ओ ए क्या हुआ सर ये ओए इसकी रेडियस होती है ओए इसकी रेडियस मतलब कितना इसकी रेडियस हो जाती है सर अंडर रूट 12 आई होप आप बोले नहीं हो अब एक बात स्टूडेंट्स अगर मैं इस ओ से इस अब कॉर्ड पर परपेंडिकुलर ड्रॉप करूं तो फिर वही बात क्या अब बायसेक्स होती हुई नहीं दिख रही है आपको इस पॉइंट पर जैसे मैं थोड़ी देर के लिए का लेता हूं एम आय होप अब बात समझ ए रही है आपको बात बड़ी इतनी सी है की सर जब आप एम निकलने चले चले जब आप चले क्या एम निकल लें तो आप क्या कहोगे सर पहले ओरिजिन से इस लाइन अब की इस लाइन अब की परपेंडिकुलर डिस्टेंस निकल सकता हूं क्या यानी मैं क्या ओम निकल सकता हूं आई होप आपको समझ ए रहा है की सर ओए हमारे पास है जो की है अंडर रूट 12 अब हम निकलना चाह रहे हैं ओम अब हम क्या निकलना चाह रहे हैं भाई अब हम निकलना चाह रहे हैं ओम 0 से इस स्ट्रेट लाइन के परपेंडिकुलर डिस्टेंस क्या होगी जीरो इन फाइव जीरो इन तू थ्री खत्म -10 का मोड प्लस 10 फॉर्मूला अपॉन में फाइव का स्क्वायर प्लस 3 का स्क्वायर 25 3 का स्क्वायर 9 25 + 9 कितना हो जाएगा आई थिंक 34 34 अंडर रूट में √34 तो ये कितना हो जाएगा सर ये हो जाएगा 10 / √34 क्या ये दोनों बातें समझ आई अब क्या आप मुझे नहीं बता सकते की सर बड़ी बेसिक सी बात है आप डिस्टेंस फॉर्मूला उसे कर लो और आपका आंसर ए जाएगा सॉरी पाइथागोरस थ्योरम उसका लो और आपका आंसर ए जाएगा कैसे ए जाएगा भाई आपको ओए पता है आपको ओम पता है तो आप एम पता नहीं कर सकते क्या हाइपोटेन्यूज के स्क्वायर में से परपेंडिकुलर सब्सट्रैक्ट कर रहा हूं तो इसके स्क्वायर में से मतलब 12 में से सब्सट्रैक्ट करना है किसे इसे तो ये हो जाएगा 10 का स्क्वायर 100 / कितना 34 अब बात समझ का रहे हो और इसका जो आप अंडर रूट लिखेंगे वही क्या होगा वही होगा आपका एम लेकिन एक मिनट सर हमें एम नहीं चाहिए था हमें तो कोड की लेंथ चाहिए थी तो इस कॉमन कोड के लेंथ क्या होगी एम का डबल एम का डबल यानी इसका डबल क्या बात समझ का रहे हो इसका डबल यानी ये क्या हो जाएगा अब तो जल्दी से निकल लो फिर अब आपको बस ये करना है की 34 12 करना फिर 100 से ट्रैक करके उसको 34 से डिवाइड करना क्या इतना कम है आपके ऊपर छोड़ डन अगर आपको आलस आता है तो मैं ही करूंगा अभी पर ये करना पड़ेगा ना कुछ-कुछ चीज तो जीवन में करनी पड़ेगी जैसे 12 34 तो फटाफट 12 34 कहते हैं देखो 12 का फोर टाइम्स कितना होता है 12 फोर टाइम्स होता है 48 48 यानी कितना 48 यानी 8 48 यानी 8 कैरी कितना है ना 12 कितना 3 36 36 + 4 आई थिंक 40 तो ये हो जाएगा कितना आई थिंक ये हो जाएगा 40 यहां ए जाएगा क्या 408 सही लिख रहा हूं क्या कोई दिक्कत तो नहीं सर एक और छोटी सी बात ध्यान से देखो सर अब इसमें से सब्सट्रैक्ट करना है आपको 100 और फिर उसे आपको डिवाइड करना है कैसे 34 से और इस पूरे एक्सप्रेशन का आपको अंत में डबल भी करना है bhuliyega मत और वही आपकी कॉमन कोड अब के लेंथ होगी अगर आप इसे थोड़ा और सिंपलीफाई करें तो ये हो जाएगा 308 ये हो जाएगा कितना भाई 308 क्या आप सबको दिख रहा है आई थिंक दिख रहा होगा यह हो जाएगा 34 और अब मेरी आपसे बस गुजारिश है आई थिंक इट्स जस्ट क्लीयरली नॉट अन वेरी डिफिकल्ट टास्क जो की आसानी से आप कर सकते हो तू से डिवाइड कर देना 17 17 से चेक कर लेना डिविजिबल है या नहीं डिवाइड हो रहा होगा तो फाइनली जो भी अंडर रूट ए रहा हो सिंपलीफाई हो रहा हो तो ठीक है नहीं ए रहा है तो ठीक है और उसे तरीके से आप आंसर लिख देंगे मेरे ख्याल से यह पार्टिकल तू डू मेरा जो मसाला था मेरा जो प्रॉब्लम था वो ये वो आपको ये सीखना है की इसे कैसे सॉल्व करना है और कमेंट की लेंथ निकलने का यही तरीका है क्या सेंटर से परपेंडिकुलर लेंथ निकल लो रेडियस निकल लो और पाइथागोरस थ्योरम लगा के अप्लाई कर दो आई थिंक यही बात हम अभी तक हर बार करते हुए आए हैं तो इस तरीके से अभी तक हमने जो चीज है सीखी वो कुछ इस तरीके से हमारा अभी तक का जो भी scenarioz बने उनका एक कंप्लीट रिवीजन हुआ अब क्या सर अब suniyega ध्यान से अब आपके आते हैं असाइनमेंट क्वेश्चन जिन्हें आप क्या कहते हैं जिन्हें आप कहते हैं सर होमवर्क क्वेश्चन तो आज का पहला होमवर्क क्वेश्चन हो द फॉलोइंग पर्स ऑफ़ सर्कल्स अरे सिचुएटेड इन द प्लेन एंड अलसो फाइंड द नंबर ऑफ़ कॉमन टांगेंट्स तो मैं क्या करूंगा सर मैं अपनी एक सिंपल सी कंडीशन अप्लाई करूंगा क्या की इनकी रेडियस और इनके सेंटर्स में इन दोनों सर्कस के सेंटर्स के बीच की डिस्टेंस और इनकी रेडियस में क्या रिलेशन है बस वैसे देख लीजिएगा वही यहां अप्लाई कर दीजिएगा आपको दिख जाएगा की ये किस तरीके से हैं किस तरीके से मतलब या तो डिसजॉइंट होंगे या फिर एक्सटर्नल टच कर रहे होंगे या फिर इंटरनल टच कर रहे होंगे या फिर सर्कल एक दूसरे के अंदर होंगे जैसे भी बने जैसे भी बने आपको किस की हेल्प से मिलेंगे उनके सेंटर्स के बीच की डिस्टेंस और उनकी रेडियस आसानी से आपको ये चीज इमेजिन करने में हेल्प करेंगे और क्लियर हम जानते हैं सर अगर ये सिनेरी है तो चार रेडी चार कॉमन टांगें ये है तो तीन कॉमन टैसेंट ये है तो एक कॉमन टैसेंट और ये है तो जीरो कमेंट बस जो भी सिनेरियो आएगा उसे अकॉर्डिंग्ली तो जल्दी से इस वाले क्वेश्चन का स्क्रीनशॉट लीजिए स्टूडेंट्स और मुझे लिखकर बताना की इस क्वेश्चन का आपने क्या आंसर निकाला था नेक्स्ट क्वेश्चन पर आए आज का जो सेकंड असाइनमेंट क्वेश्चन है ये रहा तो ये आपका सेकंड होमवर्क क्वेश्चन है जिसको आपको खुद से करना है योनि कॉन्सेप्ट्स पर बेस्ड है जो आपको सिखाया गया देखो भाई क्या लिखा हुआ है और प्लीज इन्हें करिएगा ये जानबूझकर रखे गए क्वेश्चन से ताकि आपके वो सारे कॉन्सेप्ट रिवाइज करवाए जा सके जो आपको पढ़ाया गया है ताकि आपकी प्रैक्टिस हो आपका नॉलेज बिल्ड हो आपके अंडरस्टैंडिंग चेक की जा सके सर्कल ऑफ रेडियस 5 यूनिट्स इंटरसेप्ट्स डी सर्कल इन अन सच अन वे डेट डी लेंथ ऑफ डी कॉमन कोड इस ऑफ मैक्सिमम लेंथ आप याद करिए ऐसा कुछ हमने पढ़ा है ऐसा कुछ हमने पढ़ा है मेरा यकीन करिए मैंने बहुत अच्छे से आपको पढ़ाया है की कॉमन कोड की लेंथ मैक्सिमम कब होती है आपको पता है हिंट चाहिए डायमीटर आगे सोचना खुद से फिर क्या इफ डी स्लोप ऑफ डी कॉमन कोड इस 3 / 4 तो यहां से और यहां से आप बहुत आसानी से उसे सर्कल की इक्वेशन यानी उसके सेंटर पर जा सकते हो मेरे यकीन करिए एक बहुत आसान क्वेश्चन है जो बहुत आराम से आप सोच लोग मेरा यकीन करिए इसके बाद क्या सर इसके बाद हम आते हैं आज के अगले असाइनमेंट क्वेश्चन पर आई होप आपने इसका स्क्रीनशॉट ले लिया है ये क्वेश्चन नोट डाउन किया भाई स्क्रीनशॉट मतलब आज का आपका अगला असाइनमेंट क्वेश्चन ये रहा ये आज का आपका तीसरा असाइनमेंट क्वेश्चन है जो आपको करना है क्वेश्चंस समझ लेते हैं भाई लेट तू पैरेलल लाइंस L1 और L2 तो ये क्या है भाई पैरेलल विद पॉजिटिव स्लोप्स माइंड गिव रीजंस पॉजिटिव है बहुत बड़ी बात है जिसका कुछ मतलब होना चाहिए ये टांगें है इसकी और इसकी ठीक है सर अब वह क्या का रहा है एल्बम जो है वह इस सर्कल की भी टैसेंट है सच डेट एस वैन एंड S2 लिए ऑन डिफरेंट साइट्स ऑफ L1 कुछ याद करो इस बात का कुछ तो मतलब है की S1 और S2 लिए ऑन दी डिफरेंट साइट्स ऑफ L1 मतलब इन दोनों की सेंटर्स का L1 में जब आप वैल्यू पुट करेंगे तो उनका जो प्रोडक्ट आएगा वो नेगेटिव आएगा याद ए रहा है कुछ ऐसा पड़ा था हमने बस वही कंडीशन अप्लाई करिएगा आप आसानी से L2 तक पहुंच जाएंगे मेरा यकीन करिए बहुत सारी डिटेल्स बहुत सारी इनफॉरमेशन है और आप आराम से एक क्वेश्चन सॉल्व कर लेंगे तो जल्दी से इस क्वेश्चन का स्क्रीनशॉट ले लो क्लिक किया क्या चलो इसके बाद आज का आपका अगला साइन इन क्वेश्चन ये रहा की आज का आपका नेक्स्ट असाइनमेंट या होमवर्क क्वेश्चन है जैसे आप कहेंगे क्या लिख रहा है वो डी इक्वेशन ऑफ अन सर्कल इस दिस ठीक है सर्कल लार्जेस्ट सर्कल मैं डील किया था जिसकी इक्वेशन दी गई थी जैसे टच कर रहा था और इस तरीके से वह यहां पर टच कर रहा था एक और लाइन को तो बड़ी आसानी से आप इसे सोच पाओगे अगर आप इसे विजुलाइजर करते हो प्लॉट करना ड्रा करना आराम से बन जाएगा और अब आराम से इक्वेशन निकलकर सर्कल के सेंटर को बता पाओगे यकीन करिए आसान क्वेश्चन है यह था आज का आपका अगला असाइनमेंट क्वेश्चन और आई होप आपने इसका आसानी से स्क्रीनशॉट लिया नेक्स्ट असाइनमेंट क्वेश्चन पर आते हैं ये आपका नेक्स्ट असाइनमेंट क्वेश्चन है देखो भाई क्या लिखा है फाइंड डी इक्वेशन ऑफ डी सर्कल तो आपको एक सर्कल की इक्वेशन निकालनी है जिसकी रेडियस है थ्री हु इस रेडियस इस थ्री दिस सर्कल विच टचेस दिस सर्कल इंटरनली अत थिस पॉइंट आई थिंक सर यह तो आसान है अब विजुलाइज करना आप आराम से इससे और इससे इस क्वेश्चन को वर्कआउट कर लोग आप यकीनन तौर पर कर लोग एक बार जरा ट्राई करिएगा तो ये आज का आपका अगला असाइनमेंट क्वेश्चन ये आप का लीजिए होमवर्क क्वेश्चन और क्या सर और नेक्स्ट असाइनमेंट क्वेश्चन पर आए स्क्रीनशॉट लेते चले जा रहे हैं ये आपके असाइनमेंट यानी होमवर्क क्वेश्चंस हैं इंक्रीजिंग यूनिफॉर्म विद रिस्पेक्ट तू टाइम दें आते व्हाट वैल्यू ऑफ टी विल दे टच एच अदर दिस इस बेसिकली ए क्वेश्चन ऑफ एप्लीकेशन ऑफ डेरिवेटिव्स थोड़ा वहां यह उसे होगा खैर आप बेसिक कॉमन सेंस से भी इस क्वेश्चन को ट्राई कर सकते हो एक बार ट्राई करना आसानी से वर्क आउट कर लोग तरीका ये होगा आप इनके सेंटर्स के बीच के डिस्टेंस निकलना और अभी जिस डिस्टेंस पर ये प्रेजेंट है उसे निकलना फिर ऑफ कोर्स ये क्वेश्चन बस इतना सा है डिस्टेंस और स्पीड अगर आप डिस्टेंस को स्पीड से डिवाइड करें तो आपको मिल जाएगा टाइम तो आपको इस सिनेरियो को विजुलाइज करके डिस्टेंस सोचनी है स्पीड आपको दी गई है रिलेटिव स्पीड का कॉन्सेप्ट उसे होगा और बहुत आसानी से क्वेश्चन आप कर लोग रिपीट माय स्टेटमेंट मुझे नहीं पता आपको ये पता करना है की इनके बीच में डिस्टेंस कितनी है और डिस्टेंस कवर करने के लिए वो किस स्पीड पर बढ़ रहा है वो किस स्पीड पर बढ़ रहा है दोनों की स्पीड ऐड हो जाएगी क्योंकि रिलेटिवली दोनों एक दूसरे के अपोजिट डायरेक्शन में फिजिक्स बेसिक फिजिक्स डिस्टेंस को इस स्पीड से डिवाइड करें रिलेटिव स्पीड लगाना तो ऑफ कोर्स कितना टाइम लगेगा वो आपको बहुत आसानी से दिख जाएगा मैं सच कहूं तो ये ओड का क्वेश्चन मुझे शुरू में लगा था पर एक बेसिक स्पीड टाइम और डिस्टेंस का क्वेश्चन है जहां पर आपको डिस्टेंस और स्पीड से डील करके टाइम तक पहुंचना है कर लोग क्या देखो दोनों सिनेरियो उसने बनाए हैं दो आंसर्स इसीलिए ए रहे हैं क्योंकि शायद कुछ दो रीजंस रहे होंगे आप समझेगा आप आसानी से समझ जाएंगे एक बार externalli टेस्ट करेगा एक बार इंटरनल टच करेगा इसलिए तो आंसर आएंगे और ये क्वेश्चन आपको पूछ लिया जाएगा मस्क में जहां पे एक से ज्यादा ऑप्शन करेक्ट हो आप उसे यही टच करवा के खुश हो जाओगे पर अगर वो और बढ़ता रहा तो यहां भी तो टच करेगा तब क्या होगा वह भी आपको पता करना है आसान क्वेश्चन है करिएगा केवल एक्चुअली एंजॉय सॉल्विंग दी क्वेश्चन जो आपको असाइनमेंट क्वेश्चन तो हर बार जब भी कोई क्वेश्चन दिखाया जाए आप कर लीजिएगा स्क्रीन पॉज मैं जरा है जाऊंगा स्क्रीन से एक तरफ और आप करिएगा उसे क्वेश्चन को सहूलियत से लगेगा तो मैं हिंट जरूर दूंगा ताकि आप उसे कर लें थोड़ा बहुत मदद ले लें पर ट्राई आप ही को करना है तो करेंगे की आप ट्राई बिल्कुल करेंगे यही मोटिव यही इंटेंट और यही इरादों के साथ इन क्वेश्चंस को करना जितनी भी एक्सरसाइज आएंगे अब जितने भी क्वेश्चंस लाएंगे हर तरीके क्वेश्चंस करेंगे सिंगल करेक्ट मल्टीपल करेक्ट मैट्रिक्स टाइप लिंक्ड आंसर टाइप न्यूमेरिकल आंसर टाइप सब कुछ करेंगे और बहुत अच्छे से करेंगे और यह उन सारे कॉन्सेप्ट्स को रिकॉर्ड करेंगे और आखिर में हम के एडवांस में पूछे गए प्रीवियस ईयर क्वेश्चंस भी करेंगे मतलब एक कंसोलिडेटेड तैयारी करेंगे तो बिना किसी देले की आप करते हैं शुरू बहुत अच्छी किताब है जो की आपको सच में के डबली एडवांस लेवल तक लेकर जाएगी और बहुत थारो लेकर जाएगी बहुत अच्छे से मेटिकुलुजली लेकर जाएगी शुरू करते हैं भाई ये प्लेलिस्ट जरूर से करके रख लीजिएगा बड़ा आसान सा तरीका है क्या करना है आपको आप बस इस वीडियो के डिस्क्रिप्शन में जब जाएंगे ना तो आपको क्या दिखेगा पता है इस प्लेलिस्ट का लिंक यहां लिखा होगा फुल प्लेलिस्ट वहां पर आपको क्या करना है पता है वहां लिखा होगा सिंह आगे मैथमेटिक्स फॉर आईआईटी जी मांस एंड एडवांस्ड वहीं पर आप क्लिक करेंगे तो ये पूरा प्लेलिस्ट ओपन हो जाएगा प्लस बटन दिखाई देगा वो से करके रखना ताकि आपके पास ही वीडियो हमेशा आपके अकाउंट में से रहे हैं ये रहा आज का आपका पहला क्वेश्चन आपकी स्क्रीन पर कौन बनेगा करोड़पति की तरह तो इसे कर लीजिए आप ट्राई क्वेश्चन नंबर वैन आपके सामने इस क्वेश्चन में वो क्या का रहा है मुझे जो दिख रहा है वो ये की सर एक सर्कल दे दिया है उसने अगर आपको क्वेश्चंस समझना है तो क्वेश्चंस जरूर मैं समझा दे देता हूं एक सर्कल आपको दे दिया है वो का रहा है इक्वेशन ऑफ डी सरकमसर्किल ऑफ एन इक्विलैटरल ट्रायंगल इस है तो यह जो आपका यह जो सर्कल की इक्वेशन है ना यह कोई ऑर्डिनरी या नॉर्मल सर्कल नहीं है ये एक सरकमसर्किल है किसका एक इक्विलैटरल ट्रायंगल का ठीक है सर तो क्या तो वो ये बोल रहा है अच्छा उसे ट्रायंगल की एक वर्टेक्स भी आपको बता दिए जो की क्या है वैन कमा वैन मतलब वो जो इक्विलैटरल ट्रायंगल है उसकी वैन ऑफ डी वर्ल्ड कमा वैन ठीक है सर वो फाइनली आपसे बात कर रहा है डी इक्वेशन ऑफ डी इन सर्कल ऑफ डी ट्रायंगल इस मतलब वो जो सरकमसर्किल की क्वेश्चन दे दी है और वो आपसे इन सर्कल के बारे में जानना चाह रहा है ये बात देते हुए की वैन ऑफ डी वर्टेक्स इस आते वैन कमा वैन जरा अच्छे से इसे देखिए इस क्वेश्चन को ट्राई करिए और मुझे बताइए अगर आपको हिंट चाहिए तो पहली हिंट तो मैं दूंगा वो ये की आप सोच के देखो इस सर्कल का सरकमसर्किल इन सेंटर सारी चीज एक ही जगह होती हैं सेट्रॉयड ओर तो सेंटर यह बात भूल रहे हो क्या सर ऐसा क्यों क्योंकि क्लीयरली स्टूडेंट्स ट्रायंगल है दूसरी कंडीशन या दूसरी वैन कमा वैन जो होगी भाई जो सरकमसर्किल है इसका तो क्लीयरली इस सर्कल पर लाइक सेटिस्फाई करेगा और तीसरी हेड तीसरी हिट ये होती है की इक्विलैटरल ट्रायंगल के लिए जो सरकमरेडियस होती है वो इन रेडियस का डबल होती है यह तीन अगर कंडीशंस आपने आराम से सोच कर दिमाग से अप्लाई किए द तो ये क्वेश्चन हो जाना चाहिए कैसे समझते हैं देखो बड़ी सिंपल सी बात मैं आपसे कहना चाह रहा हूं जो बात है वो ये की आपके पास पहली बात तो एक इक्विलैटरल ट्रायंगल है इक्विलैटरल ट्रायंगल में आपके पास एक सर्कल है कुछ इस तरीके से की सर वो जो सर्कल है वो आपके तीनों वर्टेक्स ऑफ़ कोर्स पास होता है यह जो सर्कल से पास होता है इस सर्कल की इक्वेशन आपको दी गई है और आपके पास एक और सर्कल है टेक्निकल आपके पास एक और सर्कल है टेक्निकल जो की आपको इक्वेशन ढूंढनी है और वह सर्कल इस तरीके से है की वह इस सर्कल इस ट्रायंगल अच्छे से समझ का रहे हैं भाई तो इस तरीके से आपको इस बाहर वाले सर्कल दे दी गई है सुंदर वाले सर्कल की इक्वेशन निकालनी है मेरी एक बात प्लीज ध्यान से सुनो और सोच कर बताओ सेंटर भी निकल लेते तो इक्विलैटरल ट्रायंगल में वो सब से होते हैं यानी की सर इन सब का जो सेंटर है वो यहां होगा और सबसे खास सबसे जरूरी बात ये की इस सर्कल का सेंटर होगा - जी - एफ तो ये जो सेंटर के cardinate है वो तो -डी कमा - फ होंगे क्या आप मेरी बात समझ पाए अगर मैं इस पॉइंट को थोड़ी देर के लिए कुछ भी का लेता हूं कुछ भी का लेता हूं लेट्स में सी ही का लेता हूं तो ये रहे आपके किसके cardinate के इस बात से कोई परेशानी तो नहीं है कोई दिक्कत है अच्छा सर एक और बात आप समझो आप का रहे हो की ये जो इक्विलैटरल ट्रायंगल है इसकी वैन ऑफ डी वर्टिस इस वैन कमा वैन तो क्या ऐसे वैन कमा वैन इस सर्कल पर लाइक करेगी की नहीं तो वैन कमा वैन है सर्कल की इक्वेशन को सेटिस्फाई करेगी तो वैन वैन तू जी प्लस तू ऍफ़ प्लस सी क्या आप मेरी बात समझ में मैं फिर से रिपीट करता हूं वैन प्लस वैन प्लस तू जी प्लस तू यानी यहां से सर शायद मैं सी की वैल्यू लिख पाऊं तो सी की वैल्यू ऐसे ही बस निकलने का मैन है तो क्या कहेंगे भाई पूरा पार्ट उसे तरफ शूट हो जाएगा तो माइंस ऑफ कितना डबल कॉमन ले लो तो कितना हो जाएगा अब क्या अब ध्यान से देखो और एक बात का जवाब दो ध्यान से देखो एक बात का जवाब दो मुझे चाहिए मुझे चाहिए इस इन सर्कल की रेडियस है रिपीट माय स्टेटमेंट मुझे चाहिए इस इन सर्कल की रेडियस ये कितना आसान है सोच के देखना आप हमेशा याद रखना जो आपकी सरकमरेडियस होती है वो डबल होती है इन रेडियस का अगर मैं सरकमरेडियस को कैपिटल आर से दिनो करूं तो इन रेडियस को मैं दिनो कर रहा हूं स्मॉल आर से मुझे इन रेडियस चाहिए तो क्या सर इससे अच्छा आप ऐसा कहते हो तो अच्छा नहीं लगता क्या की जो इन रेडियस होती है जो इन रेडियस होती है वह सरकमरेडियस का हाफ होती है और सर कम रेडियस सही से निकलेगी सरगम सर कम रेडियस क्या होगी सर कम रेडियस मतलब ये वाली रेडियस ये आपकी क्या होगी भाई कैपिटल आर और यह जो होगी यह क्या होगी भाई ये होगी स्मॉल आर तो कैपिटल आर को आप क्या कहोगे भाई कैपिटल आर को आप कहोगे सर जी स्क्वायर प्लस एक्स स्क्वायर वाइरस का हाफ सर्कल के सेंटर के भी वही होंगे तो मैं लिखूंगा एक्स - जी तो क्या हो जाएगा एक्स + g² + ए - एफ यानी कितना ए + एफ का होल स्क्वायर इस इक्वल तू रेडियस का स्क्वायर इस इक्वल तू रेडियस का स्क्वायर और रेडियस कितनी है सर अगर मैं रेडियस का स्क्वायर निकलना चाहता हूं तो रेडियस है आपकी g² + x² - सी रेडियस का स्क्वायर मतलब अंडर रूट हटा डन तो क्या लिखूंगा g² + x² - सी का हाफ का हाफ आई होप आप बात समझ का रहे हो ये आपकी सरगम रेडियस है तो इसके हाफ का स्क्वायर मतलब डिवाइडेड बाय फोर क्योंकि अंडर रूट कैंसिल लेकिन डिनॉमिनेटर में तो तू का स्क्वायर 4 होगा ना तो ये क्या दिखेगा सर ये राइट हैंड साइड पर दिखेगा माइंस सी डिवाइडेड बाय फोर इसमें एक बात और ध्यान से देखो सर आप बड़े ध्यान से देखो और बड़े ध्यान से समझो की ये जो सी है ये जो -सी है क्या इसकी भी वैल्यू लिख सकता हूं अगर माइंस यहां से उठा के यहां लगाया तो क्या -सी की कुछ वैल्यू हमें पता है सो तू टाइम्स 1 + जी + एफ - सी की जो वैल्यू है वो तू टाइम्स वैन प्लस जी प्लस एफ है ना तो उसे माइंस सी की वैल्यू तू टाइम्स वैन प्लस जी प्लस एफ रखता हूं और फोर को इधर ले आता हूं आई होप ये बातें आपको कंफ्यूज नहीं कर रहे हैं तो देखो क्या लिखेंगे हम लिखेंगे एक्स + जी का होल स्क्वायर आप चाहो तो ऐसे सिंपलीफाई कर लो के सिंपलीफाई कर लो सर क्योंकि अगर मैं अब ऑप्शंस को थोड़ा देख लूं ऑप्शंस को थोड़ा देख लूं तो कुछ अभी खास आइडिया तो लग नहीं रहा है बट ट्राई जरूर करते हैं देखो इसे हम सिंपलीफाई करने कोशिश करते हैं फोर को इधर लेकर आता हूं तो यह फोर है ना और इसे मैं भी ऐसा कैसे ही रखता हूं एक्स + जी का होल स्क्वायर लता हूं तो इसको ऐसे ही रखता हूं क्या ए + एफ का होल स्क्वायर इस इक्वल तू ध्यान से देखो भाई ये लिखा हुआ है g² + x² और माइंस सी माइंस सी मतलब सी की वैल्यू जस्ट अभी हमने निकल है -सी की वैल्यू जो हम लिख रहे हैं वो क्या है तू टाइम्स 1 + जी + एफ है ना तो यहां पर मेरे रखूंगा क्या -सी की वैल्यू जो की कितनी है तू टाइम्स 1+g+f तो 2 1 2 ये हो जाएगा 2G और ये हो जाएगा 2f ऐसा क्यों किया बस देखेंगे कुछ ना कुछ यहां पर निकल के ए रहा होगा शायद अब देखो अगर मैं चीज है आप सिंपलीफाई करता हूं अब अगर मैं फाइनली इसको सिंपलीफाई करने की तरफ बढ़ता हूं तो क्या निकल कर ए रहा है देखो भाई सर देखो एक्स स्क्वायर प्लस तू जी एक्स उसको 4 से मल्टीप्लाई करेंगे तो हो जाएगा 4x² है ना 4 एक्स स्क्वायर बन्ना आता है तो इसे बनाते हैं 4x² ठीक है सर और क्या बचेगा 4G स्क्वायर अच्छा आप देख रहे हो इधर से ये G2 जब इधर आएगा तो ये 4G स्क्वायर है अच्छा क्या होगा लिखना 3G स्क्वायर कोई डाउट तो नहीं है और क्या कर लोग सर आप और अगर मैं इसे आगे बधाई तो यहां मिलेगा मुझे 2gx 2gx को 4 से मल्टीप्लाई करेंगे तो ये होगा 8g एक्स कोई दिक्कत तो नहीं है और क्या सर यहां पर तो यहां पर कितना हो जाएगा 4y² - एक्स स्क्वायर और ये 4 मल्टीप्लाई होगा तो कितना हो जाएगा वो हो जाएगा 8 फी कोई डाउट तो नहीं है भाई और क्या बच जा रहा है सर और अगर यहां से चीजे उठाएं तो ये 2G 2 है और तू भी तो आएगा तो क्या बचेगा सर ये बचेगा -2g - 2f और ऑफकोर्स -2 = 0 आई थिंक ऐसा कुछ मुझे किसी ना किसी ऑप्शन में दिखाना चाहिए आई रिपीट माय स्टेटमेंट ऐसा कुछ मुझे किसी ना किसी ऑप्शन में दिखाना चाहिए और अगर मैं ऐसा कुछ किसी ना किसी ऑप्शन में ढूंढना देखना या फिगर आउट करना चाहूं तो ऐसा मुझे किस ऑप्शन में दिख रहा है जरा गौर से देखो आई थिंक अब अगर हम ऑप्शन सी भी मुझे बहुत दूर तक देख रहा है अब ऑप्शन देख का रहे हो की सर 4x² 4y स्क्वायर तो मिल रहा है लेकिन G2 इस तरीके से सॉर्टेड नहीं है ना ऑप्शन ए और सी में है आप देखो ऑप्शन ए और सी में बिल्कुल फॉरेक्स स्क्वायर 4y स्क्वायर जीएक्स और फी दिख रहे हैं बट बाकी चीज भी तो दिख रहे हैं तो बाकी चीज ऑप्शन ए और सी से तो नहीं मिलने वाले अब बचे हुए ऑप्शन बी और दी तो अगर मैं ऑप्शन बी और दी को सिंपलीफाई करूं तो देखो ये पार्ट तो मुझे दिख रहा है ऑप्शन बी का ये पार्ट तो मुझे दिख रहा है है तो इस वाले पार्ट को थोड़ा सिंपलीफाई कर ले तो अगर ऑप्शन बी का हिस्सा हमारे रिक्वायर्ड डिसाइड जो हमारा आंसर ए रहा है उसे मैच हो जाएगा बात बन जाएगी तो थोड़ा ध्यान से देखो देखो 1 1 और ये 1 1 तो 1 + 1 2 तो मुझे तू तो दिख रहा है और ऑफ कोर्स राइट हैंड साइड पर है ना तो तू तो मुझे दिख रहा है सर तू अगर मैं लेकर गया तो राइट हैंड साइड पर तू मिल जाएगा तो ऑफ कोर्स सर ये तो देखना आता है मैं बस चीजें मैच कर रहा हूं फटाफट ताकि मेरा आंसर अगर यही हो तो बात बन जाए अब suniyega सर ध्यान से बड़े ध्यान से सुनना ये माइंस जी ये प्लस थ्री जी तो कितना बचेगा 2G क्या हमारे पास 2G है जरा देखते हैं सर क्या हमारे पास 2G है हान सर है क्योंकि वो ऑफ कोर्स उधर जाएगा तो +2g हो जाएगा कोई दिक्कत तो नहीं साइन पॉजिटिव मिल लेता है है ना सर एक और चीज आपने नोटिस नहीं की आपने अभी क्या किया अभी आपने इससे मल्टीप्लाई नहीं किया तो बचा -3g स्क्वायर सर ये भी है कैसे है देखो जरा ध्यान से -3g स्क्वायर कैसे ये 3G स्क्वायर हो जाएगा तो माइंस थ्री स्क्वायर सर ये पार्ट तो मिला है ना सिमिलरली से बेसिस पे क्या मुझे - 2f - 2f साथ थ्री एक्स स्क्वायर मिलना चाहिए आई थिंक -2f के साथ अगर मैं इन दो चीजों को भी ढूंढ लाया तो देखो 3f - एफ कितना -2f क्योंकि यह उधर गया है तो अभी ये प्लस थ्री प्लस तू एफ दिख रहा है बट इस तरफ - 2fe ही होगा बिल्कुल सही और एक और बात ध्यान से देखो 3f एफ ये हो जाएगा - 3x² तो बिल्कुल सर ये भी आपको दिखेगा जब ये थ्री एक्स स्क्वायर हो जाएगा तो मुझे सारी टर्म्स लगेगी क्योंकि बची हुई जो टर्म्स है वो तो ऑलरेडी लेफ्ट हैंड साइड पर लिखी हुई है तो सर ज्यादा सोचने की जरूरत नहीं है आपका ऑप्शन भी आंसर होगा एक आसान सा सिंपल सा डायरेक्शन जिसमें मेरा यकीन है किसी भी स्टूडेंट को कोई डाउट नहीं होना चाहिए अगले क्वेश्चन पर मूव करते हैं स्टूडेंट्स इसे देखो क्या लिख रहा है डेट डिफरेंस बिटवीन डी रेड ऑफ डी लार्जेस्ट इन डी स्मालेस्ट सर्कल्स विच है तो सर दो चीजों पर बात हो रही है तो कैसा लार्जेस्ट और ये कैसा स्मालेस्ट इस बारे में sochiyega थोड़ा और उनकी रेड आय का डिफरेंस तो सी आर एस्क्ड डिफरेंस बिटवीन डी रेड वो जो लार्जेस्ट और स्मालेस्ट सर्कल बनेंगे उनके रेडियस का दोनों की इंडिविजुअल का डिफरेंस कैसे लार्जेस्ट केसेस स्मालेस्ट हो सकते हैं उनके जो सेंटर होंगे वो सरकम्फ्रेंसेस पर होंगे किसकी इस सर्कल की ठीक है और वो दोनों सर्कल्स आपके ए कमा बी से पास होंगे वो दोनों सर्कल्स आपके ए कमा बी से पास होंगे सर ए बी तो मेंशंड ही नहीं है ए भी तो बताया ही नहीं है तो कैसे सोचेंगे सर क्या करेंगे अच्छा मैं आपको अगर हिंट देना शुरू करूं अगर मैं पहली हिंट आपको डन तो इस क्वेश्चन को करने में पहले तो आप इस सर्कल को देखो इस सर्कल को इसलिए देखो क्योंकि आप इस सर्कल का सेंटर और रेडियस निकल सकते हो क्या जब आप यह निकल लोग तो आप खुद दो पाल के लिए इमेजिन करो ऐसे दो सर्कल्स जो की क्या रखते हो जो की सबसे बड़े और सबसे छोटे सर्कल बन पाए इस कंडीशन को फुलफिल करते हुए की उनके सेंटर इस गिवन सर्कल की सरकम्फ्रेंसेस पर हूं और ऑफ कोर्स वो पॉइंट ए कमा बी से पास हो रहे हो क्या आप मेरी बात समझ का रहे हो मैं जो बात का रहा हूं वीडियो ध्यान से सुनिए स्टूडेंट्स में से थोड़ा सा अलग तरीके से विजुलाइज करवाता हूं देखना यहां पर क्या लिख रहा है वो पहले तो इस सर्कल को देखो सर इससे देख कर मुझे जो आता है आइडिया वो ये की ये हो जाएगा -1 और ये हो जाएगा -2 -1 -2 पर आपके सर्कल का सेंटर होगा जो आपको गिवन सर्कल है उसका ठीक है सर माइंस वैन कमा - 2 पर हम एक सर्कल ले लेते हैं सेंटर डेट माइंस वैन कमा -2 तो ये एक सर्कल है जिसका सेंटर है -1 -2 पर ठीक है सर और क्या इसकी रेडियस बता सकते हो क्या वैन का स्क्वायर 4 5 - 4 यानी 5 + 49 9√3 तो इसकी रेडियस कितनी हो गई सर इसकी रेडियस हो गई थ्री इसकी रेडियस हो गई थ्री लिख लेंगे ना मैं याद है रेडियस 3 है अब सुनना ध्यान से एक पॉइंट है क्या नहीं पता तो मैं मैन लेता हूं कहीं ना कहीं लेट से देयर इस अन पॉइंट ऑन ए बी अब आप मुझे एक ऐसा छोटे से छोटा सर्कल बता सकते हो क्या एक छोटे से छोटा सर्कल जिसका सेंटर हो इस सर्कल पर जिसका सेंटर हो सर्कल की सरकम्फ्रेंसेस पर और वो कम अभी से पास हो रहा हूं सर जो ए बी से सबसे पास होगी ना लाइन सबसे पास होगी लाइन मतलब जो ए कमा बी के नेरेस्ट होगा वो स्मालेस्ट पॉसिबल सर्कल बनेगा आप मेरी बात समझ का रहे हो मतलब अगर मैं ए कमा बी से नॉर्मल ड्रॉप करो सर्कल पर अगर मैं ए बी से नॉर्मल ड्रॉ करो इस सर्कल पर तो वो सर्कल सर आपका सबसे छोटा सर्कल होगा मतलब मैं जो कहना चाह रहा हूं प्लीज ध्यान से सुनना प्लीज बहुत ध्यान से सुनना सर जो आपने नॉर्मल ड्रॉप किया है ना ये यहां पर अगर सर्कल का सेंटर हुआ और अगर वो ए बी से पास हुआ तो ये वो स्मालेस्ट सर्कल बनेगा आप मेरी बात समझ का रहे हो क्या आई थिंक आप बातें समझ तो का रहे हो और अच्छे से समझ का रहे हो मतलब ये जो सर्कल रहा होगा जो की थोड़ा सा ऑफ कोर्स अभी छोटा बना है जैसे मैं थोड़ा और बड़ा करता हूं तो ये जो सर्कल रहा होगा ना सर ये आपका वो स्मालेस्ट सर्कल होगा इससे छोटा सर्कल आप नहीं बना सकते दोनों बातें चाहिए ना उसका सेंटर इस सर्कल पर होगा और वो एकमा भी इसे पास होगा तो आप खुद सोचो की इसके अलावा कुछ और इमेजिन कर सकते द आई थिंक नहीं अब क्या अब मुझे चाहिए लार्जेस्ट सर्कल लार्जेस्ट सर्कल की भी खास बात क्या रहेगी वो ए बी पर से पास होगा और इस सर्कल की सरकम्फ्रेंसेस पे अप्लाई करेगा सर सबसे बड़ा सर्कल तो सबसे बड़ा सर्कल कौन सा होगा ऑफ कोर्स उसका एक कोऑर्डिनेट्स होगा ईकॉम अभी और ए बी से जो नॉर्मल ड्रॉ किया है सबसे दूर ये पॉइंट है ना कमा भी से तो अगर वो ए कमा बी और इसको डायमीटर ट्रीट कर दें तो बाद बन जाएगी क्या आपने ये बात समझ पाए तो सर मतलब डायमीटर नहीं ए कमा भी एक एंड और एक ये सेंटर तो अगर ए भी और इसको रेडियस मानते हो एक सर्कल बना पाया अब तो एल्डो ये सर्कल यहां पर पॉसिबल करना यू नो क्रिएट करना मुश्किल होगा बट मैं आपको एक हल्का सा आइडिया देना चाह रहा हूं की वो चीज कैसी दिख रहे होंगे मैंने थोड़ा बड़ा फिगर ले लिया बट मैं समझाना चाह रहा हूं आपसे क्यों सर कल ए बी से जा रहा होगा वो सर्कल ए कमा बी से जा रहा होगा और उसे सर्कल का जो सेंटर होगा सर्कल का जो सेंटर होगा वो ये होगा बहुत कंफ्यूज तो नहीं कर रहा हूं मैं आपको स्टूडेंट्स वह आपका लाल जस्ट पॉसिबल सर्कल होगा सबसे बड़ा सर्कल होगा क्यों क्योंकि उसका सेंटर होगा यहां इस सर्कल की सरकम्फ्रेंसेस पर और वो पास हो रहा होगा अब से तो ए बी से जो सबसे दूर पॉइंट है वो ये है और ए बी के जो सबसे पास पॉइंट है वो ये कैसे बनाया सर एक कमा बी से एक नॉर्मल ड्रॉप करके कोई तकलीफ तो नहीं है इस बात से नहीं है अब मुझे एक बात बताओ सारी बातें छोड़ो आप तो मुझे सारी बातें छोड़ के आप ये बात बताओ उसने क्या पूछा था उसने इन दोनों सर्कल्स की रेडियस का डिफरेंस पूछा था एक बात बताओ एक बात बहुत ध्यान से सोच के बताना यह जो बड़ा वाला सर्कल है जो लार्जेस्ट सर्कल इस पॉइंट से इस पॉइंट तक के डिस्टेंस बी और इसे मैं का डन लेट से दी तो आप मुझे बता सकते हो जो लार्जेस्ट सर्कल है उसकी रेडियस कितनी है लार्जेस्ट सर्कल की रेडियस है ए दी आई होप आप समझ का रहे हो कैपिटल आर से दिनो करता हूं उसे मैं कहता हूं एड स्मालेस्ट सर्कल की रेडियस कितनी है सर स्मालेस्ट सर्कल की रेडियस है बी तो स्मालेस्ट सर्कल की रेडियस कितनी है बी उसने आपसे क्या पूछा दोनों का डिफरेंस उसने आपसे पूछा दोनों का डिफरेंस तो कैपिटल आर - आर - आर मतलब आप से वो पूछ रहा है एड माइंस बी दी क्या आप मुझे बस ये देख कर आंसर बता रहे हो क्या या मैं ही बताऊं भाई ये तो बता दो सर आप कितने दिमाग से क्यों नहीं सोच का रहे हो की अगर यह है आपका एड और यह है आपका बी तो अगर आप एड में से बी सब्सट्रैक्ट करोगे तो आपका जो बचेगा वो क्या बचेगा भाई सर वो बचेगा अब और अब क्या है अब इसे नथिंग बट इस सर्कल का डायमीटर गिवन सर्कल का डायमीटर गिवन सर्कल की रेडियस जितना मुझे याद ए रहा है थ्री थी तो इसका डायमीटर कितना होगा सिक्स तो अब के लेंथ कितनी होगी सिक्स और सिक्स ही इस क्वेश्चन का क्या होगा आंसर एक बहुत बेसिक बहुत स्टैंडर्ड क्वेश्चन जो सीधे-सीधे इंशन से किया जा सकता है अगर आपने बातें अच्छे से सीखी और समझे तो इस क्वेश्चन का आंसर आप कौन सा ऑप्शन मार्क करेंगे स्टूडेंट्स मेरे ख्याल से ज्यादा सोचने की जरूरत ही नहीं थी सर इस क्वेश्चन का आंसर मार्क करेंगे ऑप्शन ए यहां तक किसी भी स्टूडेंट को डाउट है तो पूछो सर बड़ा बेसिक सा क्वेश्चन था बस यही तो कंडीशन थी क्या की हमें ये इंश्योर कर लेना था की ए कमा बी से पास होने वाला सर्कल जिसका सेंटर यहां प्लेस दें वो ये छोटे से छोटा ही होगा जो की इससे सबसे क्लोज होगा तो ये कॉम अभी का नॉर्मल जो यहां और ए बी से दूर सबसे कौन होगा सर वो ये जो सबसे दूर जा रहा होगा डायमीटर ही तो आपको क्लोजेस्ट और फादर पॉइंट देता है आप हमेशा ही बात याद रखना मसाला थी\| हमने निकल लिया आसान था आसान था की नहीं मेरा यकीन करिए ऐसे क्वेश्चंस आएंगे और आप ऐसा नहीं सोच पाएंगे तो फिर गड़बड़ होगी इस क्वेश्चन को ट्राई करो मैं जरूर हिंट देता हूं आपको इस क्वेश्चन को कैसे सोचना है भाई क्वेश्चन में पहले तो लिखा है सर्कल जिसमें आपने सर्कल को इनस्क्राइब किया विद वैन एंगल जो है वो कितने डिग्री का है 60° का है ठीक है कोई दिक्कत तो नहीं है डिस्टेंस फ्रॉम डी सेंटर ऑफ डी सर्कल तू डी नेरेस्ट वर्टेक्स है तो रोंबस की रोंबस के अंदर एक सर्कल इनस्क्राइब किया रोमांस के अंदर एक सर्कल इनस्क्राइब किया और उसे सर्कल के मतलब उसे सर्कल के सेंटर से रोंबस की जो सबसे क्लोजेस्ट वर्टेक्स है उसकी जो डिस्टेंस निकलकर कोई भी रैंडम पॉइंट में ले लूंगा और उसके अगर रोंबस के अलग-अलग होगी आपको यह बताना है आप सारी बातें छोड़ो आप सबसे पहले तो इस क्वेश्चन को इमेजिन करो मेरा आपसे बस यह कहना है की अगर इस क्वेश्चन को इमेजिन ही करना है तो हम हमारी सहूलियत के अकॉर्डिंग क्यों ना करें हम हमारे कन्वीनियंस के अकॉर्डिंग क्यों ना करें बिल्कुल करेंगे तो अगर हम हमारी सहूलियत हमारे कन्वीनियंस के अकॉर्डिंग करना चाहेंगे तो कैसे करेंगे सर सर सबसे पहले तो ये आपका क्या हो जाएगा ए एक्सिस और इसी तरीके से ये आपका क्या हो जाएगा सर ये हो जाएगा आपका एक्स एक्सिस आई होप ये बातें आप समझते हैं अच्छे से इसमें कोई आपत्ति नहीं है सर जब ये ए एक्सेस और एक्सेस बने तो मैं एक ऐसा सर्कल लूंगा जो की ऑफ कोर्स आपका कहां सेंटर होगा जो की ऑफ कोर्स आपका ओरिजिन पर सेंटर होगा ऐसा क्यों किया सर क्योंकि मुझे अपनी सहूलियत के अकॉर्डिंग चीज देखनी है ना अब आपको क्या चाहिए था सर अब आपको चाहिए था एक रोम बस तो ठीक है क्रोम बनाते हैं और उसे रोंबस के अंदर है सर्कल को इनस्क्राइब आपने किया था उसे रोंबस के अंदर आपने सर्कल को क्या किया था सर इनस्क्राइब तो लेट से यह वो इनस्क्राइब कल है जिसको हमने क्रोम किया मेरी बातें समझ सन और देख कर अच्छे से डाइजेस्ट कर का रहे हैं जो बातें मैं आपसे कहना चाह रहा हूं भाई हमने हमारे कन्वीनियंस के अकॉर्डिंग बना लिया है और हम कोई भी गड़बड़ नहीं कर रहे हैं और मैं इसे अपने बेस्ट पॉसिबल वे में क्रिएट करने की कोशिश कर रहा हूं ताकि आपको कोई तकलीफ या दिक्कत ना हो तो सर यह क्रोम पर से आई थिंक बहुत प्रॉपर रोंबस ना बनाओ शायद ये वाला पार्ट थोड़ा ज्यादा स्ट्रेच हो गया बट आई होप आप समझ का रहे हो एक रोम बस है रोंबस मतलब प्ले लिए पैरेललोग्राम जिसकी चारों साइड्स की लेंथ इक्वल होती हैं याद ए रहा है अब अगर ऐसा है अब अगर ऐसा है तो ऑफ कोर्स सर मुझे ये जानने में कोई हरजिया तकलीफ नहीं है की किसके अंदर बनाए सर यहां से देखना डी सर्कल इस इनस्क्राइब्ड इन ए रोम्बस विद वैन एंगल 60 डिग्री डी डिस्टेंस फ्रॉम डी सेंटर तू डी सर्कल तू डी नेरेस्ट वर्टेक्स इस इक्वल तू वैन मैं ये मैन लेता हूं मैं ये मैन लेता हूं की इसकी जो नेरेस्ट वर्टेक्स है वो कौन सी है इसकी जो नेरेस्ट वर्टेक्स है वो मैं मैन लेता हूं ये है है ना सुनना ध्यान से और वो का रहा है की सर्कल के सेंटर से इस रोंबस की नेरेस्ट वर्टेक्स के डिस्टेंस कितनी है वैन और अगर ये वैन है तो कहना इसे इसके cardinate जाएंगे जीरो कमा वैन और सिमेट्री से इसके क्वाड्रांट्स हो जाएंगे क्या जीरो कमा माइंस खबर इधर आपने इसके कोऑर्डिनेट्स वैन कमा जीरो और माइंस जीरो रखें नहीं ऐसा नहीं होने वाला है ये थोड़ा स्क्विज होगा है ना ये कुछ ऐसा बना होगा नेरेस्ट वर्टेक्स होगी और आपकी दूर वाली वर्टेक्स भी होगी तो ये डिस्टेंस और ये डिस्टेंस से नहीं होगी आई होप बातें डाइजेस्ट कर का रहे हो जो मैं आपको समझाना कहना है बोलना चाह रहा हूं है ना तो सर अब आप क्या कहने वाले हो क्या करने वाले हो बात इस पे करते हैं अब एक ब्लू आपको जो दिया गया था वो ये इससे लाइन को ध्यान से पढ़िए है वह का रहा था की सर वैन एंगल 60 डिग्री विद वैन एंगल 60° 60° वाले एंगल का क्या मतलब है अगर एक एंगल 60° का बन रहा है तो उससे क्या निकल कर ए रही है चीज सुनो ध्यान से अगर एक एंगल 60 डिग्री का बना तो सर सिमिट्रिकल लॉ से बहुत कुछ मजेदार सा होने वाला है मैं मैन लेता हूं ये एंगल 60 डिग्री है तो ये बताओ ये कितना होगा सर ये एंगल क्लीयरली होगा 30 डिग्री क्या आप सब इस बात से एग्री करते हो हान सर ये एंगल होगा 30 डिग्री एक कम ऐसा करूं क्या की सर इस सर्कल के सेंटर से इस पर एक परपेंडिकुलर ड्रॉप करूं आप में से किसी भी स्टूडेंट को इस बात से कोई आपत्ति जब आपने परपेंडिकुलर ड्रॉप किया तो क्या हो गया सर ध्यान से देखो जब आपने परपेंडिकुलर ड्रॉप किया तो देखो भाई ये है 30 तो कहना इसे की सर ये हो जाएगा 90° अगर ये 39 तो ये कितना सर ये हो जाएगा 60 अगर यह 60 तो ये पूरा 90 तो ये भी कितना होगा सर वापस ये होगा 30° अब suniyega आप कम की बात है ये होगा कितना 30° अब बहुत कम की बहुत जरूरी बात सुनना क्योंकि इससे आपकी क्वेश्चन खत्म कर दोगे ऐसे कैसे कर देंगे सर सुनो तो एक बात बताओ एक बात बताओ सर अगर ये है ओरिजिन इन्हें मैं कुछ कोऑर्डिनेट्स देता हूं एक बी सी और दी मुझे एक बस इतनी सी बात बताओ स्टूडेंट्स क्या यह जो है मतलब मैं इसको नाम दे देता हूं अगर आपको कोई कन्फ्यूजन है तो लेट से मैं इसे नाम देता हूं क्या यह आपको राइट एंगल ट्रायंगल नहीं दिख रहा मैं बात कर रहा हूं ट्रायंगल आप में सिर्फ ट्रायंगल आप में ध्यान से सुनना स्टूडेंट्स मुझे क्या पता नहीं है और क्या पता करना है इस बारे में आपसे थोड़ी बात करना चाह रहा हूं ऐसा क्यों क्योंकि बात समझना स्टूडेंट्स बहुत जरूरी बात है मुझे बस अगर ये पता चल जाए किसी तरीके से किस सर्कल का इस सर्कल की ये लेंथ कितनी है जो की बेसिकली क्या है रेडियस तो क्या है पता की जा सकती है सर आप दोनों पता कर सकते हो आप यह भी पता कर सकते हो और आप यह भी पता कर सकते हो क्यों पता करना है 2 मिनट थोड़ा पेशेंस तो अस्पत करना है कोई रीजन रहा होगा तो पहले तो मैं पता करने की कोशिश करता हूं लेट्स से आप बस ऐसे ही करना है बात करेंगे क्यों पता करना है अगर मैं आप पता करना चाह रहा हूं तो देखो ये है एंगल और यह है आपका उसके नीचे की साइड तो ये है बेस और आपको हाइपोटेन्यूज पता है तो आपको बेस और हाइपोटेन्यूज में रिलेशनशिप चाहिए तो कोस्थेटा तो मैं ले रहा हूं कोस 30 अब कोस 30 क्या होगा ध्यान से देखना सर कॉस्ट 30 होगा बेस अपॉन हाइपोटेन्यूज तो ये हो जाएगा आप / ओ बी तो ये कितना होगा सर आप / आप अच्छा अब की लेंथ कितनी है सर ओ बी वैन यूनिट का है आई होप ये तो बात आप समझ का रहे हो तो ob1 यूनिट क्या है तो ये तो हो जाएगा वैन तो आप की वैल्यू कितनी आएगी कोस 30 कोस 30 कितना होता है सर कोस 30 होता है अंडर रूट 3/2 सर ओ पी के ए जाने से क्या पता चल रहा है चलो मैन लो आप मुझे पता चल गया अब आप के ए जाने से जो आपको पता चल रहा है वो ये की आप ओए निकल सकते हो क्या आप निकल के बता सकते हो ओए जरा सोच के देखो की ओए आप ऐसे कैसे निकल सकते हो ने अगर आप सीधे अब और ओम में देखते तो भी निकल सकते द वैसे भी निकल सकते द इतना घुमा फिर के करने के बजाय मैं उसे तरीके पर बात करना चाह रहा हूं आपसे जिस तरीके पर मैं आपसे बात करना चाह रहा हूं वो क्या प्लीज इस बात को समझो स्टूडेंट्स जैसे की देखो इतना मत करो इतना कुछ कारी बिना भी आप सीधे निकल लेते मैंने थोड़ा सा उसे घुमा दिया मेरा कहने का मतलब है देखो यह कितना है 30 डिग्री यह 30° है ना मतलब क्यों करना है पता नहीं क्यों मेरे दिमाग में पता नहीं इतना घुमा फिर के चीजे करने को थॉट आया मैं कहता हूं सर ऐसे मत सोचो और ये पीवी लेने की जरूरत ही नहीं है आप तो ये बोलो आप तो बस ये बताओ अगर मैं ट्रायंगल आब की बात करूं ट्रायंगल आब की बात करूं तो ये 30° है क्या हान सर है अगर ये 30° है तो इतना कुछ सोचने की जरूरत ही नहीं पता नहीं क्यों मैं ऐसा करवा रहा हूं आपसे ध्यान से सुनो स्टूडेंट्स अगर ये 30° है सुनना ध्यान से तो देखो मुझे क्या पता है मुझे वो भी पता है और मुझे ओए पता करना है मुझे अब पता है और मुझे ओए पता करना है बात समझ का रहे हो क्या ये एंगल है परपेंडिकुलर और बेस तो क्या मैं 10:30 की हेल्प ले सकता हूं बिल्कुल ले सकते हो तो जैसे ही आप 10:30 की हेल्प लोग तो 10:30 क्या बोलेगा सर वो बोलेगा परपेंडिकुलर अपॉन बेस तो क्या बोलेगा वो बोलेगा वो भी अपॉन ओए कोई दिक्कत तो नहीं भाई कोई दिक्कत तो नहीं हो जाएगा सर ओए जो आप निकलना होता है वैन बाय रूट थ्री वैन बाय रूट थ्री का ऐसे प्रोकल रूट 3 तो ओए कितना ए जाएगा सर ओए ए जाएगा √3 अब जैसे ही ओए आया √3 तो कहना है की सर ए के जो cardinate होंगे जो एक-एक कोऑर्डिनेट्स होंगे वो होंगे √3 कमा जीरो पता नहीं क्यों मैं बहुत घुमा फिर के कर रहा था मैं रियली सॉरी फॉर डेट बाद में मुझे रिलाइज हुआ कितना सोचने की जरूरत नहीं है और कभी-कभी आपके साथ ही अक्सर ऐसा होगा की आप कुछ कुछ करते करते हैं पोस्चर के बाद में लगेगा अगर इतना घूमर गए ऐसे तो हम यही ए जाते ऐसा होगा सॉल्व करते-करते वक्त ऐसा होगा या जेनुइन स्वाभाविक है ये नेचुरल तो यह हमें कुछ चार पॉइंट मिल गए अब क्या अब मेरा आपसे यह पूछना है की अगर मैं इस पर एक रैंडम पॉइंट मैन लेता हूं पी इस पे कहीं भी एक रैंडम पॉइंट मैन लेता हूं पी जिसके कोऑर्डिनेट्स में मैन लेता हूं होमो के अब उसने आपसे क्या कहा है इस पी पॉइंट की ए से बी से सी से और दी से डिस्टेंस इसका स्क्वायर इसका सैम तो आप सोच के बताओ पी से जब एक ही डिस्टेंस निकलेंगे तो वो क्या आएगा सर वो होगा आई होप आप देख का रहे हो पी से ए की जो डिस्टेंस निकलूंगा वो क्या होगा ह-√3 का स्क्वायर प्लस के माइंस जीरो का स्क्वायर तो पहले तो हमें पी ए निकलता हूं तो कितना हो गया सर वो होगा ह-√3 का स्क्वायर प्लस के का स्क्वायर प्लस के माइंस वैन का स्क्वायर क्या आएगा आप ध्यान से देखो सर पीसी मतलब क्या h+√3 प्लस के स्क्वायर को ऐसे लिख देना है h+√3 का स्क्वायर प्लस के प्लस वैन का होल स्क्वायर प्लस वैन का होल स्क्वायर कोई परेशानी आई थिंक चीज इस तरीके से आपको देखनी चाहिए ये आपकी वो वैल्यू है जो वो इस क्वेश्चन में ढूंढ रहा है कुछ का रहा है तो निकलती हैं सर इस बारे में बात कर लेते हैं क्या-क्या है ध्यान से देखो भाई एक मिनट थोड़ा ध्यान से देखो बात समझना बात समझना यह हो जाएगा 2h√3 तो कैंसिल हो जाएगा अब सुनना हो जाएगा 2x² और ये हो जाएगा कितना ये थ्री के स्क्वायर भी होगा जिसपे हम बात करेंगे अभी इस बारे में अभी तो हम सीधे सीधे सॉल्व करते हैं बिना कंफ्यूज हुए और क्या सर और ध्यान से देखो के - 1 के स्क्वायर प्लस वैन स्क्वायर कितना हो जाएगा ये हो जाएगा 4K स्क्वायर मतलब आई होप यह दोनों टर्म हमने कितना मिलेगा सर यह मिलेगा आपको 1 + 1 2 कोई दिक्कत तो नहीं अब एक और बातें स्क्वायर ये कितना हो जाएगा 4² और 3 + 3 कितना हो जाएगा सिक्स तो यहां पर आपको मिलेगा सिक्स आई थिंक चीजें आसान है बिल्कुल आसान है सर अब मुझे उसे चीज की जरूरत पड़ेगी जो मैं निकल रहा था जिसकी हमें जरूरत थी suniyega ध्यान से बहुत कम की बात है बहुत कृष्ण बात है प्लीज इस बात को सुनना 4K स्क्वायर प्लस 4 स्क्वायर अब सर आप स्क्वायर क्या करोगे वो आपके सर्कल पर लाइक करने वाला कोई पॉइंट है ना मतलब सर्कल पर लाइक आने वाला मतलब ये पॉइंट ह की ओरिजिन से डिस्टेंस का स्क्वायर यानी की और अच्छी भाषा में मैं कहूं तो सर टेक्निकल सर्कल पर लाइक करने वाले किसी भी पॉइंट का उसके सेंटर से डिस्टेंस का स्क्वायर मतलब सर्कल की रेडियस का स्क्वायर सर्कल की रेडियस का स्क्वायर की रेडियस निकल सकता हूं कैसे निकलेंगे सोच कर देखो सर फिर वही बात यह है 30° है की नहीं सर अब इस 30 डिग्री में ध्यान से देखना मुझे ओए पता है मैं किसकी बात कर रहा हूं जैसे मैं इस पॉइंट को थोड़ी देर के लिए कहता हूं क्यों है ना तो मैं आपसे बात कर रहा हूं ट्रायंगल ओ के ए में ट्रायंगल ओके में ध्यान से देखना ये है 30° मुझे 30 डिग्री के नीचे वाली साइड पता है तो मुझे हाइपोटेन्यूज पता है और मुझे क्या चाहिए परपेंडिकुलर 30 डिग्री के अपोजिट परपेंडिकुलर तो परपेंडिकुलर और हाइपोटेन्यूज मतलब sinθ यानी कितना सिन 30 तो जब आप लेना चाहोगे क्या साइन 30 तो क्या मिलेगा जरा देखो भाई सिन 30 होगा इस 30 के सामने वाली साइड वो क्यों निकलना है सर तो ओक क्या आएगा ध्यान से देखो भाई ओक जो आएगा सर वो होगा आपका ओए साइन 30 ओए कितना है √3 तो √3 सिन 30 और sin30 कितना होता है सर 1 / 2 तो हो जाएगा √3/2 पर एक मिनट सर आप ओ क्यों नहीं निकल रहे द तो के का स्क्वायर कितना होगा 3 / 4 ये कितना होगा भाई 3/4 तो आप h2 + K2 की वैल्यू कितनी रिप्लेस करोगे सर आप रिप्लेस करोगे 3/4 तो ये हो जाएगा 4 3 / 4 + 8 सो दिस इस बेसिकली थ्री और 8 + 3 कितना 11 सो दिस इसे गोइंग तू बी क्लीयरली नथिंग बट 11 विच हैपेंड्स तू बी ऑप्शन बी क्या यह पूरा पार्ट आप सभी को समझ आया क्या यहां तक कोई भी डाउट परेशानी दिक्कत या तकलीफ है स्टूडेंट्स आई थिंक यह भी एक आसान सा बड़ा ही मजेदार सा डायरेक्ट सा सीधा साधा सुलझा हुआ सा क्वेश्चन था चलो अगर यहां तक चीज समझ आई क्या आगे अगर हम यहां से बड़े तो अब आता है एक और आपके सामने क्वेश्चन मैं जरूर इस क्वेश्चन को आपको samjhaunga फिर आप ट्राई करेंगे फिर हम साथ में करेंगे क्या लिखा है देखो भाई डी एरिया सर्कल्स कौन-कौन से सर्कल एक तो यह सर्कल इसको देखकर ही कुछ चमकना चाहिए और एक सर्कल इसको देख कर भी कुछ तो चमकना चाहिए जो की आपने पढ़ा है सर जीवन में सारी चीज आसान थी पर यह बहुत अजीब सा लगता है हमको अब नहीं लगना चाहिए भाई अजीब सा चलो पहले तुम समझते हो अच्छा मुझे एक बात का जवाब दो चलो इस बारे में बात करेंगे इस बारे में बाद में बात करेंगे जीरो दोनों का सेंटर जीरो कमा जीरो पर है तो दोनों कैसे सर्कल होगा सर दोनों को सेंट्रिक सर्कल हो गए हैं अगर आप मेरी बात से एग्री करते हो तो करते हो क्या करते हैं सर तो ये हो गया आपका जीरो कमा जीरो ये क्या बना रहा हूं मैं ये हो जाएगा आपका सर्कल का सेंटर पर देखूं तो इस अंदर वाले सर्कल की रेडियस कितनी है सर अंदर वाले सर्कल की रेडियस है वैन और बाहर वाले सर्कल की रेडियस कितनी है सर बाहर वाले सर्कल की रेडियस है तू क्यों फोर यानी तू का स्क्वायर ये तो अब आप देख के ही समझ जाते हो ना अब अगर यहां पर एरिया पूछा होता तो मैं कहता सर एरिया बॉन्डेड की रीजन मतलब वो इस रीजन का एरिया पूछ रहा है और मैं तो सर ये तो बचपन से करता ए रहा हूं पाई R1 स्क्वायर माइंस दिया वह तो आपको जी मांस और एडवांस की तैयारी करवा रहे हैं ना तो यहां पर कुछ डाला गया अब इसका क्या करें सर इससे कैसे जुड़ें इससे कैसे लड़े सवाल तो ये है तो मेरा आपसे कहना है क्या इसे देखकर आपको पैर ऑफ स्ट्रेट लाइंस का एक वो कॉन्सेप्ट याद आता है जहां पर हमने ये बात की थी की सर ये तो वो स्ट्रेट लाइन से जो ओरिजिन से पास होते हैं शरीर तो वो स्ट्रेट लाइन से जो ओरिजिन से पास होती हैं दोनों ओरिजिन से पास होती है अगर आप ध्यान से देखो तो सर आपको कैसे पता चल रहा है मेरा पता करने का तरीका बस इतना सा है की अगर आप देखो तो ax² + बी ए स्क्वायर प्लस तू हस्की बाकी टर्म्स है ही नहीं आपको याद ए रहा है जो मैं कहना चाह रहा हूं यह स्टूडेंट्स की अगर मैं पैर ऑफ स्ट्रेट लाइंस की बात करूं तो हमने पढ़ा है ax² + बी ए स्क्वायर सेटिस्फाई करता है तो यह जो पैर ऑफ स्ट्रेट लाइंस है ओरिजिन से पास होती है मतलब दोनों ओरिजिन से निकलती हैं दोनों ओरिजिन से गुजरते हैं मतलब ये दोनों पैर ऑफ स्ट्रेट लाइंस कुछ ऐसी निकलेंगे कुछ निकलेंगे जैसी भी निकलेंगे मुझे नहीं पता पर कुछ निकल रही होंगी एक ऐसी जा रही होगी और दूसरी वाली कुछ ऐसी जा रही होगी जैसे भी जारी होगी मुझे नहीं पता पर कुछ इस तरीके से जा रही होंगी तो आपसे एरिया बॉन्डेड बाय डी रीजन जब पूछा जाए तो इन तीनों ने किसको बुन किया इन तीनों ने जो बाउंड किया वो इसे क्या की इस तीनों ने मिलके इस रीजन को बाउंड के इसने बोला इससे ऊपर इसने बोला इससे नीचे उसने बोला इधर उसने बोला इधर तो आपसे इस रीजन का एरिया पूछा जा रहा है जो हम निकल देंगे पर ये रीजन के लिए सर सबसे बड़ी समस्या क्या है पता है मुझे पता होना की आर्क कितने का है ये आर्क कितने का है सर यह तो मुझे शायद निकलती आता है यह तो मुझे शायद निकलती आता है मैंने कुछ ऐरो ड्रा किए हैं कुछ बनाए अब मुझे आप बता दो इसका फॉर्मूला क्योंकि यह आपको मैंने बहुत शिद्दत बहुत मेहनत और बहुत अच्छे से पढ़ाया द अभी क्या आपको याद नहीं ए रहा की सर पैर ऑफ स्ट्रेट लाइंस में अगर मुझे ऐसी कोई इक्वेशन दे दी जाए तो उन दोनों के बीच का क्यूट एंगल यानी 10 थीटा इस गिवन की मोड ओवर इट्स स्क्वायर -अब / ए + बी और ऑफ कोर्स a² - अब के ऊपर आप क्या लगा लेते हो आप लगा लेते हो अंडर रूट तो यहां पर ये मैं अंडर रूट लगा दूंगा क्या आपको ये फॉर्मूला याद नहीं ए रहा है स्टूडेंट्स क्या हमने फॉर्मूला नहीं पढ़ा था x² - अब/a+b ऑफ कोर्स अगर आपको मोड नहीं लगाना है तो यहां मत लगाओ क्योंकि ये तो वैसे भी अंडर रूट वाला पार्ट है पर ए + बी के ऊपर तो जरूर मोड लगाना याद ए रहा है क्या उसको मोड लगाने की आदत है तो इस तरीके से ए जाता है तो x² - अब इनसाइड हम √ ए + बी इन साइड बिटवीन इसरो स्ट्रेट लाइंस तो सर बात करते हैं एंगल बिटवीन डीज तू स्ट्रेट लाइन आप चाहो तो इससे पैर विजय यू नो स्प्लिट भी कर सकते हो वैसे भी सोच सकते हो m1 और M2 की फॉर्म में और वहां से भी एंगल बिटवीन तू स्ट्रेट लाइन निकल सकते हो m1 - M2 / 1 + m1 M2 तन थीटा पर मैं का रहा हूं उतना जाने की वजह से रिसीव हो ना आप ध्यान से ए कितना है अंडर रूट थ्री बी कितना है √3 और 2hxy तो ह कितना है सर अगर ये इधर है तो 2h यानी 4 कितना होगा 2 2 तो ह की वैल्यू होगी -2 मतलब अगर आपको मैं बातें समझाना चाहूं तो देखो भाई ए कितना हो जाएगा √3 बी कितना हो जाएगा √3 और ह हो जाएगा आपका तकनीक के लिए माइंस तू और यहां से अगर आप तन थीटा निकलोगे तो देखोगे तन थीटा निकलते हैं तो x² मतलब -2 का स्क्वायर कितना 4 - अब यानी कितना अंडर रूट 3√3 कितना 3 इस पर क्या लगेगा अंडर रूट अपन में ए + बी ए + बी यानी √3 √3 डेट्स तू रूट थ्री सही लिख रहा हूं क्या कुछ गलत तो नहीं लिखा स्क्वायर मतलब -2 का स्क्वायर और इस तरीके से आपका यह एंगल निकल कर ए रहा है जो की ए रहा है कुछ इस तरीके से जिसे मैं लिख रहा हूं कितना 1 / 2√3 बस एक इसमें छोटा सा करेक्शन फॉर्मूले में जो मुझे लग रहा है की मैं क्यों गलती कर देता हूं ऐसी की यहां पर तू भी तो आता है भाई ये क्यों नहीं लिखा हमने यहां पर तू आता है राइट तो ये हो जाता है तन थीटा जो की होता है तू टाइम्स अंडर रूट होकर स्क्वायर माइंस ये भी अपॉन ए + बी तो यहां पर भी तू ए जाएगा सर ये ए जाता है वैन बाय रूट थ्री जो की ऑफ कोर्स आपका क्या tanθ क्या इसे देख कर मुझे थीटा की वैल्यू आप बता सकते हो तन का कौन सा एंगल तन का कौन सा एंगल आपका अंडर रूट वैन बाय रूट थ्री होता है सर मेरी ख्याल से 30° मेरे ख्याल से 30° तो ये जो एंगल है ये है 30 डिग्री और जैसे ही आपको 30° पता चला की आपको इसका आंसर स्ट्राइक कर रहा है सोच के देखो मैं क्या करूंगा सोच के देखो मुझे इस सेक्शन में अगर मैं बात करूं आपसे सर पहले तो आप इसका एरिया निकल लो फिर आप इसका एरिया निकल लेना और आई थिंक दोनों का डिफरेंस आपको आंसर दे देगा अब सुनना पूरा एरिया जो 2π का होता है होता है पाई आर स्क्वायर 30 डिग्री में तो पहले तो मुझे वो परेशन निकलना पड़ेगा जो चाहिए तो पूरा होता है जो की होता है 360 लेकिन मुझे उसके 30 डिग्री वाले परेशन मतलब मुझे इतना फ्रैक्शन चाहिए क्या बात है अब बड़े वाले अगर सर्कल की बात करें तो होगा पाई आर स्क्वायर यानी कितना 1 पाई तो ये ऐसा कुछ हो जाएगा कोई तकलीफ तो नहीं है भाई इन बातों से और सर अगर आप इसे सिंपलीफाई करें तो 30 / 360 क्या मुझे कुछ कहने की जरूरत है की थ्री का 36 12 टाइम्स क्या आपको सभी को दिख रहा है तो ये हो जाएगा 4π-π और ये हो जाएगा 35/12 जो की कितना हो जाएगा बाय बाय है और π/4 अगर मैं देखूं तो क्लीयरली कौन सा ऑप्शन ना करेंगे आप सर हम बिना सोचे समझे माफ करेंगे ऑप्शन दी आसान सा बेसिक सा क्वेश्चन है और वह ओबवियसली किस क्वाड्रेंट में पूछ रहा था वो फर्स्ट क्वाड्रेंट में पूछ रहा था मैंने इस लाइन को ठीक से नहीं पढ़ा हम रेली सॉरी फर्स्ट क्वाड्रेंट में पूछने का मतलब था की ये रहेगा आपका फर्स्ट क्वाड्रेंट तो ये दोनों लाइंस ऐसी भी तो गुजर रही होंगी तो ऐसा का ऐसे ही फोर्थ क्वाड्रेट क्वाड्रेंट में भी तो बनता पर अगर वो पूरा एरिया पूछता तो आप इसका डबल लिख रहे होते π/2 पर उसने ये वाला एरिया नहीं पूछा उसने सिर्फ फर्स्ट क्वाड्रेंट का एरिया पूछा है ये 00 है ना समझ का रहे हो क्या तो आई थिंक चीज इस तरीके से होंगे और इस तरीके से आप आंसर लिखोगे बस एक गलती मैंने जो की वो ही है की मैंने फर्स्ट क्वाड्रेंट नहीं पढ़ा और उसके बिना बस ये लिख दिया एक अच्छा ब्रेन पूरा क्वेश्चन ठीक से पड़ेगा और फिर सही आंसर निकलेगा आई होप आपको बात समझ आई इसमें कोई डाउट नहीं है चारों क्वेश्चंस मजेदार द क्वेश्चंस पहले आप ट्राई करेंगे और आप एक बार खुद से देखेंगे चीज निकल कर और वर्कआउट होती हैं तो ठीक वर्ण हम साथ में जो डेफिनेटली हर क्वेश्चन को करते ही हैं इस क्वेश्चन में क्या लिखा है पहले वो देख कर एक बार समझने की कोशिश करते हैं मेरा कहना है सर इस क्वेश्चन को ध्यान से देखो डी कंडीशन डेट ये जो कार्ड है ये जो कॉर्ड है किसकी सर इस सर्कल की वो इस सर्कल के सेंटर पर एक राइट एंगल सब्सटेंड करेगी मैं सच कहूं ना तो इस क्वेश्चन में तो आपको मेरी हेल्प लगनी ही नहीं चाहिए मैं सच कहूं तो इस क्वेश्चन में आपको मेरी सच में हेल्प नहीं लगनी चाहिए क्योंकि यह क्वेश्चन तो आप कर सकते हो इसमें जो भी दिया या पूछा गया ना ये सब आपको आता है मैं फिर से रिपीट करता हूं क्यों डी कंडीशन इस सर्कल की वह इस सर्कल के सेंटर पर इस सर्कल के सेंटर पर राइट एंगल सब्सटेंड कर रहे हैं देखो सर कुछ समझ नहीं ए रहा है तो मेरा कहना होगा की आप चीजों को विजुलाइज करिए आप क्वेश्चंस सर्कल का कहीं ना कहीं क्या होगा सेंटर होगा सर अब इस सर्कल को देखकर कोई दिक्कत तो नहीं है अब सर एक कोड है जो इस सर्कल के सेंटर पर राइट एंगल सब्सटेंड कर रही है एक कोई कार्ड है जो इस सर्कल के सेंटर पर राइट एंगल सब्सटेंड कर रही है तो क्या मैं ऐसा नहीं का सकता की सर पॉसिबली वो ये कोई कार्ड रही होगी मैन सकते हैं क्या बिल्कुल मैन लो सर हमें तो कोई आपत्तियों परेशानी नहीं है अब आप एक बात का जवाब मुझे दीजिए वैसे तो आपको दिख रहा होगा फिर भी अगर आपको स्ट्रेट लाइंस का चैप्टर याद नहीं ए रहा है तो इस लाइन को ध्यान से देखो क्या लिखा है सर एक्स कोस अल्फा + ए सिन α = दिस सर्कल मैं सब्सटेंस ए राइट एंगल डी सेंटर ऑफ डी सर्कल तो क्या कंडीशन लगनी चाहिए बिल्कुल सही पढ़ रहे हो सर आप कोई दिक्कत नहीं है अब एक बात बताओ अगर यह डिस्टेंस ए है अगर ये डिस्टेंस ए है तो क्या ना ऐसे सर ये डिस्टेंस भी ए होगी और ये डिस्टेंस भी ए होगी और अगर यह राइट एंगल ट्रायंगल है अगर ये एक राइट एंगल ट्रायंगल है तो आप क्या कहोगे अगर ये एक राइट एंगल ट्रायंगल है तो मैं बड़ी सिंपल सी बात आपसे कहूंगा सर की यहां से ये और ये ए और ए मुझे मिलाकर ये डिस्टेंस क्या देंगे सर ये जो डिस्टेंस निकल कर आएगी जो की होगा हाइपोटेन्यूज a² + a² यानी 2a² 2a² का अंडर रूट तो कितना हो जाएगा अंडर रूट 2 टाइम्स ए या आप अगर अपने बेसिक स्टैंडर्ड तरीके से लिखना चाहते हो तो आप इसे लिखोगे a√2 कोई तकलीफ तो नहीं सकते हो क्या एक छोटी सी बात का जवाब आप दे सकते हो क्या मैं क्लीयरली जानता हूं सर समझ का रहा हूं की मैन लो देख का रहे हो क्या बस ऐसे ही मेरा मैन है आपको खुद बड़ी बेसिक सी साधन सी बात समझ ए रही होगी सर की ये जो cardinate होगा ये क्या होगा ये कमा जीरो और ये कोऑर्डिनेट्स क्या रहा होगा जीरो कॉम आए बड़ी बेसिक सी बात है स्टूडेंट्स आई होप बड़ी स्माइल सीधी-सीधी सी बात है जिसमें मेरे ख्याल से तो कोई तकलीफ का दिक्कत आपको नहीं होनी चाहिए समझने को लेकर बट फिर भी मैन लो ये जो है कार्ड ये कुछ ऐसी गोद है छह कोड जरूरी है की सर आपने तो हो सकती थी जो की 90° एंगल सब्सटेंड कर सकती थी बिल्कुल कर सकती थी कोई दिक्कत नहीं ये कोर्ट कहीं भी ए सकती थी मैंने तो बस एक रैंडम सा सिनेरियो लिया है क्या कोई बात समझ आई अब एक छोटी सी बात का आप मुझे जवाब दो अगर ये एक्स एक्सिस पर है अगर ये पॉइंट एक्स एक्सिस पर है तो क्या इसका आप यहां पर एक्स koardinate निकल सकते हो मैं टेक्निकल कुछ नहीं निकलना चाह रहा हूं मैं इस कार्ड में स्कॉट की लेंथ निकलना चाह रहा हूं इस लाइन की इक्वेशन से तो मैं दो तरीके उसे कर रहा हूं मैं यहां पर एक स्क्वाड यूनिट निकल लेना चाह रहा हूं यहां पे ए कार्ड लेना चाह रहा हूं उसे लाइन की यहां से यहां तक की पॉइंट की मतलब अगर मैं इसे का रहा हूं थोड़ी देर के लिए और इसे मैं का लूं भी तो अब की डिस्टेंस निकलना चाह रहा हूं इस लाइन की तरफ से जो की मैंने यहां से निकल ए रूट तू और शायद मैं कुछ यहां पर कंडीशन अप्लाई करके आंसर लता हूं काफी कुछ मैंने का दिया है बस अगली लाइन का जवाब दे दो जो आपको पता होगा की मैं क्या कहना चाह रहा हूं मुझे बस इतना बता दो की अब आप क्या कहोगे क्या निकलोगे कैसे चीज सोचोगे क्योंकि अब तो चीज आपको पता है भाई अब तो आपको चीज पता है अब मेरे ख्याल से ज्यादा अगर मैं बता दूंगा तो ये क्वेश्चन खत्म हो जाएगा अब कैसे निकलोगे भाई अच्छा चलो मैं हिंट दे देता हूं पहली एक बार बताओ क्या इस स्ट्रेट लाइन के डिस्टेंस आप इस सर्कल के सेंटर से निकल सकते हो सर कल का सेंटर इस जीरो कमा जीरो स्ट्रेट लाइन की इक्वेशन दी गई है परपेंडिकुलर लेंथ निकलने में कोई बुराई नहीं हम निकल सकते हैं फिर एक और हिंट मैं आपको देता हूं एक बात बताओ ये जो लाइन है ये ऑफ कोर्स राइट एंगल बना रही है ये लाइन जो है ऑफ कोर्स राइटिंग बना रही है तो क्या उसे आइडिया से आप एक 90 डिग्री एंगल वाले कॉन्सेप्ट से इस सर्कल की रेडियस नहीं निकल सकते जो मैं कहना चाह रहा हूं उसे समझो एक बात बताओ बड़ी बेसिक सी बात जीरो कमा जीरो से अगर मैंने इस लाइन पर परपेंडिकुलर ड्रॉप किया तो क्या मैं परपेंडिकुलर लेंथ निकल सकता हूं बिल्कुल निकल सकते हो सर कौन सी बड़ी बात है क्या होगी जरा सोच के बताओ सर यह जो होगी वो क्या होगी जीरो से इस स्ट्रेट लाइन पे जो की कौन सी ये वाली स्ट्रेट लाइन तो देखो जीरो इन कोस अल्फा 00 सिन α 0 तो बचा क्या -पी तो माइंस पी का क्या मोड डिवाइडेड बाय अच्छा माइंस बी का मॉडल लिखूं या बी का मोड कोई फर्क पड़ता है क्या नहीं पड़ता तो लिख लेते हैं हम किसका मोड पी का मोड कोई फर्क नहीं पड़ता है ना माइंस का भी मॉडल लिख लेते तो ये था डिवाइडेड बाय किसके एक्स और ए के कॉएफिशिएंट्स के स्क्वायर अल्फा अंडर रूट वैन का अंडर रूट वैन तो ये कितना ए जाएगा मोड पी तो ये जो लेंथ ए रही है जो की क्लीयरली दिख रही थी सर मतलब आप अगर उसको दूसरी तरीके से देखते हैं ना तो भी आपको समझ ए जाता है की ये डिस्टेंस कितनी होती है ये डिस्टेंस होती है मोड पे अब मेरा कहना है क्या इसे देखकर भी आप मोड भी और ए में डिस्टेंस या रिलेशन नहीं निकल सकते चलो एक और इंटर देता हूं अब क्वेश्चन खत्म होना चाहिए और अब भी नहीं हो रहा है तो दिक्कत है एक बात बताओ यह और है क्या सर है तो अगर यह 90° है तो क्या ये 45 डिग्री होगा अरे होगा की नहीं अगर ये 45 डिग्री होगा तो आप ऐसे सोच लो इसको अगर आप कहते हो ओरिजिन यानी ओ और इस पॉइंट को लेट से आप कहते हो कुछ भी लेट से क्यों तो क्या आप ट्रायंगल एक में कुछ सोच का रहे हो ट्रायंगल ओए के में कुछ सोचने का मेरा बस बड़ा सिंपल सा मतलब है ट्रायंगल ओए के में अगर आप देखोगे तो ये एंगल कितना होगा सर ये एंगल होगा 45° एंगल होगा 45 डिग्री आप सन रहे हो आप समझ का रहे हो जो मैं कहना चाह रहा हूं मुझे बस ये पता है suniyega ध्यान से मेरे पास हाइपोटेन्यूज है मेरे पास हाइपोटेन्यूज है और मेरे पास क्या है भाई परपेंडिकुलर यह एंगल है इसके सामने की साइट परपेंडिकुलर इसके यह वाली साइड क्या हो गई हाइपोटेन्यूज 90 डिग्री के अपोजिट तो अगर परपेंडिकुलर और हाइपोटेन्यूज में रिलेशन चाहिए तो सिथेंटा तो सिन 45 से बात कर लेते हैं तो suniyega ध्यान से सिन 45 को आप क्या कहोगे सर हम साइन 45 को कहेंगे परपेंडिकुलर जो की क्या है सर मोड पी और भी बहुत सारे तरीके हो सकते द बट यही तरीका देख लो और आपका ये क्लीयरली कितना है सर ये है आपका हाइपोटेन्यूज जो की है ए सिन 45 आप मुझे बता रहे हो क्या कितना होता है सर साइन 45 होता है 1/√2 तो ये हो जाता है 1 / √2 = ए सर ए को यहां ले आते हैं ये हो जाता है √2 पी तो ये रिलेशन पर मैं बात कर रहा हूं ऐसा कुछ तो नहीं दिया है तो आपका आंसर गलत है क्या मुझे तो नहीं लग रहा है मेरा आंसर गलत है क्योंकि अगर मैं इसका स्क्वायर कर डन तो लेफ्ट हैंड साइड पर ए जाएगा ये तो सबसे ही होना चाहिए भाई सबको आया और अगर ये आया तो क्या अगला क्वेश्चन ट्राई करें क्या लिखा है समझते हैं सर्कल तो सर यह जो सर्कल है इसके किसी पॉइंट पर हमने टांगें ड्रॉ की कर लेंगे सर वह इस स्ट्रेट लाइंस से एक पॉइंट के पर ए एक्सिस ये इस क्वेश्चन का करेक्ट है ये इस क्वेश्चन का करेक्ट है इसे ठीक से पढ़ना वो इस स्ट्रेट लाइन से ए एक्सिस पर मिलती है प्लीज इसे हम इमेजिन करो इस स्ट्रेट लाइन से कोई मिलती है स्ट्रेट लाइन ए एक्सिस पर प्लीज सोच के देखो आपका आंसर ए जाएगा है और वह पूछ रहा है फिर पीके की लेंथ तो के पर मिल रही है पी की आपसे लेंथ पूछी जा रही है मेरा यकीन करिए इससे आसान क्वेश्चन आज नहीं हो सकता था सर ऐसा कैसे आसान क्वेश्चन हो गया है आप छोड़ो सारी बातें अब तो विजुलाइज कर लो कैसे लाइक करेंगे सर आप पहले तो सर्कल बना लो ये रहा सर्कल ठीक है क्या सर ये जो सर्कल है इस सर्कल के किसी पॉइंट पर आपने टैसेंट ड्रॉ की है है ना किसी पॉइंट पर आपने टैसेंट ड्रॉ किया मैंने यहां कहीं ड्रॉ की है जैसे मैंने यहां कहीं एक टैसेंट ड्रॉ की है कोई तकलीफ तो नहीं है किसी स्टूडेंट को लेट से मैंने यहां कहीं टेंशन ड्रॉ की है ना ये क्या है आपका पॉइंट पी अब वह आपसे यह का रहा है की यह जो स्ट्रेट लाइन है मैन लो यहां कहीं आपका सिलेबस में ऐसे ही रैंडम बना दे रहा हूं ना बस ये जो स्ट्रेट लाइन है ये वाली ये आपकी ये वाली स्ट्रेट लाइन है है ना और लेट से ये है आपका क्या ए-एक्सिस ये जो है आपका ये है आईएसआईएस उन्होंने आपसे ये कहा जनाब उन्होंने आपसे बस ये कहा की ये जो टैसेंट आपने ड्रॉ की ये जो टैसेंट आपने ड्रॉ की वैसे तो ये क्वेश्चन पढ़ते ही खत्म हो जाना चाहिए बट फिर भी ये जो टैसेंट आपने ड्रॉ की ये इस लाइन से कौन सी लाइन 5x - 2y + 6 = 0 से ए एक्सिस पर ये कौन सा एक्सेस है ये आपका ए एक्सिस है ए एक्सिस पर मिलती है पॉइंट के पर है तो वह पीके की लेंथ पूछ रहा है तो क्या सोच का रहे हो क्या एक्सेस पर ही मिल रही है उसी पॉइंट के ऊपर तो के के कोऑर्डिनेटर अगर निकलने हैं तो यहां पर एक्स की जगह जीरो रख दीजिए जनाब एक्स की जगह जीरो रखा एक्स की जगह जीरो रखा तो यह जीरो माइंस सिक्स माइंस सिक्स बाय माइंस तू यह पॉइंट क्या हो जाएगा सर ये हो जाएगा जीरो कैसे बात कर रहे हो बिना कैसे पी के निकल लेंगे कुछ तो भी बोलते हो की जरूरत नहीं है प्लीज इस बात को मानिए आप एक बात बताओ अपने ऐसा नहीं सिखा है क्या एक किसी एक्सटर्नल पॉइंट से किसी एक्सटर्नल पॉइंट से इस पॉइंट को मैं कुछ भी का देता हूं लेट्स से आर एक्सटर्नल पॉइंट आर के लेट से कोऑर्डिनेट्स सर्कल की कोई ना कोई इक्वेशन को मैं का देता हूं x² प्लस ए स्क्वायर नहीं कर रहे हो अगर आपको ये फॉर्मूला नहीं याद ए रहा है तो ये फॉर्मूला इतना आसान सा फॉर्मूला है भाई याद करो ये जो आर एन की लेंथ है ये जो तेनजेन की लेंथ है इसको पढ़ाया था आपको ऐसा कोई फॉर्मूला नहीं याद ए रहा है क्या ये क्या फॉर्मूला होता है सर अरे भाई यह फॉर्मूला तो यही फॉर्मूला हुआ ना की एक्स और ए की जगह आप क्या रख दीजिए आप X1 y1 रख दीजिए प्लस y1 स्क्वायर प्लस तू जी एक्स वैन प्लस कर रही है तो पीके की लेंथ को आप क्या बोलोगे भाई सर पीके की लेंथ यही होगी की इस इक्वेशन में जीरो पास कर दूंगा और उसका अंडर रूट ले लूंगा तो वो हो जाएगा अब हमें चीज क्लियर हैं और अब यहां से हम आगे बढ़ सकते हैं क्या यहां तक किसी भी स्टूडेंट को कोई परेशानी क्या यह बात आप सभी को समझ आई चलो अगर अब मैं ट्राई करता हूं नेक्स्ट क्वेश्चन हर कॉम अनादर क्वेश्चन फॉर डी दे क्या इस सर्कल के सेंटर के कोऑर्डिनेट्स आप निकल सकते हैं तो किसी पॉइंट से किसी स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस का फॉर्मूला आपको नहीं आता है क्या देखकर चलो भाई ट्राई कर लेते हैं देखो सबसे पहले तो आप क्या करोगे आपका ये एक सर्कल है और ये एक सर्कल है तो S1 - S2 आप कर लेंगे तो S1 - S2 मतलब देखो x²y² x²y² के 5x - 3X ये कितना हो जाएगा सर ये हो जाएगा 8x - 8y -7y तो -8 -7 कितना सही हो जाएगा -15 कोई दिक्कत तो नहीं है सर 1 - 25 हो जाएगा कितना हो जाएगा सर प्लस 20 क्रॉस चेक कर लेते हैं 1 - - 25 बिल्कुल सही ये आपकी क्या हो गई सर ये आपकी हो जाएगी कॉमन कोड या फिर आप इसे का सकते हो रेडिकल एक्सिस जो भी आप का रहे हो ये दोनों बातें याद है क्या सर अब आगे अगर हम पढ़े तो आगे क्या देख रहे हैं देखो भाई अब इस स्ट्रेट लाइन की इस सर्कल के सेंटर से तू एक्स इधर आएगा माइंस वैन माइंस वैन ए पॉइंट है नहीं तो 1 0 सो इक्वेशन इतना आसानी की वैन कमा जीरो की आपको इससे डिस्टेंस निकालनी है परपेंडिकुलर डिस्टेंस किसी पॉइंट की किस स्ट्रेट लाइन से कितनी आसानी है भाई 1 8 = 0 + 26 / √2 ओवर 8 का स्क्वायर 64 15 का स्क्वायर 225 तो ये कुछ एक्सप्रेशन मैंने और मैं मोड नहीं लगा रहा क्या करूंगा मोड ला के पॉजिटिव वैल्यू है 26 + 8 कितना होता है सर ये हो जाता है 34 4 + 596 + 289 289 का अंडर रूट 17 और 17 का डबल होता है 34 तो ये आपको देगा ऑप्शन ए एस दी करेक्ट आंसर किसी भी स्टूडेंट को यहां तक कोई भी आपत्ति हो तो मुझसे पूछे आई विल reexplang दिस होल क्वेश्चन तू यू ऑल सर आसान सा क्वेश्चन था कॉन्सेप्ट क्या था कॉमन कोड के बारे में सोच पाना जिसके बारे में हमने डिटेल डिस्कशन किया ऐसे ही क्वेश्चंस आएंगे पर आपको बातें पता हो तो और बातें पता हो तो क्या अगला क्वेश्चन ट्राई करें स्टूडेंट्स नेक्स्ट क्वेश्चन आपके सामने क्या लिखा है देखो भाई टांगें आर ड्रोन तू डी सर्कल तो एक ये सर्कल है सर आते डी पॉइंट वेयर इट इस मिटा सर्कल पर इस सर्कल पर कहां से कहां से सर उन पॉइंट्स पर जहां पर यह सर्कल इससे इंटरसेक्ट करता है ठीक है सर लामबीडीए कोई वेरिएबल है बात करेंगे वह पूछ रहा है वो बहुत जरूरी बात आपसे ये पूछ रहा है की वो जो पॉइंट ऑफ इंटरसेक्शन जो पॉइंट ऑफ इंटरसेक्शन क्या होगा मैं आपको कुछ याद दिलाता हूं अगर आपको याद नहीं ए रहा है तो एक क्वेश्चन कैसे सॉल्व करेंगे आप वो याद दिलाना चाह रहा हूं मैन लो आपका ये एक सर्कल है है ना मैन लो आपके पास एक और सर्कल है एक और सर्कल है ना अब अगर मैन लो आपने क्या किया आपने ये किया suniyega ध्यान से कम की बात पे आपने आपका ये सर्कल वो लामबीडीए वाला सर्कल नॉर्मल सर्कल है इस पर आपने ड्रा की पॉइंट ऑफ इंटरसेक्शन पे टांगें है ना यहां पे अपने ड्रॉ की टांगेंट्स यहां पर आपने लेट से ड्रॉ की हैंड टैसेंट में से उठा के यहां रखता हूं तो ये होगी पहली टैसेंट और सिमिलरली यहां पर आपने जो की ऑफ कोर्स आपकी कुछ ऐसी है तो इन दोनों ट्रांजिट का यह जो पॉइंट ऑफ इंटरसेक्शन आपको लॉकर्स निकलना है समझ में ए रहा है यहां से यह सर्कल है तो ये आपका सर्कल हो गया x² + ए स्क्वायर है अगर मैं सारी बातें खत्म करके ये का डन क्या आपको बस इतना सा पॉइंट याद है की सर इन दोनों का जो पॉइंट ऑफ इंटर सेक्शन होगा इन दोनों का जो पॉइंट ऑफ इंटरसेक्शन होगा उससे एक कॉमन कोड पास होगी उससे एक कॉमन पास होगी उसे कॉमन कॉर्ड की क्या इक्वेशन होगी बस मुझे बता दो उसे कॉमन कोड बस मुझे ये बता दो क्वेश्चन खत्म हो जाएगा एक तो मैं कॉन्टैक्ट ऑफ कोर्ट से सोच सकता हूं और एक रेडिकल एक्सेस के कॉन्सेप्ट से सोच सकता हूं कुछ याद ए रहा है जो मैं कहना चाह रहा हूं एक तो मैं रेडिकल एक्सेस के कॉन्सेप्ट से सोच सकता हूं और एक में कॉमन जो कॉन्टैक्ट वाला पॉइंट है उससे सोच सकता हूं बस मेरे यकीन करिए बस वो निकलती ही आपका आंसर ए जाएगा ऐसे कैसे ए जाएगा सर ऐसे कैसे सोच लेंगे जरा चीज है तो समझो आप 1 मिनट सारी बातें छोड़ो मुझे कुछ नहीं पता मुझे बताओ क्या यह जो कोड है क्या यह जो कोड आपको दिख रही है यह सर कॉमन कोड है हान सर ये कॉमन कोड है यह कॉमन कोड होने के साथ-साथ आपका रेडिकल एक्सेस है क्या यह कॉमन कोड होने के साथ-साथ मेडिकल एक्सिस है अगर यह रेडिकल एक्सेस है तो इसकी इक्वेशन माइंस लिखता हूं x² - x²y² - y² गया अब जैसे इसका नेगेटिव किया तो ये ए जाएगा लामबीडीए + 6 - है ना माइंस किया तो मैं इसको नेगेटिव कर दे रहा हूं तो ये हो जा रहा है तू लामबीडीए -8 टाइम्स ए अब सुनना यह है माइंस हुआ है -1 -1 -3 तो -1 + 3 तो 3 - 1 कितना 2 = 0 ये आपकी उसे कॉमन कोड की यानी रेडिकल एक्सेस की इक्वेशन ए रही है अब सारी बातें छोड़ो क्या आप कोड ऑफ कॉन्टैक्ट इस पर ह कॉम कैसे आपने दो टांगेंट्स ड्रॉ की है और उन पर जो पॉइंट ऑफ कॉन्टैक्ट है क्या ना बिल्कुल सही बात है तो उसे केस में क्या हो जाएगी सर उसे केस में हो जाएगी ये एचसी प्लस के ए - 1 = 0 एक तरीका कोड ऑफ कॉन्टैक्ट और एक तरीका कॉमन कोड एक तरीका रेडिकल एक्सिस का कॉमन कोड और एक तरीका कोड ऑफ कॉन्टैक्ट और अगर ये दोनों ही से स्ट्रेट लाइन की इक्वेशन को व्यक्त किया जाहिर करती हैं तो मैं कहूंगा सर इनके कॉएफिशिएंट्स का रेशों से होगा मतलब मैं क्या कहूंगा मैं कहूंगा सर लामबीडीए + 6 / ह या फिर ह / लामबीडीए + 6 विल बी इक्वल तू के अपॉन तू मेरे ख्याल से यहां पे क्वेश्चन खत्म हो गया अगर आप चीज सोच का रहे हो तो आपको सिर्फ और सिर्फ यहां से कुछ चीज सोचनी है और इसके बाद अब मैं आपको क्यों इस क्वेश्चन का सॉल्यूशन बताऊं क्योंकि यहां से तो आप सोच सकते हो स्टूडेंट्स यहां से आगे क्या सोचने की जरूरत है कैसे सोचेंगे सर कैसे निकलेंगे क्या करना होगा ऐसे सोच के देख लो बड़ी सिंपल सी बात बड़ी सिंपल सी बात स्टूडेंट्स देखो पहले तो ह को इसके इक्वल रख दो या दोनों में लामबीडीए की वैल्यू निकल लो यहां से भी लामबीडीए की वैल्यू निकालो यहां से की वैल्यू निकालो दोनों की इक्वल तू आपका ह और के में रिलेशन ए जाएगा अब मेरी बात समझ का रहे हो क्या जैसे पहले मैं लौंडा की वैल्यू यहां से निकलता हूं लेकिन इन दोनों को कंसीडर करता हूं इससे पहली और तीसरी और इसे तो ह / लामबीडीए + 6 को माइंस वैन बाय तू के इक्वल रखते हैं ह / लामबीडीए + 6 को अगर मैं माइंस वैन अपॉन तू के इक्वल रखता हूं तो ये हो जाता है 2h = - लामबीडीए -6 -6 को इधर या माइंस को इधर लाया तो लामबीडीए = 2s - 6 किसी की स्टूडेंट को कोई दिक्कत है तो पूछो अब इसी तरीके से इन दोनों को कंसीडर करते हैं के / 2 लामबीडीए -8 है ना ये कितना है सर ये है की अपॉन तू लामबीडीए -8 जो की किसके इक्वल है -1/2 के फिर वही बात ये कितना हो जाएगा 2K = 8 - 2 लामबीडीए = 8 - 2K और अगर तू से डिवाइड किया तो लामबीडीए विल बी 4 - के तू से डिवाइड किया तो लामबीडीए विल बी फोर माइंस के अब आप प्लीज इस बात को ध्यान से देखो और मुझे सोच के बताओ यहां से लामबीडीए था 2h - 6 और यहां से लामबीडीए था 4 - क्या आई हैव ट्रेड एलिमिनेटिंग लामबीडीए बेसिकली सो व्हेन यू डू सो व्हाट यू गेट इस यू गेट अन रिलेशन एंड नक विच बेसिकली बॉल्स डाउन एस देखो के कम हर सो दिस इस गोइंग तू बी 2 ह प्लस के सिक्स गोज देयर सो सिक्स वन गोज देयर 6 + 4 इस गोइंग तू बी 10 एंड अल्टीमेटली व्हेन देयर इसे अन क्वेश्चन ऑन लॉस व्हाट यू फाइनली दो यू विल प्ले ह एंड के बाय एक्स एंड ए रिस्पेक्टिवली है ना सो दिस ह एंड प्लेसिड की एक्स एंड ए एंड फाइनली और ऐवेंंचुअली व्हाट यू गेट इस दिस इस गोइंग तू बी 2X + व्हेन इफ तू ब्रिंग दिस टाइम ऑन डी लेफ्ट हैंड साइड दिस इस तू बी दिस एंड दिस क्लीयरली इसे नथिंग बट डेट लुकास विच लुकास डी इंटरसेक्शन ऑफ डी टेनिस विच टेंसेज डी टांगें विच वर्ड ड्रोन तू डी सर्कल विच सर्कल x² + y² 1 आते विच पॉइंट डी पॉइंट ऑफ इंटरसेक्शन ऑफ डीज तू सर्कल सो दिस इस हो यू विल राइट दिस इक्वेशन विच इस क्लीयरली विच ऑप्शन आय विल लीव विच इस क्लीयरली 2X + ए - 10 एंड 2X + ए - 10 इस क्लीयरली नथिंग बट ऑप्शन दी है अन्य वैन देस अन्योन हैव अन्य कन्फ्यूजन सो फार इन दिस क्वेश्चन आई थिंक दिस क्वेश्चन सिंपल इफ यू आर क्लियर विद योर एप्रोच एंड रिमेंबर दिस इस गोइंग तू बी अन कॉन्सेप्ट विच इस कांस्टेंटली गोइंग तू बी एस्क्ड वेयर यू विल बी यूजिंग डी वारी सिंपल कॉन्सेप्ट कॉमन कोड ऑफ कॉन्टैक्ट यह दोनों इक्वेशंस आपको से स्ट्रेट लाइन की तरफ लीड करेंगे इस तरीके से आपको सोचना हो वो का रहा है इक्वेशन ऑफ सर्कल विच कट दिस सर्कल है तो वो ये कहना चाह रहा है की एक ऐसा सर्कल एक ऐसा सर्कल जो इससे मतलब इस सर्कल को और इन दो लाइंस को ऑर्थोंगनली इंटरसेक्ट करता है सर सारी बातें ठीक है पर इन दोनों लाइंस को ऑर्थोगोनली इंटरसेक्ट करने वाली बात समझ नहीं ए रही है डाइजेस्ट नहीं हो रही है आप एक बात बताओ अगर कोई सर्कल किसी लाइन को ऑर्थोगोनली इंटरसेक्ट करें मतलब क्या मैं थोड़ा सा आपको चीजें प्लॉट करके दिखाने आज समझने की कोशिश करता हूं देखो भाई अगर एक सर्कल है और इस सर्कल पर अगर मैं कोई एक रैंडम सी लाइन ड्रॉ करो जैसे ये एक लाइन है तो अब बात समझ का रहे हो क्या ये लाइन है सर्कल का ऑर्थोगोनली टच या इंटरसेक्ट कर रही है बिल्कुल नहीं सर है ना लेकिन अगर मैं एक ऐसी लाइन ड्रॉ करूं जो इसके सेंटर से पास हो अगर मैं एक ऐसी लाइन ड्रॉ करूं जो इसके सेंटर से पास हो तो आप खुद इस बात को नोटिस कर पाएंगे आप खुद इस बात को जांच पाएंगे की हान सर देखो ये जो लाइन है ये इस सर्कल के साथ ऑर्जिनल है कोई लाइन किसी सर्कल पर ऑर्थोंगोनल है मतलब क्या मतलब वो लाइन उसका नॉर्मल है वो लाइन उसका क्या है भाई नॉर्मल बात समझ ए रही है क्या आई होप अगर यह लाइन आपको समझ ए गई तो आप यह क्वेश्चन ध्यान से सॉल्व कर सकते हो अब सर पहला कम तो ये है इस पार्ट को ब्रेक करना है सिंपलीफाई करना देखो भाई क्या लिखा हुआ है सर अगर इन दो स्ट्रेट लाइंस की इक्वेशंस की मैं बात करूं तो जरा सिंपलीफाई करते हैं देखो क्या लिखा हुआ है एक्स ए - 2X अगर मैं एक्स कॉमन ले लेता हूं तो क्या हो जाएगा ए - 2 कोई दिक्कत तो नहीं ये क्या हो जाएगा ए - 2X कॉमन लिया है यहां से भी अगर मैं ए - 2 बनाना चाहूं तो यहां से ए - 2 बनाने के लिए मुझे -1 कॉमन लेना पड़ेगा यहां से एक्स कॉमन लिया था तो ये हो जाएगा एक्स - 1 और इस तरीके से ये आपकी इक्वेशन फैक्ट्रीज हो जाएगी जो बहुत सुलझी हुई फॉर्मूले उसने खुद आपको दे राखी थी शैल आय से की हान सर ये दो स्ट्रेट लाइंस की इक्वेशन है कौन सी दो स्ट्रेट लाइंस अगर आप ध्यान से देखो तो एक स्ट्रेट लाइन है ए = 2 और एक स्ट्रेट लाइन एक्स = 1 और ये दोनों लाइंस इस सर्कल को जैसे हमें फिगर आउट करना है उसे सर्कल को ऑर्थोगोनली इंटरसेक्ट करते हैं सर्कल बिल्कुल सही बात है अब अगर मैं बात करूं आपसे सोच के देखो अगर इस सर्कल के दो नॉर्मल हैं अगर ये सर्कल के दो नॉर्मल हैं तो एक नॉर्मल तो कौन है सर ए = 2 ए = 2 का क्या मतलब होता है अगर मैं बात करूं ए = 2 मतलब सर एक ऐसी स्ट्रेट लाइन जो एक्स एक्सिस के पैरेलल है अब बात समझ का रहे हो और सर एक और इसका नॉर्मल है एक और इसका नॉर्मल है एक्स = 1 ये क्लीयरली एक्स एक्सिस के पैरेलल है ओह सॉरी कुछ गड़बड़ कर रहा हूं आई एम रियली सॉरी मैं उल्टा बोला आई एम रियली रियली सॉरी मैंने उल्टा बोला है ए = 2 यह नहीं होगा ए = 2 आपकी एक्स एक्सिस के पैरेलल लाइन होगी हम रिलीज सॉरी फॉर डेट तो अगर मैं एक एक्स एक्सिस के पैरेलल लाइन ड्रॉ करूं जो क्लीयरली इस सर्कल के सेंटर से पास होती है जो क्लीयरली सर्कल के सेंटर से पास होती है तो कुछ इस तरीके से होगी की आप मेरी बात समझ का रहे हो ये कौन सी स्ट्रेट लाइन होगी सर ये एक्स एक्सिस के पैरेलल एक स्ट्रेट लाइन होगी और चूंकि ये सर्कल पर नॉर्मल है तो उसके सेंटर से तो पास होनी तय है क्योंकि वो इसे ऑर्थोगोनली इंटरसेक्ट करती है दूसरी बात सर एक और स्ट्रेट लाइन है एक और स्ट्रेट लाइन है जो की इस सर्कल के सेंटर से पास होती है यानी इस पर ऑर्थोगोनली इसका नॉर्मल है एक्स = 1 तो क्लियर सी बात है सर वो लाइन भी सर्कल के सेंटर से पास होगी और अगर मैं आपसे बस इतना पूछूं की ये जो दोनों स्ट्रेट लाइंस हैं ये जो दोनों स्ट्रेट लाइंस है कहां इंटरसेक्ट करती हैं तो इन दोनों का जो पॉइंट ऑफ इंटरसेक्शन होगा सर वो होगा वैन कमा कोई डाउट कोई परेशानी आई होप आप देख का रहे हो इन दोनों लाइंस का जो पॉइंट ऑफ इंटरसेक्शन होगा वो होगा वैन कमा तू और अगर दो daigonals का पॉइंट ऑफ इंटरसेक्शन हम निकले तो डेट इस नथिंग बट डी सेंटर ऑफ डी सर्कल आई होप आप ये बात जानते हो तो वैन कमा तू क्या हुआ इस सर्कल का सेंटर अब मुझे क्या निकलना है सर मुझे सर्कल की रेडियस निकालनी है तो इस सर्कल की इक्वेशन अभी फिलहाल मैं लिखूंगा तो मैं क्या लिखूंगा एक्स - 1 का होल स्क्वायर प्लस ए माइंस तू का होल स्क्वायर इस इक्वल तू आर स्क्वायर क्या ये बात आप सभी समझ का रहे हो आई थिंक चीजें आसान सी है बड़ी बेसिक सी है अब आप एक बात पे गौर फरमाइए अगर मैसेज सिंपलीफाई थोड़ा करो तो कर लेता हूं देखो जरा ध्यान से ये क्या हो जाएगा सर ये हो जाएगा x² है ना ये वैन स्क्वायर होगा तो इस पर हम बात करेंगे यहां से मुझे मिलेगा -2x जिस पर हम बात करेंगे यहां से मिलेगा y² यहां से मिलेगा -2x और यहां से मिलेगा -4 + 1 कितना 5 तो यहां पर आपको मिलेगा 5 - r² = 0 ये हो जाती है आपके सर्कल की इक्वेशन आई होप ये बात आप समझ का रहे हो अब अगर मुझसे कभी भी लाइफ में कोई कहे की सर दो सर्कल्स ऑर्थोंगोनल है मतलब ये सर्कल हमारा जो सर्कल निकल कर ए रहा है उसके साथ औरतों के नल है तो मैं दो तो सर मैं दो सर्कल के orthognility की ये कंडीशन जानता हूं की सर तू टाइम्स जीवन जीतू प्लस F1 f2 यहां से जीवन कितना आएगा आई होप आप देख का रहे हो 2gx तो जीवन की वैल्यू कितनी है वैन अप अब देख का रहे हो जीवन की वैल्यू अगर यहां से निकलूं तो वैन है और अगर मैं F1 की वैल्यू यहां से निकल दो कितनी है जल्दी बताओ भाई F1 की जो वैल्यू है डेट इस क्लीयरली तू अगर मैं यहां से आपसे सी वैन की वैल्यू पूछूं तो C1 की वैल्यू कितनी है -4 कोई डाउट तो नहीं है भाई C1 की वैल्यू है -4 बड़ी सी बात है सिमिलरली अगर मैं यहां पर आपसे पूछता बताओ भाई जी तू की वैल्यू कितनी है सर जी तू की जो आप वैल्यू निकलोगे इसका हाफ वो होगा -1 और यहां से अगर आप f2 की वैल्यू निकले यहां से आप f2 की वैल्यू निकले तो हो जाएगा इसका हाफ यानी की माइंस तू और सी तू की वैल्यू को लेकर मेरे ख्याल से कोई टिप्पणी करने की जरूरत नहीं है वो है 5 - r² क्या कंडीशन अप्लाई करने बिल्कुल कर दीजिए -1 तो 1 - अंदर इसका डबल करना है वो मैं कर दूंगा ये हो जाएगा F1 f2 f2 है -2 तो 2 -2 तो यहां पे कितना ए जाएगा सर ये होगा -4 तो ये कितना हो जाएगा भाई -4 -4 -1 कितना -5 -5 2 कितना सारी हो जाएगा -10 तो एलएस फिर तो होगा -10 = C1 + C2 C2 कितना है सर C2 है 5 - r² 5 - r² में -4 किया तो 5 - 4 कितना हुआ तो ये हो जाएगा 1 - r² ये कितना हो जाएगा ये हो जाएगा वैन माइंस आर स्क्वायर कोई डाउट तो नहीं है कोई परेशानी तो नहीं सर अगर आप इसे सिंपलीफाई करें तो आई होप आप देख का रहे हो क्या सर 1 को यहां ले तो -11 और ये माइंस आर स्क्वायर तो r² कितना आता है सर आर स्क्वायर ए जाता है आपका 11 इस सर्कल की रेडियस हो जाती है अंडर रूट 11 और इस तरीके से आप इसकी इक्वेशन लिख सकते हो मेरे ख्याल से बहुत आसानी से बिना ज्यादा हेसीटते किए आपको कुछ नहीं करना है दोनों बातें या तो आप चाहो या तो आप चाहो तो क्या करो ध्यान से देखो भाई आपने निकल ली है r² की वैल्यू तो आप चाहो तो यहां पर r² की वैल्यू रख दो r² की वैल्यू कितनी ए रही है सर r² की वैल्यू ए रही है 11 तो 5 में से अगर 11 सूत्र किया तो कितना भाई -6 क्योंकि 5 - 11 सो थॉट्स - 6 तो x² + y² - 2X - 4y - एक्स है ना सो - 2X - 4y - एक्स विच इस गोइंग तू बी और ऑप्शन ए आई थिंक की एक बड़ा ही बेसिक सा सिंपल सा आसान सा और डायरेक्ट सा क्वेश्चन जो आप सभी स्टूडेंट्स को बहुत इजीली और डायरेक्टली और क्लीयरली समझ आया आई थिंक यहां तक तो कोई भी परेशानी या डाउट या दिक्कत नहीं है बड़ा बेसिक सा क्वेश्चन था कंडीशन शायद ट्रिक यहां से हो सकती थी की आप इस बात को समझ पाएं की दो पैर ऑफ स्ट्रेट लाइंस बेसिकली जो है वो आपकी सर्कल पर ऑर्थोंगोनल कैसे है tokanal होने का मतलब वो लाइंस उसे पर नॉर्मल है और अगर वो दोनों इसके नॉर्मल हैं मतलब वो डायगोनल है डायमीटर है और अगर वो डायमीटर है तो उनका पॉइंट ऑफ इंटरसेक्शन इस सर्कल का सेंटर है रेडियस को लेकर हम इससे बात कर सकते द इस पार्टिकुलर कंडीशन को अप्लाई करके बहुत डिफिकल्ट बहुत टू क्वेश्चन तो नहीं था अभी आई थिंक आप सभी ने इस क्वेश्चन को आंसर किया होगा अगर मैं अगले क्वेश्चन की तरफ बढूं तो अब हम ए चुके हैं केस कैटिगरी के क्वेश्चंस पर अब हम बात करेंगे मल्टीपल करेक्ट आंसर टाइप क्वेश्चंस यहां पर अब आपके एक से ज्यादा क्वेश्चंस करेक्ट होंगे जरा इस क्वेश्चन को ध्यान से देखिए और अच्छे से समझिए और अच्छे से देख कर मुझे बताओ इस क्वेश्चन का आंसर आप क्या मार्क करोगे कैसे सोचोगे क्या थॉट होगा क्या थॉट प्रक्रिया होगी क्या आप रोज होगी ज़रा सोच के बताओ फिर सारी बातें छोड़ो आप तो क्वेश्चन पर आओ क्वेश्चन क्या लिखा है वो पढ़ते हैं डी इक्वेशंस ऑफ डी टांगें तो मुझे टांगें की इक्वेशन निकालनी है मतलब एक से ज्यादा इक्वेशंस की वो बात कर रहा है शायद टॉप कोर्स शायद यहां पर मल्टीपल ऑप्शंस करेक्ट तो बिल्कुल उसे बारे में बात करेंगे ड्रोन फ्रॉम डी ओरिजिन तो सर ओरिजिन से ओरिजिन क्या होता है भाई हम जानते हैं क्या जीरो कमा जीरो से इस सर्कल पर टांगें ड्रॉ करनी है सर सारी बातें छोड़ो मैंने जीवन में इतना जरूर सिखाया की अगर कोई लाइन ओरिजिन से पास हो रही है तो ए = एमएक्स फॉर्म में होती है जिसे मैं और अच्छे तरीके से एमएक्स-y=0 फॉर्म में लिख लेता हूं इस बात से किसी को कोई आपत्ति कोई दिक्कत तो नहीं है ठीक है सर मैन लिया आपके पास अब अगर मैं इस सर्कल को देखूं तो इस सर्कल को देख के आपको क्या आइडिया लग रहा है आपको क्या आइडिया लग रहा है सर इसका ये जो जी है उसकी वैल्यू कितनी है सर इसके जी की वैल्यू है आर तू जीएक्स आई होप आप समझ का रहे हो मतलब मैं सॉरी जी की वैल्यू मतलब - जी की वैल्यू मतलब मैं क्या निकलना चाह रहा हूं उसके सेंटर के अकॉर्डिंग तो मैं डायरेक्ट सेंटर के क्वाड्रेट्स निकल देता हूं जैसे ये हमारी हैबिट या प्रोटोकॉल है उसे तरीके से इसका नेगेटिव हाफ होगा आर इसका नेगेटिव होगा हाफ होगा ह तो ये हो जाएगा आर कमा ह जो की क्लीयरली के होंगे इसके सेंटर के cordule है इनके अगर आपसे पूछा जाए इसकी रेडियस क्या होगी तो r² + x² - x² गया तो जीरो और बचा क्या r² और r² का अंडर रूट लिया तो आर तो ये हुए इसके सेंटर के क्वाड्रांट्स और ये हुई इसकी रेडियस अब एक छोटी सी बात का जवाब दो सर अगर कोई लाइन कोई लाइन किसी सर्कल की टांगें है तो क्या अगर कोई लाइन किसी सर्कल की tenjent है तो मैं बहुत जरूरी और कृष्ण बात दावे के साथ का सकता हूं की सर इस सर्कल के सेंटर से ड्रॉप किए गए परपेंडिकुलर की लेंथ इस लाइन पर इसकी रेडियस के इक्वल होगी यह बात भूले तो नहीं हो तो सर बस इसी बात पर बात करते हैं सुनेगा ध्यान से कैसे सोचेंगे देखो भाई आर एन तो हो जाएगा श्री -12 हो जाएगा m² + 1 और ये किसके इक्वल होगा वो इक्वल होगा इसकी रेडियस आर के क्या यहां तक किसी भी स्टूडेंट को कोई आपत्ति अब हमारा कंसेंट क्या है हमारा कंसर्न है हर तो हमें पता है हमें कहां तक पहुंचना है एम तक तो इससे सॉल्व करके मैं एम की वैल्यू निकालो क्वेश्चंस समझ ए रहा है आई थिंक एप्रोच आसानी से आप सभी को क्लियर हो रही है ठीक है सर इस बारे में बात कर लेते हैं कैसे सोचेंगे जरा देखो भाई अगर मुझसे पूछा जा रहा है तो मैं इसको उधर शिफ्ट कर देता हूं तो ये हो जाएगा क्या श्री-ह और उसे तरफ क्या हो जाएगा सर वो हो जाएगा आर टाइम्स अंडर रूट ओवर स्क्वायर प्लस वैन अब बड़ा सिंपल सा थॉट ऑफ दिस अंडर रूट एंड बेसिकली स्क्वायर बोथ डी साइड्स सो व्हेन यू हैपन तू स्क्वायर बोथ डी साइड्स व्हाट डू यू फाइनली गेट प्लीज हैव लुक आते दिस एवरी वैन ये हो जाएगा श्री-ह का होल स्क्वायर तो ये क्या हो जाएगा सर ये हो जाएगा m² r² + h2 -2m र इसे इक्वल तू व्हाट डू सी हैव ऑन दी राइट हैंड साइड डेट इस गोइंग तू बी आर स्क्वायर m² + r² करने से ये बन जाएगा r² तो r² m² + r² तो ये क्या हो जाएगा सर ये हो जाएगा r² एंड स्क्वायर प्लस आर स्क्वायर क्या इतनी बातें सभी स्टूडेंट्स को आसानी से क्लीयरली डायरेक्टली समझ आई कोई परेशानी कोई दिक्कत अब क्या करना है सर अब एक बड़ा सिंपल सा ऑब्जर्वेशन देखो भैया आप सब देखो r² m² m² से कैंसिल कोई दिक्कत तो नहीं है अब अगर मैं चीज सॉल्व करना चाहूं तो सर एक जरूरी ऑब्जर्वेशन देखो आप आर स्क्वायर को इधर ले आओ ले ए सकते हैं क्या तो ये हो जाएगा h² - r² ये उधर चला जाएगा 2mrh यह 2m चला जाएगा तो मैं एम को उधर ही रखता हूं मैं सिर्फ एन को उधर ही रखता हूं बाकी चीज एम के साथ और क्या है सर तू र तो मैं 2rh को यहां ले आता हूं क्या ये बात आप सभी को समझ आई अब सर इसे देखकर क्या बेसिकली मैं कहना चाह रहा हूं जब आप इसे देखोगे तो ये आती है आपकी उसे लाइन की स्लो कोई दिक्कत कोई डाउट कोई परेशानी आई थिंक चीज है सर आसान सी है चीज सली हुई थी ये आपका फाइनल डिस्कशन होकर यह रिजल्ट ए रहा है लेकिन मेरा आपसे पूछना है की सर अगर सर अगर आपकी दो टांगेंट्स आपने ड्रा किया आप ये कनक्लूड कर रहे हो देखो मेरी बात को बहुत ध्यान से सुनो यहां पे जीरो कमा जीरो पास करो जीरो जीरो तो बच के ए जा रहा है x² और क्लीयरली ये जीरो रेडियस का तो सर्कल है नहीं मतलब कुछ ना कुछ उसकी रेडियस है जो की है आर तो क्लीयरली अगर इसकी कोई नौकरी सिग्निफिकेंट लेंथ है और जीरो कमा जीरो इससे बाहर है तो जीरो कमा जीरो अगर एक एक्सटर्नल पॉइंट है जिससे मैंने इस सर्कल पर tangenced ड्रेन हैं तो कितनी drawngi सर क्लीयरली दो टेंशन ड्रॉ होंगे लेकिन जब आपने वैल्यू निकालनी चाहिए जब आपने वैल्यू निकालनी चाहिए तो आपके साथ ये गड़बड़ हो रही है की आप के अकॉर्डिंग तो स्लोप की एक ही वैल्यू ए रही है सर आपके अकॉर्डिंग तो स्लोप की एक ही वैल्यू ए रही है तो दूसरी वैल्यू कहां गई अब हम दूसरी वैल्यू की बात करने जा रहे हैं suniyega ध्यान से बस गड़बड़ ये है की जब आप स्क्वायर रूट को हटाते हैं तो चीज थोड़ी सी गड़बड़ हो जाती हैं उन्हें चीजों को रिजॉल्व करने के लिए मैं इस वाली इक्वेशन को फिर से ध्यान से देखता हूं कौन सी वाली इक्वेशन स्टूडेंट्स इस इक्वेशन को देखो आय ऍम रराइटिंग दिस होल इक्वेशन अगेन बट इन ए लिटिल डिफरेंट मैनर कैसे ध्यान से देखना एक बात सोच के बताओ स्टूडेंट्स बड़ी सिंपल सी बात अगर मैं यहां पर आपको चीज लिख कर डन अगर मैं यहां पर आपको चीज लिख डन तो ध्यान से देखो क्या निकल कर ए रहा है ये एक इक्वेशन बन रही थी जो की क्या बन रही थी से देखना यहां पर मुझे दिख रहा है 2mrh तो पहले से मैं लिख लेता हूं -2mrh प्लस क्या दिख रहा है सर प्लस दिख रहा है h2 और - r² = 0 क्या मैं इसे एम में क्वाड्रेटिक की तरह ट्रीट कर सकता हूं सर एम में अगर क्वाड्रेटिक की तरह ट्रीट करना है तो 0 m² जैसे भी तो एक टर्म चाहिए मतलब बेसिकली अगर एम में मुझे इसे क्वाड्रेटिक बनाने हैं तो एम तो यहां पर है बट दूसरी टर्म नहीं है ना आप रिलाइज करोगे की ऐसी गड़बड़ की हुई है आई टॉक अबाउट डेट टॉक अबाउट इट यू नो व्हाइट बट पहले इस बात को समझना स्टूडेंट्स बहुत ध्यान से की यहां पर प्रॉब्लम ये ए रही है प्रॉब्लम ये ए रही है की सर ये एम में क्लीन nequation बन गई और उससे मुझे एम की एक ही वैल्यू मिली बट विले डूइंग डेट हमने एम की एक और वैल्यू निकलना भूल गए उसे पॉसिबिलिटी को डिस्कस करना बोल अगर हमने शुरुआत से डिस्कस किया है की सर जनरली वो एम में क्वाड्रेटिक रहती है और जब मैं क्वाड्रेटिक रहती है यानी की दो वैल्यूज आती है वो दो टांगेंट्स आप ड्रॉ कर रहे होते हो क्या आपको ये बात याद है ना स्टूडेंट्स अगर मैं किसी एक्सटर्नल पॉइंट से तेनजेन ड्रॉ करो तो हम हमेशा से जानते आए हैं की आपकी दो टांगेंट्स ड्रॉ होंगे पर वो दूसरी टेंशन की स्लोप ढूंढने में इतनी मुश्किल है की वो ए रही है उसे बारे में अब हम बात करने वाले हैं चलो सारी बातें छोड़ अगर इसके दो रूट से अल्फा बेटा अगर इसके दो रूट से अल्फा बिता तो क्या हम जानते हैं प्रोडक्ट ऑफ डी रूट्स क्या होता है सर प्रोडक्ट ऑफ डी रूट्स होता है सी अपॉन ए यानी वो क्या हो जाएगा सी आपकी कांस्टेंट हम जो की हो जाएगी h² - r² और अपॉन ए अपॉन ए यानी कितना जीरो तो प्रोडक्ट ऑफ डी रूट्स कितना ए रहा है सर प्रोडक्ट ऑफ डी रूट्स ए रहा है इंफिनिटी प्रोडक्ट ऑफ डी रूट्स ए रहा है इंफिनिटी एंड नॉट डिफाइन अब इनमें से एक प्रोडक्ट तो एक रूट तो मैं जानता हूं ना एक रूट तो मैं जानता हूं ये है मैन लो वो अल्फा में जानता हूं तो अब अल्फा को उधर भेजा तो एनीथिंग इन इंफिनिटी विल बी इंफिनिटी तो टेक्निकल shailise से एक ही सर जो आपका दूसरा रूट है जो आपका दूसरा रूट है जो की टेक्निकल एम की आपकी एक और वैल्यू है शैल आय से की सर वो हो जाने वाली है इंफिनिटी आप मेरी बात समझ का रहे हो इंफिनिटी का मतलब क्या है होप आप सोच का रहे हो एम यानी स्लोप किसी लाइन की इंफिनिटी मतलब वो लाइन एक्स एक्सिस पर परपेंडिकुलर है यानी वो ए एक्सिस के पैरेलल है बस उसकी स्लोप कुछ ऐसी थी जो की एक साधारण ऑर्डिनरी क्वेश्चन से नहीं निकल जा सकती इसीलिए उसने उसे पॉसिबिलिटी को ही ले मिनट कर दिया और आपको एक ही स्लोप निकल कर दे पर दूसरी स्लोप के नॉट दे फिगर आउट या वो इंफिनिटी है इसलिए आपको थोड़ा सा आउट ऑफ डी बॉक्स सोचना पड़ा क्या यह बात आप सभी को डाइजेस्ट हो रही है तो आप क्या कहोगे भाई आप कहोगे सर इसकी दो इक्वेशंस ऑफ कोर्स होनी चाहिए दो स्लोप के साथ एक बार एम इंफिनिटी होगा और एक बार एंड ये होगा जब एम ये होगा तो क्या मैं स्ट्रेट लाइन की इक्वेशन लिख सकता हूं ए = एमएक्स तो स्ट्रेट लाइन के क्वेश्चन में क्या लिखूंगा मैं लिखूंगा ए = एमएक्स है ना एम मतलब क्या तू र एक्स इसे हम सिंप्लीफायर जरूर करें ए = एन इंफिनिटी को 1 / 0 एक्स लिख सकते हैं जीरो यहां गया तो ये क्या हो जाएगा सर ये ए जाएगा एक्स = 0 तो एक आपकी टैसेंट की इक्वेशन तो ये आणि है दूसरी इक्वेशन अगर मैं इसे थोड़ा सिंपलीफाई करूं तो यहां से देख कर मैं ऑप्शंस के थ्रू आइडिया लेना चाह रहा हूं की ऑप्शंस में क्या दिख रहा है की आपको एक बात तो नजर ए रही है की एक्स = 0 क्यों मिल रहा है कोई डाउट तो नहीं है तो पहला तो ये ऑप्शन करेक्ट है दूसरा इन दोनों में से अगर आप ध्यान से देखोगे तो आपको नजर आएगा सर की आप तू र यहां ले आओ तो आपको क्या दिखेगा आपको ये दिखेगा सर की एक्स में तो मल्टीप्लाई हो जा रहा है कौन h² - r² दिख रहा है क्या सभी को ये तू र ए इधर ए जाएगा जैसे मोठा के यहां ले आता हूं तो 2rh टाइम्स क्या ए = 0 और ये हो जाएगी आपकी दूसरी टेंशन की इक्वेशन जैसे हम लिख रहे हैं स्क्वायर माइंस आर स्क्वायर ये ब्रैकेट है राइट दिस इस नथिंग बट व्हाट अन ब्रैकेट तो h² - r² एक्स - 2r ही जो की क्लीयरली है आपका कौन सा ऑप्शन ऑप्शन सी क्या आप सभी स्टूडेंट्स को ये क्वेश्चन और इसका सॉल्यूशन समझ आया आई थिंक यकीनन तौर पर यह बहुत डिफिकल्ट बहुत टू बहुत मुश्किल क्वेश्चन नहीं था बस ट्रिकी पार्ट ये था की यहां पर ये बात याद रखना की सर कभी भी कोई सर्कल हमारे पास होता है तो किसी एक्सटर्नल पॉइंट से हमेशा उसे पर दो टांगेंट्स बनती है और अगर दो टांगेंट्स बनती हैं तो सर ये बात हमेशा याद रहे की अगर आपकी एक ही स्लोप ए रही है तो दूसरी स्लोप भी एक्जिस्ट करती होगी जिससे शायद थोड़ा कोई आउट ऑफ डी बॉक्स एप्रोच से सोचा जा सकता है क्या ये पूरा क्वेश्चन क्लियर है आई थिंक बहुत डिफिकल्ट नहीं था बस यह बात याद रखना है स्टूडेंट्स है ना इसी के साथ मूव करते हैं नेक्स्ट क्वेश्चन पर नेक्स्ट क्वेश्चन में क्या लिखा है देखो वो का रहा है डी कोऑर्डिनेट्स ऑफ डी सेंटर ऑफ अन सर्कल तो एक सर्कल के सेंटर के कोऑर्डिनेट्स निकले हैं जिसकी रेडियस है तू ठीक है सर विच टचेस डी पैर ऑफ दिस स्ट्रेट लाइंस है पैर ऑफ स्ट्रेट लाइन मतलब कुछ स्ट्रेट लाइन रही होंगी और एक सर्कल है जो इन दोनों को टच करता है जो इन दोनों को टच करता है और उसकी रेडियस कितनी है आर कोई बात दिक्कत तो नहीं पर सर जरूरी है क्या आर मतलब उसकी रेडियस कितनी है उसकी रेडियस है तू पर सिर्फ जरूरी है क्या की वो दोनों पैर ऑफ लाइंस को ऐसे इस वाले बाईसेक्टर में ही टच करें ऐसा नहीं हो सकता क्या की उसे सर्कल ने इन दोनों पैर ऑफ स्ट्रेट लाइंस को यहां टच किया था उसे केस में भी तो आप शायद एक सर्कल बना सकते द जिसकी रेडियस तू होती शायद मुझे नहीं पता और बना सकते द क्या तो हम दोनों पॉसिबिलिटी दोनों संभावना दोनों पर विचार करेंगे और इसीलिए क्वेश्चंस कैटिगरी में रखेंगे स्टूडेंट्स आए हो आप देख का रहे हो की मल्टीपल करेक्ट आंसर टाइप क्वेश्चंस है यहां पर आपको एक से ज्यादा ऑप्शंस मिलना तय है अब अगर सर मैं बात करूं यहां पर इस वाले पार्ट की इससे हम कैसे सॉल्व करेंगे और सारी बातें छोड़ो आप ये बताओ क्या मैं इसका कुछ कर सकता हूं x² + y² -2x+1 इस इस वाले पार्ट का जरा इसे ध्यान से देखो बस इसे अगर आप स्प्लिट कर का रहे हो इसे अगर आप दो पार्ट्स में ब्रेक कर का रहे हो तो आई थिंक आप आसानी से सोच का रहे हो की इस क्वेश्चन का क्या करना है बस इसमें छोटा सा करेक्शन कर लो स्टूडेंट्स यहां पर एक छोटी सी चीज देख लो की यहां पर आप लगा लो एक नेगेटिव साइन नेगेटिव साइन इसलिए क्योंकि इससे ये क्वेश्चंस सही हो जाएगा यहां पर ये टाइप हो है स्टूडेंट्स देखना ध्यान से वो क्या का रहा है dorites ऑफ डी सेंटर ऑफ अन सर्कल पहले तो इसको क्रैक करते हैं देखो इसको स्प्लिट करना है ना तो मैं आपसे का रहा हूं की आप इससे इससे और इससे देखो दिख रहा है a² + b² - 2ab तो क्या मैं इसे ऐसा लिख सकता हूं एक्स - 1 का होल स्क्वायर और ये क्या है आपका - y² अब आप देख का रहे हो ये a² - b² दिख रहा है क्या बिल्कुल दिख रहा है सर तो a² - b² को हम क्या लिखते हैं ए + बी यानी से हमें क्या लिखूंगा एक्स - 1 मैं लिखूंगा एक्स - 1 + ए और ए क्या हो जाएगा एक्स ये क्या हो जाएगा भाई एक्स - 1 - ए = 0 तो ये बन गई आपकी दूसरी लाइन अब इन्हें थोड़ा ध्यान से देखो एक स्ट्रेट लाइन है एक्स + ए - 1 suniyega ध्यान से एक स्ट्रेट लाइन है आपकी क्या एक स्ट्रेट लाइन है आपकी अन बना लेता हूं बस कहीं भी लिख लेता हूं एक्स + ए - 1 = 0 अब सन रहे हो क्या एक स्ट्रेट लाइन है एक्स + ए - 1 और एक क्या है सर एक स्ट्रेट लाइन है एक्स - ए - 1 = 0 एक स्ट्रेट लाइन क्या है सर एक्स - ए - 1 = 0 ऐसा क्यों किया सर आपने के ड्रॉ किया बात समझना एक छोटी सी बात देखना स्टूडेंट्स आप एक बात बताओ बड़ी सिंपल सी बात मैं आपसे पूछ रहा हूं देख के बताओ इस लाइन की स्लोप क्या है सर वही उधर गया तो एमएक्स + सी से मुझे क्लीयरली ये दिख रहा है ए = एक्स तो इसकी 45° स्लोप है और इस लाइन के स्लोप के सर ये एक्स उधर गया है तो माइंस एक्स ये 135 डिग्री नेशन है स्लोप मतलब इंक्लिनेशन की मैं बात कर रहा हूं ये 45 टॉप आपको समझ ए रहा है ये प्लस वैन है यहां पर तन थीटा माइंस वैन है एक और बात बता दो सर के इन दोनों लाइंस का पॉइंट ऑफ इंटरसेक्शन में निकल सकता हूं देखो दोनों को साथ में सॉल्व किया अगर मैंने दोनों को ऐड किया तो ए से ए गया x+x2x और ये माइंस तू माइंस तू उधर गया तो तू एक्स तू के इक्वल तो एक्स की वैल्यू वैन एक्स की वैल्यू वैन यहां है की तो 1 - 1 0 तो ए की वैल्यू जीरो तो वैन कमा जीरो यानी एक्स एक्सिस पर एक पॉइंट है एक्स एक्सिस पर एक पॉइंट है कौन वैन कमा जीरो जहां पर ये दोनों क्या कर रही होंगी इंटरसेक्ट तो ये वाली स्ट्रेट लाइन तो कुछ ऐसे ही जा रही होगी जो की क्लीयरली कितना डिग्री एंगल बना रही होगी 45 डिग्री क्या आप मेरी बात समझ का रहे हो जो की क्लीयरली कितना डिग्री एंगल बना रही होगी भाई 45 डिग्री सिमिलरली आपकी एक और स्ट्रेट लाइन जा रही होगी सर इसी से वैन कमा जीरो से जो की होगी आपकी यह वाली लाइन और यह क्लीयरली कितना डिग्री बना रही होगी ये बना रही होगी 1:35 डिग्री और मैं बहुत अच्छे से बहुत आराम से देख का रहा हूं सर की ये है 45 डिग्री एंगल और ये है 45 डिग्री एंगल ये दोनों एक दूसरे पे परपेंडिकुलर है और आपको ऐसे नहीं दिख रहा है तो इसकी स्लोप है माइंस वैन इसकी स्लोप है वैन तो 1 -1 -1 तो ये दोनों लाइंस एक दूसरे पर परपेंडिकुलर हैं और अगर यह दोनों लाइंस एक दूसरे पर परपेंडिकुलर हैं तो क्या मैं बड़ी आसान सी बात ये का सकता हूं की ये जो होगा ये इन दोनों का क्या होगा आई होप आप देख का रहे हो ये इन दोनों का एक एंगल बाईसेक्टर होगा और दूसरी बात बस ऐसे ही मेरा मैन है आपको कुछ समझने कहने या बताने का तो क्या मैं ऐसा नहीं का सकता की सर देखो आप ये जो लाइन होगी ये भी तो यहां पर एक एंगल बाईसेक्टर ही बनेगी ये जो लाइन होगी आपकी यहां पर इससे भी तो आप एक एंगल बाईसेक्टर ही कहोगे क्या आपको यहां तक किसी भी बात से कोई बात है एंगल बाईसेक्टर्स एक दूसरे पर होते हैं परपेंडिकुलर तो अगर यह एक्स-एक्स आपका एक एंगल बाईसेक्टर है तो दूसरा एंगल बाईसेक्टर आपका ए एक्सिस के पैरेलल ही होगा एक्स एक्सिस परपेंडिकुलर ही होगा और इस पॉइंट के क्वाड्रेंट हम सब जानते हैं अभी हमने निकाला पॉइंट ऑफ इंटरसेक्शन ये क्या था सर ये पॉइंट है आपका वैन कमा जीरो ये जो पॉइंट है ये है आपका वैन कमा जीरो तो क्लीयरली इस एंगल बाईसेक्टर यानी इस स्ट्रेट लाइन के इक्वेशन क्या हो जाएगी इसकी इक्वेशन हो जाएगी एक्स = 1 और इस एंगल बाईसेक्टर यानी इस स्ट्रेट लाइन की इक्वेशन हो जाएगी ए = 0 जो की क्लीयरली आपका एक्स-एस यहां तक कोई परेशानी है तो पूछो इतनी सारी बातें आप क्यों कर रहे हो सर मैं इतनी सारी बातें इसलिए कर रहा हूं क्योंकि मुझे इमेजिन करने हैं ऐसे सर्कल्स मुझे इमेजिन करने हैं ऐसे सर्कस जिनकी रेडियस हो तू जिनकी रेडियस हो तू और ऑफ कोर्स उनके जो सेंटर हो मतलब उनकी जो पॉइंट ऑफ कॉन्टैक्ट हो जो लाइन हो जो टच करें जो टैसेंट हो वो ये लाइन या ये लाइन क्या है बात आपको समझ आई क्या यह बात आप सभी को समझ आई फिर से पढ़ देता हूं डी cardinate ऑफ डी सेंटर ऑफ अन सर्कल जिसकी रेडियस है तू जो की इन दोनों लाइंस को टच करता है क्या यहां तक किसी भी स्टूडेंट को कोई आपत्ति या परेशानी किसी को भी कोई दिक्कत है तो पूछो मैं आपको एक एग्जांपल देके समझने की कोशिश करता हूं इस बात को प्लीज ध्यान से समझो आप खुद कहोगे की सर एक सर्कल तो कैसा बन सकता है तू रेडियस वाले एक सर्कल के हम बात करें तो तू रेडियस वाले अगर मैं एक सर्कल की बात करूं तो ध्यान से देखना आप प्लीज इस बात को ध्यान से देखो भाई अगर सर्कल की मैं बात करूं तो थोड़ा सबसे बड़ा कर लेते हैं आई थिंक आप बात समझ का रहे हो की यह सर्कल कैसा होगा जो की ऑफ कोर्स मैंने थोड़ा प्रॉपर्ली नहीं बनाया बट इसके सर्कल का जो सेंटर होगा वो यहां के लाइक करेगा अब बात समझ का रहे हो क्योंकि ऑफ कोर्स की एंगल बाईसेक्टर है एंगल बाईसेक्टर है और ये दोनों आपकी टांगेंट्स है तो क्लियर सी बात है इन दोनों टांगें के एंगल बाईसेक्टर पर ही सर्कल का सेंटर लाई करेगा ये बात आपको डाइजेस्ट हो रही है और सर एक बहुत जरूरी बात मुझे पता चल रही है क्या की सर ये अगर आपकी टांगें है तो उसे पर अगर आपने परपेंडिकुलर ड्रॉप किया तो इसकी रेडियस होगी जो की क्या है तू कैन आई से एक बहुत जरूरी बहुत कृष्ण बात की ना सिर्फ ये ना सिर्फ बल्कि ऐसा कैसे सर्कल यहां बन सकता था बात करेंगे इस बारे में ना सिर्फ ये बल्कि सर आप थोड़ी गड़बड़ कर रहे हो आप ऐसा क्यों नहीं सोच रहे की यही से सर्कल यहां भी तो बन सकता था ऐसा का ऐसा यही से सर्कल यहां भी तो बन सकता था यहां भी तो एक सेंटर हो सकता था जिससे अगर आपने परपेंडिकुलर ड्रॉप किया होता तो वो भी रेडियस कितनी होती है तू और क्यों आप बोल रहे हो यही का यही सर्कल यहां भी मशक्कत हो तो एक्चुअली आपके सेंटर यहां यहां कहीं और यहां कहीं हो सकते हैं अब उसे सेंटर के को-ऑर्डिनेट निकलने की जरूरत है वो ये की पहले तो है आपके पास वैन कमा जीरो आपको दिख रहा है क्या अब वैन कमा जीरो से आपकी क्या हेल्प मिलेगी इस बारे में बात करते हैं अब सारी बातें छोड़ो आप मुझे बस ये बताओ बड़ी सिंपल सी बात है सोच के बोलना स्टूडेंट्स ऑफ़ कोर्स यह जो सेंटर होगा जो यहां या यहां होगा वो एक्स एक्सिस पर होगा और अगर कोई सेंटर एक्स एक्सिस पर है तो उसके कोऑर्डिनेट्स क्या होते हैं सर उसके cardinate क्या होंगे उसका वे कॉर्डिनेट जीरो होगा तो इसका सेंटर क्या मैन सकता हूं अल्फा कमा जीरो दूसरी बात ये जो सेंटर है ये जो सेंटर है ये किस पर लाइक करते हैं एक्स = 1 पर लाइक करते हैं तो इसके जो cardinate होंगे उसका yordinate मुझे नहीं पता मैं कुछ गलत लिख रहा हूं बिल्कुल उसका yordinate मुझे नहीं पता लेकिन उसका excordinate तो मुझे पता है ना आई होप आप मेरी बात समझ का ऐसे ही यहां पर भी कुछ ए जाएगा ऐसे ही यहां पर भी कुछ ए जाएगा जिनकी मैं बात नहीं कर रहा हूं जो की हम इसी तरीके से सोच लेंगे बार-बार बात वही निकल कर आती है वही बार-बार निकल के कनक्लूड होती है की सर सर्कल के सेंटर्स हैं सर्कल की टैसेंट की परपेंडिकुलर डिस्टेंस सर्कल की रेडियस के इक्वल होती है कोई दिक्कत तो नहीं है तो अल्फा कॉमर्स जीरो से इसे टैसेंट की परपेंडिकुलर डिस्टेंस तू के इक्वल रख डन क्या रख दो सर अल्फा जीरो से एक्स + ए - 1 की डिस्टेंस है ना मैं क्या कर रहा हूं अल्फा कमा जीरो से एक्स + ए - 1 = 0 की जो डिस्टेंस है वो कितनी होगी भाई वो निकलेगी तू यूनिट्स तो निकल लेते हैं अल्फा 1 तो ये हो जाएगा अल्फा प्लस जीरो इन वैन जीरो इन वैन जीरो और फिर क्या बच रहा है सर फिर बच रहा है माइंस वैन कोई दिक्कत तो नहीं डिवाइडेड बाय वैन स्क्वायर प्लस वैन स्क्वायर मोड लगाना मत bhuliyega सर और ये कितना ए जाएगा डिस्टेंस है तू के इक्वल तो अल्फा -1 का जो मोड है वो कितना है 2√2 कितना हो जाता है सर हो जाता है प्लस माइंस तू रूट तू माइंस तू रूट ऑफ कोर्स माइंस हो रहा है वो 2√2 + 1 कौन सा कार्ड है वो 2√2 + 1 जो है वो आपका ये वाला कार्ड है और यहां पर जो ए रहा है वो है -2 √2 -1 रिपीट मैं स्टेटमेंट इस कोऑर्डिनेट्स के जो cardinate हैं इस पॉइंट के जो cardinate हैं वो है 2√2 -1 और इस सेंटर के कोऑर्डिनेटर जो होंगे वो होंगे -2√2 -1 आई होप ये बात आपको समझ आई कोई दिक्कत तो नहीं एंड ऑफ कोर्स कैन के ए का ऑर्डर आप बताओगे सर कब बताएं वो तो जीरो ही है ना क्यों जीरो है क्योंकि वो कहां पर है एक्स एक्सिस पर तो पहले तो ऐसे पॉइंट्स ढूंढ लो क्या ऐसे कोई पॉइंट हमें ऑप्शन में मिल रहे हैं 2√2+1 और 2√2 -1 से 2√2 -1 तो नहीं मिल रहा है बट तू रूट तू प्लस वैन कमा जीरो मिल रहा है तो एक आंसर तो ये तय है क्या मैं दूसरे आंसर पर बात करूं क्या अब मैं दूसरे आंसर पर बात करूं दूसरा आंसर क्या होगा दूसरा आंसर ये होगा सर की अब आप वैन कमा बेटा से इस स्ट्रेट लाइन की एक्स - ए - 1 की परपेंडिकुलर डिस्टेंस निकलेंगे तो आपके पास एक पॉइंट है वैन कमा बिता और स्ट्रेट लाइन है एक्स - ए - 1 अब कोस = 0 और इसकी भी जो परपेंडिकुलर डिस्टेंस है वो कितनी है 2 तो निकलने क्या सर बिल्कुल निकल लीजिए कैसे भाई देखो 1 1 डेट वैन माइंस वैन इन बिता थॉट्स - बेटा और क्या बच रहा है -1 मोड लगा लेना तो जरूरी है डिनॉमिनेटर में वैन का स्क्वायर 1 - 1² 1 + 1 2 यानी √2 और ये किसके इक्वल है सर ये 2 के इक्वल है वैन से वैन कैंसिल तो माइंस बिता का मोड आय थिंक बिता ही रहेगा और यहां पर ये ए जाएगा 2√2 कोई तकलीफ तो नहीं है भाई आई थिंक चीजें आसान है तो बिता कितना आएगा सर बिता की दो वैल्यूज मुझे मिलेगी जो की होंगी प्लस माइंस तू रूट तू इस बार जो कोऑर्डिनेट्स मिलेंगे इस बार जो कोऑर्डिनेट्स मिलेंगे वो क्या मिलेंगे सर एक इस पॉइंट की जो वैल्यू होगी वो तो क्या होगी सर वैन तू रूट तू आई होप आप समझ का रहे हो और इस पॉइंट के कोऑर्डिनेट्स होंगे वह -2√2 आपको किसके ऑप्शन में दिख रहे हैं आई थिंक सर ये जो दो ऑप्शंस हैं ऑप्शन बी और ऑप्शन दी यही फाइनली आपके आंसर्स होने चाहिए वीडियो ऑल अंडरस्टैंड अगेन ट्रिकी पार्ट था की आप इस पार्ट को ध्यान से देख पाएं और ये पार्ट कहां से ए रहा है ये पार्ट आपके पिछले चैप्टर से ए रहा है जो की है आपका क्या पैर ऑफ स्ट्रेट लाइंस आई थिंक चैप्टर आसान था मजेदार सा था और उसी का एप्लीकेशन ये था और आई थिंक बहुत टू बहुत डिफिकल्ट क्वेश्चन नहीं है अगर सर्कल की इक्वेशन भी पूछे जाती तो मैं का देता पर कितने सर्कल्स आते हैं टेक्निकल सर टेक्निकल हो सकता था इसके दो आंसर्स और आते हैं अगर ऑप्सनल्स में होते तो मैं उन्हें भी माफ करता हूं तो मस्क जब आपके पास एक से ज्यादा ऑप्शंस करेक्ट वाले क्वेश्चन है स्टूडेंट्स तो प्लीज थोड़ा केयरफुली दिखेगा बात बस किसी एक ऑप्शन को माफ करके खुश मत हो जाएगा की हान सर आंसर ए गया अब तो हम आगे बढ़ते हैं प्लीज थोड़ा केयरफुली रहिएगा आपकी वो एक्चुअली एबिलिटी चेक करें की आप हर पहलू को अच्छे से खोदकर khagate हैं या नहीं अब हर बारी की पर नजर डालते हैं या नहीं वो हर एक पॉइंट को एक्सप्लोर करने वाला माइंड ब्रेन एक्सपेक्ट कर रहे हैं ये बात याद रहेगी आई थिंक आप सारी लर्निंग अच्छे से सिख रहे हो की सर सारी बातें छोड़ो कुछ चीज तो मुझे दिख रही हैं कुछ चीज तो नजर ए रही हैं कौन सी चीज है देखो अपना क्वेश्चन को थोड़ा ब्रेक करो इक्वेशन पूछी जा रही है एक सर्कल की ठीक है सर विच टचेस इक्वेशन ऑफ अन सर्कल विच टच मतलब पूछी जा रही है दी जा रही है जो भी है देख लेंगे जो की ये आपके स्ट्रेट लाइन की इक्वेशन है विच टच इसे डी एक्सेस ऑफ कॉर्डिनेट इन लाइंस तो एक सर्कल है जो दोनों एक्सेस को टच करता है एक सर्कल है जो दोनों एक्सेस को टच करता है कुछ याद आया स्टूडेंट्स अगर मैं बात करूं एक ऐसे सर्कल की जो दोनों एक्सेस को टच करता है इमेजिन करो ये आप गलत से ए-एक्सिस है ये आप कर लेते क्या है भाई ए एक्सिस और एसेसरीज ये आपका है एक्स दिस इसे योर एक्स एक्सिस है ना अगर ये ए एक्सिस और एक्स एक्सिस है और अगर मैं एक सर्कल बनाऊं जो दोनों एक्सेस को टच करता है तो उसे चारों क्वाड्रेंट में बन सकता है बट अभी फिलहाल मैं अपने कन्वीनियंस के अकॉर्डिंग उसे फर्स्ट क्वाड्रेंट में बना ले रहा हूं किसी बात से किसी भी स्टूडेंट को कोई आपत्ति अब एक बात आप खुद सोचो की आपने फर्स्ट क्वाड्रेंट में क्यों बनाया है सर आप इस लाइन को देखो ना एक्स/3 + ए / 4 = 1 3 और 04 अगर ये इंटरसेक्ट फॉर्म है स्ट्रेट लाइन की जो की आपकी कहां से पास होती है सर जो की आपकी यहां से कुछ ऐसे इस तरीके से पास होती है सर की वो सर्कल को बेशक टच तो करती है लेकिन वो कैसे टच करती है वो सर्कल को इस तरीके से टच करती है की वो एक्स एक्सिस पर आपके 3 कमा जीरो पर इंटरसेक्ट लेती है और ए एक्सिस पर जीरो कमा फोर वाला इंटरसेक्ट लेती है बात से कोई आपत्ति मैं ऐसे सर्कल की इक्वेशन में जाना चाह रहा हूं जो की क्लीयरली दोनों एक्सेस को टच करता है और सर मैं भूल तो नहीं रहा हूं ना की अगर दोनों एक्सेस को अगर कोई सर्कल टच करें तो उसका एक्स कोऑर्डिनेटर उसका ए कोऑर्डिनेटर उसके सेंटर का दोनों से होते हैं और उसकी रेडियस भी वही होती है उसका मोड चूंकि फर्स्ट क्वाड्रेंट में है तो सब कुछ पॉजिटिव होगा और ये सर्कल इस स्ट्रेट लाइन को जो की एक्स-एक्सिस पर थ्री यूनिट्स का और ए एक्सिस पर पॉजिटिव फोर यूनिट्स का इस तरफ डायरेक्शन में इंटरसेप्ट मार दिए मतलब ये लाइन तो सर मेरे ख्याल से तो बड़ी आसान सी बात है बहुत जल्दी हो इसी बात है ज्यादा तकलीफ वाली बातें ही नहीं है यह बहुत डिफिकल्ट बातें ही नहीं है ऐसे कैसे नहीं है समझो बात को ये आपका क्या है सर्कल इस सर्कल को जरा गौर से देखो वो का रहा है यह आपके गिवन सर्कल की इक्वेशन है ठीक है सर मैन लिया आपके अकॉर्डिंग अगर सर ये उसे गिवन सर्कल की इक्वेशन है हु सेंट्रलाइज्ड इन डी फर्स्ट क्वाड्रेंट इस दिस तो आप सोच के बताओ आपके अकॉर्डिंग सर्कल के सेंटर के कॉर्डिनेट क्या होंगे इस सर्कल के सेंटर के कोऑर्डिनेट्स एक बार आप सोच के बताओ स्टूडेंट्स बस एक छोटे से करेक्शन के साथ जो की आप खुद देख का रहे हो ये टाइप वो ए रहा है की यहां पे सिर्फ माइंस तू सी एक्स नहीं होगा सर यहां पर इस इक्वेशन में आपको माइंस तू सी वे भी दिखेगा ऐसा क्यों का रहे हो सर प्लीज इस बात को समझो यहां पर अगर ये फर्स्ट क्वाड्रेंट में सेंटर है तो इसके सेंटर के क्वाड्रेंट क्या हो गया आप खुद समझो इसका हाफ सी और इसका हाफ सी तो सी कमा सी होना है और इसकी रेडियस भी सी होगी आई होप आप देख का रहे हो इसकी रेडियस भी क्या होगी सी ये बात आपको समझ ए रही है और अन्य वो यहां से चेक कर लो c² + c² - c² तो सी स्क्वायर कैंसिल x² जो बचेगा उसका अंडर रूट आप लेंगे तो सी ही आएगा तो इस तरीके से आपका ये कुछ बन रहा है बात जो निकल फाइनली ए रही है वह यह क्वेश्चन को सॉल्व करने के लिए मैं तो बस आपसे ये कहूंगा की सर्कल के सेंटर से जब आप परपेंडिकुलर ड्रॉप करोगे उसकी लेंथ इस लाइन पर जो होगी वो क्या होगी इसकी रेडियस के इक्वल यही एक स्टेटमेंट है जो हम कई दफा इस चैप्टर में उसे करेंगे बार-बार ऑलमोस्ट हर क्वेश्चन में कहीं ना कहीं मैं का रहा हूं आपसे सी कमा सी की इस स्ट्रेट लाइन से परपेंडिकुलर डिस्टेंस जो सी 1 / 3 यानी सी/3 सी 1 / 4 यानी सी / 4 - 1 इसका क्या लेंगे आप मोड डिवाइडेड बाय वैन बाय थ्री का स्क्वायर कितना 1 / 9 1 / 4 का स्क्वायर कितना 1 / 16 और इस तरीके से आप इसका ले लेंगे अंडर रूट और ये किसके इक्वल होगा सर इसकी रेडियस के अगर आप इतनी सारी मुश्किलों से बचाना चाहते द इसको इतना कॉम्प्लिकेट कैलकुलेशन से बचाना चाहते द तो आप इसे लिख सकते द 4X + 3y - 12 उससे मेरे ख्याल से और आसान हो जाता है बट ठीक है कर लिया तो ऐसे ही कर लेते हैं है ना अब क्या करने वाले हो सर अब मेरा कहना है देखो भाई जरा ध्यान से जरा ध्यान से देखो ये बन जा रहा है 4c ये बन जा रहा है 3c और ये बन जा रहा है -12 अगर मैं पूरा एक एलसीएम ले लूं तो क्लीयरली में क्या कर रहा हूं मैं पुरी इक्वेशन को अगर 12 से मल्टीप्लाई करूं न्यूमैरेटर में देखो अब ध्यान से तो क्या हो जा रहा है थ्री का 12 मतलब फोर टाइम्स थ्री टाइम्स और 12 12 तो ये हो जाएगा कितना 12 मतलब 12 से मल्टीप्लाई करने कुछ ऐसा होगा अभी इधर देखो अब ध्यान से बहुत ध्यान से देखना ये हो जाएगा 16 प्लस नाइन और डिनॉमिनेटर में क्या दिखेगा मतलब पता नहीं क्यों हमने इसको थोड़ा कॉम्प्लिकेट कर दिया है थोड़ा कॉम्प्लिकेट मतलब मैं का रहा हूं की ऐसे बिल्कुल ए रहा है वो ए ही रहा था एक और लाइन में करता और ए जाता और मैं का रहा हूं इससे फटाफट ऐसे ही सिंपलीफाई कर लो देखो अब क्या करोगे आप लिखोगे मतलब आप कंफ्यूज हो जाओगे इसलिए मैं थोड़ा सिंपलीफाई कर देता हूं 4X + 3y और 12 जो डिनॉमिनेटर में आएगा वो उधर जाएगा तो एक कुछ ऐसा हो जाएगा अब ये वही स्ट्रेट लाइन है ये यही स्ट्रेट लाइन है और इसकी में सी से अगर डिस्टेंस निकलूं तो मैं लिखूंगा 4c प्लस क्या 3c - आई थिंक अब ज्यादा आसान दिख रहा है और समझ ए रही है चीज और डिनॉमिनेटर में क्या थ्री और फोर आई थिंक pythagora एंटरप्राइज तो कितना ए जाएगा 5 ये बात जानते हो और ये किसके इक्वल होगा इसकी रेडियस सी के सर बड़ी बेसिक सी बात है 5 को वहां पहुंचा दो तो 5 को यहां पहुंचने हैं तो फाइव कितना हो जाएगा 5c सर एक और कम करो मोड हटा दो तो जैसे ही मोड हटाया जैसे ही मोड हटा गया तो मोड हटाने से यहां पर क्या प्लस माइंस एक और छोटा सा कम कर लो सर चार और तीन कितना होता है सात तो कितना हो जाएगा सही हो जाएगा 7c ये कितना हो जाएगा भाई 7c अब एक और कम कर लो सर ये माइंस 12 को वहां पहुंचा देते हो कितना हो जाएगा प्लस 12 कोई दिक्कत तो नहीं और 5c को यहां ले आओ तो 5c को यहां ले आए अगर 5c को यहां ले आए तो ध्यान से देखो भाई ये है 7c फिर आएगा 5c इधर तो वो हो जाएगा प्लस माइंस फाइव सी इसे इक्वल तू 12 अब आप एक बात खुद सोच के बताओ ध्यान से देखना स्टूडेंट्स कम की बात है 7c - 5c ध्यान से देखो 7c - 5c कितना हो जाएगा सो ये हो जाएगा 2c = 12 एक और बात 7c + 5c यानी कितना ट्वेल्थ इस इक्वल तू 12 तो यहां से सी की वैल्यू कितनी 6 और यहां से सी की वैल्यू वैन तो इसकी दो वैल्यूज आएंगे सर कौन-कौन सी वैन और सिक्स यानी ऑप्शन ए और दी क्या यहां तक किसी भी स्टेप में किसी भी स्टूडेंट को कोई डाउट इस क्वेश्चन को अच्छे से देख लो स्टूडेंट्स कहीं पर भी कोई डाउट हो तो अच्छे से चीज समझ लो आई थिंक ये आज का सबसे आसान क्वेश्चंस में से एक क्वेश्चन था हमने कुछ नहीं किया हमने ये जो सर्कल था ये इन दोनों को इन दोनों एक्सेस को टच करता था और इस स्ट्रेट लाइन प्रोडक्ट्स करता था तो बड़ी बेसिक सी बात हमने अप्लाई किया ऑफ कोर्स यहां पर ये टाइप जरूर इंश्योर कर लीजिएगा की आपने नोटिस किया हो की सर इस सर्कल के सेंटर से इसकी tenjent की परपेंडिकुलर लेंथ इसकी रेडियस की इक्वल होगी बस बस इतनी सी बात हमने की और वहीं से हमने सी की वैल्यूज निकल जो की i1 और 6 आई थिंक एक आसान क्वेश्चन था आसानी से सोचा जा सकता था पर बिल्कुल याद रखना ये मल्टीपल करेक्ट आंसर टाइप एक से ज्यादा करेक्ट ऑप्शंस होंगे तो आप ये सारी छोटी-छोटी बातों का ध्यान रखें है ना इस तरीके से चीज आप करेंगे क्या इस क्वेश्चन में कोई विदाउट इस क्वेश्चन के बाद आते हैं आज के अगले क्वेश्चन पर इसे जरा ध्यान से देखो और क्या लिखा है मुझे जरा समझ के बताओ बहुत अच्छा क्वेश्चन है बहुत इनसाइटफुल क्वेश्चन है अल हेल्प यू इन विजुलाइजिंग दिस क्वेश्चन की एक्जेक्टली हो क्या रहा है देखो भाई आपके पास पहले तो एक सर्कल है और एक पॉइंट है सरूप थ्री कमा वैन देखो आप इस सर्कल में पास करो थ्री प्लस वैन फोर है ना तो क्लीयरली 3 कमा वैन इस सर्कल पर लाइक करता है तो एक पॉइंट है पी जो की है √3 कमा वैन और अगर सर्कल को आप देखो तो मुझे ज्यादा कुछ सीखने कहने या बोलने की जरूरत नहीं है आप खुद कहोगे की ये जो सर्कल है सर ये जो सर्कल है ये क्लीयरली आपको कहां लाइक करता है ए एक्सिस पर मतलब ए और एक्स एक्सिस के इंटरसेक्शन ओरिजिन पर इसका सेंटर लाइक करते हैं जो आप खुद यह बात देख कर आसानी से बोल का रहे होंगे बिना ज्यादा इस चीज को घुमाए क्या मेरी इस बात से किसी भी स्टूडेंट को कोई आपत्ति नहीं है भाई तो आगे बढ़ते हैं सुनेगा ध्यान से अब आप एक बताओ √3 कमा वनसर क्लीयरली ये पॉइंट जो होगा ये यहां कहीं होगा लेट्स से ही वो पॉइंट है आपका जिसके कोऑर्डिनेट्स क्या है अंडर रूट थ्री कॉम वैन और मुझे कुछ कहने की जरूरत नहीं है इसके सेंटर के कोऑर्डिनेट्स है जीरो कमा जीरो और उसकी रेडियस है तू इतना तो सबको दिख रहा है बिल्कुल सर अब क्या बोल रहे हो ये जो पॉइंट है पी ये एक क्वार्टर ऑफ डी सर्कल मूव करता है अब एक क्वार्टर ऑफ अन सर्कल का मतलब क्या इस बात को समझना ये बहुत क्रोशिया पॉइंट है अगर मैं इस पॉइंट को इसके सेंटर से कनेक्ट करता हूं अगर मैं इस पॉइंट को उसके सेंटर से कनेक्ट करता हूं तो एक क्वार्टर क्या होता है मतलब एक सर्कल का चौथा हिस्सा एक सर्कल का चौथा कैसे पूरा होता है 360 डिग्री तो 360 का 4th यानी कितना 90 डिग्री तो ये यहां से 90° रोते करता है सर आप अपनी मर्जी से क्यों कर लेते हो चीजे आपके घर का राज है क्या मतलब आपने क्यों मैन लिया की यहां से वो इधर ही गया वह यहां से क्वार्टर जो वैन क्वार्टर लेंथ लिया जो वैन क्वार्टर डिस्टेंस ट्रेवल किया इसके आर्क पर या इसकी सरकम्फ्रेंसेस पर वो एंटीक्लाकवाइज ही ट्रेवल किया या अपने से अपने मैन से क्यों मैन लिया बिल्कुल नहीं मानेंगे गलत बात है तो मैं कहूंगा नहीं सर इसके अलावा ये भी तो संभावना थी की वो इस डायरेक्शन में चला जाता बिल्कुल पुरी बेशक संभावनाएं बात समझने की कोशिश करना स्टूडेंट्स ये पॉइंट यहां था और यहां से यह 90° ट्रेवल करके इधर ए जाता या ऐसा भी हो सकता था ये पॉइंट यहां से 90° ट्रेवल करके यहां ए जाता कहां यहां दोनों संभावनाएं पॉसिबल है फिर क्या किया उसने वो ट्रेवल करता है क्वार्टर ऑफ डी सर्कल और टांगेंशियली निकल जाता है लगा तो यहां से निकल जाएगा ऐसा या फिर यहां से निकल जाएगा कुछ ऐसा क्या बात आप समझ पाए आई थिंक क्वेश्चन यहां तक तो आसान है क्वेश्चन में यहां तक तो कोई दिक्कत नहीं होगी किसी भी अब क्या हो रहा है सर अब वह यह का रहा है प्लीज इस बात को suniyega डी इक्वेशन ऑफ डी लाइन अलांग विथ डी पॉइंट मूव्स आफ्टर लिविंग तो वो किस लाइन पर किस पथ पर मूव कर रहा है उसे लाइन की इक्वेशन आपसे पूछी जा रही है सर इस लाइन की इक्वेशन में निकल सकता हूं बहुत आसानी से कैसे आसानी से मुझे कौन सी लाइन की इक्वेशन चाहिए सर मुझे एक्चुअली मुझे एक्चुअली या तो इस लाइन की इक्वेशन चाहिए या इसकी या टेक्निकल दोनों की सर दोनों लाइंस की इक्वेशन निकलना बहुत टू टास्क नहीं है क्यों क्योंकि प्लीज इस बात को समझो स्टूडेंट्स आपने एक बात तो क्रैक की क्या सर आपको यह एंगल पता चल सकता है आपको यह एंगल पता चल सकता है फिर बस आपको यह करना है की उसमें आप इधर कुछ घुमा देना और इधर कुछ घुमा देना तो उससे आपको इन दोनों लाइंस के स्लो पता चल जाएंगी आपको बेशक इस लाइन तक की भी स्लो पता चल जाएगी और इस लाइन की भी स्लो पता चल जाएगी जो की आप बेसिक ज्यामिति से निकल लोग ऐसा मेरा मानना है मैं हिंट बता रहा हूं की आपको करना क्या है अगर इन दोनों लाइंस के स्लो का गई तो इक्वेशन तो सर अभी बहुत दूर है क्यों क्योंकि ये लाइंस की स्लोप पता चली वो भी अभी मुझे किसकी इस लाइन की और इस लाइन की मुझे तो अभी इस लाइन की और इस लाइन की भी स्लोप चाहिए जो की निकल लेंगे इस पर हम बात करेंगे लेकिन मेरा आपसे ये पूछना है सर मेरा आपसे ये पूछना है की प्लीज इस बातें गोद फॉर माय मुझे इन दोनों लाइंस की इक्वेशन निकलने के लिए ना सिर्फ स्लोप्स बल्कि वो किस पॉइंट से पास होती हैं वो भी तो पता करना पड़ेगा दोनों हिंट दोनों ब्लू आपको दे दिए आपको पता क्या करना है आपको पता करनी है इन दोनों लाइंस की स्लोप्स और दूसरा ये दोनों लाइंस किस पॉइंट से पास होते हैं मेरी हिंट आपको सलाह बस इतनी सी है जो भी डाटा जो भी इनफॉरमेशन जो भी एप्रोच इस क्वेश्चन में लगनी है वो सारी बातें यार तो फर्स्ट चैप्टर koardinate सिस्टम या सेकंड चैप्टर स्ट्रेट लाइंस में बहुत डिटेल में हमने पढ़ा है आपको सारी बातें पता है एक बार बस इस पार्ट को उसे फिगर को जो आपकी सहूलियत के लिए जो आपकी हेल्प करने के लिए ड्रॉ किया है उसे से देखो मेरा यकीन करो आप बहुत आसानी से उसे क्रैक कर लोग मेरा यकीन करिए आप बहुत आराम से उससे क्रैक कर लोग देखो मैं हेल्प करता हूं थोड़ा सा बताने की ज्यादा दूर नहीं जाना है ज्यादा दूर नहीं सोचना है आप सारी बातें छोड़ो अब तो बस ये बताओ सर आप बस ये बताओ क्या इस पॉइंट के कोऑर्डिनेट्स निकले जा सकते हैं क्यों नहीं अच्छा ऐसे सोचो ये है अंडर रूट 31 और ये है जीरो जीरो मुझे बस ये बता दो ये एंगल कितना है क्या आप में से कई सारे स्टूडेंट्स नहीं एंगल निकाला सर एंगल निकलना है ना इंक्लिनेशन निकालो कैसे स्लोप के थ्रू जाते हैं Y2 - y1 सो 1 - 0 / x2 - एक्स √3 - 0 तो ये ए रहा है 10 थीटा यहां का जो ये थीटा है यहां का जो थीटा है उसका 10 थीटा ए रहा है वैन बाय रूट थ्री उसका टेन थीटा कितना ए रहा है वो जो थीटा है ये जो थीटा है इसे हमें लिख देता हूं यहां पर थीटा और tanθ ए रहा है वैन बाय रूट थ्री मुझे बताओ 10 का कौन सा एंगल 1 / √3 होता है सर ये होता है कितना 30° आपको याद है स्टूडेंट्स ये कितना होता है 30 डिग्री आई होप आप भूल ही नहीं हो तो थीटा की वैल्यू कितनी आती है सर थीटा की वैल्यू आती है 30° आई होप इस बात से हम काफी आगे पहुंच जाएंगे तो ये जो एंगल निकाला सर ये था 30° अच्छी बात है सर मैन ली आपकी बात अब हमें आपको ले जाना चाह रहा हूं इस पॉइंट तक अब सारी बातें छोड़ो आप ये बताओ अब बस ये बताओ ये है 30 ये है 90 है की नहीं बिल्कुल है सर वैन क्वार्टर ट्रेवल किया 90 + 30 कितना 90 + 3120 ये पूरा है 180 180 में से 120 सब्सट्रैक्ट किया तो कितना सही हो जाएगा 60° अच्छा एक बात बताओ इस सर्कल की रेडियस पता है क्या सर इस सर्कल की रेडियस तू यूनिट्स हैं तो आप सोच के देखो ना ये आपका क्लीयरली कहां ए रहा है clearlization है ये आपका ए रहा है सेकंड क्वाड्रेंट ये आपका ए रहा है सेकंड क्वाड्रेंट में यहां पे एक्स नेगेटिव होगा ए पॉजिटिव होगा तो क्या मैं इसे क्या मैं इसे -2 कोस 60 यहां पे क्या ए जाएगा +2cos इसे आप लिखोगे साइन क्या हो रहा है भाई प्लस तू साइन 60 लिख सकता हूं क्या आप प्लीज इस बात को थोड़ा ध्यान से देखो मैं फिर से रिपीट करता हूं इस पॉइंट का जो एक्स कोऑर्डिनेट्स होगा नेगेटिव होगा आपने कैसे लिखा सर मैंने बेसिक सा आपका राइट एंगल ट्रायंगल में सोचा इस बारे में हमने डिटेल डिस्ट स्ट्रेट लाइंस और कोऑर्डिनेट्स सिस्टम में किया है मुझे हाइपोटेन्यूज पता है मुझे एंगल पता है तो मुझे क्या निकालनी है ये लेंथ और ये लेंथ बस ये जो लेंथ निकलोगे उसके लिए एक cardinate नेगेटिव रखना क्योंकि ये एक्सेस के लिए रेड साइड पर है और ये जो लेंथ निकलोगे वो एक्सेस की पॉजिटिव साइड पर ही है तो इसे पॉजिटिव ही रखना आप मेरी बात समझ का रहे हो तो अगर मैं थोड़ी देर के लिए मैन लूं इस पॉइंट को का देता हूं पी और थोड़ी देर के लिए इस पॉइंट को का देता हूं क्यों तो पी पॉइंट के cardinate क्या ए रहे हैं तो सर आपका हो गया पी पॉइंट के जो cardinate ए रहे हैं कोस 60 कितना होता है सर कोस 60 होता है सिन 30 जितना साइन 30 कितना होता है 1 / 2 तो -2/2 -1 तो पी का जो एक्स कोऑर्डिनेट्स आएगा वो कितना आएगा माइंस वैन सिमिलरली sin60 होता है अंडर रूट 3/2 फिर तू से तू कैंसिल तो ये हो जाएगा अंडर रूट थ्री तो ये कितना ए जाएगा सर -1 √3 ये हो जाता है पी क्या इसी आधार पर क्या इसी बेसिस पर थोड़ा दिमाग लगाया जाए तो क्या क्यों नहीं निकल सकते क्यों नहीं निकल सकते क्यों निकल सकते हैं कैसे सर सोच के देखो अगर मुझे कोई पूछे क्यों का एक cardinate तो के के कोऑर्डिनेट्स को ले के मैं क्या टिप्पणी करूंगा ध्यान से देखोगे बड़ा मजेदार सा बात है आप बताओ सर ये ये कितने डिग्री से घुमा सर ये घुमा 90° से 90 में से ये एंगल ऑलरेडी 30 था तो फिर से वही बात सर आप क्यों नहीं सोच का रहे हो की ये कितना होगा ये होगा 60° फोर्थ क्वाड्रेंट में कौन पॉजिटिव कौन नेगेटिव फोर्थ क्वाड्रेंट में expostive वही निगेटिव यानी आप के के कोऑर्डिनेट्स क्या कहोगे मैं क्योंकि कोऑर्डिनेट्स पर ही वही बात ओरिजिन से के तक के डिस्टेंस तू इसे मैं लिखूंगा तू कोस 60 पॉजिटिव और माइंस तू साइन 60 बट ऑफ कॉस्ट नेगेटिव से कोई परेशानी तो नहीं हो रही आप सभी को क्या यह बात आप सभी को समझ ए रही है तो के को लेकर हम क्या कहेंगे भाई सर के को लेके हम कहेंगे तू कोस 60 - 2sin60 है ना कोस 6p यानी वही बात वापस sin30 है ना sin30 कितना होता है 1 / 2 1 / 2 से 2 यानी वैन तो ये कितना बचेगा सर ये बचेगा फाइव कोई तकलीफ तो नहीं है आई थिंक नहीं होनी चाहिए कुछ गड़बड़ तोड़ना नहीं कर रहा हूं बस एक बार मुझे चेक कर लेने दीजिए स्टूडेंट्स एक बार बस मुझे चेक कर लेने दीजिए ये पूरा था 90° ये था 30° ये है 60° है ना ये था अगर 60 डिग्री सही लिखा है कोई दिक्कत नहीं है तो ये ए जाएगा कितना वैन बाय तू वैन बाय तू यानी ये गया उधर आप देखोगे तो सिन 60 होता है √3 / 2 2 से 2 कैंसिल तो ये बचेगा - √3 अब मेरा बस आपसे ये कहना है की आपको पी और के पॉइंट्स पता चल गए हैं पता चल गए हैं क्या आई थिंक सर बहुत बड़ी विशेष बहुत टू बात नहीं है आपको पी और के दोनों ही पॉइंट्स बड़े ही आसानी से पता चल चुके हैं अब क्योंकि आपको पी और के पता चल गए हैं तो क्या आप पी और के की क्वेश्चन नहीं लिख सकते पी और के की इक्वेशन लिखने का एक तरीका तो यह की मैं अब पी और के की स्लोप निकालो विच इस अब्सोल्युटली एंड एब्सल आइडिया विच इसे डेफिनेटली नॉट रिकमेंड सर ऐसे क्यों का रहे हो आप अभी तो आप ये का रहे द की आप इस लाइन के स्लोप निकल लो इस लाइन के स्लोप निकालो पॉइंट ऑफ कॉन्टैक्ट ए ही गया है मैं का रहा हूं आप सारी बातें छोड़ो क्या आपको ये बात याद नहीं है क्या आपको ये बात याद नहीं है की अगर एक सर्कल है और सर्कल में किसी पॉइंट पर कोई टैसेंट ड्रॉ की जाए जैसे ही आपका सर्कल है ये सर्कल की इक्वेशन क्या है सर ये है x²+y2fy प्लस सी इसे इक्वल तू जीरो और इसका एक पॉइंट ऑफ कॉन्टैक्ट है जिसको आप मैन लेते हो X1 y1 सर मुझे एक बात बताओ अगर मैं आपसे पूछूं इस टैसेंट की इक्वेशन तो स्ट्रांग के क्वेश्चन क्या होती है तो आप कहते हो टी = 0 कहते हो की नहीं तो जीरो मतलब इस टांगें की इक्वेशन को मैं कहूंगा एक्स एक्स वैन प्लस ए ए वैन प्लस जी टाइम्स एक्स + X1 प्लस एफ टाइम्स ए + y1 + सी = 0 ये आपकी टेंशन की इक्वेशन होती है भूल गए हो क्या तो सर यहां पे कौन सी बड़ी बात है आपका सर्कल क्या है x² + y² = 4 आपका सर्कल क्या है x² + y² = 4 और ये आपके पॉइंट ऑफ कॉन्टैक्ट है जिन पर आपको टांगें ड्रॉ करनी है तो इंटेलिजेंस की इक्वेशन लिखना कौन सा बड़ा मुश्किल है टू टास्क है आई थिंक नहीं है तो निकल लेते हैं सर ध्यान से देखना -1 √3 पर सर्कल सर्जन की सर्कल की इक्वेशन क्या है सर्कल के क्वेश्चन है x² + ए स्क्वायर कर रहे हो माइंस वैन कमा √3 - 1 √3 तो -1 एक्स - एक्स √3 ए तो ये हो जाएगा √3y और ये हो जाएगा -4=0 ये आपकी कौन सी tenjent है भाई ये आपकी इस टैसेंट की इक्वेशन है कोई तकलीफ तो नहीं है अच्छा उसे टेस्टोन जिनकी इक्वेशन सर वैन कमा - √3 अब एक और पॉइंट कौन सा है वैन कमा - √3 तो 1 - √3 मतलब 1 एक्स - √3 ए - √3y और फोर इधर ले तो -4 = 0 जरा ध्यान से देखो माइंस एक्स प्लस अंडर रूट 3y माइंस एक्स प्लस आई थिंक ऑप्शन बी बन सकता है अगर मैं माइंस एक्स माइंस फोर को उधर ले जाऊं तो सबसे पहले तो आप सभी को ऑप्शन बी नजर ए रहा होगा आई थिंक आप मैच कर का रहे हो दूसरा अगर मैं देखूं तो ये दिख रहा है क्या ध्यान से देखो भाई अगर मैं अंडर रूट 3y को एक तरफ रखूं √3y को एक तरफ रखूं तो ये हो जाएगा 4 - एक्स तो ऑप्शन दी तो नहीं है है ना अगर मैं अंडर रूट 3X बनाऊं तो ऑप्शन सी तो नहीं है बट ऑप्शन ए देखने की कोशिश करते हैं ए = √3x+4 नहीं बिल्कुल नहीं एक बार फिर से चेक करते हैं मैं कुछ शायद मिस कर रहा हूं थोड़ी सी रश में ऑप्शन ए ए = √3x + 4 नहीं बिल्कुल नहीं मैं एक बार फिर से चेक कर लेता हूं के के क्वाड्रांट्स जो द वो द वैन कमा माइंस अंडर रूट थ्री और अगर ये रखा तो 1 - √3 कमा ए और -4 बिल्कुल सही और इस बेस पे ऑप्शन सी देखें तो वो हो जाता है कितना ऑप्शन सी तो मुझे मुश्किल लग रहा है ऑप्शन दी जो है अंडर रूट थ्री ए = एक्स - 4 हान यह सही है अगर उधर गया तो एक्स - 4 ही तो बचेगा तो बिल्कुल सही सर पता नहीं क्यों आप ध्यान से देखते नहीं हो ऑप्शन दी आपका करेक्ट आंसर होगा और इस बेसिस पे आप दो ऑप्शंस कौन-कौन से बी और दी मार्क करोगे ट्रिकी पार्ट बस ये था की अब पी और के के क्वाड्रेटिक निकल पाते और कैसे इस प्रॉब्लम को थोड़ा ठीक से विजुलाइज करके क्या ये बहुत टू बहुत डिफिकल्ट क्वेश्चंस था आई थिंक नहीं था तो आई थिंक आप चीज सोच पाए और अगर चीज समझ पाए तो क्या नेक्स्ट क्वेश्चन पर मूव करें नेक्स्ट क्वेश्चन क्या लिख रहा है वो जरा ध्यान से देखिए अच्छे से पढ़ो ये एक अच्छा क्वेश्चन है की बहुत मजेदार क्वेश्चन है और ये बहुत इनसाइड फुल क्वेश्चन है इसे प्लीज अच्छे से ट्राई करो पहले मैं समझाऊं या ना समझाऊं एक बार आप ट्राई जरूर करो बहुत अच्छा ये क्वेश्चन है और बहुत इनसाइड फुल क्वेश्चन है जो आप सभी से होना चाहिए देखो क्या का रहा है वो क्वेश्चन समझो क्वेश्चन बहुत अच्छा है पहले तो बात समझना नॉर्मल तू दिस सर्कल क्या मतलब मतलब यह की इस सर्कल का एक नॉर्मल सा सारी बातें छोड़ो इस सर्कल का नॉर्मल मतलब मतलब आई होप आप जानते हो की सारे ही सर्कल्स के नॉर्मल्स उनके सेंटर से पास होते हैं क्योंकि उन पर नॉर्मल होते हैं सारे वो डायमीटर्स होते हैं तो अगर इस सर्कल की बात करें तो सर्कल का सेंटर है जीरो कमा जीरो मतलब वो इसके सेंटर जीरो कमा जीरो से पास हो रहा है मतलब एक स्ट्रेट लाइंस जीरो कमा जीरो से पास हो रही है तो उसकी इक्वेशन किस फॉर्म में होगी सर उसकी इक्वेशन क्लीयरली ए = एमएक्स + फॉर्म एमएक्स फॉर्म में होगी प्लस सी तो रहने दे रहा वो जो लाइन है वो जो नॉर्मल है वो इस सर्कल को डिवाइड करती है कौन से सर्कल को एक ऐसे सर्कल को जिसके सेंटर के कोऑर्डिनेट्स हैं तू कमा फोर एक ऐसे सर्कल को जिसके सेंटर के कोऑर्डिनेट्स से तू कमा फोर और इसकी रेडियस है तू इसकी रेडियस है तू है ना इस सर्कल को किसी एक प्रोपोर्शन में डिवाइड कर दिया मैं मैन लेता हूं ये जो लाइन है ए = एमएक्स ये कुछ ऐसी लाइन है ये कुछ ऐसी लाइन है ध्यान से सुनना रैंडम बना ले रहा है ना ये हमारी स्ट्रेट लाइन है ये कौन सी स्ट्रेट लाइन है सर ये आपकी वो लाइन है जो की है वही इसे इक्वल तू एमएक्स कोई तकलीफ तो नहीं है अब ऑफ कोर्स इस स्ट्रेट लाइन इस सर्कल को 200 में डिवाइड किया जो की है π-235 + स्टोर में देख का रहा हूं की बराबर तो है नहीं है इससे एक हिस्सा थोड़ा छोटा है काफी छोटा और एक हिस्सा काफी बड़ा है देखो पाई में सिर्फ तू सब्सट्रैक्ट कर दिया और यहां पाई का ट्रिपल करके तू ऐड किया तो ऑफ कोर्स एक सेक्शन badhaaoge और एक सेक्शन छोटा होगा जो की कुछ ऐसा होगा आय होप आप बात समझ का रहे हो इस सेक्शन का जो रीजन है इस इस रीजन का जो एरिया है ये जो रीजन है इसका एरिया डिवाइडेड बाय इस रीजन का जो एरिया है वो रेश्यो आपको ये दिया हुआ है कोई तकलीफ तो नहीं सर कैसे सोचेंगे इस क्वेश्चन को इस पे बात कर लेते हैं सर मेरा कहना है ध्यान से सुनो स्टूडेंट्स एक छोटी सी बात है अगर मुझे पता चल जाए ध्यान से सुनना एक छोटी सी बात क्या देखो अगर मैं इस पॉइंट को यहां कनेक्ट करूं देख रहे हो क्या आप लोग और इस पॉइंट को यहां कनेक्ट करूं ठीक है क्या और मैं मैन लूं की सर ये जो एंगल है ये है लेट से θ सर एक बात बताओ इस सर्कल की रेडियस कितनी है सर इस सर्कल की रेडियस है ये डिस्टेंस हो रही है डिस्टेंस तू हो जाएगी मैं कुछ बातें का लेता हूं जिससे मैं का देता हूं ए इसे का देता हूं बी और इसे का देता हूं सी कोई तकलीफ तो नहीं है आपकी सहूलियत के रमन पॉइंट्स को कुछ ऐसे रैंडम नाम ले लेता हूं जैसे यहां पर एक पॉइंट ले लिया दी यहां पर एक पॉइंट ले लिया आई यहां पर एक पॉइंट ले लिया एफ और यहां पर एक पॉइंट ले लिया जी क्यों लिया अभी आपको रिलाइज करवाऊंगी बस आपको समझने के लिए ताकि आपको ये क्वेश्चन अच्छे से समझ ए जाए अब एक बात बताओ बहुत सोच के समझ के बोलने स्टूडेंट्स एक छोटी सी बात बताओ इस पूरे सर्कल में अगर मैं आपसे पूछ रहा होता मानो थोड़ी देर के लिए इसका एरिया इसका एरिया सर इसका तो पूछा ही नहीं है पर मैं पूछ रहा हूं मैन लो मैं आपसे इस रीजन का एरिया पूछ रहा होता तो आप क्या कहते हैं हम कहते हैं सर पूरे सर्कल का एरिया कितना होता है वो निकल लो सर पूरे सर्कल का एरिया होता है पाई r² तो वो हो जाता पाई r² r² मतलब कितना रेडियस का स्क्वायर हो जाता है पाई आर स्क्वायर लेकिन सर पूरे सर्कल का एरिया नहीं चाहिए उसके θ पोर्शन का चाहिए किसके पूरा सर्कल होता है 360 मुझे थीटा डिग्री का चाहिए तो मुझे क्या चाहिए थीटा / 360 360 को अभी थोड़ी देर के लिए 2π कहते हैं कितना हो जाएगा थीटा / 360 अगर आपको कन्फ्यूजन है तो मैं इसे लिख देता हूं क्या थीटा / 360 तो इसका एरिया है 4 पाई लेकिन मुझे कौन सा एरिया चाहिए मुझे चाहिए थीटा / 360 डिग्री वाले रीजन कहेंगे मेरी बातें बहुत प्रॉब्लम आती तो नहीं हो रही है स्टूडेंट्स अभी हमने कौन सा एरिया निकल लिया अभी सर आपने ये वाला एरिया निकल लिया लेकिन एक मिनट सर यहां तो जाना नहीं था यहां तो जाना नहीं था मुझे तो ये वाला एरिया चाहिए अभी मुझे ये वाला एरिया चाहिए तो shailais से की सर जो एरिया आपने निकाला है जो एरिया आपने निकाला है अगर उसमें से मैं इस ट्रायंगल के एरिया को सब्सट्रैक्ट कर डन तो मुझे ये वाला एरिया बचा हो मिल जाएगा आप मेरी बातें समझ का रहे हो क्या तो ट्रायंगल का एरिया क्या होगा ट्रायंगल ध्यान से देखो स्टूडेंट्स क्या मैं जानता हूं किसी ट्रायंगल का एरिया जिसकी मुझे साइड्स की लेंथ दे दी जाए और उनके बीच का एंगल दे दिया जाए सर हाफ दोनों साइट्स का प्रोडक्ट इन साइन ऑफ डी एंगल कौन सा एंगल डी एंगल विच इस कंटेंड बाई दो तू साइट्स विच इस क्रीटेड बाय दोस्तों सेट्स तो मैं क्या कहूंगा इस ट्रायंगल एबीसी का एरिया वो कितना होगा मैं इस पूरे एरिया में से जो जस्ट अभी हमने निकाला इस पूरे एरिया में से वो ट्रायंगल का एरिया सब्सट्रैक्ट कर रहा हूं वो ट्रायंगल का एरिया सब्सट्रैक्ट कर रहा हूं तो कितना होगा सर वो होगा हाफ है ना दोनों साइड्स का प्रोडक्ट यानी 2 2 और उनका कंटेंट एंगल्स साइन थीटा तो ये जो हमने लिखा है स्टूडेंट्स ये आपका ये वाला पार्ट है आई होप बातें समझ ए रही है अच्छा सर डिनॉमिनेटर में क्या ए रहा है अब डिनॉमिनेटर की अगर मैं बात करूं तो ये तो वो आपका न्यूमैरेटर आप डिनॉमिनेटर की अगर हम बात करें तो डिनॉमिनेटर में क्या है गौर से देखो भाई सर डिनॉमिनेटर निकलना है तो आसान है ऐसे कैसे आसान है बहुत आसान है मुझे एक बात बताओ डिनॉमिनेटर क्यों आसान है की सर अगर इस पूरे सर्कल के एरिया में से अगर ये सब्सट्रैक्ट कर दें तो क्यों नहीं मुझे ये रीजन का एरिया मिल जाएगा समझ में ए रहा है जो का रहा हूं तो बिल्कुल सर ऐसा कर लो कैसा कर लो भाई मैं क्या कर रहा हूं पूरे सर्कल का एरिया वह वही बात पाई r² तो पाई r² मतलब कितना पाई r² मतलब आपका पाई आर स्क्वायर यानी पाई आर आर कितना फोर फोर का स्क्वायर और इसमें से इसमें से मैं इसका एरिया सूत्र कर रहा हूं सर इसमें से मैं इस रीजन का एरिया सब्सट्रैक्ट कर रहा हूं इस रीजन का जो एरिया है वही तो न्यूमैरेटर में लिखा है तो ये जो एरिया आप ढूंढ रहे हो जो सब्सट्रैक्ट करना है वो यहां पर न्यूमैरेटर में लिखा है तो टेक्निकल पूरे सर्कल के लिए इसके एरिया को सब्सट्रैक्ट करना है बहुत डिफिकल्ट बातें तो नहीं का रहा हूं तो जब इसे सब्सट्रैक्ट किया तो देखो भाई क्या मिलेगा सर यह मिलेगा 45 θ / 360 और माइंस किया तो यह हो जाएगा प्लस कितना समझा पाऊं ये आपका रेश्यो है लेफ्ट हैंड साइड पर ये आपका रेशों है लेफ्ट हैंड साइड पर ये आपका रेश्यो है कौन सी साइट पर लेफ्ट हैंड साइड पर और राइट हैंड साइड पर यही जो रेश्यो है यही जो प्रोपोर्शन है ये किसके इक्वल है ये π-2:3 पे +2 ये किसके इक्वल π - 2 / 3 पाई + 2 इस रेश्यो के इक्वल है इस पार्ट को जरा ध्यान से देखो इस पार्ट को जरा अच्छे से समझो और कहीं भी कोई डाउट है तो मुझे यहां से चीज आगे बढ़ाने में मदद करो क्या यहां तक किसी भी किसी भी स्टेप में कोई डाउट है अभी तक जो भी बातें हमने लिखी आई थिंक बहुत सिंपल सी शॉर्ट ट्रिक्स सी बातें लिखी बस अपनी सहूलियत के लिए सर जब आप सब चीज इंडियन में ही कर रहे हो तो क्या 360 को हमें 2pir प्लेस कर डन तो चलेगा क्या तो यहां पर मैं 360 को 2π से और यहां भी 360 को 2/7 कर दे रहा हूं आई होप इस बात से कोई आपत्ति या तकलीफ नहीं है अब सर जब आप सिंपलीफाई करना ही चाह रहे हो तो आप क्या करोगे भाई हम कहेंगे 2π का फोर पाई कितना टाइम तू टाइम्स इस तू से यह तू कैंसिल कोई दिक्कत तो नहीं कोई दिक्कत या कोई डॉक्टर सॉल्व करना चाहूं तो देखो यहां से देखना ये हो जाएगा 2 थीटा तो सर ये तो न्यूमैरेटर में है कितना तू थीटा माइंस सर ये है -2sintheta तो ऐसे लिख देते हैं -2 सिन थीटा व्हाट डू सी हैव इन डी डिनॉमिनेटर ध्यान से देखना स्टूडेंट्स डिनॉमिनेटर में जो है वो है फोर पाई डिनॉमिनेटर में जो है वो है फोर बाई माइंस यहां से देखो कितना 2 थीटा और क्या सर प्लस तू टाइम्स साइन थीटा करना है तो कैसे सोचेंगे देखो भाई क्रॉस मल्टीप्लिकेशन के लिए मेरा विचार ये होगा की 35 + 2 को यहां मल्टीप्लाई किया तो 3पाई+2 को मल्टीप्लाई किया थोड़ा केयरफुल रहेगा क्योंकि थोड़ा वल्नरेबल जोन है चीज हटेंगे मल्टीप्लाई होंगी सब्सट्रैक्ट होंगे बहुत कुछ होगा तो बस थोड़ा केयरफुली करेगा है ना 35 + 2 को आपने किया 2π-2 साइन थीटा से तो सॉरी 2π या कितना सर ये है तू थीटा माइंस तू साइन थीटा ये है तू थीटा माइंस तू साइन थीटा वैसे ही आप तू भी कॉमन ले सकते द तू कॉमन क्यों नहीं स्टूडेंट्स में तू भी कॉमन ले लेता हूं तो देखो ध्यान से यह तू और ये तू गया यहां पर बचेगा 2 और ये है जाएंगे तो मैं यहां पर लिख देता हूं कोई दिक्कत तो नहीं है आई थिंक चीज आसान है तो मैं इसे लिख लेता हूं θ - सिन थीटा है ना इससे बेहतर लैंग्वेज हम लिख लेंगे θ - ऑफ कोर्स किया सिन थीटा कोई दिक्कत तो नहीं है भाई अब आगे suniyega अब आगे suniyega इस तरफ क्या जा रहा है यह है पाई माइंस तू राइट हैंड साइड पर आपको क्या मिल रहा है राइट हैंड साइड पर आपको मिल रहा है 5 - 2 जिससे कॉर्नर multiplaye हो रहा है पर π - 2 से 2π 2pi - देता प्लस साइन थीटा है ना फिर तू से मल्टीप्लाई किया तो ये हो जाएगा 2 थीटा और यहां पर क्या हो जाएगा सर यहां पर ए जाएगा -2 सिन थीटा तो ये कितना हो जाएगा ये हो जाएगा कितना भाई 2pir सही लिख रहा हूं क्या फ्री हो जाएगा कितना पैसा मल्टीप्लाई किया तो - देता / ए गया तो पाई साइन थीटा कोई दिक्कत तो नहीं है अब - 2 से मल्टीप्लाई किया तो माइंस फोर पैट है ना और फिर दिखेगा माइंस तू साइन थीटा कोई गड़बड़ तो मैंने नहीं किए बस थोड़ा सा वल्नरेबल जोन है क्योंकि θ पाए सब कुछ आप अलग-अलग चीजों से डील करते चले जा रहे हो तो एक बार बस क्रॉस चेक कर लो की पाइप से जब मल्टीप्लाई किया तो 2pir माइंस तू से किया -45 प्लस तू थीटा माइंस तू साइन थीटा गलती होने के चांसेस हैं थोड़ा सा बस ध्यान से करना कोई भी एक प्लस माइंस साइन गड़बड़ हुआ ना आपका पूरा आंसर गलत हो जाएगा इसलिए थोड़ा सा यहीं दो मिनट लेके कर लो पूरा सॉल्व कर दोगे गलत करके फिर पूरा चेक करोगे अच्छा जहां भी लगे की सर थोड़ा सेंसेटिव जोन है वहां एक बार थोड़ा देख लो रो के हड़बड़ाहट में मत रो रैश में गलतियां होती हैं है ना सर अब आप देखो चीजों को समझो मुझे एक तो दिख रहा है की ये 2 थीटा से 2 थीटा कैंसिल कर दो सर आप सभी को दिख रहा है क्या एक बात आप सब नोटिस करो स्टूडेंट्स की देखो सर आप - 2sin थीटा से आप ये माइंस तू साइन थीटा को कैंसिल कर दो बिल्कुल कर सकते हैं और क्या कैंसिल होता दिख रहा है सर आपको और कुछ हटा हुआ कैंसिल हो तो आज chhatata हुआ दिख रहा है क्या भैया हमें सर इससे ज्यादा तो मुझे ज्यादा नजर नहीं ए रहा है बट अगर आप देखो तो चीज आइडिया सब्सट्रैक्ट करें जो होना होगा हो जाएगा बिल्कुल सर ऐसा ही कर लेते हैं है ना देखो भाई अब अगर मैं यहां से शुरू करूं तो मैं कहां लेकर आता हूं पहले तो देखो ये लिखा है थ्री थीटा बाय और ये थीटा पाए इधर आता है तो थ्री थीटा बाय में ये थीटा / ऐड किया तो ये हो जाएगा कितना फोर थीटा बाय किसी स्टूडेंट को यहां तक कोई भी अभी क्या करने वाले हैं सर अब ध्यान से सुनना सर बड़ी सिंपल सी बात है बड़ी ही सिंपल सी बात है आप ध्यान से देखो ये लिखा है -3π साइन थीटा और ये लिखा है प्लस पाई साइन थीटा ये उधर आए तो माइंस पाई साइन थीटा तो ये हो जाएगा -4 बाय साइन थीटा क्या इस बात तक तो कोई आपत्ति नहीं है अब एक छोटा सा कम और आपको कर लेना है सर कौन सा वाला ये वाला पार्ट 2 पाई स्क्वायर माइंस यहां पर आपको क्या मिलेगा सर यहां पर आपको मिलेगा 2π स्क्वायर दोनों को भी कंसीडर कर लिया और आई थिंक अब हमने सारी चीज कंसीडर करनी है बस एक छोटी सी गड़बड़ की 2² जब इधर आएगा तो वो नेगेटिव हो जाएगा और 4 इधर आएगा तो पॉजिटिव कुछ और चेंज तो नहीं दिख रहे हैं यहां तक स्टूडेंट्स इसे एक बार अच्छे से देखो अब एक जरूरी बात आप सब नोटिस करोगे क्या की सर देखो आप सब में 2/2π 2π दिख रहा है तो 2π कॉमन ले लेते हैं अब जैसे ही 2π कॉमन लिया तो यहां से 2π मतलब 2पाई से पूरे एक्सप्रेशन को डिवाइड कर रहा हूं तो 2π यहां से लिया तो ये क्या बचेगा सर ये बचेगा तू थीटा सबको दिख रहा है क्या ध्यान से देखते जाना सर 2π यहां से लिया तो 2π तो ये क्या बचा माइंस तू साइन थीटा कोई दिक्कत नहीं सर यहां से 2π ले लिया तो ये बचा कितना -5 और यहां से 2π ले लिया तो ये बचा + 2 = 0 आई थिंक ये यहां पर आपको समझ में कम करता हूं मैं क्या करता हूं मैं इस तू साइन थीटा को उसे तरफ लेकर जाता हूं ध्यान से सुनना अगर मैं माइंस तू साइन थीटा को उसे तरफ लेकर गया तो शरीर क्या हो जाएगी ये हो जाएगा प्लस तू साइन थीटा यहां है इसे मैं यहां लिख देता हूं एक छोटा सा कम और करते हैं एक छोटा सा कम और करते हैं क्या suniyega बहुत कम की बात है इस तू को इधर लेट हैं सुनना मैं क्या कर रहा हूं मैं लिख रहा हूं साइन थीटा = सिन थीटा = 2 से यहां डिवाइड किया तो ध्यान से देखना 2θ + 2 जब डिवाइड होंगे जब तू थीटा प्लस तू डिवाइड होंगे तो कितना ए जाएगा थीटा प्लस वैन तू थीटा प्लस तू जब डिवाइड होंगे तो कितना ए जाएगा थीटा + 1 या 1 + θ क्या मेरी बात समझ में और जब माइंस पाई को ये तू डिवाइड करेगा तो हो जाएगा -π/2 अब आप मेरी सारी बातें छोड़िए बस एक बड़ी बेसिक सी बात का जवाब दीजिए थीटा ऑफ कोर्स कोई पॉजिटिव वैल्यू क्योंकि वो एक सिग्निफिकेंट एंगल है राइट इन दोनों साइड्स के बीच का किन दोनों के बीच का इन दोनों के बीच का यहां पर है ना ये वाला एंगल है थीटा अब मेरा आपसे ये कहना है प्लीज इस बात को ध्यान से सुनना साइन थीटा की रेंज पर अगर मैं बात करूं तो सर वो होती है -1 से 1 तक सिन थीटा की रेंज जो होती है वो होती है -1 से 1 तक ठीक है सर अब ये है वैन प्लस समथिंग यह है वनप्लस समथिंग ऑन जा रहा है जितना भी हिस्सा बनकर बॉन्ड जा रहा है शैला से की सर वो हिस्सा जीरो होना चाहिए अब मेरी बात समझ का रहे हो मतलब सिन थीटा की मैक्सिमम वैल्यू वैन वैन के भी ऑन ये नहीं जाता है तो वैन के भी ऑन जो भी वैल्यू जीरो होनी चाहिए मतलब आप कहना चाह रहे हो सर की थीटा माइंस पाई / 2 जीरो होना चाहिए जी हान अगर थीटा माइंस पाई / 2 जीरो होना चाहिए तो थीटा कितना होना चाहिए पी / 2 और क्या आपकी ये वैल्यू उसे कंडीशन को सेटिस्फाई कर रही है देखो भाई थीटा की वैल्यू अगर मैं π/2 रखता हूं तो सिन 90 ट्रिकी पार्ट था थोड़ा सा ट्रिकी पार्ट इसलिए था की मुझे इस जोन से थोड़ा डील करने में प्रॉब्लम हो रही थी अब मेरी बात समझ का रहे हो आप यह बात डाइजेस्ट करवा रहे हो क्योंकि यहां थीटा भी था और सिन थीटा भी था और दोनों को लेकर मेरा ये एक थोड़ा सा इंशन बेस्ड एप्रोच कम कर रहा है जिसके थ्रू मैं थीटा की वैल्यू पाई / 2 ला रहा हूं कोई दिक्कत तो नहीं है अब सर थीटा की वैल्यू अगर π/2 पता चलती है अगर थीटा की वैल्यू पाई / 2 पता चलती है तो उससे क्या आई थिंक सर उससे बहुत कुछ पता चलता है उससे बहुत कुछ मतलब क्या पता चलता है जैसे ही देता की वैल्यू पाई / 2 पता चले अब ध्यान से देखना मतलब ये जो एंगल है सर ये आपका कितना है 90° एंगल अरे हान या ना बिल्कुल सही बात है सर और अगर यह 90 डिग्री एंगल है अगर यह 90° एंगल है तो मैं एक बहुत जरूरी सी बात जानता हूं सर बड़ी सिंपल सी बड़ी बेसिक सी बात सर इसको ऐसे भी सोच सकते हो या तो आप वो राइटिंग एंगल में चले जाओ बट उतनी बातें नहीं भी करनी है तो मैं एक कृष्ण पॉइंट पर यहां पर डील करना चाह रहा हूं इससे की सर यहां पर देखना बात को बात समझना आपका सर्कल का सेंटर है तू कमा फोर पर बिल्कुल सही बात सर आपकी लाइन की इक्वेशन क्या है ए = एमएक्स बिल्कुल सही बात है और इसकी रेडियस क्या है तू मतलब क्या मैं ज्यादा दिमाग लगाए बिना सर सीधे-सीधे यह कहूं क्या सीधे-सीधे यह कहूं क्या की सर 2 4 से इस सेंटर से तू कमा फोर से आई थिंक tenjent की इक्वेशन में निकलूं तो बात बन जाएगी क्या क्योंकि तू कमा फोर से अगर इस पॉइंट पर मैंने टांगें ड्रॉ की तू कमा फोर से अगर मैंने इस पॉइंट पर एक टांगें अगर ड्रॉप की मैन लो थोड़ी देर के लिए चलो छोड़ो ऐसे छोड़ो इसको ऐसे देख लो इतना घूमने की जरूरत ही नहीं है ये π/2 पता चला ना हान सर चल गया अब सुनना ध्यान से क्या मैं 2 कमा फोर से इस स्ट्रेट लाइन पर एक परपेंडिकुलर ड्रॉप कर सकता हूं क्या आप ये परपेंडिकुलर लेंथ अपने नॉलेज से निकल सकते हो सर मेरा नॉलेज यह कहता है suniyega ध्यान से ये कितना है तू ये कितना है तू और ये भी तू है मतलब ये साइड और ये साइड इक्वल लेंथ की थी मतलब ये एंगल और ये एंगल इक्वल होंगे मतलब अगर ये अल्फा होगा तो ये भी क्या होगा अल्फा suniyega कम के बाद अब प्लीज इस बात को सुनना ये अल्फा होगा तो ये भी अल्फा होगा और अगर ये आइसोसेलस राइट एंगल है तो सर मत बोलो उसको बोलो 45 डिग्री बोल सकता हूं क्या प्लीज इस बात को ध्यान से सुनना क्या मैं इस एंगल को या सिंगल को 45 डिग्री का सकता हूं क्या क्योंकि ये कितना आया आप देख का रहे हो ना ये वाला जो एंगल था कितना है फिर ये एंगल जो थीटा आप मैन के चल रहे द जिसे आप θ का रहे द ये आता है 90° ये कितना आता है भाई ये आता है 19 डिग्री अगर ये 90 डिग्री और ये तू और ये तू इक्वल है तो ये कितना हो जाता है 45 बहुत कम की बात मैं कहना चाह रहा हूं प्लीज इसे ध्यान से सुनना जैसे ये आपका पॉइंट है ए और यहां पे आपने इसे कहा एन तो मैं निकलने में एक्चुअली अभी फिलहाल इंटरेस्टेड हूं एम क्यों अभी बात करेंगे इस बारे में पर आप अपने नॉलेज से बताओ एम क्या होगा सर मेरे नॉलेज से जो एम होगा वो क्या होगा अगर मैं ट्रायंगल ए बी एम में बात करूं तो देखो ध्यान से बहुत ध्यान से मुझे हाइपोटेन्यूज पता है मुझे परपेंडिकुलर पता करना है क्योंकि एंगल है इसके सामने की साइड परपेंडिकुलर हाइपोटेन्यूज तो क्या मैं यहां पे सिन 45 से बात कर सकता हूं अगर मैं सिन 45 में डील करूं अगर मैं सिन 45 यानी क्या 1/√2 तो सिन 45 क्या होगा परपेंडिकुलर यानी क्या एम अपॉन हाइपोटेन्यूज यानी क्या तू कोई दिक्कत तो नहीं है आई थिंक तू यहां या तो तू और अंडर रूट तू ने तू को डिवाइड किया तो √2 तो ए एम की वैल्यू कितनी आएगी सर ए इनके वैल्यू आएगी √2 कोई दिक्कत तो नहीं है भाई अब अगर ए एम की वैल्यू मुझे पता चल गई है √2 जो की क्या है सर इस पॉइंट से इस लाइन पर ड्रा की गई परपेंडिकुलर लेंथ है ए = एमएक्स जैसे मैं लिख सकता हूं क्या एमएक्स - ए = 0 और इस स्ट्रेट लाइन की परपेंडिकुलर डिस्टेंस किस पॉइंट से सर 2 आई थिंक सर बस ये कर लो बात बन जाएगी और यहां पे एक और चीज आएगी जो की √1+m² डिनॉमिनेटर में आती है हमने वहां लिख दिया अगर एक छोटा सा कम नहीं करना चाहता हूं तो मैं दोनों तरफ स्क्वायर करना चाह रहा हूं सर दोनों तरफ स्क्वायर किया तो देखो भाई एक छोटा सा कम और कर लेते हैं तो आप जब इस तरफ स्क्वायर कर रहे हो मतलब मैं बस यह का रहा था आपसे की जब आप स्क्वायर करोगे तो यह भी स्क्वायर हो जाएगा और इसका भी आप स्क्वायर करोगे तो प्लीज इस बात को समझो जब आप ये स्क्वायर कर रहे हो तो यहां से 2m - 4 में से आप 2 कॉमन ले सकते हो क्या लेकिन तू का स्क्वायर यानी 4 बाहर आएगा तो अंदर क्या बचेगा अंदर बचेगा एन - 2 का होल स्क्वायर यहां पे देखना √2 का स्क्वायर कितना है तू और वैन प्लस एम का स्क्वायर का अंडर रूट कैंसिल तो ये बचा आई होप में बहुत कंफ्यूज नहीं कर रहा हूं आपको मेरा आपसे बस ये कहना है की यहां से तू इस फोर को कितना बना देगा तू आई होप आप एक क्वाड्रेटिक की तरफ बढ़ रहे हो देख का रहे हो क्या अगर इस क्वाड्रेटिक को सॉल्व किया तो देखना स्टूडेंट्स यहां से क्या निकल कर ए रहा है सर यहां से क्वाड्रेटिक को अगर सॉल्व किया तो बात सीधी सीधी सी है क्या यहां से अगर सॉल्व किया तो ये हो जाएगा m² हो जाएगा प्लस फोर माइंस कुछ तो मैंने गलत किया है बिल्कुल गलत किया है यह -4 एम होगा ना पता नहीं कहां दिमाग रहता है कभी-कभी m² + 4 - 2 से भी मल्टीप्लाई कर दें क्या कर दो भाई तो 2n² कितना हो जाएगा 8 और -4m कितना हो जाएगा -8 एम ये कितना हो जाएगा -8m और इस तरफ तो आपका क्या है m² + 1 कोई दिक्कत तो नहीं है भाई आई थिंक कोई परेशानी किसी भी स्टूडेंट को नहीं होनी चाहिए अब सर m² इधर ले तो 2m² - m² m² कितना होगा सर ये हो गया 7 सर एक छोटा सा कम कर लो स्प्लिटिंग डी मिडिल टर्म से बात कर लेते हैं स्प्लिटिंग का मिडिल टर्म पे बात कर लेते हैं बस एक बार मैं क्रॉस चेक कर रहा हूं की गड़बड़ तो हम नहीं कर रहे हैं तू कमा फोर से डिस्टेंस तू तू से तू गया तो बचा तू ये हो जाएगा m² - 8 + 7 बिल्कुल सही अब यहीं पर मैं स्प्लिटिंग का मिडिल टर्म अप्लाई कर रहा हूं आई होप ये बात समझ रहे हो तो कितना हो जाएगा सही हो जाएगा m² आई होप आप देख का रहे हो ये कितना हो जाएगा -7m - एम + 7 = 0 और अगर आप यहां से एम की वैल्यू निकले तो एक फैक्टर क्या बनेगा सर एक फैक्टर बनेगा एन - 7 आप सभी को दिख रहा है यहां से आप क्या ले लोग -1 कॉमन तो एक फैक्टर क्या बन जाएगा एन - 1 तो एम की एक वैल्यू तो आएगी क्या वैन और एक वैल्यू आएगी 7 एम की एक वैल्यू आएगी वैन और एक वैल्यू आएगी 7 और ऐसा अगर मैं देखूं तो आई थिंक मुझे कुछ कहने की जरूरत नहीं है आप सभी ने आंसर निकल लिया की एम की एक वैल्यू तो आई है वैन यानी ए = एमएक्स = 1 एक्स एक वैल्यू आई है 7 यानी ए = 7x आई होप कोई डाउट कोई परेशानी कोई दिक्कत नहीं है एक अच्छा बेहतरीन और मैं आज साहब हमारे क्वेश्चंस की कैटिगरी है लिंक्ड कंप्रीहेंशन टाइप बेस्ड क्वेश्चंस सर ये कैसे क्वेश्चंस याद ए रहा है आपको पैराग्राफ बेस्ड क्वेश्चंस जहां पर एक पैसेज है जिसमें काफी सारी इनफॉरमेशन है और उसे इनफॉरमेशन को रिड्यूस करके आपको कुछ क्वेश्चंस के आंसर निकलने होंगे तो पहले तो बस इस डाटा को ध्यान से पढ़ो अच्छे से समझो और मुझे ये बताओ की इस क्वेश्चन का आपको क्या मतलब समझ ए रहा है दो-तीन चीज है एक तो पहले तो आपको एक टांगें का दी गई है विथ इंजन ब दी गई है फिर एक सर्कल बोरा जा रहा है जिसका आई थिंक आप सेंटर हो रेडियस निकल सकते हो फिर सर एक पॉइंट पी भी है जो की एक कर्व पर है जो की हम जानते हैं ए = सिक्स है ठीक है सर फिर वो एक बात यह बोल दे रहा है ए और बी जो है वो एक ऐसे सर्कल पर लाइक करते हैं मतलब ए और बी जो है वो इस सर्कल पर लाइक करते हैं वो तो बिल्कुल तय सी बात है फिर वह एक नए फंक्शन की बात कर रहा है वह फंक्शन क्या है suniyega ध्यान से वो फंक्शन वो फंक्शन है जो की लोकस है किसका जो की लोकस है सेंटर ऑफ डी circumsir का ट्रायंगल पी अब तो जो आपने यहां से टैसेंट ड्रॉ की है इस सर्कल पर ए और बी पॉइंट पर जो मिल रही है तो वो जो पब सर ट्रायंगल बन रहा है उसका सरकमसर्किल आपको चाहिए ठीक है सर ये भी ढूंढ लेंगे अब इतना कुछ कर रहे हैं तो कुछ फॉलोइंग क्वेश्चंस आएंगे अभी उन पर हम बात करेंगे पर पहले ये कुछ बना के देखें क्या की सर बन क्या रहा है कोशिश करते हैं देखो पहले तो आप एक सर्कल से डील कर रहे हो सर एक्स माइंस 4 प्लस ए माइंस फाइव मतलब ये दो आपके रेडियस है फोर कमा 5 तो अगर इसमें सर्कल ड्रॉ करूं तो मैन लेते हैं हमारे पास एक सर्कल है जो की ऑफ कोर्स क्या है एक्स - 4 और ए -5 यानी इसके जो सेंटर के cardinate है वो है सर फोर कमा 5 सही बात है क्या बिल्कुल सही बात है इसकी रेडियस बताओ भाई सर रेडियस दिखी रही है तू है तू यूनिट से ठीक है सर बात करेंगे इस बारे में अब आपका एक कर्व है वही इसे इक्वल तू सिन एक्स ए = सिक्स जो कर्व है अब suniyega ध्यान से ऑफ कोर्स हम जान का रहे हैं हम समझ का रहे हैं ये फोर कमा फाइव है इससे तू यूनिट्स नीचे भी आया था यहां पर तू आएगा ये फोर कमा 5 है इससे 2 यूनिट्स नीचे भी आया तो ये आपकी यूनिट कितनी है फाइव तो थ्री आएगा तो सिन एक्स वाला कब तो नीचे से ही चला जाएगा ऐसे कैसे सर क्योंकि सिन एक्स की रेंज क्या है -17 तो ये तो कहां है 5 और 5 से 2 यूनिट्स नीचे भी आए तो ये कितना होगा बेसिकली थ्री तो मेरा आपसे ये कहना है की नीचे कहीं से कौन सा कर जा रहा होगा सिन एक्स वाला कर आप समझते हो की सिन एक्स वाला कर्व हमारा कैसा होता है ये आपका जो कर्व है ये सिन एक्स वाला का जा रहा होगा अब वह आपसे यह का पर आप एक पॉइंट ले रहे हो कोई सा भी लेट से रैंडम पॉइंट मैं कुछ भी मैन लेता हूं पी और इस पॉइंट से इस रैंडम पॉइंट से आपने ड्रॉ की है इस सर्कल पर टांगेंट्स आपने यहां इस पॉइंट से ड्रॉ की है क्या जनाब जो ये टैसेंट गई हैं भाई साहब ये टैसेंट ने इसको सर्कल को ए और बी पर टच किया है इंटेलिजेंस ने इस सर्कल को ए और बी पर टच किया है अब बात गौर से suniyega स्टूडेंट्स बात यह है ए और बी ऑफ कोर्स आपके सर्कल पर लाइक कर रहे हैं अब आपको कौन से सर कम सर्कल का पब का अगर आप ध्यान से देखो बहुत ध्यान से देखना स्टूडेंट्स यह जो पब है ये जो आपका ट्रायंगल बन रहा है कौन सा पब अगर मैं इन दोनों को कनेक्ट कर डन तो यह आपका जो ट्रायंगल बन रहा है अच्छा सर कम सर कल ढूंढ लूंगा मैं पर पहले मुझे ये बताओ अगर मैं इससे यहां ऐसे कनेक्ट कर डन और इसे यहां ऐसे कनेक्ट कर दो तो एक बीपी मतलब इसमें एक और पॉइंट ले ही लेता हूं एसबीपी आपको एक क्वॉड्रिलैटरल दिख रहा है क्या हम दिख तो रहा है एक और बात मुझे आप जवाब दो भाई की अगर मैं आपसे पूछूं तो आप मुझे बताओगे क्या की अगर मैं एक सर्कल ड्रॉ करूं इफ आई वर तू ड्रा अन सर्कल इन सच अन मैनर इफ आई वांट तू ड्रा अन सर्कल इन सच बेसिकली बी एंड सी टच इस बेसिकली बी एन सी सॉरी बी और ए अगर वो बी और ए से गुजरता है अगर मैं ऐसा सर्कल ड्रॉ करूं जो बी और ए से गुजरता है प्लीज इस बात को ध्यान से सुनिए बहुत कम की और बहुत कृष्ण बात है जो मैं कहना चाह रहा हूं और एक छोटा सा कम और करता हूं ध्यान से suniyega स्टूडेंट्स की मैं सी और पी को मिला देता हूं ये सब क्यों कर रहे हो अच्छा सारी बातें छोड़ो आप मुझे एक बात का जवाब दो आप प्लीज इस बात का जवाब दो बहुत ध्यान से देखकर सोच समझ के जवाब देना सर इस सर्कल के सेंटर से इसकी टांगें पर जब आपने रेडियस बनाई तो वो क्लीयरली एक परपेंडिकुलर होगा अब सुनना बहुत कम की बात बहुत कम की बात है और यही इस क्वेश्चन है सी पी बी आई रिपीट माय स्टेटमेंट कब जो है आपका ये जो सीपीवी है ये क्या एक राइट एंगल ट्रायंगल है क्या बिल्कुल है सर कब जो एक राइट एंगल ट्रायंगल है इसका एक सरकमसर्किल है इसका एक सरकमसर्किल है अच्छा बाय डी वे मैं बस ये कहना चाह रहा हूं आपसे की जो सर्कल हमने बनाया है वो सिमिट्रिकल से वो सिमेट्री caseence आप देख का रहे हो ना ये साइक्लिक क्वॉड्रिलैटरल है मतलब देखो आप ये 90 और यह 90 क्योंकि आपने रेडियस से परपेंडिकुलर ड्रॉ किया ना टेंशन होना है ये 90 है तो ये जो ए पर एंगल बन रहा है और बी पर एंगल बन रहा है इनका सैम 180 बन ही रहा है तो टोटल क्वॉड्रिलैटरल के चारों एंगल्स हम 360 होता है उसमें से 180 यहां चला गया तो बचे हुए दोनों 180 है तो मैं जानता हूं ये साइक्लिक क्वॉड्रिलैटरल है सॉरी साइक्लिक क्वॉड्रिलैटरल साइक्लिक क्वॉड्रिलैटरल का मतलब है की ये चारों पॉइंट्स encyclick होंगे मतलब ये चारों पॉइंट्स किसी सर्कल पर लाइक करेंगे मेरी बातों को बहुत गौर से सुनेगा स्टूडेंट्स बहुत बड़ा इंपॉर्टेंट सा राज छुपा है इस बात में चारों पॉइंट्स पी ए बी सी पीएसीबी अगर एक साइक्लिक क्वॉड्रिलैटरल है और अगर वो एक सर्कल पर लाइक करेंगे जैसा बनता हुआ मुझे दिख रहा है तो क्या मुझे दो-तीन बातों की आप जवाब दोगे जैसे की सर पहली बात तो ये की पी क्या है पी पर कोई रैंडम पॉइंट लेता हूं पी बेसिकली ये कर वे आपका कौन सा ए = सिन एक्स cardinate टी ले लूं तो ए cardinate क्या हो जाएगा सर वही कॉर्डिनेट हो जाएगा सिन टी तो क्या मुझे पीके आर्डिनेंस पता चल गया फोर कमा 5 अब एक बात का जवाब देना स्टूडेंट्स बहुत कृष्ण बात है बहुत कम की बात है यहीं से क्वेश्चन या तो शुरू हो जाएगा खत्म हो जाएगा अगर आप ध्यान से देख पाए तो मैं आपसे पूछना चाह रहा हूं ये जो ट्रायंगल है कब क्या ये एक राइट एंगल ट्रायंगल है हान सर क्यों नहीं दिख रहा है आपको ये रेडियस है ये टैसेंट है तो क्लियर 90 डिग्री एंगल है और अगर यह राइट एंगल ट्रायंगल है पी सी बी पी सी बी तो इसका सरकमसर्किल क्या है ये वही सरकमसर्किल है जो पीएसीबी का सर्कल है मतलब पीएसीबी का भी सरकमसर्किल और पीसीबी का circumsar एक ही होना है एनीवेज ये तो हम बात समझ ही का रहे हैं मुझे एक बात का जवाब दो अगर एक राइट एंगल ट्रायंगल है तो उसका सरकमसर्किल कहां होता है सरकम सेंटर कहां होता है इफ यू एवर हैपेंड तू एनकाउंटर राइट एंगल ट्रायंगल वेयर डोस इट्स सरकमसेंटर लाइव कैन यू तेल मी डेट सर कैसी बातें कर रहे हो कितना घुमा फिर के बात कर रहे हो आप को नहीं पता है क्या अगर एक डायमीटर है तो सर्कल के सरकम्फ्रेंसेस करता है है तो यह जो हाइपोटेन्यूज है इसका मिड पॉइंट ही तो आपका सरकमसेंटर होगा इस हाइपोटेन्यूज का मिड पॉइंट लेट्स से सर यहां पर है क्या मैं इसके मिड पॉइंट के कोऑर्डिनेट्स निकल सकता हूं क्यों नहीं निकल सकते तो इसके मिड पॉइंट के जो cardinate हमने निकलने हैं जैसे मैं का देता हूं देख लिए दी अगर अभी कोई cardinate और पॉइंट नहीं मिल रहा है तो इसके मिडिल पॉइंट के क्वांटिटी क्या होंगे सर वो होंगे टी + 4 / 2 तो दी के cardinate पर सवाल है सर अभी तो दी के कोऑर्डिनेट्स पर तो वो क्या होंगे सर पहला तो होगा क्या पहला तो होगा सर इसका क्या टी + 4 / 2 क्या आप मेरी बात समझ पाए और सर इसका व्हाइट ऑर्डिनेट क्या होगा सर वो होगा सिन टी + 5 / 2 तो ए कार्ड क्या हो जाएगा सिन टी + 5 / 2 और यही तो आपके ह कॉम के हैं मतलब अगर आप इसके सेंटर के कोऑर्डिनेट्स को कहते ह के क्योंकि हम इसी का तो लोकल ढूंढ रहे द तो यही तो आपके ह कॉम के हुए क्या आप मेरी बात समझ सन और सोच का रहे हैं तो मैं कहना चाह रहा हूं तो अगर सर आपसे सारी बातों की एक बात बड़ी सिंपल सी हम समझना चाहें तो आपने कहा की ह की जो वैल्यू आई है ह की जो वैल्यू आई है वो आप लिखते हो क्या टी प्लस फोर बाय तू जी हान मैं यही लिखता हूं और जो के की वैल्यू आई है वो क्या लिखते हो सर वो लिखते हो सिन टी प्लस फाइव बाय तू अब अगर मैं यहां से सोचना शुरू करूं अब अगर मैं यहां से सोचना शुरू करूं तो बस बात यहीं से निकल कर आएगी की आप टी को एलिमिनेट करते हुए टी को एलिमिनेट करते हुए कोई रिलेशन बना सकते हो क्योंकि हम इसी को तलाश में द जो पीएसीबी आपका जो क्वॉड्रिलैटरल है जिसमें वो ये पूछना चाह रहा है पब पब जो ट्रायंगल है इसका भी सरकमसर्किल यही है उसका भी सरकमसेंटर यही डी है तो इसे दी के लोग कस पर बात हो रही है इससे दी के भाई किस पर बात हो रही है लोकेश पर अब एक बड़ी सिंपल सी बात देखो सर अगर मैं यहां से चीज निकलना चाहूं तो क्या मैं यहां से टी की वैल्यू कुछ लिख सकता हूं देखो मैं टी की वैल्यू लिखना चाहता हूं क्या मैं टी की वैल्यू यहां से ले सकता हूं 2h-4 आई होप आपने से किसी को आपत्ति नहीं होगी अगर मैं तक वैल्यू और वह यहां रिप्लेस करता हूं तो के कितना ए जाएगा सर के ए जाएगा सिन ऑफ ध्यान से देखिएगा टी की वैल्यू कितनी 2 ह माइंस 4 प्लस 5 / 2 और लोकस पर बेस्ड क्वेश्चंस जब भी जीवन में आएंगे मैं तो एक ही कम करूंगा सर ह को एक्स से और के को ए से और प्लेस कर दूंगा तो कर दो तो जब आप फाइनली लो कस का इक्वेशन लिखते हो अभी क्वेश्चन तो पढ़ेंगे सर आपने लोकल बस ऐसे ही लिख लेते हैं बात करेंगे उसे बारे जैसे ही आप इसकी लोकल ढूंढ लेते हो किसकी उसे सरकम सेंटर की तो वो क्या ए जाती है सर ये ए जाता है तू के जैसे मैं क्या लिखूंगा 2y या ए को ए ही रखते हैं ए = सिन 2X - 4 ये कितना ए जाएगा सर ये ए जाएगा प्लस फाइव डिवाइडेड बाय तू अभी आपने कोई क्वेश्चन सॉल्व नहीं कर लिया जाना आप खुश मत hoiyega अभी तो इस क्वेश्चन की शुरुआत भर हुई है बस जिसमें आप ने उसके सरकम सेंटर का lokass फिगर आउट किया है आई थिंक चीजें आसान है बहुत टू तो नहीं है बड़े बेसिक से ऑब्जर्वेशन बेस्ड क्वेश्चन है जो वही सारी बातों पर बेस हैं जो सारी प्रॉपर्टीज हमने अभी तक पढ़ी है कहीं ना कहीं आर्डिनेंस सिस्टम यू नो स्ट्रेट लाइन पे ऑफ स्ट्रेट लाइन या सर्कल जैसे चैप्टर के अंदर ठीक है सर मैन लिया आपकी बात अब क्या करें अब मुझे जो ठीक लग रहा है जो मैं करना चाह रहा हूं ये की अब पढ़ लेते हैं क्वेश्चंस तो बात करते हैं पहले क्वेश्चन पर रेंज ऑफ ए = एफसी आप रेंज समझ रहे हो ना रेंज मतलब क्या मतलब वो जो लोकल्स है उसको उसने कहा वही इसे इक्वल तू एफसी जो लॉज ए रहा है आपका ध्यान से suniyega एफ ऑफ एक्स जो आपका फंक्शन ए रहा है वो ए रहा है साइन ऑफ 2X - 4 + 5 / 2 आपको रिलायंस भी हुआ की जो क्वेश्चन सर्कल पर बेस्ड था जिससे देखकर मैंने को-ऑर्डिनेट ज्यामिति लगाई स्ट्रेट लाइंस सरकमसर्किल काफी सारी चीज लाया वो करते-करते एक क्वेश्चन एक क्वेश्चन आपके कैलकुलस यानी फंक्शंस के रेंज और डोमेन का बन गया वेलकम तू आईआईटी जी मांस एंड एडवांस स्टूडेंट्स दिस इस व्हाट दिस एग्जाम डस्ट तू यू दिस इस एक्जेक्टली डोंट एवं रिलाइज एंड थिंग्स हैपन आउट ऑफ डी ब्लू थैंक्स जस्ट हैपन आउट ऑफ डी ब्लू ऑलवेज रिमेंबर इतने अच्छे मजेदार तरीके के क्वेश्चंस आपको मिलेंगे और आप बहुत एंजॉय करोगे इसलिए बहुत शांतिपूर्ण तरीके से हम एबीईएक्स एक एक स्टेप समझ रहे हैं ताकि ये बातें आपको याद रहे ये देखो इस तरीके से चीज बदल जाती है वो वंडर हो जाते हैं रेंज ऑफ वॉइस निकालनी है तो कैलकुलस जब हम पढ़ेंगे ना स्टूडेंट्स जब हम डिटेल में पढ़ेंगे तब हम बहुत अच्छे से बात करेंगे रेंज के बारे में पर अगर अभी फिलहाल में रेंज निकाला था तो मेरे जीवन के सच्चाई तो मुझे पता है सर साइन ऑफ एनीथिंग साइन ऑफ एनीथिंग व्हाट एवर डेट एनीथिंग इस गोइंग तू बी आई डोंट केयर अभी स्टेज जो भी एक्स की वैल्यू घूमते फिरते साइन जो होगा सर सिन एक्स फंक्शन है जो की प्रिडिक फंक्शन है और यह हमेशा -1 से वैन के बीच रहता है तो इस एक्सप्रेशन की मिनिमम वैल्यू मुझे -1 जब सिन एक्स - 1 होगा उसकी और इसकी मैक्सिमम वैल्यू जब साइन ऑफ दिस प्लस वैन होगा तब मिलेगी तो मिनिमम वैल्यू क्या होगी सोच कहता हूं इसकी मिनिमम वैल्यू -1 +5 - वनप्लस 5 मतलब कितना फोर फोर बाई तू कितना तू इसकी मैक्सिमम वैल्यू इसकी मैक्सिमम वैल्यू वैन वैन प्लस फाइव सिक्स और सिक्स बाई तू कितना थ्री तो सर ये जो होगा एक्सप्रेशन दो से थ्री तक की वैल्यूज ये क्या होगी सर इस फंक्शन की रेंज किसकी रेंज की कोई दिक्कत तो नहीं है कोई परेशानी तो नहीं है तो सर कोई ऐसा ऑप्शन जो मुझे मिल जाए तो कमा 3 जैसा तो मैं उसे मार्क कर दो खुश हो जाऊं आप सभी स्टूडेंट्स को क्या सबसे पहले तो ये ऑप्शन दी समझ आया की क्यों हमने मार्क किया किसी भी स्टूडेंट को कोई आपत्ति है तो पूछ ले आई विल reclairify री एक्सप्लेन डी होल आई थिंक सर आसानी से अब क्या अब आते हैं अगले क्वेश्चन पर वह पूछ रहा है इस फंक्शन का पीरियड किस फंक्शन का पीरियड ए = एफसी का बात करेंगे जब हम सेट रिलेशन एंड फंक्शंस चैप्टर पढ़ेंगे तब हम पीरियड के बारे में बात करेंगे की किसी फंक्शन का पीरियड क्या होता है अभी फिलहाल बस आपकी सहूलियत के मैं आपसे ये कहना चाह रहा हूं की अगर मैन लो मुझे जीवन में कभी कोई फंक्शन मिले कभी कोई फंक्शन मिले और मुझसे कोई कहे की ये जो फंक्शन है ये प्रिडिक फंक्शन है प्रिडिक समझते हो ना स्टूडेंट जो फंक्शन अपने आप को बार-बार रिपीट करता है अभी ऐसे ही समझ लो और अगर ये प्रिडिक फंक्शन में विथ पीरियड टी अगर इस प्रिडिक फंक्शन की periadicity है टी की ये टी यूनिट डिस्टेंस के बाद वापस अपना से पार्ट रिपीट करता है तो मैं ये कहूंगा की आप से कभी भी कोई कहे ना एफ ऑफ एक्स जहां पे ये जीरो नहीं होना चाहिए जहां पर एक यानी होना चाहिए जीरो अगर कोई कहे एफ ऑफ एक्स के बारे में आपका क्या ओपिनियन है तो मैं कहूंगा भाई एफ ऑफ ए एक्स बी एक प्रिडिक फंक्शन होगा मैं नहीं संकोच बिना किसी hectation के कहूंगा की एफ ऑफ ए एक्स बी एक रिकॉर्डिंग फंक्शन का बशर्ते इस चीज के साथ की एफ ऑफ एक्स जो फंक्शन है उसका पीरियड जो होगा वो होगा टी / मोड ए ऑफ एक्स जो फंक्शन था जिसका पीरियड हुआ करता था टी एफ ए एक्स बी एक प्रिडिक फंक्शन ही रहेगा जिसका पीरियड हो जाएगा पी / मोड ए क्यों हो रहा है कैसे हो रहा है बात करेंगे डिटेल डिस्कशन जब हम सेट्स रिलेशन और फंक्शंस पढ़ेंगे बट अभी बस इस बात का याद रखिएगा अब अगर मैं वापस यहां पर आऊं तो सर आप खुद देख का रहे हो ये लिखा हुआ है सिन 2X - 4 और इसमें प्लस फाइव तू करने से कोई फर्क नहीं पड़ता ये भी याद रखना तो ये जो एफ ऑफ एक्स फंक्शन है इसमें जो प्लस है Y2 कर दिया इससे कोई फर्क नहीं पड़ता है अभी भी प्रिडिक है मैं जानता हूं सर साइन एक्स एक पर और एक फंक्शन होता है जिसका पीरियड होता है 2π सिन एक्स अपने आप को 2π के बाद रिपीट करता है ये फंक्शन ऐसा चलता है ये बात जो है ये जीरो से 2π तक की है फिर वापस ये 4π - 2π से 0 ऐसे रिपीट होती है तो सिन एक्स एक पीरियोटिक फंक्शन जिसका पीरियड होता है 2π सुनेगा ध्यान से मैं आपसे कहना चाह रहा हूं की सिन एक्स एक पर एक फंक्शन है जिसका पीरियड है 2π क्या आप मुझे सिन 2X का पीरियड बता सकते हो सर क्यों निकल रहे हो आपको जो 2X - 4 + 5 / 2 का भी रेट पूछा है 1 सेकंड क्या आप मुझे सिन 2X का पीरियड बता सकते हो सर अगर साइन एक्स का पीरियड है 2π तो सिन 2X का पीरियड क्या होगा इसका पीरियड डिवाइडेड बाय इसकी पॉजिटिव वैल्यू तो 2π/2 और 2π/2 कितना होगा फाइव तो सिन 2X का पीरियड होगा बाय सिन 2X का अगर पीरियड हुआ पाए तो हम ये बात वहीं डिस्कस करेंगे अभी बस इस ए फैक्ट समझ लीजिए की अब 2X में कुछ ऐड करो या सब्सट्रैक्ट इसका पीरियड वही रहेगा इस फंक्शन में कुछ ऐड करो या डिवाइड इसका पीरियड वही रहेगा तो जो सिन 2X का पीरियड है वही उसे फंक्शन का पीरियड रहेगा क्यों इस बारे में सर्च रिलेशन ऑफ फंक्शन बात कर लेंगे तो मैं क्या कहना चाह रहा हूं भाई मैं कहना चाह रहा हूं की इस फंक्शन का जो पीरियड निकल कर आएगा वो कितना होगा सर वो होगा क्या आपने अगले क्वेश्चन का भी आंसर मार्क कर दिया आई थिंक सर जब पूछा गया की इस फंक्शन का पीरियड कितना है तो मैं मार्क करूंगा ऑप्शन सी बहुत अच्छे से इन सारी चीजों को देख लीजिए बहुत आसानी से बहुत आराम से हर एक बात को समझ लीजिए मैं नहीं चाहूंगा की आप कहीं भी किसी भी बात को लेकर कंफ्यूज हो रहे हैं अगला सवाल विच ऑफ डी फॉलोइंग इसे ट्रू विच ऑफ डी फॉलोइंग इसे ट्रू एफ ऑफ एक्स 4 रियल रूट्स एफ ऑफ एक्स वैन रियल रूट है ना नॉन ऑफ डीज या एफ ऑफ इन्वर्स एक्स की रेंज जो है वह कुछ बता रहा है मैं पहले तो ऑप्शन ए और ऑप्शन बी को चेक कर लेता हूं हम पहले तो चेक कर लेते हैं ऑप्शन ए और ऑप्शन बी है ना एक को चेक कर लेंगे तो बात पता चल ही जाएगी एक बात बताओ रियल रूट्स का मतलब क्या है ये आपका एफ ऑफ एक्स है एफ ऑफ एक्स के सिन 2X - 4 + 5 / 2 एफ ऑफ एक्स के सर वो है साइन ऑफ अब वैसे तो ये पढ़ते ही भर आपको क्लिक हो जाना चाहिए बट फिर भी अगर नहीं हो रहा है तो मैं पूछूंगा आपसे यही लिखा है ना साइन ऑफ 2X - 4 + 5 / 2 एक बात बताओ रूट्स का मतलब क्या रूट्स का मतलब ये एक एक्स की कितनी वैल्यूज पर जीरो होता है कितने रूट्स हैं का मतलब है एक्स की कितनी वैल्यू पर ये जीरो होता है तो वो पूछना चाह रहा है की सर आपका जो एक्सप्रेशन है साइन तू एक्स माइंस फोर ये तू तो उधर गया इस इक्वल तू व्हाट माइंस फाइव तो आप खुद सोच के बताओ सर सिन ऑफ 2X - 4 - 5 के इक्वल कब होगा सर कैसी बात कर रहे हो आपको पता नहीं है क्या बड़ी बेसिक से साधारण से सिंपल सी बात है आप किसी भी वैल्यू का जब साइन लेते हैं आप जब किसी भी वैल्यू का साइन लेते हैं तो वो वैल्यू तो माइंस वैन से वैन के बीच होती है है तो इसकी मिनिमम वैल्यू तो -1 है ना तो राइट हैंड साइड से आपका क्या रिलेशन है बस में यह समझाना चाह रहा हूं की यह है -17 के बीच और ये माइंस फाइव कितना पीछे जा रहा है तो क्लीयरली कोई स्कोप ही नहीं है दोनों की इक्वल होने का तो बात खत्म हो गई सर इक्वल एक रूट दो रूट चार रूट छोड़ो एक भी रूट नहीं आएगा मतलब ऐसी कोई भी वैल्यू नहीं है जहां पर ये एक्सप्रेशन आपका जीरो हो बड़ी बेसिक सी बात है ठीक है सर ये बात आपकी मैन ली तो अभी इसमें एक और बात थी क्या बात थी सर इसमें एक और बात ये भी चलो ठीक है शायद ऑप्शन सी ट्रू ऑप्शन सी निकलती हैं अब उसने देखो आपको cardinate से सेट्स रिलेशन फंक्शन में इन्वर्स फंक्शन तक लेके चला गया वो रेंज और डोमेन तो छोड़ो आप है ना प्रायोरिटी फंक्शन भी चेक कर लिया आप समझ रहे हो इस तरीके से एक ही क्वेश्चन के थ्रू कितने सारे चैप्टर का कितना डिटेल नॉलेज वो आपका चेक कर रहा होगा ये यही एग्जाम है तो इस क्वेश्चन में क्या बोल रहा है वो रेंज ऑफ एक्स तो एफ इन्वर्स ऑफ एक्स निकलने हैं ठीक है सर आपका फिक्स किया है तो ये बताओ सर यह था मेरे ख्याल से एफ ऑफ एक्स तो एफ-16 निकलने की प्रक्रिया बड़ी आसान सी है आप क्या लिखते हो आप लिखते हो ए = जो आपका एफ ऑफ एक्स है जो की है सिन 2X - 4 + 5 / 2 अब आप ये करोगे की अभी जो ए एक्स की टर्म्स में एक्सप्रेस है आप कोशिश करोगे एक्स को ए की टर्म्स में एक्सप्रेस करने का तो पहले तू को इधर ले तो ये हो जाएगा 2y फिर फाइव को इधर ले तो हो जाएगा -5 = व्हाट इसे इक्वल तू सिन ऑफ 2X - 4 अब मैं सिन इन्वर्स ले लेता हूं तो ये हो जाएगा sin⁻¹ ऑफ क्या 2y -5 है ना फिर क्या बचेगा इस तरह साइन इन्वर्स सिन इनफेक्ट मैंने 5 कर देगा तो ये हो जाएगा 2X - 4 तो थोड़ा और सिंपलीफिकेशन के भी गुंजाइश है -4 इधर ले तो 2X कितना हो जाएगा सर 2X हो जाएगा आपका कितना 2X आपका हो जाएगा -4 इधर आया तो कितना सिन इन्वर्स ऑफ 2y - 5 + 4 एक छोटा सा कम और कर लीजिए और यहां मतलब तू को हटाए और यहां ले आई तो ये है आपका एक्स अब इन्वर्स फंक्शंस की परिभाषाएं ये कहती हैं की अब जब आपने फाइनली एक्स को ए के टर्म्स में एक्सप्रेस कर ही दिया है तो आप एक्स की जगह लिख दीजिए f⁻¹ एक्स और ए को रिप्लेस कर दो एक्स से तो यहां मैं लिखूंगा एक्स की जगह क्या f⁻¹ एक्स और ए को रिप्लेस करूंगा एक्स से तो एफ-1 एक्स की परिभाषा होगी साइन इन्वर्स ऑफ 2X - 5 प्लस 4 डिवाइडेड बाय व्हाट डिवाइडेड बाय तू सो दिस इस हो यू आर गोइंग तू राइट निकलने का तरीका जो हमारा होगा वो बड़ा आसान सा यही होगा की सर पहले तो आप ये पता करो की ये सिन इन्वर्स ऑफ इसकी रेंज क्या होती है ये तो बड़ा आसान है सर कैसे आसान है बात समझो सिन इन्वर्स ऑफ 2X - 5 की रेंज अगर मैं बात करूं तो सिन इन्वर्स ऑफ एनीथिंग फिर चाहे वो 2X - 5 ही क्यों ना हो ये हमेशा सर आपके माइंस π/2 से π/2 के बीच होता है कोई दिक्कत तो नहीं सर अब एक कम करते हैं दोनों तरफ तीनों तरफ फोर ऐड कर देते हैं तो मैंने यहां भी फोर ऐड किया जाएगा क्या माइंस पाई बाय तू प्लस फोर यहां पर कितना हो जाएगा पाई / 2 + 4 एक छोटा सा कम और करते हैं सर तीनों तरफ तू से डिवाइड कर देते हैं ऐसा कुछ पहुंचना चाह रहा तो जैसे ही यहां तू से डिवाइड किया तो ये कितना हो जाएगा 2 / √2 2 कितना 4 यहां तू से डिवाइड किया तो ये कितना हो जाएगा भाई ये हो जाएगा 2 आई होप आप समझ का रहे हो यहां तू से डिवाइड किया तो ये हो जाएगा तू यहां पर ये पूरा एक्सप्रेशन तू से डिवाइड हो जाएगा और से बात यहां होगी हो जाएगा π/4 + 2 तो कितना हो जाएगा π/4 + 2 तो क्लियर सी बात है सर इसकी जो रेंज है वो है -π/4 + 2 से +5/4 + 2 तक ऐसा कोई ऑप्शन मुझे दिख रहा है क्या देख लेते हैं भाई इसकी जो रेंज है वो -π/4 + 2 से लेकर π/4+2 तक है बिल्कुल सही बात और इसके लिए नान ऑफ डीज पर जाने के बजाय मैं मार्क कर दूंगा ऑप्शन सी बिल्कुल टच करता है कम की बातें इससे बहुत कुछ पता चलेगा आपको ठीक है फर्स्ट क्वाड्रेंट इसने कुछ और मजेदार बात का दी मेरा कहना है बस इतना देखकर इतना पढ़कर आप इसे थोड़ा ड्रा करने की कोशिश करो सुनेगा ध्यान से आपके पास क्वाड्रांट्स है है ना तो आपके पास ये रहा क्या भाई ये रहा आपका सुरवी एक्सेस ये रहा सर आपका क्या ए और ऐसा ही यही कही क्या होगा सर यहीं कहीं होगा आपका लेट्स से ये एक्स एक्सिस देयर इस अन सर्कल जो एक्स एक्सेस को टच करता है एक सर्कल है सर जो की एक्स-एक्सिस को टच करता है ठीक है सर मैंने की सर्कल बनाया जो की एक्स-एक्सिस को टच कर रहा था अब सुनना ध्यान से यह सर्कल की रेडियस है वैन यूनिट और ये एक्स एक्सिस को पॉइंट ए पर टच करता है कम की बातें सुनना अगर ये जो पॉइंट है ये है ए और इस सर्कल की रेडियस है वैन यूनिट तो कैन आई से इसका अगर ये जो सेंटर है यहां से इसकी जो एक ही डिस्टेंस है वो 1 यूनिट होगी तो इसके सेंटर का एक्स कोऑर्डिनेट्स मुझे नहीं पता पर क्या ए cardinate मुझे वैन पता है आप मेरी बातें समझ का रहे हैं क्या आई होप आप यह बात समझ का रहे हो इससे मैं यहीं पर रहने देता हूं बस मैं तो आपसे ये कहता हूं की यहां से जब मैंने कोई परपेंडिकुलर ड्रॉप किया यहां से जब मैंने कोई परपेंडिकुलर ड्रॉप किया तो उसकी लेंथ कितनी होगी भाई वैन आई होप ये बात आप हमेशा याद रखेंगे इस क्वेश्चन के कॉन्टेक्स्ट में तो बिल्कुल हान तो यहां से एक परपेंडिकुलर ड्रॉप किया जा रहा है भाई और इस परपेंडिकुलर के लेंथ ऑफ कोर्स कितनी होनी चाहिए वैन कोई तकलीफ यहां तक तो कोई दिक्कत नहीं है भाई ये परपेंडिकुलर इसकी लेंथ है वैन अब suniyega ध्यान से ओरिजिन ओ तू डी सर्कल ऑफ कोर्स बनाई गई है ओरिजिन से जो मेरे ख्याल से कुछ ऐसी जा रही होगी और यहां पर भी अगर आप से मैं बात करूं तो आप कहोगे की ये जो परपेंडिकुलर लेंथ है यह भी वापस से कितने डिस्टेंस की है यह भी वैन लेंथ जैसा जैसा वह का रहा है वैसा वैसा बनाते जा रहे हैं सर और tagent जो है ओरिजिन है वो सर्कल को टी पर टच करती है वो टी पर टच करती है तो सर ये जहां टच करती है वो पॉइंट है टी एक और बात सुनना एक और बात वो का रहा है सर एक पॉइंट पी है इस तरीके से की ट्रायंगल ओ आप ट्रायंगल ओ ए पी एक राइट एंगल डी ट्रायंगल है ठीक है सर बात करेंगे एक पॉइंट पी हमें लेना है इस तरीके से की ये जो है आपका ओरिजिन ये जो है आपका ए और आपका कहीं ना कहीं आपको मिलेगा पी वो एक राइट एंगल ट्रायंगल होना चाहिए अब मैं पी कहां लूं अब मैं पी किस तरीके से लूं मेरा ये सवाल है मेरा कहना है अपनी सहूलियत के लिए अपने कन्वीनियंस के लिए मुझे जहां पी लेना ठीक लग रहा है वो ऐसा है की अगर मैं यहां से थोड़ा आगे बधाई अगर मैं यहां से इसे थोड़ा आगे बधाई तो लिसन मैंने इसे थोड़ा आगे बढ़ाया तो क्या मैं इसे यहां मिला डन तो क्या हो जाएगा फायदा से तो मुझे यहां पर एक पॉइंट पी मानने में साहू रहेगी और आसानी रहेगी और मैंने तो सिर्फ वही किया जो उसने कहा उसने कहा आपका जो ट्रायंगल आप है पी इस तरीके से लेना की आपका ट्रायंगल आप कैसा बन जाए एक राइट एंगल ट्रायंगल तो हान सर मैंने आपकी बात मैन ली जैसा जैसा क्वेश्चन मैं कहा मैंने तो वैसा किया इसके अलावा इससे ज्यादा मैंने दिमाग नहीं लगाया ये आपका एक पॉइंट ऑफ कॉन्टैक्ट था और ये आपका पॉइंट ऑफ फंटा है बहुत अच्छी बात है तो यहां तक चीज दिख रही है क्या पहले तो इस पार्ट को ध्यान से देख लो बस एक छोटी सी बात हम मिस कर गए राइट एंगल डी ट्रायंगल है ना तो ए वर्टेक्स पर वो राइट एंगल है तो ये आपकी ए वर्टेक्स यहां पर राइट एंगल है और आप जो ट्रायंगल है उसकी पेरिमीटर है 8 यूनिट्स आई रिपीट माय स्टेटमेंट ये जो ट्रायंगल ओ आप है इसकी पेरिमीटर है 8 यूनिट्स आई होप आप ये बात याद रखोगे अब यहां तक अगर चीज समझ ए गई है तो मूव करते हैं क्वेश्चंस पर पहला क्वेश्चन पढ़िए और अब आप सोचिए की चीज कैसे निकालनी है अब आपको चीज सोचनी हो सबसे पहला सवाल है की पी के की लेंथ क्या होगी मेरा कहना है पहली बात तो के क्या है हम मिस कर रहे हैं क्या सर पीके बड़ा सिंपल सा टर्म है जो के आपका क्या था सर्कल का सेंटर तो ये रहा के आई थिंक ये तो वो इस पीके की लेंथ पूछ रहा है मैं अभी थोड़ी देर के लिए अपनी सहूलियत के लिए क्या पीके की लेंथ को एक्स मैन सकता हूं सर यह जो पीके की लेंथ है क्या मैं इसे थोड़ी देर के लिए एक्स का सकता हूं आई थिंक कोई बुराई नहीं है अगर इसे मैंने कहा अगर इसे मैंने कहा एक्स तो थोड़ी देर के लिए सोच के देखो बड़ी सिंपल सी बात आपसे मैं पूछ रहा हूं सीटीपी राइट एंगल ट्रायंगल है हान सर राइट एंगल ट्रायंगल है प्लीज ध्यान से सुनना इस राइट एंगल ट्रायंगल में क्या आप मुझे बता सकते हो की टीपी पीके और क्यूट पर मैं अगर मैं हाइपोटेन्यूज मतलब राइट एंगल ट्रायंगल में पाइथागोरस थ्योरम लगाऊं तो देखो भाई वैन मुझे पता है ये मैंने एक्स मैन लिया है तो क्या पॉइंट निकल सकता हूं ये 90 डिग्री है ये हाइपोटेन्यूज है तो हाइपोटेन्यूज के स्क्वायर में से अगर ये बेस मैं सब्सट्रैक्ट कर डन स्क्वायर तो x² - 1 और इसका अगर मैं अंडर रूट ले लूं तो क्या मुझे पॉइंट मिल जाए तो पॉइंट की जो लेंथ होगी वो कितनी होगी सर वो होगी √x² - 1 अगर मुझे पॉइंट की लेंथ मिल गई है अब मैं आपको यह सवाल दे रहा हूं की अगर आप टी ओ और ओए पता कर ले बहुत हद तक मैंने हेल्प किया स्टूडेंट्स की आप बता दो अगर आप कर लो कुछ कुछ आपके सामने है तो इस ट्रायंगल की सरकम्फ्रेंसेस जो की आपको 8 दी गई है उससे आप एक्स की वैल्यू निकल के पीके बहुत इजीली निकल सकते हो सर मैं सारी चीज निकल लूंगा सब कुछ कर लूंगा पर मुझे ये बात समझ नहीं ए रही है आप कैसे जाओगे वहां तक जो बात हमसे ओटी और ओए को लेकर पूछे जा रही है ओटी और ओए कैसे निकलूं मेरा आपसे कहना है आप इन दो ट्रायंगल को ध्यान से देखो बहुत ध्यान से देखना है इन दो ट्रायंगल इसको मैं लिख देता हूं ताकि आप कंफ्यूज ना हो आप पहले तो देखो टी के पी ट्रायंगल को अगर मैं आपसे पूछूं आप पहले तो देखो टीसीपी ट्रायंगल को और इसी तरीके से अगर मैं एक ट्रायंगल आपको डन तो आप देखो ट्रायंगल आप को सर ऐसा क्यों का रहे हो प्लीज इस बात को सुनना ध्यान से बहुत ध्यान से सुनना अच्छा एक बात बताओ स्टूडेंट्स ये जो टी के पी ट्रायंगल है और ये एंगल है बिल्कुल सही बात है सर टीपीक्यू में यह एंगल पी है और एंगल आओ मतलब आप ट्रायंगल में भी एंगल पी है आप मेरी बात समझ का रहे हो मतलब अगर ये एंगल पी इस टीपी में भी है और ओपा में भी है तो पी एंगल्स से एक 90° एंगल से तो केन आई से ये जो बचा हुआ एंगल है के ये जो बचा हुआ एंगल है के ये सर क्लीयरली ये सर क्लीयरली इस एंगल ओ के इक्वल होगा मतलब क्या सर मतलब आप समझो ना एंगल एंगल क्या कहना चाह रहा हूं एंगल एंगल सिमिलरिटी यानी ये दोनों ट्रायंगल टीसीपी और आपके कौन से दो ट्रायंगल आप ये दोनों ट्रायंगल आपके हो गए क्या भाई ये दोनों ट्रायंगल से आपके हो गए सिमिलर हो गए क्या अब एक बात सोच के देखो अब आप एक बात सोच के देखो मैं यहां से ले रहा हूं इस ट्रायंगल में कौन से ट्रायंगल में आपका ओपा ट्रायंगल में ले रहा हूं ओए आपको दिख रहा है क्या बिल्कुल ले लीजिए सर यहां से आपने लिया ओके और इसी ट्रायंगल में मैं एक और साइड ले रहा हूं जैसे मैं का रहा हूं आप मैंने क्या लिया एक साइड मैंने ले ली है दोनों साइड मैंने क्या किया इस राइट एंगल ट्रायंगल में baseekly अगर मैं इसे एक एंगल मैन हूं तो मैंने बेस और हाइट मैंने क्या लिया है बेस और हाइट कोई दिक्कत तो नहीं सर इसी तरीके से अच्छा ये इसके लिए बेस और किसके सी बात सोचे देखो ये वाला जो एंगल है ये वाला जो एंगल है इसका बेस और इसकी हाइट मैंने ली है तो सिमिलरली इस वाले एंगल के लिए इसका बेस क्या होगा तक और इसकी हाइट क्या होगी पीके कोई दिक्कत नहीं तो मैं क्या क्या ले लूंगा सर मैं यहां से ले लूंगा ओए के अपॉन में ओए इसका बेस था तो इस वाले एंगल में ये बेस क्या होगा तक क्या ये बात आपको समझ ए रही है पापी बात समझ का रहे हो ये क्या हो जाएगा सर ये हो जाएगा तक क्या सारे स्टूडेंट्स यहां तक कंफर्टेबल है सिमिलरली बड़ी ही बेसिक सी बात है सर यहां पर हम रियली सॉरी मैंने पीके बोल दिया था क्या जब यहां पे परपेंडिकुलर था आपका आप तो इस वाले ट्रायंगल में इसके अपोजिट साइड कौन है पॉइंट तो आपका ये हो जाएगा पॉइंट कोई तकलीफ कोई डाउट किसी भी स्टूडेंट को किसी भी स्टूडेंट को कोई परेशानी अब गौर से देखो स्टूडेंट्स बहुत कम की बातें हम करेंगे जिन पर आपको डिस्कशन को ध्यान से देखना है बहुत ध्यान से देखना इस वाले पार्ट को सर आपको ओए पता है क्या अभी तो नहीं पता सर आपको टी के पता है क्या सर मुझे शायद तक पता है शायद मुझे टी क्यों पता है तो तक कितना हुआ वैन तो मैं ओए को लिखना चाह रहा हूं अब ओए को मैं क्या लिखूंगा स्टूडेंट्स ध्यान से देखो आप ओए को लिखोगे आपकी बड़ी बेसिक सी बात आप सर आपको आप पता है क्या देखो भाई आप क्या है a³ + पीके तो कितनी ए जाएगी एक्स + 1 तो आप कितनी हो जाएगी सर आप हो जाएगा आपका एक्स + 1 और सर इसी के साथ-साथ में आपसे पॉइंट पूछूं तो आपको पॉइंट पता है क्या आई थिंक मैंने पॉइंट निकाला था पॉइंट कितना है सर सर पॉइंट है अंडर रूट ओवर पॉइंट है अंडररूट ओवर x² - 1 तो अगर आप ध्यान से देखो तो हम ओए तक तो पहुंच गए क्या हम ओए तक पहुंच गए भाई अब एक बात का जवाब आप मुझे दे दो अगर ये एक पॉइंट है एक्सटर्नल जहां से मैंने सर्कल पर टांगें ड्रॉ की है तो क्या मैं ऐसा नहीं कहूंगा की ओए और ओ टी इक्वल लेंथ के होंगे किसी एक्सटर्नल पॉइंट से ड्रॉ की गई tenjins की लेंथ किसी सर्कल पर इक्वल होती हैं याद है क्या तो सर अगर यही ओए है तो मैं ज्यादा दिमाग लगाए बिना क्या इसे ही ओटी का दूंगा इतना क्यों कर रहे हो सर ये सब करने से हो क्या रहा है बस थोड़ा सा पेशेंस रखें अब अगर मैं आपसे बात करूं अब अगर मैं आपसे बात करूं किस बारे में इस ट्रायंगल आप के पैरामीटर के पेरिमीटर की तो ट्रायंगल आप का पैरामीटर मेरे नॉलेज से क्या होगा सुनना ओटी प्लस टी पी प्लस कोई दिक्कत तो नहीं है तो पैसे तो क्या लिख लिया क्या आप जो की क्या है 1 + एक्स तो यहां पर 1 + एक्स और हमने ऐड किया फिर क्या ऐड किया सर फिर इसमें बस एक छोटा सा बच रहा है पॉइंट जो की क्या है अंडर रूट ओवर x² - 1 जो की है बस आप इसे सॉल्व कर लीजिए आई थिंक आप एक्स की वैल्यू निकलोगे जैसे क्वेश्चन काफी सारा आपका सॉल्व हो जाएगा मैं सोच रहा हूं इसको सॉल्व करने का कोई आसान तरीका क्योंकि यह बन रहा है बहुत खतरनाक सा एक्सप्रेशन सर कैसे करें अब कुछ तो सोचना पड़ेगा मेरे दिमाग में जो थॉट ए रहा है वो ये कम करते हैं मैं अंडर रूट x² - 1 से पूरे एक्सप्रेशन को मल्टीप्लाई करता हूं एलएस और rhsas में फायदा क्या होगा उससे सर नहीं बता पर करते हैं कुछ तो होगा तो देखते हैं अंडर रूट x² - 1 से मैंने दोनों साइड एलएस और रहा को मल्टीप्लाई किया देखो यहां पे मल्टीप्लाई करेंगे तो क्या होगा यहां पर मल्टीप्लाई करेंगे तो बचेगा 2X + 1 तू अंदर मल्टीप्लाई कर दिया तो 2X + 2 गड़बड़ आएगी यहां पर की वनप्लस एक्स में मल्टीप्लाई हो जाएगा अंडर रूट ओवर x² - 1 जिसकी हम बात करेंगे इसके साथ-साथ एक छोटी सी चीज हो रही होगी की जब यहां मल्टीप्लाई होगा तो ये बन जाएगा अंडर रूट्स एंड रूट कैंसिल और x² - 1 तो यहां क्या बचेगा सर आपका ये बचेगा सिर्फ x² - 1 जैसे हमने लिख लेता हूं यहां x² - 1 है ना और एक छोटी सी बात सिर्फ लेफ्ट हैंड साइड ही नहीं सर राइट हैंड साइड पर भी आपको मल्टीप्लाई करना होगा बिल्कुल तो राइट हैंड साइड पर जब मल्टीप्लाई होगा तो क्या होगा आते टाइम्स अंडर रूट ओवर x² - 1 ये क्या हो जाएगा सर ये हो जाएगा 8 टाइम्स 8 टाइम्स अंडर रूट ओवर एक्स स्क्वायर माइंस कोई दिक्कत अब क्या करें तो क्या मिलेगा देखो भाई थोड़ा ट्रिक ही तो हो रहा है पर मेरे ख्याल से इससे आसान तरीका मुझे तो नजर नहीं ए रहा है आपको आए तो आप बताना ये क्या बन जा रहे हैं सर ये बन रहा है x² है ना प्लस तू एक्स जिसका हम कुछ नहीं कर सकते और 2 - 1 + 1 और जब इससे वहां शिफ्ट किया इसे वहां शिफ्ट किया तो देखो यहां से वनप्लस एक्स यानी कितना मुझे मिल जाएगा अंडर रूट ओवर x² - 1 कोई दिक्कत तो नहीं है अब क्या करेंगे सर देखो भाई बड़ी सिंपल सी बात हमने ऐसे उठा के वहां रखा मैंने कुछ गड़बड़ किया हान गड़बड़ किया सर आपका ये एक्स + 9 नहीं रहेगा जब इसको उठा के हमारे रखोगे तो 8 - 1 ये 7 हो जाएगा और ये माइंस एक्स हो जाएगा तो टेक्निकल ये हो जाएगा 7 - एक्स कोई दिक्कत तो नहीं अब क्या करने का मैन हो रहा है सर आपका अब मैं कहूंगा की सर एक कम कर लेते हैं आई होप आपको दिख रहा है की मैं सोच रहा हूं दोनों तरफ स्क्वायर करें ऑफ कोर्स इस तरह स्क्वायर करेंगे तो प्रॉब्लम और बढ़ जाएगी क्योंकि पावर फोर तक जाएगा और ये भी बढ़ेगा पर मेरे ख्याल से और कोई आसान तरीका दिख भी नहीं रहा है ना तो सिंपलीफाई करने के लिए दोनों साइड का स्क्वायर करते हैं दोनों साइड्स का स्क्वायर किया तो सर यहां पर जब स्क्वायर करोगे तो ये हो जाएगा और यहां पर स्क्वायर करने से या अंडर रूट है जाएगा है ना कोई दिक्कत तो नहीं है कुछ ऐसी चीज हमें दिखेंगे अब इसे सिंपलीफाई करते हैं देखो भाई ध्यान से ये है ए + बी + सी का होल स्क्वायर ये होता है a² + b² + c² + 2ab + 2bc+2z तो a² x² का स्क्वायर एक्स तू दी पावर 4 2X का स्क्वायर कितना हो जाएगा सर ये हो जाएगा 4x² वैन का स्क्वायर लेकिन अगर आप राइट हैंड साइड को देखें तो राइट हैंड साइड पर भी काफी सारी चीज होनी है जैसे ध्यान से देखो सर 7 - एक्स का जब आप स्क्वायर एक्सपेंड करोगे तो वो कितना बनेगा ध्यान से देखो सर 7 माइंस एक्स का स्क्वायर जब आप एक्सपेंड करोगे तो ये बनेगा सेवन का स्क्वायर 49 - एक्स यानी क्या प्लस एक्स स्क्वायर 7 2 14 तो ये हो जाएगा -14x ये कितना हो जाएगा -14x और सर इससे आपको मल्टीप्लाई करना है x² - 1 को इससे आपको मल्टीप्लाई करना है x² - 1 को तो इससे हम मल्टीप्लाई करेंगे किसको एक्स स्क्वायर माइंस वैन को क्या मैं यही लिख डन थोड़ी मेहनत हमारी बच जाएगी तो इस एक्सप्रेशन को मैं यही लिख रहा हूं घबराना मत है ना ज्यादा घबराना मत स्टूडेंट ध्यान से देखा पहले तो ये पूरा एक्सप्रेशन x² से मल्टीप्लाई होगा तो क्या बचेगा सर मिलेगा 49x² फिर क्या मिलेगा एक्स स्क्वायर एक्स - 14x एक्स तो ये कितना हो जाएगा -14x³ और जैसे ही माइंस वैन से मल्टीप्लाई किया तो ये हो जाएगा -49 ये कितना बचेगा प्लस 14 एक्स तो ये पूरा हटके आपका क्या बचेगा प्लस 14 आई होप आपको यहां तक चीज समझ ए रही हैं अब हम करेंगे कुछ चीज हमारे फीवर में हो रही हैं क्या जिससे हमारी लाइफ थोड़ी आसान हो जाए एक अच्छा कम हुआ सर जीवन में यह हटा बहुत खुशी हुई मुझे देखकर और कुछ ऐसा खास हटा हुआ नहीं दिख रहे तो जो होगा आप देखेंगे भाई शुरुआत करते हैं क्यूबिक पार्ट से तो यह माइंस 14 एक्स कब है 4x³ है तो ये जब उसे तरफ जाएगा ध्यान से माइंस 14 x³ जब इधर आया लेफ्ट हैंड साइड पर तो माइंस 14 प्लस फोर कितना हो जाएगा आपका 18 एक्स कब ये हो जाएगा आपका 18x³. ठीक है सर और हम अगर आगे बड़े और हम आगे बड़े तो अब x² वाली टर्म्स पर बात करते हैं तो देखो भाई ध्यान से 4x² + 2x² कितना 6x² और 6x² में ये उधर गया है तो कितना हो जाएगा सारी हो जाएगा 7x² 7x² में से आपको 49 सब्सट्रैक्ट करना है है ना क्योंकि 49 भी उधर जाएगा तो 7x² में से अगर 49 सूत्र किया तो 7 - 49 या 49 - 7 कितना आई थिंक 42 आप सभी को दिख रहा है क्या तो ये ए जाता है 42x² बट ऑफ कोर्स विद डी नेगेटिव साइन है ना चार और दो छह और यहां से 49 - 48 तो 48 उधर गया है - 42x² तो यहां पर मुझे क्या दिखेगा यहां पर मुझे दिखेगा माइंस 42x² अब क्या सर अब अगर हम आगे देखें तो अब कौन-कौन सी टर्म्स बच रही है हमने इसे कंसीडर कर लिया हमने x² वाली टर्म्स कंसीडर कर लिया अब एक्स वाली टर्म्स की बात की जाए तो सर वहां पर सीधा सीधा बस 4X लिखा हुआ है 14x उधर गया तो 4X को सब्सट्रैक्ट किया तो कितना हुआ -10x तो मुझे एक और टर्म मिलेगी क्या -10x और क्या देख का रहे हो सर आप और अगर मैं ध्यान से देखूं तो इधर से अब सीधी सीधी सी एक्सप्रेशन है ये है वैन और ये है -49 ये उधर गया तो 49 + 1 थॉट्स 50 तो यहां पर मुझे दिखेगा प्लस = 0 यह हमें एक एक्सप्रेशन मिला क्या इस बात से किसी भी स्टूडेंट को कोई आपत्ति बस इस तरीके से चीज हमको दिखी अब इससे क्या करेंगे सर यह सोचने वाली बात है इससे अगर मैं चीज निकलूं तो मुझे यह दिख रहा है क्या की सर तू कॉमन ले सकते हैं ले लेते हैं सर तो तू कॉमन लिया तो ये बच रहा है 9x³ ये बच रहा है 21x² ये हो जाता है -5x और ये हो जाता है प्लस 25 = 0 कुछ इस तरीके से हमें चीज मिल रही है क्या इस बात से किसी भी स्टूडेंट को कोई आपत्ति तो ये बना एक क्यूबिक एक्सप्रेशन जो अब हमें फाइनली सॉल्व करके एक्स की वैल्यू फिर आउट करनी है एक्स की के वैल्यू फिगर आउट करनी है क्योंकि अगर मैं क्वेश्चंस को पढ़ूं तो लेंथ ऑफ पीके पूछ रहा है जब वो पीके की लेंथ आपसे मांगेगा तो आपको एक्स चाहिए होगा इसके बाद जब आप अगला क्वेश्चन करोगे तो पूछ रहा है इक्वेशन ऑफ सर्कल सी तो सी सर्कल की इक्वेशन निकलने के लिए मैं क्या चाहिए मुझे के के कोऑर्डिनेट्स चाहिए तो के के कोऑर्डिनेट्स में हमने निकल लिया था की इसका वे कोऑर्डिनेटर तो वैन होगा लेकिन नेक्स्ट क्वाड्रेंट क्या होगा सर एक्स कोऑर्डिनेट्स ओए होगा वो मुझे पता करना है और ओए पता करना है मतलब क्या ओए हम निकल चुके हैं ये यानी मुझे वापस एक्स पता करना है मतलब एक्स पता चल जाए और जीवन आसान हो जाएगा इसका थर्ड क्वेश्चन भी देखो इक्वेशन ऑफ टांगें ओटी अब अगर ओटी की भी इक्वेशन मुझे जानी है अगर मुझे ओटी की भी क्वेश्चन जानी है तो मुझे तक koardinate तो चाहिए ही चाहिए ना अब बात समझ का रहे हो क्योंकि अगर टी के क्वाड्रेंट पता चल गए तो मैं सर्कल की इक्वेशन से होती की टैसेंट की इक्वेशन लिख सकता हूं या टी के अकॉर्डिंग पता चल गया वो पता ही है तो बात खत्म हो जाएगी और तक कोऑर्डिनेट्स पता करने के लिए भी वापस बात वही आती है की यहां से मुझे कुछ ना कुछ एक्स में निकलना पड़ेगा तो प्रॉब्लम बस इतनी सी है की सर अब आप सॉल्व करो इस क्यूबिक इक्वेशन को और यहां से निकालो किसकी वैल्यू भाई यहां से आपको निकालनी है एक्स की वैल्यू अब एक्स की वैल्यू कैसे सोचेंगे क्या करेंगे इस पर ही सवाल है मेरा कहना है यह क्यूबिक इक्वेशन तो क्या इसको हम कुछ अपनी यू नो इंशन बेस्ड एप्रोच से सॉल्व कर सकते हैं जैसे अगर मैंने पहले ट्राई किया मैन लो माइंस वैन पहले ट्राई करते हैं वैन पहले मैं क्या ट्राई करता हूं वैन तो देखो वैन ट्री किया तो क्या मिलेगा सर जैसे ही आपने वैन ट्री किया तो देखना वैन वैन और वैन अब सुनना ऐसा करने से फाइव सब्सट्रैक्ट करते ही ये बन गया 20 21 में से 9 सब्सट्रैक्ट करते ही सर ये तो कुछ और ही बन गया इससे तो बात नहीं बनी क्योंकि 21 में से ना ही सब्र करती है कुछ और बना नहीं बात बन रही है जीरो तो नहीं हो रहा है आई होप आप देख का रहे हो की 21 - 9 वहां पे केक वाली है सर माइंस वैन ट्री करते हैं -1 अगर मैंने ट्राई किया तो देखना क्या हो रहा है अगर एक्स की वैल्यू में माइंस वैन ट्री करता हूं तो देखो भाई ये हो जाएगा -1 यानी कितना हो जाएगा -9 यहां पर माइंस वैन ट्री किया तो ये तो रहेगा -21 मेरी अब बात समझ का रहे हो यहां पर यह क्या रहेगा प्लस 25 अब आप बहुत ध्यान से इस बात को सुनना बहुत ध्यान से क्योंकि इससे कुछ कम की बात निकल कर आने वाली है ये ए जाता है -30 ए जाता है प्लस 30 ये ए जाता है -30 ए जाता है प्लस 30 तो -1 ने इसको सेटिस्फाई किया है अगर माइंस वैन ने इसको सेटिस्फाई किया तो मैं कहूंगा सर की एक्स - 1 इसका एक फैक्टर होगा मैं कहूंगा की एक्स - 1 इसका एक फैक्टर होगा किसका suniyega ध्यान से अब इस पार्ट को सॉल्व करने का तरीका क्या होगा ये लिखा हुआ है 9x³ - सुनेगा ध्यान से ये लिखा है 9x³ - 21x² फिर क्या लिखा है -5x + 25 अब बात समझ का रहे हो और यहां से मुझे एक्स - 1 एक वैल्यू मिलती है एक्स की क्या वैल्यू -1 बट आई होप आपको नज़र ए रहा है की एक्स की वैल्यू -1 हमारे किसी कम की नहीं है क्योंकि यहां पे एक्स की वैल्यू -1 नहीं हो सकती ना ये तो एक लेंथ है तो ये हमारे लिए एक डाइजेस्टिबल वैल्यू नहीं है क्या आप मेरी बात समझ का रहे हो तो मुझे कोई एक और वैल्यू लेनी होगी अब उसे वैल्यू की तरफ अगर मैं बढ़ाना चाहूं तो मैं कैसे सोचूंगा मैं सोचूंगा की मैं इसको यहां पर फैक्ट्रीज कर लूं एक फैक्टर तो मिला एक्स - 1 दूसरा फैक्टर ढूंढने की कोशिश करते हैं दूसरा फैक्टर जब ढूंढने गए तो ध्यान से देखो स्टूडेंट्स क्या निकल के ए रहा है मैं पूरे इस एक्स - 1 एक्सप्रेशन को मल्टीप्लाई करता हूं x² से ये डिवीज़न आपने सिखा है स्टूडेंट्स उसे पे बात करेंगे x² को एक्स - 1 से मल्टीप्लाई किया और आदर पंत तो 9X कब भी हटाना चाहता हूं ना तो मैं कम करता हूं इसको मल्टीप्लाई करता हूं 9x² से तो 9 x² एक्स्ट्रा कितना हो जाएगा 9x³ लेकिन 9x² -1 तो ये हो जाएगा कितना - 9x² क्या आप मेरी बात समझ पाए आई होप आप मेरी बात समझ का रहे हो और अब आप क्या करेंगे अब यहां पर आप माइंस करेंगे और इसे प्लस कर देंगे तो ये कुछ निकल कर आएगा ये कैंसिल हो जाएगा 21 में से 9 सब्सट्रैक्ट किया जैसे ही ट्वेंटी वैन में से 9 सब्सट्रैक्ट किया तो कितना भाई 21 - 120 और 20 में से आते सब्सट्रैक्ट किया तो 12 तो ये हो जाएगा -12x² है ना और यहां पे क्या बच जा रहा है ये बच रहा है माइंस फाइव एक्स प्लस 25 अब क्या सर अब देखो भाई अगली टर्म की मैं अगर बात करूं तो मुझे चाहिए -12x² तो क्या मैं -12x से मल्टीप्लाई करूं करिए सर तो -12x से एक्स को मल्टीप्लाई किया तो ये हो जाएगा -12x² - 12x है -1 को मल्टीप्लाई किया तो कितना हो जाएगा प्लस 12x आप मेरी बात समझ का रहे होंगे -12x है माइंस वैन को मल्टीप्लाई किया तो प्लस 12 और फिर से वापस चीज की तो इस बार वापस हमने सब्सट्रैक्ट किया तो ये हो जाएगा क्या पॉजिटिव और ये हो जाएगा नेगेटिव अपनी बात समझ का रहे हो भाई तो क्लियर सी बात है सर ये कैंसिल और बच क्या रहा है बट जो रहा है वही है तो ये हो जाएगा -17x+25 ये क्या बचा रहा है सर माइंस 17x + 25 क्या बात आप सभी को समझ ए रही है अब क्या सर ध्यान से सुनते जाएगा अब अगर मैं आपसे बात करूं तो ध्यान से नोटिस करना स्टूडेंट्स -17x + 25 चाहिए तो यहां चाहिए -17x तो क्या मैं यहां पर सीधे सीधे बड़ी सिंपल सी बात ले ए सकता हूं की मैं यहां पर मल्टीप्लाई कर डन इसे किस से मैं यहां पर ऐसे मल्टीप्लाई कर डन -17 से मैं यहां पर इसे मल्टीप्लाई कर डन किस माइंस 17 से कोई दिक्कत तो नहीं भाई हमने सीधे सीधे देखा की एक्स - 1 है ना एक्स - 1 एक्स - 1 होना चाहिए था क्योंकि आपका जो फैक्टर होगा वो होगा एक्स - और माइंस वैन ने सेटिस्फाई किया हम रिलीज सॉरी फॉर डेट आई रियली रिपीट माय वर्ड मैं इसको थोड़ा सा आपको थ्योरम समझा देता हूं की सर अगर एफ ऑफ एक्स में अगर एफ ऑफ एक्स में अगर आप रखते हो अगर आप एक्स की जगह रखते हो ए अगर आपका एक्स = ए इसे सेटिस्फाई करता है तो एक्स - ए आपका एक फैक्टर होता है एफ ऑफ एक्स का तो मैंने एक छोटी सी गड़बड़ कर दी है आपका एक्स की वैल्यू वैन नहीं थी आपके एक्स की वैल्यू थी माइंस वैन आई रिपीट माय स्टेटमेंट एक्स की जो वैल्यू थी वो थी -1 तो एक्स की वैल्यू -1 का मतलब है की एक्स + 1 इसका वैन ऑफ डी फैक्टर्स होगा हम रियली सॉरी फॉर डेट और एक्स + 1 अगर इसका फैक्टर होगा तो यहां पे मैं एक फैक्टर कौन लूंगा x+1 बस छोटा सा करेक्शन हो गया इस बात पे स्टूडेंट्स इसे प्लीज ठीक कर लीजिएगा अगर आप कंफ्यूज हो गए हैं तो और इस बात पर अब हम बात करेंगे प्लीज ध्यान से सुनते जाएगा बहुत फोकस में अंडर में समझना थोड़ा लांगती हो रहा है क्योंकि इस पार्ट को सॉल्व करना थोड़ा ट्रिकी हो चला है है ना 9X कब चाहिए तो मैं यहां पर 9x² से मल्टीप्लाई करता हूं तो कितना हो जाएगा सही हो जाएगा 9x³ और 9x² 1 तो कितना हो जाएगा + 9x² अब कोस सब्सट्रैक्ट करेंगे तो यह दोनों साइन क्या हो जाएंगे नेगेटिव तो यह नेगेटिव और यह नेगेटिव जैसे ही आपने ऐसा किया तो यह कैंसिल और यह हो जाएगा कितना - 30x². और इसे ले आए नीचे तो माइंस फाइव एक्स और ये कितना + 25 अब एक छोटा सा कम हम ये करेंगे मुझे चाहिए -30x² तो यहां पे एक्स है तो मैं क्या ले लूंगा -30x - 30x लेते ही मुझे क्या नजर आएगा गौर से देखिए भाई ये हो जाएगा -30 x² और - 30x से मल्टीप्लाई किया तो ये हो जाएगा -30x लेकिन सर जैसे ही नेगेटिव साइन से आप मल्टीप्लाई करेंगे तो ये दोनों हो जाएंगे पॉजिटिव ये दोनों हो जाएंगे पॉजिटिव और उससे क्या मिलेगा ध्यान से देखो ये पार्ट कैंसिल और ये बचेगा प्लस 25x है ना और ये प्लस 25 आई होप आप देख का रहे हो की यहां पर मैं क्या ले लूंगा यहां पर मैं ले लूंगा 25 तो 25 लेते ही ये कितना हो जाएगा सर ये हो जाएगा 25x + 25 क्या आप मेरी बातें समझ का रहे हो स्टूडेंट्स और यहां से फाइनल कंक्लुजन की तरह जब आप बढ़ेंगे तो ये ऑफकोर्स कितना बचेगा जीरो तो मुझे जितना याद है वो ये की सर डिविडेंड इस इक्वल्स तू डिवीज़न इन क्वेश्चन प्लस रिमाइंडर कर चुके हैं की एक्स की वैल्यू -1 तो हमारे लिए एक एक्सेप्टेबल सॉल्यूशन नहीं है तो मैं इस वाले फैक्टर को सॉल्व करके एक्स की वैल्यू तक पहुंचने की कोशिश करूंगा जो की क्या है 9x² - 30x+25=0 अंडरस्टैंड थोड़ा लेंडी है बट इसके बाद कई सारी चीज सिख रहे हो की क्वेश्चंस कैसे घुमा कर पूछे जाते हैं अब अगर इस पर बात की जाए तो मुझे इस माइंस 30 को ऐसे दो पार्ट्स में स्प्लिट करना है की प्रोडक्ट मिले मुझे 25 9 मुझे उसका प्रोडक्ट मिले 25 9 25 बेसिकली क्या है 25 है 5 5 और ये क्या है 3 3 क्या मैं इसे 15 और 15 के पार्ट्स में ब्रेक कर सकता हूं क्योंकि देखो 5 3 = 15 5 3 = 15 आई होप मेरी बात समझ पाए और मैं इसी तरीके से इन क्वेश्चंस को सोचता हूं मैं क्या लिखूंगा 9x² - 15x - 15x + 25 टेक्निकल पता है एक बात कहूं ये परफेक्ट स्क्वायर है अगर आप ठीक से देख पाओ तो ऐसा करने की जरूरत ही नहीं थी सच कहूं तो क्यों सर ध्यान से देखो आप आपको क्लिक नहीं कर रहे हैं 9x² मतलब क्या 3X का होल स्क्वायर मतलब क्या मतलब 5 का स्क्वायर और ये कितना है 30 नहीं है 30 नहीं है ये है तू इन फाइव आय होप आप देख का रहे हो ए स्क्वायर प्लस बी स्क्वायर माइंस तू अब अन्य हो जाएगा सर वो आएगी 5 / 3 अब जब चूंकि मुझे पता चल गई है एक्स की वैल्यू 5 / 3 जो मैं इस सारी चीजों के दौरान निकलना चाह रहा था तो जैसे ही आती है एक्स की वैल्यू 5 / 3 आई होप आपके सारे सवालों के जवाब आपको मिल गए आपको आपके सारे सवालों के जवाब मिलने लगे क्योंकि सर पहला क्वेश्चन तो यही था ना हमने जो चेक किया था क्या देखा था सर पहला क्वेश्चन जो था वो आई थिंक यही था क्या ध्यान से देखो भाई लेंथ ऑफ पीके और लेंथ ऑफ पीके क्या हो जाएगा 5/3 तो सर पहले क्वेश्चन का तो आंसर ए जाएगा ऑप्शन सी कोई दिक्कत तो नहीं है अब अगर मैं दूसरे क्वेश्चन की तरफ बढूं तो दूसरा क्वेश्चन क्या हुआ उसको सोचते हैं तो दूसरा क्वेश्चन जहां तक मुझे याद ए रहा है उसे सर्कल की इक्वेशन है इस सर्कल की इक्वेशन में सबसे जरूरी बात मुझे जो निकालनी है वो ये की सर सर्कल की इक्वेशन में रेडियस तो पता है वैन है अगर के के कोऑर्डिनेट्स पता चल जाए तो बात खत्म हो जाएगी क्योंकि कोऑर्डिनेट्स पता करने के लिए मुझे चाहिए ओए क्योंकि क्यों का जो एक्स कोऑर्डिनेट्स है वो एक-एक इक्वल है अब ओए कितना है सर ओए ये रहा अब यहां पर रखता हूं मैं 5/3 एक्स की जगह 5/3 तो suniyega ध्यान से जब मैं ओए निकलना चाह रहा हूं तो ओए निकलने के लिए क्या करूंगा मैं लिखूंगा एक्स + 1 यानी 5 / 3 + 1 / 25 / 9 - 1 कोई दिक्कत तो नहीं है इससे जब आप सिंपलीफाई करेंगे तो पंच और तीन आठ तो ये हो जाएगा 8/3 ये क्या हो जाएगा सर 5 और 38 / 3 और यहां पर देखा तो 25 - 9 यानी 16 तो ये हो जाएगा 16 / 9 और 16 / 9 मतलब 4 / 3 क्या आप मेरी बात समझ पाए सर थ्री से थ्री कैंसिल 8 / 4 कितना होता है 2 8 / 4 अगर होता है तू मतलब इसका एक्स कोऑर्डिनेट्स हुआ तू इसका ए koardinate बता सकते हो क्या सर के की एक्स एक्सिस से डिस्टेंस है वैन यूनिट तो इसका ए cardinate हुआ वैन तो टेक्निकल ये एक ऐसा सर्कल है जिसके सेंटर है तू कमा वैन पर और जिसकी रेडियस है वैन तू कमा 1 और 1 रेडियस आई रिपीट माय स्टेटमेंट तू कमा वैन सेंटर और वैन रेडियस तू कमा वैन सेंटर और वैन रेडियस वाला ऑप्शन मुझे ढूंढने की जरूरत नहीं है क्योंकि ऐसा मुझे ऑप्शन ए में दिख रहा है किसी भी स्टूडेंट को कोई तकलीफ या परेशानी हो तो कुछ आई विल रे एक्सप्लेन दिस पार्ट नौ कॉम डी फाइनल क्वेश्चन फॉर दिस चैप्टर और फॉर दिस सॉरी फॉर दिस चैप्टर बिल्कुल ही फॉर दिस पार्टिकुलर लेक्चर जिसमें लिंक्ड कंप्रेशन टाइप इस कंप्रेशन का ये फाइनल क्वेश्चन है डी इक्वेशन ऑफ टांगें ओटी तो मुझे ओटी की इक्वेशन चाहिए ये बात मैं आपके ऊपर छोड़ता हूं की आप मुझे ओटी की इक्वेशन कैसे निकल कर देंगे ओटी के लिए आपको क्या चाहिए मेरा यकीन करिए ओटी निकलने के लिए बहुत ही आसान सा तरीका आपके पास है ओटी निकलने के लिए क्या आप आप की स्लोप निकल सकते हो क्या सर आप की स्लोप क्यों निकलवा रहे हो क्योंकि जो आप की स्लोप होगी वही ओटी की स्लोप होगी और अगर आप की स्लो पता चल जाए तो ओटी की स्लोप निकलने में कोई दिक्कत नहीं है क्योंकि चाहे ओटी हो या आप दोनों ओरिजिन से पास होती हैं अब आप क्यों ढूंढ रहे हो क्योंकि आप मुझे दिख रहा है कैसे दिख रहा है सर ये है वैन और यह है 5/3 आई थिंक यह है वैन और यह तो क्या मैं पी का ध्यान से देखो भाई पी का एक्स कोऑर्डिनेट्स बता सकता हूं सर पी का एक्स कोऑर्डिनेट्स वही होगा जो ओए का मतलब के का एक्स कोऑर्डिनेट्स है क्योंकि सब से लाइन पर लाइक कर रहे हैं ना तो ये भी तू होगा ये भी तू होगा ये भी होगा तो इसका एक्स कोऑर्डिनेट्स तो क्या हो गया 2 सर इसका ए koardinate पता करना है कितना आसान है ये डिस्टेंस है वैन ये डिस्टेंस है 5 / 3 तो 5 / 3 + 1 5 / 3 + 1 करके देख लो भाई यहां पे कर दे रहा हूं घबराना मत 5 / 3 + 1 दिख रहा है ना 5 + 3 8 / 3 तो ये कितना ए रहा है 8/3 अब मेरा आपसे ये कहना है अगर मैं आप की इक्वेशन निकलूं तो क्या ओटी की इक्वेशन अपने आप नहीं निकलेगी तो तू कमा 8 / 3 और 0 से पास होने वाली लाइन जीरो से पास होने वाली लाइन कौन सी लाइन भाई 2 8 / 3 और 0 से पास होने वाली स्ट्रेट लाइन कैसे निकलेंगे मेरे ख्याल से बहुत मुश्किल बात नहीं है आप खुद सोच लो एक्स - X1 यानी एक्स कोई तकलीफ अब क्या सर आई थिंक अब तो चीज आसान है आप देखो यहां से 2 ने 8 को डिवाइड किया कितना टाइम्स फोर टाइम्स तो ये कितना हो जाएगा सारी हो जाएगा 4/3 ये कितना हो जाएगा सर ये हो जाएगा 4/3 तो ए = 4 / 3 टाइम्स एक्स वही इसे इक्वल तू नहीं दिख रहा है पर मेरा यकीन है आप सभी को दिख रहा होगा ऑप्शन ए जो की ऐसा ही है इस क्वेश्चन में जो शायद सारे स्टूडेंट्स को सबसे मुश्किल पार्ट लगा होगा वो एक्स की वैल्यू का निकलना क्योंकि इसने हमारी क्या कहेंगे धज्जियां उदा दी कैसे भाई क्योंकि पहले तो हमने क्या किया पहले तो हमने जीवन को आसान बनाने के लिए एक क्वेश्चन बनाई पर वो खुश नहीं द इस बात से उन्होंने ऐसे क्वाड्रेटिक किया लीनियर तक नहीं रोका उन्होंने गुमाया उन्होंने इसे पहले तो अनरूट में लाया है तो अंदुरूट को हमने सुझाव से sujhate पहुंचे तो अंडर रूट सुधारते सुधारते घबराहट तो तभी बढ़ गई थी जब व्यक्ति बाहर 4 ए गई थी पर उन्होंने हम पे थोड़ी यू नो बेहतर टीवी बढ़ती और उन्होंने एक्स ^ 4 हमारा हवा दिया बहुत खुशी हुई साहब इस बात से जो आपने हमारे ऊपर इतनी क्या कहेंगे भाई दया राखी अब जब एक्स कब में भी इक्वेशन नहीं तो हम तो इसे देखकर ही परेशान हो गए पर इसे देखकर परेशान मत होना जनरली ये ऐसी इक्वेशन होती है जिसका रूट वैन या माइंस वैन होता है अब इसको माइंस वैन ने सेटिस्फाई किया ना मैंने एक्स = -1 यानी एक्स + 1 फैक्टर होगा यानी एक्स + 1 इसे प्रॉपर्ली डिवाइड करेगा तो एक्स + 1 से मैंने इसको डिवाइड करवाया पॉलिनॉमियल का डिवीज़न आय होप ये देख कर आपको कुछ याद आया होगा नॉस्टैल्जिया अब जब डिवाइड किया तो हम जानते हैं सर जो डिविडेंड होता है वह डिवीज़न और क्वेश्चन के प्रोडक्ट को अगर आप रिमाइंडर से ऐड कर दें तो डिवाइड के इक्वल होता है तो दिस विल बी इक्वल तू दिस इन दिस इससे तो बात बन नहीं रही थी क्योंकि एक्स की वैल्यू माइंस वैन हो नहीं सकती से नेगेटिव कैसे होगी डिस्टेंस फर्स्ट क्वाड्रेंट में है नीचे तो जा नहीं सकते द तो सीधी सीधी सी बात है सर एक्स की फिर आपको ये वाली वैल्यू लेनी चाहिए जो यहां से निकल कर आएगी जो की एक परफेक्ट स्क्वायर था बस इसी बात से हमने यहां पर चीज सोचनी चाहिए निकल और एक्स की वैल्यू हुई 5 / 3 और 5 / 3 आते ही एक कंप्रेशन पर बेस्ड क्वेश्चंस हैं जिस पर मल्टीपल क्वेश्चंस होंगे और उन सारे क्वेश्चंस को हमें सॉल्व करना है जो की इस पार्टिकुलर थॉट या प्रक्रिया या डाटा पर बेस्ड होंगे क्या दिया है तू सर्कल्स इंटरसेक्ट ऑर्थोगोनली सर अगर दो सर्कल्स और सेकेंडरी इंटरसेक्ट करते हैं तो जितना मैं जानता हूं मैन लो अगर यह एक पहला सर्कल है और एक और सर्कल है सर जो इसे लेट्स से ऑर्थोगोनली इंटरसेक्ट कर रहा है है ना तो मैं बड़ी बेसिक सी छोटी-छोटी सिंपल सी बातें जो जानता हूं ये की लेट से यहां पर इसे orthagonal इंटरसेक्ट कर रहा है ना मैन लेते हैं अब व्हाट्सएप करने के लिए इंटरसेप्ट करने का मतलब क्या है सर इस पर जो ये मैंने टांगें ड्रॉ किए कुछ इस तरीके से है ना और जो हमने इस पर ये जो टैसेंट दरो के हम रिलीज हो रही फॉर दिस इसे तो मैं यही रखता हूं थोड़ा सा इसे जरूर रोते कर दूंगा आई थिंक यह इसकी टेंशन होगी मैन ली जाए क्या यह बिल्कुल टांगें आपको ड्रा करनी है सर जो की होगी ऑफ कोर्स यहां पर थोड़ी सी ऑफ कॉस्ट रिलेटेड होगी और उसे जड़ों जो है मैं हम इसका 90°c बिगाड़ देंगे उससे ना मैं आपको समझने में थोड़ी प्रॉब्लम फेस करूंगा तो मैं इसका एक कम करता हूं उसको थोड़ा और थोड़ा बनाता हूं जब औरतों को analysity पर क्वेश्चन आए ना तो वह फिगर सही नाम बनेगा तो आपको प्रॉब्लम हो जाएगी उसे क्वेश्चन को समझने में तो इस शॉप को मैं थोड़ा और अंदर ही शिफ्ट करता हूं ताकि मैं आपको समझा पाऊं होगा या नहीं यह एक लाइन आएगी और एक लाइन है तो बहुत दूर चले जाएंगे है ना एक लाइन यह आएगी और एक लाइन हो जाए अगर ईश्वर ने हमारा साथ दिया तो बस कोशिश हमारी पुरी रहेगी की चीज सही चले है मतलब इंजन और काफी हद तक मुझे लग रहा है एक 90 डिग्री हैंगर कोई दिक्कत तो नहीं अब ऑफ कोर्स ये दोनों ट्राएंगल्स ऑर्थोंगोनल हैं अच्छा इनकी कॉमन कोड भी दिख रही है क्या हान सर इनकी कॉमन कोड भी दिख रही है जो की ऑफ कोर्स ये हो जाएगी तो ये रही इनकी क्या भाई ये रही इनकी कॉमन कोड कोई दिक्कत तो नहीं है यहां तक किसी भी स्टूडेंट को मेरे ख्याल से तो नहीं होनी चाहिए तो ये हमने बना लिए इनकी कॉमन कोड सर अब क्या अब बात करते हैं कुछ चीजों पर कैसे ध्यान से देखो उसने क्या कहा उसने कहा ये दो सर्कल से इंटरसेक्ट कर रहे हैं orthoonely और इनकी जो कॉमन कोड है उसकी लेंथ है 24/5 तो यहां पर ये जो आपको कॉमन कार्ड दिख रही है ना इसकी लेंथ है कितनी 24/5 तो इससे अगर मैं एक हूं रिम में बी कहूं तो व्हाट गिवन आई एम टोल्ड डेट अब 24/5 एवरीवन विथ मी सो फार अन्य कन्फ्यूशंस ऑन डेट नौ कमिंग तू डी नेक्स्ट पार्ट ऑफ दिस क्वेश्चन डी रेडियस ऑफ वैन ऑफ डी सर्कल इसे 3 नौ व्हाट आई विल लेट डी रेडियस ऑफ वैन ऑफ डी सर्कल प्लीज ट्री एंड अंडरस्टैंड प्लीज ट्री एंड अंडरस्टैंड व्हेन तू सर्कल्स आर आर्गनल व्हेन एवर तू यू नीड तू अंडरस्टैंड इट बाय योरसेल्फ प्लीज अंडरस्टैंड दिस आप एक बात बताओ क्या आपको ये वाला एंगल 90 डिग्री दिख रहा है हान सर ये तय है 90° होना क्यों क्योंकि ये सर्कल्स और नल है तो इनके पॉइंट ऑफ कॉन्टैक्ट पर पॉइंट ऑफ इंटरसेक्शन पर इफ आई हैपन तू ड्रॉट 10 जींस दैन दे आर गोइंग तू बी परपेंडिकुलर ऑन इ आदर एंगल मेड बिटवीन डेम इस गोइंग तू बी 90° एंड थॉट्स वही दे आर कॉल्ड ऑर्जिनल ऑन एच आदर टुवर्ड्स इ आदर विद रिस्पेक्ट तू एच आदर अब एक बात पर अब जोर दीजिए एक बात को ध्यान से समझने की कोशिश करिए अगर मैं इन दोनों सर्कल्स के सेंटर से आपसे milvaun इन दोनों सर्कल्स के सेंटर से आपको milvau तो अब क्या ऑब्जर्व करेंगे क्या कंक्लुजन पाएंगे उसे देखते हैं क्या यहां तक कोई डाउट है अच्छा सर एक सर्कल की रेडियस मुझे दे दी गई है जो की कितनी है एक सर्कल की रेडियस मुझे बता दी गई है थ्री ठीक है इससे बात करेंगे पर पहले क्वेश्चन पढ़ते हैं डी रेडियस ऑफ डी आदर सर्कल तो मुझे दूसरे सर्कल की भी रेडियस निकालनी है निकल लेंगे अगली बात डी एंगल बिटवीन डायरेक्ट कॉमन tangences तो सर इन पर डायरेक्ट कॉमन टांगें भी ड्रॉ हो रही है और उनके बीच का एंगल निकलना है ठीक है सर वो भी पहुंच जाएंगे और तीसरी बात डी लेंथ ऑफ डी डायरेक्ट कॉमन tangentis मतलब सर डायरेक्ट कॉमन टांगें से भी तो डील करना है अभी तो अगर डायरेक्ट कॉमन टांगें से हमें डील करना ही है तो कैन आई से की सर डायरेक्ट कॉमन टेंट कुछ ऐसी सी दिख रही होगी आई थिंक यह आपकी एक डायरेक्ट कमेंट टैसेंट आप सभी को दिख रही है क्या और मैं एक बहुत कृष्ण बहुत जरूरी सी बात हमेशा से जानता हूं सर की जो डायरेक्ट कॉमन टांगें होती है ना यह जो आपकी डायरेक्ट कॉमन टांगें होती है सर यह आपके दोनों सर्कल के सेंटर्स को मिलने वाली लाइन से कहीं ना कहीं जाके टच होती है रिपीट माय स्टेटमेंट जैसे मैन लो इस सर्कल का सेंटर का सेंटर यहां कहीं होगा तो इनको चिल्लाने वाली लाइन से कहीं ना कहीं आगे जाकर मिलती है यह हमेशा से हम सीखते और ऑब्जर्व करते आए हैं इन सारी बातों में किसी भी बात से आप में से किसी भी स्टूडेंट को कोई परेशानी है तो पूछो फिर हम बढ़ेंगे आगे मेरे ख्याल से यहां तक तो किसी भी स्टूडेंट को कोई तकलीफ नहीं है जल्दी से सोच के बताएं स्टूडेंट्स अब सर एक कम करते हैं चूंकि ये डायरेक्ट कॉमन टांगें है और मैं मैन लेता हूं इस सर्कल की रेडियस सेंटर यहां कहीं होगा जैसे उसको मैं बना लेता हूं C1 और इस सर्कल का सेंटर यहां कहीं होगा जैसे मैं बना लेता हूं लेट्स सी तू है ना कहीं ना कहीं बना लेते हैं अब एक बात बताओ स्टूडेंट्स बड़ी साधारण सुलझी और सिंपल सी बात बताओ अगर आपको C1 और C2 को लेकर कन्फ्यूजन है तुमने थोड़ा अलग बना लूंगा पर आप नोटिस करो की आपको डायरेक्ट कमेंट टैसेंट के लेंथ चाहिए आप समझ का रहे हो डायरेक्ट कॉमन टांगें की लेंथ निकलने का मतलब क्या है दूसरी बात मुझे आपसे इस बात को निकलवाने में इंटरेस्ट है इस बात को जानने में इंटरेस्ट है की आप इन डायरेक्ट कमेंट्स के बीच का एंगल कैसे देंगे ये भी पूछा जा रहा है और दूसरे सर्कल की रेडियस तीनों बात पर बात करेंगे है और पहले तो मैं इन दोनों सर्कस को थोड़ा और प्रेसीजली प्लॉट करने की कोशिश करता हूं क्या आप यह बात जान समझ या देख का रहे हो की ऑफ कोर्स सर इसके सेंटर से इस पर परपेंडिकुलर ड्रॉप करेंगे टैसेंट पर तो इसकी रेडियस और इसके सेंटर से इस पर परपेंडिकुलर ड्रॉप करेंगे तो इसकी tengent कोई दिक्कत तो नहीं ये तो बड़ी बेसिक सी बात है और ऑफ कोर्स ये दोनों एंगल्स क्या होंगे सर ये दोनों एंगल आपके 1990 डिग्री के होंगे कोई तकलीफ तो नहीं है और क्लियर सी बात है सर अगर मैं इसे अपना सेंटर मानूं सेंटर मानो तो ये जो टैसेंट है यह 90° अगर है रिपीट में स्टेटमेंट यह लाइन इस सर्कल की टैसेंट पर 90° है तो ये भी इसकी रेडियस ही हुई मतलब यह भी एक रेडियस और ये भी एक रेडियस और अगर दोनों इंटरसेक्ट करेंगे मतलब बेसिकली जो डायगोनल सेंटर सेट कर रहे हैं तो वो सर्कल का सेंटर होना तय है सिमिलरली ये भी एक रेडियस और ये भी एक रेडियस और दोनों अगर इंटरसेक्ट कर रही हैं तो डायगोनल ही है और वो हो जाएगा इस सर्कल का सेंटर आप मेरी बातें समझ का रहे हैं या नहीं मैं फिर से समझता हूं गौर से suniyega इससे यू नो सर्कल का सेंटर कहने का रीजन ये है अब समझो यह रेडियस है मतलब ये एक डायमीटर है और यह रेडियस है मतलब ये दोनों भी डायमीटर है और डायमीटर्स का पॉइंट ऑफ इंटरसेक्शन ही तो आपका क्या होता है सर्कल का सेंटर अब बात समझो इस टैसेंट पर अपने परपेंडिकुलर ड्रॉप किया रेडियस इस टैसेंट पर आपने परपेंडिकुलर ड्रॉप किया है रेडियस इस टैसेंट पर परपेंडिकुलर ड्रॉप किया रेडियस इस टैसेंट पर परपेंडिकुलर ड्रॉप किया रेडियस टू कोर्स की रेडियस है है ना अब बात सुनना है स्टूडेंट्स क्या मैं ये का सकता हूं की सर अगर मैं इसे कहूं R1 तो आप इसे कहोगे r2 ऑफ़ कोर्स सर आप इसे भी r2 का सकते हैं और आप इस लेंथ को भी R1 का सकते हैं मुझे कोई आपत्ति परेशानी तकलीफ दिक्कत नहीं है और इन सबसे पहले जरूरी बात की अब की लेंथ मुझे पता है 24/5 और मुझे ये भी बहुत अच्छे से पता है की ये एंगल 90 डिग्री होगा एंगल 90 डिग्री होगा ही होगा और सर्कल के सेंटर से कॉर्ड पर परपेंडिकुलर ड्रॉप किया जाए तो अगर मैं इसे थोड़ी देर के लिए का लू के सेंटर से ड्रॉप करके परपेंडिकुलर से बहुत दिक्कत तो नहीं है इन सारी बातों से स्टूडेंट्स आई थिंक अब चीज क्लियर है अब अगर मैं डायरेक्ट कॉमन टांगेंट्स के बारे में बात करना शुरू करूं तो एक बात और आप देख रहे होंगे स्टूडेंट्स की सर अगर आपने यहां से भी एक टैसेंट दरो की है अगर आपने यहां से भी एक टैसेंट अगर ड्रा की होती तो पहले तो उसका पॉइंट ऑफ इंटरसेक्शन यहां होता और वो इन दोनों को भी क्या कर रही होती टच तो मेरे ख्याल से चीज सिमिट्रिकली कुछ ऐसी जा रही होती बट मैंने इसका इंक्लिनेशन बहुत अजीब बनाया है और मैं इसे ठीक से नहीं बना पाया हूं है ना रोते किया तो इससे तो मैं हटा देता हूं और मैं दूसरी टांगें बनाने की कोशिश करता हूं जो की ऑफ कोर्स में यहां से लेता हूं उसे टच करती हुई जो किसी यहां टच करती हुई यहां से पास तो करेगी लेकिन इसे थोड़ा नीचे ले ले तो कोई बुराई है क्या डॉटेड लाइन ले लेता हूं ताकि हम बहुत प्रेसीजली बना नहीं पाएंगे वो कुछ इस तरीके से आपकी टच करती हुई यहां से जा रही होगी मैंने बस एक पंजेंट बना दिया आई होप ये बात बहुत ज्यादा बॉर्डर नहीं कर रही है आपको इन दोनों टांगें के बीच का एंगल चाहिए देखो क्वेश्चन पढ़ो आपको इन दोनों डायरेक्ट कमेंट के बीच का एंगल चाहिए क्या मेरी आप यह बात समझ का रहे हो की अगर ये एंगल आपको चाहिए क्योंकि आप थोड़ी देर के लिए अगर मानते हो ठीक-ठाक तो ऑफ कोर्स में इस एंगल को क्या कहूंगा देता / 2 कोई तकलीफ कोई परेशानी कोई दिक्कत बिल्कुल सर स्वाभाविक है एंगल थीटा बाय तू होगा क्योंकि यह जो लाइन है यह क्लियर एंगल बाईसेक्टर की तरह बिहेव करेंगे मेरे पास बहुत सारी इनफॉरमेशन बहुत सारा डाटा है बहुत सारी जानकारी है अब एक बात का जवाब दो मैं थोड़ी देर के लिए जो एम लेंथ है ट्रायंगल C1 ए एम की बात कर रहा हूं मैं ट्रायंगल सी वैन ए एम की बात करना अच्छा बाय डी वे मैं एक और बात मैन ले रहा हूं की जो R1 है R1 कितना है आपका जो R1 है वो आपको दिया हुआ है और कार्बन की वैल्यू कितनी आप मैन ले रहे हो सर R1 की वैल्यू है यह जो रेडियस 3 दी गई है तो इसको मैं मैन लेता हूं थ्री अब एक बात बताओ अगर मैं ट्रायंगल ए c1m में बात करूंगा आपसे मैं किस ट्रायंगल में बात कर रहा हूं मैं ट्रायंगल ए सी वैन एम में अगर बात करूं आपसे है ना अगर आपको कोई परेशानी हो तो क्या मैं इससे कोई एंगल का सकता हूं इस एंगल को मैं नाम नहीं दे रहा हूं बस ऐसे ही एक एंगल पर बात कर रहा हूं सुनना ध्यान से है ना ऐसा मैं एंगल क्या ही लेता हूं ताकि आप कंफ्यूज ना हो और इसे मैं का लेता हूं की एंगल फाइव अब आप एक बात बताओ अगर मैं सिन फाइंड निकालो अगर मैं सिन फाइंड निकलू है ना तो सिन 5 क्या आएगा आप खुद सोच के बताओ सर जब आप निकलेंगे सिन 5 तो सोच के देखो सिन 5 मतलब परपेंडिकुलर अपॉन hypotaneous तो एम अपॉन R1 तो कितना हो जाएगा है तो यह है 90 डिग्री क्या हो जाएगा ये एंगल फाइव\| देखो ये है फाइव ये है तो ये एंगल कितना होगा 90 - 5 यह पूरा 90 है तो क्या करूंगा आई होप मैं आपको सारी बातें अच्छे से समझा का रहा हूं बिना किसी बात की तकलीफ है परेशानी अब एक छोटा सा कम करते हैं छोटा सा कम ये अगर मैं इसका और इसका स्क्वायर करके ऐड करूं तो आप एलएस पर देखो साइन स्क्वायर फाइव प्लस कोस स्क्वायर फाइव वैन तो ये क्या मिलेगा वैन ठीक है सर और राइट हैंड साइड पर क्या मिलेगा m² / r1² a² / r² मतलब आपको क्या मिलेगा एलएस फिर साइन स्क्वायर फाइव प्लस कोस स्क्वायर बट राइट हैंड साइड पर प्लस क्या मुझे एम का स्क्वायर पता है सर ए एम का स्क्वायर तो नहीं पता है लेकिन एम जो है वो अब का हाफ है एम जो है टेक्निकल वो अब का हाफ है डेट इस नथिंग बट हाफ ऑफ अब तो यहां से कुछ हेल्प मिल सकती है क्या यहां तक पहुंचने में देखो ध्यान से मैं फिर से रिपीट करता हूं एम जो है वो अब का हाफ है अब का हाफ मतलब 24/5 का हाफ यानी 12/5 तो ए बी अगर 24/5 है तो एम हो जाएगा 12 / 5 a.m कितना हो जाएगा भाई 12/5 तो मैं एम का स्क्वायर कॉमन ले लेता हूं ए एम का स्क्वायर कॉमन ले लेता हूं तो ये हो जाएगा 1 / r1² + 1 / r² तो कितना हो जाएगा सर ये हो जाएगा r2² + r1² बहुत कृष्ण सी बात है आप दूसरे सर्कल की रेडियस निकलने के काफी करीब है एम इसे नथिंग बट अब इसे दी कॉमन कोड और कॉमन कोड 24/5 इसका हाफ है 12 / 5 तो ये हो जाएगा 12 / 5 12 5 को इधर ले तो ये हो जाएगा 5/12 का स्क्वायर तो 5² / 12² = 9 / r2² पता करना है तो 9 r2² कोई तकलीफ कोई डाउट कोई परेशानी तो नहीं है स्टूडेंट्स सर इसे सॉल्व करिए चीज शायद आसान हो जाएंगे और इसे ही सॉल्व करना होगा शायद r2² निकलने के लिए क्या यहां तक कोई तकलीफ चलो भाई अब सर फाइव अपॉन फाइव स्क्वायर अपॉन 12 स्क्वायर को भी मैं लिख सकता हूं कितना 25 बाय 144 सिंपलीफिकेशन तो हो जाएगा 25 9 अब क्या गुंजाइश को दिख रहा है सर एक स्कोप अब मुझे कुछ नहीं दिख रहा अब तो आपको क्रॉस मल्टीप्लिकेशन करनी पड़ेगा\| 25 का 9 टाइम्स कितना होता है सो 25 का 10 टाइम्स होता है 250 तो 9 टाइम्स होगा 225 आप समझ का रहे हो तो ये हो जाएगा कितना ये हो जाएगा 225 दिस इस गोइंग तू बी 225 अब ये यहां ए रहा है ये वहां जा रहा है कर सकते हैं क्या ऐसा कर लेते हैं देखते हैं चीज कैसे वर्क आउट होते हैं तो अगर समझा जाए तो ये हो जाएगा 225 टाइम्स r2² है ना ये हो जाएगा मैंने 225 में से वैन फोर सब्सट्रैक्ट कर लेते 145 में से 144 सब्सट्रैक्ट अगर मैंने किया तो थोड़ा सा ट्रिकी पर इसको ऐसे सोचना 144 में कितना ऐड करें की 200 ए जाए देखो 144 में सिक्स ऐड किया तो कितना आई थिंक 50 और 50 ऐड किया तो 56 तो कितना ए जाएगा 200 56 में अगर 25 ऐड कर लें मैं क्या करना चाह रहा हूं 56 में 25 56 में फाइव ऐड किया तो 56 में 5 विकेट गया तो कितना 61 61 में 28 गया तो क्या 81 तो ये ए जाएगा कितना ये ए जाएगा 81 r2² कितना r2² किसी भी स्टूडेंट को कोई दिक्कत है तो पूछो हो गई मैंने कुछ गड़बड़ तो नहीं की बस ऐसे एक बार क्रॉस चेक कर लो चीजें आई थिंक सही है आसान है अब इधर क्या लिखा हुआ है इधर लिखा है 144 9 ये लिखा हुआ है 144 9 अब क्या करें सर अब सर एक छोटा सा कम आप और कर लीजिए 81 को वहां लेकर गए तो 81 को वहां लेकर गए तो डिनॉमिनेटर में तो ये हो जाएगा 9 / 81 यह कितना हो जाएगा 9 / 81 जो की होगा कितना अंडर रूट होगा 12/3 और 12/3 सर्कल की जो रेडियस आती है डेट्स आउट तू बी फोर आई अंडरस्टैंड शायद मैंने कहीं ना कहीं इसको ट्रिकी बना दिया है मुश्किल बना दिया है कैलकुलेशन के टर्म्स में बट अब जो आया वो आया तो रेडियस जो सेकंड सर्कल की आई है वो कितनी है भाई 4 यानी आप कौन सा ऑप्शन मार्क करेंगे आप मार्क करेंगे ऑप्शन बी क्या यहां तक तो किसी स्टूडेंट को कोई दिक्कत नहीं रही यहां तक चीज अच्छे से समझ ए गई हैं तो आगे बड़े आई थिंक सर ये तो चीज आसान थी इसमें तो कोई ज्यादा ऐसी बहुत बड़ी विशेष तकलीफ नहीं हुई होगी मेरा ऐसा मानना है किसी भी स्टूडेंट को अब क्या करना है सर अब क्या सोचना है उसे बारे में बात करते हैं अब थोड़ा सा वापस आते हैं इस बात पर और इस फिगर पर जहां हम हेल्प लेंगे इस स्टेटमेंट से की सर देखो अगर मैं C1 C2 के पैरेलल एक लाइन ड्रॉ करता हूं फिर ड्रॉ ए लाइन पैरेलल तू सेवन सी तू तो ये लाइन ऐसी कुछ जा रही होगी नौ आई हैव ड्रॉन लाइन पैरेलल तू सी वैन सी तू मतलब यह लाइन और यह लाइन पैरेलल है अगर यह लाइन और यह लाइन पैरेलल है और ये कॉमन लाइन है तो अगर ये एंगल थीटा / 2 था तो shailise है की सर ये एंगल भी थीटा / 2 होगा यहां मैं सिन थीटा / 2 आपसे पूछूं तो ये क्या होगा सर सिथेंटा / 2 होगा दिस अपॉन दिस आप समझ का रहे हो स्टूडेंट्स और दिस अपॉन दिस की jadhoj है झांसी ये जो लेंथ है ये क्या है सर r2 - R1 आप इस लेंथ पर बात करो ये जो लेंथ है ये लेंथ कितनी है सर पुरी r2 - R1 क्योंकि ये आपका क्लियर ही क्या बन रहा है ये आपका क्लीयरली पैरेललोग्राम है जिस बारे में डिटेल डिस्कशन किया है तो ये लेंथ हो जा रही है आपकी r2 - R1 मैं किस पर बात कर रहा हूं सिन थीटा / 2 पर हम बात कर रहे हैं साइन थीटा बाय तू जो की होता है परपेंडिकुलर अपन हाइपोटेन्यूज तो परपेंडिकुलर हुआ r2 - R1 तो हाइपोटेन्यूज क्या होगा हाइपोटेन्यूज आपका ये जो की है C1 C2 के पैरेललोग्राम है तो जो ये लेंथ होगी वही ये लेंथ होगी तो ये हो जाएगा 7 C2 के बीच की डिस्टेंस ये हो जा रही है C1 C2 के बीच के डिस्टेंस सर मेरा जीवन बहुत आसान है क्योंकि मुझे r2 पता है मुझे R1 पता है लेकिन मैं C1 C2 कैसे निकलूं क्या आप मेरी सी वैन सी तू को निकलने में मदद कर सकते हैं मेरा कहना है आप C1 C2 निकल चुके हो बस को दिख नहीं रहा है ऐसा कैसे कुछ भी बोलते रहते हो अच्छा बताओ R1 कितना है सर R1 आया था थ्री अच्छा बताओ r2 कितना है सर r2 आया था 4 अब देखो ना C1 C1 ए C2 एक राइट एंगल ट्रायंगल है क्या हान सर है दिख रहा है क्या C1 C2 जब हमने निकाला पाइथागोरस थ्योरम से तो ये आया फाइव अगर C1 C2 आया है फाइव सुनेगा ध्यान से C1 C2 अगर आया है फाइव तो एक्चुअली आप क्वेश्चन सॉल्व कर चुके हो कैसे सर ध्यान से देखो साइन थीटा बाय तू आर तू माइंस आर वैन माइंस 24/25 अभी मैं इसे ऐसा ही लिख ले रहा हूं बाद में सिंपलीफाई कर लूंगा मत घबराओ है ना कोस थीटा / 2 ऐसा कैसे निकाला आपने मैं आपको याद दिलाता हूं मैं बड़ी बेसिक सी प्रॉपर्टी उसे कर रहा हूं की sin² थीटा / 2 + कोस स्क्वायर θ / 2 क्या होता है वैन होता है बस में नहीं यही किया है सिन थीटा / 2 पता था जो की कितना था 1/5 1/5 का स्क्वायर 1/25 उधर गया तो 1 - 1 / 25 तो 25 - 1 / 25 एंड 24/25 और उसका अंडर रूट तो ये क्या हो गया कोस थीटा / 2 क्यों निकल रहे हो सर एक बात बताओ एक बात बताओ सिन 2theta कितना होता है सर होता है 2sinθcosθ तो सिन थीटा कितना होगा सो सिन थीटा होगा 2sinθ/2cosθ/2 तो साइन थीटा हम निकलना चाह रहे हैं ठीक है सर निकालिए साइन θ जब आप निकलना चाहेंगे जनाब तो सिन थीटा क्या होगा सर साइन थीटा होगा तू टाइम्स साइन थीटा बाय तू विच इसे दिस अब सुनना ध्यान से 24 को क्या मैं 46 लिख सकता हूं 4 6 तो 2/5 बाहर आएगा अंदर बचेगा √6 तो 2/5 बाहर आएगा और अंदर बचेगा अंडर रूट 6 तो कितना ए जाएगा 4/5 तो थीटा कितना होगा सर थीटा होगा सिन इन्वर्स ध्यान से देखना थीटा होगा सिन ^ -1 4 टाइम्स √6/5 यह आपका टेक्निकल उन दोनों डायरेक्ट कॉमन टांगेंट्स के बीच का एंगल है याद करो यह थीटा ही तो है उन दोनों डायरेक्ट कॉमन टांगेंट्स के बीच का एंगल याद ए रहा है क्या तो ये कितना ए जा रहा है सर ये ए जा रहा है फोर टाइम्स रूट 6/5 क्या ये फोर टाइम्स रूट 6/5 ए रहा है देख लेते हैं सर बिल्कुल सही बात है पर कुछ गड़बड़ है शायद मैंने गड़बड़ की हान मैंने कैलकुलेशन में गड़बड़ की जो मुझे अभी जस्ट रिलीज हुआ सर बेवकूफी तो आप करते हो ना मुझे वो अंडर रूट में लगा पता नहीं क्यों तो हो जाएगा 25 और ये रहा आपका सीधा सीधा ऑप्शन मैं बहुत अच्छे से आपको सारी चीज क्लियर कर दूंगा अगर अभी तक कोई भी परेशानी है तो भाई मेरा मानना है यहां तक तो कोई दिक्कत नहीं है अब सर जब पूछ रहा है आपसे इस डायरेक्ट कमेंट टैसेंट की लेंथ तो वो कैसे निकलोगे मैं तो सिर्फ फॉर्मूला भूल गया हूं मैंने तो फॉर्मूला नहीं रखा मुझे तो फॉर्मूले याद नहीं रहते तो फॉर्मूला नहीं रहता तो कोई बात नहीं मुझसे क्या पूछ रहा है वो डायरेक्ट कॉमन टांगें की लेंथ है ना मैं इस पॉइंट को कहता हूं लेट से पी और इस पॉइंट को लेट्स से मैं कहता हूं के आपसे ये पीके की लेंथ पूछी जा रही है चलो इस पॉइंट को हम का लेते हैं एल बस ऐसे ही का देते हैं तो ट्रायंगल पीकेएल में आपसे पीके पूछा जा रहा है ट्रायंगल पीकेएल में आपसे पीके पूछा जा रहा है है ना आई रिपीट माय स्टेटमेंट ट्रायंगल पीकेएल में आपसे पूछा जा रहा है पीके पीके तक मैं पहुंच जाऊंगा सर लेकिन क्या मैं कहूं pq² + p² सॉरी pq² [संगीत] मुझे पता है क्या सर pq² पता है क्या [संगीत] सॉरी 4 - 3 कितना है वैन तो ये कितना ए गया है वैन अबू कल अगर आता है वैन तो ये हो गया वैन वैन को मद्र शिफ्ट कर रहा हूं तो पीके का स्क्वायर जो होगा वो पी एल स्क्वायर में निकल रहा हूं अब p² क्या होगा सर सर पल का स्क्वायर आप खुद देखो क्या C1 C2 के डिस्टेंस के इक्वल है की नहीं बस ये दोनों बातें अप्लाई कर लो ये क्वेश्चन भी खत्म हो जाएगा देखो बड़ी सिंपल सी बात मैं का रहा हूं पी एल मतलब क्या सर C1 C2 पैरेललोग्राम है ना तो सी वैन सी तू कितना आया था फाइव तो ये हो जाएगा सर ये हो जाएगा 5 का स्क्वायर मतलब 25 माइंस का स्क्वायर तो पीके का जो स्क्वायर ए रहा है वह ए रहा है 24 का स्क्वायर तो पीके कितना आएगा वो ए जाएगा अंडर रूट 24 मतलब 6 4 तो ये ए जाएगा 2 रूट 6 इस गोइंग तू बी तू रूट सिक्स और इस तरीके से आप इसका भी आंसर मार्क करेंगे तो 2√6 जो है आपके डिस्टेंस आणि चाहिए जो की क्लीयरली आपका ऑप्शन होगा सी क्या किसी भी स्टेप में किसी भी क्वेश्चन किसी भी पार्ट में आपको कहीं कोई परेशानी सबसे पहला डाटा सर्कल एस ऑफ रेडियस वैन है ना तो एक सर्कल है जिसकी रेडियस है वैन यूनिट इनस्क्राइब्ड इन एंड इक्विलैटरल ट्रायंगल पिक तो सबसे अच्छा कम तो यही होगा सर की पहले मैं एक इक्विलैटरल ट्रायंगल पिक बना लूं और उसमें एक सर्कल को सब्सक्राइब कर रहे हैं तो जब आप सर्कल इनस्क्राइब करते हैं तो आई थिंक चीजें कुछ इस तरीके से हॉपफुली होनी चाहिए थोड़ा सा इसे और बड़ा कर लेंगे तो मेरे ख्याल से यह सर्कल हो गया आपका इनस्क्राइब क्या आपको अच्छे से चीज दिख रही है बिल्कुल सर्कल है इसकी रेडियस की रेडियस है डेट इक्विलैटरल ट्रायंगल इस पिक तो इसकी वर्टिसेज है पी के और आर अब क्या सर अब अगर मैं आगे बढ़ता हूं डी पॉइंट ऑफ कॉन्टैक्ट ऑफ पिक एंड आरपी है ना तो पीके के साथ जो पॉइंट ऑफ कॉन्टैक्ट है वो है दी के आर के साथ जो है वह है जो लाइन पीके है जो लाइन पीके है उसकी इक्वेशन क्या है सर इसकी इक्वेशन प्लस ए माइंस सिक्स इस इक्वल तू जीरो आई होप यही है √3x + ए - एक्स है ना अभी क्या और पॉइंट दी के cardinate हुआ आपको दे दे रहे हैं पॉइंट दी के cardinate का रहे हैं थ्री टाइम्स अंडर रूट 3/2 और 3/2 सो थ्री टाइम्स रूट 3/2 आपके पॉइंट दी के जो cardinate हैं ये पॉइंट दी है इसके कोऑर्डिनेट्स वो आपको का रहा है वो है दी पॉइंट के कोऑर्डिनेट्स की हम बात कर रहे हैं राइट ये है थ्री टाइम्स रूट 3/2 और ऑफ कोर्स 3/2 डीज आर नथिंग बट डी कोऑर्डिनेट्स ऑफ पॉइंट दी ये इस तरीके से चीज दे रही हैं फरदार इट इसे गिवन डेट डी ओरिजिन ओरिजिन अत डी सेंटर ऑफ सर्कल जो आपका सेंटर है आर ऑन डी से साइड ऑफलाइन पीके उसने ये कहा इस बात का जो भी मतलब है मुझे नहीं पता बट इस सर्कल का जो सेंटर है मैन लो यहां कहीं है ये और ओरिजिन ये जो लाइन पीके है इसके एक ही तरफ है इसका ये मतलब हुआ की अगर यह पॉइंट पु के इस तरफ है तो यहीं कहीं यहीं कहीं आपका ओरिजिन भी होगा याद रहेगी ये बात कुछ कारण होगा कोई ना कोई रीजन होगा जिसकी हमें हेल्प लगेगी इस बात को उसे करने में तो ये हमारा डाटा है बस एक छोटी सी बात की अगर आप परपेंडिकुलर ड्रॉप करते तो वो जो रेडियस होती इन रेडियस उसकी लेंथ होती है वो याद रहेगा अब देखते हैं आज का पहला क्वेश्चन सबसे पहला क्वेश्चन इस सर्कल की इक्वेशन निकालनी है अगर मुझे सर्कल की इक्वेशन निकालनी है तो सर सर्कल की रेडियस तो पता है सर्कल के सेंटर के कोऑर्डिनेट्स पता करने हैं पहला क्वेश्चन suniyega अगला सवाल क्या है भाई सर अगला सवाल है एंड एफ आर गिवन बाय तो सर उसने आपको दी तो दे रखा है उसने आपको दी तो दे रखा है अब आपको ए और एफ निकलना है ठीक है सर सोचेंगे कैसे देखा जा सकता है और आज का तीसरा क्वेश्चन था इक्वेशन ऑफ डी साइट कर इन आरपी आर तो ये जो आपकी दो साइड से हैं कौन-कौन सी के आर आरपी इनकी भी आपको और इनकी क्वेश्चंस अब देख रहे हो सर्कल के साथ-साथ उसमें स्ट्रेट लाइंस को भी मर्ज कर दिया और इस तरीके से पूरा क्वेश्चन आपके सामने ए चुका है तो जरा ये वाला क्वेश्चन सॉल्व करो स्टूडेंट्स ये देख कर की सर इस सर्कल के सेंटर के कोऑर्डिनेट्स हम कैसे निकलेंगे इन केस इफ यू नीड अन्य हिंट तो मैं कहूंगा की आप पहले तो सर्कल के सेंटर से परपेंडिकुलर ड्रॉप करिए उससे क्या होगा सर उससे जो ये परपेंडिकुलर बनेगा शायद वहां से आप कुछ हेल्प लेकर कुछ आइडिया लेकर चीजें वर्कआउट कर पाएं चीज सोच पाए ऐसे कैसे सुनिए ध्यान से मैन लो मैं सर्कल के सेंटर को कहता हूं सी का सकता हूं क्या अब हम ये का रहे हैं आपसे जनाब की सी से आप ड्रॉप करिए परपेंडिकुलर ठीक है सर सी से हमने ड्रॉप किया क्या परपेंडिकुलर और यह जो आपने परपेंडिकुलर ड्रॉप किया है ये कुछ ऐसा सा है बिल्कुल सही बात है सर अब मुझे सी तक पहुंचना है अच्छा सी तक पहुंचने से पहले क्या आप मुझे पीके की स्लोप बता सकते हो क्यों सर बस ऐसे ही मुझे जानना है तो सर आपका हो गया पी की स्लोप यहां से निकल जा सकती है बिल्कुल तो जरा ध्यान से देखो अगर मैं अंडर रूट 3X को वहां पहुंचना हूं आई होप यही इक्वेशन थी आपकी अंडर रूट 3x + ए - एक्स बिल्कुल तो अगर मैं अंडर रूट 3X + ए - 6 में से √3x को उसे तरफ पहुंचा हूं तो क्या मैं इसकी स्लोप निकल पाऊंगा देखो भाई आई थिंक ये अंडर रूट 3X उधर गया तो -√3 तो shailise है ये जो लाइन पीके है इसकी जो स्लोप आएगी वो कितनी ए जाएगी सर वो ए जाएगी माइंस अंडर रूट थ्री सर आपकी बात मैन ली हमने अगर लाइन पीके की स्लोप आती है -√3 तो क्या आप सीडी की स्लोप निकल सकते हो कौन सी बड़ी बात है सर पीके पर परपेंडिकुलर है सीधी तो क्या सीडी की स्लोप निकलना बहुत टू टास्क के बिल्कुल भी नहीं सर तो सीडी की स्लोप क्या होगी आई थिंक सीडी है ना सो सीडी की स्लोप में कहूंगा सर इस लाइन की स्लोप का नेगेटिव रिसिप्रोकल यानी -√3 का अंडर रूट सॉरी माइंस अंडर रूट थ्री का नेगेटिव अंडर रूट 3 और उसका रिसिप्रोकल 1 / √3 तो सीडी लाइन जो है उसकी स्लोप आती है वैन बाय रूट थ्री सारी बातें छोड़ो सर क्या आप मुझे एक छोटी सी बात बताओगे क्या आप मुझे एक बड़ी बेसिक सी बात बताओगे suniyega ध्यान से बहुत कम की बात है मुझे यह पता है की यह जो दी पॉइंट है इसकी cardinate है और दी पॉइंट से वैन यूनिट डिस्टेंस पर आपका सी पॉइंट सिचुएटेड है आप सुनना ध्यान से तो अगर मैं सीडी की इक्वेशन निकल लूं दी पॉइंट के रिस्पेक्ट में पैरामेट्रिक फॉर्म में याद ए रहा है तो उससे मैं डिस्टेंस उसे करके सी के कोऑर्डिनेट्स निकल सकता हूं क्या थोड़ी सी याद दिलाता हूं आपको कुछ चीजों की याद करो स्टूडेंट्स हमने कुछ ऐसा पढ़ रखा है जीवन में की सर अगर आपके पास एक स्ट्रेट लाइन इफ यू हैव अन स्ट्रेट लाइन है ना इस स्प्रेड लाइन पर ये एक पॉइंट है जिसके cardridge है X1 - y1 आपको बस शर्तें ये पता है की इसकी स्लोप यानी इसका टैसेंट एक्स-एक्सिस के साथ जो आप बना रहे हो यानी क्या इसका इंक्लिनेशन वो टेन थीटा कितना है तो आपको पॉजिटिव एक्स के यहां तो इंक्लिनेशन भी पता है जो की है लेट्स से कितना थीटा अब अगर इससे आर डिस्टेंस पर कोई पॉइंट सिचुएटेड है अगर इससे आर डिस्टेंस पर अगर इससे आर डिस्टेंस पर कोई पॉइंट सिचुएटेड इसके कोऑर्डिनेट्स आप मैन लेते हो ह के तो आप वह कैसे लिखते हैं जरा इस बारे में बात कर लेते हैं भाई ध्यान से देखिएगा याद करिए सर आप लिखते हो ह-x1 आप लिखते हो अपॉन कितना अपॉन कोस थीटा = के-51 अपॉन साइन थीटा बल्कि इसके पीछे की तरफ भी तो देख सकती है तो जरूरी नहीं है मुझे सिर्फ ये cardinate मिले मुझे ये cardinate भी मिल सकता है और इसीलिए हम आर के आगे प्लस माइंस साइन लगाते हैं क्योंकि मुझे नहीं पता वो आगे जाकर या पीछे जाकर मिल रही है से प्रॉब्लम यहां आने वाली है मुझे नहीं पता की दी से मुझे पीछे जाना है आगे दी से मुझे पीछे जाना है आगे मैंने तो बस एक सिनेरियो बनाया है तो मैं कैसे निकलूंगा suniyega ध्यान से मैं कहूंगा आप सबसे बहुत ध्यान से सुनना मैन लो मैं सी के कोऑर्डिनेट्स वही बात मैन लेता हूं लेट्स से ह कॉम के अपनी सहूलियत के तो मैं क्या कहूंगा भाई मैं कहूंगा ह-3√3 / 2 मैं क्या कहूंगा मैं कहूंगा ह-3 रूट 3/2 यहां आएगा √3 देख कर क्या आपको इसकी स्लोप यस का इंक्लिनेशन पता चल रहा है सर वैन √3 होता है 10:30 तो टेक्निकल यह जो लाइन सीडी है यह पॉजिटिव एक्स के साथ 30 डिग्री एंगल बना रहे हैं तो यहां क्या लिख लेंगे सर यहां लिख लेंगे कॉस्ट 30 क्या ये बातें आपको समझ ए रही है सिमिलरली के-3/2k - 3 / 2 आई होप आप बातें समझ का रहे हो और ये कितना होगा सर ये होगा सिन 30 बस इस बात का आपको ध्यान रहे इल्म रहे की सर यहां पर जो आर है जो डिस्टेंस है वो वैन नहीं है उससे आप लीजिएगा प्लस माइंस में क्योंकि आपको नहीं पता की आप किस डायरेक्शन में गए तो मैंने दोनों पॉसिबिलिटीज डिस्कस की यहां तक कोई परेशानी अब हम एक बार प्लस वैन लेकर और एक बार माइंस वैन लेकर रूट्स सॉरी इस सी के कोऑर्डिनेट्स निकलेंगे और जब आप सी के कोऑर्डिनेट्स निकलेंगे तो आपको ये नजर आएगा सर की क तो मुझे दो कार्ड यूनिट्स मिल जाएंगे तो दो कोऑर्डिनेट्स मिल जाए तो कोई बहुत बड़ी बात है क्या कोई प्रॉब्लम है क्या उसने कहा तो था की सर जो आपका ओरिजिनल है और जो सर्कल का सेंटर है वो लाइन पीके की से साइड पर लाइक करते हैं तो शायद इस बात इन दोनों में से एक सेंटर के कोऑर्डिनेट्स को इल्यूमिनेट कर दो और दूसरा हमारा आंसर होगा पहले क्वेश्चन क्या समझ ए रहा है आपको तो बिल्कुल सर आगे बढ़ते हैं यहां से पहले प्लस वैन लेते हैं और जैसे ही प्लस वैन लिया तो suniyega ध्यान से ये जो लिखा हुआ है यह जो लिखा हुआ है कितना ह-3√3/2 अब डिनॉमिनेटर में क्या है कोस 30 और इस इक्वल तू क्या है वैन कोस 30 कितना होता है सर कोस 30 होता है √3 / 2 तो मैं यहां क्या लिख लेता हूं √3 / 2 सिमिलरली अगर मैं आपसे बात करूं तो एक और क्या लिखा हुआ है के - 3 / 2 डिनॉमिनेटर में क्या है सिन 30 सिन 30 है 1 / 2 तो 1 / 2 उधर गया तो कितना 1/2 कोई दिक्कत तो नहीं है सर जब इससे वहां पहुंचा तो ये हो जाएगा 4 टाइम्स रूट 3/2 से 4 डबल कैंसिल तो ये हो जाएगा 2√3 तो ह की वैल्यू आती है तू रूट थ्री और के की वैल्यू आती है यहां से 3/2 उधर गया तो 4 / 2 4 / 2 यानी 2 तो 1 सेंटर के कोऑर्डिनेट्स तो आएंगे तू रूट थ्री तू है ना इसी तरीके से अगर मैं सोचूं सर तो क्या एक और संभावना है एक और पॉसिबिलिटी है सर सेंटर के कार्ड के जो आपने सोच ही नहीं ह-3 रूट 3/2 / जो कोस 30 है याने के अंडर रूट 3/2 है वो -1 से मल्टीप्लाई हो तो ये हो जाता - √3 / 2 क्या मेरी बात आप समझ में सिमिलरली ये जो है के-3/2 इसके अपॉन में जो sin30 है वो माइंस वैन से मल्टीप्लाई हो तो वो हो जाता है -1/2 क्या बड़ी बेसिक सी बात है आप समझ का रहे हो और सर अगर आपने इससे थोड़ा एक्सप्लोर किया तो आप क्या कहोगे देखो जरा - ये इधर आया तो ये हो जाएगा ऐड तो थ्री में से वैन सब्सट्रैक्ट हुआ थ्री में से मैं सब्सट्रैक्ट हुआ तो तू और तू से तू कैंसिल तो कितना आता है ह ह आता है अंडर रूट थ्री कोई दिक्कत तो नहीं सिमिलर सिमिलरली सर ये जब उधर गया है तो 3 - 1 2 2/21 तो के की वैल्यू कितनी आती है सर के की वैल्यू आती है वैन तो एक सेंटर का कार्ड हमें कहता है √3 कमा वैन और एक कहता है 2√3 कमा 2 आप कौन सा चेंगे आप कौन कर लेंगे इस पर बात करते हैं आप याद करो एक बड़ी बेसिक सी प्रॉपर्टी हमने पढ़ी है क्या की अगर दो पॉइंट्स अगर दो पॉइंट्स किसी एक स्ट्रेट लाइन के जैसे ही आपकी स्ट्रेट लाइन है एक्स + बी ए + सी = 0 अगर दो पॉइंट्स किसी स्ट्रेट लाइन के एक ही तरफ अप्लाई करें जैसे की एक पॉइंट है क्या X1 y1 और एक और पॉइंट है इसके इस तरफ जो की क्या है एक्स तू Y2 तो सर बात बड़ी साधारण सी ये है की जब आप X1 y1 इसमें और X2 y1 X2 Y2 इसमें पास करेंगे तो दोनों ही या तो पॉजिटिव आएंगे या दोनों ही नेगेटिव आएंगे क्योंकि दोनों एक ही तरफ हैं पर अगर दो-दो ही पॉजिटिव हो या दोनों ही नेगेटिव हो तो उनका प्रोडक्ट यानी क्या यानी की सर ए X1 प्लस बी ए वैन प्लस सी वैन टाइम्स ओरिजिन यानी जीरो जीरो रख के देख लूं मैं की उसका साइन क्या है वही साइन इन cardinate से आना चाहिए की आप मेरी बात समझ पाए तो ये तरीका ज्यादा अच्छा है फटाफट हो जाएगा मेरे ख्याल से तो देखो सर पहले तो हमने इस स्ट्रेट लाइन में जीरो कमा जीरो पास किया तो जीरो इन जीरो तो ये बचा माइंस एक्स माइंस एक्स है ना नेगेटिव तो इस लाइन के इधर जो भी पॉइंट्स लाइक करते हैं इस लाइन के इधर जो भी पॉइंट्स लाइक करते हैं वो क्या होंगे सर नेगेटिव आपको कोऑर्डिनेट्स देंगे नेगेटिव वैल्यू देंगे इसमें रखने पर तो अब मुझे सी के वो कोऑर्डिनेट्स चुना है जिन्हें सी के मुझे वो cardinate चुना है जिन्हें उसे लाइन में अगर पास करें तो नेगेटिव वैल्यू है तो 2√3 पास करते हैं 2√3 अगर पास किया तो ध्यान से देखना तो बताओ दूसरा वाला आंसर होगा इन केस अगर आपको वेरीफाई करना है तो वेरीफाई कर लेते हैं देखो भी √3 कमा वैन रखकर √3 कमा वैन पास किया तो √3 √3 √3 3 + 1 4 - 6 - तो आप कौन सा वाला कौन सा वाला आप ऑप्शन एक्सेप्ट या सिलेक्ट करेंगे सर आपके लिए जो सेंटर के कार्ड होंगे वो होंगे √3 कमा वैन ये नहीं और जैसे ही मुझे पता चलता है √3 कमा वैन हमारे सेंटर के कोऑर्डिनेट्स हैं और सर ये सर्कल की रेडियस है वैन तो क्या मैं इसका आंसर लिख सकता हूं सर √3 वैन से मुझे क्लीयरली दिख रहा है ऑप्शन दी आपको नहीं दिख रहा है तो मुझसे पूछो मैं फिर से एक्सप्लेन करूंगा मुझे ऑनेस्टली और जल्दी से बताओ क्या आपको ऑप्शन दी में कोई दिक्कत सर बड़ी आसान सी बात है रूट 3 कमा वैन इसके सेंटर के कॉर्डिनेट इसकी रेडियस वैन तो कौन सी बड़ी बात है ये क्वेश्चन तो आसान था आई थिंक बहुत आसान क्वेश्चन था अब अगले सवाल में उसने पूछा आपसे आई और एफ के cordonts अब पॉइंट ए और एफ तक मुझे पहुंचना है ठीक है सर पॉइंट ए और एफ तक जब आपको पहुंचना है तो ए और एफ तक पहुंचने के लिए आपको कहा जाना है आई होप यह हम जान चुके हैं की सी के cardinate हैं सर सी के कार्डिनल्स आप निकल चुके हो ये हैं क्या ये है √3 वैन अब जब आई तक आपको पहुंचना है मतलब क्या मतलब यहां से एक परपेंडिकुलर ड्रॉप करना है और एफ तक पहुंचना है मतलब यहां से एक परपेंडिकुलर ड्रॉप करना है तो आपको इन पॉइंट्स तक पहुंचना है अगेन आपके पास काफी सारी इनफॉरमेशन है जरा दिमाग लगाइए और मुझे ये बताइए की आप ए और एफ तक कैसे पहुंचेंगे अच्छा सारी बातें छोड़ो बड़ा बेसिक सा कॉन्सेप्ट है बड़ी ही बेसिक चीज ज्यामिति लगाकर क्या मैं आपसे अगर पूछूं की पीढ़ी की लेंथ कितनी होगी तो आप बता सकते हो क्या मैं आपको हिंट देने की कोशिश कर रहा हूं स्टूडेंट्स मैं आपसे पूछना चाह रहा हूं की अगर मैं आपसे पूछूं पद की लेंथ निकल कर बताओ तो क्या आप मुझे पद की लेंथ निकल के बता सकते हो कोशिश करो उससे क्या होगा पद की लेंथ निकलने के बाद अगेन मैं आपसे पूछूंगा की पीके की जो इक्वेशन है वो आप निकल के बताओ मुझे पैरामीटर एक फॉर्म में रिपीट माय स्टेटमेंट मैं आपसे पूछता हूं की पद की जो लेंथ आपने निकल है बस यह लेंथ निकलने के बाद आप कोशिश करो आपकी स्ट्रेट लाइन पीके मैं हिंट देना चाह रहा हूं ताकि ये क्वेश्चन आप खुद से ट्राई करें पीके की इक्वेशन निकालिए पैरामेडिक फॉर्म में जैसा जस्ट अभी हमने निकाला है आपका इस वाले तरीके से सन का रहे हो सोच का रहे हो स्टूडेंट्स आई एम जस्ट गिविंग यू सैम टाइम्स सो डेट यू लर्न योरसेल्फ ताकि आपकी प्रैक्टिस करनी की आदत डालें अब आप यह सोच के देखो की अब क्या मैं ए और एफ तक क्या मैं ए और एफ तक किसी ना किसी तरीके से पहुंच सकता हूं कोई थॉट कोई आइडिया कोई भी हमारे दिमाग में कोई बात जिससे हम चीज निकल पाए जिससे हम चीज सोच पाए देखो बड़ी सिंपल सी बात है ऐसे सोचो मैं सबसे पहले तो कहां तक पहुंचना चाह रहा हूं मैं सबसे पहले पहुंचना चाह रहा हूं देखो हमारे पास क्या है सुना है बहुत सारी प्रॉपर्टीज है ज्यामिति में टेक्निकल एक एक आइडिया आपको देने की कोशिश करता हूं अगर अभी क्लिक नहीं किया तो सोच के देख अगर मैं आपसे पूछूं की आप मुझे बताओ ये लेंथ कितनी है वैन हान सर ये लेंथ वैन है बस एक छोटा सा शॉर्ट है क्योंकि यह इक्विलैटरल ट्रायंगल है तो ऑफ कोर्स यह एक मीडियम की तरफ बिहेव कर रही होगी यह सर हम जानते हैं एक मेडियन की तरह है ना यह क्लीयरली सर हम जान रहे हैं समझ का रहे हैं की यह क्लियर एक मेडियन की तरह बिहेव अब अगर ये एक मीडियम की तरह बिहेव कर रही है तो ये एंगल बाईसेक्टर परपेंडिकुलर बाईसेक्टर और एल्टीट्यूड तीनों की तरफ बिहेव करेगी तो ये अगर एंगल था 60° तो शैलेश है की सर ये एंगल हो जाएगा 30° क्या आप मेरी बात से एग्री करते हो मैं आपसे ये बोल लेना पूछना है बताना चाह रहा हूं स्टूडेंट्स की आपको ये डिस्टेंस पता है वैन तो क्या आप यह पीढ़ी की डिस्टेंस निकल सकते हो सोच के देखो निकल सकते हो क्या देखो ध्यान से सर ऐसे ही मैंने वैन की हेल्प से पद की डिस्टेंस निकले बड़ा ही आसान सा तरीका है सोच के देखो भाई या फिर अगर आपको वो नहीं निकाला है बट ने निकल लो चलो ना कोई ज्यादा बड़ा फर्क नहीं पड़ता है देखना अगर ये मुझे पता है लेंथ cd1 है अब देखो राइट एंगल ट्रायंगल है ना ये 90 डिग्री है 30 डिग्री है तो मुझे ये परपेंडिकुलर पता है और मुझे बेस चाहिए मुझे परपेंडिकुलर पता है और मुझे बेस्ट चाहिए बेसिक आपकी ज्यामिति बेसिक ट्रिगो तो मैं कहूंगा सर अगर ये वैन है और ये चाहिए जहां पर 10:30 अगर मैं उसे कर लूं तो बात बन जाएगी तो अगर मैं उसे करता हूं तन 30 है ना तो 10:30 का मैं कितना कहूंगा ध्यान से देखो सर 1030 जो होगा आपका वो होगा ध्यान से देखो इस ट्रायंगल में किस में डीपी में हो जाएगा सीडी/पद सीडी कितना है वैन तो 1 / पद आप निकल रहे हो आप क्या निकल रहे हो 1 / पद बेसिक सी बात है पद को यहां पहुंचा 10:30 कितना होता है सर 10:30 होता है 1 / √3 तो 1/1/√3 यानी कितना रूट थ्री तो आप एक बात और कनक्लूड कर पाए सर ये जो पद है ये जो पद है इसकी लेंथ है अंडर रूट थ्री ये क्यों निकाला सर आपने हमें समझ नहीं आया इससे फायदा क्या हुआ 1 सेकंड थोड़ा सा पेशेंट रखिए इससे फायदा ये हो रहा है की वापस गौर से सुनना वापस से सुनना क्या मैं इस स्ट्रेट लाइन पीके की इक्वेशन इस स्ट्रेट लाइन पी के की इक्वेशन क्या मैं अपनी वही इस फॉर्म में पैरामेट्रिक फॉर्म में लिख सकता हूं ऐसा क्यों सर सुनना एप्रोच लॉन्ग जो पर्सपेक्टिव जो लॉन्ग टर्म में आपको चीज दिखाना चाह रहा हूं सोचना अगर मैं इस लाइन पीके की इक्वेशन इस लाइन पीके की इक्वेशन अगर मैं लिखूं किस तरीके से इस तरीके से लिखूंगी सर दी पॉइंट से पैरामीटर फॉर्म में लिखूं तो उससे अगर मैं अंडर रूट थ्री ऊपर गया तो पी मिल जाएगा और √3 नीचे गया तो के मिल जाएगा आप समझ रहे हो मैं क्या कहना चाह रहा हूं सुनना ध्यान से अगर पी मिल गया अगर पी मिल गया तो सोच के देखो ये मेडियन है ये पी ए मेडियन है तो 2:1 के रेश्यो में आपका सेट्रॉयड डिवाइड कर रहे हैं जो की इन सेंटर सरकमसेंटर अलसो सेंटर सब है तो इसके cardinate और इसके कोऑर्डिनेट्स पर 2:1 वाला रेश्यो लगाकर आई थिंक यहां पहुंच जाऊंगा और ना सिर्फ पी मिलेगा बल्कि सर के भी मिल जाएगा और अगर के और सी पता है तो वही तू वैन वाला रेश्यो लगाकर मैं एफ तक भी पहुंच जाऊंगा तो आई थिंक आई और एफ निकल जाएंगे बस यही हिंट आपको समझने या देने की मैं कोशिश कर रहा हूं तो सबसे पहला तो क्या सर आपने डीपी निकल लिया बहुत खुशी की बात है अब क्या करेंगे आई होप आप यह बात रिलाइज कर का रहे हो की सर ये मेडियन भी है तो डीपी और डीसी इक्वल लेंथ के होंगे क्योंकि इक्विलैटरल ट्रायंगल है ये मेडियन परपेंडिकुलर बाईसेक्टर एंगल बाईसेक्टर एल्टीट्यूड सब बन जाएगा यहां पर है ना चीज बहुत शॉर्ट्स थीं तो अगर मैं आपसे पूछूं की सर दी से आप √3 यूनिट शिफ्ट करिए तो आप आगे भी जाएंगे पीछे भी जाएंगे वही प्लस माइंस एंड √3 तो एक से आपको पी और मिल जाएगा और एक से आपको के cardinate मिल जाएगा आप मेरी बात समझ का रहे हैं क्या तो बिल्कुल सर निकलती हैं ध्यान से सुनना अब जब आपने निकलना चाहा पॉड तो देखो क्या हो रहा होगा ध्यान से देखना भाई बड़ी सिंपल सी बात है पहले तो मैं इस लाइन की इस लाइन की इक्वेशन लिखना चाह रहा हूं इस लाइन की इक्वेशन लिखने में मुझे पता चला था की पीके की स्लोप कितनी थी माइंस अंडर रूट थ्री पीके की स्लो कितनी थी - √3 मतलब टांगें जो tanθ था वह था -√3 क्या आप मुझे -√3 को थोड़ा सिंपलीफाई करने में हेल्प कर सकते हो जैसे कोई कहे आपसे की तन थीटा जो है वो है माइंस अंडर रूट थ्री मैं ऐसा बोल सकता हूं क्या की सर 10 थीटा जो होगा वो होगा माइंस 1060 क्या मैं इसे 1060 बोल सकता हूं अभी फिलहाल आपको समझने की 60 को ऑफ कोर्स आप क्या लिख सकते हो 16 को आप लिख सकते हो क्या π/3 है ना अगर आप रेडियंस में जाना चाह रहे हैं लेकिन सर यहां माइंस है तो मैं ऐसा कर सकता हूं क्या की तन थीटा = ध्यान से देखना 10 अगर मैं पाई में से सब्सट्रैक्ट करता हूं पाई / 3 को अगर मैं पाई में से सब्सट्रैक्ट करता हूं π/3 को ऐसा क्यों सर ध्यान से देखो भाई यहां कहीं होता है 5 में से सब्सट्रैक्ट करते हैं सेकंड क्वाड्रेंट में साइन और कोसेक पॉजिटिव होते हैं तो पाई की वजह से 10 तो 10 रहेगा लेकिन ये क्या हो जाएगा -10 π/3 और ये मुझे मिल रहा है लेकिन जब ऐसा आप करेंगे तो आपको जो कंक्लुजन मिलेगा वो क्या की सर जो आपका 10 थीटा होगा वो होगा 10 कितना -π यानी 2π/3 तो क्या आप देख का रहे हो पाई / 3 देता जो है वह कितना है 2π/3 सर ऐसे कंपेयर कर सकते हैं क्या देता और इस एंगल को अभी फिलहाल आप मैन लीजिए कर सकते हैं जब त्रिकोण पढ़ेंगे तब हम समझेंगे की अभी मैंने क्यों किया तो थीटा कितना ए रहा है 2π/3 इसका इंक्लिनेशन ए रहा है 2π/3 तो अब अगर मैं इस पद एसपीडी या डीसी इस इस इक्वेशन पीके को पैरामेट्रिक फॉर्म में लिखना चाह रहा हूं तो थोड़ी देर के लिए मैं मैन लेता हूं जैसे की मैन लो पीके कोऑर्डिनेट्स हैं इसको हम आगे और ऑफ कोर्स उसी तरीके से के के भी निकल जाएंगे तो सुनना ध्यान से बहुत कम की बात ध्यान से कहना चाह रहा हूं क्योंकि मैं पहुंचना चाह रहा हूं पी और के तक मैं पहुंचना चाह रहा हूं पी और के तक तो मैं क्या कहूंगा ध्यान से मैं कहूंगा ह माइंस वही बात आपके किसके अकॉर्डिंग 3 रूट 3/2 और 3/2 तो कितना ए जाएगा ह माइंस अपॉन कोस 120 25/3 120 होता है 60 का डबल 120 कोई तकलीफ तो नहीं है अगर आपको तकलीफ है तो आप इसे लिख सकते हो 2π/3 बट मेरे ख्याल से ऐसा भी लिखेंगे तो कोई बहुत बड़ी विशेष बात नहीं है बट अगर आपको अच्छा नहीं लग रहा है आप इसे लिख लो क्या 2π/3 कोई डाउट तो नहीं है विल बी इक्वल तू विल बी इक्वल तू के माइंस 3 / 2 अपॉन कितना अपॉन sin25/3 एंड डेट इसे इक्वल तू अब जब आप पैरामेट्रिक फॉर्म में लिखना चाह रहे हो तो मैं प्लस माइंस एंड √3 ले रहा हूं मैन लेते हैं प्लस अंडर रूट 3 से पी मिल जाएगा और माइंस अंडर रूट थ्री से मुझे क्यों मिल जाएगा क्या ये आपको एप्रोच समझ ए रही है ए रही है अब सुनना ध्यान से अब जब यहां पे मैंने चीज वर्कआउट करने की कोशिश की तो क्या निकल कर ए रहा है गौर से देखोगे आई थिंक बहुत टू बहुत मुश्किल बात है नहीं पर पहले तो मैं कॉस्ट तू बाय बाय थ्री और साइन तू बाय बाय थ्री सिंपलीफाई कर लेना चाह रहा हूं क्योंकि हमने त्रिकोण नहीं पढ़ा है तो ये पार्ट शायद आपके लिए मुश्किल हो तो इससे मैं थोड़ा सिंपलीफाई कर डन क्या मैं कोस 2π/3 क्या मैं कॉस्ट उपाय अमरेली सॉरी 25 + 3π/3 को कुछ ऐसा लिख सकता हूं प्लीज ध्यान से देखना कॉस्ट बाय थ्री स्टूडेंट्स उसे मैं लिख सकता हूं क्या सिन 2π/3 को साइन माइंस एग्री करते हो कोस 2π/3 को π - π/3 और सिन 2π/3 को π-π/3 लिखा इससे क्या है देखो पाई के कारण कॉस्ट तो कॉस्ट ही रहेगा लेकिन π - मतलब सेकंड क्वाड्रेंट और वहां कोस होता है नेगेटिव बहुत बढ़िया यहां पर भी पाई के कारण साइन तो साइन रेश्यो ही रहेगा लेकिन π- सेकंड क्वाड्रेंट और यहां ये पॉजिटिव होता है तो ये हो जाएगा sinπ/3 कोई तकलीफ तो नहीं है सर ऐसा करने से क्या ऐसा करने से देखो कोस 60 कोस 60 यानी सिन 30 आई होप आप समझते हो यानी ये कितना हो जाएगा -1/2 कोई दिक्कत तो नहीं यहां मिल जाएगा अंडर रूट 3/2 कोई दिक्कत कोई डाउट कोई परेशानी तो इस तरीके से आपको कोस 2π/3 और सिन 2π/3 दिस वैन नथिंग बट व्हाट डीज व्हाट डी वैल्यूज ऑफ डीज वर डी वैल्यूज ऑफ कोस ये चूंकि अभी आपने त्रिकोण नहीं पढ़ा है इसलिए मेरा मानना है की बस आपको कोई परेशानी ना हो इस स्टेप को समझने में इसलिए मैंने इतना डिटेल एक्सप्लैनेशन दिया वर्ण जब हम ट्रिगो पढ़ लेंगे तो मैं कहूंगा की ये सारा कम आप अपने ब्रेन में कैलकुलेट करेंगे इतनी कैलकुलेशन आप ऐसे नहीं करेंगे अब क्या सर अब अगर मैं सोचता हूं इस वैल्यू को रखने का क्या कॉस्ट तू बाय बाय 3 - 1 / 2 और √3 / 2 तो कैसे सोचेंगे बात कर लेते हैं इस बारे में ध्यान से देखिए स्टूडेंट्स यहां पे ये वैल्यूज हम रखेंगे राइट तो रख के देखिए और निकालिए क्या ए रहा है तो सर सबसे पहले तो कोस 2π/3 जो हमने निकाला कितना सर ये निकल रहा है हमारा -1 / 2 है ना तो यहां रखा - 1 / 2 अब दोनों आएगा उसे पर बात कर लेते हैं एक बार प्लस रूट थ्री एक बार माइंस रूट 3 पर पहले से मैं सॉल्व कर लेता हूं ह माइंस कितना 3√3 बाय तू माइंस वैन बाय तू और ये जो ए रहा है सिन 2π/3/2 इटली एंड दिस इसे गोइंग तू बी प्लस माइंस अंडर रूट थ्री आई होप यही ए रही है दो चीज यहीं इस तरीके से दो चीज ए रही हैं माइंस वैन बाय तू और √3 / 2 अब सर अगर आप चीज निकलना चाहो तो मैं दोनों पहलू एक-एक करके एक्सप्लोर करूंगा पर पहले मैं थोड़ा सिंपलीफिकेशन कर लेता हूं ह की वैल्यू अगर हम यहां से निकल ले तो ह की वैल्यू क्या ए रही होगी सर ह की जो वैल्यू ए रही होगी वो अगर माइंस वैन बाय तू मल्टीप्लाई किया है ना माइंस वैन बाय मल्टीप्लाई किया तो ये हो जाएगा -+√3 / 2 और क्या √3 से √3 कैंसिल मतलब 3 बन जाएगा तो हो जाएगा 3/2 तो ये हो जाएगा प्लस माइंस 3/2 और इसमें आकर ऐड हो रहा होगा कौन थ्री बाय तू आई होप आपको ये बातें नज़र ए रहे हैं कोई तकलीफ कोई परेशानी तो सर एक बार वो ऊपर वाले साइन माइंस प्लस और प्लस प्लस उसे कर लीजिए एक बार प्लस प्लस और माइंस प्लस उसे कर लीजिए तो जैसे ही ऐसा किया तो मेरे ख्याल से मुझे जो cardinate मिलेंगे पहली बार जब हमने माइंस प्लस उसे किया तो अंडर रूट 3/2 को सब्सट्रैक्ट किया 3√3/2 में से तो ए जाएगा 2√3 / 2 आई होप आप समझ रहे हो तू से तू कैंसिल तो कितना बचा √3 तो एक cardinate क्या मिला अंडर रूट थ्री इसका ए कोड आप समझ का रहे हो ना नेगेटिव साइन उसे किया तो 3√3 में से √3 का है तो तू रूट थ्री तो तू साइड तू कैंसिल तो बचा √3 यहां पर प्लस प्लस उसे तो 3/2 3/2 तो आई थिंक थ्री क्योंकि ये हो जाएगा तू टाइम्स 3/2 तो तू से तू कैंसिल तो बचेगा थ्री तो एक cardinate तो ए जाएगा √3 अब दूसरा कॉर्डिनेट अब मैं उसे करूंगा यहां भी प्लस साइन यहां भी प्लस साइन तो रूट थ्री रूट थ्री प्लस वैन फोर रूट थ्री बाई तू उसको तू से कैंसिल किया तो ये हो जाएगा कितना सही हो जाएगा तू रूट थ्री कोई दिक्कत तो नहीं है अच्छा इसके बाद क्या सर इसके बाद अगर मैं आगे बात करूं आपसे तो फिर क्या सर यहां पे - साइन उसे किया तो 3/2 में से 3 / 2 सब्सट्रैक्ट किया तो ये ए जाता है जीरो तो सर एक तो आएगा √3 और एक आएगा 2√3 कॉमर्स जीरो एक आएगा √3 और एक आएगा 2√3 अब यहां जो उसे करना चाहें है ना ये है थ्री रूट थ्री बाय तू है ना जहां जैसा उसे करना चाहे आपकी कहीं भी लिख लीजिए कोई भी बहुत मेजर या बड़ा डिफरेंस नहीं ए रहा है तो एक जगह तो मैं देख लेता हूं क्या लेट से अंडर रूट थ्री कॉमर्स थ्री और एक जगह हम लिख लेते हैं 2 रूट 3 जीरो आई थिंक चीज अभी सॉर्टेड जैसा भी आपको लिखना है कोई बहुत बड़ा मेजर फर्क या डिफरेंस नहीं ए रहा है सर अभी तो आपने निकाला पी और के जब आपने निकल लिया है पी और के तो क्या आप अब ए और एफ निकल सकते हैं मेरा यकीन करिए आसान है ऐसे कैसे आसानी से बात समझो आपकी जो ये पे है ये जो पे है ये मेडियन हुई ना और मीडियम यानी की सेट्रॉयड सी मेडियन को 2:1 के रेश्यो में डिवाइड करता है अब मेरी बात समझ का रहे हो और अगर मुझे ए के कोऑर्डिनेट्स चाहिए अगर मुझे ए के कोऑर्डिनेट्स चाहिए तो मैं क्या कहूंगा मैं अगर इसे थोड़ी देर के लिए मैन लेता हूं आपको समझने के लिए X1 कमा y1 चलो हम मानते चले जा रहे हैं ह कॉम के तो मैं इसे लिख लेता हूं लेट से ह कॉम के तो आप क्या कहोगे सर ये रेशों जो है पीसी इस तू सी आई वो है तू इस तू वैन तो आप क्या कहोगे सर ध्यान से देखना तू ह प्लस करेंगे बहुत इजीली निकल सकते हो देखो कैसे सोचेंगे सर यहां से 3 मल्टीप्लाई हुआ तो 2h जो होगा वो कितना ए जाएगा वो ए जाएगा 3√3 और उसमें से एक रूट थ्री सब्सट्रैक्ट हुआ तो ए जाएगा 2√3 आई रिपीट माय स्टेटमेंट जैसे ही आपने क्या किया तो 2√3 और 2 से 2 कैंसिल तो ह की वैल्यू कितनी आती है √3 तो एक्स की वैल्यू कितनी आती है रूट थ्री है ना सिमिलरली यहां पर देखो भाई ये थ्री गया थ्री से थ्री कैंसिल आई होप आई रिपीट माय स्टेटमेंट थ्री गया थ्री से थ्री कैंसिल आई होप आप देख का रहे हो है ना और जैसे ही 3 से 3 आय होप ये थ्री माइंस थ्री जीरो तो के की वैल्यू कितनी जीरो तो ये ए जाएगा अंडर रूट थ्री जीरो कमा जीरो और आई होप अब मुझे कुछ कहने की जरूरत नहीं है याद आएगा अंडर रूट थ्री कमा जीरो और आपको दिख रहा है अगर के आई इस लाइन पर है तो 2√3 और √3 तो ये आपको दिख रहा होगा क्या है आपको दिख रहा होगा क्या है फिर भी अगर नहीं दिख रहा है तो मैं हेल्प करता हूं सुनेगा ध्यान से हम तो ने एफ तक पहुंचने हैं चलो अगर आपको यह दिख रहा होगा ना तो इससे और इससे भी आप इस तक ए सकते हो मेरा ये कहना है बट फिर भी मैन लो आपको ये नहीं दिख रहा है जो की इससे दोनों को देख कर देखना चाहिए क्योंकि ये लाइन का yordinate फिक्स है ना तो इसका भी वो एक ऑर्डिनेट फिक्स्ड है जीरो और आप डिस्टेंस देख लो 2√3 से √3 तो √3 डिस्टेंस तो रूट 3 से √3 डिस्टेंस और माइंस की तो क्या आपको नजर ए रहा है मैं कहां ले जाना चाह रहा हूं ने आपको ऐसे देखना है तो ऐसे देख लो सोच के देखो आपके पास सर कैफ एक वही बात वापस मैं इसको क्या मैन लेता हूं एफ के क्वाड्रेंट को अगर अभी फिलहाल सॉल्व करना है जहां भी मुझे वेरिएबल लग रहा है उसे मैन ले रहा हूं ह कॉम आके फिर वही बात स्टूडेंट्स आपके क्यों एफ को सी ने 2:1 के रेश्यो में डिवाइड किया है आपके कैफ को सी ने 2:1 के रेश्यो में डिवाइड किया तो फिर से वही बात सुना दिया 2:1 तो 2h आई होप समझ का रहे हो 2h+2√3 यहां पर क्या लिखूंगा मैं सर यहां पर मैं लिखूंगा 2h+2√3 / 3 = अगेन कौन √3 क्या आप मेरी बातें समझ का रहे हो भाई एंड सिमिलरली सिमिलरली क्या भाई तू के प्लस जीरो डिवाइडेड बाय तू प्लस वैन आई थिंक अब आप चीजे सोच का रहे हो जो मैं कहना चाह रहा हूं की आपको स्ट्राइक कर रहे हैं हम कहां जा रहे हैं देखो भाई ये ए जाएगा 3 यहां पर तो 3 रूट 3 रूट 3 में से 2√3 तो √3 और √3 को तू ने डिवाइड किया तो कितना ए जाएगा √3 / 2 आई होप आप समझ का रहे हो मैं जो कहना चाह रहा हूं वो है 2h ये हो जाएगा 3√3 में से 2√3 का है तो √3 तो आपको दिख रहा है क्या ह की जो वैल्यू है वो है अंडर रूट 3/2 मैं ऐसे स्टेप बाय स्टेप लिख रहा हूं पर यह सारी कैलकुलेशन आपको ब्रेड में करनी है ताकि आप टाइम बचाए यहां से मुझे कुछ कहने की जरूरत है क्या की सर के की जो वैल्यू होगी वो कितनी होगी सर की की जो वैल्यू हो जाएगी वो हो जाएगी 3/2 रूट 3/2 और 3/2 जरा सोच के देखो √3 / 2 और 3/2 क्यों होगा क्योंकि आर डेफिनेटली क्या रहा होगा जीरो तो एफ के कोऑर्डिनेट्स भी मुझे पता चल गए जो की क्या है सर एफ के कोऑर्डिनेटर जो हमें पता चले वो चले √3 / 2 कमा 3/2 मैं यही आपसे कहना चाह रहा था की आप चाहें तो अब मेडियन वाला सेट्रॉयड वाला तू इस तू वैन का कॉन्सेप्ट मत लगाओ इसका और इसका मिड पॉइंट है इसका और इसका मिड पॉइंट है एफ वैसे सोच लो की आप मेरी बात समझ का रहे हो भाई ये सारी डिस्टेंस अंडर रूट थ्री लैटरल ट्रायंगल इन इंडिया ने क्या यहां तक चीजे समझ आई सर इससे क्या इससे आई थिंक मुझे ए और एफ के कोऑर्डिनेट्स मिले जीरो और √3/2 है ना तो अगर यहां से मैं आगे देखूं अगर यहां से मैं चीज देखूं तो कौन सा ऑप्शन दिखता है भाई अगर यहां से देखें सर तो मेरे ख्याल से सर सीधा-सीधा आप ऑप्शन ए मार्क करके खुश रहिए क्या किसी भी स्टूडेंट को इस बात से कोई आपत्ति अगर मैं ऑप्शन ए मार्क करूं तो आई थिंक आप सभी को दिख रहा है सर अंडर रूट 3/2 3/2√3 कमा जीरो ये हमारे दो कोऑर्डिनेट्स हैं अब क्वेश्चन क्या बचा है अब क्वेश्चन में क्या पार्ट बचा है अब बचा है सर कर और आरपी की इक्वेशन बताओ मेरा कहना है अगर ये दोनों कर लिए ना तो इससे आसान क्वेश्चन कुछ नहीं है अब आपसे क्या पूछ रहा है वो कर और आरपी की इक्वेशन आई थिंक सर अब क्या बचा है अब तो जीवन आसान है के आर और आरपी की इक्वेशन आप क्यों नहीं लिख सकते आपके पास सब कुछ तो है अरे सर आपके पास सब कुछ है मतलब मैं क्यों बताऊं आपको करपी की इक्वेशन कैसे निकालनी है आपको क्या नहीं पता है आई थिंक यह तो अब आसान है सर कर मतलब क्या के के और आर के अकॉर्डिंग बताएं क्यों नहीं निकल सकते और से आरती के भी कोऑर्डिनेट्स आरोप के पता है क्यों नहीं निकल सकते मैन लो मुझे कर की इक्वेशन लिखनी है तो कर क्या है 2√3 कॉमर्स जीरो कमा जीरो और आरपी क्या है √3 और 0 क्या सीधे-सीधे लिखें 2√300 से बनने वाली इक्वेशन किस पूछ रहा हूं भाई 2√3 कमा जीरो और एक पॉइंट है क्या जीरो कमा जीरो इन से बनने वाली स्ट्रेट लाइन की इक्वेशन आप क्या लिखेंगे भाई सर इनसे बनने वाली स्ट्रेट लाइन की इक्वेशन जो मैंने सीखिए फिर वही बात ए - 0 = Y2 - y1 सो Y2 -5 0 - 0 आप समझ रहे हो तो ये क्लीयरली सर एक लाइन है जो की एक्स एक्सिस के पैरेलल है आप समझना बात को मैंने कुछ गड़बड़ तो नहीं की बस एक बार मैं नोटिस कर लेता हूं मैं कुछ तो गलत का रहा हूं गलत तो नहीं का रहा हूं क्योंकि ये लाइन क्लीयरली क्या है सर ये लाइन आपकी जो है इसका व्हाइट ऑर्डिनेट जीरो है तो एक्स एक्सिस के पैरेलल बस मैं एक बार चेक कर लेता हूं मैं कुछ गलत तो नहीं करो 2√3 कमा जीरो और जीरो कमा जीरो तो मैं क्या कर रहा हूं ध्यान से देखना ए - 0 = Y2 - y1 / X2 - X1 अब समझ का रहे हो अरे रिपीट मैं स्टेटमेंट Y2 - y1 यानी जीरो माइंस जीरो तो ये हो जाता है जीरो अपॉन X2 - X1 तो ये हो जाएगा कितना -2√3 और ऐसे दौरान बस ये ध्यान रखिएगा की आप एक छोटा सा कम और कर लें क्या सर अब छोटा सा कम ये कर लें एक्स - है ना आप यहां पे लिख लीजिए तो ये कितना हो जाएगा एक्स - कितना एक्स - आई थिंक मैंने सही लिखा है बिल्कुल सही तो ये हो जाएगा एक्स - कितना 2√3 क्या आपको सारी बातें समझ ए रही हैं बिना किसी परेशानी दिक्कत या तकलीफ के आई थिंक चीज आसान है अगर आपको चीज दिख रही हैं तो सर बिल्कुल सीधे-सीधे चीज दिख रही है 0 / एनीथिंग और इन एनीथिंग इस जीरो तो ये जा रहा है ए = 0 ये क्या जा रहा है सर ए = 0 कोई दिक्कत तो नहीं है भाई ये कौन सी स्ट्रेट लाइन है ए = 0 मतलब ये क्लीयरली कौन सा एक्सिस है एक्स एक्सिस मेरा कहना है सर इतना घुमा फिर के सोचने से अच्छा होता आप यही देख लो अब देखो ना एक्स एक्सिस पे लाइक करने वाले पॉइंट है मतलब ये एक्चुअली फ्लिप हो के ऐसा बन्ना था अगर आप ठीक है जैसा भी ए रहा है वो ए रहा है है ना जीरो कमा जीरो जीरो कमा जीरो के बाद ए रहा है √3 कमा जीरो और ये ए रहा है 2√3 कमा जीरो तो इन सबका yordinate जीरो है और एक्स koardinate चेंज होता चला जा रहा है तो ये क्लीयरली इक्वेशन क्या ए रही है ये इक्वेशन ए रही है ए = 0 मैं भी थोड़ा सा एक सेकंड के लिए कंफ्यूज हुआ क्योंकि हमने जो बनाया है और उसमें थोड़ा सिंक नहीं है तो ये रही आपकी इक्वेशन तो ये ए गया आपका ए = 0 सिमिलरली यहां से सोचो तो वैसे तो देख कर ही क्लिक हो जाएगा बट फिर भी अगर आपको मानना है तो निकालो 3√3 से सॉरी √3 से 0 3 0 सो √3 एक पॉइंट है और एक पॉइंट है 0 इस स्ट्रेट लाइन की इक्वेशन के बारे में आपका क्या विचार है suniyega ध्यान से ए - y1 है ना तो ये वाला ले लो = Y2 - y1 सो 3 / X2 - X1 √3 और एक्स - 0 तो ये हो जाएगा एक्स आई थिंक सर ये तो और आसान है √3y = 3 एक्स जैसा भी आपको दिखे ऐसा कोई एक ऑप्शन होगा या फिर आप इसे और सिंपलीफाइड मैनर में लिखना चाहते तो आप इसे ऐसे लिख लेते कैसे भाई की सर थ्री को √3 √3 लिख लेते हैं तो ए कितना हो जाता ए हो जाता अंडर रूट 3X डिस्क को हैव इन डी मोस्ट सिंपलीफाइड वर्जन तो एक तो होगा ए = √3x और एक होगा ए = 0 क्या आप मेरी बात समझ पाए तो इस तरीके से आपकी ये दो स्ट्रेट लाइंस की इक्वेशन आएंगे जो की मुझे दिखाना चाहिए इन दोनों ऑप्शंस में मतलब किसी एक ऑप्शन में रहकर तो यहां पर एक तो आपका है ए = √3x और एक है ए = 0 जिसको लेकर मेरे ख्याल से किसी भी स्टूडेंट को कोई आपत्ति नहीं होनी चाहिए आपको यकीन हो रहा है कितना डिटेल है ना कितना मतलब थारो आई वुड से और कितना मतलब कंप्रिहेंसिव कितना मतलब सब कुछ उसमें दे दिया उसने हर चीज पूछ ली सारा ओवर ऑल कंप्रिहेंसिव नॉलेज आपका चेक किया उसने सर्कल स्ट्रेट लाइंस एवरीथिंग देयर असिस्ट इन दिस हो थिंग्स आर गोइंग तू बी एस्क्ड तो इस क्वेश्चन को करने में तो वक्त लगेगा तो तीन क्वेश्चंस भी तो होंगे तीन क्वेश्चंस भी तो हुए और दिल क्या यार सैम गुड मार्क्स तो आपके इस तरीके से आपको चीज सोचनी है आई होप आपको समझ ए रहा है डिफिकल्ट नहीं है क्वेश्चंस कब जब आपने कॉन्सेप्ट्स पढ़कर रखे हो जब आपने बातें अच्छे से सिख कर राखी हूं है ना एक जरूरी बात स्टूडेंट्स याद रखिएगा की आपको क्या करके रखना है जब आप इस वीडियो के डिस्क्रिप्शन में जब जाओगे तो आपको क्या मिलेगा वहां पर आपको एक लिंक दिखाई देगा क्या लिखा हुआ मिलेगा फुल प्लेलिस्ट लिंक है ना तो वहां पर लिखा होगा सिंग आगे मैथमेटिक्स फॉर आईआईटी जी मांस एंड एडवांस्ड उसे प्लेलिस्ट को ओपन करोगे तो से प्लेलिस्ट या ऐसा कोई प्लस साइन दिखाई देगा उससे ये प्लेलिस्ट आप अपने अकाउंट में से करके रख लीजिए इससे आप हमेशा सिंक में रहोगे आपको हमेशा यू नो देखते रहेंगे की कितने वीडियो ए चुके हैं कितने रिलीज हो चुके हैं आपने कहां तक देख लिया है आपने कहां तक देख लिया और कितने बाकी हैं ताकि आपका कोई भी बैकलॉग क्रिएट ना हो आप हमेशा अपडेटेड रहो यू ऑलवेज स्टे प्राइस है वो डिटेल में हम ये कर रहे हैं ये बेसिकली में हम डिस्कस करते हैं की इंगेज का जो पैकेज है यह मैथमेटिक्स का जो पैकेज है 500 में डिवाइड है cardinate ज्यामिति कैलकुलस वेक्टर एंड 3D और ऑफ कोर्स आपका क्या अलजेब्रा और ट्रिगोनोमेट्री हम ऑलमोस्ट इसी सेंस में इसी इसी ट्रेडीशन में इसी इसी यू नो मैनर में इन सारी बुक्स को कवर करेंगे और हम बिल्कुल ये अशर करने रहे हैं आपको की हम हर चैप्टर हर बुक के हर चैप्टर के हर कॉन्सेप्ट को बहुत डिटेल में पढ़ेंगे मतलब उसके हर कॉन्सेप्ट को पढ़ेंगे उसमें दिए गए इलस्ट्रेशंस को जो की एग्जांपल के आपके क्वेश्चन से वो करेंगे उससे चैप्टर के ऑन एंड होने पर उसके एक्सरसाइज करेंगे और यकीनन तौर पर आप निश्चित रहिए की उसमें जो आर्काइव दिए हैं जो जी एडवांस के प्रीवियस इयर्स क्वेश्चंस दिए हैं वो भी करेंगे तो जो भी आपका सर्कस चैप्टर चल रहा है इसकी एक्सरसाइज हम कर रहे हैं फिर हम के एडवांस के प्रीवियस इयर क्वेश्चन होंगे और इस तरीके से हर चैप्टर को बहुत डिटेल में मैटर में पड़ेंगे आय होप आप ये समझ का रहे हो एक ही क्वेश्चन में टेक्निकल 4 क्वेश्चंस और वह अलग क्वेश्चंस और ऐसा बिल्कुल नहीं है जो बचपन में आप क्या करते द की किसी एक को किसी एक से मैच कर दिया हो सकता है किसी एक की मैपिंग या मैचिंग एक से ज्यादा एलिमेंट से हो रही हो तो इस तरीके से आपको चीज सोचनी है ऐसा हो सकता है ना और यही ज एडवांस्ड करवाइए आपसे तो एक अच्छा क्वेश्चन है सारे क्वेश्चंस एक-एक करके हम करेंगे शुरुआत करते हैं फर्स्ट क्वेश्चन से वो क्या का रहा है बढ़िया सर वो का रहा है ये जो दो सर्कल्स हैं एक दूसरे को टच करते हैं मेरा ख्याल है की अगर आपको याद है बातें तो ये बातें bhuliyega मत की अगर सर दो सर्कल्स एक दूसरे को टच करें तो हमने तो बड़ी आसान सी एक जरूरी बात ये सीखी है सर क्या की सर देखो इस सर्कल की सेंटर को लेकर आपका क्या विचार है इस सर्कल के सेंटर को लेकर मैं आपसे कहूंगा इसके कोऑर्डिनेट्स होंगे -1 0 और इसके सेंटर के कार्ड 0 - 0 - A2 कमा जीरो ठीक है सर तो इसका सेंटर जो हो जाएगा वो होगा -1 0 और इसका जो सेंटर होगा वो होगा -a2 कमा जीरो आई थिंक चीज आसान फिर क्या सर फिर वो ये का रहा है की ऑफ कोर्स ये दोनों एक दूसरे को टच करते हैं तो बताओ इसकी और इसकी रेडियस क्या होगी सर सीधी सीधी सी बात है a1² - बी √ ए तू स्क्वायर माइंस सर्कल की जो रेडियस लिखोगे वो क्या होगी सर वो होगी ऑफ कोर्स a1² - बी एन अंडररूट और यहां जब आप लिखोगे इसके रेडियस तो वो कितनी होगी सर वो होगी a2² - बी अब मेरा विचार है की सर जब आप सब जानते ही हो तो क्या आप यह कनक्लूड नहीं करोगे की देखो सर इन दोनों को अगर मैं करंट करूं तो इन सर्कल्स के सेंटर्स के बीच की डिस्टेंस इनकी रेडियस के सैम के इक्वल होगी क्योंकि ये दोनों एक दूसरे को टच करते हैं क्या ये बात आप समझ पाए आई होप आप ये बड़ा बेसिक सा डायरेक्ट शॉट सीधा-सीधा कॉन्सेप्ट समझ पाए बस इस बेसिस पे आप सॉल्व करेंगे इस क्वेश्चन का पहला पार्ट क्या यहां तक किसी भी स्टूडेंट को कोई तकलीफ कोई दिक्कत है तो पूछो भाई मेरे ख्याल से तो सब चीज आसान है बहुत टू बहुत मुश्किल बहुत बड़ी बात नहीं अच्छा सर आप एक सिनेरियो क्यों बोल रहे हो एक सिनेरियो आप ये नहीं बोल रहे हो क्या उसने आपसे बोला था की वो externalli टेस्ट ऐसा भी तो हो सकता था की सर यह दोनों सर्कस एक दूसरे को इंटरनली टच कर रहे होते और अगर यह दोनों एक दूसरे को इंटरनल टच कर रहे होते तो कहना इसे की इंटरनल इस तरीके से टच कर रहे होते बिल्कुल सही बात है सर और इंटरनल अगर टच कर रहे होते तो मेरा कहना है मैं उसे केस में इसको इस तरीके से चेक कर लेता की सर इनकी जो कमेंट टैसेंट होती है ना इनकी जो कॉमन टैसेंट होती उसे पर जो आप इनके सेंटर से जो परपेंडिकुलर ड्रॉप करते इनके सेंटर से जो आप परपेंडिकुलर ड्रॉप करते द वो उनकी रेडियस के इक्वल होता और क्या वही बात मैं यहां पर नहीं सोच सकता था क्या वही बात मैं यहां पर नहीं सोच सकता था की सर सेंटर्स के बीच की डिस्टेंस और रेडियस का सैम शायद यहां एप्लीकेबल ना हो बट एक कॉमन थॉट जो दोनों सीनरी में अप्लाई होगा वो ये की अगर दो सर्कल से एक दूसरे को एक्सटर्नल या इंटरनल टच करें तो उनकी कॉमन टांगें अब कॉमन टांगें को यहां पर मैं क्या कहूंगा वो ना सिर्फ इन दोनों की क्या है भाई कॉमन टांगें बल्कि वह इन दोनों की क्या है जल्दी से बता दो मुझे रेडिकल एक्सेस भी है सर इन दोनों की ना सिर्फ वो कॉमन टांगें है बल्कि रेडिकल एक्सेस भी है ना ये आपका सर्कल है S1 और यह आपका सर्कल है S2 तो इन दोनों सर्कल्स की क्या रेडिकल एक्सेस की इक्वेशन आपको लिखते आती है बिल्कुल आती है सर वो क्या होती है S1 - S2 = 0 बात करते हैं इस बारे में जब आपने निकलना चाहा s1-s2=0 तो देखो इसमें से इसे सब्सट्रैक्ट करना है x²y² तो हमेशा है ही जाते हैं अब क्या बच रहा है तू ए वैन एक्स माइंस तू ए तू एक्स और बी से बी कैंसिल तो क्या बच रहा है अभी अब हमारे पास बच रहा है 2a1x - 2a 2X है ना मैं क्या कहना चाह रहा हूं जो रेडिकल एक्सिस की इक्वेशन ए रही है वो ए रही है तू a1x - 2 a2x अगर आप ध्यान से देखोगे तो आप पाओगे की सर 2 और 2 हटा दो तो A1 और A2 कॉमन लिया तो ये हो जाएगा A1 एम रियली सॉरी वर्ण वो एक ही सर्कल बन जाता तो आपके पास संभावना क्या निकल कर ए रही है एक्स = 0 मतलब आप ये देख का रहे हो सर की इनका जो रेडिकल एक्सेस है इनका जो रेडल एक्सिस है वेदर हो एवर दे यू नो टच एच आदर इन केस ऑफ एक्सटर्नल और इंटरनल इट इस गोइंग तू बी एक्स इस इक्वल्स तू जीरो विच इस नथिंग बट वही एक्सेस एक्स = 0 का क्या मतलब इट इस क्लीयरली डी ए एक्सिस आई होप यू ऑल आर गेटिंग डेट अब मेरा बस आपसे ये कहना है की ये जो cardinate है ये जो कॉर्डिनेट है इससे इस सेंटर से इसकी परपेंडिकुलर डिस्टेंस निकालिए सिमिलरली इस कोऑर्डिनेट्स की परपेंडिकुलर डिस्टेंस निकालिए से इस sinariyon में करिए मतलब इन दोनों सर्च करके सेंटर से इनकी परपेंडिकुलर डिस्ट्रेस निकल लिए और वो किसके इक्वल रख दीजिए सर वो इक्वल रख दीजिए इनकी किसकी रेडियस के इक्वल कोई तकलीफ कोई बात कोई परेशानी इस बात से आई थिंक बस इतनी सी बात है की सर यह जो आपने रेडिकल एक्सेस निकाला है ये दोनों सर्कल्स को टच करें और दोनों सर्कस को टच करने का मतलब है की सर्कल के सेंटर से सर्कल के सेंटर से इन रेडिकल एक्सिस यानी कॉमन टैसेंट की परपेंडिकुलर डिस्टेंस आपकी क्लीयरली किस के इक्वल होनी चाहिए इसकी रेडियस के बस यही phenomenom अप्लाई करेंगे आई होप आपको हिंट समझ आई और आपने इस क्वेश्चन को अटेम्प्ट किया ट्राई किया क्या कहना चाह रहा हूं माइंस ए जीरो से क्या एक्स = 0 के डिस्टेंस आई होप ये आप खुद से वर्बल इंट्यूटिवली भी सोच सकते हो जब आप - A1 0 से डिस्टेंस निकल रहे हो तो क्लीयरली क्या होगी आप खुद सोचो ये क्या है -7 कमा जीरो सर ये एक्स एक्सिस पे लाइक करने वाला कोई पॉइंट है इससे सिर्फ ए-एक्सिस के डिस्टेंस कितनी होगी सर A1 मोड ले लोग ना क्योंकि डिस्टेंस है तो डिस्टेंस कितनी होगी मोड ए वैन तो डिस्टेंस कितनी होगी सर वो होगी मोड A1 आप चाहो तो डिस्टेंस फ्रॉम ना लगा लेना मैं का रहा हूं ऐसे ही कर लो और ये जो डिस्टेंस है वो किसके इक्वल है सर उनकी akording के डिस्टेंस इसकी रेडियस यानी की a1² - बी के इक्वल है एक और कंडीशन आपको मिलेगी कौन सी दूसरी वाली ये वाली आई होप आप समझ का रहे हो दूसरी कंडीशन क्या मिलेगी की सर आपकी जो इस सर्कल की रेडियस है इस सर्कल की जो रेडियस है यहां पर भी वही बात हो रही है ना रेडियस है और परपेंडिकुलर डिस्टेंस है ना तो इस सर्कल की जो रेडियस है वो क्या है सर माइंस ए तू सॉरी सेंटर से इस लाइन की परपेंडिकुलर डिस्टेंस कितनी है मोड ऑफ - A2 यानी क्या मोड ऑफ ए तू और ये किसके इक्वल होगी सर ये इसकी रेडियस के इक्वल होगी यानी कितनी अंडर रूट ओवर ए तू स्क्वायर माइंस बी आप से बात यहां भी अप्लाई करो इसके सेंटर के कोऑर्डिनेटर इसके सेंटर के cardinate है परपेंडिकुलर डिस्टेंस वही एक्स एक्सिस पर है और वो रेडियस के इक्वल होगी वही बात यहां से भी निकल कर आएगी जो वहां से निकल कर ए रही है एक ही कंक्लुजन पर आप ए जाओगे फाइनली इन दोनों का स्क्वायर अगर किया तो क्या मिलेगा सर लेफ्ट हैंड साइड पर दिखा तो ये मिल जाएगा a1² = a1² - बी आई थिंक बात सही है क्या ये बात आप सभी को समझ ए रही है आई थिंक इसी तरीके से क्या आप दूसरा कंक्लुजन निकल का रहे हो की सर दूसरा कंक्लुजन जो निकलेगा फाइनली वो क्या होगा सर वहां से मैं निकल लूंगा फाइनली दूसरा कंक्लुजन तो कितना हो जाएगा सर्व हो जाएगा a2² = देख का रहे हो की अब हम किस तरफ बढ़ रहे हैं अब हम किस तरफ पहुंच रहे हैं और हम किस तरफ चीजें कनक्लूड कर रहे हैं बहुत मुश्किल बहुत टू ये बातें नहीं है भाई क्या आप मुझे बता का रहे हो की आप क्या कंक्लुजन दोगे सर बी को लेकर मेरे ख्याल से आप खुद देख का रहे हो ये और ये कैंसिल तो बी की वैल्यू जीरो यहां से ये और ये कैंसिल तो बी की वैल्यू जीरो किसी भी केस से का लो तो बी की वैल्यू जीरो होनी चाहिए बी की वैल्यू जीरो होनी चाहिए और जब आपसे कोई कहे की बी की वैल्यू जीरो होनी चाहिए जब आपसे कोई कहे बी की वैल्यू जीरो होनी चाहिए तो देखो ध्यान से A1 A2 और बी क्या हो सकता है मेरा कंसर्न है बी जीरो इन सब में बी जीरो मुझे आर वाले केस में दिख रहा है A1 और A2 से फर्क नहीं पड़ता बस बी जीरो हो उतना काफी है तो सर सीधी सीधी सी बात है उसे केस में बस हमारे पास जो आंसर होगा वो क्या होगा सर a1 मतलब आप ऐसे इमेजिन करो आपके पास टेक्निकल ए एक्सिस है और यहां पर दो सर्कल टच कर रहे हैं ए-एक्सिस को तो या तो ऐसे कर रहे होंगे या फिर वैसे कर रहे होंगे तो क्लीयरली सर इस केस में बी नहीं होगा बस इतनी सी बात होगी बी नहीं होगा मतलब आपका ये जीरो होगा मतलब आपका जो फर्स्ट ए पार्ट है इसकी मैपिंग आप किस करोगे ए की मैपिंग होगी आर से मेरे ख्याल से एक आसान अच्छा डायरेक्ट और सीधा सीधा सा क्वेश्चन था जिसमें कोई दिक्कत या तकलीफ नहीं हुई और आप railise कर का रहे हो किस तरीके से चीजों को सोचना है सिर्फ एक पहलू पर बात नहीं करनी है की आप बस का दो सर एक्सटर्नल टच करते हैं आपको हर एक पहलू सोचना है और उससे फाइनली कंक्लुजन पर आना है याद रहेगी ये बात सर अगर ये बात याद रहेगी तो क्या अब हम ट्राई करें आज का नेक्स्ट क्वेश्चन बी क्वेश्चन में क्या दिया है वह का रहा है f2 सर्कल्स स्क्वायर प्लस ए स्क्वायर प्लस तू ए वैन एक्स प्लस बी सर ऐसा ही कुछ है और यह भी ऐसा ही कुछ है लेकिन इस बार कंडीशन क्या है दे टच एच आदर अगर वो का रहा है ये जो सर्कल है दे टच एच आदर तो फिर ये ट्रिपलेट A1 a2b क्या होगा इस बार बस आप इस बात पे ध्यान दीजिए स्टूडेंट्स की एक बात मैं करके जो मैंने थोड़ा सा रुक कर सोचा और देखा क्योंकि मुझे लगा शायद कोई टाइप होगा लेकिन एक फर्क है की देखो यहां 2 1 A1 तू ए वैन एक्स है और ये 21x है लेकिन यहां प्लीज ध्यान से देखो 2a 2X है और ये 282 वही आप मेरी बात समझ का रहे हो तो इस बार भी कुछ है लेकिन इस बार थोड़ा सा mindyut सा बड़ा ही बारीक सा चेंज है जिसे आप सभी देख का रहे होंगे और मैं वापस अगर बात करूं आपसे वही कंडीशन की जो हम पढ़ते हुए आए हैं की सर अगर दो सर्कल्स मुझे नहीं पता एक्सटर्नल इंटरनल पर अगर वो एक दूसरे को टच करें तो क्या नजर आता है सर सर ये बात तय है की उनके रेडिकल एक्सिस को निकालो और उससे ही चीज हम डील करेंगे तो क्या मैं फिर से यही बात करूं की सर रेडिकल एक्सेस के बारे में बात करते हैं चलो रेडिकल एक्सरसाइज के बारे में बात करते हैं कौन से दो सर्कल्स हैं एक सर्कल जो दिया हुआ है आपको वो दिया है x² + y² + बस यही फर्क ध्यान से देखना इस क्वेश्चन और इस क्वेश्चन में की यहां पे एक्स ही था और यहां पर ए ही था तो 21x2 a1x + बी है ना तो ये है तू टाइम्स a1x + बी = 0 और आपका जो दूसरा सर्कल है डेट इस 282 है ना तो ये है x² + y² + 2 + बी = 0 मैं फिर से आपसे वही कहूंगा मुझे नहीं पता ये दोनों एक दूसरे को externalli या इंटरनल टेस्ट करता है व्हाट आर व्हाट ऑल आय नो इस डेट एच आदर एंड दें तू सर्कल्स हैपन हैपन तू टच एच आदर व्हाट डू यू बेसिकली नो रेडिकल एक्सिस देयर रेडिकल एक्सरसाइज गोइंग तू बी डेट कॉमन टेंस इट एंड हैपेंस तू बी डेट कॉमन टेंसेज और डेट इट विल हैव उसकी जो डिस्टेंस होगी फ्रॉम डी सेंटर्स ऑफ डी सर्कल्स विल बी इक्वल तू देयर रिस्पेक्ट व्हाट - A1 बी कोई तकलीफ तो नहीं है बस ऐसे ही निकल लेते हैं इस सर्कल के सेंटर के अकॉर्डिंग ध्यान से देखना जीरो कमा - A2 है ना मैंने -ए बी क्यों लिखा पता नहीं कहां ध्यान है भाई -7 0 रेडियस के लिए बी के हेल्प लगेगी रेडियस के लिए भी की हेल्प की मतलब अगर रेडियस इन केस निकालनी होंगी तो क्या कहता है सर रेडियस के लिए मैं कहता बड़ी सिंपल सी बात की देखो भाई ये जब आप रखोगे क्या A1 तो ये हो जाएगा a1² और ऑफ कोर्स क्या -बी आप समझ का रहे हो भाई और सिमिलरली सर यहां से जब आप निकलती हो तो ये कितना ए जाता है सर ये ए जाता a2² है ना माइंस बी इस बार थोड़ा ध्यान से देखिएगा चीजे थोड़ी सी बदलेगी सर रेडिकल एक्सेस निकलने का बिल्कुल निकल लो रेडी का एक्सेस क्या ए रहा है सर रेडिकल एक्सरसाइज को लेकर मेरा विचार है दिस माइंस दिस x²y² x²y² है तो ये बचा क्या ये बचा 2 a1x है ना बी में से बी चला जाएगा और यहां बचेगा क्या -2 a2y आप समझ का रहे हो सर एक बात आप ध्यान से देखो 22 गया लेकिन ये बचेगा a1x है ना माइंस ए तू ए = 0 अब फिर वही बात मेरा तो कहना ये है की सर इस सर्कल के सेंटर से इस कॉमन टैसेंट की परपेंडिकुलर डिस्टेंस इसकी रेडियस के बाल और इस सर्कल के सेंटर से इस कॉमन टांगें की परपेंडिकुलर डिस्टेंस इसलिए इक्वल होनी चाहिए बड़ा बेसिक सा मसला है बिल्कुल सर यही कॉन्सेप्ट हम अप्लाई करेंगे और सोचेंगे क्योंकि यही हमने सिखा है और यही हम जानते हैं तो जब मैंने निकल माइंस A1 0 से इसकी परपेंडिकुलर डिस्टेंस तो देखो भाई तो ये हो जाएगा - a1² कोई दिक्कत तो नहीं फिर क्या सर माइंस A2 0 क्लियर सी बात है सर ये हो जाएगा जीरो और कुछ बचा नहीं है तो इस पर हम लगा देंगे मोड और मोड लगाने से अच्छा मैं कहूंगा देखो a1² पॉजिटिव ही है तो माइंस साइन ऐसे नेगेटिव ही बनाएगा तो मोड लगाने से क्या होगा इसे नेगेटिव है जाएगा आई होप आप समझ का रहे हो डिनॉमिनेटर में कितना आएगा सर डिनॉमिनेटर में आप इनके कॉएफिशिएंट का स्क्वायर का पूरा अंडर रूट लेते हैं तो वह अंडर रूट ओवर ए वैन स्क्वायर प्लस ए तू स्क्वायर होगा जो की आपका ए रहा था क्या a1² - बी कोई दिक्कत तो नहीं है क्या इसी आधार पर मैं यहां पर भी सोच सकता हूं बिल्कुल सोच सकते हो सर देखो जीरो इन ए वैन जीरो माइंस ए तू प्लस a2² और ओ पॉजिटिव ही रहेगा तो मोड लगाना है देस नॉट मेकिंग सेंस तो ये हो जाएगा प्लस a2² / वही बात इन दोनों के कॉएफिशिएंट्स के सैम का स्क्वायर तो वो हो जाएगा a1² + a2² और ये किसके इक्वल होगा इसकी रेडियस जो की क्या है अंडर रूट ओवर क्या ए तू स्क्वायर माइंस बी आई थिंक सर जब आप इन पार्ट्स को सॉल्व करेंगे तो शायद आप कुछ निकल पाएं बात करते हैं इस बारे में और देखते हैं की हम क्या निकल का रहे हैं मेरा कहना है की सर अगर दोनों तरफ स्क्वायर करें अगर दोनों साइड्स का स्क्वायर करें तो क्या बन रहा है गौर से देखो स्टूडेंट अगर दोनों तरफ स्क्वायर किया तो ये हो जाएगा A1 मतलब सम में उधर भेज देता हूं तो ये हो जाएगा a1² का स्क्वायर यानी a12 दी पावर फोर आप समझ का रहे हो और यहां जो मिलेगा इसका स्क्वायर क्या दे देगा सर ये दे देगा आपको A1 का स्क्वायर माइंस बी आप समझ का रहे हो स्टूडेंट्स और ये जो है इसका स्क्वायर क्या देगा ये दे देगा आपको a1² मैं पहले इसी को थोड़ा सिंपलीफाई करता हूं फिर इसी बेसिस पे उसे हम जनरलाइज कर सकते हैं तो देखो इससे जब सिंपलीफाई करने की कोशिश की तो ध्यान से देखना क्या मिल रहा है लेफ्ट हैंड साइड पर लेफ्ट हैंड साइड पर A1 तू दी पावर फोर है बिल्कुल है सर तो लेफ्ट हैंड साइड पर सी हैव विच विल गेट कैंसर की दिस थिंग बट मैं अभी फिलहाल आपको समझने के लिए लिख लेता हूं एंड दें व्हाट एल्स देखो भाई a1² a2² सो दिस इस गोइंग तू बी A1 A2 का होल स्क्वायर या फिर आप लिख लीजिए a1² एंड ऑफ कोर्स a2² व्हाट एल्स आर सी हैव इन हर और अगर आप ध्यान से देखें तो -बी a1² और -बी a2² क्या आप मेरी बात समझ का रहे हैं माइंस बी a1² और माइंस बी a² तो मैं यहां पर लिख लेता हूं क्या यहां पर लिख लेता हूं -बी एवं स्क्वायर माइंस बी a² सर अगर आप इसे सिंपलीफाई करें तो आपको एक रिजल्ट जरूर मिल रहा होगा वो क्या सर देखो आप इससे इसको कैंसिल होता है देख का रहे हो और ये क्लीयरली ए रहा है जीरो के और अगर कुछ सिंपलीफिकेशन आप देखना चाहो तो आप चाहो तो इसे उठाकर इधर ले आओ आप चाहो तो इसे उठाकर इधर ले आओ तो जैसे ही इसे उठाकर इधर ले हम जैसे इसे उठाकर इधर ले तो मैं कहता हूं ये a1² और 2² को इस तरफ लता हूं है ना तो क्या हो जाएगा ये हो जाएगा - a1² और a2² अब आप देखो ये सब प्लस हो जाएंगे ये सब क्या हो जाएंगे प्लस और अगर बी कॉमन ले लिया तो मैं फाइनली बी की वैल्यू लिखता हूं बी की वैल्यू जो लिखूंगा वो होगा a1² + a2² / a² a2² क्या है सर शायद ऐसा का ऐसा ठीक बिल्कुल ऐसा कंक्लुजन शायद आप यहां से का रहे होते कैसे समझो ये हो जाता है a2² ये वापस वही बात आती है फिर ये a2² a2² A2 की पावर 4 बना था क्योंकि आप इसको कैंसिल कर रहे होते और ये भी मतलब ये स्क्वायर कर रहे होते आप समझ का रहे हो जो से वहां किया तो वो ए तू की पावर 4 यहां के तू के पावर फोर से कैंसिल हो जाता था अब जो बचत वो क्या वही a1² a2² फिर क्या बन जाता a1² बी यहां से a2² है चुका था तो a2² बी फिर वही बात है निकल कर आती जितना मुझे लग रहा है तो शायद आप वहां से भी इसी कंक्लुजन की तरफ पहुंच रहे होते जो की है की सर बी कुछ इस तरीके से डिज़ाइन हो बी कुछ इस तरीके से डिज़ाइन हो की वो इस कंडीशन को फुल करें अगर आपका बी इस कंडीशन को फुल फिल कर रहा है तो आप कहेंगे की सर यही आपका फाइनल आंसर होना चाहिए अगर बी आपका इस कंडीशन को फुल फिल कर रहा है तो मेरे ख्याल से सर ढूंढने की कोशिश करते हैं की ऐसा कौन सा ऑप्शन है जो इस कंडीशन को फुल कर रहा है इस कंडीशन को आर एस जो आपके ऑप्शन हैं दोनों का स्क्वायर का सैम डिवाइडेड बाय दोनों के स्क्वायर का प्रोडक्ट है ना तो देखते हैं भाई अगर मैं बात करूं इसकी प्लेट की बात करें A1 a2b की तो A1 A2 मतलब इनके पहले ये दो स्टेटमेंट हमको ने समझ में ए रहा है तो इनके स्क्वेयर्स का सैम क्या है ना सुनना ध्यान से पहली बात तो यहां पे दोनों से हैं फिर भी कर ही लेते हैं फटाफट देखो 2² 4 2 का स्क्वायर 4 + 4 कितना आते तो 4 + 4 कितना है 8 आप समझ पाए तू का स्क्वायर 16 और 8 / 16 कितना होता है 1 / 2 8 / 16 कितना होता है सर 1 / 2 क्या बात आपको समझ आई और ये क्लीयरली हमको दिख रहा है बी के इक्वल नहीं है सॉरी मतलब 2 के इक्वल नहीं है क्या ये बात आपको समझ आई तो यहां से मैं कहूंगा की सर ये तो हमारी कंडीशन को सेटिस्फाई नहीं कर रहा है बस एक बार और से कर लेते अगर मैंने कोई गलती की हो तो बस इसे थोड़ा मैं यहां पर देख लेता हूं क्योंकि मुझे कुछ शायद गड़बड़ लग रही है क्योंकि हमने थोड़ा सा ध्यान से नहीं की ये चीज है ना A1 तू डी पावर 4 ये लिखा हुआ है यहां पे a1² और a1² है जो की आपको ए डी पावर फोर देगा अब आपके जो बचेंगे वो क्या वो बचेगा a1² a2² और क्या मिलेगा -बी a1² - बी a2² बिल्कुल सही बात है है ना और अगर यहां से देखें तो ए वैन तू दी पावर फोर आपका कैंसिल हो चुका था तो बच जो रहा था वो क्या बच रहा था a1² a2² बिल्कुल सही बात और जो बचा था -बी स्क्वायर और प्लस a2² बिल्कुल सही बात और उससे हमने जब निकाला तो मैंने कुछ गड़बड़ किया है बिल्कुल मैं बेवकूफी तो करता हूं पता नहीं क्यों मैं क्यों बेवकूफी करता हूं है ना जब आप देखोगे सर तो देखो आप बी कॉमन लो तो बी जैसे ही आप कॉमन लोग तो पता नहीं क्यों आप ऐसी गलती करते हो है ना मतलब ऐसे सिली एरर्स आप प्लीज लाइफ में कभी मत करिए थोड़ा सा फोकस मैनर में चीज करिए आउट ऑफ डी ब्लू पता नहीं कैसे चीजे मैंने बस ऐसे लिख दी बस ये बहुत बहुत अजीब सा ब्लड है क्योंकि इससे आपके सारे आंसर की गलत हो जाएंगे है ना आप समझ का रहे हो बस ये सब कुछ सही करके आप गलत कर देते हो ये प्रॉब्लम है इस तरीके के है ना तो बी कितना ए जाएगा सर a1² / a1² + a2² अब सोचते हैं अब अगर मैं बात करूं अब मैं अगर सारे ऑप्शंस की एक-एक करके बात करूंगा ना तो दोनों के स्क्वायर का प्रोडक्ट अपॉन दोनों का स्क्वायर इसका प्रोडक्ट पी तो लिखोगे लिखोगे क्या और हो सकते हैं देख लेते हैं सर क्यों हो सकता है क्या फिर वही बात इन दोनों के स्क्वायर का प्रोडक्ट इन दोनों के स्क्वायर्स का प्रोडक्ट कितना होगा देखो भाई आई होप आपको न्यूमैरेटर में क्या लिखना है प्रोडक्ट राइट और डिनॉमिनेटर में सैम तो न्यूमैरेटर में लिखा प्रोडक्ट तो कितना हो जाएगा सर 1² 1² 1 तो 1 1 ये कितना ए रहा है वैन कोई दिक्कत तो नहीं अब क्या सर डिनॉमिनेटर में दोनों के स्क्वायर्स का सैन तो 1 + 1 कितना 2 और ये ए रहा है 1 / 2 1 / 2 से मैच हो रहा है हो रहा है सर तो पी बल्कि सर आपका कब भी इसको फुलफिल कर रहा है किसी भी स्टूडेंट को कोई तकलीफ सर ऑप्शन आर तो आंसर नहीं हो सकता क्यों सर क्योंकि आप इनके स्क्वायर का प्रोडक्ट करोगे और उससे किसी से डिवाइड करोगे तो कटाई जीरो आने का स्कोप चांस नहीं दिख रहा आप देख लो अब करके देख लो तो आर तो नहीं है इसके बारे में क्या ख्याल है सर जब वैन कमा 1 पैर रखते हैं तो 1 / 2 आता है थ्री बाय फाइव नहीं तो एसबीआई रिजेक्ट हो सकता है तो के लिए क्या होना जाएंगे पी और के चीज समझ ए रही है आई थिंक आसान है अब आते हैं थर्ड क्वेश्चन पर वो क्या का रहा है इस स्ट्रेट लाइन स्ट्रेट लाइन a1x - b1y + b² ये स्ट्रेट लाइन है वो इस सर्कल को टच करती है एक आसान सा सर्कल है अगर आप ध्यान से देखो तो अगर कोई स्ट्रेट लाइन किसी सर्कल को टच करती है तो क्या आप मुझे बताएंगे सर इससे आसान तो क्वेश्चन आज का होगा नहीं देखो भाई सर ये तो आज का सबसे आसान क्वेश्चन है इसमें तो मेरे ख्याल से ज्यादा मेहनत नहीं लगेगी सर ये तो आसान है ना ये तो किया है हमने कैसे किया है सर आप सिंपल सी बात सोचो पहले तो आप सर्कल बताओ सर्कल क्या है सर आपका सर्कल है x² + y² और क्या है बस यही हमारे कम के फैक्टर्स हैं जो की हैं माइंस ए तू एक्स माइंस पी ए ध्यान से सुनना - a2x - बी ए = 0 अच्छा आई होप यहां तक कोई दिक्कत नहीं है अच्छा इसके साथ साथ लाइन की इक्वेशन क्या है अब समझ रहे हो डेरिवेशन बेस्ड क्वेश्चन आएंगे हो यहां पर नंबर दे देता तो सिर्फ वह ऑप्शन देखता पर इस तरीके से उसने ये दे दिया तो एक ही चीज के लिए मल्टीपल ऑप्शंस करेक्ट हो सकते हैं उसने ऐसी संभावनाएं बनाएं और ऐसे ऐसे इंटेलिजेंट क्वेश्चंस को आपसे पूछेगा रत्न वाले क्वेश्चंस नहीं पूछेगा की बस फॉर्मूला रात लिया और अप्लाई कर दिया और आंसर निकलना है नहीं जनाब कैलकुलेटर कर लेगा ये कम तो है ना आप तो दिमाग लगाओ आप इंसान हो तो देखो क्या हम इवॉल्व इसलिए हुए ना ह्यूमंस हैं हम हम आगे क्यों बढ़ पाए हम प्रोग्रेस क्यों कर पाए हम बाकी एनिमल्स बाकी स्पीशीज में सबसे आगे क्यों निकल पाए बिकॉज सी कॉल सर दिस नोट स्मार्ट टेस्ट ऑल दो हमसे भी स्मार्टर एनिमल्स हैं डॉल्फिन हमसे भी ज्यादा अच्छा ब्रेन होता है उसका पर फिर भी हम इवॉल्व हो पाए हम आगे बढ़ पाए तो अपने ब्रेन का उसे करिए और जिंदगी को अच्छा बनाया है ना एम्टी को प्रॉस्पर करने में हेल्प करिए तो देखो भाई क्या करना है यहां पर यहां पर अगर मैं बात करूं तो ये आपकी स्ट्रेट लाइन है a1x - बी ए + b² है ना स्ट्रेट लाइन क्या है सर a1x - बी ए एम रियली सॉरी प्लस बी स्क्वायर बिल्कुल कम है ही नहीं मेरी लाइफ में इन दोनों के नेगेटिव हो जाएगा सर ये हो जाएगा a2/2 ये हो जाएगा बी / 2 तो ये हुआ इसका सेंटर इसकी रेडियस निकल लो वो हो जाएगा सर ए तू स्क्वायर बाय फॉर प्लस बी स्क्वायर बाय फोर माइंस जीरो है ही नहीं तो बात खत्म और इसका क्या ले लीजिए आप अंडर रूट ये इसकी क्या हो गई सर ये इसकी हो जाती है रेडियस कोई दिक्कत तो नहीं है अब यह आपके प्रेसिडेंट है इस सेंटर से इस इंजन पर जब आप परपेंडिकुलर ड्रॉप करेंगे तो क्या वो रेडियस होगा बात खत्म बस इतनी सी बात आपसे कही जा रही है तो सर ट्राई करते हैं ना माइंस बी तो माइंस बी इन माइंस बी बाय तू थॉट्स - b² / 2 और ये क्या लिखा है प्लस बी स्क्वायर और इस पर आप लगाइए मोड ये हुआ आपका न्यूमैरेटर डिस्टेंस फॉर्मूला वाला A1 A2 + b1 B2 मतलब एक्स वैन प्लस बी ए प्लस सी आई होप आपको याद ए रहा है डिवाइडेड बाय अंडर रूट ओवर ए स्क्वायर प्लस बी स्क्वायर तो a1² + b² ये कितना हो जाएगा अंडर रूट ओवर क्या a1² है ना विच विल बी इक्वल तू दिस रेडियस जो की मैं यहीं लिख डन ताकि मेरा डबल से लिखने का कम खत्म हो जाएगा अब आप करो सर स्क्वायर अब क्या करेंगे स्क्वायर कर देते हैं तो जब स्क्वायर किया जब स्क्वायर किया तो सबसे पहले तो एक अच्छा कम मुझे ये दिख रहा है की यहां से 4 और यहां से फोर कॉमन ले लो सर ये 4 और ये फोर कॉमन लिया तो यहां पे तो प्लस लग जाएगा और बाहर ए जाएगा तू बाहर क्योंकि अंडर रूट से बाहर ए रहा है मैं इस पुरी इक्वेशन को तू से मल्टीप्लाई और तू से डिवाइड करता हूं क्यों कर रहे हो सर देखो जैसे ही तू से मल्टीप्लाई किया जैसे ही तू से मल्टीप्लाई किया तो ये है जाएगा ये है जाएगा और यहां पे क्या ए जाएगा सर तू कोई दिक्कत तो नहीं और तू से डिवाइड भी तो करना है पर तू से जब आपने डिवाइड किया तो यह वैन बाय तू है गए आई होप कोई परेशानी नहीं है सर एक और कम 2b² में से b² सूत्र किया तो ये कितना बचेगा सर ये बचेगा 202 तो ये हो जाएगा कितना b² कोई दिक्कत तो नहीं भाई आई थिंक चीजें आसान है अब सर स्क्वायर कर दो दोनों तरफ है ना अब अगर मैंने दोनों तरफ स्क्वायर किया तो मुझे कुछ सिंपलीफाइड सा दिखाना चाहिए तो यहां स्क्वायर किया तो ये कितना हो जाएगा सही हो जाएगा a2² + b² और इससे भी इधर ले आए तो ये कितना हो जाएगा सही हो जाएगा a1² + b² विच इस इक्वल तू विच इसे इक्वल तू मोड ऑफ मोड ऑफ क्या मोड ऑफ ए वैन ए तू प्लस ये बी स्क्वायर और इसका मैं क्या ले लूंगा सर स्क्वायर क्यों ले लिया आपने क्योंकि ऑफ कोर्स ये अंडर रूट में हटा चुका हूं ना अब जैसे हम कर लेते हैं सिंपलीफाई जब आप इसे सिंपलीफाई करोगे तो ध्यान से देखो भाई सिंपलीफाई करते वक्त क्या दिखेगा और इस मोड का ये स्क्वायर है राइट तो मोड का कोई फर्क पड़ता नहीं है यहां पर देखो ये ए जाएगा a2² और एवं स्क्वायर सभी को दिख रहा है तो क्या देखेगा a1² और a2² कोई दिक्कत तो नहीं और क्या सर देख रहे हो और ध्यान से देखो अब दिखेगा a2² b² और क्या दिखेगा a1²b² आप समझ रहे हो ना तो ये लिख लेता हूं मैं क्या दिखेगा सर एक तो दिखेगा a1²b² और एक आपको दिखेगा a2² b² और अगर आप एक टर्म और देखोगे तो वो दिखेगा b² b² जो की होगा बी तू दी पावर 4 ठीक है सर और क्या और अगर हम इस तरफ देखें तो A1 A2 + b² का स्क्वायर जो की कितना हो जाएगा सर ध्यान से देखो इसका जब स्क्वायर होगा तो वो होगा a1² a2² अब बी स्क्वायर का स्क्वायर यानी बी तू डी पावर 4 सही लिख रहा हूं क्या बिल्कुल सही और अगली जो टर्म आएगी वो होगी प्लस तू एवरीवन वो क्या होगा +2 A1 A2 b² अब ध्यान से देखो भाई सर a1² मुझे दिख रहा था लिखते वक्त और ये भी मुझे दिख रहा था लिखते वक्त अब ध्यान से देखो इस टर्म को उधर लता हूं तो ये हो जाता है a1 जीरो एक बात तो ये नजर ए रही है सर यहां से जो कंक्लुजन नज़र ए रहा है a² + b² - 2ab तो ये क्या हो जाएगा सर ये ए जाएगा A1 - A2 का होल स्क्वायर है ना क्या आप मेरी बात समझ पाए और यहां से क्या मिल रहा है सर यहां से मिल रहा है बी = 0 यहां से क्या मिल रहा है सर यहां से मिल रहा है A1 जो है वो A2 के इक्वल हो तो या तो क्या हो A1 और A2 इक्वल हो या फिर क्या हो बी 0 दोनों में से कोई सी भी बात मुझे मिलेगी क्योंकि आप इसका और इसका प्रोडक्ट जीरो है मतलब या तो ये जीरो हो या ये जीरो अब मेरी बात समझ का रहे हो जहां पर भी A1 या A2 इक्वल होंगे या 20 ऑप्शन होगा अब A1 और ए तू इक्वल कहां-कहां है सर A1 और A2 इक्वल है आपके कंक्लुजन है बस ऐसे एक बात थोड़ा सा कम से कर लेते हैं कहीं हमने कुछ गलती तो नहीं की बस एक बार थोड़ा सा पार्ट क्रॉप चेक कर लेने में कोई बुराई नहीं है तो सर ये आज ए रहा था आपका कितना देखो इसमें यह ए जाता है तू ए वैन ए तू बी स्क्वायर है ना 2a1a2b2 और यहां पर क्या था ये ये था a1² और ये था a2² बिल्कुल सही बात है सर और यहां से जब आपने कर तो इस तरफ लेकर आए तो सब में आपको b² कॉमन दिखा तो b² आपने कॉमन लिया और जो बच गया वो आपका क्या बचा वो आपका बचा a1² + a2² - 2a1a2 है ना और अगर इस कंडीशन को देखा जाए अगर इस कंडीशन पर बात की जाए तो आई थिंक यहां से मैं कनक्लूड कर का रहा हूं की a1² - a2² - 2a1a2 है ना मतलब A1 - A2 का होल स्क्वायर और ये बी स्क्वायर तो ऑफ कोर्स A1 और A2 इक्वल हो और बी जीरो तो या तो यह हो तो एवं और A2 इक्वल है क्या सर A1 और A2 जो इक्वल हमने देखे वो क्लीयरली ऑप्शन पी ऑप्शन के में देखें और ऑप्शन एस में भी देखें और बी जो जीरो है वो आपको ऑप्शन आर में है तो मैं क्या कहूंगा सर मैं सी ऑप्शन के लिए जो कंक्लुजन दूंगा जो सी ऑप्शन के लिए कंक्लुजन दूंगा ध्यान से सुनना स्टूडेंट सी टेक्निकल सारे ही ऑप्शंस को अपने अंदर लिए बैठा है क्योंकि आप देखो ना आपको A1 और A2 आई होप आपका A1 और A2 है बिल्कुल A1 और A2 इक्वल है आपके पी के और एस ऑप्शन तो या तो ऐसा हो या फिर कैसा हो जाएगा या फिर ऐसा हो जाए की आपका जो बी की वैल्यू है वो जीरो हो जाए तो इन तीन ने A1 और A2 इक्वल कर दिया और इसने बी को जीरो कर दिया बस इसको एक और अलग तरीके से मैं चेक कर लेना चाह रहा हूं जैसे मैन लो पी है आपका 222 अगर पी है आपका 22 तो क्या टू-टू 2 यहां पर भी आपको दिख रहा है आई थिंक क्या बुराई होगी इसमें मेरे ख्याल से तो कोई बुराई नहीं होनी चाहिए की तू का स्क्वायर 42 का स्क्वायर 4 + 48 बस एक बार चेक करो क्योंकि चारों सही है इसके और यहां पर क्या जा रहा है ये ए जा रहा है आपका 2 2 है ना बिल्कुल सही बात है 8 - 8 0 ए रहा है बिल्कुल सही बात तो बिल्कुल सही बात है की वो चारों ही ऑप्शंस आपके ट्रू होंगे और चारों ही ऑप्शंस ट्रू होंगे मतलब आप क्या कहोगे आप कहोगे पी के आर और एस कोई दिक्कत तो नहीं है अब आते हैं हम आज के हमारे दी ऑप्शन पर अब दी ऑप्शन में क्या लिखा है वो देखते हैं सर दी को जब आप ध्यान से देखेंगे तो दी को तो मेरे ख्याल से आप सब सॉल्व कर सकते हो क्योंकि दी बड़ा ही आसान सा वैसा ही क्वेश्चन है जो अभी तक हम करते हैं जो हमने लास्ट पिछला क्वेश्चन किया बस उसी तरीके की चीज अगर आप सोच पाओ तो ऐसे कैसे सर अरे भाई ऑप्शन मतलब दी में तो क्या है दी तो खत्म हो जाना चाहिए ना पद्धति है देखो वो क्या का रहा है ये जो लाइन है अब लाइन में वैल्यू दे दी है वो इस सर्कल को टच कर रही है तो सर ये तो आसान है सर्कल के पहले तो ये बताओ सर्कल बड़ा सुलझा हुआ है सर्कल के सेंटर के कोऑर्डिनेट्स हैं A1 A2 और सर्कल की रेडियस है रिपीट माय स्टेटमेंट सर्कल के सेंटर के cardinate है A1 A2 और उसकी रेडियस है भी और दूसरी बात आपकी जो स्ट्रेट लाइन है वो क्या है 3x 6 + 4y - 4 आपके जो स्ट्रेट लाइन है वह है 3x + 4y - ए = 0 डिस्टेंस की तो A1 A2 से सो 3 A1 ये कितना हो जाएगा 3a1 प्लस आप मोड लेंगे डिवाइडेड बाय फिर वही बात 3 और 4 इनके स्क्वायर का जब आप सैम करके अंडर रूट लेंगे तो pythagora इंटरप्रेट फाइव आने वाला है और ये पुरी बातें किसके इक्वल है सर ये पुरी बातें इसकी रेडियस बी के इक्वल है सर यहां से आप एक बहुत मजेदार चीज लिख सकते हो की फाइव को यहां से उठा के उधर ही पहुंचा दो तो ये कितना हो जाएगा 3 हो जाएगा फाइव बी ये कितना हो जाएगा फाइव बी ऑर्डर तू गेट आउट ऑफ दिस मोड बेसिकली व्हाट आई एम गोइंग तू डू सेट्स बट डेट इस व्हाट आई बेसिकली डोंट रिकमेंड आप यही वैल्यूज रख के देख लो ना क्यों स्क्वायर करना है तीन टर्म्स का स्क्वायर कितना लंबा और लेंडी जाने वाला है तो आप क्या बस इस कंडीशन को सेटिस्फाई कर सकते हो चारों ऑप्शंस आपको पता है स्टूडेंट्स चारों ऑप्शंस कंडीशन में रखो जहां पर आपकी कंडीशन है 3a1 + कोस 4a 2 - 4 / 5 जब वो अंदर लेके गए तो 5 बी अब ये कौन सा ऑप्शन ऐसी कंडीशन फुल फुल करता है हमें ये देख लेना है है ना अब अगर मैं देख लूं यहां पर तो कौन-कौन से ऑप्शन है आपके पास ये सारे ऑप्शन हैं 222 अगर ट्राई किया है है तो 22 अगर मैंने ट्राई किया पहला जो करेक्ट ऑप्शन मुझे लग रहा है वो पी तो है वैन वैन बाय तू हो गया यानी देख लेते हैं जल्दी से वैन वैन बाय तू ट्राई किया है ना तो वैन वैन बाय तू वैन और वैन तो तीनों चार सात माइंस चार तीन है ना 5 1 / 2 ऑप्शन है क्योंकि वैन वैन थ्री बाय फाइव मुझे दिख रहा है तो ऑप्शन एस भी ट्राई करते हैं वैन वैन थ्री बाय फाइव सो वैन वही ट्राई किया जस्ट अभी तो कितना ए रहा है 4 से 4 कैंसिल द बच्चा 3 और ये कितना है 3/5 तो 3/5 में से 5 से 5 कैंसिल तो थ्री बचा तो 3 = 3 होता है तो ऑफ कोर्स सर इसका एक और ऑप्शन होगा दी का कौन सा एस लेकिन एक और ऑप्शन बच रहा है सर ये तो नहीं हो सकता क्यों देखो ध्यान से ये जीरो होती है जीरो हो गया अब यहां पे देखो 2 3 तो 6 और 0 इक्वल नहीं है तो सर ये नहीं होगा तो इस तरीके से आप कनक्लूड करोगे इस क्वेश्चन के चारों ऑप्शन की सर इस तरीके से इनकी मैपिंग होगी और इस तरीके से इसे सॉल्व करोगे मेरा मानना है की आपको कहीं पर भी किसी भी स्टेप में कोई डाउट हेसिटेशन या फ्लकचुएशन नहीं है चीज आसान थी स्मूथ थी और आसानी से क्लियर से मैटर में आपको समझ आई है बिना ज्यादा घबरा बिना ज्यादा परेशान हुए बिना ज्यादा सोचे समझे आई थिंक कोई डाउट कोई परेशानी नहीं है भाई क्या चीज है क्लियर है आई थिंक सर अच्छा क्वेश्चन था बहुत कुछ हमने सिखा बहुत कुछ हमने देखा बहुत सारी बातों पर हमने गौर किया इन सारी बातों में इन सारी बातों की सारा कंक्लुजन ये है की आपको चैप्टर अच्छे से पढ़ना है वो का रहा है दो सर्कल्स हैं उन सर्कल्स की इक्वेशन है और यह जो सर्कस है इनके बारे में अगर मैं बात करूं x² + y² - 2X और x² + Y2 + y² + 2X = 0 है ना और एक सर्कल जो है आपका वो है x² + y² - 2X = 0 कुछ बातों के मेरे अब जवाब दीजिए पहला सवाल इसका सेंटर क्या होगा सर इसका सेंटर होगा -1 0 इसका सेंटर क्या होगा तो आप अब सोच पाते हैं सर इसकी रेडियस क्या होगी वैन का स्क्वायर माइंस जीरो वैन का स्क्वायर हो जाएगी वैन इसकी रेडियस भी क्या होगी वैन इनसे कुछ कंक्लुजंस कुछ मतलब कुछ रीजंस निकल सकते हैं क्या सर कोशिश करते हैं देखो भाई ये रहा होगा क्या सर आपका आईएसआईएस ये रहा होगा आपका आईएसआईएस जो की काफी हद तक हमने स्ट्रेट मैन्युअल में बनाया है और ये रहेगा आपका क्या सर एक्स है ना ये क्या रहेगा सर आपका एक्स कोई दिक्कत तो नहीं है अब मेरा सवाल है सर आपके पास दो सर्कल्स हैं एक सर्कल जिसका सेंटर है वैन कमा जीरो पर यानी एक्स एक्सिस पर उसकी रेडियस है वैन क्या मैं का सकता हूं की वो एक्स इसको टच कर रहा है बिल्कुल बेहतरीन सा शॉर्ट ट्रिक्स सुलझा हुआ सर्कल है जो की यहां पर ऐसा कुछ बन रहा होगा आई थिंक चीज बिल्कुल सर इसका सेंटर है माइंस वैन कमा जीरो पर इसका सेंटर है बराबर ही है ना लगभग बराबर ही है और ये ऑफ कोर्स आपका कहां हो रहा होगा सर ये आपका यहां बन रहा होगा कहां पर -1 0 पर इन दोनों बातों से कोई आपत्ति है आपको ये ऑफ कोर्स आपका कौन सा वाला सर्कल है ये वाला और ये आपका ये वाला सर्कल कोई दिक्कत तो नहीं है आई थिंक नहीं होनी चाहिए अब जब आपसे कोई कुछ पूछ रहा है क्या पूछ रहा है गौर से देखो वो पूछ रहा है एक पी पॉइंट है लामबीडीए लामबीडीए अगर ये पॉइंट है लामबीडीए लामबीडीए जो की है बेसिकली एक वेरिएबल पॉइंट तो आप क्या निष्कर्ष निकलती हो सर अगर पी एक पॉइंट है लामबीडीए लामबीडीए जो की है वेरिएबल पॉइंट तो आप खुद सोचना लैंडर लमदा मतलब क्या मतलब ए = एक्स स्ट्रेट लाइन पर लाइक कर रहा है ये कहां लाइक कर रहा है सर ये ए = एक्स आपके जो स्ट्रेट लाइन है ये बाहर लाइट कर रहा है आप मेरी बातें सोच समझ सिख और सन का रहे हैं क्या बिल्कुल सर तो ए = स्ट्रेट लाइन हम ड्रॉ करते हैं जो की ऑफ कोर्स आपकी कुछ इस तरीके से जा रही हो की जितना मेरा ये मानना है इसे थोड़ा सा कोर्स इंक्लाइंड जरूर करना होगा और इस तरीके से जा रही हो अब आप खुद थोड़ा सा चीजों को सोच कर देख कर समझ कर बताओ सर इस सर्कल में यहां कहीं ये होगा क्या सेंटर जो की होगा वैन कमा जीरो और ये जो वैन कमा जीरो होगा ये ऑफ कोर्स आपकी लाइन को यहां कहां वैन कमा वैन पर इंटरसेक्ट कर रहा होगा की आप मेरी बात समझ का रहे हो कुछ कहने की जरूरत है और इस सर्कल का सेंटर क्या है -1 कमा जीरो तो ये यहां से कहां ये यहां से -1 -1 है ना वहां पर क्या कर रहा होगा इंटरसेक्ट क्योंकि चीज बहुत semetrical है ये ए = एक्स स्ट्रेट लाइन है तो इस पर लाइक करने वाले पॉइंट की एक्स और ए koardinate दोनों ही से होते हैं तो इस तरीके से गुजर रहे होंगे मतलब कार्ड को मैं कहूं तो ये होगा -1 -1 आपका ओरिजिन है और ये ऑफ कोर्स पॉइंट क्या होगा आपका वैन कमा कोई दिक्कत कोई तकलीफ कोई परेशानी सर इस तरीके से चीज हो रही होंगी और जो लामबीडीए लामबीडीए है वो आपका इस स्ट्रेट लाइन पर लाइक करता है तो लामबीडीए कमा लमदा की जो उसने बात की है वो इस स्ट्रेट लाइन पर लाइक करने वाले सारे पॉइंट्स हैं आसान है चीज बिल्कुल सर अब सॉल्व करेंगे क्वेश्चन पहला पार्ट उसे पॉइंट पी की ये कंडीशन की वो C1 के अंदर हो लेकिन C2 के बाहर हो सर अगर वो C1 के अंदर हो C1 के अंदर हो C2 के बारे में एक बार देख लेता हूं C1 और C2 क्या है C1 आपका है -2x वाला है ना तो ये टेक्निकल जो है आपका यह सी वैन है और यह कंडीशन की पी जो आपका पॉइंट है वह C1 के अंदर हो C1 के अंदर होना मतलब क्या shailise यहां हो क्या परिभाषा है और यहां होने के लिए लामबीडीए क्या होगा सर वो माइंस वैन से जीरो तक जाएगा तो लामबीडीए क्या होगा -1 से 0 तक जाएगा और माइंस में से जीरो तक किस ऑप्शन में जा रहा है आर में तो पी के लिए मैं क्या मार्क करूंगा मैं ए के लिए क्या मार्क करूंगा की सर एक ही मैपिंग आप कर दीजिए आर से कोई दिक्कत तो नहीं है इनसाइड सी तू बट आउटसाइड सी वैन तो सी तू के अंदर होना चाहिए सर C2 के अंदर होना चाहिए मतलब आई थिंक ये वाला हिस्सा कोई दिक्कत तो नहीं अब जब वो सी तू के अंदर है तो ऑफ कोर्स जीरो कमा जीरो से वैन कमा वैन इन लामबीडीए क्या है जीरो से वैन तक तो लामबीडीए क्या है जीरो से वैन तक लामबीडीए अगर जीरो से वैन तक है मतलब कौन सा ऑप्शन ऐसा ऑप्शन तो बी के लिए आप क्या बात करेंगे वीकली ऑफ मार्क करेंगे एस कोई दिक्कत तो नहीं है आई थिंक सर आसानी चीज है वो C1 के बाहर हो और C2 के भी बाहर हो पी जो है वो C1 के और C2 के दोनों के बाहर हो सर अगर पी C1 और C2 दोनों के बाहर तो C1 के बाहर होना मतलब ये है और C2 के बाहर होना मतलब ये हिस्सा इसमें क्या देख का रहे हो सर ये माइंस इंफिनिटी से माइंस वैन तक ए रहा है और ये वैन से इंफिनिटी तक जा रहा है तो माइंस इंफिनिटी से माइंस वैन और वैन से इंफिनिटी तो मैं क्या कहूंगा सर माइंस इंफिनिटी से माइंस वैन और वैन से इंफिनिटी यानी सी जो है आपका वो क्वेश्चन आप होगा सर सी जो है आपका वो मैप होगा के से और इस बेसिस पर आप का सकते हो की दी के समय आप होगा क्योंकि वैन तू वैन napenge फिर भी आपको देखना है तो वो का रहा है पी देस नॉट लाइक इनसाइड सी तू पी C2 में लाइन ना करें पी C2 में लाइन ना करें उसका आंसर वहीं से भी देख सकते द पर फिर भी अगर C2 में लाइन ना करें मतलब ये हिस्सा ना हो ये हिस्सा ना हो मतलब ये हिस्सा ना हो मतलब सर या तो ये हो या फिर ये हो तो मैं क्या कहूंगा माइंस इंफिनिटी से जीरो और वैन से इंफिनिटी सो माइंस इंफिनिटी से जीरो और वैन से इंफिनिटी और माइंस इंफिनिटी से जीरो मैं कुछ गलत बोल रहा हूं क्या मैं शायद कुछ गलत बोल रहा हूं फिर से देखते हैं कुछ गलत तो मैं कर रहा हूं क्योंकि आर एस के ए गया अब होना चाहिए इनसाइड सी तू ये आपका C2 है तो C2 के अंदर नहीं होना चाहिए और C2 के अंदर नहीं होना चाहिए मतलब या तो यहां हो है ना C2 यही है ना आपका या फिर यहां हो और यहां पर क्या है ये माइंस इंफिनिटी से जीरो और ऑफ कोर्स वैन से इंफिनिटी तो - इंफिनिटी से जीरो आई थिंक यहां पर ये होना चाहिए था जीरो और ये होना चाहिए था वैन से इंफिनिटी बस ये एक करेक्शन होना चाहिए था जो मुझे फुल हो रहा है बस इसी पार्ट्स में हम कंटिन्यू करते हैं बस इस वाले पार्ट को देखो भाई मैं बी से वेरीफाई करता हूं पर इस वाले पार्ट को देखो ये जीरो से वैन तक ही होना चाहिए एक बार आप लोग भी देख लो इसे अच्छे से एक करेक्शन मुझे दिख रहा है मैंने शायद सी वैन और C2 गलत मैन लिया देख लेते हैं एक बार C2 जो है आपका वो प्लस वाला है है ना C2 जो है आपका प्लस वाला है तो वो है आपका C2 तो बस ये करेक्शन है ज्यादा आंसर्स इफेक्ट नहीं होंगे मेरे ख्याल से ज्यादा आंसर्स अफेक्ट नहीं होंगे या सारी आंसर्स अफेक्ट हो गए एक्चुअली तो बस करेक्ट कर लो ये है आपका C2 ये है सी वैन बहुत बड़ा फर्क नहीं आएगा बस ऐसे समझ लो की अगर पहले मैं एक फटाफट वीकली देखी ही लेता हूं की पहला जो आपका था पहला क्या था सर पीर लिस इनसाइड सी वैन बट आउटसाइड C2 तो C1 के जो अंदर है मतलब ये दोनों के तो आंसर स्विफ्ट हो जाएंगे होप आप समझ रहे हो क्योंकि ये इसके लिए था इसके लिए था आई होप आप समझ रहे हो तो ए और बी के लिए जो हमने लिखा है ए और बी के लिए जो मैंने आर और एस लिखा है की हम उसको एस और लिख डन आई होप आप समझ का रहे हो कुछ बड़ा फर्क नहीं है बस वो फ्लिप हो जाएंगे है ना सी ऑप्शन सी लिस आउटसाइड सी वैन तिलाई से आउटसाइड सी वैन एंड आउटसाइड सिरोही सर ये तो बदलेगा ही नहीं है तो वही बात है सी वैन और सी तू से बाहर है वो इंटरचेंज कर दिया कोई फर्क नहीं रहा है दी ऑप्शन देख लेना पी देस नॉट लाइक इनसाइड सी तू तो पी जो है वो C2 के अंदर ना हो और फिर से देख लेते C2 के अंदर ना हो मतलब क्या सर सी के अंदर ना हो मतलब आप कहोगे की सर यहां से स्कोडा यहां से इसको हटाओ C2 के अंदर नहीं चाहिए ना आपको तो C2 के अंदर नहीं चाहिए मतलब कौन सा जॉन चाहिए आपको वो वाला जोंस चाहिए आपको ये वाला चाहिए है ना आपको ये वाला जोन चाहिए और ये क्या है सर ये क्लीयरली सी तू के बाहर है यहां से और यहां से है ना C2 के अंदर वाला हिस्सा ही है तो इसे हटा दो तो बिल्कुल सर C2 के बाहर वाले से चाहिए मुझे क्या चाहिए पी जो है वो C2 के अंदर लाइन नहीं करता है तो C2 के बाहर मतलब माइंस इंफिनिटी से माइंस वैन और जीरो से इंफिनिटी और ऐसा ऑप्शन मुझे दिख रहा है माइंस में से -1 0 से इंफिनिटी तो दी के लिए आप मार्क करोगे पी बस ये छोटा सा करेक्शन था मैंने वो C1 C2 की नमिंग गलत कर दी जो की वापस गलती इस वजह से हुई की आप सर क्वेश्चंस ठीक से नहीं पढ़ते हो और मेरी इस गलती से आपको ये बात रिलाइज हो रही होगी की क्वेश्चंस ठीक से पढ़ना बहुत जरूरी है भाई क्वेश्चंस मतलब गलत मत पढ़ो है ना तो से कप से कप पर अगर हम बात करें तो से कप आपका है ऑप्शन दी अच्छा क्वेश्चन था आगे बढ़ते हैं नेक्स्ट क्वेश्चन आपका आता है न्यूमेरिकल वैल्यू टाइप क्वेश्चन तो ये कैसा क्वेश्चन है सर ये उसे कैटिगरी का क्वेश्चन है जहां पर आपको इसे सॉल्व करके एक न्यूमेरिकल वैल्यू फीड करनी है बहुत सारे तरीके हो सकते हैं इस क्वेश्चन को सॉल्व करने के एक तरीका जो मेरे दिमाग में ए रहा है एक तरीका तो आए होप आप समझ का रहे हो ज्योमैट्रिकली आप सोचते हो और मैं का रहा हूं देखो क्या का रहा है वो एक रियल नंबर एक्स और ए है जो इस कंडीशन को सेटिस्फाई करते हैं जो की इस कंडीशन को सेटिस्फाई करते हैं फिर वो का रहा है दें डी मिनिमम वैल्यू ऑफ दिस दैन डी मिनिमम वैल्यू ऑफ दिस दिस इस नथिंग बट व्हाट अब खुद सोच के देखो इस सर्कल की इस सर्कल की इस सेंटर से डिस्टेंस है इस सर्कल पर लाइक करने वाले पॉइंट की सेंटर से डिस्टेंस आपको निकालनी है मेरा यकीन करिए एक बहुत ही आसान बहुत ही बेसिक बहुत ही सिंपल सा क्वेश्चन है दो तरीके पहला तरीका सर मैं कैसे सोच सकता हूं समझना बात को ये एक सर्कल है इस सर्कल की खास बात क्या है सर्कल की खास बात है इसका सेंटर है माइंस फाइव इस सर्कल का सेंटर है 5 कमा ऑफकोर्स कितना 12 बहुत बढ़िया सर और वो क्या का रहा है इसकी रेडियस कितनी है सर इसकी रेडियस है 14 इसकी रेडियस कितनी है भाई इसकी रेडियस है 40 अब आप एक बात समझो मैन लो वो आपसे पूछ रहा है अंडर रूट ओवर x² + y² तो एक्स कमा ए क्या है सर एक्स कमा ए इस नथिंग बट टेक्निकल बट डी पॉइंट लाइन aneware ऑन डी फिल्म सर्कल फेदर ऑफ डी सर्कल इसकी पेरिमीटर पर लाइक करने वाले सारे सर्कल्स का पॉइंट्स का उन्होंने लोग का सही एक क्वेश्चन जैसे मैं बता सकता हूं की एक्स की सर का फिनिश पे लाइक करें अब मैन लो आपको समझने के लिए मैं कहता हूं की सर लेट से 0 यहां पर है अब आप बताओ अब आप बताओ इस एक्सप्रेशन का मतलब क्या है सर इस एक्सप्रेशन का मतलब है जो सर्कल पर लाइक कर रहा है कोई पॉइंट पी उसकी ओरिजिन से डिस्टेंस अब देखो ना एक्स-0 का स्क्वायर + ए - 0² इनसाइड एंड अंडर रूट तो आपसे इसकी मिनिमम वैल्यू पूछी जा रही है मतलब इस सर्कल की सरकम्फ्रेंसेस पर लाइक करने वाले किसी भी पॉइंट की इस पॉइंट जीरो कमा जीरो से मिनिमम डिस्टेंस तो मेरा कहना होगा सर बड़ा ही आसान सा तरीका होगा ज्यादा जीवन में दिमाग मत लगाइए क्योंकि अगर मैं इस सर्कल के सेंटर से इस पॉइंट के डिस्टेंस तो वह बिल्कुल नॉर्मल बन जा रहा है और नॉर्मल जो बन जा रहा है यह आपकी ऑफ कोर्स सर्कल किया रहेगी टैसेंट तो आप अगर ये डिस्टेंस बता दोगे तो मिनिमम होगी क्योंकि इसके अलावा पी यहां के अलावा पी कहीं और लिया होता तो ये और बढ़ती है और बढ़ती ऐसे करते हैं बढ़ती चली जाती है और ये डिस्टेंस आपकी मैक्सिमम होती है आप समझ का रहे हो एम रियली सॉरी ये नहीं यहां से बढ़ते अगर इसके जस्ट पीछे जाते हैं डायमीटर में तो वो डिस्टेंस आपकी मैक्सिमम होती पर हमें मैक्सिमम डिस्टेंस नहीं पूछे हमसे क्या पूछी है मिनिमम डिस्टेंस तो मैं कहूंगा दिस इसे 0 से -5 कमा 12 की डिस्टेंस निकल दो और उसमें से इस सर्कल की रेडियस सूत्र कर दो क्योंकि वही तो ये डिस्टेंस हो गया आप समझ का रहे हो मैं क्या कहना चाह रहा हूं मैं कहना चाह रहा हूं आपसे की जीरो कमा जीरो से इस सर्कल के सेंटर के डिस्टेंस निकल दो और उसमें से इस सर्कल की रेडियस सब्सट्रैक्ट कर दो तो क्या बन जाएगी आपकी ये डिस्टेंस जो की क्या होगी मिनिमम डिस्टेंस क्या आपको ये मिनिमम डिस्टेंस का कॉन्सेप्ट अच्छे से डाइजेस्ट हो रहा है आई थिंक सर बिल्कुल ऑब्जर्वेशन से दिया जा सकता है तो फाइव कमा 12 से इसकी डिस्टेंस आप खुद देखोगे इतनी ए रही है बहुत ध्यान से देखें अच्छा एक और बात सर आपको कैसे पता 0 सर्कल के बाहर रिपीट माय स्टेटमेंट आपको कैसे पता जीरो कमा जीरो सर्कल के बाहर है ऐसा भी सर्कल के अंदर उसे सर्कल पर हूं तो तीनों बातें चेक कर लेते हैं मैंने बस एक बात मैन ली है इसे है ना अब इस बात को मानने से समझते हैं इस चीज से देखना जीरो कमा जीरो पहले तो ये चेक कर लेते हैं की कहां पर है देखो भाई 0 पास किया तो ये जीरो और ये जीरो तो फाइव का स्क्वायर कितना है 25 12 का स्क्वायर 144 144 + 25 169 और 169 है ना अब 169 में से 14² अगर सब्सट्रैक्ट किया तो 16913 का स्क्वायर होता है 13² में से 14 का स्क्वायर करेंगे तो नेगेटिव आएगा 13 के स्क्वायर में से 14 का स्क्वायर करेंगे तो नेगेटिव आएगा मतलब मैं देख का रहा हूं सर की 0 जो आपने गलती से बाहर मैन लिया था एक्चुअली सर्कल के अंदर है मैं मैन लेता हूं जीरो कमा जीरो ये है है ना मैं बस ऐसे ही कुछ भी मैन लेता हूं जैसे माइंस फाइव 12 गया है ना तो मैं ज्यामिति के लिए ठीक तरीके से मारना चाहूं तो ये आपका नेगेटिव है ये पॉजिटिव है तो मैन लो जीरो कमा जीरो यहां के कहीं भी मैन लो बस मैं मैन लेता हूं यहां कहीं बस ऐसे ही मैन लेता हूं कोई दिक्कत तो नहीं है भाई ये इसके अंदर हमने माना जीरो कमा जीरो अब आप खुद बताओ इस बार जो डिस्टेंस होगी मिनिमम डिस्टेंस हो गया होगी आप खुद सोचो सर अगर जीरो कमा जीरो से पास होता हुआ मैं डायमीटर बनाऊं 0 से पास होता है डायमीटर बनाओ तो ऑफ कोर्स पी क्या है आपका एक्स तो ये जो पॉइंट है और यहां से जीरो कमा जीरो के डिस्टेंस है मैक्सिमम होगी और क्या ना इसे जस्ट इसका जो दूसरा वाला अपोजिट आपका वर्टेक्स है मतलब सॉरी जो एक्सट्रीम है किसका इस डायमीटर का यहां से डिस्टेंस मिनिमम होगी तो ये जो डिस्टेंस है ये ऑफ कोर्स आपका पॉइंट है ये आपका पॉइंट है कौन ओरिजिन जीरो कमा जीरो और यहां पर आप ले आए हो इस पॉइंट को तो ये डिस्टेंस आपको निकालनी है क्या डिस्टेंस निकलना है बहुत टू टास्क के नहीं है सर क्यों क्यों नहीं है क्योंकि सर इस सर्कल की रेडियस निकलेंगे जो की हमें पता है उसमें से माइंस फाइव कॉमन 12 से इस जीरो कमा जीरो की डिस्टेंस को सब्सट्रैक्ट कर देंगे तो क्या बचेगा तो मुझे ये वाली डिस्टेंस मिल जाएगी क्या आपको बॉर्डर्स ऑपरेटिंग के स्टैंडर्ड ऑपरेटिंग प्रक्रिया से समझ ए रही है बिल्कुल सर बस ये चेक कर लेना जीरो कमा जीरो उसके अंदर है या बाहर है उसे पर है ना तो जीरो कमा जीरो से इस सर्कल पर लाइक करने वाली किसी भी पॉइंट की मिनिमम डिस्टेंस जो निकालनी है वो क्या होगी पहले तो हमने सर्कल की क्या निकल रेडियस जो की मुझे पता है कितनी 14 उसमें से क्या माइंस करूंगा क्लीयरली सर 0 से इसकी डिस्टेंस जो की हम निकल चुके हैं जस्ट अभी आप देखो फाइव का स्क्वायर 12 का स्क्वायर 25 प्लस 144 169 169 का अंडर रूट 13 तो 14 - 1 कितना 13 की वैल्यू कितनी होती है वैन इसका आंसर होगा वैन बहुत डिफिकल्ट था ये क्वेश्चन मेरा हमेशा से मानना रहा है अगर आप चीज ठीक से विजुलाइज कर लें तो क्वेश्चंस आसान है तो क्वेश्चंस सच में बहुत आसान है क्या ये क्वेश्चन और इसका सॉल्यूशन समझ आया अगर जीरो कमा जीरो बाहर होता तो भी मैं यही कहता फिर मैं उसे केस में इस डिस्टेंस में से जो फाइव कमा 12 ये जो माइंस फाइव 12 है इसकी जो डिस्टेंस थी उसे पॉइंट से जीरो कमा जीरो से उसे डिस्टेंस में से रेडियस अट्रैक्ट करते हैं यहां उल्टा किया रेडियस में से डिस्टेंस सूत्र की क्योंकि ये अंदर है बाहर होता तो उल्टा कर देता की आप बात समझ का रहे हो बहुत टू बहुत मुश्किल बहुत बहुत ज्यादा दिक्कत वाली बात है तो नहीं है अगर यहां तक चीज है क्लियर है दो सर्कल है जिनकी externalli टांगें है है ना वो टच कर रहे हैं और जो लाइन पी ए बी और पी ए दश बी है वह उनकी कमेंट है जहां पर ऑफ कोर्स स्मॉलर सर्कल को और बी और बी दश पर वो बड़े सर्कल को टच करते हैं वो ये कहना चाह रहा है की दो सर्कल जो है वो externalli एक दूसरे को टच करना है दे आर टचिंग externalli इ आदर तो पहली बात तो दो तुर्क दो सर्कल्स एक दूसरे को externalli टच करना है तो अगर दो सर्कल जो एक दूसरे को externalli टच कर रहे हैं अगर मैं वो ड्रॉ करता हूं तो मैन लो ये आपका पहला सर्कल है और इसी का एक और दूसरा सर्कल बना लेता हूं लेट्स से मैं ये वाला है ना और ये दोनों सर्कल्स एक दूसरे को ऑफ कोर्स एक्स्ट्रा ली टच कर रहे हैं राइट ये दोनों सर्कल्स एक दूसरे को externalli टच अगर कर रहे हैं तो ये इस तरह का कोई नारिया रहा होगा अब वो का रहा है इनकी आप बना दो सर tagent इनकी अगर आप बनाते हो tangej तो आई थिंक टैसेंट तो ये हो जाएगी है ना इसे मैं थोड़ा सा रोते कर लेता हूं और ऊपर शिफ्ट कर देता हूं तो सर ये रहेगी आपकी हॉपफुली वैन ऑफ डी टेंसेंट तू डी सर्कल है ना आई थिंक सर यह एक टैसेंट बन जा रही है और इसी तरीके से आपके पास एक और टैसेंट है एक और टैसेंट कौन सी है सर अगर मैं इसे थोड़ा छोटा बना लूं तो चलेगा क्या क्योंकि मुझे लग रहा है मुझे थोड़ा स्पेस और चाहिए होगा तो मैं थोड़ा सा ऐसे प्रेस साइज भी बना लेता हूं एक सर्कल को थोड़ा छोटा बना लेते हैं है ना तो एक सर्कल थोड़ा छोटा बनाते हैं और दूसरा सर्कल ऑफ कोर्स थोड़ा बड़ा बना लेते हैं ताकि चीज हमारी आसानी से हमें दिख जाए बनती हुई है ना तो यह हमारे दोनों सर्कल्स आपके बन रहे हैं जिनको मैं कॉमन ग्राउंड पर लाना चाह रहा हूं इस तरीके से अब देखना भाई अब आपको कहना है सर इनकी टांगें सब बना रहे हो तो इनकी जब टांगेंट्स आप बना रहे हो तो एक ट्रेन तो ये हो जाती है है ना एक टैसेंट तो सिर्फ और आपकी ये हो जाती है जैसे अगर आप थोड़ा शिफ्ट डाउन करें तो दिस इस बिहेविंग ऑफ डी टेनिस और सिमिलरली एक और टैसेंट जब आप बनाएंगे एक और टैसेंट जब आप बनाएंगे तो वो टैसेंट कुछ ऐसे जा रही होगी जो की क्लीयरली आपकी कुछ इस तरीके से इसे टच कर रही होगी इसमें थोड़ा सा लेफ्ट करो सर अगर आप एक और लाइन ड्रॉ करते हैं एक और लाइन इस तरीके से ड्रॉ करते हैं जो इन दोनों साइकिल्स के सेंटर से पास हो रही हो तो ये लाइन की खास बात क्या होगी की ये लाइन ऑफ कोर्स इन दोनों टांगेंट्स का एंगल बाईसेक्टर होने के साथ साथ इन दोनों सर्कस के सेंटर से पास होगी ये बात हम जान देख समझ और सिख का रहे हैं आई होप आप सब भी इस बात से एग्री करते हो की ऑफ कोर्स यह दोनों सर्कस के सेंटर से पास हो रही होगी अब हम बात करते हैं जो क्वेश्चन में सिनेरियो दिए गए इसे मैं मैन लेता हूं C1 और इसमें मैन लेता हूं C2 अब उसने क्या कहा था उसने बोला आपकी जो लाइन पब और पी ए दश बी दश है वो कॉमन टांगेंट्स है स्मॉलर सर्कल को ए दश पर तो स्मॉलर सर्कल को वो ए और ए दश पर और बड़े सर्कल का वो बिगड़ सर्कल को वो बी और बी दश पर कनेक्ट करती है कौन सी आपकी tagent पी अब और यहां तक की कहानी है अब वो क्या का रहा है का और अब इक्वल है लेंथ फोर के बहुत ध्यान से suniyega एक बहुत जरूरी उसने बात कही आपसे जरूरी बात क्या कही की जो आपके का और अब हैं सर वो kaunsitali इक्वल लेंथ के हैं मतलब ये जो पाल लेंथ है ये जो लेंथ है सर ये लेंथ कितनी है ये क्लीयरली लेंथ है फोर यूनिट्स और इसी तरीके से वो आपसे कहना चाह रहे हैं की ये जो ए बी लेंथ है ये जो ए बी लेंथ है ये भी कितनी है 4 यूनिट्स और अब कोर्स अगर ऐसा है तो मैं बिल्कुल बिना है कहूंगा की पी ए दश और ए बी दश ए दश बी दश थी इक्वल लेंथ के होंगे क्योंकि ये भी फोर यूनिट्स होगा ये भी फोरेंस होगा क्योंकि चीज बिल्कुल से मैट्रिक है यहां तक तो कोई परेशानी नहीं है अच्छा सर यहां तक कोई परेशानी नहीं है तो अब क्या अब मेरा आपसे ये कहना है जरा ध्यान से देखोगे और कुछ चीजों के जवाब दो वो पूछ रहा है डी स्क्वायर ऑफ डी रेडियस ऑफ डी स्मॉलर सर्कल मतलब जो आपका छोटा वाला सर्कल है उसकी रेडियस का उसकी रेडियस का अगर मैं स्क्वायर कर डन तो उसकी कितनी वैल्यू होगी वो आपसे पूछ रहा है सर कैसे सोचेंगे क्या करेंगे कोई थॉट कोई आइडिया कोई दिमाग में आपके पास मेरा कहना है आप ऊपर वाले ट्रायंगल को ध्यान से देखो और कोशिश करो की क्या आप किसी तरीके से इसमें मतलब ऊपर वाले ट्रायंगल में अगर मैं यहां पर आपकी रेडियस और कनेक्ट कर डन जैसे की मैन लो ये है ना तो मैं एक वैन को कनेक्ट कर देता हूं ऑफ कोर्स और मैं bc2 को भी कनेक्ट कर देता हूं है ना कोई दिक्कत तो नहीं और ऑफ कोर्स यह क्या हो जा रही है सर आपकी रेडियस अब मेरा जो आपसे कहना है अगर आप इन दो ट्रायंगल उसको गौर से देखें कौन से दो ट्रायंगल की बात करें सर आप बात कर रहा हूं आपसे पीएसी वैन की मैं बात कर रहा हूं और मैं आपसे बात कर रहा हूं पी बी सी वैन की ऐसा क्यों सर आप प्लीज इस बात को देखो भाई ये जो एंगल है ऑफ़ कोर्स रेडियस में परपेंडिकुलर ड्रॉप किया है मतलब रेडियस जो है आपकी टेंशन पर परपेंडिकुलर होती है और इन बातों के साथ ये बात होता है की ये कॉमन एंगल है तो ये एंगल ये एंगल और सिमिलरली ये एंगल और ये एंगल इन दोनों ट्रायंगल इसमें से है तो इनके ये वाले एंगल्स तो से होना तय है और एंगल एंगल से milaity से की एन ऐसे की सर pa1 जो आपका ट्रायंगल है वो pbc1 के सिमिलर होगा इस बात को आज अन फैक्ट याद रखना प्रूफ करने की जरूरत नहीं होनी चाहिए मतलब इससे जब भी आप देखें तो आपको याद रहे की हान सर ये एक बात है जो आप हमेशा याद रखें क्या यहां तक कोई दिक्कत स्टूडेंट्स आय होप यहां तक कोई परेशानी नहीं है कोई दिक्कत नहीं है कोई भी दिक्कत नहीं है अब अगर मैं आपसे दो चीजों का रेश्यो पूछूं दो चीजों का रेश्यो अगर मैं पूछूं इन दोनों ट्रायंगल के अगर मैं बात करूं तो देखो इस ट्रायंगल में मैं अगर इसकी रेडियस और इस लेंथ पर आपसे बात करूं तो इसकी रेडी कितनी है सर इसकी रेडियस को अगर मैं थोड़ी देर के लिए का लूं R1 और इसकी रेडियस को थोड़ी देर के लिए मैं का लूं r2 तो आपसे जो मैं पूछना चाह रहा हूं वो ध्यान से देखो भाई एक बड़ा सिंपल सा सवाल है आप बताओ ये एक वैन इस वाले ट्रायंगल की बात कर रहा हूं तो ये एक वैन है ये क्या है सर आपका ac1 और इसकी ये जो परपेंडिकुलर है इसको हम इसके बेस यानी की इससे आप से डिवाइड कर रहा हूं तो अगर ये सिमिलर है तो सिमिलरिटी के अकॉर्डिंग रेश्यो विजय bc2 और बीपी भी बराबर होंगे बिल्कुल सर तो आपके bc2 और बीपी का रेश्यो भी वही होगा इससे अगर मैं किसी एक नए कंक्लुजंस की तरफ पहुंचे तो मैं क्या कहूंगा सर एक वैन है आपके सर्कल की रेडियस यानी R1 है ना और आप कितनी है सीरियल लेंथ है फोर कोई दिक्कत तो नहीं सिमिलरली सर bc2 कितना है bc2 आपके दूसरे सर्कल की रेडियस है जो की क्या है r2 और बीपी कितना है सर बीपी अगर आप निकलोगे तो 4 + 4 कितना ए जा रहा है 8 क्या इससे आप R1 और r2 का रेश्यो निकल का रहे हो सर एक बहुत मजेदार चीज पता चल रही है R1 और r2 का रेश्यो ए रहा है 1 / 2 क्या इस बात से कोई दिक्कत क्या इस बात से कोई परेशानी यहां तक कोई डाउट है तो पूछो स्टूडेंट्स आई थिंक चीजें अभी तक तो बड़ी सॉर्टेड और सुलझी हुई सी हैं अब अगर मैं आपसे पूछूं अगर आप किसी भी एक राइट एंगल ट्रायंगल में पाइथागोरस थ्योरम के पर्सपेक्ट से सोचें किसी भी एक राइट एंगल ट्रायंगल में अगर आप पाइथागोरस थ्योरम के पर्सपेक्टिव से सोचे या दोनों में सोच लें तो आप एक बहुत जरूरी बात निकल पाओगे क्या बात निकल पाओगे अच्छा सुनना ध्यान से मैं यहां पर पहले तो ये का सकता हूं क्या की अगर इस स्मॉलर सर्कल की रेडियस को मैन लो मैं का डन थोड़ी देर के लिए R1 r2 जो है वो रहने दो है ना छोड़ो आप इस ट्रायंगल देखो ए पी सी वैन सर एक जरूरी ब्लू है की आपका apc1 एक राइट एंगल ट्रायंगल है ये राइट एंगल डी ट्रायंगल है तो पीसी वैन को मैं क्या कहूंगा सोच के देखो पीसी वैन आपका हाइपोटेन्यूज है तो जो इसका स्क्वायर होगा वो इसके और इसके स्क्वायर के सैम का एडिशन होगा तो मैं पीसी वैन को क्या लिखूंगा सोच के बताओ सर pc1 को आप लिखोगे अंडर रूट ओवर है ना R1 का स्क्वायर R1 का स्क्वायर प्लस ऑफ कोर्स क्या आप का स्क्वायर जो की कितना है 4 यानी कितना 16 कोई तकलीफ तो नहीं है इसी तरीके से क्या मैं इस ट्रायंगल में भी सोच सकता हूं पीसी तू बी में तो उसमें भी ये है हाइपोटेन्यूज ये है परपेंडिकुलर और ये है बेस तो यहां पर अगर मैं निकलूं तो क्या मैं का सकता हूं जो पी सी तू हाइपोटेन्यूज है वो होगा अंडर रूट ओवर r2 का स्क्वायर और आर तू के स्क्वायर बोल सकता हूं आप सन रहे हो जो मैं का रहा हूं मैं का रहा हूं आपसे की अगर मैं आपसे पूछूं पी सी तू तो pc2 क्या आएगा फिर से देखो ध्यान से पी सी तू जो आएगा वो आपका क्या आएगा ध्यान से देखो r2 यहां पहुंचा है तू यहां पहुंचा है तो ये हो जाएगा तू आर वैन तो पहले मैं लिख रहा हूं क्या r2 का स्क्वायर यानी तू आर वैन का स्क्वायर तो क्या मैं कर सकता हूं और क्या करना चाह रहे हो सर आप और मैं एक और जरूरी बात करना चाह रहा हूं की इसमें से अब ये वाला पार्ट ऐड कर रहा हूं तो यानी पी बी पी भी क्या है 4 + 48 का स्क्वायर कितना 64 तो मैं यहां पर ये 64 लिखना चाह रहा हूं अब आप थोड़ा सा और ध्यान दिए थोड़ा सा और अगर गौर से देखें तो आप कनक्लूड कर पाओगे सर ऐसे कैसे कंट्रोल कर पाएंगे अब सुनिए तो जरा ध्यान से आप एक बात बताओ पीसी वैन अपॉन पीसी तू कुछ है क्या आपको पता है क्या की पीसी वैन अपॉन पीसी तू कितना है सर मुझे तो पता है आपको पता हो या नहीं मुझे नहीं पता आप मुझे बस पीसी 1 / pc2 बता दो आपका क्वेश्चन सॉल्व हो जाएगा ऐसे कैसे सॉल्व हो जाएगा ऐसा आप खुद देखो ना आपको पता है पीसी वैन अपॉन पीसी तू आपको बहुत अच्छे से पता है बहुत सारे तरीके से पता है आप जिस तरीके से निकलना चाहते हो उसे तरीके से पता है और ये रेश्यो नहीं दिख रहा है तो एक और तरीका है आपके पास खुद सोच के देखो सर बड़ी सिंपल सी बात है देखो ये pc2 है है और यह पीसी वैन है इन दोनों में डिफरेंस कितना है रिपीट माय स्टेटमेंट अगर मैं पीसी तू में से पीसी वैन सब्सट्रैक्ट करूं तो सर इसकी रेडियस है आर मतलब R1 और उसकी रेडियस है r2 तो अगर मैं pc2 या तो ऐसे कर लो या डिवाइड कर लो दोनों एक ही बातें कहीं नहीं जा रहे क्वेश्चन आपसे मिल रहा है ऑलरेडी प्रूफ कर चुके हो और आप प्रूफ कर चुके हो की ये जो है इसमें डबल का रिलेशन है अब देखो ना रेडियस का रिलेशन 1 / 2 है तो क्लीयरली हम ये रिलेशनशिप कर चुके हैं की ये देखो ये लेंथ है फोर और ये लेंथ है 4 और टेक्निकल पुरी लेंथ है 8 तो 4:8 यानी 1:2 का रिलेशन है तो अब इनकी इन साइड्स में भी 1:2 का रिलेशन होगा एंड अगर हमें पीसी वैन अपॉन पीसी तू लिखूं तो आप क्या कहोगे सर वो आप कहोगे वैन बाय तू अगर मैं आपसे पूछूं पीसी 1 / pc2 तो वो आपका हो गया 1 / 2 बस इस वैन बाय तू के इक्वल आप कैसे रख दो 1 / 2 के इक्वल आप इसे और इसे रख दो आपका R1 की वैल्यू ए जाएगी आप बात समझ में या फिर अगर ऐसे नहीं जच रहा है तो आप ऐसे सोचो C1 और C2 हेलो अगर मैं पूछूं आपसे pc2 - pc1 तो यह है पीसी तू और यह है पीसी वैन तो डिफरेंस कितना है C1 C2 और C1 C2 क्या है सर R1 + r2 और R1 और r2 आप जानते हो तो आई होप यहां से भी आप निकल सकते हो वही बात वापस आपको चीज सोचनी जैसा आपको ठीक लगे तो अगर मैंने यहां से करना चाहता तो ये आता है पीसी 1 / pc21/2 वो हो जाएगा r1² + 16 ये कितना हो जाएगा सर ये हो जाएगा आपका क्या r1² + 16 में अंडर रूट लगा रहा हूं थोड़ा सा पेशेंस रखिएगा और डिनॉमिनेटर में pc2 तो 4 r2² + 64 ये कितना हो जाएगा फोर आर तू स्कुएर + 64 अब जो आप एंपावरमेंट लगाएंगे उसका कम आई थिंक यहां पे स्क्वायर लगा के कर सकते हैं और अब जब हमने इसे सॉल्व करने की कोशिश की तो वैन से मल्टीप्लाई होंगे ये तो ये हो जाएगा 4r2² + 64 है और तू का स्क्वायर 4 से मल्टीप्लाई कर रहा हूं क्योंकि मुझे कुछ तो फीलिंग ए रही है गड़बड़ सी कुछ तो मैं गलत नहीं कर रहा हूं ना भाई क्योंकि यहां से हम रेश्यो तक ही पहुंचेंगे वापस और बात गड़बड़ हो जाएगी तो मेरे ख्याल से ये तरीका हेल्प नहीं करेगा क्योंकि आप समझो वापस अब वापस R1 और r2 में रेश्यो निकल रहे हो इससे कोई फायदा नहीं होगा इसके बजाय जो तरीका दूसरा हम डिस्कस कर रहे द वो क्या कर रहे द ध्यान से देखो स्टूडेंट जो दूसरा तरीका हमने डिस्कस किया था वो था पीसी तू में से अगर आप पीसी बन सकता है रिपीट में स्टेटमेंट सर अगर आप पीसी तू में से पीसी वैन सूत्र बताओ अगर आप pc2 निकल रहे द ये फोर आर वैन स्क्वायर प्लस 64 ये कितना है सर 4r1 स्क्वायर प्लस 64 अंडर रूट में ये आपका पीसी है माइंस पी सी वैन पीसी वैन कितना है सर वो है r1² + 64 √ में r1² + 64 अब मेरा आपसे कहना है tc2 - पीसी 1 कितना है सर पीसी तू माइंस पी सी वैन है आपका क्या R1 + r2 ये है R1 + r2 suniyega ध्यान से और ध्यान से देखो भाई r2 कितना है तू आर वैन तो R1 +2r1 यानी थ्री आर वैन तो ये कितना ए जाएगा सर लेफ्ट हैंड साइड पे दिस इस गोइंग तू बी थ्री आर वैन तो कैन आई रिप्लेस दिस एक्सप्रेशन बाय थ्री आर वैन एस सी ऑल कैन तो pc2 - पीसी वैन गेट्स रिप्लेसिड बाय 3 टाइम्स आर वैन नौ हर कॉम डी मोस्ट ट्रिकी पार्ट अब आपको R1 का स्क्वायर निकलना है जिसको निकलने के लिए मुझे थोड़ी मेहनत तो करनी पड़ेगी थोड़ी चीज तो सोचनी पड़ेगी पर एक मेहनत आपकी आसान हो सकती है अगर आप चीज ध्यान से देखें तो कैसे देख लें सर आई होप आपको ये पार्ट दिख रहा है एक बात सोच के देखो स्टूडेंट्स आपने बस थोड़ा ध्यान नहीं दिया क्या यहां से मैं फोर कॉमन ले सकता हूं आई होप मैंने कुछ गलत तो नहीं लिखा है R1 + 64 क्यों लिख लिया है बस यही तो गलतियां हम करते हैं ये 64 नहीं है है ना ये कितना है अब आप में से कहीं सारे स्टूडेंट्स ने नोटिस किया अगर मैं यहां से फोर कॉमन लेता हूं रिपीट मैं स्टेटमेंट अगर मैं यहां से फोर कॉमन लेता हूं तो आप देखो ये कितना है फोर और ये 16 तो फोर कॉमन लेते ही अंदर बचेगा वैन और 64 का फोर कितना 16 टाइम्स लेकिन अंडर रूट के बाहर अगर फोर आया अगर अंडर रूट के बाहर 4 आया तो कितना दे देगा आपको 2 तो ये है तू टाइम्स और ये है वैन आप समझ रहे हो ये identifically आईडेंटिकल तू टाइम्स एक्स - एक्स कितना होता है एक्स ये हो जाएगा 3r1=√ r1² + 16 क्या आप मेरी बात समझ में सिर्फ एक छोटा सा कम कर लीजिए आपका लाइफ सॉर्ट आउट हो जाएगा अगर आप दोनों तरफ स्क्वायर कर दें तो थ्री आर वैन का स्क्वायर कितना 9 R1 का स्क्वायर ये कितना हो जाएगा सर ये हो जाएगा आपका r1² और ये हो जाएगा प्लस 16 r1² इधर आया तो ये हो जाता है 8 R1 है ना स्क्वायर = 16 तो r1² कितना सर अरविंद स्क्वायर ए रहा है जिसमें कोई बहुत मुश्किल बहुत डिफिकल्ट बातें हमसे नहीं कही गई थी और ये आसानी से हम निकल पाए सोच पाए और देख पाए तो R1 का स्क्वायर कितना है सर तू और इस बेसिस पे आप मार्क करोगे कौन सा ऑप्शन तो नहीं है न्यूमेरिकल आंसर है क्वेश्चन तो यहां पर आप फीड करोगे तुम अब यहां पे पॉसिबिलिटी बहुत कम होती है आप सोच रहे हो न्यूमेरिकल आंसर टाइप अभी यहां पे ऑप्शन नहीं है तो आपको आंसर लाना ही है और उसे वहां लिख कर आना है इनपुट आपकी कैटिगरी दी जाएगी वहां पे कीबोर्ड से आपको लिखकर सबमिट करना होगा आंसर तो सोच रहे हो इस तरीके से भी आपको सोचना है चीजें ये क्वेश्चन अगर समझ आया आई थिंक टू नहीं था ये बात याद रखना की जब भी ऐसा कुछ हो तो ये बात एप्लीकेबल होती है इसके बाद अगर मैं आपसे दूसरी बात करूं तो ये रहा आपका एक और क्वेश्चन यहां पर वो का रहा है आपसे की सर जो क्यूट एंगल है बिटवीन दिस लाइन एंड दिस सर्कल तो नाइन कोस थीटा की वैल्यू बताओ मेरा आपसे कहना है बस आप यह सिनेरियो इमेजिन करो की जब आपको कोई सर्कल दे दिया जाए और कोई स्ट्रेट लाइन दे दी जाए तो वो स्ट्रेट लाइन और वह सर्कल में क्या रिलेशन होने वाला है और उन दोनों के बीच का जो क्यूट एंगल है किस-किस का सर इन लाइन का इस सर्कल के साथ इस लाइन का इस सर्कल के साथ क्यूट एंगल का मतलब क्या है पहले तो ये समझते हैं पहले तो एक बात बताओ सर कुछ जरूरी बात बताओ की मुझे कैसे पता चलेगा की ये लाइन इसकी टैसेंट है या यह लाइन ऐसे दो पॉइंट में इंटरसेक्ट कर रही है या ये लाइन इससे दूर है कैसे पता चलेगा तो मेरा तरीका आसान सा सर ये है पहले तो यह पता करो की इस सर्कल के सेंटर के cardinates क्या है सर वो हो जाएंगे आपके 2 - 1 इस सर्कल की रेडियस बता सकते हो क्या निकल लेते हैं सर आई होप आपको दिख रहा है तू कमा मैंने सुना है ना 4 + 15 - 49√3 तो ये कुछ ए रहा है अब अगर मैं 2 - 1 से इसकी परपेंडिकुलर डिस्टेंस निकलता हूं तो कितना हो जाएगा 3 2 - 4 कितना +4 और ये कितना -5 / 3 और 4 कितना होता है थ्री और फोर थ्री का स्क्वायर 94 का स्क्वायर 16 वो हो जाता है 25 उसका अंडर रूट फाइव और उसपे कोर्स आप मोड लगाते हो छह और चार 10 में से पंच गए तो पंच और पंच को पंच मिनट डिवाइड किया तो एक यानी जो परपेंडिकुलर डिस्टेंस ए रही है इस लाइसेंस के सेंटर की वो ए रही है वैन अब suniyega ध्यान से sunhariyo क्या है सेंटर रेडियस और सेंटर से लाइन की परपेंडिकुलर डिस्टेंस जो सेंटर की सेंटर से लाइन की जो परपेंडिकुलर डिस्टेंस है वो रेडियस से कम है क्या आप यह सिनेरियो खुद से विजुलाइज कर सकते हो की सर यह आपका सर्कल हुआ ये क्या हुआ सर आपका सर्कल अब इस सर्कल में आप खुद नोटिस करोगे की इस सर्कल के रेडियस से अगर एक स्ट्रेट लाइन के परपेंडिकुलर डिस्टेंस छोटी है मतलब वो जो लाइन रही होगी सर वो जो लाइन रही होगी ऑफ कोर्स आपकी इससे 2 पॉइंट्स पर इंटरसेक्ट कर रही होगी बिल्कुल कर रही होगी क्या आप मेरी इस बात को समझ का रहे हो क्या आप मेरी इस बात को अच्छे से समझ का रहे हो मैं उसको ऐसा बना ले रहा हूं ताकि मैं रेडियस को परपेंडिकुलर दिखा पाऊं है ना तो ऐसे वो जा रही होगी आई होप इस बात से किसी भी स्टूडेंट को कोई आपत्ति कहीं से कहीं तक नहीं है है ना अब मेरा यह कहना है की मैन लो सर ये आपकी एक दूसरे स्टेट लाइन हो गई बिल्कुल हो गई अब ये जो दोनों अंदर मैंने बनाई है रेडियस क्या इनको लेकर आपको चीज समझ ए रही है ऑफ कोर्स रेडियस है तो इसको मैं थ्री का डन क्या और इसे भी थ्री का डन और उसके सेंटर के अकॉर्डिंग -1 बिल्कुल सही बात है सर और आपकी स्ट्रेट लाइन है जिसकी इक्वेशन क्या है जिसकी इक्वेशन है 3X - 4y = 0 अब आपने की सर अगर इससे इस पर परपेंडिकुलर ड्रॉप किया जाए है ना इस पॉइंट से रिपीट माय स्टेटमेंट इस पॉइंट से जो आपने ड्रा किया है स्टूडेंट्स ये जो सर्कल का सेंटर है इससे इस लाइन पर अगर एक परपेंडिकुलर ड्रॉप किया जाए तो उसकी परपेंडिकुलर लेंथ कितनी होगी सर वो परपेंडिकुलर लेंथ हमने निकल थी कितनी वैन तो ये जो परपेंडिकुलर लेंथ है ये कितनी हो जाएगी वैन क्या यहां तक चीज समझ ए रही है पूरा सिनेरियो आपके सामने बना कर रख दिया है अब आपको बस ये क्वेश्चंस सॉल्व करना है अगर आपको कन्फ्यूजन है तो बस ये सिनेरियो पूरा आपके सामने है अब जब वो आपसे पूछ रहा है क्या पूछ रहा है भाई वो पूछ रहा है क्यूट एंगल क्यूट एंगल बिटवीन डी लाइन एंड डी सर्कल इस तो किसी भी लाइन और किसी भी सर्कल के बीच के क्यूट एंगल से आपका क्या मतलब है किसी भी लाइन और किसी भी सर्कल के बीच के क्यूट एंगल से मेरा तात्पर्य है लाइन और सर्कल के बीच एंगल मतलब है की सर इस पॉइंट पर जहां पर इस लाइन ने सर्कल को इंटरसेक्ट किया जहां पर इस लाइन ने सर्कल को इंटरसेक्ट किया वहां पर आप ड्रॉ करिए एक टैसेंट वहां पर आप ड्रॉ करिए एक टैसेंट आई रिपीट माय स्टेटमेंट वहां पर हमने किया ड्रा किया भाई एक ट्रांजिट अब इस tagent है जो सर्कल और लाइन का पॉइंट सेशन यहां भी बना सकते द कोई फर्क नहीं पड़ता वही आंसर आएगा इस टैसेंट का जो की इस लाइन और सर्कल का पॉइंट ऑफ इंटरसेक्शन है और ये लाइन जो है इनके बीच जो क्यूट एंगल है इनके बीच जो क्यूट एंगल है वो है थीटा और आपसे पूछी जा रही है नाइन कोस थीटा की वैल्यू क्या क्वेश्चन यहां तक समझ आया किसी भी पार्ट में कोई तकलीफ आई थिंक नहीं होनी चाहिए सर आप का रहे हो की यह थीटा है बिल्कुल जी हान मैं आपसे का रहा हूं की ये थीटा है तो एक बात बताओ सर वर्टिकल अपोजिट एंगल के कॉन्सेप्ट से क्या मैं का सकता हूं की ये एंगल भी थीटा होगा आई होप अगर ये थीटा है तो ये भी थीटा होगा बिल्कुल सही बात है सर अच्छा सर आपको ये बात नहीं पता है क्या ध्यान से सोचो की सर सर्कल की रेडियस से सर्कल की टांगें पर अगर लाइन ड्रॉ की जाए मतलब सर्कल की रेडियस जो होती है वो सर्कल की टेंशन पर कितना डिग्री एंगल क्रिएट करती है 90° सर ये थीटा है ये 90 है तो जो बच जा रहा है यानी ये जो एंगल है ये जो एंगल है ये कितना होगा सर ये जो एंगल है ये रहा होगा 90 - θ आप सन रहे हो क्या आप मेरी बातें समझ का रहे हो या नहीं आई थिंक आप समझ का रहे हो सर आप फिर से थोड़ा कंफ्यूज हो रहे हो आप ऐसे क्यों नहीं सोच रहे हो की सर ये है 90 और यह है 90 - θ तो ऑफ कोर्स यह एंगल कितना रहा होगा सर ये एंगल वापस अगर मैं निकलूं तो ये जो एंगल रहा होगा ये रहा होगा θ आई थिंक आप एंगल्स में तो कंफ्यूज नहीं हो गए फिर से शुरू करते हैं देखो ये है इस स्ट्रेट लाइन और इस सर्कल के बीच एंगल कैसे सर ये स्ट्रेट लाइन सर्कल को इस पॉइंट पर इंटरसेप्ट करती है वहां मैंने टांगें ड्रॉ की तो इस tagent और इस स्ट्रीट लाइन के बीच में क्यूट एंगल थीटा है वर्टिकली अपोजिट एंगल तो ये भी θ ये 90 डिग्री है तो ये 90 - θ ये 90 ये 90 - θ है तो उसका थीटा होना है अब अगर मैं आपसे पूछूं की आप कोस्थेटा मुझे निकल कर बताओ तो कोस थीटा में निकल सकता हूं कैसे निकल सकता हूं सोच के देखना है स्टूडेंट्स बड़ी बेसिक सी बात है सिंपल सी बात है सर आपके पास आपके पास परपेंडिकुलर है कितना वैन आपके पास रेडियस है कितनी 3 अगर आपके पास परपेंडिकुलर और रेडियस सॉरी परपेंडिकुलर और हाइपोटेन्यूज है परपेंडिकुलर और हाइपोटेन्यूज है तो आप क्यों नहीं निकल सकते बिल्कुल निकल सकते हैं आई रिपीट माय स्टेटमेंट अगर आपके पास परपेंडिकुलर और हाइपोटेन्यूज है तो आप क्यों नहीं निकल सकते अगर परपेंडिकुलर और hypotaneous है suniyega ध्यान से क्या मैं इस एंगल पर थोड़ी देर आपसे बात करूं शरीर कितना है 90 - θ अगर मैं यहां पर इस एंगल का साइन रेश्यो लेता हूं मैं आपसे क्या पूछ रहा हूं अगर मैं इस एंगल का साइन रेश्यो लेता हूं अगर मैं लेता हूं आपसे साइन 90 - θ तो इस एंगल का साइन रेश्यो लेना मतलब इसके अपोजिट परपेंडिकुलर और इसका hypotaneous वो कितना हो जाएगा सर वो हो जाएगा 1 / 3 किसी भी स्टूडेंट को कोई आपत्ति तो नहीं है सर अभी तक तो नहीं है और जब आपने 1/3 से सिन 90 - θ तो सिन ऑफ 90 - θ कितना हो जाएगा सो ये हो जाएगा कोस थीटा ये हो जाएगा कोस थीटा जो की कितना ए रहा है 1 / 3 आपसे कोस्थेटा नहीं पूछा सर आपसे नाइन कोस थीटा पूछा है तो 9/3 कितना हो जाएगा 3 हो जाएगा 3 और इस क्वेश्चन का आप मार्क करेंगे आंसर कितना आसान इतना बेसिक इतना सिंपल क्वेश्चन है आप कंफ्यूज ना हो शांति से बस इस क्वेश्चन को फॉलो करें एक-एक लाइन का अपना-अपना मतलब है अगर आपको बेसिक पता है अगर आपने हर कॉन्सेप्ट को ठीक से पढ़ाया तो ये क्वेश्चन कहीं नहीं गया बस हमेशा कोऑर्डिनेट्स ज्यामिति में जो सबसे ज्यादा जरूरी है वो है विजुलाइजेशन या फिर आप उसे कहें इमेजिनेशन अब क्वेश्चंस को ठीक से बनाया देखिए समझिए और आपको चीज क्लियर होंगी बहुत दूर नहीं जाएंगे और हान इन सब के साथ जो सबसे ज्यादा जरूरी स्किल है वो है ज्यामिति आपको ज्यामिति के बेसिक्स पता होने चाहिए तभी जाकर आप चीज सोच पाएंगे आई होप ये क्वेश्चन अच्छा आसान और बेसिक सा क्वेश्चन था जो बहुत डिटेल में आप सभी को समझ आया अगर ये क्वेश्चन क्लियर है तो क्या मैं मूव करूं आज के अगले क्वेश्चन पर ये भी आपका कौन सी कैटिगरी का क्वेश्चन है न्यूमेरिकल वैल्यू टाइप क्वेश्चन है जहां पर ऐसे सॉल्व करके आपको इसकी क्या निकालनी है आपको इसकी न्यूमेरिकल वैल्यू निकालनी है कोई दिक्कत तो नहीं है यहां तक अब सर क्वेश्चंस समझते हैं पहले तो वो का रहा है दो C1 और C2 बोथ पास थ्रू डी पॉइंट वैन बाय इसका मतलब है यह दोनों आपके दो पॉइंट्स पर इंटरसेक्ट करते हैं जब भी ऐसा दिखे तो ये कंक्लुजन तो आएगा तो दो पॉइंट पर जब ये दो सर्कल सेंटर सेट करें तो एक सर्कल तो आपका ये रहा और दूसरा सर्कल आपका ये रहा है ना तो ये दो सर्कल्स हैं और वो का रहा है की सर ये जो दोनों सर्कल्स हैं ये वैन कमा तू और तू कमा वैन है ना तो मैं इसे मैन लेता हूं लेट्स से क्या तू कमा वैन है ना और इसे मैं मैन लेता हूं वैन कमा तू कोई फर्क नहीं पड़ता कहीं भी कुछ भी मैन लीजिए आपने तो बस अपनी सहेली से मैन लिया वैन कमा तू क्या था एक तो ये जो पॉइंट है ये आपका ऑफ कोर्स ए है और ये जो पॉइंट था ये कोर्स आपका क्या है ए ये जो पॉइंट है ना अब क्या का रहा है वो सुनेगा ध्यान से ये एक लाइन को टच करते हैं एक लाइन है 4X - 2y = 9 उससे बी और डी पर टच करते हैं आप सोच के बताओ सर अगर एक लाइन को टच करते हैं मतलब वो लाइन इनकी क्या है कॉमन टेंट इट तो उसे वो बी और दी पर टच करते हैं तो मैं एक लाइन बनाता हूं बी और दी जो इसकी क्या हो गई है ना वो इन दोनों के क्या हो जाएगी सर वो इन दोनों के लिए हो जाने वाली है tagent आई थिंक ये जो लाइन है आपकी ये क्या हो गई इन दोनों की कमेंट्री और ये दोनों जो सर्कल्स हैं ये दोनों सर्कल्स हैं उसे कहां कहां टच करते हैं उसे टच करते हैं बी और दी पर तो ये आपके दो पॉइंट्स हो गए कौन-कौन से भाई ये पॉइंट मैन लेते हैं बी और ये पॉइंट हम मैन लेते हैं और ये जो स्ट्रेट लाइंस है सर आपकी ये स्ट्रेट लाइंस जो हैं ये स्ट्रेट टेक्निकल उसकी इक्वेशन है 4X - 2y - 9 = 0 ठीक है सर अच्छी बात है अब क्या निकल कर ए रहा है suniyega ध्यान से डी पॉसिबल cardinate ऑफ अन पॉइंट सी सी क्या है सच डेट डी क्वॉड्रिलैटरल कौन सा एबीसीडी इसे अन पैरेललोग्राम अब वो पूछ रहा है आपसे की ए बी सी दी अब ए बी सी दी अगर मैं बनाना चाहूं तो देखो भाई मुझे ए पता है बी पता है दी पता है अब मुझे सी चाहिए तो टेक्निकल अगर मैं आपको कुछ चीजों पर बात करूं तो सर आप एक कम करो आप ए से बी को मिलवा दो तो ये रहा ए और बी का मिलन है ना आप सर ए से दी को मिलवा दो ये रहा ऐसे दी का मिला अब आप खुद सोच के बताओ सर जब आपने ए बी और ए दी को मिलाया तो आपको क्या नहीं दिख रहा है की अगर आप इसे पैरेललोग्राम बनाना चाहते हैं तो सी को आप कहीं ऐसी जगह लोग तो यहां पर ये पैरेललोग्राम बन रहा हूं तो आई थिंक ये पैरेललोग्राम कुछ इस तरीके से इस तरीके से थोड़ा सा इसे मैं डिलीट कर दूंगा कुछ इस तरीके से ये एक आपका पैरेललोग्राम बन रहा होगा है ना मैंने थोड़ा सा सिर्फ पुश कर दिया ऊपर की तरफ बट आई होप आप मेरी बात समझ का रहे हो और ये आपका पैरेललोग्राम बनेगा और ये जो पॉइंट आएगा आपका यहां पर ये पॉइंट आपका कौन सा होगा सी कोई दिक्कत कोई परेशानी तो नहीं है अब इस सी के कोऑर्डिनेट्स उसने कहे क्या है सर सी के कोऑर्डिनेट्स हैं ए गौमा भी और वो फाइनली आपसे पूछ रहा है मोड अब की वैल्यू जो हम निकल लेंगे पर क्या यहां तक आपको सारी बातें समझ आई यहां तक ये पूरा क्वेश्चन क्लियर है क्या अगर यहां तक कोई भी डाउट है तो पूछो वर्ण फिर हम आगे बढ़ेंगे एक बात जो हमने हमेशा से सीखी है जो हमेशा से हमने देखते समझते आए हैं शुरुआत से की सर ये बात तो होता है क्या सर की ये जो आपकी कॉमन कोड होगी ये जो आपकी कॉमन कोड यहां पर बनेगी वो कॉमन कॉर्ड आपके इस पैरेललोग्राम के ऑफ कोर्स यहां से इसे गुजरेगी है ना मैं स्ट्रेट लाइन बनाना चाह रहा था तो अब कोस ये जो कॉमन कोड आपकी होगी वो ऑफ कोर्स आपके इस तरीके से बन रही होगी की वो क्लीयरली आपके पैरेललोग्राम को ना सर टच करें बल्कि उसके इस पॉइंट से पास हो रही होगी है ना तो मैं इसे थोड़ा सा ऐसे ही बना लेता हूं क्योंकि हमने चीज बहुत मतलब परफेक्शन से नहीं बनाया पैरेललोग्राम बिल्कुल परफेक्ट पैरेललोग्राम तो मैं क्या करता हूं वैसे थोड़ा सा इस तरीके से बना लेता हूं मैं क्या कर लेता हूं एक पैरेललोग्राम का मुझे डायगोनल चाहिए तो मैं डायगोनल लेता हूं जो की ऑफ कोर्स आपके इस पॉइंट से पास हो रहा होगा और यहां पर आकर मिल रहा होगा तो ये क्या हो गया आपका डायग्नल क्या यहां तक चीज क्लियर है इसे देखो और समझो अब क्या करना है उसे पर बात करते हैं दो ऑब्जर्वेशन से क्वेश्चन खत्म हो जाएगा पहला ऑब्जर्वेशन आप सोच के देखो आप ये बात रिलाइज करोगे की सर ये जो आपका पैरेललोग्राम बन रहा है इसमें ए और सी का मिड पॉइंट है ये आपका यहां पर ये वर्टेक्स ये e21 है ये ए है और इसका जो मिड पॉइंट है वो है एम अगर मैं इसे थोड़ा सा अगर अच्छे तरीके से बताऊं तो प्लीज कंफ्यूज मत होना ये जो ए पॉइंट है ये तू कमा वैन आपका ये वाला पॉइंट है ये पॉइंट है आपका तू कमा वैन है ना घबरा कम है ये आपका पॉइंट है ए और जिसकी मैं बात कर रहा हूं वो है आपका मिड पॉइंट जो की है आप कमेंट पॉइंट जो की आपका लाइक करेगा यहां और इसे मैं दिनो करता हूं एम से और वो क्लीयरली आपकी कमेंट इंजन पे लाइक करेगा ये प्रॉपर्टीज है जो हमने पढ़ी है बहुत अच्छे से डिस्कस किए हैं सारी बातें है ना तो पहले तो मैं इस बात पर बात कर लेता हूं की सर जो एम है किसका ए और सी का तो मिड पॉइंट क्या होगा सर वो हो जाएगा ए प्लस वैन बाय तू कमा बी प्लस तू बाय तू और अगर यह मिड पॉइंट है जो की है ऑफ कोर्स आप के अकॉर्डिंग क्या एम और एम अगर इस लाइन पर लाइक करता है तो क्या इस लाइन को ये पॉइंट उसे लाइन का सेटिस्फाई करेगा बिल्कुल तो तू से फोर कैंसिल तो क्या बचा ए + 1 का डबल समझ रहे हो क्या तू से तू कैंसिल तो क्या बचा माइंस ऑफ बी + 2 और 1 क्या दिख रहा है माइंस 9 = 0 यहां से अगर मैं इसे सिंपलीफाई करूं तो देखो क्या मिल रहा है सर आपको मिल रहा है 2a यहां से मिल रहा है माइंस बी तू माइंस तू जीरो और ये हो जा रहा है -9=0 आपको ए और बी पे ए के क्वेश्चन तो मिल गई सब सही चल रहा है 2a - बी - 9 है ना कुछ गड़बड़ तो नहीं की हमने नहीं की सर आप सुनाओ मिढ्वाइंट तो निकल लिया अब आप सोच के देखो अगर मैं आपसे पूछता यहां पर ऑफ कोर्स ये पॉइंट था ये आपका ये कौन सा पॉइंट था ये पॉइंट अगर मैं आपसे पूछूं ए लाइन की इक्वेशन तो अब कोर्स ए लाइन आपकी जो कॉमन कोड है उसी पर सी लाइक करता है ये पैरेललोग्राम ऐसे ही बनता है सिमिट्रिकली तो आप बात समझो सर ए की इक्वेशन क्या होगी सर ए आई की इक्वेशन होगी वैन कमा तू और तू कमा उनसे पास होने वाला है तो ए - y1 है ना ए - 2 = है ना दोनों का डिफरेंस 2 - 1 1 - 2 - 1 तो ये हो जाएगा -1 टाइम्स ये क्या हो जाएगा एक्स - 1 आई होप आप मेरी बात समझ का रहे हो ये आपकी लाइन की इक्वेशन है जो की है आपकी इक्वेशन की इक्वेशन जो है वो आपकी ये ए रही है इसे मैं सिंपलीफाई कर लूं क्या - एक्स इधर ले तो ये हो जाएगा एक्स + ए ये माइंस तू है और ये है प्लस वैन प्लस वैन इधर आए तो -2 +1 -1 अब आप एक बार ध्यान से देखो सर ये जो ए लाइन है वही ए टाइम है वही ऐसी है तो सी मतलब ए बी इस पर लाइक करता है तो ए इसको सेटिस्फाई करेगा बिल्कुल करेगा सर तो यहां से क्या ए जाएगा सर यहां से ए जाएगा ए + बी - 1 = 0 क्या आपको ये दोनों स्ट्रेट लाइंस की इक्वेशन समझ ए रही है पहली जो आती है आपकी क्या 2a -बी + 9 = 0 - 9 = 0 और दूसरी लाइन जो आपकी क्लीयरली ए रही है ए + बी - 1 = 0 कोई दिक्कत तो नहीं है बस ऐसे ही एक बार क्रॉस चेक कर लो हमने कहीं कोई गड़बड़ तो नहीं कर दी मुझे बस एक गड़बड़ जो लग रही है बस इस इक्वेशन को मैं देखूं ध्यान से तो देखो भाई ये है ए - 2 ये है आपका - एक्स है ना माइंस वैन ही ए रहा था ना और ये कितना ए जा रहा है +1 तो +1 इधर आएगा तो माइंस वैन होगा ना ये पता नहीं क्यों हम एक गड़बड़ कर रहा है तो -2 -1 कितना होगा सर ये हो जाएगा -3 ये अगर हुआ -3 तो यहां भी आप -3 लिखिए कोई दिक्कत तो नहीं भाई इसे आप लिख दीजिए -3 अब क्या सर अगर मैं इन दोनों को ऐड कर डन अगर मैं इन दोनों इक्वेशंस को ऐड करूं तो आपको एक रिज़र्व मिलेगा सर ध्यान से देखो बी से बी कैंसिल तू ए प्लस ए कितना सही हो जाएगा 3a और -9 -3 कितना -12 है ना हो आप समझ का रहे हो इंस्टॉल करें तो प्लस 12 तो एक ही वैल्यू आती है कितनी 4 आप सभी को समझ ए रहा है क्या सर एक ही वैल्यू फॉर या तो यहां रख दो या यहां रखा तो 4 - 3 कितना सर फोर माइंस थ्री वैन बी यू वैन उधर गया तो माइंस वैन तो बी की वैल्यू कितनी आती है -1 तो अगर आपसे ए बी के कोई cardinate पूछता ना तो आप उसे कहते 4 - 1 क्या आप ये बात समझ पाए और फोर और माइंस वैन का जब आप प्रोडक्ट लेंगे तो वो कितना ए जाएगा -4 और उसका जब आप मोड लेंगे तो वो हो जाएगा 4 तो इस क्वेश्चन का आंसर आप क्या मार्क करेंगे भाई इस क्वेश्चन का आंसर आप इनपुट आंसर फीड करके आएंगे जो की होगा फोर वो का रहा है tangenced ऑन टांगें ड्रॉ की गई है फ्रॉम डी पॉइंट पी वैन कमा 8 ठीक है सर तू डी सर्कल तो ये एक सर्कल है सर्कल आते पॉइंट्स ए एंड बी दें डी इक्वेशन ऑफ डी सरकमसर्किल ऑफ डी ट्रायंगल पब तो पी ए और बी वाला जो ट्रायंगल है उसके सरकम सर्कल की इक्वेशंस या इक्वेशन आपसे वो पूछ रहा है ये कब का क्वेश्चन है सर ये आईआईटी जी एडवांस्ड ने 2009 में पूछा था और इस क्वेश्चन को हम कैसे करेंगे मेरा कहना है सर जब भी चीज है ट्रिकी हो आपके पास सबसे अच्छा तरीका होना चाहिए उसे विजुलाइज करना क्वाड्रेंट ज्यामिति में डी बेस्ट एप्रोच वुड बी विजुलाइजिंग इमेजिंग यू नो ड्राइंग डी प्रॉब्लम ऑन डी शीट तो ड्रॉ करते हैं देखते हैं क्या चीजे निकल कर आती हैं मेरा कहना है सबसे पहले तो सर मुझे एक सर्कल दिख रहा है x² + y² यहां से देखो इस सर्कल के सेंटर के क्वाड्रेट्स थ्री कमा तू और रेडियस की अगर बात करें तो थ्री का स्क्वायर 9 तू का स्क्वायर 4 9 + 4 कितना होता है सर नाइन प्लस फोर होता है 13 है ना 13 में से अगर 11 ऐड किया तो 13 में 11 ऐड किया ए जाएगा सर क्लीयरली 24 तो √24 इस गोइंग तू बी दी रेडियस आई होप आप बात समझ का रहे हो ये बड़े बेसिक से कॉन्सेप्ट्स हैं तो पहले तो हम सर्कल से ही शुरुआत करते हैं सर अगर मैं सर्कल पर आपसे बात करूं तो सर्कल को लेकर आप क्या-क्या ओपिनियन होगा की आपके पास एक सर्कल है बिल्कुल है सर इस सर्कल पर आपने किसी एक्सटर्नल पॉइंट पी से टांगें ड्रॉ की है बिल्कुल ड्रा की है तो एक टांगें टू कार्स आपने कुछ इधर ड्रॉ की होगी है ना जैसे में कुछ ऐसी बना देता हूं यहां से इसे थोड़ा सा डाउनलोड ही लेकर जाएंगे हम तो इस तरीके से जस्ट गिव मी अन सेकेंडरी वैन ये आपके पास एक टैसेंट होगी जो की आपकी यह जा रही होगी और सिमिलरली इसी तरीके से एक एक्सटर्नल पॉइंट वही पीस है उसने एक और टेंशन ड्रॉ की है और इससे वो इंडियन पॉइंट को का रहा है क्या ए और बी और ये जो एक्सटर्नल पॉइंट है आपका पेज इसके cardinate सब के अकॉर्डिंग क्या है सर क्लीयरली वैन कमा 8 कोई कन्फ्यूजन स्टूडेंट्स यहां तक मेरे ख्याल से तो डाउट या परेशानी नहीं होनी चाहिए अब एक और बात पे गौर कीजिएगा अगर मैं यहां से आपसे बात करूं तो ये देखो क्या लिख रहा है सर वो ये है थ्री कमा तू दिस इसे गोइंग तू बी थ्री कमा तू सो डी सेंटर हर इसे गोइंग तू बी थ्री कमा तू ये मैन लेते हैं आपका सेंटर है और सेंटर से भी मैं इस पी पॉइंट को कनेक्ट करता हूं बस एक छोटा सा कम और करेंगे क्या यहां से मैं परपेंडिकुलर ड्रॉप कर सकता हूं क्या यहां सर्कल के सेंटर से मैं परपेंडिकुलर ड्रॉप कर सकता हूं और सी के कोऑर्डिनेट्स क्लीयरली क्या है थ्री कमा तू अब क्या आप मेरी यह बात फिगर आउट करने में हेल्प करेंगे की सर यह जो ए और बी है यानी पब जो आपका ट्रायंगल बन रहा है जिस ट्रायंगल विच आर गेटिंग फ्रॉम हर पब इस पी ए बी ट्रायंगल बन रहा है इसके सरकम सर्कल की इक्वेशन क्या होगी कैसे सोचेंगे जरा बताइए स्टूडेंट्स मेरा कहना है इसे तरीका तो आसान है बस चीजें थोड़ा सा दिमाग से सोचने की जरूरत है थोड़ा इसे और बड़ा कर लेते हैं आई थिंक चीज है सर कुछ ऐसी आपको देखनी चाहिए आई एम रियली सॉरी क्या हो रहा होगा अगर आप ध्यान से देखें तो एक्जेक्टली चीज कैसे वर्कआउट हो रही होंगी बस थोड़ा सा ऐसे ठीक से प्लॉट कर लेते हैं मेरे ख्याल से ये काफी हद तक बन चुका है बिल्कुल बस लो की ये वो आपका circumsar है अब लगभग हर एक चीज आपके सामने ड्रून है एक हिंट और अगर एक्स्ट्रा चाहिए तो मैं और दूंगा उससे आपका एक क्वेश्चन खत्म होना चाहिए बस इसमें एक बात का ध्यान रखिएगा स्टूडेंट्स ये जो सर्कल होगा ना इसके लिए ये पॉइंट साइक्लिक होंगे मैं इसको और ठीक से ड्रॉ करता हूं ताकि आपको मैं समझा पाऊं ये जो सर्कल आपने बनाया है ना इसके लिए ये पॉइंट साइक्लिक होंगे कौन से पॉइंट साइक्लिक होंगे स्टूडेंट्स ध्यान से देखिएगा आपका वो जो सर्कल है वो सी बी पी और ए से होता हुआ जाएगा आपने ये बात समझ का रहे हैं आपका जो सर्कल है वो सी बी पी और ए सबसे होता हो जाएगा मैंने बहुत प्रॉपर्ली ड्रॉ नहीं कर पाया हूं आय होप आप मेरा रीजन समझ का रहे हो बट वो क्लीयरली इन सारे पॉइंट से पास होगा यू स्टे अन शॉर्ट ऑफ इट क्यों पास होगा अल प्रूफ इट मैथमेटिक्स पहले मैं प्रूफ कर देता हूं suniyega ध्यान से अब बस इस बात पे खंडों की सर बड़ी बेसिक सी बात है ये एंगल कितना है सर ये एंगल 90 डिग्री है अच्छा ये बताओ ये एंगल कितना है सर ये भी 90° है अब ये और ये 90 डिग्री है मेरे रिपीट माय स्टेटमेंट ये और ये 90° है तो क्लीयरली एसीपी बी ए सी बी पी ये एक क्वॉड्रिलैटरल है जो की साइक्लिक क्वॉड्रिलैटरल क्यों क्योंकि सर इनका और इनका 180 होना आता है क्योंकि टोटल सैम 360 होता है उसमें से ये दोनों अपना 180 हिस्सा लेके चले गए तो इन दोनों अपोजिट एंगल्स का भी सैम वैनिटी होगा और अगर किसी क्वॉड्रिलैटरल में अपोजिट अपोजिट एंगल्स का सैम है ना अपोजिट एंगल्स का सैम 180 हो तो वो हमारा एक conscitelic क्वाड्रेंट क्वॉड्रिलैटरल होता है सर kansaikali क्वॉड्रिलैटरल का क्या मतलब होता है इसका मतलब होगा की यहां पर आप एक ऐसा सर्कल ढूंढ पाओगे जिस पर ये चारों पॉइंट्स लाइक करें या बेहतर लैंग्वेज में सर इन चारों पॉइंट से एक सर्कल पास होगा की आप मेरी बात समझ का रहे हो बहुत ट्रिक की पार्ट है और यहीं से क्वेश्चन खत्म होगा बहुत प्रॉपर्ली मैंने चीज है ड्रॉ कर पाया बट मैं आपको चीजें समझने सीखने या बताने की कोशिश कर रहा हूं डेट यू विल बी एबल तू ड्रा अन सर्कल इन दिस मैनर आप एक डेफिनेटली सर्कल बना पाएंगे जो का पी बी से जाएगा अब मेरे बस एक सवाल का जवाब दे दीजिए क्वेश्चन खत्म हो जाएगा सवाल है सवाल है suniyega ध्यान से क्वेश्चन खत में क्योंकि अब बस आपको सॉल्व करना है इसे अगर मैं पब अगर मैं ट्रायंगल पब का circumcircled ढूंढ रहा था तो क्या वो सरकमसर्किल यही है जो क्वॉड्रिलैटरल पीएसीबी का भी circumsar है जरा ध्यान से देखिए आपको आपका आंसर मिल गया है अब बस आपको एक क्वेश्चन सॉल्व करना है जो की आप बहुत आसानी से कर सकते हो ऐसे कैसे आसानी से कर सकते हैं सर देखो भाई बहुत आसान सी बात थी अगर आपने इस बात को ऑब्जर्व किया होता तो आपको हिंट दी गई थी सोचेगा ध्यान से सर आपको क्यों नहीं दिख रहा है की जो पीसी है जो पीसी है वो इस सर्कल का ध्यान से सुनना मैं आपको तीनों बातें करता हूं तो आप इन सब को देखो आप देख लो ट्रायंगल एक को या फिर आप देख लो ट्रायंगल ई को या फिर आप देख लो ट्रायंगल पब को या फिर आप चाहे तो देख लीजिए और कोई ट्रायंगल बच रहा है क्या आई थिंक और कोई भी आप देखना चहल से आप इसको भी मैन लो तो कोई दिक्कत वाली बात नहीं है है ना आप ट्रगल ये भी मैन लेते हैं या फिर आप क्वॉड्रिलैटरल मैन लो कौन सा भाई आप मैन लो ए सी बी पी है ना यह जो सारे आपके शेप्स हैं तीन ट्राएंगल्स और ये चौथा आपका इन सब का जो सरकमस सर्कल है वो से है जो की है ये ये जो है इन चारों का ठीक है सर आपके हमने बात मैन ली लेकिन आप बात करना क्या चाह रहे हो प्ले कर कहा जाना चाह रहे हो कहना क्या चाह रहे हो मैं बस आपसे बड़ी आसान सी बात कहना चाह रहा हूं प्लीज इस बात पे गौर करिए जो ऐप्स अगर एक ट्रायंगल हमने माना और इसका सरकमसर्किल ढूंढने की कोशिश की तो क्या मैं का सकता हूं वही आपके रिक्वायर्ड ट्रायंगल का भी या इन दूसरे ट्रायंगल या underlater का भी सरकमसर्किल है बिल्कुल सही बात अब आप सीधे-सीधे सोचो भाई कितना आसान सा क्वेश्चन है सर पीसी एक स्ट्रेट लाइन है कॉर्ड है सर्कल की जो की सर्कल की सरकम्फ्रेंसेस पर किसी पॉइंट पर 90° एंगल सब्सटेंड करती है तो पीसी इस नॉट जस्ट अन कॉल इट इस गोइंग तू बी वैन ऑफ डी डायमीटर्स ऑफ डी सर्कल यस ओर नो एंड इफ इट इस वैन ऑफ डी डायमीटर्स ऑफ डी सर्कल शैल आय से की सर अगर मुझे डायमीटर के extrimas के cardinate पता है तो क्या मैं सर्कल की इक्वेशन लिख सकता हूं क्या अब आपको हिंट क्लियर हुई की मैं क्या कहना चाह रहा हूं इतनी देर से सर ये तो बड़ा आसान है कैसे भाई जल्दी बताओ सर सर कल का इक्वेशन जो लिखेंगे वो होगा एक्स - X1 प्लस ए माइंस ए वैन यू विल गेट दी आंसर तू दिस क्वेश्चन यू ऑल रिलाइज हो इजी दी इस क्वेश्चंस आर वैन तू यू फिगर आउट डी होल यू नो अंडरस्टैंडिंग और ये क्लेरिटी यू गेट बाय विजुलाइजिंग डी प्रॉब्लम बाय इमेजिन इन डी प्रॉब्लम ऑन डी शीट सर अगर इसे थोड़ा ध्यान से देखा जाए तो इसे क्या सिंपलीफाई करने का स्कोप है क्यों नहीं है सिंपलीफाई अगर हमने किया तो ये बन जा रहा है क्या सर ये बन जाता है x² ध्यान से देखना - 3X - एक्स कितना हो जाएगा -4x अच्छा पहले यहां से भी Y2 मिल जाएगा क्या बिल्कुल मिल जाएगा अब सन ना ध्यान से जो मैं का रहा था -3x - एक्स तो ये कितना हो जाएगा -4x अच्छा यहां से -2y -8 ए तो ये कितना हो जाएगा -91 एक और ध्यान से देखो -3 -1 + 3 + 3 सोली से दिस इसे गोइंग तू बी आंसर दिस इस डी रिक्वायर्ड सर्कल जो की सरकमसर्किल है पब का है ना और अगर ऐसा हम देखना चाहे तो x² + y² - 4X - 10y +19 इस क्लीयरली इस क्लीयरली गिवन विद विच ऑप्शन सी क्लीयरली आपको दिख रहा है ऑप्शन बी में किसी भी स्टूडेंट को इस क्वेश्चन को समझने में कोई तकलीफ कोई परेशानी आई थिंक ये आसान सा बेहतरीन सा डायरेक्ट सा सैंपल सा क्वेश्चन था जो हर एक स्टूडेंट ने क्रैक किया आई होप ये पुरी स्टैंडर्ड ऑपरेटिंग प्रोसीजर जो हमने मोड ऑफ ऑपरेटिंग उसे किया जो एक बेसिक से मेकैनिज्म हमने उसे किया हमारे उन सारे कॉन्सेप्ट्स को रिकॉल करते हैं जो सारी बातें हमने पढ़ी है आई थिंक इसमें ऐसा कुछ भी नहीं था जो हमने नहीं पढ़ा था तो आई थिंक सर चीजे आसान है और मैं आपको बस यही समझाना चाह रहा हूं की जी एडवांस्ड स्टफ एग्जाम नहीं है अगर आपको बेसिक पता है तो अगर आपने हर कॉन्सेप्ट को करने से पड़ा है चलिए इसके साथ मूव करते हैं नेक्स्ट क्वेश्चन पर और नेक्स्ट क्वेश्चन को देखते हैं क्या लिखा हुआ है पढ़िए जरा ध्यान से एक सर्कल है जो की पास हो रहा है माइंस वैन कमा जीरो से और टच कर रहा है ए एक्सिस को जीरो कमा तू पर तो वो और किस पॉइंट से पास होगा इस बारे में सवाल किए जा रहे हैं कैसे सोचेंगे सर अगेन मुझे तो कुछ ज्यादा समझ नहीं आता है जब तक मैं चीज विजुलाइज ना कर लूं तो सर ये रहा आपका क्या ये रहा आपका आईएसआईएस और सिमिलरली एरर आपका क्या यह रहा आपका एक्स-एक्सिस कोई तकलीफ तो नहीं है भाई किसी भी स्टूडेंट को है तो पूछो सर यह ए एक्सेस और एक्सेस ए गया तो क्या मैं एक बेसिक सा सिनेरियो ले रहा हूं है ना मैं एक बहुत ही बेसिक बहुत ही सिंपल सा डायरेक्ट ले रहा हूं बाय डी वे अभी हम कौन सी कैटिगरी के क्वेश्चंस कर रहे हैं अभी सर आप सिंगल करेक्ट आंसर टाइप क्वेश्चंस कर रहे हो है ना अब suniyega ध्यान से कम की बात को अब अगर मैं आपसे पूछूं सर फिर वही बात एक ऐसा सर्कल जो की क्या कर रहा है वही एक्सेस को जीरो कमा तू पर टच कर रहा है जीरो कमा मतलब यहां पर जीरो कमा तू पर टच करना मतलब क्या shailise यहां के टेस्ट करना ऐसा कहीं वह यहां कहीं टच कर रहा है किस टच कर रहा है इट इस टचिंग डी ए एक्सिस जीरो कमा तू एंड दुआ आई नो डी रेस्ट ऑफ डी डिटेल्स डू आय नो बाकी चीज सर क्या जरूरी है अगर कोई सर्कल जीरो कमा 2 पर टच करें तो इसी तरह वो बेशक को इस तरफ भी हो सकता था तो इसी सर्कल का आप रिप्लिका भी ढूंढ सकते द सर इसी सर्कल का रिप्लिका भी आपको मिल जाता वो शायद यहां पर बन रहा होता इस बात से तो किसी स्टूडेंट को कोई आपत्ति नहीं चांसेस द सर बराबर चांसेस द ऐसा हो सकता था की ये आपका ऐसा कुछ बन रहा होता है ना अब मेरा आपसे पूछना है सर इस क्वेश्चन में आप कैसे सोचोगे मैं आपको हिंट देता हूं और अगेन लीव इट अप तू यू की आप एक बार बताएं फिर हम साथ में बेशक डिस्कस करेंगे जब इसने से ए एक्सिस पर टच किया यहां व्हेन इट टच जीरो माइंस वैन जीरो से अब आपको अच्छी क्लेरिटी मिलेगी डेट विच वैन आर यू गोइंग तू चूस ये पास होता है -1 0 तो ये पॉइंट होगा -1 0 -1 0 कूद बी एनीथिंग वो या तो ये हो सकता है कुछ भी हो सकता है बस स्टैंड हम पॉइंट आप मैन लेते हैं हमारी समझ के लिए है ना तो पहले तो मैं ये वाला पार्ट हटा दे रहा हूं की ऑफ सर ये ट्रायंगल आप नहीं चेंगे क्योंकि ए एक्सिस को अगर उसने टच किया है तो फिर उसे तरफ से कैसे टच करेगा अगर वो माइंस उनको कमा जीरो से पास हो रहा है तो आई थिंक दिस इसे डी पॉसिबिलिटी वेयर ऑल आर लुकिंग फॉर अब चूंकि ये माइंस वैन कमा जीरो से पास हो रहे हैं तो लेट्स से डेट दिस पॉइंट इस डेट मिंस वैन कमा जीरो एंड दिस सर्कल डेफिनेटली हेस अन सेंटर आई होप आई हैव मेड यू अंडरस्टैंड डी प्रॉब्लम प्रिटी वेल इन अन प्रिटी वेल मैनर अब आप मुझे बताएंगे की इस क्वेश्चन को सॉल्व करने के लिए आपके पास क्या आइडिया है आपके पास क्या थॉट है आपके पास क्या एप्रोच है एक आइडिया एक हैंड एक थॉट आप हमेशा अपने पास कैरी करते हो की सर बड़ी बेसिक्स सी बात है अगर ये सर्कल बाय एक्सेस को जीरो कमा तू पर टच करता है तो कुछ तो मुझे पता है आपको क्या पता है प्लीज इस बात को सुनिए जैसे की अगर मैं आपसे कहूं चीजें फिर से विजुलाइज करिए क्या इस सर्कल के सेंटर के कोऑर्डिनेट्स के लिए मैं यहां से कुछ ढूंढ सकता हूं देखो भाई अगर मैंने यहां से इस पर एक परपेंडिकुलर ड्रॉप किया कुछ इस तरीके से तो क्या आपको नजर ए रहा है सर ये जो पॉइंट है क्लीयरली क्या है सर ये है जीरो कमा तू और ये है सर्कल का सेंटर बिल्कुल सर सही बात है सर्कल का सेंटर तो अगर ऐसा कुछ है की यह है जीरो कमा तू और ये है सर्कल का सेंटर तो क्या इस सर्कल का सेंटर के एक्स या yordinate में से कोई एक क्वाड्रेंट में बता सकता हूं क्या सर कोशिश किए जा सकती है अब बात समझो ये ये कौन सी लाइन है ये ए = स्ट्रेट लाइन है ना तो shailise की सेंटर का भी जो ए koardinate होगा वो क्या होगा तू आप इस सभी इस बात से एग्री करते हैं क्या कोई आपत्ति कोई परेशानी सर अभी तक तो नहीं है अब क्या सोचना होगा अच्छा अब आप इसका एक्स कोऑर्डिनेट्स क्या मैन सकते हो सर कुछ भी मैन सकते हैं कोई दिक्कत नहीं है तो लेट्स से इसका एक्स cardinate है ह इसका cardinate क्या माना जा सकता है सर इसका एक्स cardinate माना जा सकता है लेट से ह किसी भी स्टूडेंट को इस बात से दिक्कत सर ह ही क्यों माना कुछ भी मैन सकते हो देस नॉट मेक अन्य डिफरेंट इट ऑल आई होप यहां तक कोई परेशानी नहीं है है ना अब अगर मैंने इसका एक्स कॉर्डिनेट ह माना है अच्छा बाय डी वे suniyega ध्यान से ये डिस्टेंस कितनी है क्या मुझे बता सकते हो सर यही तो वो डिस्टेंस है ह आप देखो ना ये जो डिस्टेंस है अगर मैं इसे डिस्टेंस की बात करूं तो डिस्टेंस की टर्म्स में ये यही तो डिस्टेंस है ह क्योंकि सर्कल के सेंटर से जो डिस्टेंस है वो ह है बाय डी वे कोऑर्डिनेट्स ज्यामिति के कॉन्टेक्स्ट में अगर वो ह है तो मुझे डिस्टेंस लेना होगा क्योंकि मुझे नहीं पता वो नेगेटिव है पॉजिटिव बट एनीवे इस बार है वो नेगेटिव में बट फिर भी हम तो अपने कन्वीनियंस के अकॉर्डिंग जब रेडियस की बात करें जब डिस्टेंस की बात करेंगे तो मोड ह लेंगे सर अभी आपने बोला रेडियस की बात करें आप ह को रेडियस क्यों ले रहे हो आप क्यों नहीं लोग ह को रेडियस ये सर्कल का सेंटर ये सर्कल की सरकम्फ्रेंसेस तो सर्कल के सेंटर से सर्कल के सरकम्फ्रेंसेस की जो डिस्टेंस होती है वही तो उसकी रेडियस होती है तो क्या मैं कुछ लिख सकता हूं बिल्कुल लिख सकते हो सर तो मैं सर्कल की इक्वेशन लिखूंगा सर्कल की इक्वेशन लिखने का मेरा क्या तरीका होगा सर सर्कल की इक्वेशन लिखने का मेरा तरीका होगा की सर्कल के सेंटर के जो कोऑर्डिनेट्स हैं आई होप आप जानते हो यह है और सर्कल की रेडियस कितनी है सर वो हम जानते हैं वो क्या है ह तो क्या सर्कल के सेंटर के कोऑर्डिनेट्स और सर्कल के रेडियस देख कर सर्कल की इक्वेशन नहीं लिखे जा सकती क्यों नहीं लिखे जा सकती देखो भाई हम क्या लिखेंगे हम लिखेंगे बाय डी वे ये कब का क्वेश्चन है ये आईआईटी जी एडवांस में 2011 में पूछा था जब जेली एडवांस्ड आईआईटी जी 2000 मतलब आईआईटी जी के नाम से जानी जाती थी और जो आपकी के डबल आई मांस होती है आईआईटी मिंस जो होती है ना तो अगर हम बात करें इसकी तो सर सर्कल की इक्वेशन मेरे लिए तो बहुत टफ तास नहीं है मैं लिखूंगा ह एक्स - ह का होल स्क्वायर प्लस ए माइंस सिंपलीफाई दिस वेदर यू आल कैन बट आई वुड रिकमेंड की इसको सिंपलीफाई करने से पहले एक छोटा सा थॉट लो दिस सर्कल पासेस थ्रू अन सर्टेन पॉइंट विच इस नथिंग बट व्हाट -1 0 ये सर्कल पास होता है एक खास पॉइंट्स है जैसे आप कहते हो -1 0 तो shailise एक ही सर माइंस वैन कमा जीरो से सेटिस्फाई करेंगे और आय होप आप देख का रहे हो जैसे आप माइंस एंड कमा जीरो पास karvaoge तो विल गेट तू नो दी वैल्यू ऑफ ह एंड यू नो पास दी वैल्यू ऑफ ह आई थिंक चीज आसान हो जाएगी बस यही मैं आपसे कहना चाह रहा हूं तो देखो भाई सिंपलीफाई करने की कोशिश करते हैं पहले पास करेंगे -1 कमा जीरो तो -1 अगर यहां रखा तो हो जाएगा -1 - ह - कॉमन लिया तो स्क्वायर के कारण पॉजिटिव 1 + ह का स्क्वायर जो की कितना होगा हो जाएगा अल्टीमेटली सर h² गेट डी सेट स्क्वायर गया तो उधर और ये हो जाता है ह कितना -5/2 तो ह इस गोइंग तू बी माइंस फाइव बाय तू यहां पर सर जो ये ह के cardinate है ना इसका एक्सपोर्ट इन मिलेगा वो होगा -5/2 तो क्या अब मैं सबसे जरूरी बात सर्कल का सेंटर बता सकता हूं सर जो ऑप्शन बी में लिखा हुआ है ना वही सर्कल का सेंटर है बट ये आंसर नहीं है उसने क्या पूछा था था सर्कल पासेस थ्रू विच पॉइंट सर्कल आपका किस पॉइंट से पास होता है आपको ये बताना है तो मेरे ख्याल से अब सब चूंकि हमें पता है की ह की वैल्यू क्या है मैं यहां रख सकता हूं क्योंकि मुझे पता है इसके वैल्यू क्या है तो वो मैं यहां रख सकता हूं तो जैसे ही मैं ह की वैल्यू यहां रखता हूं व्हाट दस दिस इक्वेशन बिकम सो दिस इक्वेशन bikams कितना ह की वैल्यू जब आप -5/2 रखते हैं तो ये हो जाता है एक्स यहां रखा तो एक्स प्लस 5/2 का होल स्क्वायर बाय तू का स्क्वायर लिख सकते हो विच इस आई थिंक से चीज है ना अब अगर आपसे कोई पूछे इनमें से कौन सा पॉइंट इस इक्वेशन को सेटिस्फाई करता है तो क्या आप देखकर चेक करके नहीं बता सकते क्या सीधे-सीधे हम ट्राई कर लें सर दो-तीन चीज जैसे अगर मैंने ट्राई किया -3/2 कॉमन जीरो एक-एक करके हम एक-एक ऑप्शन ट्राई करेंगे स्टूडेंट्स तो जैसे ही हमने ट्राई किया -3/2 कमा जीरो तो देखो क्या ए रहा है यहां रखा -3/2 सुनेगा ध्यान से कम की बात है ना 5 / 2 - 3 / 2 तो 5 - 3 कितना 2 तो 2/2 कितना वैन वैन का स्क्वायर 1 याद रखिएगा जीरो तो ये कितना हो गया 2² 1 + 4 5 आई थिंक फाइव इसके इक्वल तो नहीं है ऑप्शन तू ट्राई किया ऑप्शन बी ट्राई किया सर ये तो सर्कल के सेंटर के कोऑर्डिनेट्स हैं ये तो सर्कल के सेंटर के cardinate तो ने से सेटिस्फाई नहीं करेंगे तो ऑप्शन ए एंड ऑप्शन बी आर डिस्कार्डेड व्हाट अबाउट ऑप्शन सी ट्राई करते हैं ऑप्शन सी जब ट्राई किया तो ये वापस - 3 / 2 - 3 / 2 + 5 / 2 वापस 2/2 2/2 मैंने 1 का स्क्वायर 1 ये 1 यहां रखा कितना 5/2 अब सुनेगा ध्यान से अब 5 / 2 में से तू सब्सट्रैक्ट करिए तो 5 / 2 में से अगर तू सब्सट्रैक्ट किया तो 5 / 2 - 5 - 4 / 2 यानी कितना हो जाएगा 1 / 2 1 / 2 यानी कितना 1/2 का स्क्वायर यानी 1/4 तो ये हो जा रहा है वैन बाय फोर प्लस वैन है और वैन बाय फोर प्लस कब डिस्कार्ड इट तो सी आर लेफ्ट विद ऑप्शन दी बट वेरीफाई करना तो हमारा कम है सर हो सकता है हमारी इक्वेशन ही गलत आई हो आपको क्या पता चेक करिए तो ऑप्शन दी देखा -4 0 जब माइंस फोर रखा यहां पर तो कितना ए रहा है सर 5/2 मैसेज 4 आपको सब्सट्रैक्ट करना है तो 5/2 में से अब में से अलग से लिख देता हूं तो यहां पे मैं सब्सट्रैक्ट कर रहा हूं क्या 5 / 2 - 4 ये कितना हो जाएगा सर 2 की 2 4 = 8 में से 5 से ट्रैक किया तो कितना 3 तो ये हो जाएगा -3/2 अब जब आप -3/2 का स्क्वायर करेंगे तो कितना हो जाएगा सर वो हो जाएगा 9/4 है ना प्लस ध्यान से देखो भाई यहां रखा जीरो यहां रखा कितना जीरो जीरो रखा तो 2² कितना 4 अब ध्यान से देखो भाई 4 और 4 16 और 9:25 और 25/4 यहां भी दिया हुआ है तो आई थिंक बिल्कुल सही सर आपका ऑप्शन दी behadchak की संकोच होकर आप मार्क करिए सही है तो आप कौन सा ऑप्शन मार्क करेंगे ऑप्शन दी वीडियो ऑल अंडरस्टैंड दिस होल क्वेश्चन की सर कितने आसान कितने बेसिक कितने सिंपल डायरेक्ट स्ट्रेट फॉरवर्ड क्वेश्चंस पूछे हैं जी एडवांस्ड है जिन्हें आप इतनी आसानी से सोच सकते हैं बिल्कुल आसानी से सोच सकते हैं और अगर इतनी आसानी से सोच सकते हैं तो क्या नेक्स्ट क्वेश्चन ट्राई करें ये के डबल आई एडवांस नहीं 2012 में पूछा है अभी हम कौन सी कैटिगरी के क्वेश्चंस करें सिंगल करेक्ट आंसर टाइप क्वेश्चन इस क्वेश्चन को देख कर आपको क्या समझ ए रहा है अगेन सर मेरे लाइफ में तो चीज है तभी सुलझी हुई रहती हैं जब मैं चीज विजुलाइज कर लूं जब मैं चीज अच्छे से वर्कआउट कर लूं ड्रॉ करके तो सर इस प्रॉब्लम को ड्रा करने के लिए क्या बोलेंगे देखो भाई डी लोकस ऑफ डी मिड पॉइंट ऑफ डी कोड ऑफ कॉन्टैक्ट डी कोड ऑफ कॉन्टैक्ट डी मोमेंट यू रीड डी स्टंप डी गोद ऑफ कॉन्टैक्ट आय थिंक समथिंग मस्ट ट्री योर ब्रेक गोद ऑफ कॉन्टैक्ट मतलब क्या आई थिंक कुछ तो स्ट्राइक कर ही रहा होगा ना कोड ऑफ कॉन्टैक्ट्स उनके सर ये एक बड़ी इंपॉर्टेंट टर्मिनोलॉजी है ऑफ टेंसेज ड्रॉन फ्रॉम डी पॉइंट फ्लाइंग ऑन डी स्ट्रेट लाइन दिस तू दिस सर्कल व्हेन यू लुक आते दिस सर्कल डू यू रिकॉग्नाइज्ड समथिंग सर ये एक ऐसा सर्कल है जिसके सेंटर के कोऑर्डिनेट्स से जीरो कमा जीरो और इसकी रेडियस A3 आई होप ये तो आप देख का रहे हो तो एक ओरिजिन पर ले ओरिजिन पर सेंटर रखने वाला सर्कल और उसे पर मैंने एक स्ट्रेट लाइन ये एक स्ट्रेट लाइन है ना लेट ड्रॉप थिस इस स्ट्रेट लाइंस स्ट्रेट लाइन पर लाइक करने वाले किसी पॉइंट से मैंने यहां पर कोड ऑफ कॉन्टैक्ट ड्रॉप की है ना तो लोकस ऑफ डी मिड पॉइंट ऑफ डी कार्ड ऑफ कॉन्टैक्ट सर टेक्निकल ये क्वेश्चन ये है इस लाइन पर कोई भी पॉइंट आप उठाओ वहां से इस पर tenjent ड्रॉ करो उन टेनिस का वही कोड ऑफ कॉन्टैक्ट बनेगा तो वो जो कोड ऑफ कॉन्टैक्ट बनेगा वो जो कोड ऑफ कॉन्टैक्ट का कोड ऑफ कॉन्टैक्ट का जो मिढ्वाइंट है उसका लोकस क्या होगा सर लोकस के बारे में क्योंकि ये पॉइंट वेरिएबल है इस लाइन पर लाइक कर रहा है उससे आपने इस पर टेंडर ड्रॉ किया आई होप आप समझ का रहे हो आई होप आप विजुलाइज कर का रहे हो तुम्हें एक बेसिक सा एक स्टैंडर्ड लेकर बताता हूं जैसे की मैन लो मैंने यहां पर एक रैंडम पॉइंट लिया लेट्स टेक लेट्स से मैंने यहां पर एक रैंडम पॉइंट लिया अब जैसे ही सर आपने यहां पर एक रैंडम पॉइंट लिया है तो लेट से आपने यहां से तो एक टांगें ऑफ कोर्स यहां ए जाएगी ऐसे और दूसरी तंग कर सकती यहां पर ऐसे और क्लीयरली आप खुद सोच रहे होंगे की गोद ऑफ कॉन्टैक्ट जब आप बनाएंगे तो ये हो जाएगी आपकी क्या कोड ऑफ कॉन्टैक्ट इस कोड ऑफ कॉन्टैक्ट का लुकास मिढ्वाइंट का मुझे लोकस निकलना है तो उसे मैं का लेता हूं ओके मैं का लेता हूं होमो के है ना और सर्कल के सेंटर के बारे में ज्यादा बातें करने की जरूरत नहीं है जो केस जीरो कमा जीरो इन केस मुझसे सर्कल की रेडियस पूछी जाती तो मैं क्या कहता थ्री कोई तकलीफ कोई आपत्ति नहीं है सर अब एक बात का जवाब दो आप अब बड़ी बेसिक सी बात का जवाब दो इस कोड ऑफ कॉन्टैक्ट पर ये जो ये जो tenjent रॉक की गई है इस लाइन पर लाइक करने वाले पॉइंट से उसके cardinate क्या है सर अगर उसके एक्स koardinate को लेट्स से मैं थोड़ी देर के लिए का लूं कुछ भी है ना जैसे मैं का लेता हूं टी अगर उसके excordinate को मैं का लूं टी तो उसका ए कार्ड एजेंट क्या बनेगा आप खुद सोच के देखना इसका जो ए कार्ड आएगा आप 4X को ये रखो फाइव बाय को उधर भेजो तो 20 + 5 / 4 एक्स की वैल्यू कितनी ए जाएगी सॉरी 5 मुझे ए की वैल्यू चाहिए हम रिलीज सॉरी फॉर डेट तो मैं क्या करता हूं 5y उधर पहुंचना हूं 20 इधर लता हूं मैं क्या कर रहा हूं सुनेगा ध्यान से मैं लिख रहा हूं 4X से ही 20 को इधर लेकर आए और फाइव ए को उधर पहुंचा है लेकिन मैंने सिर्फ ए को उधर रखा और उसके डिनॉमिनेटर में क्या लिया है 5 आप देख का रहे हैं जब एक्स कोऑर्डिनेट्स अपने माना है टी तो यहां पर रख दीजिए टी तो ए cardinate क्या हो जाएगा सर ए cardinate हो जाएगा 40 - 20 / 5 ये आपकी इस स्ट्रेट लाइन पर लाइक करने वाले किसी भी रैंडम पॉइंट के क्या हो जाएंगे cardinate ये सब बातें क्यों कर रहे हो सर अब आपके पास दो तरीके हैं या दो हिंट हैं इस बात को कहने के मैं आपसे सीधे सीधे कहता हूं suniyega और जवाब दीजिएगा और सोच के दीजिएगा अगर मुझे किसी को मिड पॉइंट के कोऑर्डिनेट्स बताएं तो क्या उसे सर्कल की इक्वेशन लिखी जा सकती है आई रिपीट माय स्टेटमेंट इफ आई हैव डी cardinates ऑफ डी मिड पॉइंट ऑफ अन कार्ड ऑफ अन सर्कल कैन आई राइट डी इक्वेशन ऑफ डेट कार्ड ए बी किसी एक्सटर्नल पॉइंट से अगर कोड ऑफ कॉन्टैक्ट फ्रॉम अन्य एक्सटर्नल पॉइंट सो मेरे पास दो तरीके हैं एक तो यहां से और एक यहां से बस अब आपको सोचना है की वो दोनों इक्वेशंस क्या होंगे और उन्होंने कंपेयर करके आपको ह और के में इक्वेशन बनाना है जहां से आप चीजें सोच पाएं एक तरीका जो हमारे दिमाग में ए रहा है वो बड़ा सिंपल सा की सर अगर ये कॉर्ड अगर ये बेसिक रिकॉर्ड जो की बायसेक्स हो रही है ह कॉम के पर क्या याद है आपको अपने पढ़ा था अगर मुझे कार्ड का मिड पॉइंट पर था हो तो मैं उसे कार्ड की क्वेश्चन लिख सकता हूं तो सर अगर मुझे ये चोर्ड बायसेक्स होती मतलब मिड पॉइंट ह कमा के ऊपर दिख रहा है होप यू आर ऑल राइजिंग ह कॉम के दिस इसे वेल इट इस गेटिंग बाईसेक्टर राइट सो ये जो कॉर्ड है नाम दिस है ना स्कॉट को अगर मैं नाम दे डन लेट से अब तो हम जानते हैं सर बड़ी बेसिक्स इक्वेशन क्या टी = X1 और टी आपको याद ए रहा है सर ह को हम के के लिए अगर मैं टी निकलूं ह के लिए अगर मिट्टी निकालो तो दिस इसे गोइंग तू बी कितना सर एचसी आप क्या लिखोगे उसे एचसी + के ए - सिन दिस इस टी और S1 क्या होगा सर S1 का मतलब होता है S1 का मतलब होता है आप ह कॉम के इस सर्कल की इक्वेशन में पास कर दीजिए है ना तो क्या हो जाएगा h² + K2 - 9 सो दिस इस गोइंग तू बी h² + K2 -9 ये किसकी इक्वेशन है ये उसे कार्ड अब की इक्वेशन है सो इक्वेशन ऑफ अब कट और वो भी किस तरीके से निकलना है हमने निकाला टी = S1 तकनीक सो जो की हमने पढ़ा है सर अब अगर इस बारे में बात करें तो आप कहोगे ये और ये कैंसिल आउट हो जा रहा है सो दिस कट एंड आउट तू बी कितना एचसी प्लस के स्क्वायर ए डिफरेंट रूट बट डू सी हैव अन्य आदर एप्रोच आज वेल सर सी डू हैव हमारे पास क्या एक और तरीका है एक और तरीका है जनाब बस थोड़ा ध्यान से देखिएगा एक और तरीका क्या कहता है ध्यान से देखो भाई सर अगर मुझे पता है अगर मुझे पता है की किसी एक्सटर्नल पॉइंट से मैंने दो टांगेंट्स ड्रॉ किए हैं और उन टेंसेज का जो पॉइंट ऑफ कॉन्टैक्ट है उन पॉइंट ऑफ कॉन्टैक्ट से मैंने एक कोड ऑफ कॉन्टैक्ट नहीं है सो दिस इसे अलसो नॉन आज कार्ड ऑफ कॉन्टैक्ट तू डी सर्कल विद रिस्पेक्ट तू दिस पॉइंट दिस पॉइंट लेट्स कॉल इट पी फॉर ऑयल दिस पॉइंट पी तो सर क्या इसके रिस्पेक्ट में भी आपको इसकी इक्वेशन लिखते आती है मुझे तो आती है क्या आपको आती है हमने पढ़ी है भाई अगर कार्ड ऑफ कॉन्टैक्ट की इक्वेशन लिखनी है विथ रिस्पेक्ट एंड पॉइंट पी तो टी = 0 से दी जाती है और ग = 0 मतलब फिर वही बात अब आपके कौन-कौन हो जाएंगे दिस मतलब ये हो जाएगा ए और ये हो जाएगा बी है ना तो आप क्या लिखोगे ए एक्स + बी ए - 9 = 0 तो बिल्कुल लिखते हैं जब टी पास किया तो ये हो जाएगा टक्स जब टी पास किया तो कितना हो जाएगा टक्स तो यही इक्वेशन दूसरे लॉजिक से क्या हो जाएगी मैं ऐसे यहां लिख लेता हूं है ना दूसरे लोग t=0 के लॉजिक से वही इक्वेशन है ना क्या हो जाएगी टक्स + 40 - 20 / 5 टाइम्स y20 के साथ भी एक और चीज आने वाली क्या -9 तो दिस इस गोइंग तू बी से इक्वेशन डी से क्वार्ट का एक क्वेश्चन वही कोड अब के क्वेश्चन ये है और वही कॉर्ड अब की इक्वेशन है अब करना क्या चाह रहे हो आप निकलना क्या चाह रहे हो सर रिलेशन क्या है मेरा कहना है मेरे पास कुछ वैरियेबल्स हैं जैसे एक्सी यहां पर है टी तो मैं इससे निजात पाना चाह रहा हूं मैं चाह रहा हूं की मेरे पास जो रिलेशन है वो ह कमा के में ही हो क्योंकि वही तो लोकस होगी वही तो ह कमा के में एक रिलेशन बनाएगा और वही फाइनली आपका आंसर होगा जहां पर आप ह और केक हो अल्टीमेटली एक्स और ए सर प्लीज करेंगे डू यू ऑल गेट बट एक लॉजिक जो मेरा कहता है सर पहले से थोड़ा सिंपलीफाई कर लेते हैं जब मैं सिंपलीफाई करना चाहूंगा तो लेट्स मल्टीप्लाई डी होल इक्वेशन बाय फाइव सो दिस bikams कितना 5dx है ना प्लस कितना है सर 40 - 20 टाइम्स ए और 9 5 कितना हो जाएगा सही हो जाएगा 45 = 0 आई होप आप जानते हो जब मुझसे कभी भी कोई कहेगा की सर ये इक्वेशन और ये इक्वेशन रिप्रेजेंट डी से लाइन डी से स्ट्रेट लाइन एंड डेट केस आई वुड सिंपली सजेस्ट की सर इनके एक्स और ए और कांस्टेंट टर्म्स के रेश्यो से होंगे यानी की क्या पहला कंक्लुजन यहां पे एक्स का कॉएफिशिएंट है ह और यहां पर एक्स का कॉएफिशिएंट फाइव टी डू यू ऑल गेट डेट सिमिलरली सर अगर आप आगे बात करें तो अगला क्या मिल जाता है आपको नेक्स्ट जो मिलता है यहां से जो ए का कॉएफिशिएंट मिलता है डेट इस के सो दिस इस गोइंग तू बी के और अगर वहां से आप ए का कॉएफिशिएंट देखें तो डेट इस गोइंग तू बी फोर टी -20 कोई डाउट तो नहीं है सर सिमिलरली सिमिलरली अगर मैं और आगे आपसे बात करूं है ना और आगे अगर आपसे बात करूं तो क्या कांस्टेंट टर्म्स जो मिल रही यहां पर मिल रहा है कितना सर यहां पर आपको मिल रहा है - x² - K2 वो है -45 तो नेगेटिव कैंसिल करके हमें लिख दे रहा हूं सीधे-सीधे क्या इट्स ² प्लस के स्क्वायर डिवाइडेड बाय 45 क्या किसी भी स्टूडेंट को कोई आपत्ति ये जो इक्वेशन हमने बनाई ये जो रिलेशन हमने लिखा इससे इसमें लिखी गई किसी भी बात से आई थिंक सर अब आपको कोशिश बस ये करनी है की आप बड़ा बेसिक सा फंडा निकालो बेसिक सा फंडा क्या अब देखो हमारे पास ह के और टीम में इक्वेशन है मुझे रिलेशन चाहिए तो मैं क्या करूंगा पता है मैं यहां से टी की कोई वैल्यू निकलूंगा किस तरीके से गौर से suniyega स्टूडेंट्स में इन दोनों को इक्वेट करके इससे और इससे इक्वेट करके मैं टी की कोई ना कोई वैल्यू निकलूंगा और फिर इससे और इससे इक्वेट करके वो टी की निकल गई वैल्यू यहां पर रिप्लेस कर दूंगा तो मुझे सिर्फ और सिर्फ ह और के में एक रिलेशन मिलेगा डू यू ऑल गेट डेट कोशिश करते हैं सर पहले मैं ह / 5 टी और h2 प्ले के स्क्वायर अपॉन 45 क्विक वेट करता हूं तो बड़ा बेसिक सा रिलेशन जो आप देख पाओगे क्या यहां से आपको पहले तो दिख रहा क्या है सर यहां से दिख रहा है ह / 5D और ये किसके इक्वल है सर ये है इक्वल है h² + K2 / 45 आई होप ये सब आप नोटिस कर का रहे हो अब सर ध्यान से देखो आप अगर नोटिस कर रहे हो तो फाइव का 45 कितना ये हो जाता है नाइन टाइम्स है ना ये हो जाता है नाइन टाइम्स अब क्या सर मैं टीको वहां ले जाता हूं नाइन को वहां ले जाता हूं और इसे यहां ले आता हूं तो ध्यान से देखना ये हो जाएगा 9h / h² + K2 ये हो जाएगा 9h / कितना x² + K2 बाय डी वे दिस इसे दी वैल्यू ऑफ नथिंग बट टी सो यहां से सर हमने टी की वैल्यू निकल वेस्ट आउट तू बी दिस अब प्लीज इस बात पे गौर फरमाइए स्टूडेंट्स मैं आपसे जो कहना चाह रहा हूं वो ये की अब आप इन दोनों को एक वेट करिए कर लेते हैं सर तो देखो ध्यान से अगर मैं वेट करता हूं तो सीधे सीधे लिखना चाह रहा हूं देखो ये हो जाएगा 45 की 45 के लिख देता हूं इसे फिर से तो यहां पर क्या ए जाएगा सही ए जाएगा 45 के इस इक्वल तू 45 के = h² + k² टाइम्स 14 - 20 है ना तो कितना हो जाएगा ये हो जाएगा h² + K2 टाइम्स 14 - 20 नौ डी ओनली थिंग सी नीड तू डू 40 - 20 में ये जो मुझे टी दिख रहा है दिस टी नीड्स तू बी रिप्लेसिड की दिस एंड पॉसिबल विल गेट दी रिजल्ट और रिलेशनशिप बिटवीन नक है ना तो सर कोशिश करते हैं देखो भाई आपके पास बहुत सारे तरीके मेरा कहना है टी की वैल्यू जो आपने निकल है ये सीधे सीधे वहां रिप्लेस कर दीजिए जो होना होगा हो जाएगा है ना तो मैं यहां पर तो लिखता हूं 45 के टी की वैल्यू रिप्लेस कर रहे हो तो टी की वैल्यू कितनी है सर वो है 9h / h2 प्लस के स्क्वायर प्लस के स्क्वायर भी मल्टीप्लाई किया तो ये हो जाएगा 20 h² + K2 आई होप डेट दज इन बदर अन्योन टॉक नौ व्हेन यू फाइनली गेट रीड ऑफ दिस पार्ट आई थिंक दिस हो दिस गैस सैंपल फाइट तो अब क्या मिल रहा है ऐसा 9 4 कितना होता है 36 सो दिस इस गोइंग तू बी 36 ह दिस इस गोइंग तू बी 36 ह और 36 ह को इधर ले या 36 ह को वही रखा 45 के को उधर ले गए 20 को इधर ले तो लेफ्ट हैंड साइड पर मुझे दिख रहा है 20 टाइम्स इट्स स्क्वायर प्लस के स्क्वायर एंड ऑन दी राइट हैंड साइड व्हाट डू सी सी 9 4 36 ह माइंस कितना 45 के कोई तकलीफ किसी भी स्टूडेंट को सर आपको सिंपलीफाई करने का कोई स्कोप देख का रहे हो क्या मुझे तो ज्यादा नहीं दिख रही सिंपलीफाई होने की और अगर मैं ऑप्शंस में देखूं तो आई होप आप रिलाइज कर का रहे हो ह और के को आप किस से रिप्लेस करोगे एक्स और ए से और जब आप ह और के को एक्स और ए सर प्लेस करते हो तो आपका एक तो 20 में x² + y² दिखाना चाहिए और 36 ह और 45 ए देखना चाहिए और ऐसा मुझे जो दिख रहा है वो क्लीयरली ऑप्शन ए दिख रहा है सर आपको ऑप्शन ए क्यों दिख रहा है क्योंकि देखो ना आप 29 h2 + K2 = 36 ह - 45k वो जब इस तरफ ए रहा है तो आपको साइन चेंज हो जा रहे हैं और ये क्लीयरली हमें कौन सा ऑप्शन सेटिस्फाई करता हूं दिख रहा है ऑप्शन ए आईआईटी जी एडवांस्ड 2012 का ये सवाल जो की क्लीयरली एक आसान सा बेहतरीन सा शॉर्ट ट्रिक्स ऑफ क्वेश्चन है वो का रहा है एक लाइन एमएक्स + 1 ए = एक्स + 1 दिस लाइन अच्छा सर ये किस तरह के क्वेश्चन है अभी भी आपके सिंगल करेक्ट आंसर टाइप क्वेश्चंस ही चल रहे हैं राइट वो का रहा है ये जो लाइन है ए = एमएक्स + 1 है ना सो दिस इसे योर लाइन ए = एमएक्स + 1 ये इंटरसेक्ट करती है कैसे इस सर्कल सर्कल को देख कर कुछ ओपिनियन आपके दिमाग में सर देख रहे हो आप यह जो है ये है आपका क्या सेंटर के cardinate आप देख का रहे हो ना आपके जो हैं दोनों चीज और 25 के लिए क्या है उसकी मतलब फाइव जो है उसकी रेडियस बहुत आसान सी बेसिक्स सी बात है अब वो का रहा है की ये जो लाइन है वो इस सर्कल को पी और के पर इंटरसेक्ट करती है ठीक है सर डी मिड पॉइंट ऑफ डी लाइन सेगमेंट पीके हज एक्स कोऑर्डिनेट्स -3 कमा फाइव अब कुछ बातें गड़बड़ हो रही है मेरा कहना है इस 2 मिनट ड्रॉ कर लेंगे क्या बिल्कुल ड्रा करिए हमेशा वेरियस करिए ज्यामिति अगर मैथ्स में कहीं जगह आपको प्रॉब्लम्स इमेजिन करके उसको यू नो प्लॉट करके ज्यादा आसानी से आप देख पाओगे तो सर क्या करें पहले तो एक सर्कल बना लो सर सर्कल कैसा है थ्री कमा - 2 पर इसका सेंटर है और फाइव इसकी रेडियस है ठीक है सर सी हैव अन सर्कल बेसिकली जिसके सेंटर के कार्ड के अगर मैं बात करूं आपसे तो वो है क्लीयरली थ्री कमा - 2 आई होप आप सभी इस बात से एग्री करते हो और सर इसकी रेडियस है फाइव विच सी ऑल आर क्लीयरली मीटिंग अब अगर मैं एक स्ट्रेट लाइन की बात करूं इफ आई टॉक अबाउट अन स्ट्रेट लाइन विच इंटरसेप्ट्स डी सर्कल तो क्या इस स्ट्रेट लाइन की इक्वेशन मुझे पता है सर स्लोप को हटाकर बाकी बातें मुझे पता है ओके थॉट्स एक्सेप्टेड नौ व्हाट आर सी टोल्ड की सर ये जो लाइन है जब इसे इंटरसेक्ट करती है तो इस कॉर्ड का जो मिड पॉइंट आता है इस कॉल्ड का जो मिडिल पॉइंट आता है डेट मिड पॉइंट का जो एक्स कोऑर्डिनेटर डेट इस -3 कमा फाइव इसका जो एक्स कोऑर्डिनेट्स है वो है -3 / 5 कमा फाइव -3/5 एक बात बताओ स्टूडेंट्स एक बात सोच के बताओ बाय सैम सेंस बाय यूजिंग सैम नॉलेज बाय यूजिंग सैम अंडरस्टैंडिंग ऑफ स्ट्रेट लाइंस इफ दिस पॉइंट हैपेंस तू ले ऑन दिस स्ट्रेट लाइन कैन आई कैन clueed की सर कैन आई कनक्लूड इस पॉइंट का जो एक्स cardinate है वो इस लाइन को सेटिस्फाई करेगा और उसकी हेल्प से क्या मैं इस पॉइंट का ए cardinate निकल सकता हूं आप मेरी बात समझ का रहे हो मैं क्या करना चाह रहा हूं मैं बस इतना सा कहना चाह रहा हूं की सर अगर इस पॉइंट का एक्स कोऑर्डिनेट्स यहां पास करूं क्योंकि ये पॉइंट इस लाइन पर लाइक करता है तो इस लाइन में पास किया तो एक्स की जगह क्या रखा -3 / 5 तो ये हो जाएगा -3n / 5 सो ये हो जाएगा 1 - 3m / 5 तो इसी पॉइंट का जो ए cardinate होगा वो होगा 1 - 3m / 5 आई होप यू ऑल एग्री विद डेट सो दिस इस गोइंग तू बी दी कोऑर्डिनेट्स ऑफ दिस पॉइंट जो की है क्या मिड पॉइंट किसका है स्क्वाड का सो दिस इस दी मिड पॉइंट ऑफ दिस कट आई होप यू ऑल एग्री विद डेट कोई कन्फ्यूजन कोई डाउट तो नहीं है अच्छा सर हमसे कहा गया था ये जो लाइन है ये सर्कल को पी और के पर इंटरसेक्ट करती है तो मैं इन पॉइंट्स का नाम दे देता हूं लेट्स कॉल दिस पॉइंट पी दिस कॉल दिस पॉइंट के है ना अरे सर्कल के सेंटर को बस अपनी सहूलियत के लिए मैं का देता हूं सी एन और ये जो मिड पॉइंट है इसे मैं अभी अपनी कन्वीनियंस के लिए का देता हूं एम किसी भी स्टूडेंट को कोई आपत्ति इस बात से यहां तक तो नहीं होनी चाहिए अब आप थोड़ा और मैं यह स्टूडेंट थोड़ा सा पेशेंस के साथ इस चीज को देखिए की अगर मैं इसको कनेक्ट करता हूं तो हम जानते हैं सर सर्कल के सेंटर्स अगर आपने कार्ड के मिड पॉइंट को मिलाया है तो वो क्लीयरली परपेंडिकुलर ही रहता है बिल्कुल सही बात और उसी से मैं देख पता हूं की ये जो बायसेक्स होती हुई गोद आपको दिख रही है अच्छा एक छोटा सा ऑब्जर्वेशन और मैं देना या लेना या पूछना चाहूंगा किस तरह आपने पी से सी को मिलाया होता वैसे तो जरूरत नहीं है इसी से बात बन जाएगी आप मत मिलाओ सर आप बस इतना देखो दो बातें स्टूडेंट्स प्लीज दो बातों पर गौर फरमाइए हिंट जो आपको मैं देना चाह रहा हूं वो ये है की अगर मैं आपसे पूछूं इस लाइन की स्लोप तो आप बता सकते हो क्या अगर मैं आपसे पूछूं इस लाइन के स्लोप तो आप बता सकते हो क्या सर मुझे ना अगर मैं थोड़ा ध्यान से देखूं तो इस लाइन के स्लोप को देखने के दो तरीके हैं कम लाइन की स्लोप को देखने के दो तरीके एक नजरिया 1 पर्सपेक्टिव ये की कम जो लाइन है वो पीके लाइन पर परपेंडिकुलर है कुछ याद आया दूसरा नजरिया की मेरे पास सी और एम के cardinate हैं आपके पास दोनों हिंट है दोनों तरीके हैं अब जरा मुझे आप थोड़ा आराम दीजिए और आप करिए स्क्रीन पीओएस और इस क्वेश्चन को करिए ट्राई मुझे तो आराम बस आप स्क्रीन पीओएस करके ही दे सकते हैं क्योंकि रियलिटी में तो मैं आराम पे हूं नहीं है ना रियलिटी में तो हम लगातार क्वेश्चंस सॉल्व करें जा रहे हैं तो सर इस क्वेश्चन को ध्यान से देख रहे हैं मुझे चाहिए पीके स्ट्रेट लाइन की स्लोप तो सर वो तो दी हुई है ना आपको इतनी भाई सर वो दी हुई है तो पहले तो मैं आपको क्या कहना चाह रहा हूं की जो पीके लाइन है उसे पर परपेंडिकुलर है कौन सर कम की स्लोप जो मैं देख का रहा हूं वही इसे इक्वल तू एमएक्स + सी फॉर्म में वो है एन तो shailise स्लोप ऑफ कम इस गोइंग तू बी माइंस वैन अपॉन है अगर दो लाइंस परपेंडिकुलर हो तो बड़ा ही बेसिक सारा रिलेशनशिप है जो आप सब जानते होंगे अब आप कहना क्या चाह रहे हो सर मैं आपसे कहना चाह रहा हूं की ये तरीका था लाइन पर परपेंडिकुलर होने के कारण आपके पास क्या तरीका है मेरा कहना है सर सी और मेक और एम दोनों के कार्ड Y2 - y1 अपॉन X2 - X1 भी तो एक तरीका हो सकता है बिल्कुल हो सकता है तो वही ट्राई करते हैं Y2 - y1 सो - 2 - 3 / 5 सो दिस इसे गोइंग तू बी माइंस तू प्लस थ्री बाय फाइव आई एम रियली सॉरी मैंने ये X2 - X1 किया है जितना मैं देख का रहा हूं मैंने बहुत गलत किया है मैंने Y2 - X1 कर दिया जो की बिल्कुल गलत है आई एम रियली सॉरी फॉर डेट है ना मैं फिर से करता हूं रेली सॉरी फॉर था -2 -1 - इसका प्लस हो जाएगा 3m / 5 - 2 - 1 सो ये हो जाएगा -3 -3 और ये जो नेगेटिव होगा ए प्लस हो जाएगा तो ये हो जाएगा -3 + 3m / 5 सो न्यू > में क्या मिलेगा सर -3 प्लस 3 कोई तकलीफ तो नहीं लेकिन सर डिनॉमिनेटर में क्या मिलेगा अगर डिनॉमिनेटर को देखें तो ये हो जाएगा X2 - X1 सो 3 - 3 / 5 ये हो जाएगा 3 + 3 / 5 तो ये कितना हो जाएगा सर ये हो जाएगा 3 + 3 / 5 आई थिंक ये आपकी एक स्लोप आती है लेकिन सर ये स्लोप आपने जो निकल है ये स्लोप भी तो यही स्लोप है बिल्कुल सही बात तो जो आप यहां से शुरू निकल रहे द वो है -1/एम तो बिल्कुल सर ये जो है एक इसकी इक्वल होगी ये जो है इक्वल होगी -1/m³ आई थिंक सर आपके पास एक क्वेश्चन है जो आपको सॉल्व करनी है बिल्कुल तो ट्राई करेंगे और पहले इस वाले पार्ट को जरा सिंपलीफाई कर लें तो 5 3 कितना -15 प्लस कितना 3 एम डिवीज़न में फाइव आएगा यहां भी फाइव आएगा सैम नॉट बाय डेट 5 3 कितना 15 + 3 कितना 18 डू यू ऑल एग्री विथ डेट आई 3 15 + 3 18 है ना एक छोटा सा कम और कर लेते हैं सर यहां दिख रहा है माइंस वैन बाय एम थोड़ा और सिंपलीफाई कर लेते हैं क्योंकि स्कोप है कैसा सर देखो अब थ्री यहां से और यहां से कॉमन ले सकते हो आप तो जैसे ही ऐसा करोगे तो देखो ये तरीका वैन टाइम ये थ्री का फाइव टाइम्स और ये थ्री का सिक्स टाइम्स डू यू ऑल गेट डायट अब जैसे ही आप ऐसा करते हो तो सुनना ध्यान से क्या मैं एक कम कर सकता हूं अगर ये नेगेटिव साइन यहां मल्टीप्लाई किया तो यहां से नेगेटिव हटा ये ऑफकोर्स वैन है आपका है ना तो यहां पर से भी नेगेटिव हटा ये पॉजिटिव हो जाएगा और ये पॉजिटिव क्या हो जाएगा नेगेटिव तो सर अब जो इक्वेशन बन कर ए रही है ये एम यहां पर मल्टीप्लाई होता है तो ये हो जाएगा कितना 5 एम यहां मल्टीप्लाई ए जाएगा तो दिस इसे गोइंग तू बी फाइव एम यहां ये एम होता है तो दिस इस गोइंग तू बी m² कोई तकलीफ तो नहीं है और बड़ी आसान सी बात है सिक्स यहां मल्टीप्लाई होता है दिस इसे गोइंग तू बी सिक्स इसे वहां शिफ्ट किया तो ये हो जाएगा m² - 5 एम ये हो जाएगा m² - 5m + 6 = 0 सिर्फ क्वाड्रेटिक और यह तो हम सॉल्व करना जानते हैं क्वाड्रेटिक सॉल्व करने का तरीका जो मेरे दिमाग में ए रहा है सर बड़ा ही बेसिक सा क्या मैं कहूंगा सर आप देखो स्प्लिटिंग का मिडिल टर्म देख का रहे हो क्या -2m -3m और वो है सिक्स अगर आपको नहीं दिख रहा है तो मैं लिख देता हूं वैसे देखना है क्या है ना m² - 2m -3 एम + 6 अब देख का रहे हो माइंस फाइव माइंस तू माइंस थ्री कितना होता है +6 अगर प्रोडक्ट की बात करेगा तो एक फैक्टर बनेगा एन - 2 और अगर यहां भी एन - 2 चाहिए तो -3 कॉमन लेंगे यहां से एम कॉमन लिया था तो दूसरा फैक्टर हो जाएगा एन - 3 और अगर आप यहां से रूट्स निकलने हैं तो एम की जो वैल्यू आएगी वो होंगी 2 या 3 या दोनों और 2 3 पर बात हो रही है तो ऑप्शंस को देखो भाई एम जो है टेक्निकल वो 2 से 3 के बीच ले करता है ऑप्शन ए तो गलत है ऑप्शन सी भी गलत है ऑप्शन दी सही होता लेकिन यह तो कुछ तो भी लिखा है ऑप्शन दी तो मतलब मैथमेटिकली ही गलत है आप मेरी बात समझ का रहे हो ना और एनीवे वो 2 तक जा नहीं यहां पे टेक्निकल वो माइंस थ्री होना चाहिए था ना वो टाइप है तो आई वुड से यू शुड गो विद ऑप्शन क्या आप सब इस बात से एग्री करते हो की हान सर ऑप्शन बी आपका करेक्ट आंसर होगा की आप सभी को ये बात समझ ए रही है तो आई होप ऑप्शन भी आप क्यों का रहे हो सर ऑप्शन बी में बस इसलिए कहना चाह रहा हूं स्टूडेंट्स क्योंकि ऑप्शन भी आप ध्यान से देखो क्यों है क्योंकि एम क्या है तू और थ्री तो बिल्कुल सर तू और थ्री अब देख का रहे हो 2 ईयर हान थ्री हान कहीं ए जाएगा फोर से पहले तो बिल्कुल सर ऑप्शन बी हम बचका बिना ज्यादा सोचे समझे मार्क करेंगे ये क्वेश्चन तब सॉल्व होंगे जब आप इन चीजों को इस तरीके से लाइक करेंगे मीनल घबरा बिना डरे है और उनकी दो वैल्यूज आना तय है क्यों सर क्योंकि इस पॉइंट से स्कॉयर्ड रिस्पेक्ट में उसे ऑर्थो सेंटर की मिरर इमेज उसे ऑर्थो सेंटर की मिरर इमेज इन साइज के रिस्पेक्ट में हमेशा सर्कल पर लाइक करेंगी कौन से सर्कल सरकमसर्किल गेट डेट बस रीजन या कहना बस इतना सा है आय होप आप देख का रहे हो चीज की चूंकि ये आपका अलसो सेंटर है तो ऑर्थो सेंटर्स का जो अर्थरो सेंटर होगा उसके लिए ये स्ट्रेट लाइंस जो हैं यह अगर मैं इनको देखूं तो इस स्ट्रेट लाइन के उसे सेंटर की मिरर इमेज देखो तो वो यहां होगी इसके रिस्पेक्ट में देखो तो यहां हो गया इसके रिस्पेक्ट में देखो तो यहां होगी तो ये जो तीन एज हैं तीन जो आपकी साइड्स हैं ट्रायंगल की इनके रिस्पेक्ट में अगर आप मिरर इमेज निकले तो वो आपकी हमेशा इन सर्कल पर लाइक करेंगी किस पर सर कम सर पर किसके ट्रायंगल ये लाइन अगर आपने याद रखी तो ये क्वेश्चन आज का वैन ऑफ डी इजीएस क्वेश्चन है आप अगर सोच पाए तो कैसे सर सबसे पहले तो यह बताओ क्या मुझे और सेंटर दिया हुआ है बिल्कुल दिया हुआ है क्या मुझे इसकी इक्वेशन दी गई है इसकी जो जो तीन साइड्स हैं उनकी इक्वेशन दी गई है क्या सर थोड़ा सोचें तो मिल सकती है और अगर ये बातें मुझे पता है अगर मुझे बातें पता है तो सर कुछ तो सोचा जा सकता है बात समझने की कोशिश करो भाई बात इसलिए समझने की कोशिश करो की सर देखो एक ट्रायंगल है जिसकी दो साइड से एक साइड तो है एक्स एक्सिस अगर कोई साइड एक्स एक्सिस पर लाइक करें तो उसकी इक्वेशन क्या होती है ए = 0 दूसरी साइड क्या है दूसरी साइड आपकी है एक्स + ए + 1 = 0 है ना अब प्लीज इस बात को ध्यान से सोच रहा हूं वो का रहा है आपसे की जो आपका और तो सेंटर है जो आपका और तो सेंटर है वो है वैन कमा वैन यह कम की बात है अब वह का रहा है इक्वेशन ऑफ डी सर्कल पासिंग थ्रू डी वर्टिकल ट्रायंगल ठीक है सर सारी बातें करते हैं पहले एक बात बताओ पहले एक बात बताओ ऑर्थो सेंटर के कोऑर्डिनेट्स के हैं सर वो है वैन कमा वैन अगर ये आपका ऑर्थो सेंटर है तो क्या मैं इसकी इमेज निकलूं ए = 0 के रिस्पेक्ट में सर अगर आप किसी भी यू नो पॉइंट की अगर इस लाइन के रिस्पेक्ट में जो की ऑफ कोर्स आपका क्या है ए = 0 ए = 0 यानी एक्स एक्सिस है ना तो किसी भी पॉइंट की अगर आप एक्स एक्सिस के रिस्पेक्ट में मिरर इमेज निकलती हैं तो उसका एक्स कोऑर्डिनेट्स तो वही रहता है बस उसका ए कोऑर्डिनेट्स उतनी ही डिस्टेंस पर नेगेटिव पॉइंट क्या ही कर रहा होता है तो shailise वो जो सरकम सर्कल बनेगा डेट वुड बी पासिंग थ्रू दिस पॉइंट सो दिस इसे गोइंग तू बी डेट फर्स्ट पॉइंट डू यू नो एग्री विद डेट अन्य कन्फ्यूजन सो फार सिमिलरली आपके पास एक और स्ट्रेट लाइन है सर एक्स + ए + 1 सो दिस एंड आदर्श फेक लाइन देयर इस वैन मोर स्ट्रेट लाइन सी डू हैव जिसकी इक्वेशन में मैन लेता हूं लेट से कुछ ऐसी है है ना तो कहना है सर इसी ह पॉइंट का जो मिरर होगा जो मिरर इमेज होगी वो क्या होगी सर सन मिरर इमेज निकलती आती है की आपको मेरे ख्याल से मैंने सिखा है सर लाइफ में की अगर मेरे पास कोई पॉइंट है मेरे पास अगर कोई पॉइंट है लेट से X1 और इसकी न्यूज़ है मेरा इमेज चाहिए किस लाइन के रिस्पेक्ट में मिरर इमेज जो है उसके कोऑर्डिनेट्स हैं ह कॉम के तो आप लिखते हो ह - X1 अपॉन इसका एक्स का कॉएफिशिएंट ए = के - y1/ए का क्वेश्चन विच इस बी = मिरर इमेज इन अगर फीड के अकॉर्डिंग साइन उसे करता है लेकिन मिरर इमेज लेनी है तो - 2 टाइम्स क्लीयरली ए X1 + dy1 + सी तो ये हो जाएगा ए X1 + बी ए वैन प्लस सी और डिनॉमिनेटर में अंडर रूट ओवर ए स्क्वायर प्लस बी स्क्वायर नहीं लिखना है मैं बार-बार का रहा हूं डिनॉमिनेटर में सिर्फ आपको लिखना है a² + b² और इस फॉर्मूले के अब से आप ये फीट के सॉरी मेरा इमेज के कार्ड निकलती हो बस वही एप्रोच हमें यहां पर उसे करनी है सर देखो भाई बहुत ध्यान से देखना सर इसके अगर आप मिरर इमेज निकलती हो यहां पे जिसको आप कहते हो थोड़ी देर के लिए लेट से ह कॉम के तो आप क्या कहोगे मैं कहूंगा ह माइंस वैन मैं कहूंगा ह माइंस वैन अपॉन वहां पर एक्स का ऑपरेशन क्या है वैन इस इक्वल तो -2 टाइम्स - 2 टाइम्स ध्यान से देखो आपका क्या X1 सो 1 1 थॉट्स वैन आप समझ रहे हो प्लस बी ए वैन सो वैन इन वैन लेट्स अगेन वैन प्लस सी वो सी वापस कितना है वैन और इससे आप डिवाइड करते हो किस से वैन स्क्वायर प्लस वैन स्क्वायर यानी कितना तू से अंडर रूट मत लगाना गलती मत करना है ना सर अब अगर आप इसे सॉल्व करना चाहो तो मेरे ख्याल से मैं फटाफट से सॉल्व करने की कोशिश करता हूं मुझे जो सबसे खास बात नजर ए रही है वो सर ये और ये कैंसिल हो जा रहा है तो ये बच रहा है -3 है नहीं तो ह का जो माइंस वैन उधर गया -3 के पास तो हो जाएगा प्लस वैन माइंस थ्री प्लस वैन माइंस तू ह की वैल्यू कितनी आती है -2 से बात के लिए भी होगी क्योंकि वही बात वापस -3 -3 - 3 + 1 - 2 तो सर इसके जो अर्थ और सेंटर के इस लाइन के रिस्पेक्ट में की आपकी क्या है सर एक्स + ए + 1 = 0 और ये लाइन क्या थी सर आपकी एक्स एक्सिस जो की है ए = 0 इसके रिस्पेक्ट में जो मिरर इमेज ए रही है वो क्या है सर वो है -2 -2 कोई तकलीफ तो नहीं है मैं ये कहना चाह रहा हूं की ये पॉइंट ए और लेट्स से ये पॉइंट बी ये दोनों ही आपके उसे सरकमस सर्कल से पास होंगे तीसरा पॉइंट शिलाई से ये जो तीसरा पॉइंट होगा एल्डो मैंने बहुत प्रॉपर्ली चीजे बनाई नहीं है पर इन दोनों vertises का जो पॉइंट ऑफ इंटरसेक्शन होगा इन दोनों vertises का जो पॉइंट ऑफ इंटरसेक्शन होगा क्या मैं ये का सकता हूं की वो भी आपके क्लीयरली उसे circumsarkar से पास होगा आप मेरी बातें समझ सन और देख का रहे मैंने बहुत रैंडम कुछ भी पॉइंट बना दिया कहीं भी बना दिया पर मैं आपको बस ये कहना चाह रहा हूं की आपका जो इन दोनों वर्ड इन दोनों एज का जो पॉइंट ऑफ इंटरसेक्शन है क्या वो एक वर्टेक्स होगी और क्या वह circumsar कल और क्या वो सरकमसर्किल उसे वर्टेक्स से भी पास होगा आप खुद सोचो आपने एक पॉइंट निकल लिया ए आपने एक पॉइंट निकल लिया बी ये आपने निकाला इस लाइन के रिस्पेक्ट में बी आपने निकाला इस लाइन के रिस्पेक्ट में ये दोनों जहां इंटरसेक्ट करती हैं उससे भी तो वो पॉइंट वो सर्कल पास होगा तो आपको एक और पॉइंट मिलने वाला है सी जो की क्या होगा इनका पॉइंट ऑफ इंटरसेक्शन तो ए पॉइंट जो हमने निकाला उसको मैं क्या मैन लेता हूं भाई वो था वैन कमा -1 वैसे ही लिख लेता हूं वैन कमा -1 वैसे ही बना दे रहा हूं है ना दी पॉइंट जो आपने निकाला वो क्या था - 2 - 2 और जो सी पॉइंट आप निकलेंगे वो क्या हो गया इन दोनों का इंटरसेक्शन यानी ए की वैल्यू जीरो यहां पास कर दीजिए जैसे ही आप ए की वैल्यू जीरो पास करते हैं तो एक्स की वैल्यू कितनी ए जाती है -1 तो ये हो जाएगा -1 डिप्लोमा जीरो दिस सी कोऑर्डिनेट्स गोइंग तू बी माइंस सर्कल की इक्वेशन चाहिए जो इन तीन पॉइंट से पास होता है दो तरीके दो तरीके पहला तरीका आप ऑप्शंस में ढूंढ लीजिए ऐसा जो भी ऑप्शन है जो इन तीनों पॉइंट्स को सेटिस्फाई करें -1 -1 -2 कमा -2 और ऑफ कोर्स क्या आ -1 0 अगर इन तीनों पॉइंट्स को कोई भी ऑप्शन पास कर रहे द तो आपका आंसर होगा या फिर कोई और तरीका है देयर इस अंदर एप्रोच एक जो तरीका हमने सिखा है सर्कल्स चैप्टर के अंदर जहां पर हम बात करेंगे सर की मैन लेते हैं सर्कल की इक्वेशन है क्या एक्स स्क्वायर प्लस ए स्क्वायर प्लस इस इक्वल्स तू जीरो अब मैं जानता हूं सर की सर कल जो है वह इस पॉइंट से पास होता है तो यह सेटिस्फाई करेगा तो एक्स और ए की जगह वैन और माइंस वैन रखा तो ये हो जाएगा वैन और वैन कितना तू प्लस ये कितना हो जाएगा 2G ये कितना हो जाएगा -2f+c है ना अगर आप इसे थोड़ा सॉल्व करें थोड़ा भी इससे सिंपलीफाई करें तो क्या मिल जा रहा है सर 2G - 2f ना प्लस सी प्लस 2 = 0 ये आपकी पहली इक्वेशन ए रही है जब आप -1 0 रखेंगे तो देखो भाई -1 0 रखते ही क्या मिल जाएगा माइंस वैन कमा जीरो रखते ही -1 यहां रखा तो वैन और ये जीरो है ना -1 रखते ही ये कितना हो जाएगा सही हो जाएगा - 2G हो जाएगा जीरो प्लस मुझे दिखेगा सी और यहां क्या दिखेगा प्लस वैन इसे इक्वल तू जीरो माइंस तू बी तो आप रखेंगे बिल्कुल रखिए सर जैसे ही ये रखा तो 2 का स्क्वायर रूट तू का स्क्वायर 4 + 48 तो सी प्लस आते तो मिलना आता है बिल्कुल सर आप मेरी बात समझ का रहे हैं अब ये हो जाएगा - 2G 2G 2 डेट माइनस 4G थॉट्स - 4G और क्या सर - 4f तो आपको एक और टर्म मिलेगी वो क्या होगी -4f मेरा आपसे ये कहना है आप एक बार आप एक बार इसे और इसे सॉल्व कर लीजिए आप एक बार इसे और इसे सॉल्व कर लीजिए और आप एक बार इसे और इससे भी सॉल्व कर लीजिए आप कुछ चीज मिल जाएंगी कौन-कौन जीएफ और सी की वैल्यू वो यहां रख दीजिएगा आपके सर्कल के क्वेश्चन ए जाएगी कैसे करेंगे सर दो मिनट लगेगा तो बात करिए लेते हैं देखो मैंने क्या किया मैंने पहले तो इन दोनों को सॉल्व करने की कोशिश की किसे इसे और इसे है ना इन दोनों को जब हमने सॉल्व करने की कोशिश की तो एक छोटा सा कम करते हैं सर इसका करते हैं इसका जब डबल किया तो कितना हो जाएगा देखो भाई इसका जब डबल क्या तो हो जाएगा 4G - 4f है ना प्लस तू सी प्लस 4 = 0 और ये इक्वेशन ऑलरेडी लिखी हुई है क्या -4g-4f है ना प्लस सी प्लस 8 अब सर अगर आप इन दोनों को ऐड कर दें अगर आप इन दोनों को ऐड ही करते हैं तो आपको क्या दिखाई देगा बड़े ध्यान से देखो अभी इन दोनों को अगर मैंने ऐड किया तो ये दोनों टर्म्स कैंसिल तो ये बच के ए जा रहा है सर ये बच रहा है -8f प्लस 3c + 12 = 0 तो मुझे एफ और सी में एक इक्वेशन मिली क्या मेरे पास a4c में एक और इक्वेशन है अभी तो नहीं है अभी तक तो नहीं है है ना अब suniyega ध्यान से यहां से मुझे जी और सी में क्वेश्चन मिल रही है यहां से मुझे जी और सी में एक क्वेश्चन मिल रही है क्या मैं कोशिश कर सकता हूं सर ध्यान से देखना समझने की कोशिश करना इन दोनों में से भी मैं जी को ilminate कर डन तो मुझे a4c में एक और एक क्वेश्चन मिल जाएगी वो ऑलरेडी है उन दोनों को सॉल्व करके क्वेश्चन मिलेगा सुनार से इसका भी मैं डबल करता हूं इसका भी मैं डबल करता हूं बट नेगेटिव डबल मतलब -2 से मल्टीप्लाई करता हूं - 2 से मल्टीप्लाई किया तो ये हो जाएगा कितना 4G है ना आप मेरी बात समझ का रहे हो फिर क्या माइंस तू से मल्टीप्लाई किया तो -2c -2 से मल्टीप्लाई किया तो -2 अब जो मैंने इन दोनों को किया तो आप खुद देखो बड़े ध्यान से ये पार्ट है गया तो बच क्या रहा है सर बच रहा है -4f तो यहां मैं लिखता हूं क्या -4f suniyega ध्यान से ये माइंस तू सी प्लस सी सो ये हो जाएगा सर कितना -सी -2 + 8 ये हो जाएगा प्लस सिक्स इसे इक्वल तू जीरो अब एक छोटा सा कम कर लीजिए जैसे यहां से या तो सी को या एफ को iliminate करने की कोशिश करिए मैं सी को एलिमिनेट करने की कोशिश करता हूं तो मैं एक कम करता हूं इस इक्वेशन को थ्री से मल्टीप्लाई करता हूं तो देखो भाई 4 3 कितना सब हो जाएगा 12 ये तो क्लीयरली मुझे थ्री दिखे रहा है और 6 3 कितना सही हो जाएगा 18 ये कितना हो जाएगा सर 18 आई होप आप मेरी बात तो सही कर रहे हो अब जब आपने इन दोनों को ऐड किया जब आपने इन दोनों को ऐड किया तो गोद से देखोगे -12 -8 ये कितना हो जाएगा सर ये हो जा रहा है -20 एफ कोई दिक्कत तो नहीं है ये पार्ट कैंसिल हो जा रहा है 12 + 18 देखो 12 और 8 20 + 10 कितना 13 सो 30 उधर गए तो -30 यहां से सर आप एफ की वैल्यू पाते हो माइंस से माइंस कैंसिल कोई तकली इस बात से तो यह जाता है 3/2 अब जब आप की वैल्यू जहां भी आपको पास करना है कहीं भी पास कर दो कोई बहुत बड़ी विशेष बात नहीं है तो एफ की वैल्यू 3/2 यहां पास कर देता हूं है ना यहां भी कर सकते हो यहां मुझे ज्यादा आसान लग रहा है तो जैसे ही 3/2 यहां पास किया तो देखो तू से कैंसिल किया तू से यह कैंसिल किया तो कितना 6 आप सन रहे हो क्या मैंने 3/2 यहां पास किया बस ध्यान से देखना आप बहुत अच्छे जोन में हो इस क्वेश्चन को सॉल्व करने के एफ की वैल्यू 3/2 राखी तो हो जाएगा -12 3 / 2 बस हम कोई कैलकुलेशन क्या है ना करते हैं इस समय थोड़ा कोशिश कर रहा हूं यहां से देखते जाएगा भाई तू का 12 कितना है ये हो जाएगा सिक्स टाइम्स तो ये हो जाएगा सिक्स टाइम्स है ना ये हो जाएगा -36 कितना भाई सर ये क्लीयरली आपको दिख रहा होगा ये हो जा रहा है माइंस 18 और ये है प्लस 18 तो -18 प्लस 18 आपका क्या बना देगा इसको जीरो और थ्री सी की वैल्यू जीरो है मतलब सी की वैल्यू कितनी ए जाती है जीरो अब सी की वैल्यू जीरो whitewap कहीं रख दीजिए अगर मैंने सी की वैल्यू जीरो यहां राखी तो वैन उधर गए तो -1 - से - कैंसिल तो जी की वैल्यू हो जाएगी 1 / 2 तो जी की वैल्यू कितनी आती है 1 / 2 आई थिंक यू आर वेरी क्लोज तू डी आंसर नौ क्योंकि जीएफ और सी अगर ए चुके हैं तो सर्कल की इक्वेशन लिखना इस नॉट अवेलेबल डिफिकल्ट टास्क आप लिखेंगे x² + y² है ना + 2G एक्स तो देखो तू से ये तू कैंसिल अब समझ का रहे हो तो कितना बचेगा ये बचेगा सिर्फ एक्स आई होप आप कंफ्यूज नहीं हो 2G यानी 2 1 / 2 तो ये बचा एक्स + 2fy तो तू से तू कैंसिल तो ये बचेगा कितना 3y और थ्री है नहीं तो ये कितना हो जाएगा जीरो आप मेरी बात समझ रहे हो ना c0 है सो x² + y² + एक्स + 3y² + y² + एक्स + 3y और मुझे कुछ कहने सुनने या बोलने की जरूरत नहीं है आप कौन सा ऑप्शन मार्क करेंगे बी सर इतनी मेहनत करने के अलावा एक और आसान तरीका था आप इसे सॉरी इससे इससे हम रियली सॉरी आप कैसे इससे और इससे सेटिस्फाई करवा देते तो भी आपका आंसर यही आता इस तरीके से भी बेशक इसे सोचा जा सकता था इसमें कोई बुराई नहीं होती तब भी आंसर ए जाता क्या यहां तक कोई कन्फ्यूजन डाउट या परेशानी कोई दिक्कत तो नहीं है एक बार इस पार्ट को अच्छे से देख लीजिए अच्छे से समझ लीजिए और देख के बताओ मुझे की कितना निकल कर ए रहा है या कोई गड़बड़ तो आपको नहीं लग रही किसी जगह तो दोनों तरीके आपको समझ आई या तो ये तरीका जो ऑफ कोर्स थोड़ा ट्रिक है बट मजेदार है या फिर वो वैल्यूज ट्राई कर लीजिए ऑप्शंस में उसमें भी वक्त लगना है क्योंकि चार ऑप्शन में तीन-तीन वैल्यूज और तीनों ही सेटिस्फाई करें तो हम आंसर मार्क करेंगे कोई दिक्कत तो नहीं है चलो भाई आगे बढ़ते हैं पहली बात तो ये जी एडवांस्ड ने 2013 में एक क्वेश्चन पूछा था तो बिल्कुल ये अच्छा क्वेश्चन है जैसे हम समझेंगे और एक और जरूरी बात की ये किस कैटिगरी का क्वेश्चन है सर ये कैटिगरी है आपके पास मल्टीपल करेक्ट आंसर टाइप क्वेश्चन तो अब हम ए चुके हैं एक से ज्यादा ऑप्शंस करेक्ट की कैटिगरी में क्वेश्चन समझ सबसे पहला मोटिवेट होना चाहिए तो का रहा है सर्कल या सर्कल्स हमें नहीं पता टचिंग डी एक्स एक्सिस आते डिस्टेंस ऑफ थ्री यूनिट्स फ्रॉम द ओरिजिन ओरिजिन से 3 यूनिट के डिस्टेंस पे सर्कल है जो एक सक्सेस को टच कर रहा है हेविंग एन इंटरसेक्ट ऑफ लेंथ तू बाय रूट 7 ऑन डी ए एक्सिस हेविंग डी लेंथ ऑफ डी इंटरसेप्ट तू टाइम्स रूट 7 सर सारी बातें करेंगे मेक बेसिक सैन्य आपको देने की कोशिश करता हूं की देखो मैन लो ऐसा कभी है ये आप गलत से ए एक्सिस है लेट्स हैव का क्या है भाई एक्सिस है ना और इसी तरीके से आपके पास यहां पर है एक्स एक्सिस ये मैन लो मैंने बनाया है एक्स एक्सिस और इसी तरीके से मैं बना लेता हूं ए एक्सिस भी क्या प्रॉब्लम हो रही है ना तो ये हो जाता है क्या भाई आपका ये हो जाता है आपका सर ए एक्सिस कोई दिक्कत तो नहीं सर अब एक छोटी सी बात मेरी आपसे एक कहानी है बहुत ध्यान से suniyega सर एक एक रैंडम सा सर्कल बना रहा हूं मैं बहुत ज्यादा सेंसरी क्रिएट नहीं कर रहा हूं एक रैंडम सा सर्कल हम बनाना चाह रहे हैं जो इस तरीके से है की सर वो एक्स एक्सिस पर तो आपको जो टच कर रहा है वो कहां टच कर रहा है थ्री कमा जीरो पर एक्स एक्सिस पे थ्री कमा जीरो पर यही बात है प्लीज इस बात को सुनना यही बात इस तरीके से भी हो सकती थी की ये सेकंड क्वाड्रेंट में हो रहा है तो देखो अभी भी ओरिजिनल 3 यूनिट डिस्टेंस को टच कर रहा है और इंटरसेक्ट कर रहा है लेकिन ऐसा यहां भी तो हो सकता था सर देखो माइंस थ्री कमा जीरो से थ्री यूनिट डिस्टेंस कोशिश करना और वह इंटरसेप्ट और ऐसा कम आपका फोर्थ क्वाड्रेंट में भी हो सकता था मैं तो बस एक सिनेरियो लेकर चल रहा हूं की मैन लो फर्स्ट क्वाड्रेंट में हो रहा होता तो क्या होता है हम बात कर लेंगे फिर जो भी बनेगा आप समझ रहे हो मेरी बात हम जो भी सिनेरियो बन रहा होगा वैसे बाते कर लेंगे बट फिलहाल मैं ऐसा मैन लूं क्या की फर्स्ट क्वार्टर में हम सॉल्व कर लें ऐसा का ऐसा दो जगह तीन जगह और सोच लेंगे ठीक है सर आपकी बात मैन ली तो ये पॉइंट क्या रहा सर ये टच कर रहा है एक्स एक्सिस को ओरिजिन से थ्री यूनिट डिस्टेंस में तो अभी फिलहाल हमारे केस में ये पॉइंट हो जाएगा थ्री कमा जीरो कोई परेशानी तो नहीं है नहीं है सर और क्या कहना चाह रहे हो ये जो इंटरसेप्ट है ये जो इंटरसेक्ट बन रहा है इसकी लेंथ कितनी ए रही है सर इस इंटरसेक्ट की जो लेंथ है सर यह जो इंटरसेक्ट बनाना आपका इसकी लेंथ आती है तू टाइम्स अंडर रूट 7 बहुत बढ़िया अब मुझे एक बात बताओ सर सर्कल का सेंटर कहीं होगा क्या बिल्कुल होगा तो लेट्स से ये सर्कल का सेंटर है अब मैं एक बहुत खास बात जो आप सभी को सोच कर बताना चाह रहा हूं की देखो ये क्या हो जाएगा आपका परपेंडिकुलर और ऑफ कोर्स ये आपका क्या हो जाएगा यहां पर एक परपेंडिकुलर ये आपकी क्या हो जाएगी इसकी रेडियस भी क्या इस समय रेडियस का सकता हूं बिना ज्यादा आपत्ति लिए किसी भी बात की तो सर ये हो जाएगी आपके सर्कल की रेडियस बाय डी वे इसकी जब आपने रेडियस निकल तो रेडियस कितनी आई सर इसकी रेडियस को लेके मेरा ओपिनियन है की अभी तो मुझे नहीं पता पर ये जो लेंथ है ये लेंथ रेडियस होगी के क्योंकि सर्कल के सेंटर से अपने इसकी टैसेंट पर ड्रॉप किया है बहुत बढ़िया ये आपका क्लीयरली है सर ये आपका एक्स एक्सिस है और ये आपका क्लीयरली एक्सेस है अभी अभी बस हम क्वेश्चंस समझ रहे हैं सी आर नॉट सॉल्विंग डी क्वेश्चन वेयर अंडरस्टैंडिंग डी क्वेश्चन एंड इट इसे वेरी वेरी वेल सेट डेट डी मोमेंट यू अंडरस्टैंड डी प्रॉब्लम हाफ ऑफ डी प्रॉब्लम इस सॉल्ट सो इन्वेस्ट सैम टाइम एंड अंडरस्टैंडिंग डी प्रॉब्लम वेल आप प्रॉब्लम समझिए आपको दिख जाएगा पता है प्रॉब्लम समझते समझते दिख जाएगा की क्या-क्या दिया है क्या पूछा और कैसे निकले क्लिक कर जाएगा अगर आपने सारे बेसिक से अच्छे से पढ़ रहे हैं तो मेरा कहना है सर मुझे एक हिंट और दिखाई दे रही है शायद आप नोटिस नहीं कर रहे हो बट वो हिंट है कुछ ऐसे ही की अगर आप इन दोनों को कनेक्ट कर लेते अगर आपने इन दोनों को कनेक्ट किया था मैं सिर्फ फिर से कनेक्ट कर देता हूं मेरा आपसे कहना है की अगर आपने इन दोनों को कनेक्ट किया होता तो आपको एक बड़ी मजेदार से बात दिख रही होती है क्या बात दिख रही होती अब suniyega ध्यान से सर आप यह नहीं जानते हो क्या की ये जो सर्कल के सेंटर से अपने परपेंडिकुलर ड्रॉप किया है वो इस कॉर्ड को बायसेक्स करेगा 2 रूट 7 तो यह जो एक साइड है इसकी लेंथ कितनी है सर ये है अंडर रूट 7 क्या आप मेरी बात समझ का रहे हो कोई तकलीफ तो नहीं है सर कोई दिक्कत नहीं है अच्छा इस सेंटर के कोऑर्डिनेट्स में मुझे एक cardinate तो पता है अभी फिलहाल 3 इसका व्हाइट cardinate नहीं पता है ठीक है सर अच्छी बात है क्या आप मुझे ये बता सकते हो सोच के बोलेगा क्या आप मुझे बता सकते हो की सर या तो मुझे ये लेंथ दे दो या फिर आप मुझे ये लेंथ दे दो कुछ तो बताओ सर कुछ तो हेल्प करो इसके थ्रू तो ठीक है सर वो निकल लेंगे मैं फिर से रिपीट करता हूं क्या आप मुझे या तो ये लेंथ या ये लेंथ दे सकते हो शायद उससे मैं कहीं तक पहुंच जाऊं तो बस आपको बस वहीं से सोचना है और वहीं से क्वेश्चन को और काउंट करना है काफी सारी चीज हमने सोच लिया फिर जैसा भी हमारे कार्ड में रहेंगे जैसे भी हम रेडियस और सेंटर के अकॉर्डिंग निकलेंगे रेडियस तो से रहने वाली है उसे यहां यहां और यहां सेंटर को शिफ्ट करके सोच लेंगे तो काफी सारा डाटा आपके सामने है ये हर एक चीज हमने सोचना सीखी है बस अब आपको सोचकर मुझे बताना है की सर हम कैसे इस पॉइंट फिर आउट करेंगे की क्या इनफॉरमेशन हमारे पास है कैसे निकलेंगे भाई अन्य वैन हेविंग अन्य थॉट्स अन्य अन्य एप्रोच अन्य आइडिया की सर कैसे सोचेंगे देखो आपको एक चीज पता है अगर आप सोच का हो तो आपने नोटिस किया स्क्रीन पॉज करके की हान सर अगर मैं ये डिस्टेंस निकलूं और ये डिस्टेंस निकल लूं तो क्या आपको नहीं दिख रहा है दोनों इक्वल हैं क्योंकि ये एक्टिंग है और अगर ये रेक्टेंगल है तो जितनी ये डिस्टेंस है उतनी ही डिस्टेंस है आपने थ्री लिखा भी तो था आपको हिंट दी गई थी भाई तो ये जो लेंथ है ये कितनी है सर ये है थ्री आप प्लीज इस बात को ध्यान से सोचिए ये आपके पास एक राइट एंगल तू ट्रायंगल है दिस इस दी राइट एंगल डी ट्रायंगल यू हैव ऐसे में कहूं लेट्स सी इसे मैं कहूं लेट्स से ए और इसे मैं कहूं लेट से बी तो ये जो ए बी सी आपका ट्रायंगल है और राइट एंगल जो ट्रायंगल है जिसमें आपको परपेंडिकुलर और बेस पता है की आप हाइपोटेन्यूज नहीं निकल सकते कोशिश करते हैं सर अगर मैंने हाइपोटेन्यूज निकलना चाहा तो कितना ए रहा है गौर से देखो √7 का स्क्वायर 16 का अंडर रूट 4 तो यह जो लेंथ ए रही है एक वो कितनी ए रही है 4 सर अगर एक इसकी रेडियस है तो अगर मैं इस पॉइंट को थोड़ी देर के लिए का लूं दी तो क्या एक और एड = लेंथ के नहीं होंगे होंगे ना क्योंकि एक और एड दोनों ही रेडियस है तो अगर ये फोर है तो ये भी फोर होगा ये फोर होगा तो ये भी फोर होगा यानी ओ से बी तक के डिस्टेंस तो कैन आई से इसके सेंटर के अकॉर्डिंग इट्स हो जाएंगे थ्री कमा 4 आप सन रहे हो जो मैं का रहा हूं इसके सेंटर के आर्डिनेंस हो जाएंगे थ्री कमा फोर और इसकी रेडियस होगी 4 पहली बात दूसरी बात अगर मैंने इसे इस क्वाड्रेंट में माना होता तो ऐसा सेकंड क्वाड्रेंट में माना होता तो आप सोच का रहे हो सेकंड बस अब आपको सेंटर से खेलना है रेडियस तो 4 फिक्स है तो आपको अब सर्कल में करना है थ्री कमा फोर सेंटर पे है ना तो एक सर्कल तो क्या बनेगा सर 3 कमा फोर सेंटर लेते हुए जिसकी रेडियस क्या है फोर है ना एक क्या बन जाएगा सर एक्स नेगेटिव हो जाएगा ए अभी भी पॉजिटिव रहेगा सेकंड क्वाड्रेंट वहां पे रेडियस 4 एक ही हो सकता है की आपका थर्ड क्वाड्रेंट में आप चले जाओ एक्स और ए दोनों नेगेटिव बट रेडी है 4 ये क्या बन सकता है सर आपके पास कैसे नारिया बन जाए की आप कौन से क्वाड्रेंट में हो फोर्थ क्वाड्रेंट में जहां एक्स तो पॉजिटिव हो लेकिन ए नेगेटिव हो तो ट्रेडिशनल बीपी 4 क्या आप चारों से नारिया में सर्कल की इक्वेशन लिख सकते हो क्यों नहीं लिख सकते सर कोशिश करते हैं यहां जब लिखेंगे आप तो क्या लिखेंगे भाई एक्स - 3 का होल स्क्वायर प्लस ए माइंस अपना 16 इससे थोड़ा सिंपलीफाई करते हैं और इसी बेसिस पर इनको भी हम सोच लेते हैं मेरा कहना है बस यहां पे प्लस माइंस लगा दो यहां पे प्लस माइंस लगा दो जो जो केसेस बन रहा है वो आपके आंसर होंगे आप समझ का रहे हो आप समझ का रहे हो क्या मेरी बात प्लीज ध्यान से सुनिए बड़ी सिंपल सी बात है स्टूडेंट्स देखो भाई पहला कंक्लुजन suniyega है ना जब आप इसे सिंपलीफाई करेंगे तो क्या बनेगा सर यहां से मिलेगा x² यहां से मिलेगा Y2 ठीक है क्या बिल्कुल सर अब क्या मिलेगा देखो थ्री का स्क्वायर 94² 16 से 16 कैंसिल तो बचेगा क्या नहीं तो जो कांस्टेंट टर्म है वो भी फिक्स होगी क्या 9 बहुत ध्यान से सुनना अब जो बात मिलने वाली है वो क्या -3 अब सीधी सीधी सी बात है सर ये क्या चाहिए तो यहां पर क्या मिलेगा सर यहां पे मिलेगा -6x आप समझ रहे हो क्या यह क्या चाहिए अब बाकी चीज सोच सकते हो आई होप इसे देख कर बाकी चीज सो सकते हो बार-बार पुरी प्रक्रिया दोहराने की जरूरत नहीं है जैसे मेरी बात सुनना आपको एक और चाहिए ना एक क्वेश्चन तो अगली इक्वेशन क्या होगी सोच के देखो सर एक्सेस स्क्वायर और y² तो मिलना तय है ना माइंस थ्री कमा फोर तो कैन आई से उसके लिए मुझे लेना होगा प्लस 66 और -8x और वापस क्या प्लस साइन आप समझ रहे हो ना क्योंकि 34 और 9 तो रेडियस अगर फोर है तो 3 से 9 कैंसिल हो जाएगा क्योंकि थ्री का स्क्वायर बचेगा 4 अब मेरी बात समझ का रहे हो सेंटर चाहिए -3 - 4 तो इन दोनों को पॉजिटिव कर देंगे तो वो सर्कल के क्वेश्चन क्या हो जाएगी सर वो हो जाएगा x² + y² + 6 + 8x प्लस नाइन इसे इक्वल तू जीरो सर एक्स पॉजिटिव ए नेगेटिव तो उसके लिए मैं क्या करूंगा x² + y² - 6x + 8x + 9 टेक्निकल इसका सारा पार्ट्स से रहा बस सेंटर्स को चेंज करने के बजाय बार दोनों नेगेटिव एक बार दोनों पॉजिटिव एक बार एक्स पॉजिटिव ए नेगेटिव एक्स नेगेटिव ए पॉजिटिव मैंने यहां एक एक्स लिख दिया सॉरी इसको आप क्या लिखेंगे आते फाइव आई थिंक क्या चीजे आसान है अब देख लो भाई कौन-कौन से ऑप्शंस हैं क्या आपको अच्छे से चीज समझ आई क्या आपको अच्छे से ये चारों इक्वेशंस बनते हुए देखें अब हम देख लेते हैं ऑप्शंस के मुझे तो सिर्फ 8 और 9 मिनट इंटरेस्ट है 6 8 और 9 और सारे कंपटीशन चलेंगे 9 पॉजिटिव रखे और रेट के नेगेटिव पॉजिटिव कुछ भी कांबिनेशन चलेंगे\| तो बस अब आप देख लो और मुझे बता दो इन में से देख कर कौन सा आंसर आप मार्क कर रहे हो सर बहुत सारी चीज तो बाय डिफॉल्ट हो जा रही है क्योंकि 7 से हमें कुछ लेना देना नहीं है तो ऑप्शन बी और ऑप्शन दी तो है गए अब ऑप्शन ए और ऑप्शन सी को आप देखें तो हान सर नाइन पॉजिटिव है बढ़िया और सिक्स और एट के कुछ भी कांबिनेशंस बन ही सकते द तो ए और सी दोनों हम ले लेंगे क्या आपको एक क्वेश्चन और इसका सॉल्यूशन समझ आया एंड यू डू अंडरस्टैंड ये एग्जाम बहुत टू एग्जाम नहीं है कितनी आसान एग्जाम है पर हाउस में रखा है इसे कौन सी कैटिगरी में एमएस के में एमएस के मल्टीपल सिलेक्ट मतलब एक से ज्यादा ऑप्शन तो यहां पर हो सकता है आप सिर्फ ये देख के खुश हो जाती है ये वाला सोच ही नहीं पाते तो सिनेरियो सोचना होगा की यहां यहां कहीं भी वो बन सकता था वो टू तो नहीं था एक क्वेश्चन अगर ये क्वेश्चन बहुत टू नहीं था तो मुझे अब थोड़ा यकीन फेथ और दिला एक और क्वेश्चन सॉल्व करके जो की आप करने वाले हैं कैसे करेंगे सर इस क्वेश्चन को ध्यान से देखिए भाई मैं अगेन हिंट दूंगा मैं हेल्प करूंगा ट्राई आपको ही करना है अगर आपको अभी लग रहा है की नहीं सर आप रुक जाओ हम ट्राई करना चाहेंगे ज एडवांस का 2014 में पूछा गया क्वेश्चन खुद से आपके समझे बिना तो बिल्कुल आप स्क्रीन पॉज करके स्वतंत्र आजाए मुझे आप प्ले और पॉज अपनी शहर के मुताबिक कर सकते हैं पर उसके बाद प्ले होने के बाद मुझे क्या करना है मुझे ही डिसाइड करना है पर प्ले और पॉज आप कर सकते हैं इसका फायदा ही होगा की आप क्वेश्चंस खुद से अटेंड करिए पहले आप खुद से ट्राई करिए सर इस क्वेश्चन में क्या करना है देखो भाई एक सर्कल है एस जो पास होता है ठीक है सर देख लेंगे इस बात का क्या मतलब है तू दिस इन दिस सर्कल इन दोनों सर्कल्स को आप गौर से देखें सर देखो ध्यान से इन दोनों सर्कल्स को अगर मैं देखूं तो इसका सेंटर है क्या वैन कमा जीरो इसका सेंटर है जीरो कमा जीरो इसकी रेडियस है 4 और इसकी रेडियस है वैन ऑप्शन का रहे हो अगर मैं थोड़ा डायरेक्टली ही अच्छे तरीके से प्लॉट करना चाहूं तो मैं आपसे सारी बातें छोड़ के पहले इनको थोड़ा प्लॉट करना चाह रहा हूं तो मैं प्लॉट करते वक्त क्या करूंगा सर मैं कहूंगा एक्स एक्सेस है बिल्कुल क्या मिलेगा सर एक्स एक्सेस में ये एक्स एक्सेस हो गया इसी तरीके से आपके पास क्या होगा आईएसआईएस है ना वो एक्स इसको भी हम थोड़ा ठीक कर लेते हैं तो यह हो जाएगा यह हो गया एक सर्कल है सेंटर पर सिचुएटेड ओरिजिन पर सिचुएटेड जिसकी रेडियस है वैन तो ये तो बड़ा आसान सा सर्कल है जो हम बना लेंगे इजीली इसमें तो किसी भी स्टूडेंट को कोई तकलीफ नहीं हुई होगी सर एक और सर्कल है जो की कहां पर सिचुएटेड है वैन कमा जीरो पर इसका सेंटर है वैन कमा जीरो मतलब यहां पर आप समझ का रहे हो ना यहां पर और इसकी रेडियस कितनी है सर इसकी रेडियस है फोर इसकी रेडियस कितनी है 4 अगर इसकी रेडियस 4 इयर्स का सेंटर यहां है आप समझ का रहे हो तो ये सर्कल कैसा जाएगा सर ये सर्कल थोड़ा बड़ा हो जाएगा ये सर्कल थोड़ा सा बड़ा हो जाएगा क्योंकि इसका सेंटर ही यही है इसका सेंटर ही यही है और ये कुछ इस तरीके से बन रहा होगा जितना मैं सोच का रहा हूं अब मेरा बस आपसे ये कहना है आय होप ये दोनों बातें आप समझ पाए ये वाला सर्कल और ये वाला सर्कल मेरा बस आपसे ये पूछना है की आप मुझे ये जानने में हेल्प कर सकते हो की अगर मुझे एक ऐसा सर्कल की इक्वेशन चाहिए जो की इन दोनों सर्कस पर ऑर्थोंगोनल है सर जितना मुझे आता है सर ये सारी बातें छोड़ो आप तो ये सारी बातें छोड़ो आप तो सीधे पॉइंट पर आओ मैं जितना लाइफ में जानता हूं सर जितनी बातें मैंने समझिए अगर एक सर्कल की नहीं दो सर्कल्स पर ऑर्थोंगोनल है तो वो उन दोनों सर्कस के वो उन सर्कल उन दोनों सर्कल के बाय डी वे बस ऐसे ही डिस्टेंस मुझे निकालनी है ये है वैन है ना ये है वैन और ये होगा वैन तो ये हो गई डिस्टेंस कितनी तू थ्री फोर और थोड़ा लंबा जाएगा थोड़ा और आगे जाएगा तो आप चाहते हैं तो इसे थोड़ा और बड़ा बना सकते द बट ऐसे ही बस एक हल्का सा मैं आइडिया लेना चाह रहा हूं अब मेरी इस बात को ध्यान से गौर से suniyega स्टूडेंट्स प्लीज इस बात को भूल गए अब बस एक कॉन्सेप्ट पर लिए जो हमने डिस्कस किया था सर की अगर आपके पास तीन सर्कल्स हैं S1 S2 और S3 अगर आपसे कोई कहे की S1 और S2 आपस में परपेंडिकुलर है और S1 और S3 आपस में परपेंडिकुलर है मतलब जो एस वैन सर्कल है वो एस वैन वो S1 जो सर्कल है वो S2 S3 पर orthagonal लें तो हम ये हमेशा से जानते हैं सर की S1 जो सर्कल है S1 जो सर्कल है इसका जो सेंटर होगा याद करो स्टूडेंट से प्रॉपर्टी बहुत अच्छे से पढ़ थी S1 जो सर्कल है इसका जो सेंटर होगा उसका जो लोकस होगा वो हमेशा इसके और इसके रेडिकल एक्सरसाइज पर होगा अब भूल तो नहीं रहे हो हमने रेडिकल एक्सरसाइज बहुत डिटेल में ये कॉन्सेप्ट बार-बार बहुत बार डिस्कस किया था की सर किन्ही भी दो सर्कल्स का जो रेडिकल एक्सिस होता है उसे पर उसे सर्कल का सेंटर लाइक करता है जो इन गिवन सर्कल पर ऑर्थोंगोनल हो इन पर परपेंडिकुलर हो आपको कुछ क्लिक किया अगर नहीं किया तो सारी बातें छोड़ दो सारी बातें छोड़ दो क्या मैं इन दोनों सर्कल्स पर एक ऐसा सर्कल ढूंढ सकता हूं जो परपेंडिकुलर है मैं ढूंढो सर बात की बात है पर मैं यह जरूर पता कर सकता हूं की एक ऐसा सर्कल एक ऐसा सर्कल जिसका सेंटर मुझे अगर पता करना है एक ऐसा सर्कल जो की इन दोनों पर औरतों का सेंटर इन दोनों के रेडिकल एक्सिस पर लाइक करेगा क्या आपको याद है ये बात और सर अगर ये सर्कल है मैन लो S1 और ये सर्कल है S2 तो इनका रेडिकल एक्सिस इस गिवन बाय आई थिंक मुझे बताने की जरूरत नहीं है s1-s2=0 आई होप यू ऑल हैव नॉट फॉरगेट इन दिस और अगर आपने भुला नहीं है तो सर इसी बात से कहानी को आगे बढ़ते हैं और देखते हैं चीज कैसे सोचते हैं सर S1 क्या आएगा देखो भाई ध्यान से एस वैन जो आएगा इससे सिंपलीफाई कर लेते हैं तो ये हो जाएगा x² + 1 है ना -2x + y² है ना और ये 16 को भी इधर ले आए तो ये हो जाएगा कितना वैन में हटा लेता हूं तो 16 - 1 कितना हो जाएगा सर यह हो जाएगा x² - 2X + y² और यहां पर दिखाता वैन 16 इधर आया तो वैन के कारण 16 कितना हो गया भाई - 15 कोई दिक्कत तो नहीं इस इक्वल तू जीरो ये आपका S1 है s2o जो भी आपको माना है और आपका जो दूसरा सर्कल है वो है x² + y² - 1 = 0 और जवाब s1-s2 लिखेंगे तो मुझे कुछ सीखने की जरूरत नहीं है की सर x²y² कैंसिल है ना अब क्या बचेगा -2x - 15 - 1 यानी -15 + 1 तो ये कितना हो जाएगा सर ये क्लीयरली हो जाएगा -15 + 1 यानी -49 है ना माइंस 14 और सर यहां से अगर आप एक्स की वैल्यू निकले मतलब सीधी-सीधी अगर मैं क्वेश्चन लिखूं तो हो जाती है एक्स + 7 = 0 ये क्या हुआ सर आपका ये हो जाता है आपका रेडिकल एक्सरसाइज ये क्या बन जाते हैं सर आपका रेडिकल एक्सरसाइज रेडिकल एक्सेस का मतलब है वो सर्कल जो की इन दोनों पर ऑर्जिनल है उसका सेंटर इस स्ट्रेट लाइन पर लाइक करेगा जैसे मैं और आसान भाषा में लिख सकता हूं एक्स = -7 और इसे देखकर आप खुद अंदाजा लगा का रहे होंगे यह वह स्ट्रेट लाइन है जो एक्सेस के पैरेलल है एक ऐसी स्ट्रेट लाइन है जो ए एक्सिस के पैरेलल है ठीक है सर तो कैसे स्ट्रेट लाइन जो एक्सेस के पैरेलल है अब मैं इस पर बात करना चाह रहा हूं आपसे एक ऐसी लाइन जो एक्सेस के पैरेलल हो तो उसका एक्स cardinate फिक्स हो गया ना अब अगर इसका एक्स कॉर्डिनेट फिक्स्ड हो गया है तो बस आप थोड़ा सा दिमाग लगाकर सोच के बता दो की अगर वो जीरो कमा 1 से पास होता है अगर वो जीरो कमा 1 से पास होता है तो एक कंडीशन तो यहां से दूसरी कंडीशन इस बात से की सर इन दोनों को अगर वो औरतों का नल है तो वहां से इन दोनों के थ्रू आप उसकी रेडियस निकालो आपका आंसर ए जाएगा मैं क्या कहना चाह रहा हूं स्टूडेंट्स आपको दो कंडीशंस मैंने देने की कोशिश की है आपको दो बड़ी बेसिक सी आसान सी डायरेक्ट सी बात मैंने कहने की कोशिश की है इन दोनों कंडीशंस को अगर आप अप्लाई कर लें तो इस क्वेश्चन को आप आंसर बहुत आसानी से निकल देंगे पहले तो अगर ऑफ कोर्स हम जान गए हैं की सर ये जो लाइन है इस पर वो लाइक करता है और यहां से अगर मैं जीरो कमा वैन के डिस्टेंस निकालो ये जो पॉइंट है आपका एक्स = -7 इसके साथ अगर मैं एक yordinate कुछ भी एक वेरिएबल मैन लो लामबीडीए के कुछ भी वहां से अगर जीरो कमा वैन के डिस्टेंस निकालो तो वो आपकी एक रेडियस हो जाएगी क्या आप मेरी बात समझ पाए लेकिन सर रेडियस निकलने का तो एक और तरीका हम जानते हैं कौन सा सर की सर अगर मैं इससे अगर इसकी टांगें पर अगर परपेंडिकुलर ड्रॉप करूं तो वो भी तो रेडियस होती है सर आप सन रहे हो क्या मैं फिर से दोहराता हूं आपसे कम की जरूरी बात बहुत बेसिक सी इंपॉर्टेंट सी बात सुनना ध्यान से पहले तो ये सुना एक्स = -7 अगर मैं मैन लूं की इस सर्कल के सेंटर का एक्स कोऑर्डिनेट्स है ये -7 क्योंकि वो इसी लाइन पर लाइक कर रहा है तो yordinate मैन लेता हूं लामबीडीए तो जो इन पर ऑर्थोंगोनल है ना उसके सेंटर के अकॉर्डिंग में मैन लेता हूं -7 कोई तकलीफ तो नहीं है सर अब चूंकि वो सर्कल जो है वो आपका जीरो कमा वैन से पास होता है तो ऑफ कोर्स इन दोनों के बीच की जो डिस्टेंस होगी उसकी रेडियस होगी तो डिस्टेंस निकलती हैं -7 - 0 7 का स्क्वायर तो कितना हो जाएगा भाई 149 कोई तकलीफ तो नहीं है अरे पेट में स्टेटमेंट -7 -7 का स्क्वायर कितना 49 + लामबीडीए - 1² तो लामबीडीए - 1² भी हम कर देते हैं और इसका क्या लेंगे आप अंडर रूट ये आप क्या अकॉर्डिंग क्या होती है सर रेडियस मैन लिया आपकी बात ये आप के अकॉर्डिंग क्या होती है रेडियस है लेकिन एक और तरीका हो सकता है रेडियस निकलने का कैसा प्लीज उसे बात को सुनो ये सर्कल का सेंटर है ये सर्कल का सेंटर है क्या मैं इस सर्कल की टांगें की इक्वेशन निकल सकता हूं सर क्या मैं इस सर्कल की टैसेंट की इक्वेशन निकल सकता हूं कोई थॉट कोई एप्रोच से हम सर्कल की टेंशन की इक्वेशन निकल लें इस क्वेश्चन में अगला थॉट जो निकल कर आता है इस बात को ध्यान से सुनो सर जब सर्कल्स ऑर्थोगोनली इंटरसेक्ट कर रहे हैं ना आप इस बात को समझो जब सर्कल्स ऑथर के लिए इंटरसेक्ट करेंगे तो होगा क्या जैसे ही केस सर्कल अफसर और एक और सर्कल में मैन लेता हूं यहीं कहीं जैसे की जैसे ये रहा बस मुझे ऑर्थोगोनली इसको इंटरसेक्ट करवाना है तो क्या यहां पर ऑर्थोगोनली कर रहे होंगे जैसे एक टैसेंट ये हो जाएगी एक ये हो जाएगी थोड़ा सा और करीब लेट हैं तो एक ये टैसेंट हो रही होगी की थोड़ा और करीब लेट हैं हॉपफुली अब हो जाना चाहिए नहीं आई थिंक अब बन जाएगा क्या एक टैसेंट ये रही एक टैसेंट ये रही थोड़ा और अंदर लाना होगा शायद में से और बढ़ा देना एक आइडियल डिस्टेंस होगी जहां पे ये ऑर्थोंगोनल इंटरसेक्ट कर रहे होंगे ना मैं बस बना लेता हूं की ये ऑर्थोगोनली यहां पर इंटरसेक्ट कर रहे हैं ना मैन लो एक टैसेंट तो ये रही आपकी सन रहे हैं की आप लोग आई होप आप समझ का रहे हो बस मैं आपको ये पॉइंट समझाना चाह रहा हूं की एक टैसेंट रहेगी खास बात ये है की ये दोनों एक दूसरे पर क्या है भाई परपेंडिकुलर ये दोनों इंडियन सेक्स दूसरे पर के मैन लो थोड़ी देर के लिए की बस क्या है भाई परपेंडिकुलर क्या आप इस बात से एग्री करते हो मैन लो बस है ना बस ऐसे ही कहना चाह रहा हूं वो बनाने में बहुत मेहनत लग जाएगी इसलिए अब जब भी ऐसा हो कभी भी ऐसा हो की अगर दोस्त सर्कल्स आपके ऑर्थोंगोनल हैं तो आप खुद ये बात ऑब्जर्व करेंगे आप खुद ये बात ऑब्जर्व करेंगे की सर जब इस सर्कल के सेंटर से इस सर्कल के सेंटर से वो मुझे बनाना ही होगा उसके बिना मैं नहीं समझा पाऊंगा तो मुझे थोड़ा सा वक्त आप दीजिए और मैं इसे थोड़ा सा properally ड्रॉ करके आपको प्रूफ करता हूं क्योंकि उससे ना आप बहुत अच्छे से बात समझ पाओगे आपको ये बात हमेशा याद रहेगी और उससे से ये बात आप हमेशा अप्लाई कर पाओगे मेरा जो कहना है इसे ध्यान से सुनिए मेरा आपसे मैसेज इतना सा कहना है की पहली बात तो इन सर्कल को ऐसे डिस्टेंस पर रखना होगा की ओर सेनल हो जाए आई थिंक अब हो रहे हैं क्या हॉपफुली एक और दो होना चाहिए है ना जैसे की ये अगर टैसेंट है मैन लो एक टैसेंट तो आपकी है है ना बात करेंगे इस सर्कल के सेंटर से जब इसकी tenjent पर इस सर्कल की tenjent है इस पर जब परपेंडिकुलर ड्रा किया तो ये हो जाएगा अब आप कम की बात सुनो बहुत कम की बात इस सर्कल के सेंटर से जब किसी दूसरे सर्कल पर इस सर्कल के सेंटर से जब किसी दूसरे सर्कल पर अपने tenjent ड्रॉ की तो वो सर्कल की रेडियस हुई ऐसा क्यों हो रहा है सर क्योंकि यह दोनों टांगें परपेंडिकुलर है तो इन्होंने काफी सारी चीज हमारे फीवर में लड़ी क्या बात है फीवर मिला दी सुनना मैं आपसे ये कहना चाह रहा हूं ये अगर सर्कल S1 और ये सर्कल है S2 है ना तो मैं आपसे ये कहना चाह रहा हूं सर्कल S1 के सेंटर से है ना मैं क्या कहना चाह रहा हूं टैसेंट है ना ड्रोन ड्रोन सी है ना तू विच सर्कल तू डी सर्कल कौन सा वाला सर तू डी सर्कल S2 विल बी इक्वल तू क्या सर लेंथ ऑफ डी टांगें जो आपने tagent ड्रा की है उसकी लेंथ विल बी इक्वल तू डी रेडियस ऑफ डी सर्कल S1 ड्यूल गेट माय पॉइंट आप खुद सोचो सर ऐसा होना तय है ये तो तय बात है ना यह क्यों तय बात है क्योंकि आप समझो ये आपका सर्कल है S1 सेंटर है सी इससे मैंने एक परपेंडिकुलर ड्रॉप किया इसकी टांगें पर उसकी रेडियस हुई तो ये जो है पॉइंट लेट्स से सी और ये पॉइंट है ए तो का क्या हुई सर रेडियस हुई आई होप आप मेरी बात समझ का रहे हो अब इस बात को इस नज़रिया से देखते हैं किस नज़रिया से किस सर एक पॉइंट है एक पॉइंट है जहां से मैंने इस दूसरे सर्कल पर टांगें ड्रॉ की है एक पॉइंट है जहां से मैंने इस सर्कल पर टांगें ड्रॉ किया तो टैसेंट की लेंथ क्या होगी सर हम जानते हैं की अगर सी के कोऑर्डिनेट्स हैं लेट्स से क्या X1 y1 और सर्कल S2 की कुछ इक्वेशन है कुछ भी है लेट्स से x² + y² + 2G एक्स + 2fy प्लस सी है ना अगर इसकी इक्वेशन तो हम जानते हैं अंडर रूट S2 अंडर रूट S2 मतलब अंडर रूट S2 मतलब सर इस S2 में आप एक्स और ए की जगह X1 y1 रख दो एक्स और ए की जगह आप X1 y1 रख दो तो ये हो जाएगा x1² + y1² + 2G X1 + 2fy वैन प्लस सी सर ये जो आपने वैल्यू निकल है ना ये जो आपने वैल्यू निकल है ना यही आपके इस पॉइंट से ड्रॉ कीजिए सर्कल पर टैसेंट के लेंथ होगी और जो की टेक्निकल अभी हमारे लिए है इस सर्कल की रेडियस आई होप ये सारी बातें बहुत कृष्ण स्टेटमेंट है यही स्टेटमेंट हमारे इस आईआईटी जी एडवांस के 2014 के क्वेश्चंस को सॉल्व करने में हेल्प करेगा तो इस कॉन्सेप्ट को ऐसा फैक्ट के ठीक है बहुत कम का फैक्ट अब अगर वापस आए सर इस कॉन्सेप्ट को यहां अप्लाई करने पर तो कैसे सोचना होगा ज़रा सोचते हैं सर थॉट बड़ा आसान सा है मेरे पास इस सर्कल के सेंटर के कोऑर्डिनेट्स हैं जो की हमने क्या माने द से जो की हमने माने द -7 कमा लामबीडीए और इससे जो इस सर्कल पर ऑर्थोंगोनल सर्कल है सर इस पर तो दो हैं ये और ये कोई सा भी ले लो तो कोई सा भी ले लो मुझे छोटा और अच्छा लग रहा है ये लो है ना तो इस पर ड्रा की गई टैसेंट की लेंथ इस सर्कल पर ड्रा की टेंशन के लेंथ कहां से -7 कमा ले लामबीडीए से कितनी होगी भाई जल्दी बताओ सर S1 का अंडर रूट तो एक्स की जगह रखा -7 -7 का स्क्वायर 49 ए की जगह खाली हो जाएगा लामबीडीए स्क्वायर सो 49 प्लस स्क्वायर माइंस वैन आप समझ का रहे हो एक रेडियस आई थी ये और ये जो आपके अकॉर्डिंग एक रेडियस आई है ये ये दोनों रेडियस है तो उसी से सर्कल की अब अगर मैं इस इक्वेशन को सॉल्व करना चाहूं तुझे दिमाग लगाते हैं पहले तो अंडर रूट से अंडर रूट वाले पार्ट कैंसिल तो जैसे ये वाले पार्ट हटाए तो वर दिया सर ये तो बहुत अच्छी बात हो गई आई थिंक सर आप एक और बात नोटिस करो ये 49 से 49 कैंसिल बिल्कुल सही बात है सर अब अगर आप चीज सिंपलीफाई करना चाहो तो मेरे ख्याल से मैं कुछ तो सोच ही लूंगा मैं कुछ तो सोच ही लूंगा ना तो सोचते हैं सर ये कितना ए जाएगा सही हो जाएगा लामबीडीए स्क्वायर प्लस वैन माइंस स्क्वायर माइंस वैन आई होप यू आर ग्रीन विद डेट सर हटा दो बिल्कुल हटा देंगे सर तो हटा दिया अब अगर मैं देखना चाहूं तो -2 λ को वहां और वैन को यहां लाओ क्या आप देख का रहे हो तो जैसे ही माइंस तू लामबीडीए को वहां पहुंचा है तो तू लामबीडीए उधर गया वैन इधर है तो इसको लामबीडीए की वैल्यू कितनी आती है सर लामबीडीए की वैल्यू ए जाती है वैन लामबीडीए की वैल्यू वैन मतलब इस सेंटर के कोऑर्डिनेट्स कितने हैं सर इसके सेंटर के cardinate आए -7 कमा वैन इस सर्कल के सेंटर के कोऑर्डिनेट्स आए -7 कमा वैन यानी आप एक ऑप्शन तो मार्क कर दोगे लेकिन अभी सर चूंकि ये आपके किस कैटिगरी के क्वेश्चन है ये मल्टीपल करेक्ट आंसर टाइप क्वेश्चंस है तो रेडियस भी तो निकालनी होगी सर इस सर्कल का सेंटर है माइंस सेवन कमा वैन और इसके पास होने का एक पॉइंट है जीरो कमा 1 तो गैस की रेडियस नहीं निकल सकते ये वैन वैन फिक्स्ड है तो रेडियस कितनी ए जाएगी सर 7 आप डिस्टेंस फॉर्मूला अप्लाई कर लीजिए आप रेडियस निकल लीजिए उसे केस में अब बहुत आसानी से इस क्वेश्चन का आंसर बहुत इजीली अपने लॉजिक से अपने दिमाग से बता पाएंगे ज्यादा सोचे समझे बिना तो shailise इसके आंसर्स हो जाएंगे ऑप्शन बी और ऑप्शन सी बहुत टू क्वेश्चन तो नहीं था बस ये प्रॉपर्टीज जरूर याद रखिएगा स्टूडेंट्स ये बहुत कृष्ण प्रॉपर्टी अगर दो सर्कल्स और जनरली इंटरसेक्ट करें तो इस एक सर्कल के सेंटर से इस पर ड्रा के टैसेंट की लेंथ दूसरे सर्कल पर वो इसी सर्कल की रेडियस की इक्वल होगी और आई होप ये आप डायरेक्टली आसानी से देख ही का रहे मतलब इट इस वेरी एवीडेंट इट इस क्लीयरली यू नो अपेरेंट किए होता है यह तो ट्रू है ना अगर हम बड़े बेसिक सारे जानते हैं तो अब हम शुरुआत कर रहे हैं लेंडर कंप्रेशन टाइप क्वेश्चन से यानी की ऐसे क्वेश्चंस जो की एक पैसेज के फॉर्म में दिए जाएंगे और इस कंप्रीहेंशन पर बेस्ड कुछ क्वेश्चंस होंगे आई होप आप समझ का रहे हो तो पहले तो पैसेज को क्रैक कर लेते हैं की इसमें दिया क्या क्वेश्चन बोल रहा है बाय डी वे यह आईआईटी जी एडवांस्ड 2012 का क्वेश्चन है क्या बोल रहा है वो समझते हैं एक टैसेंट पॉइंट ड्रा की पॉइंट से पॉइंट सर्कल इस बारे में आपके क्या विचार है क्या थॉट ए सकता है मेरा कहना है ज्यादा सोचने की जरूरत नहीं है बड़ी बेसिक सी बात है आप तो पहले ये बताओ अगर मेरे पास एक सर्कल है x² + y² = 4 और इसके किसी पॉइंट पर मुझे tagent ड्रॉ करनी है तो क्या है बहुत टू टास्क है नहीं सर आप टी = 0 से वो टेंशन की इक्वेशन बिना सोचे समझे बहुत डायरेक्टली लिख सकते हो हमने सिखा है तो अंडर रूट थ्री कमा 1 के लिए जो t=0 लिखूंगा वो मैं क्या लिखूंगा अंडर रूट 3X + 1 4 = 0 दिस इसे डेट टांगे विच टांगें वो जो ट्स आपकी पीटी तो किसकी इक्वेशन है सर यह आपकी टांगें पॉइंट की इक्वेशन कोई तकलीफ तो नहीं है अब suniyega ध्यान से इस पॉइंट पर एक परपेंडिकुलर लाइन है सर इस पॉइंट पर अगर एक लाइन परपेंडिकुलर है तो मैं इस लाइन जो एल जो पास हो रही है एल है ना तो वो जो लाइन एल है क्या उसकी स्लोप बता सकता हूं सर उसकी स्लोप निकलने का तरीका ऐसा है की ये हो जाएगा ए = एमएक्स + सी तो एम कितना हो जाएगा -√3 तो जो लाइन एल है जो की इस पॉइंट पर परपेंडिकुलर है उसकी अगर आप स्लोप निकलना चाहें तो उसकी स्लोप आप क्या कहेंगे -√3 का नेगेटिव रिसिप्रोकल तो हो जाएगा वैन बाय रूट थ्री तो वो जो लाइन एल है उसकी स्लोप ए जाएगी वैन बाय अंडर रूट थ्री अच्छी बात है सर तो क्या तो मेरा बस आपसे ये कहना है प्लीज इस बात को सोच के बताइए वो जो पॉइंट परपेंडिकुलर है आपकी लाइन कौन सी एल है ना उसे पे बस इतना सा बताओ की अगर वो टैसेंट है वो टैसेंट है किसके लिए वो टैसेंट है आपके इस सर्कल के लिए एक लाइन जिसकी स्लो मुझे अभी पता है बस उसका इंटरसेप्ट नहीं पता है एमएक्स + सी फॉर्म में अगर बात करें तो वो मैं सोच सकता हूं ए = एमएक्स + सी अगर वो टैसेंट है इसके लिए तो क्या सोचा जा सकता है सी निकलने के लिए आई होप सारे स्टूडेंट्स के दिमाग में थॉट्स लाइक किया होगा सर इस सर्कल के इस इस सर्कल के सेंटर से परपेंडिकुलर डिस्टेंस इस लाइन एल पर जो होगी वो ही इस सर्कल की रेडियस के इक्वल होगी क्या आपको कुछ बातें याद है तो बिल्कुल सर इस बारे में भी बात कर ही लेते हैं अगर इस लाइन की इक्वेशन लिखना चाहूं की तो मैं लिखूंगा ए = एमएक्स एमएक्स यानी एम कितना है अंडर रूट 3 है ना एक्स + सी सी मुझे अभी नहीं पता है ठीक है सर अब जब मैं कहना चाहूंगा की सर्कल के सेंटर के कोऑर्डिनेट्स के होंगे तो सर इस सर्कल के सेंटर के कार्ड होंगे आपके 3 कमा जीरो और इस सर्कल की रेडियस कितनी होगी सर वो होगी वैन आई होप ये बात समझ का रहे हो तो इन केस आई एम अस तू फिगर आउट डी क्वेश्चन ऑफ डी लाइन स्ट्रेट लाइन एल व्हाट आय गोइंग तू डू आज सिंपली फिगर आउट डी परपेंडिकुलर डिस्टेंस ऑफ दिस स्ट्रेट लाइन फ्रॉम दिस पॉइंट एंड अल पुट डेट इक्वल तू डेट डिस्टेंस = रेडियस आई हो सी ऑल गेट तो सर ऐसा अगर सकते हो तो क्या निकल कर आता है इस बारे में बात करते हैं ध्यान से देखो भाई परपेंडिकुलर डिस्टेंस है ना थ्री कमा जीरो से अगर इस स्ट्रेट लाइन को और बेहतर फॉर्म में लिखना चाहें शो कैन आई ब्रिंग दिस वही ऑन डी आदर साइड आई थिंक लिए जा सकता है अब सुनेंगे क्या ध्यान से थ्री इन वैन बाय रूट थ्री थ्री तू वैन बाय रूट थ्री आई होप समझते हो ax1 सो ए X1 तो 3 1 / √3 3 4 √3 √3 एक रूट 3 कैंसिल हो जाएगा √3 आप समझ का रहे हो फिर माइंस वैन इन जीरो जीरो प्लस सी ये तो आता है आपका न्यूमैरेटर डिवाइडेड बाय अंडर रूट ओवर इनके एक्स और ए के कॉएफिशिएंट्स के स्क्वायर सैम ध्यान से देखो इसका स्क्वायर 1/2 इसका स्क्वायर 1 / 3 + 1 1 / 3 + 1 कितना हो जाएगा 4/3 और 4 / 3 का क्या लेना है अंडर रूट आय रिपीट माय स्टेटमेंट इसका स्क्वायर 1/3 इसका स्क्वायर कितना हो जाएगा 4/3 4/3 का अंडर रूट तो फोर का अंडर रूट कितना होता है सर फोर का अंडर रूट होता है यहां पे लिखना चाहो तो लिख सकते हो 4/3 तो 4√ हो जाएगा 2 और वो जो अंडर रूट 3 है क्या मैं उसे यहां पर लिख सकता हूं मेरे जैसा मैन हो रहा है की सर ये जो डिस्टेंस है इसकी रेडियस के इक्वल होगी जो की है वैन अगर आप यहां से निकलना चाहो तो बेशक ये कहोगे सर ये जो 2/3 है ये जो 2/3 है उसको आप चाहते तो इसी तरह लिख लेते तो वैन से जब मल्टीप्लाई किया तो ये हो जाएगा 2/3 2 / √3 तू बी मोर प्रेसिस तू बी मोर करेक्ट अब सर सीधा सीधा सा फंडा है व्हेन यू वांट तू गेट ऑफ दिस मोड आप इस तरफ प्लस माइंस लगाते हो बिल्कुल सही बात है तो जब आप प्लस माइंस लगाते हैं तो क्या फर्क आएगा ध्यान से देखोगे भाई दो बात निकल करेंगे अंडर रूट 3 + सी विल बी इक्वल तू प्लस माइंस तू रूट थ्री है ना तो सी विल बी इक्वल तू प्लस माइंस 2 / √3 - √3 यहां से आपकी सी की वैल्यू हो जाएगी सी की वैल्यू ए चुकी है एम पता है तो क्या स्ट्रेट लाइन मिल जाएगी मिल जाएगी पर पूछा क्या है सर वो तो निकालो आप अब क्वेश्चन तो दिखाओ हमको तो क्वेश्चन देखते हैं अभी पहला क्वेश्चन है पॉसिबल इक्वेशन ऑफ एल एस तो बिल्कुल यही पूछा है दूसरा क्वेश्चन है अन कॉमन लेंथ ऑफ डी तू सर्कल्स और कमेंट टांगें ऑफ डी तू सर्कल बात करेंगे पहले तो हम एल की इक्वेशन पे वर्कआउट करने की कोशिश करते हैं तो एल की जो इक्वेशन मेरे लिए ए रही है मैं सी की वैल्यू सिंपलीफाई करना चाहता हूं तो कर सकता हूं वो कितनी ए जाएगी देखो जब प्लस साइन लेंगे तो हो जाएगा 2 - √3 से √3 ए गया तो कितना हो जाएगा थ्री देखो पापा समझ का रहे हो मैं फिर से लिखता हूं जब मैं प्लस साइन लेता हूं तो क्या होगा जब माइंस साइन लेता हूं तो क्या होगा दोनों पर बात कर लेते हैं जब प्लस साइन लिया तो ध्यान से देखो 2 - 3 + 2 - 3 एक और केस देखें तो -2 -3 तो उसे केस में कितना मिलेगा सर वो मिलेगा -5/√3 क्या आप मेरी बातें समझ पाएं सर अगर आपके पास आप सारी ही वैल्यूज ए चुके हैं तो आप ट्राई कर लो सी की वैल्यू है ना यहां रखता हूं मैं सी की दोनों वैल्यूज आई एम पुटिंग सी की है दोनों अलग-अलग वैल्यूज वैन आफ्टर वैन आता है तो पहले ये रखा तो कितना है सर 1 / √3 एक्स सी की वैल्यू कितना रखना है माइंस वैन बाय रूट थ्री आई होप हम सही कर रहे हैं है ना + 2 - 3 बिल्कुल माइंस वैन बाय थ्री आना था तो ये हो जाएगा -1 / √3 - ए = 0 ऑप्शंस किस फॉर्म में दिए जरा वो देख लेंगे ताकि हमारा कम आसान हो जाएगा सर वो जो आपका कांस्टेंट है उसे तरफ जाएगी तो पुरी इक्वेशन को अंडर रूट 3 से मल्टीप्लाई करता हूं तो देखो ये हो जाएगा एक्स -1 - ए टाइम्स ए टाइम्स √3 = 0 एंड सॉरी जीरो नहीं वैन तो एक इक्वेशन तो आपकी सर ये आने वाली है एक इक्वेशन तो बेशक आपकी ये होगी एक और इक्वेशन बन सकती है क्या देखते हैं भाई अगर -5/√3 रखा तो सीधा-सीधा सॉल्यूशन है देखो ध्यान से माइंस फाइव बाय रूट थ्री अगर यहां रखा तो देखो सीधी सीधी सी बात है यहां पर ए जाएगा -5/√3 तो आपकी क्या लिखोगे सर आप लिखोगे 1/3 है ना तो ये कितना है सर ये है वैन बाय अंडर रूट 3X से मल्टीप्लाई किया जैसे ये √3 से मल्टीप्लाई किया तो ये हो जाएगा सिर्फ एक्स ये बचेगा -5 और ये बचेगा - ए √3 ये कितना ए जाएगा - ए √3 = 0 इससे थोड़ा और सिंपलीफाई करिए सर 5 को उधर भेज दो आप तो ये क्या बन जाएगा सर ये बचेगा एक्स - 5√3 = 5 सो दिस इसे गोइंग तू बी दी अंदर एक्शन व्हाइट हैव गो इट तो दो इक्वेशंस आपको टेक्निकल मिलेंगे आप इससे यहां से भी सिंपलीफाई करके लिख सकते द जो भी तरीका आपको सही लगता है तो या तो ये या फिर इन दोनों में से कोई ऑप्शन आपका मौजूद होना चाहिए आंसर्स में ठीक है क्या तो देख लेते हैं भाई क्या ये दोनों मुझे दिख रहे हैं पहले तो देखते हैं एक्स - ए √3 = 1 सो एक्स - ए √3=1 ऑप्शन ए में है और एक्स - ए √3=5 ऑप्शन दी में है और इस बेसिस पे देखना चाहें तो हम दो ऑप्शंस मार्क कर सकते हैं भाई कौन-कौन से ऑप्शन दी नहीं मार्क करेंगे क्योंकि यहां पर प्लस है और मेरे आंसर में माइंस ए रहे हैं तो हम एक ही ऑप्शन मार्क करेंगे जो की क्लेरिटी क्या होगा आपका ऑप्शन ए लिए ऑल विद मी सो फास्ट स्टूडेंट्स किसी भी स्टूडेंट को कोई डाउट आई होप अगर यहां माइंस होता तो मैं दी भी मार्क करता बट दी में यही गड़बड़ है अब अगला सवाल क्या निकल कर ए रहा है सर अगले सवाल को ध्यान से देखते हैं और कॉमन टांगें तू डी तू सर्कल्स अब दो सर्कल्स की कॉमन टैसेंट पर अगर सवाल आता है तो पहले तो सर्कल्स को देखो सर एक तो है आपका 0% जिसकी रेडियस कितनी है तू और एक है आपका थ्री कमा 0% जिसकी रेडियस 7 मैं थोड़ा इनको प्लॉट करना चाह रहा हूं है ना सर ध्यान से देख रहा हूं वैन सर्कल यू हैव वैन सर्कल यू हैव सेंटर आते जीरो कमा जीरो सेंटर आते 0 विच हज रेडियस वैन है ना उसकी रेडियस कितनी है सर रेडियस है वैन यूनिट आई एम रियली सॉरी तकलीफ तो नहीं है बिल्कुल सर कोई दिक्कत नहीं सिमिलरली सी हैव अंदर सर्कल सी हैव नॉर्थ सर्कल विच इस सेंटर्ड इट थ्री कमा जीरो विच इसे सेंट्रल इट थ्री कमा जीरो इसकी रेडियस कितनी है वैन सुना गया सर आई होप आप ये सोच का रहे हैं की ये कितनी डिस्टेंस है तू अगर ये 0 है और अगर मैं इसे एक्स एक्सेस की तरह ट्रीट करूं तो ये पॉइंट क्या होगा सर ये पॉइंट होगा आपका क्लीयरली तू कमा जीरो आप मेरी बातें समझ का रहे हो अगर मैं इसे एक्स की तरह ट्रीट करूं तो ये आपका एक्स एक्सिस है आप समझ रहे हो ना और ये क्लीयरली आपका क्या है ये आपका क्लीयरली एक्सेस है कोई तकलीफ तो नहीं है अब suniyega ध्यान से बड़ी ही बेसिक सी और सिंपल सी बात इस सर्कल को लेकर के ओपिनियन थ्री कमा जीरो जिसके रेडियस है वैन थ्री कमा जीरो तू कमा जीरो से एक यूनिट ए गया 7 तो शैल आय से जो आपका दूसरा सर्कल बनेगा जो आपका दूसरा सर्कल बनेगा वो कुछ ऐसा बनेगा हम रियली सॉरी जो आपका दूसरा सर्कल बनेगा वो कुछ ऐसा बनेगा जिसको क्लीयरली आप पाएंगे की वो सिचुएटेड सो था सर्कल बताओ तो मैं कहूंगा सर इस बार आपको तीन कॉमन टांगेंट्स देख सकती हैं एक तो हो सकता है उनकी क्या ये वाली लाइन सर इस लाइन को लेकर तो मैं बहुत अच्छे से बात कर चुका हूं ये लाइन ना सिर्फ इनकी कॉमन टांगें है बल्कि अभी फिलहाल तो मुझे ये इनकी क्या बनती है दिख रही है रेडिकल एक्सिस भी बनता हुआ दिख रहा है अगर आप ध्यान से देख पाएं तो बिल्कुल सही बात सर एक बात तो ये होगी क्या इसके अलावा भी कोई और कॉमन टांगें देख सकती थी ध्यान से देखो भाई अगर इसके अलावा आपने देखा होता तो सर क्या मैं ये नहीं का सकता की ये जो लाइन है यह जो लाइन है यह भी तो उसकी कमेंट है यह जो लाइन है सर यह भी तो इसकी एक कॉमन टांगें हुई की नहीं भाई बिल्कुल होगी सर क्या इस बात से किसी भी स्टूडेंट को कोई आपत्ति मेरे ख्याल से तो नहीं होनी चाहिए की इन दोनों कॉमन टेनिस को अगर एक्सटेंड किया जाए इन दोनों कॉमन टांगेंट्स को एक्सटेंड किया जाए तो ये इन दोनों के सेंटर को मिलने वाली लाइन को कहीं ना कहीं इंटरसेक्ट कर रही होंगी तो हमें या तो यह या फिर इन दोनों की इक्वेशन निकालनी है बहुत टू बहुत मुश्किल क्वेश्चन तो नहीं है अब एक बात बताओ बड़ी ही बेसिक सी बात है इसकी इक्वेशन लिखना टू टास्क नहीं है क्यों सर क्योंकि ये तो आपकी यह तो आपकी रेडिकल एक्सरसाइज भी निकल जा सकती है तो अगर इससे निकलना हो तो और अगर ऐसा ढूंढना होता तो फटाफट ऑप्शंस को देख लेते हैं और देख लेते हैं की इससे निकल पाएगी की नहीं तो जरा इसे हम सिंपलीफाई कर ले ये कितना हो जाएगा देखो जरा सर ये हो जाएगा x² है ना प्लस ये कितना है y² फिर यहां से क्या मिलेगा देखो यहां से मिलेगा -6x ये कितना मिलेगा -6x और प्लस होगा नाइन में से वंस अट्रैक्ट होगा टाइम में से मैं सब्सट्रैक्ट होगा तो प्लस 8 = 0 तो ये आपका एक सर्कल और ये आपका एक सर्कल है अब आप सीधे सीधे सोच पाओ स्टूडेंट्स अगर मैंने इसमें से इसे सब्सट्रैक्ट किया तो x²y² गए -6x+8 और यह फोर इधर ए गया तो माइंस फोर तो माइंस और माइंस वाले ने +4 - 68 + 8 + 4 यानी -6x+12 आपकी लाइन के आगे भाई आपकी लाइन ए गई क्या फिर से suniyega आपकी जो लाइन आएगी वो होगी क्या - 6x + 12 मेरा ये कहना है की सर ये तो आप ऐसे भी सोच सकते हो इतना खुद देखो ये पॉइंट क्या होगा ये पॉइंट होगा 2 0 आप सन का रहे हो तो 2 0 से पास होने वाली ए एक्सिस के पैरेलल लाइन उसकी इक्वेशन क्या होगी सर वो भी एसेसरीज इक्वेटर तू आई होप आप देख का रहे हो यहां से भी अगर आप सोचो जी की अगर मैं इसे जीरो के इक्वल रख डन है ना अगर मैं इसे यहां से जीरो के इक्वल रख डन जो की जीरो के इक्वल है ये राइट तो यहां से ले लिए दिख रहा है की एक्स की वैल्यू तू ए जाएगी जो की यही है क्या एक्स की वैल्यू तू हमें किसी ऑप्शन में दिया गया है क्या एक्स की वैल्यू तू मुझे किसी भी ऑप्शन में दिख रहा है सर ए = 2 दे दिया उसने एक्स = 4 दिया तो ये तो बात नहीं बन रही है अब शायद मुझे इसी बात की लगी की इसके कोऑर्डिनेट्स में ढूंढ पाओ मेरा आपको सजेशन बस इतना सा है की अगर इन दोनों के कोऑर्डिनेट्स आप ढूंढना चाह रहे हो तो क्या आपके दिमाग में थॉट आता है की सर यह तो आसान है या आसान इसलिए है की अगर मैं इन दोनों टांगें का पॉइंट ऑफ इंटरसेक्शन ढूंढ रहा हूं इन दोनों टांगें का पॉइंट ऑफ इंटरसेक्शन ढूंढ रहा हूं तो बात बन जाएगी दोनों वहां इंटरसेक्ट कर रहे हैं सर हमने पढ़ा है की इसके जो cardinate होंगे ना वो कुछ नहीं होंगे वो वही होंगे जो क्या होंगे 0 और 2 0 को 2:21 के रेश्यो में एक्सटर्नल ये जो इंटरसेक्ट करें याद है ना C1 और C2 को R1 इस तू आर तू के रेश्यो में एक्सटर्नल इंटरसेक्ट करता है ये पॉइंट याद ए रहा है क्या बस यही बात है तो सुनेगा ध्यान से मैं क्या करूंगा मैं करूंगा वैन इन जीरो अपॉन तू माइंस वैन तो सिक्स अपॉन वैन यानी कितना सिक्स तो इसका एक cardinate क्या मिला 6 डिड ऑल गेट डट आगे बढ़ते हैं सर ध्यान से देखिएगा अब अगर मैं आगे बढ़ता हूं तो नेक्स्ट पॉइंट 2 0 0 1 0 0 तो 0 + 0 / 101 ठीक है सर ये cardinate आया सिक्स कमा जीरो तो इससे आप क्या निकल लोग इससे क्या फायदा हो जाएगा इससे आप इस tagent की इक्वेशन कैसे निकलोगे मेरा कहना है मुझे पता चल गया ना की ये लाइन कहां से पास होती है और मुझे इसका एक पॉइंट पता चला है मुझे दोनों तो पता नहीं चले तो क्या मैं इस स्ट्रेट लाइन के क्वेश्चन को लिखना चाहो तो मैं इसे लिख सकता हूं क्या ए - y1 = एम एक्स - X1 आई होप आप लिख सकते हो अगर इसकी इक्वेशन को मैं लिखता हूं ए - 5 = एम टाइम से एक्स - एक्स तो मैं सिर्फ क्या लिख सकता हूं ए है ना ये ए जाएगा एमएक्स उधर यार अगर ए को उधर ले गए तो ये मिलेगा आपको एमएक्स यहां से देखना - ए - 6m = 0 है अब इस बात पर हमने कई दफा यह बात की है सर की सर किसी से भी इससे या इससे जिससे आपको ठीक लगे उसे सर्कल के सेंटर से इस डायरेक्ट कॉमन टांगें की जो परपेंडिकुलर लेंथ होगी वो इसकी रेडियस के इक्वल होगी बस इतनी सी बात थी जो आपको सोचनी थी तो सर यही कॉन्सेप्ट अप्लाई करके हम अक्सर अपने जीवन की समस्या suljhaenge तो ठीक है सर देखते हैं यह है जीरो कमा जीरो इससे इस tagent की अगर परपेंडिकुलर लेंथ निकले तो इसकी रेडियस तू के इक्वल होनी चाहिए बिल्कुल सर तो जीरो इन एम जीरो जीरो इन माइंस वैन जीरो तो बचा क्या -6m तो मैं ले लूंगा मोड ऑफ -6m जैसे मैं चाहूं तो मोड ऑफ सेक्शन लिख लो क्या या सिक्स टाइम्स मोड ऑफ एम लिख लूं क्या आई होप आप मेरी बात समझ का रहे हो डिवाइडेड बाय आपकी लाइन के एक्स और ए के कॉएफिशिएंट के स्क्वायर का अंडर रूट तो हो जाएगा m² + 1 जिस से हम हर बार डील करते आए हैं इसका अंडर रूट इस इक्वल तू रेडियस वो क्या है तू अब अगर आप सर सिंपलीफाई करना चाहें तो एक बात आप ध्यान से देख का रहे हो ये तू सेक्स कितना टाइम हो जाएगा 39 स्क्वायर प्लस वैन वहां चला जाएगा 3m² और ये हो जाएगा m² + 1 m² को इधर ले आते हैं भाई तो ये हो जाएगा 2m² = व्हाट वैन कोई दिक्कत तो नहीं है भाई तो m² की जो दो वैल्यूज मुझे मिलेंगे और अगर एम की जो दो वैल्यूज बनेगी वो मिलेगी प्लस माइंस वैन बाय रूट तू सर एम की दो वैल्यू उसकी मिली अरे मिलेगी भाई क्यों नहीं एक आपकी ये डायरेक्ट कॉमन टैसेंट एक ये भी तो है तो उसमें दोनों दे दिया और दोनों कहां से पास होती हैं सेक्स कमा जीरो से अब आई थिंक इक्वेशन ये है ए = एम टाइम एक्स - 6 अगर प्लस वैन बाय रूट तू रक तो 1 / √2 यहां आया तो √2 इधर चला गया तो ये क्या हो जाएगा √2 ए उसको मैंने उधर शिफ्ट किया तो ये हो जाएगा - और ये हो जाएगा एक्स अब समझ रहे हो ना आप समझ रहे हो नहीं स्टूडेंट आई होप आप मेरी बात डाइजेस्ट कर का रहे हो मैं कुछ गड़बड़ तो नहीं कर रहा हूं बस एक बार चेक कर लेते हैं √2 है तो √2 ए ये वहां शिफ्ट हो और ये बचेगा क्या बिल्कुल -6 तो एक इक्वेशन तो ये मिलेगी कोई तकलीफ तो नहीं है सर अगर आपने माइंस वैन बाय रूट तू रखा होता जैसे ही माइंस वैन बाय रूट तू रखते तो -√2 यहां था तो -√2 बाय वहां पहुंच जाता तो हो जाता एक्स प्लस रूट तू टाइम्स ए और माइंस सिक्स से तो आपको कोई छेड़खानी करनी है तो सर ये वो दो लाइंस हैं जो शायद आपको ऐसा इक्वेशन मिल रही होती इनमें से अब आप देख लो आपका कौन सा ऑप्शन मैच हो रहा है बस वही आपका करेक्ट आंसर होना चाहिए अगर ढूंढना चाहा तो एक्स - √2y - 1 सेकंड ऑप्शन ये है तो एक्स + 2y = 6 एक्स + 2y = 6 बिल्कुल सर हमारे पास एक उपलब्ध ऐसी स्प्रे लाइन है जो की क्लीयरली हमारे कहां दी गई है ऑप्शन दी में तो क्या इस क्वेश्चन को लेकर किसी भी स्टूडेंट को कोई आपत्ति कोई परेशानी अच्छे से सॉल्यूशन आपने दोनों देख लिए दोनों बातें समझ ए गई क्या यह आपके लिंक्ड कंप्रेशन टाइप क्वेश्चंस हैं जहां पर आपको एक पैसेज पर बेस्ड क्वेश्चंस पूछे जाते हैं और ऑफ कोर्स यह पूछा गया है जी एडवांस्ड ने 2018 में क्या पूछा है ज़रा क्वेश्चन समझिए क्वेश्चन में लिख रहा है वो एक एक्सी प्लेन में आपका एक सर्कल है एक सर्कल है x² + y² = 4 ठीक है सर सो सी हैव अन सर्कल बेसिकली और इस सर्कल के बारे में उनका ये ओपिनियन है एक ले लिया आपका क्या बन जाएगा ए-एक्सिस और इसी तरीके से आपका यहां पर होगा एक एक्स एक्सिस है ना अब उनका आपसे यह कहना है की सर आपके ये जो एक्स एक्सेस पर एक सर्कल आप बना रहे हो एक सर्कल इस तरीके से आप बना रहे हो सर की इस सर्कल का जो है ना सेंटर जो है वह जीरो पर सिचुएटेड है और जो आपके सर्कल की रेडियस है सर इस सर्कल की रेडियस है तू यूनिट है ना ये ऑफ कोर्स आपका जीरो है जैसे कहने की जरूरत नहीं है तू कमा जीरो होगा है ना यह आपका हो जाएगा -2 0 और ये हो जाएगा आपका क्लीयरली जीरो कमा तू और आई थिंक ये हो जाएगा आपका कितना 0 - 2 एंड आई नीड नॉट टेल यू दिस इस क्लीयरली दी ओरिजिन आई होप ये बातें आपको समझ ए रही है वो आपसे ये का रहा है ध्यान से suniyega यहां तक चीजे क्लियर है ये क्लीयरली आप देख का रहे हो ना ये रेडियस कितनी है सर ये रेडियस है आपकी 2² और ये सर्कल के सेंटर जो की ओरिजिन पर प्लेस है कोई तकलीफ तो नहीं बातों को समझने में क्वेश्चन में क्या पूछ रहा है उसे पर आते हैं वो का रहा है एवं I2 और F1 f2 आपकी दो कोट्स हैं इस सर्कल की जो की पी नोट वैन कमा वैन से पास होते हैं और वो एक्स एक्सिस और ए एक्सिस के पैरेलल है रिस्पेक्टिवली तो मैं एक्स-एक्सिस के पैरेलल एक लाइन ड्रॉ करने की कोशिश करता हूं जो वैन कमा वैन से पास हो रही होगी है ना तो फाइव ट्री तू ड्रा अन है ना जो वैन कमा वैन से पास हो रही होगी तो शैल आई से 1 मिनट पॉइंट तो ये हो जाएगा जितना मैं सोच का रहा हूं है ना तो एक स्ट्रेट लाइन हम बनाते हैं एक्स एक्सिस के पैरेलल आई एम ड्राइंग ए स्ट्रेट लाइन पैरेलल तू एक्स एक्सिस जो की ये रही एंड सिमिलरली आई एम गोइंग तू ड्रा अंदर स्ट्रेट लाइन विच पैरेलल तू बाय एक्सेस विच इस गोइंग तू बी सैम व्हाट लाइक दिस अन्य वैन हेविंग अन्य क्वेरीज अन्य कन्फ्यूजन सो फार आई थिंक दिस इस वेरी मच सॉर्टेड सोफा और ये जो पॉइंट ऑफ इंटरसेक्शन है सर ये वो पॉइंट है कौन सा पिनो दिस इसे डेट पॉइंट पी नॉट जो की रियली क्या है वैन कमा वैन कोई तकलीफ तो नहीं है भाई और हम बड़ी बेसिक सी साधारण सी जरूर इसी बात ये जान का रहे हैं की ये जो आपकी दो कार्ड्स बन रही हैं उन कार्ड्स के नाम क्या दे रहा है वो एक्स एक्सिस के पैरेलल को तो एवं A2 और ए एक्सिस वाली F1 f2 है ना तो ये जो है आपकी ये पॉइंट हो जाएगा A1 और इस पॉइंट को का रहा है A2 और ये पॉइंट आपका हो जाएगा F1 और ये पॉइंट को वो का रहा है f2 आई थिंक चीज आसान है स्मूथ है कोई दिक्कत नहीं अब वो क्या कर रहा है आगे सुनते हैं जीतू जी जीवन जी तू बी डी कोड ऑफ एस अगेन पासिंग थ्रू p0 एंड हेविंग स्लोप -1 अगर अगर p0 से एक लाइन पास हो रही है मैं बड़ी बेसिक सी बात कहना चाह रहा हूं इस बात का प्लीज अच्छे से मतलब समझना किस बात का सर इस बात का की जीवन जी तू आपकी एक है जो एस से पास हो रही है उसकी स्लोप है माइंस वैन इस बात का क्या मतलब है इस पे सोचो देखो आप इमेजिन करो एक स्ट्रेट लाइन जो वैन कमा वैन से पास हो रही है पी नॉट क्या है एंड कमा वैन और स्लोप है -1 तो मैं क्या करूंगा ए नॉट ए - y1 इस इक्वल तो म क्या है -1 एक्स - X1 तो 1 से पास होने वाली स्लोप माइंस वैन कैरी करने वाली स्ट्रेट लाइन अगर इसे थोड़ा फिगर आउट करने की कोशिश कर रहा हूं तो मुझे क्या मिलता है सर ए - 1 है ना ए - 1 = - एक्स + 1 ड्यूल एग्री विद डेट तो शैल आय कनक्लूड सर ये हो जाएगी एक्स + ए = 2 इंटरसेक्ट फॉर्म में क्या होगी सर एक्स / 2 + ए / 2 = 1 सो दिस इसे डेट स्ट्रेट लाइन जो जीवन जी तू है आपकी वो ये वाली स्ट्रेट लाइन है जो एक्सपायर 2 + ए / 2 = 1 है क्या आप मेरी बात समझ का रहे हो तो ये वो स्ट्रेट लाइन है जो की आपकी एक्स / 2 + ए / 2 = 1 वाली स्ट्रेट लाइन है मतलब मैं ये कहूंगा सर प्लीज इस बात को सुनना ध्यान से की आप एक स्ट्रेट लाइन ड्रॉ कर रहे हो तो कमा जीरो कमा वैन से तो पास हो ही रही क्योंकि उसे ग्लोब माइंस वैन तो आप टेक्निकल कैसी स्ट्रेट लाइन ड्रॉ कर रहे हो सर जो की यहां से गुजरते हुए इससे गुजरते हुए इससे स एक्सटेंड हो रही है आप मेरी बातों से एग्री कर रहे हो भाई तो ये आपकी टाइमिंग जो की क्लीयरली हमें विजिबल है सर जो की बड़ी बेसिक सी साधारण सी बात है जो हमें खुद समझ ए रही है यहां सर ये माइंस वैन स्लोप है ना 135° एंगल पॉजिटिव एक्सिस है क्योंकि वैन होता तो 45 होता और वैन कमा वैन से पास हो रही है तो ये स्ट्रेट लाइन और ये कौन सी ये कौन से वो पॉइंट्स का रहा है ये का रहा है जीवन जी तू तो ये पॉइंट का वो का देता है जीवन और इस पॉइंट को का देता है G2 आई होप यू आर नॉट कंफ्यूज सो फार विथ वाटेवर इनफॉरमेशन डेट हज बिन conveyan ऑन डी यू नो एक्सी अभी वो क्या बोलता है सर अगर मैं अब इससे आगे बढ़ता हूं तो कम की बात पर आते हैं वो कुछ पूछना शुरू करता है आपसे लेट डी टांगेंट्स तू एस आते एवं एंड अन्य तू मीत आते A3 अब आपने एवं और e2 पर मिला है तो A3 और F1 तो पर जो मिला है वह टैसेंट बनाई वह है G3 सर इन तीनों के बारे में बात करते हैं suniyega ध्यान से मैं आपको ये तीन पॉइंट्स डिफाइन करना चाह रहा हूं की सर आपने क्या किया आपने एवं और e2 पर एक टांगें बनाई आसान है कैसे आसानी से अगर आपने E1 से टैसेंट E1 से एक टैसेंट बनाई तो मैं इतना समझ का रहा हूं सर चीज बिल्कुल सिमिट्रिकल होंगी चीज बिल्कुल sametrical होंगी और क्लीयरली वो आपके एक्स एक्सिस पर इंटरसेक्ट कर रही होंगी जो आप एवं I2 वाली टैसेंट बनाओगे ना सर वो क्लीयरली आपकी एक्स एक्सेस पर इंटरसेक्ट कर रही होंगी क्योंकि सिमिट्रिकल चीज है ये बिल्कुल एंगल यहीं पर बायसेक्स हो रहा होगा राइट इसी तरीके से जब आप कहां पर F1 और f2 पर बनाओगे तो एक्सेस सॉरी मैं उल्टा बोल गया एवं और e2 वाली आपकी जो टांगें है वो एक्सेस पर और जीवन और G2 वाली जो टैसेंट है वो आपके एक्स एक्सिस पर इंटरसेक्ट कर रही होंगी एंड डाउट इडली है ना तो जो जीवन और जी 2 पर मैंने टांगेंट्स बनाई है ऑफ कोर्स यह आपकी कौन सी ट्रेन है सर यह आपकी जीवन वाली टैसेंट है जो की थोड़ा दूर जा रही है पता नहीं क्यों भाई ऐसा क्यों हो रहा है यहां पर जीवन और जीतू मैं कंफ्यूज हो रहा हूं मैं का रहा हूं F1 और f2 वाली जो टैसेंट है वो आप देख का रहे हो ये F1 वाली टैसेंट होगी क्या आप सभी नोटिस कर रहे हो ये आपकी F1 वाली टैसेंट होगी और सिमिलरली ये आपकी f2 वाली टैसेंट हो जाएगी जो की ये आपकी हो गई f2 वाली ट्रेन है तो सर ये F1 f2 वाली tenjent के लिए आपके वो एक्सेस प्रिंटर सेट करेगी और मैं देख का रहा हूं उनकी लैंग्वेज के अकॉर्डिंग एवं और ई-2 वाली टैसेंट आपके ए एक्सिस पर इंटरसेक्ट करेगी जैसे आप A3 कहोगे F1 और f2 वाली इंटरसेक्ट करेगी जिससे आप f3 कहोगे और एक और बात उसने कहे की G1 और G2 वाली जो टैसेंट है जो जीवन और जी तू वाली टैसेंट है तो जीवन और जी 2 पर भी अगर मैं टेंशन ड्रा करूं तो सर ये तो बनाना आसान है तो जीतू पर अगर आपने बनाई तो क्लीयरली सर आपके ए एक्सिस के पैरेलल हो जाएगी और जब आप जीवन पर tenjent बनाएंगे तो क्लीयरली सर आपकी एक्स एक्सिस के पैरेलल जाएगी आई होप आप इन बातों से एग्री करते हो ये वाली जो आपकी टांगें होगी क्लीयरली सर आपके एक्स एक्सेस के पैरेलल होना तय है ना क्या ये सारी बातों से आप एग्री कर रहे हो और इतना मुझे इल्म है की सर जो जीवन जीतू पर जो टैसेंट बन रही है सर वो तो क्लीयरली अगर ये ये जो जीवन है यह जो लाइन है ये एक्स एक्सिस के पैरेलल और रे लाइन ए एक्सिस के पैरेलल है तो जो G3 है उसके कोऑर्डिनेट्स मुझे तू कमा तू कहने में कोई परेशानी नहीं है क्योंकि तू से यहां से और तू से यहां से पास हो रही है तो तू कमा तू ता है बहुत कॉम्प्लिकेटेड तो नहीं हो गया बहुत ज्यादा दिक्कत तो नहीं दे रहा है फिर आई होप आप अभी तक बहुत ज्यादा डिस्टर्ब नहीं हूं इन सारे बनाया है इस तरीके से गया की आप बस इसे देख कर ही वेट कर दें मैं इसके बाद ये क्वेश्चन आसान है बस ये बनाना trikiyan मुश्किल तो नहीं है पर यह कन्ज्यूरिंग है की सर अब क्या तो अब जरा क्वेश्चन पढ़ते हैं और देखते हैं की चीज क्या बोली गई वो पूछ रहा है दें और G3 ये जो आपकी तीन पॉइंट्स हैं कौन-कौन से आपके I3 है ना और यहां पर है जी थ्री और ये f3 तो E3 f3 और G3 के बारे में वो पूछ रहा है वो किस कर्व पर लाइक करेंगे डिले ऑन विच कर्व वो इनमें से किस कर्व पर लाइक करेंगे ये हमें सोच लेना है तो कहां से शुरुआत करें कैसे करें क्या करें सर मुझे तो कुछ समझ नहीं ए रहा है मेरा कहना है की अगर मैं एक सीधे थॉट्स से शुरुआत करूं तो मुझे पता है सर अगर वैन कमा वैन पर एक कोड मैंने बनाई है जो की पैरेलल है एक एक्सेस के तो क्या मैं वो पॉइंट ऑफ इंटरसेक्शन निकल सकूं अब थोड़ा चीजों को सोच के देखना स्टूडेंट्स में क्या कहना चाह रहा हूं मैं आपसे कहने जा रहा हूं बड़ी साधारण सी बात सुनेगा ध्यान से है ना इस वैन कमा वैन से इस वैन कमा वैन से जो आपने टेंशन जो आपने ये पहली बार तो कार्ड ड्रॉ की है ओके लेट्स हैव एक्सेस के पैरेलल तो इसकी इक्वेशन क्या है सर इसकी इक्वेशन है एक्स = 1 यह जो कोड आपने ड्रा किया इक्वेशन क्या है सर इसकी इक्वेशन है ए = 1 अच्छा सर सर्कल के क्वेश्चन है x² + y² = 4 क्या मैं एक्स = 1 और इसे सॉल्व कर सकता हूं पॉइंट ऑफ इंटरसेक्शन निकलने के लिए कौन-कौन से सर एक तो f2 और एक F1 निकलने के लिए आई होप ये बिल्कुल किया जा सकता है तो वैन रखा है एक्स की जगह तो ये हो गया वैन वैन उधर गया तो 4 - 1 3 कौन-कौन से पॉइंट्स हैं जो आपने एक्स = 1 / डी वे ये एक्स = 1 नहीं है एक्स = 1 ये है है ना तो जो आपने पॉइंट निकले कौन-कौन से f2 और F1 तो F1 और f2 के कोऑर्डिनेट्स क्या मिलेंगे सर F1 के जो कोऑर्डिनेट्स होंगे है ना F1 के जो cardinate होंगे वो होंगे अंडररूट थ्री कमा वैन आई होप आई एम नॉट बोथरिंग योर मैं के उल्टा बोल रहा हूं एक्स कोऑर्डिनेट्स वैन होगा ना भाई हम रियली सॉरी फॉर डेट योग वैन कमा √3 और f2 के कोऑर्डिनेट्स कितने होंगे सर एफ तू आपको दिख ही रहा होगा f2 के कार्ड आप कहोगे वैन कमा -√3 बस इसी बेस पे क्या आप एवं और वी2 निकल सकते हो कौन सी बड़ी बात है सर जब आप एवं A2 निकालो जी तो E1 और e2 मेरे ख्याल से तो आप कुछ कहना बहुत मुश्किल नहीं होगा क्योंकि सिर्फ वही बात सर ये वाइज = 1 वो रखोगे तो अंडर रूट थ्री ए जाएगा इसका ए कार्ड नेट फिक्स्ड है एक्स कोऑर्डिनेट्स क्या होगा अंडर रूट थ्री कमा वैन और माइंस अंडर रूट 3 कमा वैन सर ये सब क्यों निकल रहे हो इस बात को प्लीज ध्यान से सुनो ये हो जाएगा ऑफ कोर्स माइंस अंडर रूट 3 कमा वैन ये हो जाएगा प्लस रूट 3 कॉमन अब आप एक बात बताओ ये एक सर्कल है क्या का रहे हो कुछ समझ नहीं आया सर अरे अगर X1 y1 पर मुझे सर्कल पर टांगें ड्रॉ करनी है जो की X1 सर्कल पर लाइक करता है तो तेनजिंग की इक्वेशन क्या होती है सर वो होती है टी = 0 यानी की क्या F1 और f2 वाली जो टैसेंट है कौन सी टेंशन है सर ये f2 वाली टैसेंट और ये F1 वाली टैसेंट जो है उनकी इक्वेशन क्या होगी सर वो हो जाएगी ये एक्स X1 y1 = 4 है ना तो मैं क्या लिखूंगा एक्स एक्स में 1 एक्स एक्स + 1 और इस फोर को इधर ले आए तो -4 = 0 आपकी एक टैसेंट ये हो गई इसी तरीके से इसको पास किया ये जो f2 वाली टैसेंट ये जो लिखना चाह रहा हूं मैं इसकी इक्वेशन अगर मैं लिखना चाहता हूं तो ये है आपकी F1 f3 के अंदर इक्वेशन है ना अगर मैं आपके सामने लिखना चाहूं f2 इफ थ्री की इक्वेशन तो f2 f3 की इक्वेशन को आप क्या लिखेंगे ध्यान से देखिए सर f2 f3 की इक्वेशन को आप लिखेंगे देखो भाई ये f2 है आपका है ना तो एक्स एक्स वैन यानी एक्स + ए ए वैन यानी कितना माइंस एंड √3y है ना और ये माइंस फोर तो इधर ए ही जाएगा अब आपसे बस एक गुजारिश है आप प्लीज इस बात को ध्यान से देखिए आप प्लीज इस बात को ध्यान से देखिए सर इन दोनों इक्वेशंस को ऐड किया जाए अगर इन दोनों इक्वेशंस को ऐड किया जाए तो इससे ये कैंसिल - 8 उधर गया तो 2X अगर -8 की इक्वल है तो एक्स के इक्वल आएगा सर एक्स इक्वल आएगा आपका टेक्निकल फोर के है ना तो f3 का जो एक्स कोऑर्डिनेट्स होगा वो कितना होगा 4 और जब मैं का रहा था की एक्स एक्स तो लाइक करेगा आप खुद देख लो आप खुद देख लो ये मैथमेटिक्स प्रूफ किया अगर एक्स की वैल्यू 4 रखता हूं तो फोर माइंस फोर जीरो तो ए की वैल्यू कितनी आती है जीरो तो ये ए जाता है फोर कमा और चूंकि चीज है semetrical है क्योंकि चीज semetrical है तो क्या आप E3 के अकॉर्डिंग से बात है स्टूडेंट्स फिर वही बात देखो आप रखो ये कितना हो जाएगा अंडर रूट 3X + ए - 4 = 0 ये किसकी इक्वेशन है सर ये है आपकी इक्वेशन E1 E3 की दिस इसे दी इक्वेशन ऑफ एवं आई थ्री सिमिलरली अगर मैं थ्री की क्वेश्चन निकलूंगा तो कितनी ए जाएगी सर वो ए जाएगी माइंस अंडर रूट 3X + ए - 4 = 0 / 2y = + 8 तो ए की वैल्यू 4 ए की वैल्यू 4 रखा तो एक्स की वैल्यू जीरो तो ये कितना ए जाएगा सर ये ए जाएगा 0 4 डू यू ऑल गेट डू यू ऑल गेट डेट नौ व्हाट आर सी एस्क्ड हमसे पूछा जा रहा है f3 G3 और E3 किस कर्व लाइक करते हैं तो मेरा कहना है या तो हम कुछ सोच लें और अगर हम नहीं सोच का रहे हैं तो मैं कहूंगा सर की आपके पास ऑप्शन में ट्राई कर लो ऑप्शन ए में ट्राई कर लो एक्स + ए = 4 फोर प्लस जीरो फोर तू प्लस फोर फोर मतलब हान सर आपका ऑप्शन ए आंसर है क्या ऑप्शन बी आंसर हो सकता है रख के देख लो फोर कमा जीरो सेटिस्फाई कर रहा है तू कमा 2 और 2 का स्क्वायर नहीं हो रहा है नहीं करेगा तो आप देखिए तो नहीं कर रहा है और ये भी नहीं कर रहा है फोर कमा जीरो जीरो कमा फोर नहीं कर रहा है आप देख का रहे हो तो क्लीयरली सर आंसर होगा ऑप्शन है ये इतना आसान इतना ए या इतना सिंपल क्वेश्चन था पर ट्रिक की पार्ट क्या था इसे अच्छे से विजुलाइज करना मैं हर बार आपसे बस यही कहूंगा की इन क्वेश्चंस को सॉल्व करने का सबसे अच्छा तरीका होगा विजुलाइज करना ड्रॉ करना प्लॉट करना अच्छे से ड्रा करो अच्छे से बनाओ आप कुछ चीज दिखे और जितना बारीकी में जितना सफाई से बनाओगे आई थिंक चीजें उतनी आसानी से दिखेंगे मेरी काफी सारी हेल्प शायद इस बात ने कर दी की चीज सिमेट्री के लिए दिख रही थी बनती है तो कुछ चीज मैंने यहां से प्रिडिक्ट की वो यहां मैं ऑलरेडी बना चुका था मेरे कॉमन सेंस है मेरे नॉर्मल बेसिक सेंस है और वो यहां से वैलिडेट हो गई की देखो सर ये एक्सेस पे इंटरसेक्ट करेंगे क्योंकि सिमिट्रिकल है ना चीजे बहुत अब तक सॉर्टेड है और ऐसा आप भी देखोगे आप भी समझोगे आई थिंक आप बातें समझ पाए अगर ये क्वेश्चन आप सभी को समझ आया तो क्या आप नेक्स्ट क्वेश्चन ट्राई करेंगे नेक्स्ट क्वेश्चन में क्या का रहा है suniyega ध्यान से वो का रहा है पी एक पॉइंट है सर्कल पर जिसके दोनों कोऑर्डिनेट्स पॉजिटिव है अब अगर सर्कल पर कोई एक पॉइंट है जिसके दोनों कोऑर्डिनेट्स पॉजिटिव है मतलब वो कहां पर बात कर रहा है वो आपसे इस जोन की बात कर रहा है क्योंकि सर फर्स्ट क्वाड्रेंट में ही दोनों क्वाड्रेंट पॉजिटिव होंगे ये बात से तो आप एग्री कर रहे हो तो यहां पर है ना फिर वो का रहा है एंड दें डी मिड पॉइंट ऑफ डी लाइन सेगमेंट में मस्ट फ्लाई ऑन विच कर्व अब आप बात समझना आप तो सारी बातें छोड़ो आप तो बड़ा बेसिक सा थॉट लगाओ आप बड़ा बेसिक सा थॉट ये लगाओ की सर आपका ही ऑफ कोर्स ए एस एस है मैं यहां पर थोड़ा अलग बना देता हूं क्योंकि वहां पे ऑलरेडी चीज बहुत बुरे तरीके से ना तो वहां और कुछ बनाया है तो आप एक क्लास छोड़ के चले जाएंगे ऐसा मेरा मानना है है ना ये क्या हो जाएगा सर ये हो जाएगा एक्स एक्सिस आप कहोगे की सर अब रहने दो आप है ना आप कुछ तो भी कर लेते हो आपकी मनमानी चलती है कौन सा डायलॉग है तू लकी शासन चल रहा है यहां पर है तानाशाही तो इस बात को समझो वैसे तो हमें फर्स्ट क्वाड्रेंट से ही मतलब है हमारे कम के ही नहीं है सेकंड थर्ड और फोर्थ एनीवे बट फिर भी बात कर लेते हैं वो का रहा है पी एक पॉइंट है लेट्स से एक पॉइंट है इसकी रेडियस कितनी है सर ये जो सर्कल है इसकी रेडियस के लिए कितनी है सर इसकी रेडियस है तो क्या मैं सर्कल पर लाइक करने वाले किसी पॉइंट के क्वाड्रेंट पैरामीट्री फॉर्म में rcostheta और साइन थीटा यानी रेडियस आर यानी तू कोस थीटा और तू साइन थीटा हमने तो हमारे तरीके से कर ही लिया है आप आपके तरीके से बता दो यह आपका सर्कल था बोले तो नहीं हो अब वह का रहा है लेट टीवी डी पॉइंट ऑन डी सर्कल जिसके दोनों koardinate पॉजिटिव है तो आपको इससे इसके बीच उसे रखना है टैसेंट ड्रॉ करनी है सर मेरा जो ज्ञान है मेरा जितना गणित है वो ये कहता है अगर कभी भी किसी पॉइंट पर tenjent ड्रॉ की जाए तो ये तो ऑफ कोर्स आपकी वो टैसेंट होगी सर लेकिन मेरा यह मानना है हमेशा से की ये टैसेंट की इक्वेशन बड़ी आसानी से जीरो तो क्या लिखूंगा तू कोस थीटा सिन थीटा वो बचेगा सर कोस्थेटा सिन थीटा डू यू ऑल एग्री विद डेट नौ व्हाट डू यू मिन बाय दिस पार्टिकुलर थिंग्स आर ये बताओ अब तो वो का रहा है ये जो टैसेंट है आपकी ये एम और एन पे इंटरसेक्ट करती है आपके कोऑर्डिनेट्स को लेट्स दिस पॉइंट इसे एन एंड दिस पॉइंट इसे एन अगर मैं ध्यान से सीधी सोचो अगर मैं ए की वैल्यू जीरो रख देता हूं ए की वैल्यू जीरो रख देता हूं तो एक्स koardinate कितना ए जाएगा 2 / कोस थीटा तो एम का कोऑर्डिनेट्स कितना होगा 2 / कोस थीटा और इसी बेसिस पे कॉमन सेंस लगा लो आप तो एन के कोऑर्डिनेट्स क्या हो जाएंगे सर वो जाएंगे जीरो कमा 2 सिथेंटा कोई तकलीफ तो नहीं है भाई आई थिंक कोई परेशानी नहीं होनी चाहिए किसी भी स्टूडेंट को अब क्या करना है सर वो पूछ रहा है इस लाइन सेगमेंट में का जो मिडिल पॉइंट होगा वह कौन से कल पर लाइक करेगा तो पहले तो मैं में का मिढ्वाइंट निकलता हूं तो मैं में का मिढ्वाइंट कैसे निकलूंगा सर में कमेंट पॉइंट कैसे निकलेंगे तू अपॉन कोस थीटा + 0 / 2 आप समझ रहे हो ना मैं फिर से कहता हूं तू अपॉन कोस थीटा प्लस जीरो अपॉन तू तो तू से तू कैंसिल तो इसका जो मिड पॉइंट होगा लेट्स से वो ये है ना हमें इसके कोऑर्डिनेटिंग निकालनी है जिसको मैं का देता हूं ह कॉम के जैसे मैं का देता हूं के तो ह की वैल्यू कितनी आएगी सर ह की वैल्यू आएगी तू अपॉन कोस θ / 2 तो 2 से 2 कैंसिल तो इसकी वैल्यू आएगी 1 / कोस थीटा सिमिलरली बड़ा ही बेसिक semilarity है के की वैल्यू क्या ए जाएगी 1 / सिन थीटा क्या आप ये सब सोच का रहे हो ये आपके आईआईटी जी के क्वेश्चंस हैं स्टूडेंट्स है ना आप suniyega ध्यान से क्या कोस थीटा को 1 / ह और सिन थीटा को वैन अपॉन के लिख सकते हैं लिख सकते हो वैन अपॉन के सर मुझे एक बड़ी खास बात याद है की कोस स्क्वायर थीटा प्लस साइन स्क्वायर थीटा वैन होता है की नहीं तो हम क्या करेंगे भाई हम लिखेंगे 1 / x² + 1 = 1 अगर सिंपलीफाई करें तो हो जाएगा स्क्वायर के स्क्वायर और अगर मैं एक्स और ए से रिप्लेस कर डन तो कुछ तो मुझे देखना चाहिए क्या देखना चाहिए सर मुझे देखना चाहिए ऑप्शन दी एक्स स्क्वायर + y² = एक और अच्छा क्वेश्चन एक और बेहतरीन सा क्लेरिटी वाला क्वेश्चन अच्छे से समझ ए रही है और आप काफी एंजॉय कर रहे हैं इन लेक्चरर्स को और बहुत ही रिटेल मिनर में समझ का रहे हैं की चीज हम कैसे कर रहे हैं इस प्लेलिस्ट को से करके रखो स्टूडेंट हमने क्या किया है हम सिंग बुक कर रहे हैं यह जो आपकी बुक है जिसके थ्रू आपको डीटेल्ड मैनर में कहां की तैयारी करवाई जा रही है सर आपको आईआईटी जी मांस और ज एडवांस की तैयारी करवाई जा रही है तो आप दोनों एग्जाम को क्रैक करने वाले हैं या दोनों एग्जाम की तैयारी बेसिकली कर रहे हैं क्रैक तो आप करेंगे तो आईआईटी जी मांस और आप किस एग्जाम की तैयारी कर रहे हैं एग्जाम्स की तैयारी आप इसलिए कर रहे हैं स्टूडेंट्स क्योंकि बेशक आपकी यह जो बुक है यह पंच हसन में डिवाइडेड है आपके पूरे 11th एंड 12th के मैथमेटिक्स के सिलेबस को कवर करते हो मतलब इन्होंने इस बुक को पंच हसन में बनता है पंच अलग-अलग बॉक्स में डिवाइड किया है इस पूरे पैकेज को कौन से पार्ट्स पार्ट्स पहले तो हम पढ़ रहे हैं कोऑर्डिनेट्स ज्यामिति से तो हमने शुरुआत की है कोऑर्डिनेट्स ज्यामिति से जिसमें आपके सारी चीज ए रही है इसके बाद हम क्या करेंगे सर इसके बाद हम के कैलकुलस जो की आपका डर इंटीग्रेशन हिसाब होता है फिर क्या कर लेंगे सर फिर हम कर लेंगे एक्टर्स एंड 3D जिससे बच्चे बहुत डरते हैं तो पहले हमने इजी पार्ट उठाया है फिर इसके बाद के दो पार्ट्स स्टफ है जिनसे बच्चों को बहुत दर लगता है पर मैं आपको यकीन dilvaunga की देखो चीज आसान है और उसके बाद क्या करेंगे सर उसके बाद हम उठा लेंगे अलजेब्रा वही सारे आपके टॉपिक्स और आखिरी में हम कर लेंगे ट्रक और जो की मेरा यकीन है आप सभी को अच्छी लगती होगी है ना तो जनरली स्टूडेंट्स इसे और इसे आसान पाते हैं तीन को टू बना लेते हैं मेरा कहना है पांचो ही आसान है कैसे देखते जाएंगे हम आपको कैसे करवा रहे हैं तैयारी क्या हमारी स्ट्रेटजी क्या हमारी प्लानिंग है इस पर बात कर लेते हैं स्ट्रेटजी बड़ी आसान सी है सर स्ट्रीट्स कैसे आसान सी है हम हर एक बुक के हर एक चैप्टर को उठा रहे हैं फिर उसे चैप्टर में क्या कर रहे हैं हर एक कॉन्सेप्ट को पढ़ रहे हैं उसे कॉन्सेप्ट को कंसोलिडेटेड करने के इलस्ट्रेशंस एंड एग्जांपल्स हम कर रहे हैं और फिर ऑफ कोर्स चैप्टर के कंप्लीशन पर उसे चैप्टर के हो जाने पर उसकी हम एक्सरसाइज भी कर रहे हैं भाई साहब और क्या सर और क्या-क्या करवा रहे हो इसके बाद जब वह चैप्टर खत्म हो रहा होता है तो उसके एंड में आपको जी आर का ही यानी आपको जी एडवांस्ड के प्रीवियस इयर्स क्वेश्चंस भी दिए होते हैं अभी तक पूछे गए हम वो भी कर रहे हैं साहब तो इस तरीके से आपकी पुरी तैयारी हो रही है बहुत ही डिटेल कंप्रिहेंसिव और थारो तो क्या नहीं हो रहा है आपका क्या छूट रहा है सब तो करवाया जा रहा है सब तो हो रहा है एम एक इक्वेशन है या सेट है कैसा है जहां पर वह इस तरह की कंडीशन को फुलफिल कर रहा है x² + y² लेस दें इक्वल तू आर स्क्वायर मतलब मैं कहना चाह रहा हूं की एक्स और ए कुछ पॉइंट्स हैं अंग प्लेन एक्स ए में लाइक करने वाले हैं इस तरीके से हैं वो की वो इस सर्कल के अंदर लाई करते हैं कब क्योंकि लेस दें इक्वल तू लगा हुआ है इस सर्कल के अंदर किस सर्कल के अंदर जिसका सेंटर है जीरो कमा जीरो और जिसकी रेडियस है आर याद रहेगा ठीक है सर इसके बारे में बात करेंगे फिर क्या फिर सुनते हैं वह का रहा है कंसीडर डी ज्यामितीय प्रोग्रेशन ए एन जो की है 1 / 2 ^ एन - 1 एन की वैल्यू 123 ऐसे चली जा रही सर ज्यामितीय प्रोग्रेशन कैसे हुआ ये बात करते हैं ज्यामितीय प्रोग्रेशन ऐसे हुआ ये की अगर आप अन में अगर अलग-अलग वैल्यूज आप ट्राई करते हो जैसे एमएफ पर वैल्यू ट्री करते हो जैसे आप A1 निकलती हो तो ये टेक्निकल क्या है 1 / 2 ^ एन - 1 है ना आप जनरलाइज कैसे कर रहे हो सर आप ए एन लिख रहे हो 1 / 2 ^ एन - 1 तो जरा बताओ A1 कैसे लिखा होगा एन की जगह वैन तो वैन माइंस तू की पावर जीरो वैन तो A1 कितना आता वैन अच्छा सर A2 बताओ कितना होगा ये सेकंड टर्म है ज्यामितीय प्रोग्रेशन का तो इनकी जगह आप रख देते तू 2 - 1 तो 2 की पावर वैन तू थ्री हो जाता है 1 / 2 अच्छा सर ऐसे ही A3 निकल रहे होते तो A3 में क्या हो रहा था 3 - 1 2 यानी 1 / 2² यानी 1 / 4 आपको क्लिक हुआ क्या है वैन वैन बाय तू वैन बाय फोर वैन बाय आते सी या ज्यामिति प्रोविजन होता ठीक है सर और क्या और वह का रहा है s0 और एस नोट इस इक्वल्स तू जीरो है ना यह कुछ है यह क्या है सर सन जो है वह डिनोट करता है सैम ऑफ फर्स्ट अच्छा उसने पहले तो आपको एक बात का दी की ये तो बात याद रखना आप एस नोट इस इक्वल्स तू जीरो समझेंगे इसको सैम ऑफ डी फर्स्ट एंड टर्म्स ऑफ दिस प्रोग्रेशन मतलब अगर मैं इसकी फर्स्ट एंड टर्म्स का प्रोग्रेशन निकालो सब निकालो तो सीधी-सीधी सी बात है पहले मैं अगर सन पर बात करूं तो वो क्या होगी सोच के बताओ सन मतलब क्या सन मतलब A1 + A2 + A3 से लेके ए एंड तक A1 + A2 + A3 कहां तक चली जा रही है सर ये बात ये बात चली जा रही है ए एंड तक तो अगर मैं सन निकलना चाहूं तो सन क्या होगा सर ये सारी टर्म्स जो की क्या है सर 1 + 1 / 2 + 1 / 4 और कहां तक जा रही है सर ए एंड तक ए एन कितना है 1 / 2 ^ एन - 1 अब आप इस बात पर गौर फरमाइए क्या आपको ये ज्यामिति प्रोविजन दिख रहा है आप सोचो ज्यामितीय प्रोग्रेशन क्या होता है सर नेक्स्ट टर्म आप अटैं या अचीव करते हो प्रीवियस ये से वैल्यू से मल्टीप्लाई करके सो वैन को वैन बाय तू से मल्टीप्लाई किया 1/ 2 को 1 / 2 से मल्टीप्लाई किया 1/ फोर को वैन बाय तू मल्टीप्लाई तो 1/ 2 आपका कॉमन रेश्यो है यहां पर आपका कॉमन रेश्यो क्या है सर कॉमन रेशों है आपका अमरेली सॉरी कॉमन रेशों कितना है सर आपका 1 / 2 है ना और फर्स्ट टर्म क्या है सर जो फर्स्ट टर्म है वो है आपकी आगे से आप जिसे नोट करते हो वो है वैन और अगर मैं किसी जीपी की एंड टर्म्स के सैम की बात करूं किसी ज्यामितीय प्रोग्रेशन के antrmas के सब की बात करूं तो जो फॉर्मूला हमने पढ़ा है वो होता है ए टाइम्स 1 - आर तू दी पावर एन / आर रिपीट माय स्टेटमेंट यह एक ज्यामितीय प्रोग्रेशन की एन टर्म्स का सैम का फॉर्मूला होता है एंड अगर मैं यहां पर ऐसे लिखना चाहूं तो सन को क्या लिखूंगा ध्यान से देखो भाई ऐसे यहां लिखना चाह रहा हूं बस ध्यान से नोटिस करना और चीज समझते जाना जब आप सन लिखना चाहेंगे सर तो आप सन को क्या लिखोगे सन आपका आएगा ए कितना है वैन लिखने की जरूरत नहीं है 1 - आर तू दी पावर एन आर मतलब 1 / 2 ^ एन तो यहां पर कितना ए जाएगा सर ये ए जाएगा 1 / 2 ^ एन डिवाइडेड बाय फन क्यों निकल रहे हो पूछेगा तो अभी निकल के रख लेते हैं अब सन को मैं सिंपलीफाई कर सकता हूं कोशिश करते हैं ध्यान से देखो भाई इस कोशिश में मैं क्या करने वाला हूं बात समझता हूं बहुत ध्यान से सुनना अगर मैं इसे ऐसा ही लिखा रहने डन तो मुझे क्या दिख रहा है मुझे कम की बात दिख रही है मैं थोड़ा सिंपलीफाई कर रहा हूं देखो इसको ये हो जाएगा 2 - 1 अब देख का रहे हो क्या तो ये हो जाएगा 1 / 2 डिनॉमिनेटर में 1 / 2 न्यूमैरेटर में गया तो तू से मल्टीप्लाई हो रहा है तू से मल्टीप्लाई हो रहा है ये टर्म तो अगर मैं सन सिंपलीफाइड लैंग्वेज में लिखना चाहूं तो सन क्या ए रहा है ध्यान से देखना एस एन ए रहा है तू टाइम्स क्या 1 - 1 / 2 आई होप आप ये देख का रहे हो जो मैं कहना चाह रहा हूं तू ऊपर गया वैन माइंस वैन की पावर एन तो वैन रहने वाला है अपॉन तू डू का रहे हैं तो ये कुछ आपको मिला ये क्या मिला सर आपको सन क्या इस बात से कोई आप बताइए आप चाहो तो इससे सिंपलीफाई कर लो कैसे सेट तू को यहां मल्टीप्लाई करो तो 2 को यहां पे मल्टीप्लाई कर दो आप कर सकते हो तो जैसे ही ऐसा किया तो सन का एक और वर्जन या वेरिएंट क्या मिल सकता है 2 को अंदर मल्टीप्लाई किया तो तू अब देखो ये 2 की पावर वैन अपॉन तू की पावर एन तू डिनॉमिनेटर में ए जाएगा तो 2 के पावर माइंस वैन हो जाएगा तो हो जाएगा 1 / 2 डी पावर एन - 1 तो ये आपका एसएनआर रहा है ऐसा ले लो ऐसा ले लो कोई बड़ा फर्क नहीं पद रहा दिस इसे गोइंग तू बी दी एस एन कोई दिक्कत तो नहीं है सर ठीक है सर बात करेंगे इस बारे में फिर वो आपसे ये का रहा है suniyega ध्यान से न जो आपका सर्कल है वह एक ऐसा सर्कल है जिसका यह सेंटर है और यह रेडियस है अभी जस्ट हमने देखा यह आपका अन है और ऐसे नाम निकल चुके हैं दे आर टॉकिंग अबाउट एस ऑफ एन - 1 है ना वो किसकी बात कर रहा है सन-1 की हमने क्या निकाला है हमने सन निकाला है क्या बस एक बार प्रक्रिया कर लेते हैं स्टूडेंट्स मैंने जो निकाला है वो सन निकाला है क्योंकि अन आपका क्या था 1 / 2 ^ एन - 1 ये एन जो आपका था वो था 1 / 2 ^ एन - 1 और अगर मैं सन निकल रहा हूं हमसे क्या बोल रहा है वो सन-1 है ना हमसे वो बोल रहा है और मैंने निकाला है सन तो बिल्कुल जब आपने सन निकल लिया है सर तो सन-1 निकलना कौन सी बड़ी बात है बस एन की जगह एन - 1 रख देंगे आय होप यू वाला गेटिंग डेट आई रिपीट माय स्टेटमेंट आपने अगर ऐसा निकल लिया सर A1 A2 A3 से लेके एन टर्म्स तक कसम है और वह हमने निकल लिया बड़ा आसान तरीका 1 - आर तू दी पावर है ना अपॉन 1 - आर बिल्कुल सही बात बस इसी बात को सिंपलीफाई करके हमने संस लिखा है खैर उसने जो क्वेश्चन में कहा है आपसे की जो आपका न है वो एक ऐसा सर्कल है जिसके सेंटर है और ये रेडियस है और दी एन आपके वो सारे सर्कल्स हैं जिनका ये सेंटर है और ये रेडियस है पहले तो इनको देख के घबराना मत पहले इस क्वेश्चन को अच्छे से आप पढ़ लो अच्छे से समझ लो अच्छे से देख लो कहीं भी कोई भी डाउट हो किसी भी स्टेप में कोई भी कन्फ्यूजन हो तो अच्छे से पूछ लो आई विल रे क्लेरिफाई मैं आपसे दो बातें लिखना चाह रहा हूं मैं पहले थोड़ा सा का इनको ठीक से डिफाइन करना चाह रहा हूं केन आपके लिए क्या है सर सर ध्यान से देखना अगर आपसे मैं बात करूं न की तो न वो सारे सर्कल्स हैं जिनका सेंटर जिनके सेंटर की अगर मैं आपसे बात करूं जिनका सेंटर है सन-1 कमा जीरो सर सन क्या था वो था 2 - 1 / 2 बार एंड -1 है ना न का जो सेंटर है सर वह है सर वह है सन-1 0 ये आपको दिख रहा है सन-100 बिल्कुल दिख रहा है सर अब अगर मैं सन -1 ढूंढता हूं तो वो कितना है 2 - 1 / ध्यान से सुनना स्टूडेंट्स वो है 2 - 1 / 2 ^ ध्यान से देखना सन में होता है एन - 1 तो सन-एन में क्या होगा एन - 1 - 1 तो ये कितना हो जाएगा सर ये हो जाएगा एन - 2 बहुत डिफिकल्ट तो नहीं हो रहा है स्टूडेंट्स लेक्चर आय होप आप आसानी से समझ का रहे हो और इसी तरीके से अगर मैं बात करूं तो इसका ए koardinate कितना है जीरो तो ये आया आपका इसके सर्कल का सेंटर का कॉटन कोई दिक्कत तो नहीं है अच्छा सर न में एक और बात है सर न में उसकी रेडियस भी दी गई है ए जो की क्या है वैन अपॉन टोटल बार एन - 1 तो इसकी जो रेडियस है ये तो क्या रहा सर आपके न का सेंटर यह है आपका सेंटर के कोऑर्डिनेटर पूछे जाए तो रेडियस आप क्या कहोगे सर रेडियस तो साफ-साफ दिखे रहिए वो है 1 / 2 ^ एन - 2 1 मिनट क्रॉस चेक कर लेते हैं 1 / 2 डी पावर एन - 2 नहीं एंड -1 है ना अब जस्ट थोड़ा सा कोशिश रहेगा क्योंकि बहुत ही अच्छे कॉम्प्लिकेटेड से वैरियेबल्स आपको दिए गए जिनसे आप डील करते वक्त शायद कुछ गलती कर दें तो ये रही बात से इनकी सर कम की बात की है तो न की भी बात कर लें क्या बात कर रहे हो सर दी एन में मुझे दिया जा रहा है उसका सेंटर के कोऑर्डिनेट्स सन - 1 सन -1 तो इनके इस मुझसे कभी पूछा जाएगा दी एंड पर सवाल तो मैं दी एन के सेंटर के cardinate क्या कहूंगा सन -1 तो सर जब कम में आपका सेंटर था 2 - 1 / 2 - 2 तो यहां होगा 2 - 1 / एन - 2 - 1 - 2 और इस बार भी रेडियस तो वही है इस बार भी न हो या दी एंड रेडियस तो वही है कितनी सर ए एन जो की है आपकी कितनी जो की है 1 / 2 डी पावर एन - 1 तो ठीक है सर ये डाटा आपने क्रैक किया ये इनफॉरमेशन हमारे पास कम की है की ये आपका न है और ये आपका दी है कोई दिक्कत तो नहीं है आई होप कोई परेशानी नहीं है अच्छा बस ऐसे ही मेरा मैन हो रहा है की मैं का इंडियन को थोड़ा समझ लूं तो अगर मैं न को समझना चाहूं तो न में मैं क्या कहूंगा ऐसे ही बस थोड़ा सा ध्यान से देखो आप इन सर्कल्स की थोड़ी सी बात समझो इन सर्कल्स को थोड़ा ऑब्जर्व करो जैसे अगर मैं आपसे पूछता है की C1 क्या होगा तो C1 को लेकर आप क्या कहोगे C2 को लेके क्या कहोगे C3 को लेके क्या कहोगे जरा ये सोच के देखो जैसे मुझे C1 निकलना है तो C1 निकलने के लिए मैं क्या करूंगा ये आपके बेसिकली सी एन है राइट ये आपके सी एन है ध्यान से देखो जैसे ही आपने C1 निकलना चाहा बहुत ध्यान से देखो भाई ये आप का कितना ए रहा है ये आपका ए रहा है एन - 2 डिनॉमिनेटर में तो मुझे बता सकते हो क्या की C1 कितना आएगा बड़ी बेसिक इसी बात है इनकी जगह क्या रख दूंगा सर इनकी जगह रख दोगे आप वैन तो जब आप सेंटर निकलोगे तो सेंटर कितना आएगा देखो भाई तू माइंस तू माइंस वैन तू की पावर -1 तो ऊपर गया तो तू माइंस तू जीरो है ना तो C1 के सेंटर के कोऑर्डिनेट्स 0 कमा जीरो यहां पर भी वैन रखा तो वैन माइंस वैन जीरो तू की पावर जीरो वैन अपॉन वैन वैन तो C1 एक ऐसा सेंटर है जिसके सेंटर मतलब C1 एक ऐसा सर्कल है जिसके सेंटर है 0 और उसकी रेडियस है वैन सर ऐसे ही C2 निकलती हैं C2 को लेकर क्या बोलोगे एन की जगह तू रख दो एन की जगह तू रखा तो 2 - 2 0 2 की पावर जीरो वैन तू माइंस वैन तो इसका सेंटर कहां मिलेगा इसका सेंटर मिलेगा वैन कमा जीरो पर सर इसकी रेडियस बता सकते हो क्या इसकी रेडियस कितनी मिलेगी सर वो मिलेगी 1 मिनट ये तो ए एन है ना तो वो तो वही मिलेगा सेंटर पे थोड़ा आपको सोचना है जैसे की मैं पूछता है आपसे सी थ्री तो C3 कितना होता है C3 के लिए सोच लो अब आपको पैटर्न दिख जाएगा और वही मैं समझाना है दिखाना है कहना है बोलना चाह रहा हूं C3 जब आप निकलोगे तो एन की जगह रखो थ्री सन रहे हो आप 3 - 2 1 / 2 सूत्र / 2 तो इसका सेंटर कितना हो जाएगा सर इसका सेंटर हो जाएगा 3/2 कमा जीरो अब बातें समझ का रहे हो क्या और ऑफ कोर्स इसका जो रेडियस होगा वो होगा वैन अपॉन तू तू डी पावर 3 - 2 3 - 2 एंड 2 तो 2 की पावर 2 यानी 1 / 4 आई होप ये बात है आपको कंफ्यूज नहीं करनी है आप आसानी से चीज देख का रहे हो कंफ्यूज हुए बिना अब अगर ये सारी बातें क्लियर हैं अगर ये सारी बातें क्लियर है तो मेरा मानना है की अब सर क्वेश्चंस ट्राई करते हैं पहला क्वेश्चन बहुत ध्यान से देखना बहुत अच्छा क्वेश्चन है बहुत इनसाइड फुल क्वेश्चन है और बहुत मजेदार ये क्वेश्चन जी एडवांस्ड का 2021 का बहुत प्यारा बहुत बहुत मजेदार मतलब बहुत बारीक बहुत मेटिकुलोस ये क्वेश्चन है मतलब बहुत टू नहीं है बट स्टूडेंट्स देख के घबरा जाते हैं क्योंकि उन्हें टर्म्स हुआ तो अजीब लग रही होती है नंबर्स बहुत अजीब लग जाता है पहले देखो क्या लिख रहा है कंसीडर एम विद आर 1025 1325 / 513 है ना एम जो है एम क्या था सर आपका ये जिसमें आर कितना देर है 1025 / 513 जो स्मॉल आर की वैल्यू वो ले रहा है वो है 1025 / 513 ठीक है सर अब क्या करना है अगर यह आपकी रेडियस है तो सर 512 होता है 1024 दो की पावर्स में होते हैं आपको याद है तो तू की पावर नाइन होता है 512 तू की पावर 10 होता है 1024 सुनना ध्यान से 512 का डबल होता है 1024 513 का डबल होता है 1026 खाली भी तू से हल्का सा कम है इसकी रेडियस तू से कम है ये क्या लिखा है सर ये लिखा है r² और ये है रेडियस तो रेडियस जो है वो तू से कम है अब सुनना ध्यान से बात को समझना स्टूडेंट्स यहीं पर आपके सारे इस क्वेश्चन का करेक्ट छुपा हुआ है ये आपका एक्स ए है ना वो एक्सेस तो क्या ही बनाते हो सर आप है ना माशाल्लाह और यह क्या है ना तो एक सर्कल वो हमने लिया है ना और अगर इस सर्कल को हमने यहां पर कुछ इस तरीके से बना पाए तोड़ने को भी अरेंज कर सकते हैं हम आई थिंक यह आपका सर्कल है जिसकी रेडियस कितनी है कोई दिक्कत तो नहीं है नहीं है कोऑर्डिनेट्स हैं जो इस कंडीशन को फुलफिल कर रहे हैं की आपके वो सारे पॉइंट्स वो जो इस सर्कल के अंदर लाइक करें तो अभी हम जिसकी बात करें वो जोन की बात कर रहे हैं जो आपका कैपिटल एम है वो ये सारे पॉइंट्स हैं जो इस इस हिस्से में लाइक कर रहे हैं इसके अंदर क्या मेरी बात समझ का रहे हो और यहां पर क्लीयरली उसने इक्वल का साइन भी लगाया है ठीक है सर तो टेक्निकल एम मतलब इसके अंदर लाइक करने वाले सारे पॉइंट्स जो की क्लीयरली आपका ये पॉइंट है जस्ट थोड़ा आगे कहीं ना कहीं क्या रहा होगा तू है ना यहीं कहीं बीच में कहीं जा रहा होगा वैन और यहीं कहीं पर क्या है सर जीरो कोई दिक्कत तो नहीं कोई परेशानी तो उसे जस्ट पहले आपका यह पॉइंट गुजारा है ना यह कौन सा सर्कल है सर ये आपका ये वाला सर्कल है x² + ए स्क्वायर लेस दें इक्वल तू का मतलब है सर्कल के अंदर और आर मतलब आपकी कोई दिक्कत नहीं और जीरो कमा जीरो पर आपके सेंटर है ठीक है सर आगे बढ़ते हैं क्वेश्चन को नेक्स्ट पार्ट क्या दिया है क्वेश्चन का ध्यान से सुनते जाएगा इस क्वेश्चन का नेक्स्ट पार्ट है लेट के बी डी नंबर ऑफ ऑल दो सर्कल्स न तो सर न की बात हो रही है मैं क की बात करूंगा पर पहले न की बात हो रही है है ना तो सी एन डेट आर इनसाइड एन तो मैं थोड़ा सीन को समझना चाह रहा हूं सर न क्या है सर मुझे जितना याद ए रहा है न ये है जीरो जीरो वैन जीरो थ्री बाई तू कमा जीरो तो जो जीरो पर था उसकी रेडियस थी वैन जो 00 पर था उसकी रेडियस थी वैन तो जो जीरो पर था उसकी रेडियस थी वैन तो सर मेरे ख्याल से वो एक ऐसा सर्कल रहा होगा जितना मैं समझ का रहा हूं है ना उसकी वीडियो से कर वैन थी तो वो कुछ इस तरीके से बनने वाला सर्कल रहा होगा फिर क्या सर यह था आपका C1 है ना फिर C2 क्या था देखो सर सर C2 था वैन कमा जीरो जिसकी रेडियस थी वैन बाय तू सी तू था आपका वैन कमा जीरो परसेंट जिसकी रेडियस थी वैन बाय तू वैन बाय तू मतलब क्या आई थिंक वैन का हाफ आई थिंक वैन का हाफ तो इसको छोटा करो सर ये तो रेडियस 1 हो गई है ना अगर यह पुरी डिस्टेंस आपकी वैन है तो इतना डायमीटर होगा उसका यहां कहीं कोई सर्कल रहा होगा ये आपका C2 है आपको एक बहुत मजेदार सा पैटर्न दिखेगा स्टूडेंट अगर मैं आपसे सी 3 पूछता सर बाय डी वे वैन का हाफ उसकी रेडियस है तो ये वैन है तो ये क्या रहा होगा ये रहा होगा 1.5 यानी 3/2 तो ये जो पॉइंट रहा होगा ये आपका रहा होगा क्या यह रहा होगा 3/2 अब सर अगले पॉइंट की अगर मैं आपसे बात करूं है ना अब अगर मैं बात करूं आपसे C3 की तो सी थ्री क्या रहा 3/2 0 और 1/4 उसकी रेडियस 3/2 और उसे वैन बाय फोर उसकी रेडियस अब सन रहे हो क्या 3/4 3/2 कमा जीरो यानी उसका सेंटर और 1/4 यानी आधे की भी आधी उसकी रेडियस अब ये समझ रहे हो ये आधा है इसका भी आधा है ना मतलब वो यहां कहीं से गुजर रहा होगा वो यहां कहीं से ऐसा गुजर रहा होगा 3/2% और ऐसा कुछ आपको चीज दिख रही हैं कैसे बन रही होगी मैं अब और अंदर जाना नहीं चाह रहा हूं बट ये छोटे होते चले जा रहे हैं मतलब जितना मतलब क्योंकि ज्यामितीय प्रोग्रेशन है और आपकी 1 / 2 की पावर्स हैं तो C1 आपका सबसे बड़ा था है ना फिर आपका ये C2 रहा फिर आपका सी थ्री रहा फिर एक और सी फोर आएगा फिर सी फाइव आएगा फिर सिक्स आएगा ऐसे बनते चले जाएंगे है तो वह आपसे क्या पूछ रहा है वह ऐसे C1 C2 C3 कितने बन रहे हैं इसके अंदर वो बात कर रहे हैं देखो अब क्वेश्चंस समझो क्वेश्चन में क्या लिखा है देखो ध्यान से क्वेश्चन में लिखा है सर क्वेश्चन में लिखा है की वो नंबर है वह सारे कम C1 C2 C3 जो बना सकते हो किसके अंदर इसके अंदर इस एम वाली इस कंडीशन के अंदर क्या है बात आपको समझ आई के बारे में बात कर लेंगे सर 1 मिनट के बारे में बात कर लेंगे पहले एक और वो बात का रहा है लेट एल बी डी मैक्सिमम पॉसिबल नंबर ऑफ सर्कल्स अमंग डीज के सर्कल सच डेट नो तू सर्कल्स इंटरसेक्ट मतलब आपसे वो ये कहना चाह रहा है के तो हम निकल लेंगे के क्या होगा निकल लेंगे तो निकलेंगे अभी जीवन तो बहुत मुश्किल है तो ये भी मुश्किल है तो के तो क्या होंगे की इसके अंदर ऐसे कितने C1 C2 C3 पर अब बैठ के मैं girunga थोड़ी ऐसे बनाऊंगा थोड़ी की सर अगला वाला यहां पे आएगा फिर अगला वाला यहां पे आएगा फिर अगला वाला यहां पे आएगा ऐसे कितने आएंगे ऐसे जोड़ने हैं थोड़ी बैठेंगे हम इतना तो मेहनत नहीं करना है सर हमको हम तो मैथ्स जानते हैं सब और मैथ्स ऐसे आंसर लगा देता है बस आप समझ लो प्रॉब्लम को और मैथ्स प्रेसिस आंसर लता है अनुमान होगा ऑब्जर्वेशन होगा की सर ऐसा करते जाओ तो कब तक बनाऊंगा मैं इसे कोई दे लेके थोड़ी जाऊंगा एग्जाम में जो बनाने बैठूंगा तो मुझे सोचना होगा तो मैथ्स आपकी हेल्प करेगा पर उसके बारे में उसने एल के बारे में एकदम कही के तो ये हो गया एल क्या है एल वैन सर्कल्स की वो बात कर रहा है जो इंटरसेक्ट नहीं कर रहे हैं जैसे की देखो आप C1 C2 इंटरसेक्ट कर रहे हैं लेकिन C1 और C3 इंटरसेक्ट नहीं कर रहा है तो ऐसे किस तरह के कितने सर्कल से जो इंटरसेक्ट नहीं कर रहे हैं उन पर भी हमें बात करनी है वो एल देगा तो के और एल निकलना है कोई तकलीफ तो नहीं है सर के और एल निकलने का मेरा फंड आई होप ये क्वेश्चन आपको समझ ए चुका है के और एल निकलने का मेरा फाइनल फंड आपको बस इतना सा है बस ये सोचना ये हिंट है और उससे क्वेश्चन आप खत्म करोगे की आपको जब कोई कहे की कोई सर्कल किसी सर्कल के अंदर है रिपीट माय स्टेटमेंट जब भी जीवन में आपसे कोई कहे की कोई सर्कल किसी सर्कल के अंदर है तो आप तो बस इतना सा जानते हो कहना सर की उनके सेंटर्स के बीच की डिस्टेंस उनकी रेडी यस या दिए के डिफरेंस से हमेशा कम होती है मैं क्या कहना चाह रहा हूं आपसे मैन लो मेरे पास एक कम करके एक सर्कल हुआ एक सर्कल हुआ सीन कोई सा भी होगा आगे जाकर तो अगर दो सर्कल से एक दूसरे के अंदर मैंने जितना जाना है जीवन में ये की अगर दो सर्कल्स एक दूसरे के अंदर है यह एक सर्कल है एक सर्कल है मैन लो तो हम क्या जानते हैं सर हम यह जानते हैं की इनके सेंटर्स के बीच के डिस्टेंस मैन लो इसका सेंटर यहां कहीं है मैं थोड़ा और उसके सब बना देता हूं ताकि आप समझ पाओ चीजों को जैसे इसका ये सेंटर है यहां और इसका सेंटर यहां तो इनके सेंटर्स के बीच की डिस्टेंस जो है वो हमेशा इनके रेडियस के डिफरेंस से क्या होती है कम इनकी रेडियस का ये डिफरेंस है ये एक रेडियस है और एक की रेडियस ये है तो इनके डिफरेंस से हमेशा कम होती है कौन इनके सेंटर्स के बीच की डिफरेंस जीवन और अगर यह सेंटर है C2 इस बड़े वाले सर्कल की रेडियस थी आर वैन और छोटे वाले सर्कल की रेडियस है r2 तो इनके बीच की डिस्टेंस किस-किस की C1 और C2 के बीच की डिस्टेंस जो है हमेशा कम होती है इनकी रेड आय में बड़ी वाली को R1 मैन लेता हूं छोटी वाली को R1 मैन लेता हूं तो ये कंडीशन आपकी अप्लाई होती चली जाती है आप खुद सोचो ये जो डिस्टेंस है ऑफ कोर्स आप सोचो ना ये R1 है और R1 में से r2 भी सब्सट्रैक्ट किया यह पूरा R1 है इसमें से r2 भी घटाया तो कितना हिस्सा बच जा रहा है सर आपका ये तो बची रहे और ये भी बच रहा है तो ये वो एक्स्ट्रा पार्ट है जो दोनों रेडियस के डिफरेंस को दोनों सेंटर्स के बीच की डिफरेंस से ज्यादा करता है फिर से दोहराता हूं प्लीज इस बात को नोटिस करना यही क्रैकर्स क्वेश्चन का आप समझो बात को इन दोनों सेंटर्स के बीच की डिस्टेंस है तो सर आप खुद सोचो ना ये पुरी रेडियस इसमें से यह रेडियस हटा भी दी तो आपके पास इतना और इतना हिस्सा बचेगा तो आप का रहे हो इतने से तो यह ज्यादा ही है हान ज्यादा है क्योंकि इतना ज्यादा है आप मेरी बात समझ का रहे हो तो जब भी कोई सर्कल अपने अंदर किसी दूसरे सर्कल को समाहित करता है तो कुछ ऐसा सिनेरियो होगा की उनके सेंटर्स के बीच के डिस्टेंस उनके रेडियर के डिफरेंस के से हमेशा छोटी हो गई आई होप आप ये बात समझ में अब अगर मुझसे कोई का रहा है अगर मुझसे कोई का रहा है की आपके एम के अंदर हैं कितने न तो मैं कहूंगा मैं कहूंगा सर की एम और न अच्छा न क्या है सर कम के बारे में कुछ टिप्पणी करने की जरूरत तो नहीं है भाई ये रहा आपका न है और हम आपका यह रहा तो एम और न यह आपका एम और यह फिलहाल एक सर्कल की तरह ट्रीट करूं तो सर्कल क्या है मेरे पास x² + ए स्क्वायर की बात करूं तो न क्या है सर न के बारे में भी ध्यान से देख लो न आपका जो सर्कल है न आपका जो सर्कल है सीन कहां गया भाई काफी कॉम्प्लिकेटेड हो रहा है क्या अभी सीन ये रहा है ना इसके सेंटर के कॉर्डिनेट है ये और ये रही इसके रेडियस कोई दिक्कत नहीं तो सेंटर्स के बीच के डिस्टेंस तो डिस्टेंस फॉर्मूला लगा लो इसकी 0 से यही डिस्टेंस होगी आप समझ रहे हो X2 - X1 का स्क्वायर किया इसमें से जीरो सब जीरो में से जीरो इसके स्क्वायर का अंडर रूट लेंगे तो स्क्वायर से √ कैंसिल तो यही बचेगा तो दोनों के सेंटर्स के बीच की डिस्टेंस यानी की न और एम के बीच की डिस्टेंस यही आने वाली है जो की क्या होगी सर जो की होगी आपकी तू माइंस वैन अपॉन तू तू डी पावर एन - 2 क्या आप मेरी बात समझ पाए ये तो हुई आपकी दोनों के सर्कल्स के सेंटर्स के बीच की डिस्टेंस और ये ये सर छोटी होगी ये किस छोटी होगी इनकी रेड आई के डिफरेंस और बड़ा वाला सर्कल को हमने बड़ा वाला आपका सर्कल यही वाला है सॉरी वो एम वाला है और एम की रेडियस कितनी है सर जो एम की रेडियस हमने निकल वैन जीरो तू फाइव अपॉन 513 है ना तो इनकी रेडियस है सर आपकी 1025 / 513 और इसमें से आप क्या सब क्रैक कर ले रहे हो सर इसमें से दूसरे वाले सर्कल यानी आप न की रेडियस और न की रेडियस कितनी है सर 1 / 2 ^ एन - 1 तो ये कितना है 1 / 2 ^ एन - 1 अब मैं इसे सॉल्व कर लूं तो यह मुझे एन की वैल्यू निकलने हेल्प कर देगा और जो एन की वैल्यू ऑफ निकलेंगे वही तो वह नंबर ऑफ न होंगे जो इस एन के अंदर लाइक कर रहे होंगे अब समझ का रहे हो आई होप आप समझ का रहे हो मेरी बात है ना ये क्या है सर ये बेसिकली डिस्टेंस है ये बेसिकली डिस्टेंस है किस-किस के बीच आपके सी एन और एन के सेंटर्स के बीच है ना डिस्टेंस बिटवीन पता चल जाएगा इससे की सर कितने न आपके यहां लाइक कर रहे हैं क्या यह पार्ट अब सॉल्व करना शुरू करें डायरेक्टली अब इस बात पर आएं सर कोशिश करते हैं की क्या करें एक तरीका कांस्टेंट टर्म्स को इस तरफ शायद बात बन जाए और वही एक अच्छा तरीका होगा या फिर कुछ कंपैरिजन तो कंपैरिजन के लिए यहां एन - 2a एंड -2 है तो वो तो बात हमें अच्छी ज्यादा लग नहीं रही है है ना तो कोशिश करते हैं क्या की मैं कम करता हूं सर एन वाली टर्म्स को उधर और कांस्टेंट टर्म्स को इधर है ना तो ध्यान से देखना बहुत ध्यान से देखना है है ना क्या मैं ऐसा लिख सकता हूं देखो हम ये जानते हैं ये क्या है तू से छोटा है है ना तो उसको इधर ले तो कितना हो जाएगा सर 2 - 1 0 2 5 / 4 कितना 513 साइन < ही रहेगा ये उधर गए तो कितना हो जाएगा 1 / 2 माइंस वैन माइंस वैन कोई आपत्ति किसी भी स्टूडेंट को यहां तक जो हमने किया उससे लेकर कोई परेशानी तो नहीं है अब suniyega 5131026 ड्यूल एग्री विद डेट आई रिपीट माय स्टेटमेंट 513 का डबल 2026 है ना तो ये हो जाएगा 1026 - 1 0 2 5 यानी वैन तो कैन आई से की सर ये तो आपका दिख रहा है 1 / 513 यह तो आपका दिख रहा है वैन अपॉन 513 से इसमें भी कुछ सोच के आइडिया लगा सकते हैं क्या बस इस पे बात करेंगे स्टूडेंट्स अब हम मेरा यह मानना है मेरा ये कहना है की सर इनसे आप कुछ निष्कर्ष क्यों नहीं निकल रहे हो इसको आप थोड़ा ध्यान से देखो इसको प्लीज आप थोड़ा ध्यान से देखो यह क्या है - 2 और ये क्या है तू की पावर एन - 1 इन दोनों को आप थोड़ा ध्यान से देखो आप कुछ का पाओगे कुछ समझ पाओगे कुछ सोच पाओगे क्या का पाएंगे सर हमें तो नहीं समझ ए रहा है अच्छा छोड़ो सारी बातें एन - 1 क्या है जैसे मैन लो मैन लो इसको भूल जाओ अभी थोड़ी क्लियर बस यहां पे कुछ होता है बस इसे भूल जाओ इसे देखो जैसे मैन लो इनकी जगह होता है 5 इनकी जगह होता है फाइव तो ये होता है 5 और ये होता है फाइव ये फाइव होता मतलब 2 की पावर थ्री और ये फाइव होता है मतलब तू की पावर फोर तो वैन अपॉन तू की पावर थ्री पावर फोर और जब इनमें आप डिफरेंस निकलोगे तो आप बड़ी बेसिक सी साधन सी बात पाओगे चाहे तो आप अलजेब्राइक ही सॉल्व करके निकल लो बट ऐसे देख लो अगर मैं कहना चाह रहा हूं तो मैं जो कहना चाह रहा हूं वो ये है जैसे आपके पास लिखा है जैसे 1 / 2 का कब है ना और यहां लिखा होगा 1 / 2 ^ 4 ऐसा कोई एग्जांपल मैं आपको लेके समझाना चाह रहा हूं जैसे की देखो यहां से सुनना बस एक आइडिया देना चाह रहा हूं की जब भी लाइफ में ऐसा ना हो एनकाउंटर करो तो डायरेक्टली सोचना की मैं जो का रहा हूं ये अप्लाई करना देखो आप क्या करोगे आप लिखोगे या फिर तू की पावर 4 - 2q आप लिखोगे तू की पावर 4 - 2q हम वहां भी कर सकते हैं पर मैं कहना चाह रहा हूं प्लीज यहां से देख लो डिनॉमिनेटर में क्या हो जाएगा 2³ 2 की पावर फोर सैम भी कर सकते हैं बट मैं ऐसे ही रख रहा हूं अब आप 2 कब कॉमन ले लो सर तू कब अगर आपने कॉमन लिया तो क्या बचेगा सर 2 - 1 और डिनॉमिनेटर में 2³ 2 की पावर 4 अब आप ये कैंसिल होता है देख पाओगे ये ए जाएगा वैन बाय तू की पावर 4 आप मेरी बात समझ का रहे हो यह हमेशा हो गए तय है यह फिक्स्ड आई होप आप बात समझ का रहे हो आई होप अब बात बहुत अच्छे से समझ का रहे हो अगर ये बात आप समझ का रहे हो तो यही कम हम वहां करेंगे लेकिन वहां शायद आप कन्फ्यूज्ड हो जाएं इसलिए मैंने इसे थोड़ा सा इस तरीके से नंबर्स के थ्रू समझने की कोशिश की आई होप अभी तक चीज बहुत प्रॉब्लम है अभी तक चीज समझ ए रही है अब suniyega ध्यान से इससे यहां और इसे यहां पर इन दोनों को आपस में मल्टीप्लाई किया suniyega 2 की पावर एन - 1 है ना तो न्यूमैरेटर में क्या मिला सर न्यूमैरेटर में मिला 2 की पावर एन - 1 - 2 की पावर एन - 2 की पावर एन - 2 जैसे मैं ऐसे ही रख लेता हूं एन - 1 2 और सर ये जो वैल्यू है ये जो वैल्यू है ग्रेटर दें है 513 से तो दिस इसे ग्रेटर दें वैन अपॉन 513 अभी क्या करना चाह रहे हो सर अब इस पार्ट को ध्यान से देखो बेसिकली बनने वाला है 1 / 2 की पावर एन - 1 अभी जो लोग लिख दिया उससे और फिर भी अगर आपको कन्फ्यूजन है तो क्या मैं पर एन - 2 को क्या मैं इस तू की पावर एन - 2 को ऐसा लिख सकता हूं ध्यान से देखो एन - 2 को एन - 1 - 1 कैसा लिखूं तो आपको कोई आपत्ति या फिर ऐसा लिखेंगे हो जाएगा एन - 2 क्या इस एन - 1 को यह जो पावर लिखिए 2 की पावर एन - 1 है ना ये क्या लिखिए 2 की पावर एन - 1 जो लिखा हुआ है क्या इससे मैं ऐसा लिख सकता हूं सुनेगा ध्यान से 2 की पावर एन - 2 + 1 आई होप इससे कोई फर्क नहीं पद रहा है अब जब आप ऐसा लिख रहे हो तो अगर मैं इससे इससे एन - 2 + 1 में स्प्लिट करना चाहूं तो क्या मैं ऐसा लिख सकता हूं suniyega यहां से आप मेरी बात याद कर का रहे हो तो एन - 2 + 1 को मैं जब लिखना चाह रहा हूं तो मैं लिखना चाह रहा हूं इसे इन तू की पावर वैन क्या आप मेरी बात समझ का रहे हो अब मैं क्या करना चाह रहा हूं मैं 2 की पावर एन - 2 कॉमन ले ले रहा हूं तू की पावर एन - 2 जैसे ही कॉमन लिया तो यहां से कॉमन गया तो यहां पे क्या बचेगा और यहां से सर जैसे ही आपने देखा तो 2 की पावर एन - 2 से 2 की पावर एन - 2 गया और न्यूमैरेटर में तू में से वैन सब्सट्रैक्ट या तो कितना बचा सिर्फ वैन तो ये क्या हो जाएगा 1 / 2 की पावर एन - 1 ये कितना हो जाएगा है अब क्या करना है सर अब सुनो ध्यान से सोच के देखो आप खुद बताओ अभी भी नहीं लग रहा है तो ऐसे सोचो सर इससे यहां और इससे यहां मल्टीप्लाई ऐसे ही कर सकते हैं क्या नहीं है रिसिप्रोकल लेना है ना तो साइन चेंज हो जाएगा तो साइन जब चेंज होगा तो ये हो जाएगा 2 की पावर एन - 1 आई वांट तू बी लेस दें 513 आप समझ का रहे हो ना ऐसे प्रोकल ले रहे हो तो एक क्वालिटी का साइन चेंज हो जाता है अब क्या करना है सर अब अगर मैं आगे आपसे बात पूछूं तो आप खुद देखो मुझे तू की पावर बताओ कोई ऐसी जो 513 से छोटी हो सर 2 की पावर 10 जो होती है तू की पावर 10 जो होती है वो होती है वैन जीरो तू फोर तो तू की पावर नाइन क्या होती है 512 तू की पावर नाइन होती है आपका 10 सॉरी - 1 की वैल्यू ले लेता हूं 10 इन्हें एन की वैल्यू कितनी ले लेता हूं 11 अब जैसे माइंस वैन की वैल्यू आपने 10 ली तो तू की पावर 1024 क्या 1024 513 से छोटा होता है नहीं तो सर आप ये नहीं लेंगे तो आप क्या लेंगे सर आप एन - 1 की कुछ और वैल्यू लेंगे 10 नहीं ले सकते तो क्या लेंगे 10 नहीं ले सकते तो आप लीजिए 9 क्यों सर अब जब आप नाइन लेंगे एंड माइंस वैन की वैल्यू तो तू की पावर नाइन कितना होता है 512 तो एन - 1 अगर 9 है तो एन कितना है 10 एन कितना है स्टूडेंट्स सर एन है 10 अगर एन है 10 अगर इन है 10 इसका मतलब यह हुआ सर इसका मतलब यह हुआ की आपने पहले क्वेश्चन के पहले पार्ट को सॉल्व 10 तक के सर्कल्स उसे एम के अंदर लाइक करेंगे मतलब जो हमने सर्कल्स बनाए द ना जो हमने सर्कल्स बनाए द यहां पर यह जो सर्कल बनाए द कहां तक बात जाएगी उससे छोटे होते होते कहां तक जाएंगे उसे c10 तक जाएंगे जो की बेशक आपका इसके अंदर ही रहेगा मतलब हुआ आई थिंक आप के की वैल्यू ढूंढ ले आपके की वैल्यू ढूंढ ले आपने के की वैल्यू निकल ली अब सर जब आपने के की वैल्यू निकल लिए तो क्या आप हमारी एल निकलने में हेल्प करेंगे एल की वैल्यू क्या होगी मेरे ख्याल से एल तो और आसान कम है सर एल में क्या था एल के बारे में वो क्या का रहा है ज्यादा से ज्यादा जो एक दूसरे को इंटरसेक्ट नहीं करते मैं कोई भी दो सर्कल आपस में एक इंटरसेक्ट ना करें तो सुनना ध्यान से बड़ी बेसिक सी बात का रहा हूं मैंने जो ऑब्जर्व किया वो ये की सर C1 C3 C5 C7 और C9 आपस में इंटर सेट नहीं कर रहे द है ना एक छोड़ के कर रहे द और वैसे ही C2 C4 c6 c8 और c10 आपस में इंटर से नहीं का रहे द तो जो मैक्सिमम नंबर ऑफ सर्कल से जो उसमें कोई भी दो आपस में एक दूसरे को इंटरसेक्ट ना करें वो या तो इनमें से कोई पंच उठा लो एन में से कोई पंच उठा लो तो दोनों ही केस में पंच हैं ये पंच भी ऐसा एक पैर है जिम कोई भी दो आपस में इंटरसेक्ट नहीं कर रहे द और ये पंच भी उन सर्कल्स का पहले दिन में कोई भी दो आपस में इंटरसेक्ट नहीं कर रहे द अपनी बात समझ का रहे हो तो कहना इसे सर जहां के की वैल्यू आपने निकल 10 कैसे निकल सर वो c10 के थ्रू और जहां एल की वैल्यू आप निकल रहे हो क्या पंच इसके थ्रू एल की वैल्यू कितनी निकल रही हो पंच तो जब आपको के और एल की वैल्यू पता चल गई है तो क्या अब आप आंसर मार्क नहीं कर सकते अगर आपको के और एल की वैल्यू पता चल गई है तो सर अब तो आप आंसर बता सकते हो ना जैसे देखो मैं क्या करता हूं मैं के की वैल्यू रखता हूं 10 है तो यहां रखा 10 प्लस 10 कितना होता है 525 ये भी गलत है है ना के की वैल्यू रखता हूं 10 तो ये कितना हुआ 20 प्लस एल की वैल्यू कितनी है तो ये होता है 35 ये भी गलत है है ना के की वैल्यू रखते हैं 10 यानी कितना 30 एल की वैल्यू रखते हैं 5 या 10 और 30 + 10 = 40 बिल्कुल सही तो अब पहले क्वेश्चन का आंसर निकल पाए जो की है ऑप्शन दी आप इस तरीके से मार्क करेंगे बहुत टू बहुत डिफिकल्ट क्वेश्चन तो नहीं था बड़ा बेसिक सा कॉन्सेप्ट था जहां पर हमने सीक्वेंस सीरीज और भी कई सारी चीजों को उसे किया अब अगर मैं इससे आगे बढूं और इस क्वेश्चन पर तो इस क्वेश्चन में क्या लिख रहा है वो एक-एक करके इस पार्ट को समझते हैं यहां पे उसने क्या ले लिया यहां पर उसने दी एन ले लिया फिर वही बात इस बार एम कुछ लिया है और नंबर ऑफ ऑल दो सर्कल्स पूछ रहा है जो एम के अंदर है फिर वही कॉन्सेप्ट फिर वही कॉन्सेप्ट सर क्या कॉन्सेप्ट की आप जब निकलोगे डंक डंक और एन के सेंटर्स के बीच के डिस्टेंस जब आप इनके सेंटर्स के बीच के डिस्टेंस निकलोगे इन दोनों के बीच की तो वो होनी चाहिए कम किस इन दोनों के रेड आय के डिफरेंस इन दोनों के डिस्टेंस जो सेंटर्स के बीच की डिस्टेंस है वो इन दोनों के रेड है किसके न और एम के रेड आय से छोटी होनी चाहिए बस यही कॉन्सेप्ट लगाना है कोई तकलीफ बट इस बार एम की रेड आई चेंज हो गई इस बार एम की रेड आई है ये और दी एन क्या था सर दी एन में जब मैं एम से निकलूंगा अच्छा बाय डी वे एम का सेंटर क्या था पहले तो मैं डीएम तक पहुंचता हूं मुझे न कहीं मिल जाए मैंने निकल रखा है क्या बिल्कुल निकल रखा है सर अब suniyega ध्यान से डीएनए अच्छा सर एम का सेंटर क्या है सर एम का सेंटर जो है वो है जीरो कमा जीरो है ना और इस बार सेकंड क्वेश्चन के लिए जो दी गई है आपको वो कितनी दी गई है एम की जो रेडियस आपको दी गई है एम के जो रेडियस है वो कितनी है सर वो है तू की पावर 19 9 - 1 है ना सुनेगा ध्यान से यहां पर जो रेडियस है वह की पावर 198 टाइम्स अंडर रूट 2 यह आपको रेडियस दी है फिर वही बात इन दोनों सेंटर्स के बीच की जो डिस्टेंस है वो इन दोनों रेड आय के डिफरेंस से हमेशा कम होगी अगर न यानी डीएनए जो सर्कल है ये इसके अंदर लाइक कर रहा है तो ऐसे कितने न होंगे वो हमें फिगर आउट करना है कोई दिक्कत तो नहीं कोशिश करते हैं सर आप सोचो इसका एक्स और ए koardinate से है जीरो कमा जीरो से डिस्टेंस वो आने वाली है ध्यान से देखना √2 टाइम्स सॉरी आपके अंदर वाला सर्कल तो ये क्या है तू की पावर 19 9 - 1 है ना ये कितना है सर 2 की पावर 198 इसका कितना टाइम्स इसका अंडर रूट 2 टाइम्स करोगे इसमें से आप सूत्र करोगे इसकी रेडियस जो की है 1 / 2 बार एंड माइंस वैन इसकी कंफर्म कर लिया बस कुछ भी गलती की तो पूरा पार्ट गलत हो जाएगा न विचार इनसाइड एंड एम विद आरजे नंबर ऑफ ऑल डी सर्कल बिल्कुल सही बात 199 हमारा जो पार्ट है वह है 199 - 1 / 198 बिल्कुल सही अंडर रूट तू ये एक्सप्रेशन है और मेरा वापस आपसे ये कहना है की इस एक्सप्रेशन को भी आप लोग बहुत आसानी से कर सकते हो बिना ज्यादा क्वेश्चन को फ्लकचुएट किए लेकिन सर इस बार कैसे सोचेंगे इस बार क्या करना होगा इस बार क्या थॉट प्रेस प्रक्रिया होगी इस बार कैसे दिमाग लगाना है जरा आइडिया दीजिए मेरा कहना है की आप तो जा सकता है बट कैसे वो सोचते हैं सर कोई तरीका कोई आइडिया कोई भी एप्रोच आपके दिमाग में ए रही हो तो बताओ कैसे सोचेंगे स्टूडेंट हेविंग अन्य थॉट्स की इसको कैसे वर्कआउट किया जा सकता है क्या किसी ने स्क्रीन पॉज करके उसको सोचा इसको निकलने की कोशिश की से वैसा ही है लगभग बस इस बार कुछ थोड़े से महंगी उससे छोटी मोती चेंज हो रहे हैं यहां पर मैं जो सोचना चाह रहा हूं ये की सर एक कम कर लेते हैं ये वाली जो टर्म्स है ये वाली जो टर्म है क्योंकि अंडर रूट तू वाला पार्ट और ये अंडर रूट तू वाले पार्ट में शायद कुछ मैं कर पाऊं तो देखो मैं क्या करने वाला हूं या फिर मैं ऐसे उधर पहुंचा दो बाकी सारी चीजों को इधर लाओ जैसा आपको ठीक लगे वैसा कर लीजिए या तो आप ये 2 ^ एन - 2 वाला जो पार्ट है इसे वहां शिफ्ट कर दीजिए ठीक है ऐसा कर लेते कोई बात बहुत खास फर्क पढ़ना नहीं है तो √2 का टाइम दिस मैंने वहां शिफ्ट किया कोई दिक्कत तो नहीं भाई कोई परेशानी तो नहीं जैसे ही वहां शिफ्ट किया तो मैं सुधार लिख देता हूं तो मैं क्या लिख रहा हूं अंडर रूट तू अपॉन ऑफ कॉस्ट कितना एन - 2 अब इस दौरान क्या-क्या हुआ उसे बारे में बात करते हैं ऑफ कोर्स आपका ग्रेटर दें साइन अब बाकी सारी चीज मैं इधर लेके आया है बहुत ध्यान से सुनना 2 टाइम्स अंडर रूट तू और ये भी है अंडर रूट तू ओवरऑल कॉमन कॉमन है तो √2 जो ओवरऑल कॉमन आया है उसमें से देखो ये तो इधर ही था तो ये रहा तू माइंस क्या माइंस हुआ ये वाला पार्ट ये कौन सा पार्ट है सर जिसमें लिखा हुआ है 2 की पावर 199 - 1 / 8 किसी भी स्टूडेंट को इस बात से कोई आपत्ति और क्या हो जा रहा है सर ये वाली जब तुम ए रही है तो ये हो जा रहा है प्लस वैन माइंस वैन कोई दिक्कत तो नहीं है अब एक छोटा सा कम देखो तू की पावर 198 यहां मल्टीप्लाई हो रहा है माइंस साइन अंदर पूरा ओवरऑल स्प्रेड कर रहा हूं मैं है ना तो क्या हो रहा है है वह देखना 2 की पावर 198 जैसे ही मल्टीप्लाई हुआ तू से तो कहना ऐसे हो जाएगा 2 की पावर 199 ये हो जाएगा - ये हो जाएगा प्लस और इस पूरे का एलसीएम हो जाएगा 2 ^ 19 का क्या फायदा है आपको दिख रहा है तू की पावर 199 से 2 की पावर 199 गो कैंसल्ड सो फाइनली सी आर लेफ्ट विद वैन अपॉन तू डी पावर 198 सो सी आर लेफ्ट विद वैन अपॉन तू डी पावर 198 में से अलग से लिख देता हूं है ना तो दिस इसे 1 / 2 की पावर 198 और उससे मल्टीप्लाई हो रहा है √2 तो ये एक्सप्रेशन मुझे कुछ ऐसा दिख रहा है यहां तक कोई आपत्ति किसी भी स्टूडेंट को कोई परेशानी है तो पूछो अब मैं जो छोटा सा कम एक और करने जा रहा हूं वो ये की सर ये जो पावर आपको दिख रही है तू की कौन सी भाई ये वाली जो पावर है क्या मैं इसे वहां शिफ्ट करूं क्योंकि मैं एन वाली टर्म से अब डील करना चाह रहा हूं एक साथ तो मैं क्या करना चाह रहा हूं मैं बड़ी बेसिक सी ऑपरेशन इससे मैं यहीं रखता हूं और मैं इसे वहां शिफ्ट कर देता हूं इससे मैं बाहर रखना चाह रहा हूं और अब बात करो लास्ट इसी के लास्ट क्वेश्चन किया था डिनॉमिनेटर में तू की पावर एन - 2 - 2 न्यूमैरेटर में अंडर रूट तू है तो कुछ सोचा जा सकता है क्या याद करो वही ऑपरेशन और अगली स्टेप आप आसानी से सोच सकते हो इस थॉट के साथ जो हमने जस्ट अभी डिस्कस किया है जहां पर हमने बात की है की सर आप सीधे सीधे कॉमन लेकर सोच सकते हो मैं क्या कहना चाह रहा हूं उसे बात को ध्यान से देखो मैं आपसे कहना चाह रहा हूं बड़ी सिंपल सी बात इससे वहां शिफ्ट करो पहले तो तो ये क्या है सर 1 / 2 ^ एन - 1 है ना यहां गया तो ये हो जाएगा 1 / 2 ^ एन - 1 कोई आपत्ति तो नहीं है भाई अब इसमें से क्या करना है वो समझना आप समझो बात को ये यहां मल्टीप्लाई कर दो ये यहां मल्टीप्लाई कर दो चलेगा कोई दिक्कत नहीं है बट इतना सोचने के लिए बट ठीक है चलो ऐसे करना है तो ऐसे कर लो ये वाला साइन मैं थोड़ा इधर लगा लूं तो कोई दिक्कत तो नहीं है ना अब तू की पावर एन - 1 यहां मल्टीप्लाई हुआ तो ये हो जाएगा 2 की पावर एन - 1 10 तो ये कितना हो जाएगा ये हो जाएगा ऑफ कोर्स आपका 2 ^ एन - 2 पूरा एक्सप्रेशन आपका कमेंट एक्सप्रेशन मिल जाएगा जिसमें 2K पावर एंड माइंस वैन हो तू की पावर एन - 2 4 रहे हैं और ऑफ कोर्स यहां पे क्या लगेगा ये माइंस साइन कोई दिक्कत तो नहीं अब बड़ी बेसिक सी बात स्टूडेंट्स क्या तू की पावर एन - 1 को क्या तू की पावर एन - 1 को मैं लिख सकता हूं गौर से देखना 2 की पावर एन - 2 + 1 - 1 ही है आप देख का रहे हो तो और जो ये प्लस वैन आपने लिखा है सर इसके लिए यहां पर एक तू की पावर वैन ऐसे लिख सकता हूं कोई बड़ी बात नहीं है ऐसा क्यों कर रहे हो ऐसा देखने से देखो 2 की पावर एन - 2 तू की पावर एन - 2 यहां से कॉमन लिया और वो इस तू की पावर एन - 2 से कैंसिल किया क्या आप ज्यादा कंफ्यूज तो नहीं हो रहे हैं भाई ये चीज आपको समझ ए रही हैं आसानी से और तू की पावर वैन का मतलब क्या होता है सर 2 की पावर वैन का मतलब होता है बेसिकली तू क्या आपको अब कुछ चीज नजर ए रहे हैं बहुत ही सिंपल सी शॉर्टकट्स ये सिंपलीफाइड से वैसे और थोड़ा अच्छे से लिखना चाहूं है ना मैं थोड़ा अच्छे से लिख देता हूं ताकि आप कुछ दिख जाए √2 -1 न्यूमैरेटर डिवाइडेड बाय तू की पावर एन माइंस विद दिस पार्टिकुलर एक्सप्रेशन नौ अगेन यू दिस चांस तू लुक आते दिस पार्टिकुलर एक्सप्रेशन एंड तेल मी व्हाट नीड्स तू बी डैन थैंक्स आर क्यूट सिंपल थिंग्स आर वेरी सिंपल बस हिंट यही है की एक कम करो सारी चीजों को अंडर रूट तू में ले जाओ आई रिपीट माय स्टेटमेंट अगर मैं हर एक चीज को अंडर रूट तू में ले आया मतलब इसे भी अंडर रूट तू में ले आए ऐसे भी अंडर रूट 2 में लिया है तो क्या हम कुछ सोच पाएंगे देखो आई एम नॉट टचिंग दिस पार्ट मतलब इसे एक तरफ रखते हैं इसे यहां लेट हैं तो तू की पावर एन - उनको यहां लेकर आए हैं ना हम रियली सॉरी तू की पावर एन माइंस साइड है ना अब क्या मैं ये कर सकता हूं 2 की पावर 198 को 2 की पावर 198 को मैं सब कुछ अंडर रूट तू में चाह रहा हूं इसलिए क्या मैं लिख सकता हूं अच्छा बाय डी वे पहले देता हूं 198 से बाद में बात करेंगे तू को मैं क्या अंडर रूट तू का स्क्वायर लिख सकता हूं और इसकी पावर क्या है 198 तो आई थिंक ऐसा लिख सकता हूं मैं क्या इस बात में आपको बॉर्डर तो नहीं किया से ये जो तू लिखा हुआ है क्या इसे भी अंडर रूट तू √2 का स्क्वायर लिख सकता हूं क्योंकि मेरे पास सारी चीज फिर अंडर रूट तू में ए जाएंगे ऐसा क्यों कर रहे हो सर आप खुद रिलाइज करोगे उससे फायदा क्या होगा यहां तक तो कोई दिक्कत नहीं है भाई अब क्या करने वाले हो सर अब देखो हम चीजों को थोड़ा एडजस्ट करते हैं चीजों को थोड़ा सेट करते हैं है ना इससे कीप ऑन था साइड ओनली बट इस पार्ट को एमआईएम तू सॉल्व ध्यान से देखना सर एक चीज तो आप नोटिस करो √2 √2 बेस से तो पावर ऐड होगी तो 2n - 2 में वैन ऐड किया तो 2n - 1 तो न्यूमैरेटर में आई विल हैव अंडररूट तू की पावर 2 की पावर कितना तू एंड माइंस वैन 98 सॉरी दिस इस गोइंग तू बी 396 यह पावर वहां जाएगी तो 2n - 1 - 396 दिस इस गोइंग तू बी माइंस 397 सो दिस इसे गोइंग तू बी ध्यान से देखना स्टूडेंट्स अंडर रूट तू √2 तू दी पावर 2 जस्ट ऑफ दिस क्वेश्चन ये कितना था 200 - लिख देता हूं ये कितना है दिस वाज 200 एक्सप्रेशन वेरी केयरफुली आप बहुत अच्छा बेसिक आइडिया लगा पाओगे इसे देख करके सर अब तो क्वेश्चन आसान है अब तो मैं क्रैक कर सकता हूं हिंट है यू लुक आते डेट पार्ट 2√2 -1 अगर आपसे मैं पूछूं तू रूट तू माइंस वैन को लेके आपका क्या विचार है तो आप क्या कहोगे बड़ी बेसिक सी बात है स्टूडेंट सोच के देखो अंडर रूट तू कितना होता है सर अंडर रूट 2 इस 1.4 का डबल सर 1.4 का डबल 2.8 2.8 में से वैन सब्सट्रैक्ट किया 1.4 का डबल कितना आई थिंक 2.2 इस एप्रोक्सीमेटली कितना 1.8 है ना दिस इस dropsimatli 1.8 ये जो पार्ट है 1.8 है ना एप्रोक्सीमेटली हमसे है ना मुझे ये चाहिए ये जो एक्सप्रेशन है ये जो एक्सप्रेशन है जो की क्या है वैसे एप्रोक्सीमेटली लिख रहा हूं ना ये जो एक्सप्रेशन जो की है 1.4 ये जो पावर है ये कितनी है 2n -397 अब आप खुद सोच के देखो 1.4 की पावर 2 भी हुई 1.4 की पावर तू भी हुई तो यह बियोंड चला जाएगा आप समझ का रहे हो मेरी बात को लेट तू पॉइंट समथिंग हो जाएगा मतलब 14 का पावर आप सोच के देखो ना 196 तो एप्रोक्सीमेटली 1.4 का पावर कितना हो जाएगा 1.96 मतलब तू क्या है पास पास अराउंड 1.8 तक ही जाना चाह रहे हैं वो भी एक एप्रोक्सीमेटेड गैस पे मैं कम करूं तो मैं बहुत ही एप्रोक्सीमेटेड साइड है लेना चाह रहा हूं की क्या मैं ऐसा का सकता हूं की 2n - 397 12 मैक्सिमम वैन तक हो क्योंकि ये सब intesers हैं क्योंकि नंबर ऑफ सर्कल्स हैं ना ये तो ये तू नहीं हो सकता मतलब 2n - ये 2n - 397 तू तक भी नहीं जा सकता सुनना क्यों का रहा हूं क्योंकि अगर ये 2 तक गया अगर ये पावर तू तक गई तो 1.4 की पावर 1.4 की पावर तू 1.8 के भी ऑन चली जाएगी मेरी बात समझ का रहे हो आई कनॉट एवं हैव दिस हज तू और तू से छोटा इंटिगर कौन होता है सर वैन तू से छोटा अगर वैन होता है तो मैं कहूंगा की ये जो पावर सर आपने ली थी क्या 2n -3 197 दिस विल बी इधर इक्वल तू वैन और लेस दें वैन क्योंकि तू तो नहीं हो का रहा है वह पावर 2 तक तो नहीं जा रही है वर्ण इधर गड़बड़ हो जा रही है इससे सिंपलीफाई करते हैं तो ये हो जाएगा 2n हम सॉरी ऐसे अपरोक्ष आपको लेने होंगे जब क्वेश्चंस इतने ट्रिकी होंगे 398 सो दिस इस गोइंग तू बी थ्री 98 तो 398 का हाफ आई थिंक सर 400 का आधा 200 कम है तो 199 तो एन कितना होना चाहिए सर एन शुड बी लेस दें 199 एंड शुड बी लेट्स दें इक्वल तू 199 और नंबर ऑफ जो मुझे मिले सर्कल्स वो कितने हैं सर बी याने सर जब आपकी ये रेडियस होगी तो इसके अंदर जो इस कैटिगरी के सर्कल्स लाइक करेंगे ना वो इतने लाइक करेंगे यानी कौन सा ऑप्शन आप मार्क कर देंगे आप मार्क कर देंगे ऑप्शन बी क्या सारे स्टूडेंट्स को एक क्वेश्चंस समझ आए आई टोटली अंडरस्टैंड एंड एक्सेप्ट ये क्वेश्चन अच्छे से दे varsum रियली गुड कैटिगरी क्वेश्चंस की सर आर एस एक डायमीटर है ठीक है किसका एक सर्कल का जोगी है बहुत मजेदार सा सर्कल क्यों सर क्योंकि इसका सेंटर है ओरिजिन पर और इसकी रेडियस है वैन तो ज्यादा सोचने की जरूरत नहीं है अब वेयर एस इस डी पॉइंट कमा जीरो तो सर इस सर्कल पर एक पॉइंट है कौन वैन कमा जीरो आई होप यू आर गेटिंग डेट सर आपने ऐसा क्यों कहा इस सर्कल पर alproof डेट वैन कमा जीरो वैन जीरो तो ये हो जाता है वैन तो बिल्कुल सर स्लाइस ऑन दिस सर्कल लेट पी अन वेरिएबल पॉइंट तो सर दो तो कौन द आर एस जो की था एक डायमीटर उसमें हमारे पास एक पॉइंट कौन था एस और अब वो ले रहे हैं aquariable पॉइंट आदर दें आरएनएस तो आर और एस जो आपके डायमीटर के एक्सट्रीम है उनके अलावा दे आर टेकिंग अनदर पॉइंट पी इस तरीके से इस सर्कल पर ऑफ कोर्स क्या का रहा है वो टैसेंट और टैसेंट तू डी सर्कल आते सर बहुत बड़ा हो रहा है ये क्वेश्चन बहुत ज्यादा नहीं बार ए रहे हो मेरा क्या नाम रुक जाओ देखो इतनी सारी इनफॉरमेशन आप रिटेल नहीं कर पाओगे तो सर क्या तरीका है ड्रॉ करिए जो जो पढ़ते जा रहे हो उसे प्लॉट करो यू नो पुट इट इन योर एक्सप्रेशंस आप अपनी भाषा में उसे ड्रॉ करिए डी बेस्ट थिंग यू कैन डू इस ड्राई इट ऑफ कोर्स फिर से पूरा पढ़ने में अब टाइम वेस्ट कर रहे होंगे तो सर एक कम करते हैं ना एक-एक लाइन पढ़ते जाते हैं प्लॉट करते हैं क्योंकि काफी सारी इनफॉरमेशन इट्स बटोर डेट सी ड्रॉयड एंड दें विजुलाइज ओ इमेजिन डेट व्हाट इस गोइंग तू हैपन और व्हाट ऑल आर थिंग्स विच आर हैपनिंग सो डेट सी कैन वेरी इजीली और कन्वेनिएंट सॉल्व दिस क्वेश्चन एंड हान बटोर और एफिशिएंट मैनर तो देखो भाई क्या लिखा हुआ है सर पहले तो एक सर्कल है x² + y² = 1 ऑल राइट है लेट्स प्लॉट डेट सर्कल हमारे पास क्या होगा हमारे पास होगा आपका ए एक्सिस राइट और इसी तरीके से आपके पास क्या होगा सर आपके पास होगा एक्स एंड आप इस तरीके से ए एक्सेस और एक्सेस आपके पास होंगे बिल्कुल होंगे सर तो ये रहे आपके ए एक्सिस और एक्स कोई दिक्कत तो नहीं है भाई मेरे ख्याल से तो कोई परेशानी किसी भी स्टूडेंट को नहीं होनी चाहिए तो यह रहे आपके वॉइस एक्सेस करवाना चाहता हूं तो आप आसानी से देख पाएं चीज एंड देयर कॉम दी सर्कल x² + y² = 1 कोई दिक्कत तो नहीं है भाई आई थिंक चीज आसान है और चीज आपको अच्छे से समझ ए रही हैं सेंटर डेट ओरिजिन एंड इट हज इट्स रेडियस वैन है ना तो सर ये रहा आपका ओरिजिन जो की ऑफ कोर्स ओ जिससे आप का दें जीरो कमा जीरो अब क्या बात करना चाह रहे हो सर अब वो ये बोलना चाह रहा है एस एक पॉइंट है वैन कमा जीरो है ना एस जो है वो एक पॉइंट है वैन कमा जीरो तो आर एस जो आपका डायमीटर है उसमें जो एस पॉइंट है डेट इस दिस विच इसे वैन कमा जीरो और सर अगर एस वैन कमा जीरो है एंड दिस आर इस गोइंग हैपेंस तू बी दी मतलब आर एस जो है आपका फेस्ट तू बी डी डायमीटर ऑफ डी सर्कल तो शैल आय से जो आर होगा वो यहीं होगा और आर के cardinate में क्या कहूंगा सर वो मैं कहूंगा माइंस वैन कमा जीरो डू यू ऑल गेट की आरएसएस क्यों है क्योंकि एस ये है और स्ट्रेस डायमीटर है अब क्या करना है सर अब अगर मैं आगे बढ़ता हूं तो मुझे अगला ब्लू क्या मिलता है अगली इनफॉरमेशन जब मैं प्राप्त होती है वो ये एक पी एक वेरिएबल पॉइंट है ना हमने tenjans ड्रा किए हैं सर्कल पर कहां-कहां एस और पी पर एस और पीपल arandum पॉइंट पी ऑन aneware ऑन डी सर्कल एंड मेक सर्कल पर एक पॉइंट बी मैन लेता हूं तो सर एक रैंडम पॉइंट जब आपने पी माना तो लेट से ये वो लेट्स रन पॉइंट है पी अब अगर मैं पैरामीटर एक फॉर्म में इस पॉइंट पी को मानना चाहो तो हम जानते हैं सर आर कोस थीटा आर सिन थीटा सर्कल की रेडियस है वैन तो आर की वैल्यू हो जाएगी वैन तो रैंडम पॉइंट पी के अगर पैरामीटर एक फॉर्म में हम क्वाड्रांट्स मानना चाहे तो क्या मैं इसे वैन टाइम्स कोस थीटा और वैन टाइम साइन थीटा लिखूं तो आपको कोई आपत्ति ऑपरेशन ही तो नहीं होगी आई होप यहां तक सारी चीज स्मूथ है अब वो ये कहना चाह रहा है प्लीज इस बात को सर आप दो टेंशन ड्रॉ करते हो कहां कहां आप एस पर और पी पर टेंसेज ड्रा करते हो तो ठीक है हम एस और पीपल टेंसेज ड्रा कर लेते हैं आई एम गोइंग तू ड्रा वैन ऑफ डी टांगेंट्स आते फेर आते आते दिस पॉइंट एस है ना विच इस क्लीयरली नथिंग बट व्हाट दिस इस क्लीयरली पैरेलल तू आपका क्या एक्स एक्सिस है ना ये क्लीयरली सर आपके एक्सेसिव के पैरेलल होगी ये बात से आप सब एग्री करते हो सर एक और टैसेंट आपको ड्रॉ करनी है जो की आप कहां पर ड्रा करेंगे सिर्फ जो की आप ड्रॉ करेंगे पॉइंट पी पर इस बात से किसी भी स्टूडेंट को कोई परेशानी सर आप एक और टैसेंट ड्रॉ करेंगे जो की आप ड्रॉ करेंगे पॉइंट पेपर सो इस तरीके से आपने पी पर और एस पर सर्कल की तेनजिन स्ट्रोक की है और ये दोनों टांगें वो का रहा है कहां मिलती है इंटरसेक्ट आते पॉइंट के सो देयर पॉइंट ऑफ इंटरसेक्शन इस गोइंग तू बी के ये जो पॉइंट है सर ये पॉइंट दिस इस के इसे डेट ओके विथ एवरीवन सर आप खुद सोच के देखो अगर एस पॉइंट पर आपने tagent ड्रॉ की है सो डू यू नो ड्राई की आय एम रियली सॉरी एक्स नहीं दिस इसे गोइंग तू बी पैरेलल विद ए एक्सेस तो ये ए एक्सिस के साथ पैरेलल होगी अब अगर ये एक्सेस के साथ पैरेलल है तो के का डिनर तो मैं जानता हूं वैन ही होगा क्योंकि एस विश्वास हो रही है जो की वाइरस हम निकल लेंगे कुछ ना कुछ करके देख लेंगे समझ लेंगे यहां तक किसी भी स्टूडेंट को कोई आपत्ति कोई परेशानी अगर मैं इस तेनजिंग की क्वेश्चन आपसे पूछूं तो आप क्या कहेंगे मैं कहूंगा सर वही एक्सेस के पैरेलल है तो इसकी इक्वेशन क्या हो जाएगी एक्स = 1 किसी भी स्टूडेंट को कोई दिक्कत आई थिंक आसान है सर आप लिख सकते हो आप चाहो तो कैसे सर्कल क्या है सर कल है x² + y² = 0 सर एक पॉइंट है जहां पर मैंने सर्कल की टांगें ड्रॉ की है जो की पॉइंट पी जो है वो सर्कल पर लाइक करता है डी टांगें इस गिवन की ग = 0 तो टी = 0 का क्या मतलब हुआ एक्स कोस थीटा तो मैं इसे क्या लिखूंगा मैं ऐसे लिखूंगा एक्स कोस थीटा प्लस ए ए वैन यानी ए साइन थीटा है ना आई होप ये आपको याद है जीरो नहीं है हम रियली सॉरी फॉर दिस दिस वैसे बेसिकली वैन तो ये हो जाएगा एक्स कोस थीटा प्लस फाइव साइन थीटा इसे इक्वल तू वैन दिस इसे दी इक्वेशन ऑफ दिस टेंट इट कोई तकलीफ तो नहीं है सर अगर आप सीधे सीधे पॉइंट ऑफ इंटरसेक्शन तक ही बात लेकर आना चाह रहे हो तो पॉइंट ऑफ इंटरसेक्शन को लेकर आपका क्या ओपिनियन होगा मैं कहूंगा बड़ी सिंपल सी बात है ये एक स्ट्रेट लाइन है यह आपकी पी के स्प्रे लाइन है जिसकी इक्वेशन हमने सर्कल की टैसेंट फॉर्म में इस तरीके से दी मेरा कहना है मुझे स्ट्रेट लाइन पीके मैं जो पॉइंट के है उसका एक cardinate पता है तो अगर इसका एक्स कोऑर्डिनेट्स यहां पास किया तो क्या के पॉइंट का मुझे ए कार्ड पता चल जाएगा बेशक पता चल जाएगा तो अगर एक्स की वैल्यू वैन पास करता हूं आई होप आप देख का रहे हो वैन पास किया तो कोस थीटा यहां गया तो दिस इसे गोइंग तू बी 1 - कोस थीटा / सिन थीटा तो ये कितना हो जाएगा माइंस कोस थीटा अपॉन कितना अपॉन साइन थीटा स्मूथ है आसान है सो अभी तक तो चीज जैसी जैसी दी गई है वर्कआउट किए हैं अब आगे क्या बोल रहा है वो ध्यान से देखिए अभी डी नॉर्मल तू डी सर्कल सर आपने पी पॉइंट पर सर्कल का एक नॉर्मल ड्रॉप है ठीक है सर बात करते हैं हमने इस पी पॉइंट पर सर्कल का एक नॉर्मल ड्रॉप किया है आई होप आप जानते हो की सर सारे सर्कल के नॉर्मल्स उनके सेंटर से पास होते हैं सो ये जो नॉर्मल आपने ड्रा किया है पी पर इस सर्कल का वो क्लियर ही है और नॉर्मल की इक्वेशन को लेकर मेरा ज्यादा कुछ कहना है नहीं सर इसका सर्कल का सेंटर है जीरो कमा जीरो दूसरा पॉइंट है कोस θ कमा सिन थीटा तो ये जो रहा आपका ये है नॉर्मल तू दिस सर्कल आते पी व्हाट आई विल बी टोल्ड ये जो नॉर्मल है आपका सर्कल का पी पर वो इंटरसेक्ट करता है ऑनलाइन ड्रोन थ्रू क्यों पैरेलल तू आर एस लाइन का क्या मतलब है यह थोड़ा ध्यान से देखना है मैं फिर से रिपीट करता हूं आपने क्या किया आपने क्यों ये जो फिर से suniyega आपका जो पॉइंट है देखो ध्यान से यहां पर पहले देखो यहां पर हमारे पास ये पॉइंट है ध्यान से उसने क्या किया आपने फिर से सुनना आपका जो नॉर्मल है पी वो पी से पी पॉइंट पर जो आपने नॉर्मल ड्रॉ किया है वो मिलता है सर्कल से कहां पर नॉर्मल तू जो सर्कल आते पी इंटरसेक्ट ए लाइन ड्रॉन थ्रू थ्रू के पैरेलल तू आर एस पैरेलल तू आर एस नौ फॉरगेट एवरीथिंग कैन यू हेल्प मी इन राइटिंग और ड्राइंग आलिंगन पैरेलल तू आर एस अब देखो ये रस है डेट इस क्लीयरली नथिंग बट सक्सेस तो एक लाइन एक्स एक्सिस के पैरेलल तो मैं एक लाइन एक्स एक्सिस के पैरेलल ड्रॉ करता हूं कुछ पासेस थ्रू के आय होप आपको ये बात समझ ए रही है तो अगर हमने ये करने की कोशिश की जो बेसिकली एक लाइन है जो की एक्स के पैरेलल है और जो की क्लीयरली आपकी कहां से पास होती है अगर वो के से पास होती है सो दिस इसे विथ डेट लाइन ऐसे में थोड़ा ऐसे ड्रॉ कर लूं तो आपको कोई आपत्ति तो नहीं है उन्होंने कहा आपसे ये जो के से पास होने वाली लाइन है जो की रस के पैरेलल है वो इस नॉर्मल को जो आपने पी पॉइंट से ड्रॉ किया है सर्कल पर उसे ए पर इंटरसेक्ट करती है तो ये जो पॉइंट है आपका दिस दिस पॉइंट हर ओवर हर इस नथिंग बट आई तो ये जो पॉइंट है आपका दिस इस ए एवरीवन विद मी सो फार कोई डाउट तो नहीं है स्टूडेंट्स कोई परेशानी तो नहीं अब क्या अब जब वो पूछ रहा है आपसे क्या इनफॉरमेशन पूछ रहा है दें डी लो कास्ट ऑफ आई तो लोकस्ट जो ए का होगा उसको कौन से पॉइंट सेटिस्फाई करेंगे कौन से पॉइंट्स क्यों भाई क्योंकि ये कौन सा कैटिगरी का क्वेश्चन है मल्टीपल करेक्ट आंसर टाइप क्वेश्चन आप सारी बातें भूलिए आप क्या मुझे सबसे पहले ये फिगर आउट करने में हेल्प कर सकते हैं की ये जो नॉर्मल सर आपने ड्रॉ किया है और ये जो केवी की इक्वेशन आपने ड्रॉ की है ये दोनों चीजें आप कैसे बनाएंगे उससे हमें शायद ए पॉइंट को पॉइंट ऑफ इंटरसेक्शन के थ्रू निकल पाऊं मैं जो कहना चाह रहा हूं इसे फिर से सुनिए स्टूडेंट्स क्या सबसे पहले तो आप मुझे आप की इक्वेशन बता सकते हैं सर कौन सा बड़ा टास्क है आप यानी नॉर्मल आप नॉर्मल के थ्रू निकल दो या ऐसा नहीं समझ ए रहे तो देखो पीके cardinate से ओके कोऑर्डिनेट्स हैं तो 0 0 और कोस्थेटा सिथेंटा से पास होने वाली स्ट्रेट लाइन की इक्वेशन आई होप बहुत डिफिकल्टी नहीं है सो ना ध्यान से ए - y1 यानी ए = एन एम नहीं पता तो देखना Y2 - y1 अपॉन X2 - X1 तो सिन थीटा / कोस थीटा तन थीटा एक्स - X1 सो एक्स - 0 सो ए = 10 थीटा एक्स या ए = एक्स तन थीटा जो भी आप कहना चाहो तो ये जो आपका आप ए रहा है आप की इक्वेशन क्या है दिस इसे वही इसे इक्वल तू तन थीटा टाइम्स एक्स डू यू ऑल एग्री विद डेट आई होप नो वैन इस कन्फ्यूज्ड विथ दिस मच इनफॉरमेशन कन्वेट सो फार नौ व्हाट डू यू उपाय ऑन ऑन डी इक्वेशन ऑफ के आप के की इक्वेशन को लेकर क्या ओपिनियन रखते हो आई वुड से ये तो बड़ा आसान आंसर टास्क के कैसे अब बात समझो ना के जो है आपकी लाइन के ए जो आपकी स्ट्रेट लाइन है डेट इस पैरेलल तू एक्स एक्सिस जो की आर एस के पैरेलल था विच वाज नथिंग बट एक्स-एक्सिस पर लाइक करने वाली लाइन तो मैं कहूंगा कोई लाइन एक्स के पैरेलल है तो उसका ए koardinate उसकी जो वही जो एक्स एक्सिस से डिस्टेंस है ना ए एक्सिस पर जो कवर कर रहा है डिस्टेंस वो कितनी है 1 - कोस थीटा / सिन थीटा तो shailise ये जो के ए लाइन है इसकी इक्वेशन नथिंग बट वही इसे इक्वल तू वैन माइंस कोस थीटा / सिन थीटा करना होगा अगर मैं चीज विजुलाइज ना करूं तो ये जोन ही नहीं पाएगा जिसे हम ढूंढना चाह रहे हैं अब सारी बातें छोड़ो इफ आय विश तू गेट दी यू नो cardinate ऑफ आई व्हाट मी गोइंग तू डू आप खुद सोच के बताओ सर अगर आपको ए के cardinate चाहिए तो आप क्या करोगे बेसिकली फाइंड डी पॉइंट ऑफ इंटरसेक्शन ऑफ के एंड ओ पी नथिंग बट व्हाट आई तो सर जो आपका के ए और आप का पॉइंट ऑफ इंटरसेक्शन है वो ए है क्या बात आपको समझ ए रही है शायद इससे मैं ए के इसको डिफाइन कर पाऊं बिल्कुल सही बात है सर तो हम किसका पॉइंट ऑफ इंटरसेक्शन निकलने की जरूरत है हम के ए और आप का पॉइंट ऑफ इंटरसेक्शन निकलने वाले हैं तो कोशिश करते हैं सर सी आर फाइंडिंग के यू आर आप तो आप क्या था सर आप किस स्ट्रेट लाइन है ए = एक्स टाइम अपॉन साइन थीटा आई थिंक सर वैन ऑफ डी वैरियेबल्स कैन बी लिमिटेड और इससे आप शायद एक्शन निकल पाओ क्या थॉट है क्या विचार है सर देखो ध्यान से यहां भी हुआ है यहां भी हुआ है तो ये वाला पार्ट और ये वाला पार्ट एक करते हैं तो तन थीटा को मैं लिख देता हूं क्या वैसे लिख देता हूं सिन थीटा / कोस θ टाइम्स ऑफ कोर्स एक्स यहां पर क्या लिखा है सर यहां पर लिखा है 1 - कोस थीटा 1 साइन थीटा क्या आप मेरी बातों को समझ का रहे हैं आई थिंक मैंने क्या किया ए की वैल्यू यहां से यहां से इक्वल रख दी क्योंकि वो पॉइंट ऑफ इंटरसेक्शन हम निकलना चाहें इन दोनों इक्वेशंस को हम सॉल्व करना चाह रहे हैं बस मेरा आपसे ये कहना है जरा इस इक्वेशन को ध्यान से देखिए क्योंकि इसे आप बहुत आसानी से सॉल्व करके मुझे पॉइंट ऑफ इंटरसेक्शन दे सकते हैं और बता सकते हैं की वो पॉइंट क्या आएगा आप बड़े आराम से बड़े ही इजीली वो पॉइंट ऑफ इंटरसेक्शन निकल सकते हैं बिना ज्यादा मेहनत किए अन्य वैन हेविंग अन्य थॉट्स की सर कैसे सोचेंगे क्या करेंगे किसी के दिमाग में कोई आइडिया आप से मेरा कहना है सर एक्स की वैल्यू निकलती हैं तो एक्स की वैल्यू निकलने के लिए क्या करेंगे सेंड दिस सिन थीटा / कोस थीटा 2 डी आदर साइड ट्राई करते हैं तो देखो क्या होने वाले हैं सर एक्स को यहीं रखा है तो दिस इसे गोइंग तू बी वैन माइंस कोस थीटा है ना और अगर यहां पर देखा डिनॉमिनेटर में तो ये कितना हो रहा है सर ये हो जा रहा है सिन थीटा है ना क्या हो रहा है ये सिन थीटा यहां पर ए रहा है दिस इस गोइंग तू बी साइन स्क्वायर थीटा सर इसे थोड़ा और सिंपलीफाई करते हैं बिल्कुल करेंगे तो अगर इससे थोड़ा और सिंपलीफाई मैंने किया या फिर अगर मैं चाहूं अगर मैं चाहूं और कनक्लूड करना चाहूं तो क्या मैं ये एक्स की वैल्यू यहां रख सकता हूं उससे मुझे ए कार्ड मिल जाए बट एनीवे आई ऍम ट्राईंग तू गेट रीड ऑफ व्हाट आई एम ट्राईंग तू गेट अराउंड ऑफ दिस थिंग इन दोनों में फाइनली रिलेशन निकलना था तो मैं अगर यहां से कुछ चीज उसे करना चाहूं तो क्या कर सकता हूं सर ध्यान से देखो मैं कहूंगा पहले ध्यान से देखिएगा मैं बड़ी बेसिक सी बात आपसे कहूंगा की पहले तो मैं लिख लेता हूं 1 / sin² थीटा वो कितना हो जाएगा कोस एक्स स्क्वायर थीटा और फिर मैं लिखूंगा अमरेली सॉरी ये होगा कोस्थेटा / sin² थीटा सो था डोस नॉट मेक मच ऑफ डी सेंस अगर इसके बजाय मैं यहां पर अगर कुछ और सोचता तो कुछ और तरीका क्या हो सकता था आई थिंक पैरामेट्रिक फॉर्म के अलावा कोई और तरीका तो बात नहीं बना रहे ना नहीं बना रहे सर तो मेरे ख्याल से यही करना ज्यादा ठीक आइडिया होगा इसी तरीके से हम इसको सोचना होगा एक तरीका मेरे दिमाग में यहां से ए रहा है suniyega ध्यान से क्या की सर अगर सोच रहे द कैसे यहां से तन थीटा निकलता जो की क्या होता है ए / एक्स अगर तन थीटा पता चल जाता है तो क्या उससे मैं कोस्थेटा और सिन थीटा निकल सकता हूं और वो जो वैल्यूज आएंगे उन्हें यहां पर प्लेस कर देंगे तो शायद एक्स और ए में एक रिलेशन हो जाए दिस इस अनादर थॉट हम हेविंग आप ऐसे भी कर सकते हो बट मुझे लग नहीं रहा की हम इससे ज्यादा चीज सोच पाएंगे और ऑल टाइम में से कॉम्प्लिकेट कर रहे हैं और घुमा रहे हैं सो आय ऍम स्किपिंग दिस पार्ट मैं इस पार्ट को हटा रहा हूं और मैं इस पार्ट को इस तरीके से सोच रहा हूं ध्यान से देखिएगा स्टूडेंट्स मैं इस पार्ट को हटाकर मैं तन थीटा से चीज निकलने की कोशिश कर रहा हूं देखो यहां से क्या निकल सकते हो सर यहां से लिख सकते हो तन थीटा जो होगा वो होगा ए / एक्स अब अगर मैं बेसिक सी बात आपसे करो अगर मैं एक बहुत ही बेसिक आइडिया आपको डन तो आप क्या कहोगे सर मैं आपसे कहूंगा की देखो बड़ी बेसिक सी बात है ये एक 90 डिग्री का आपका क्या है राइट एंगल मैन लो थोड़ी देर के लिए 19 डिग्री का राइट एंगल ट्रायंगल तन थीटा क्या होता है परपेंडिकुलर अपॉन बात समझ रहे हो तो आपका हाइपोटेन्यूज कितना हो जाएगा x² + y² बहुत ध्यान से सुनना कम की बात है उससे क्या उससे अगर मैं ये वाली इनफॉरमेशन ऑफ कोर्स ये आपने क्या माना था θ तो क्या यहां से आप मुझे दो चीज बता सकते हो सर साइन थीटा कितना होगा सिन थीटा होगा आपका परपेंडिकुलर अपॉन हाइपोटेन्यूज तो वो हो जाएगा ए / कितना मतलब आप चाहते तो यहां पे भी इन दोनों को क्या कर सकते द मेरे ख्याल से न्यूमैरेटर और डिनॉमिनेटर दोनों को डिवाइड भी करते हैं तो भी वहां कोस थीटा तो लगता ही लगता है तो निकल ही लेते हैं है ना तो सिन थीटा कितना होता है सिन थीटा हो जाता है ए / √ ओवर क्या x² + y² एंड आय होप आई नीड नॉट टेल यू कोस थीटा क्या हो जाएगा सर कोस थीटा आपका हो जाएगा एक्स अपॉन अंडर रूट ओवर x² + y² ड्यूल एग्री विद डी आई थिंक सर अब चीजे आसानी से दिख रही हैं कुछ होती हुई अब मेरा आपसे बस ये कहना है की ये जो वैल्यूज आपने निकल हैं सिन थीटा और कोस थीटा की लैट्रिन प्लेस डेम हर तो मुझे जो वैल्यू मिलेगी ए की तो वही इसे गोइंग तू बी वैन माइंस कोस थीटा विच इस व्हाट एक्स अपॉन अंडर रूट ओवर व्हाट एक्स स्क्वायर + y² कोई कन्फ्यूजन तो नहीं है सिंपलीफाई दिस फादर व्हाट आई गोइंग तू डू मल्टीप्लाई दिस हर तो मुझे क्या मिलेगा सर मुझे मिलेगा ए = मुझे जो मिलेगा वो होगा ए = क्या √ ओवर x² + y² - एक्स इसके अपॉन में ही आएगा और इसके भी अपॉन में आएगा विच विल गेट कैंसल्ड बाय एच अदर तो नॉर्मेलिटी में क्या बचेगा ए क्या ये बात आप सभी समझ का रहे हो कुछ और तो चीज आपको नहीं दिख रहे हैं आई थिंक यहां पर एक्स से देखना था तो आई थिंक अगर आप ऐसे सिंपलीफाई करेंगे तो आप देखेंगे y² + एक्स है ना यहां से ये ए और ए इधर ए जाएगा तो दिस इसे गोइंग तू बी ए स्क्वायर प्लस एक्स तो जो लो कस आपको मिलेगा फाइनली वो क्या होगा वो आपको मिलेगा ऐवेंंचुअली क्या भाई ए स्क्वायर करने हैं और मुझे बताना है की इनमें से ऐसा कौन सा ऑप्शन है जो इसे सेटिस्फाई कर रहा होगा तो एक-एक करके ट्राई करते हैं पहले ऑप्शन ए ट्री करेंगे जब ऑप्शन ए ट्राई किया तो देखो भाई क्या ए रहा है ध्यान से देखो ऑप्शन ए रहा है क्या 1/3 और 1/√3 है ना तो देखो ये एक्स की वैल्यू वैन बाय थ्री हम रखते हैं और ए की वैल्यू कितनी है 1 / √3 तो ए स्क्वायर कितना हो जाएगा 1/3 क्या यह इसके पहले देखो जरा बहुत ध्यान से वैन बाय थ्री का स्क्वायर कितना 1 / 9 + y² कितना 1/3 और इसका हमें क्या निकलना है अंडर रूट आय थिंक ये हो जा रहा है आपका 2/3 तो लेफ्ट हैंड साइड पर ये आर हेविंग तू बाय थ्री ये कितना है ध्यान से देखो सर बड़ी बेसिक्स की बात मैं आपसे कहना चाह रहा हूं क्या इसे मैं 3/9 लिख सकता हूं मेरा बस मैन है क्या मैं इसे थ्री बाय नहीं लिख सकता हूं उससे क्या हो जाएगा सही हो जाएगा 4/9 और 4/9 का अंडर रूट कितना सर वह 2/3 तो सर ऑप्शन ए तो आपका सेटिस्फाई कर रहा है ऑप्शन ए आपका क्लीयरली इस लोकस को सेटिस्फाई कर रहा है मतलब ऑप्शन ए तो लाइक करेगा थोड़े बहुत ऑप्शन दी ट्राई करते हैं सर 1 / 4 1 / 2 सो दिस इस गोइंग तू बी वैन बाय फोर ए स्क्वायर वैन बाय तू वैन बाय तू का स्क्वायर 1/4 कोई दिक्कत तो नहीं अब देखो ध्यान से राइट हैंड साइड पर एक्स कितना है सर 1 / 4 1 / 4 का स्क्वायर 1/16 ए कितना है 1 / 2 1 / 2 का स्क्वायर 1/4 अब समझ का रहे हो y² मतलब 1 / 4 मैं 4 से मल्टीप्लाई करता हूं तो ये हो जाएगा 4 / 16 आप सन रहे हो क्या सर क्लीयरली में देख का रहा हूं 4 + 1 हो जाएगा फाइव फोर प्लस वैन हो जाएगा 5/16 तो वो मेरे ख्याल से इसके इक्वल नहीं है है ना तो सर ऑप्शन बी तो आपका मैच नहीं हो रहा है मतलब ऑप्शन भी इस लोकाज को सेटिस्फाई नहीं कर रहा है व्हाट अबाउट ऑप्शन सी 1/3 और -1/√3 आप प्लीज इस बात को समझना इट दस नॉट मैटर अगर आप माइंस ऑन माय रूट 3 भी लेते हैं क्योंकि जहां जहां ए है वो ए स्क्वायर है तो आई थिंक सर ऑप्शन ऑप्शन सी सेटिस्फाई करेंगे क्योंकि ऑप्शन ए एंड ऑप्शन सी आर आईडेंटिकल सर नहीं है आईडेंटिकल आप देखो ना ये नेगेटिव साइन है तो इट देस नॉट मैटर क्यों सर क्योंकि यहां भी Y2 यहां भी y² है तो अन्य वे वो वही बन जाएगा आप मेरी बात से एग्री करते हो क्या ऑप्शन दी देखते हैं 1 / 4 - 1 / 2 सर वही बात वापस अगर ऑप्शन बी नगरी नहीं किया तो ऑप्शन डीबी सेटिस्फाई नहीं करेगा क्योंकि ऑप्शन दी में भी वे पार्ट नेगेटिव है और यहां पर भी ए और ए नेगेटिव है तो shailise है सर ज्यादा सोचने की जरूरत नहीं है आप माफ करेंगे ऑप्शन ए और सी बिना ज्यादा हेसिटेशन के वीडियो ऑल अंडरस्टैंड हो इजी दिस क्वेश्चन वैसे एंड इफ यू आर क्लियर इफ यू आर क्लियर विद योर एप्रोच और आप अगर इसे ठीक से प्लॉट करते हैं ठीक से ड्रॉ करते हैं और एक सही माइंड सेट से इस क्वेश्चन को एप्रोच करते हैं तो इट इसे नॉट गोइंग तू बी अन वेरी डिफिकल्ट क्वेश्चन बट चीजें इस तरीके से बहुत बहुत क्लीयरली बहुत बहुत क्लीयरली आपको इमेजिन करनी होंगी वर्ण अब कुछ का कुछ कर रहे हो आई थिंक बाकी चीज त्रिकों ने हमारी हेल्प कर दी क्योंकि पैरामीटर एक फॉर्म बहुत बहुत कन्वेनिएंट फॉर्म में जो हमारी हमेशा हेल्प करती है नौ देर कॉम एंड आदर कैटिगरी ऑफ क्वेश्चन जहां अब हम बात करेंगे किसकी यहां पर हम बात करेंगे मैट्रिक्स मैच टाइप क्वेश्चंस की मैट्रिक्स मैच टाइप मतलब मैचिंग करनी है अब इस क्वेश्चन को अगर मैं आपसे बात करूं तो ये क्वेश्चन जो आपके इंगेज बुक में दिया है अब ये सर्कल चैप्टर में उन्होंने पैराबोला लेप्स और हाइपरबोला के भी क्वेश्चंस दे दिए हैं तो अभी हम इसका बी सी और दी पार्ट नहीं करेंगे बेटा आप चिंता मत करिए जब हम पैराबोला एलिप्स और हाइपरबोला पढ़ रहे होंगे तब बेशक में ये पार्ट्स करवाऊंगी आई विल रिमेंबर दिस दिस इस माय रिस्पांसिबिलिटी यू जस्ट वेरी अबाउट दिस पार्ट आते दिस स्टेज आज के लेक्चर में आप सिर्फ इस पार्ट को देखो की अगर मुझे एक सर्कल की इक्वेशन चाहिए तो इनमें से कौन सा ऑप्शन मैच करता है अब चूंकि मैं चीज है जानता हूं मैंने चीज सॉल्व की है मुझे पता है तो मैं सीधे आपको आंसर बताता हूं और क्यों है वो आंसर वो बताता हूं क्योंकि हम पूरा क्वेश्चन सॉल्व नहीं कर रहे हैं मतलब एप्रोच क्या होती है इस क्वेश्चन को सॉल्व करने की मैं एक एक पार्ट सॉल्व करता हूं पांचो को सॉल्व करता है और फिर देखता की किसकी किस नैपिंग होनी है बट क्योंकि आपने ये तीन चैप्टर नहीं पढ़े हैं अभी इन्हें हम पढ़ने वाले हैं सर्कल के बाद हम पैराबोला फिर हाइपर बोला ही पड़ेंगे तो तब आपको ये चीज है डिस्कस करूंगा और हमेशा के मैं वापस करूंगा तो मैं जानता हूं आंसर क्या है तो मैं आपसे सीधे आंसर डिस्कस कर रहा हूं की पी के इसका आंसर है देखो भाई डी लोकस ऑफ डी पॉइंट ह कमा के फॉर विच डी लाइन एचसी + की = 1 टच दिस सर्कल है ना तो एक सर्कल है आपका क्या एक सर्कल है सर आपके पास जो की है x² + y² = 4 और वो ये कहना चाह रहा है इस सर्कल को ये लाइन कौन सी ह एक्स + के ए एचसी + के ए = 1 टच करती तो मैंने तो जीवन में बस इतना सिखा है सर ये जब भी कोई लाइन किसी सर्कल के लिए टांगें की तरह बिहेव करें तो सर्कल के सेंटर से उसे सर्कल के सेंटर से उसे लाइन की परपेंडिकुलर डिस्टेंस उसकी रेडियस की इक्वल होनी चाहिए तो क्या स्क्वायर डिवाइडेड बाय h2 + K2 इनसाइड एंड अंडर रूट ये जो है सर ये किसके इक्वल है ये क्लीयरली होगा इसकी रेडियस के जो की कितना है तू किसी भी स्टूडेंट को इस बात से कोई आपत्ति मेरे ख्याल से तो नहीं होनी चाहिए तो जब आप इसे सिंपलीफाई करेंगे तो क्या दिखेगा ब्रिंग दिस हर और 2 को यहां लेट हैं तो ये हो जाएगा 1 / 2 है ना और वैन बाय तू की इक्वल क्या हुआ अंडर रूट ओवर x² + K2 ये कितना हो जाएगा अंडर रूट ओवर h2 प्लस के स्क्वायर अब फाइनली अगर मैं कंक्लुजन पर आऊं तो मैं क्या कहूंगा मैं कहूंगा सर स्क्वायर कर देते हैं तो जो एक्स स्क्वायर प्लस के स्क्वायर है अंत में आप ह और के को एक्स ए सिर्फ प्लेस करते हैं तो दिस इसे गोइंग तू बी x² + y² = 1 / 4 और जितना मेरा नॉलेज कहता है सर ये जो इक्वेशन है ये क्लियर सर्कल की इक्वेशन है डू यू ऑल रिलाइज एन एक्सेप्ट डट की सर ये क्लीयरली आपकी एक सर्कल की इक्वेशन है और इस केस में आप मार्क करेंगे ए के लिए पी सर बाकी ऑप्शंस का क्या ये हम डिस्कस करेंगे आने वाले चैप्टर के दौरान बट अभी फिलहाल बस के पार्ट्स मुझे ये पार्ट हम कब पढ़ेंगे जब हम क्या कर रहे होंगे आपके parabolis और हाइपर बोला जो लास्ट क्वेश्चन था आपका मैट्रिक्स मैच टाइप क्वेश्चन लेकिन इससे पहले आपको उसमें दिए गए एक पैसेज एक-एक कंप्रेशन कुछ डायरेक्शन कुछ सेट ऑफ रूल्स से डील करना है तो मैं पहले तो इस क्वेश्चन को समझने की कोशिश कर रहा हूं की सर्कल मुझे कौन-कौन सी दिए गए हैं तो सर सबसे पहले मुझे दो सर्कल दिए जा रहे हैं कौन-कौन से अगर आप इन सर्कल्स को गौर से देखें तो सर इस सर्कल के बारे में मेरा सीधा ओपिनियन बन जा रहा है सेंटर डेट ओरिजिन रेडियस 3 सेंटर डेट थ्री कमा फोर रेडियस 4 तो सर चीज ऐसे ही प्लॉट की जाए तो अच्छा है और फिर हम चीज विजुलाइज इमेजिन या क्रिएट करेंगे है ना तो सबसे पहले आई नीड तू दे विल विद डी ए एक्सिस है ना तो ये रहा ए एक्सिस यहां के आपके पास यह है ए एक्सिस है ना इसी तरीके से सर आपको डील करनी होगी एक्स एक्सेस से तो ये हो जाएगा आपका क्या उसे आने वाला है एक्स-एक्सिस एंड आई बिलीव दिस फॉर्चूनेटली हज गोटन क्रीटेड आज अन कंप्लीट प्रॉपर स्ट्रेट लाइन अब सर आपके पास एक सर्कल है जिसके सेंटर हैं ओरिजिन पर तो सर ये आपका एक सर्कल है आप इस बात को मानते हो क्या बिल्कुल मानते हैं सर और हमारे पास एक और सर्कल है ना ये सर्कल कौन सा है सर ये सर्कल है आपका सेंटर आते ओरिजिन जीरो कमा जीरो और दूसरा सर्कल जो आपको दे रखा है वो कौन सा है सर जो दूसरा सर्कल आपको दे रखा है डेट इसे सी तू जो की सेंटर है थ्री कमा फोर पर और वो क्लीयरली आपका फोर रेडियस कैरी किए हुए अब आप बात समझना मेरी बहुत बात जरूरी है जो समझना ये 4% ए एक्सिस का जो है सेंटर का वो है 4 और इसकी रेडियस भी है 4 क्या आप क्या समझ का रहे हो सर जब भी ऐसा होगा तो मैं एक और बड़ी इंपॉर्टेंट बात कहूंगा आपसे की जैसे ये तो क्लीयरली जीरो कमा जीरो इसकी रेडियस कितनी है इसकी रेडियस है आपकी थ्री है ना इसकी रेडियस है आपकी थ्री यूनिट्स आपके पास एक और सर्कल है सर एक और सर्कल है जो की थ्री कमा 4% है तो अगर ये थ्री है आई होप आप मेरी बात समझ का रहे हो अगर ये थ्री है है ना तो इससे फोर यूनिट्स डिस्टेंस तो 3 4 यहां कहीं होगा और ये थोड़ा सा इससे बड़ा होगा है ना तो मैं यहां कहीं से मानूंगा और ये इस तरीके से आपका बन रहा होगा क्या आप मेरी इन सारी बातों से चीज क्लीयरली समझ का रहे हैं जो मैं कहना चाह रहा हूं अच्छा जरूरी है क्या है सर्कल इसको यहां टच करें मेरा मानना है ये एक्स एक्सिस को तो टच करेगा सर ऐसा क्यों का रहे हो ये एक्स एक्सिस को बेशक टच करेगा स्टूडेंट्स क्यों क्योंकि आप इस बात को समझो ये एक्स एक्सिस को इसलिए टच करेगा क्योंकि पॉइंट सन लो इसका cardinate जो है वो है थ्री कमा फोर सर्कल और इसके रेडियस क्या है 4 तो अगर थ्री कमा फोर तो फोर अगर इसकी रेडियस है तो ये क्लीयरली एक्स एक्सिस को टच करेगा क्योंकि वो एक्सेस के रिस्पेक्ट में जो डिस्टेंस है वो भी फोर यूनिट है और रेडियस भी फोर ये बड़ा बेसिक सा ऑब्जर्वेशन अगर आपको मैं चीज समझाना चाहूं तो ये सर्कल है सेंटर्ड ओरिजिन जीरो कमा जीरो पर और इसकी रेडियस है थ्री यूनिट्स और ये सर्कल सेंटर है आपका थ्री कमा फोर पर क्या इन सारी बातों से आपको कोई आपत्ति अब मेरा आपसे कहना है की सर अगर इन दोनों को मिलने वाली इन दोनों के सेंटर्स को मिलने वाली एक स्ट्रेट लाइन अगर मैं ड्रॉ करूं इन दोनों के सेंटर्स को मिलने वाली अगर मैं स्ट्रेट लाइन ड्रॉ करूं आई थिंक ये वो स्ट्रेट लाइन होगी क्यों कर रहे हो सर बस थोड़ा सा पेशेंस रखिए ये बहुत कम आने वाली है हमारी ये वाली स्ट्रेट लाइन है ना तो ये हमारी वो स्ट्रेट लाइन रही ये क्यों कम आने वाली है बस इस बात को गौर से देखो समझो बात को अब हम बात करेंगे तीसरे सर्कल की ये आपके C1 और C2 बन चुके हैं अब सुने का रहे हैं यहां से वो का रहे हैं की ये जो दोनों सर्कल से एक्स और ए दो पॉइंट्स पर इंटर सेट करते हैं ठीक है सर इन दोनों ने दो पॉइंट्स पर इंटरसेक्ट किया अब एक तीसरा सर्कल है C3 जिसके ना तो सेंटर के आर्डिनेंस बताएं ना ही रेडियस लेकिन वो कुछ कंडीशंस को फुल फुल करता है कैसे सर ध्यान से देखना सेंटर ऑफ सी थ्री स्कूल लीनियर विद डी सेंटर ऑफ C1 एंड C2 मैंने इसलिए इन दोनों के सेंटर्स को मिलाया क्योंकि आपका जो सर्कल वैन का जो सेंटर है और सर्कल तू का जो सेंटर है उसी लाइन पर उन दोनों को मिलने वाली स्ट्रेट लाइन पर ही आपके थर्ड सर्कल का भी सेंटर लाइक करता है ठीक है सर और क्या और मेरा आपसे ये कहना है की सर ये जो सर्कल है ये जो सर्कल है C1 C2 वो दोनों C3 के अंदर है मतलब C3 एक बड़ा सर्कल होगा C3 बड़ा सर्कल होगा जो इन दोनों को अपने अंदर समाहित किए हुए हैं और तीसरा ब्लू क्या है तीसरा ब्लू है की C3 C1 को और C2 वीसी एम सॉरी C3 C1 और C2 को टच करता है इन पर अब आप एक बात बताओ सर C3 सर्कल इन दोनों के बाहर बना हुआ है बिल्कुल सही बात है और अगर इन दोनों के बाहर बना हुआ है तो कैन आई से सर की वो सर्कल अगर इन दोनों को टच करता है तो वो इसी तरीके से टच कर सकता है अब मैं इन दोनों को इस तरीके से टच करवाने की कोशिश कर रहा हूं जो की आई होप आप समझ का रहे हो हम इस तरीके से कर रहे होंगे है ना बस में से थोड़ा सा प्रॉपर्ली बना डन तो ये सर सर्कल आपका इस तरीके से टच कर रहा होगा इन दोनों को क्या इस बात से किसी भी स्टूडेंट को कोई आपत्ति कोई परेशानी आई थिंक सर यह सर्कल आप किस तरीके से टच हो रहे होंगे अगर मैं इसे थोड़ा सा और इधर पुष्कर हूं तो थोड़ा सा इसे पीछे या नीचे पुश करने की जरूरत है एंड आई थिंक बहुत ज्यादा अच्छे से ड्रॉ नहीं कर का रहे हैं क्योंकि हम बहुत ही प्रेसीजली चीज नहीं बना पाएंगे बट मेरा यकीन है आप चीज समझ अच्छे से का रहे हो बिना ज्यादा फ्लकचुएट हुए हैं मैं बस इस सर्कल को ड्रॉ कर लूं ताकि आपको समझ अच्छे से ए जाए आई थिंक अब आप मैन का रहे हो की सर हान ये दोनों टच कर रहे हैं है ना इस बात को क्योंकि यह बोर्ड मुझे बहुत ज्यादा फ्लैक्सिबिलिटी नहीं दे रहा है इन दोनों सर्कल को टच करवाने के लिए तो मैं मैन लेता हूं ये वो सन रही हूं क्या यहां तक बात समझ आई रिपीट माय स्टेटमेंट यह आपका सर्कल था सेवन विच वास सेंटर आते ओरिजिन है ना विच वास सेंटर डेट के लिए हम जान का रहे हैं इन दोनों को टच करने वाला एक और सर्कल है जिसका सेंटर जीरो कमा जीरो थ्री कमा फोर को मिलने वाली लाइन है कैन आई से वो इस बाहर वाले सर्कल सी थ्री के लिए भी डायमीटर की तरह कम करेगी क्योंकि इसका सेंटर्स लाइन पर लाइक करता है की आप बात समझ का रहे हो ऑफ कोर्स आपका सर्कल वैन है और ये आपका सर्कल तू है जिनके सेंटर्स को मैं C1 और C2 से दिनो कर रहा हूं कोई आपत्ति तो नहीं अब क्या सर अब आप क्या पहले तो ये सिनेरियो आपको समझ ए रहा है क्योंकि C3 ने C1 और C2 को अपने अंदर कंटेन किया हुआ था और क्लीयरली हम समझ का रहे द सर क्या हम ये समझ का रहे द की C3 जो है वो इन दोनों के बाहर है और इन दोनों को इंटरनल टच कर पाएगा इसी तरीके से अब अगर मैं थोड़ा सी थ्री के बारे में जानना या समझना या देखना चाहूं तो मेरा क्या ओपिनियन होगा प्लीज इस बात को समझना आप पहले तो मुझे बताओ क्या हम ये देख का रहे हैं की सर ये जो लाइन है ये जो लाइन है है ना इसे जहां वो टच कर रहा है वो कहां का टच कर रहा है वो भी देख लेते हैं सर ये C3 C1 को तो एम पर है ना तो सर C1 को एम पर यह जो पॉइंट है ये एन है मैं बस आपसे ये कनक्लूड करना चाह रहा हूं की अगर मैं एम एन को डायमीटर का रहा हूं C2 C1 C3 तीनों का तो अगर मैं मतलब ये वाला पार्ट C1 का है ये डायमीटर C2 का है और ये पूरा में जो है वो C3 का है मेरा बस ये ओपिनियन है की अगर मैं एम एन को फेंक आउट कर लूं तो क्या मैं C3 की रेडियस और सेंटर तक पहुंच सकता हूं कोशिश की जा सकती है कम की बातें सुनना है स्टूडेंट्स मैं C3 के सेंटर के कोऑर्डिनेट्स और रेडियस वगैरा आउट करना चाह रहा हूं देखो ध्यान से सुनना एक बात बताओ एक बात बताओ C1 जो सर्कल था उसकी रेडियस क्या थी सर C1 जो सर्कल था उसकी रेडियस थी थ्री और C2 की थी 4 तो केन आई से अगर एम से C1 की जो डिस्टेंस है वो कितनी है थ्री यूनिट्स अपने ये बात समझ का रहे हो है ना सर C1 से C2 की डिस्टेंस कितनी है सर वो वही है जो जीरो कमा जीरो की थ्री कमा फोर से होगी तो 3² + 4² तो C1 से C2 की डिस्टेंस कितनी हो जाएगी सर वो हो जाएगी फाइव अगर आप यहां कंफ्यूज हो रहे हैं तो मैं यहां लेता हूं आपका एम से C1 के डिस्टेंस कितनी अगर वह है आपके लिए थ्री क्योंकि वो उसकी रेडियस है सर C1 से C2 की डिस्टेंस कितनी है सर वो हमने डिस्टेंस फॉर्मूला उसे किया तो वो ए जाती है फाइव और एक बात बताओ अब C2 से एन की डिस्टेंस बता सकते हो क्या सर C2 से इनके डिस्टेंस तो बड़ी आसान है फोर क्योंकि ये इसकी रेडियस है ना तो वो कितनी है फोर सी तू सर्कल की जो रेडियस है वो कितनी है c2n जो की कितनी है 4 अब आप एक बात बताओ अगर मैं mc1 C1 C2 और c2n को ऐड कर लूं तो ये कितना ए रहा है सर 3 + 588 + 4 12 है ना तो मुझे जो में की लेंथ मिल रही है मुझे जो में की लेंथ मिल रही है वो है 12 जो की आपके सर्कल C3 का डायमीटर है तो shailise है की सर यहां से मैंने पता किया की जो सर्कल सी थ्री की रेडियस है है ना किसकी रेडियस भाई आपके सर्कल C3 की वो कितनी ए रही है सर वो ए रही है सिक्स क्या आप सब इस बात से एग्री करते हैं जो की हमें क्वेश्चन में कैसे किस दिनो की गई है स्मॉल आर से अगर आप देख का रहे हो तो सर मैं स्मॉल आर तो निकल पाया इस बात से किसी भी स्टूडेंट को कोई परेशानी यहां तक तो नहीं है सर अब क्या करना चाह रहे हो अब अगर मैं इसके सेंटर तक पहुंचना चाहूं इस सर्कल के सेंटर तक पहुंचना चाहूं तो कैसे pahunchunga इस बात का जवाब दीजिए सर बात आसान सी है बात आसान सी क्यों इस बात को सुनना आप मुझे ये बताओ सोच कर बताओ सिंपल सी बात है बहुत ध्यान से देखना जीरो कमा जीरो से थ्री कमा फोर के डिस्टेंस कितनी है अच्छा सारी बातें छोड़ ये में टोटल डिस्टेंस कितनी है सर में टोटल डिस्टेंस है 12 अब सुनना ध्यान से मैं थोड़ा सा एक रफ आपको चीजें दिखा रहा हूं ये जो डिस्टेंस है कितनी है 12 आप समझ रहे हो क्या तो इसकी हाफ डिस्टेंस क्या होगी सर वो होगी सिक्स ये भी आप समझ रहे हो क्या तो शैला आई से यहां हाफ डिस्टेंस पर आपका कहीं ना कहीं सेंटर लाइक करेगा अब सुनना सिक्स में से सिक्स में से ये जो डिस्टेंस है ओरिजिन से ये जो डिस्टेंस है वो थ्री है तो shailise ओरिजिन से ऊपर की तरफ कहीं ओरिजिन से ऊपर की तरफ मतलब फर्स्ट क्वाड्रेंट में ओरिजिन से थ्री यूनिट डिस्टेंस पर इस स्ट्रेट लाइन पर जो पॉइंट होंगे वही इसके सेंटर होंगे आई होप आप मेरा ऑब्जर्वेशन समझ का रहे हो जो आपको मैं दिखाना समझाना है कहना चाह रहा हूं तो बस आपसे मेरा ये कहना है की जो सर्कल सी थ्री का सेंटर होगा जो सर्कल सी थ्री का सेंटर होगा जैसे का रहे हो होमो के उसके अगर मैं कोऑर्डिनेट्स जानना चाह रहा हूं तो मैं उसके लिए एक अच्छा ओपिनियन यह रखता हूं सर की क्लियर सी बात है क्लियर सी बात है बहुत appearantent है की सर यह जो स्ट्रेट लाइन है कौन सी एम एन जिसकी इक्वेशन क्या है C2 और C1 से पास होने वाली स्ट्रेट लाइन तो उसकी इक्वेशन क्या लिख सकता हूं गौर से देखिएगा 340 सर कुछ कहने की जरूरत नहीं है देखो ए - 0 मतलब अब देख के आपको बोल देना चाहिए बट फिर भी चलो है ना इस इक्वल तू Y2 - 5 1 / x2 - एक्स 1 तो 4 / 3 टाइम्स एक्स सो ए = 4 / 3 टाइम्स एक्स क्या बोलना चाह रहा हूं मैं ए = 4 / 3 तो कहना ऐसे थ्री ए = 4X ये आपके एक स्ट्रेट लाइन है वही इस इक्वल्स तू यह जो 4/3 है की नहीं अब आप एक बात बताओ सर आप यहां पर यहां पर इस इस C1 से अब थ्री यूनिट ऊपर जाना चाह रहे हो आप मेरी बात समझ रहे हो आप C1 से 3 यूनिट ऊपर जाना अगर मैं इसको थोड़ा अलग तरीके से समझाऊं ना तो मैं ऐसे कहूंगा ध्यान से देखो भाई आपके पास एक राइट एंगल ट्रायंगल है सर आपके पास एक बेसिकली राइट एंगेल्ड ट्रायंगल है ध्यान से सुनना है स्टूडेंट्स ये आपका ओरिजिन है यहां से आपको थ्री यूनिट्स डिस्टेंस ऊपर जाना है यहां से आपको 3 यूनिट्स डिस्टेंस ऊपर जाना है क्या इतनी बात आप समझ का रहे हो आपको 3 यूनिट डिस्टेंस ऊपर जाते हुए ये करना है प्लीज इस बात को suniyega इस लाइन की स्लोप यानी इस लाइन का पॉजिटिव एक्स-एक्सिस के साथ जो स्लोप है वो है आपका थीटा है ना और इस ट्रायंगल में मैं जानता हूं की जो मेरा तन थीटा ए रहा है वो कितना है 4/3 अब अगर मैं इस पॉइंट तक जाना चाह रहा हूं तो मैं इस पॉइंट का कौन सा एक्स कॉर्ड इन इट और इस पॉइंट का मैं ए koardinate जानना चाह रहा हूं आई थिंक आप समझ गए हो मैं क्या कहना चाह रहा हूं सर हमारे पास छोटे न्यू से हमारे पास 3 यूनिट डिस्टेंस तो नहीं है क्यों नहीं है सर बहुत आसान सी बात है देखो भाई बड़ी सिंपल सी बात में आपसे कहना चाह रहा हूं सोच के देखो बहुत ध्यान से सुनना तन थीटा कितना है सर तन थीटा है 4 / 3 तन थीटा कितना होता है इसका तन थीटा होता है आपका परपेंडिकुलर अपॉन 20 है ना लेकिन ये डिस्टेंस है आपकी थ्री यूनिट्स ये डिस्टेंस आपकी 300 से और ये फोर और थ्री आपके क्या है रेश्योस ये 4/3 कितना है रेश्यो तो क्या मैं परपेंडिकुलर को थोड़ी देर के लिए सिर्फ 4K है ना परपेंडिकुलर को हमें थोड़ी देर के लिए फोर के और बेस को क्या मैं थ्री के मैन सकता हूं देखो आप इस बात को समझ का रहे हो जो मैं कहना चाह रहा हूं कोई दिक्कत तो नहीं तो आप यहां से क्या कनक्लूड करोगे अगर मैं सर फोर के और थ्री के मैन का रहा हूं तो ध्यान से सुनना इसका आपके बात को सर क्या मैं बड़ी आसान सी बात ये का सकता हूं की सर इस राइट एंगल ट्रायंगल में आप अगर पाइथागोरस थ्योरम लगाओ तो 3 के स्क्वायर प्लस 4K स्क्वायर कितना हो जाएगा 5k मतलब थ्री के और फोर के वाले राइट एंगल ट्रायंगल में हाइपोटेन्यूज कितना होगा सर वो होगा ओके और हमें कितना सर्कल सी थ्री का सेंटर क्या आप मेरी बात समझ का रहे हो तो फाइव के किसके इक्वल है 3 के अगर फाइव के थ्री के इक्वल है तो के कितना हो जाएगा सर के हो जाएगा 3/5 आई होप आप मेरी बात समझ का रहे हो अगर के हो गया है 3/5 इफ गेट अन हो तू बी थ्री बाय फाइव फोर के की वैल्यू कितनी हो जाएगी सर 4K की वैल्यू हो जाएगी 4 3 12 / 5 है ना और सर थ्री के की वैल्यू कितनी ए जाएगी आई होप आप देख का रहे हो 3 / 5 है ना तो ये हो जाएगा 9/5 तो दिस टर्न्स आउट तू बी 9/5 तो क्या मैं का सकता हूं जो सर्कल थ्री है उसके सेंटर के कोऑर्डिनेट्स 9/5 कमा 12/5 हो जाएंगे तो इसके जो सेंटर के कोऑर्डिनेट्स हैं दे आर 9 / 5 कमा 12 / 5 ये किसकी वैल्यू है वो ह कमा क हमने ऐसे क्यों निकाला मेरे ख्याल से आप अपने कन्वेंशनल कोऑर्डिनेट्स सिस्टम में पड़े गए रूल्स यानी वो जो आपने पैरामेट्रिक फॉर्म में सिखा था एक फॉर्मूला उससे भी निकल सकते द पर वो फॉर्मूला बना ही नहीं है तो मेरे पास आपको रिकॉल या reconlect करवाना चाह रहा था की चीज ऐसी होती इनके साथ भूल जाओ चीजें तो आप अपनी ट्रेडिशनल अप्रोचों हमेशा अपने पास रख सकते हो जिनसे आप इस तरीके से आंसर निकल सकते हो तो दिस इस हो यू गेट 9 / 5 एंड 12 / 5 विच आर दी सेंटर्स है ना तो सर हमने एक चीज तो फिगर आउट कर ली आपका 9/5 और 12/5 और इसकी रेडियस मतलब हमने सर्कल सी थ्री क्रैक कर लिया टेक्निकल सी क्रांत डी सर्कल सी थ्री जिसके सेंटर है 9/5 और उसकी रेडियस है कितनी सिक्स दिस इनफॉरमेशन सी हाव गार्डन सो फार आई होप यहां तक कोई डाउट नहीं है सर यहां तक चीज है क्लियर है अब क्या करना चाह रहे हो अब जब मुझे पता चल गया है C3 अब वो क्या का रहा है सुना गया था लेट डी लाइन थ्रू एक्स एंड ए अब एक्स और ए क्या द सर आपके एक्स और ए अभी तो नहीं पता इंटरसेक्ट C3 आते znw है ना अगर मैं एक्स और ए की बात करूं तो एक्स और ए यह रहेगा कहना जो सर्कल्स एक्स और ए पर इंटरसेक्ट कर रहे द तो आपके जो सर्कल्स हैं वो मैन लो आपके एक्स और ए पर इंटरसेक्ट कर रहे हैं है ना अब सर सर लाइन जो आपकी एक्स और ए के थ्रू जा रही है उसको लेकर तो मेरा ओपिनियन बहुत शॉर्ट है मेरा ओपिनियन बहुत सॉर्टेड इसलिए है क्योंकि मुझे ज्यादा कुछ आपको कहने या समझने की जरूरत ही नहीं है मैं कहूंगा सर आपसे की आप देखो जरा ध्यान से आप देखो जरा ध्यान से की यह आपकी लाइन है ये आपकी लाइन एक्साइज ए रही है ऑफ कोर्स आपका कौन सा पॉइंट है एक्स और ये आपका कौन सा पॉइंट है ए और ये एक्स ए से जो लाइन जा रही है दिस इस नथिंग दिस इस नथिंग डी कॉमन सो आई कूद हैव कॉल्ड दिस एक्स ए इधर डी कॉमन कोड बट मेरा यकीन है कई सारे स्टूडेंट्स ने इसे आईडेंटिफाई किया होगा सी कैन कॉल इट अन कॉमन कोड इधर और कॉमन कोड और व्हाट और रेडी कल एक्सेस क्या यह सारी परिभाषाएं भूले तो नहीं सर इससे क्या करना चाह रहे हो अगर यह कॉमन कोड नहीं है मेरे ख्याल से तो नहीं है क्यों सर क्योंकि जो ये कॉमन कॉर्ड या रेडिकल एक्सेस है जो स्ट्रेट लाइन एक्स ए है वो C1 और C2 का रेडिकल एक्सेस है अगर वो C1 और C2 का रेडिकल एक्सिस है तो इसकी इक्वेशन तो मैं लिख सकता हूं देखो आपके पास एवं है आपके पास C2 भी है C1 क्या है सर नाइन वाला और ये 34 और 16 वाला मतलब C1 कितना है सर C1 आपका है x² + y² = 9 है ना और C2 क्या है सिर्फ C2 जो है आपका वो है एक्स - 3 + 5 - 4 है ना माइंस थ्री और फोर ही है ना थ्री कमा फोर बिल्कुल सही तो मैं ऐसे लिखूंगा x² + y² मेरी बातों को अब suniyega ध्यान से आप ने सर यहां पर देखो एक्स - 3 और एक्स - 4 का स्क्वायर किया आप suniyega यहां से थ्री का स्क्वायर 16 से 16 कैंसिल तो बचेगा क्या +9 तो यहां पर आपको क्या मिलेगा 9 आई होप इस बात से आप सब सहमत हैं और यहां पर आपको मिलेगा प्लस नाइन इसे इक्वल तू जीरो सर ये क्यों निकाला क्योंकि मैं कॉमन कोर्ट तक पहुंचना चाह रहा हूं और कॉमन कोड क्या होती है सर कॉमन कोड होती है आपकी दोनों सर्कस के आप उनको क्या कर दो सूत्र तो अगर आपने सी वैन में से C2 सब्सट्रैक्ट किया है सर्कल के क्वेश्चन सब्सट्रैक्ट किया तो ये और ये ये और ये पार्ट गए है ना वहां नेगेटिव होगा तो हो जाएगा 6x + 8x ये बात सबको समझ ए रही है सर ध्यान से ध्यान से देखना ये यहां पर ए रहा है माइंस नाइन वो भी वहां माइंस नहीं तो माइंस साइन माइंस साइन कितना माइंस 18 = 0 ये आपकी क्या है ये आपकी रेडिकल एक्सिस या कॉमन कोड एंड वो जो आपकी स्ट्रेट लाइन है कौन सी एक्स ए इसकी इक्वेशन है आप चाहें तो आप इसे थोड़ा सिंपलीफाई कर सकते हो क्योंकि सब में से मुझे तू कॉमन आता है वो दिख रहा है क्या आपको दिख रहा है तो सर आप सिंपलीफाइड इनिक्वेशन क्या लिखोगे आप लिखोगे 3x इस बात से कोई डाउट इतना पार्ट अच्छे से देखो समझो मुझे बताओ अगर यहां तक कोई भी डाउट कोई भी परेशानी या कोई भी दिक्कत है तो अब क्या थॉट ए रहा है सर आपका अब क्या करना है अब हम कोशिश करते हैं क्वेश्चन पढ़ के समझने की की हमसे क्या-क्या चीज निकलने को कही गई हैं अभी तक इतनी चीज मुझे दी गई थी अगर आप देख का हो तो अभी हमने क्वेश्चंस तो सॉल्व करना शुरू ही नहीं किया है अभी तो हमने इसका साइंस इसका मीनिंग निकाला है की सर लेट डी लाइन थ्रू एक्स एंड ए इंटरसेक्ट सी थ्री इट्स znw तो C3 को ये स और व पर इंटरसेक्ट कर रही है है ना तो मैं ये बात भी निकल लेता हूं की C3 को उसने कहां का इंटरसेप्ट किया C3 को उसने स और व पर इंटरेस्ट है अगर मैं C1 और C3 की कॉमन टैसेंट की बात कर रहा हूं तो सर ये C1 और ये सी थ्री कैन आई से सर क्योंकि जो कॉमन टांगें है वो कुछ ऐसी जा रही होगी दिस इस गोइंग तू बी डेट कॉमन टांगें यस और नो स्टूडेंट्स गिव मी अन रिस्पांस क्या यहां तक चीजे समझ ए रही है सो दिस इस डेट कॉमन टांगें जिसके बारे में हम बात करेंगे अभी नहीं कर रहे बात करेंगे क्योंकि वो ये कहना चाह रहा है वो जो कॉमन टांगें है वो इस पैराबोला के साथ कमेंट है अब पैराबोला भी हमने चैप्टर पढ़ा नहीं है बट पैराबोला के साथ वो कमेंट लेना चाह रहे तो मैं बस थोड़ी सी लिबर्टी लूंगा इस वाले जोन में जहां पर मैं आपसे कहूंगा की आप मुझे थोड़ा सा रिस्पांस दीजिए की मैं पेरा बोला की यहां पर एक टांगें बना पाऊं मैं बनाऊंगा और मैं डिस्कस करूंगा आपसे की इसे कैसे करते हैं बट इससे पहले वाला पार्ट आप सॉल्व कर सकते हो पेरा बोला के साथ वो जो कॉमन टांगें है उसके अलावा जितना भी पार्ट है वो अब आसानी से इस पार्ट को देख कर जोन में आकर आराम से कर सकते हो सर कैसे करना होगा ये रहे आपके क्वेश्चंस पहले तो उसने पूछा आपसे तू ह प्लस के वैल्यू निकल सकते हैं क्या कोशिश की जा सकती है तू ह प्लस के इसलिए निकाला जा सकता है क्योंकि मेरे पास है 9/15 है ना 9/15 और 12/5 आपके हैं रिपीट माय स्टेटमेंट 9 15 तो 2h प्लस के मतलब इसका डबल प्लस के ये आपसे पूछा जा रहा है और अगर आप 2h + के के बारे में कोई टिप्पणी करें तो क्या आप मुझे बता रहे हो 2h+ के की वैल्यू कितनी ए रही है सर देखो ध्यान से 9 का डबल 18 और 12 कितना 20 और 10 30 और 30 को पंच से डिवाइड किया तो कितना अच्छा है तो 2h+k कितना आता है सिक्स और तू ह प्लस के आता है सिक्स तो मैं वैन की मैपिंग कैसे कर दूंगा पी से तो 1 के लिए तो रिजर्व्ड है कौन सा वैन के लिए रिजर्व्ड है नौ देयर आस्किंग यू लेंथ ऑफ जो अपॉन लेंथ ऑफ एक्सी तो सर मुझे zedblue और एक्सी से बात करनी है तो स व एक्स ए तक अगर पहुंचना है तो आई थिंक आप रिलाइज कर रहे हो ये है आपका जो और ये है आपका एक्सी अब ये कैसे सोचेंगे यह सोचने का तरीका जो मेरे पास होगा वह यह की सर जो तक पहुंचने के लिए आप शायद C3 की हेल्प ले रहे होंगे बिल्कुल सही बात है और एक्स ए तक पहुंचने के लिए शायद में शायद में शायद में कोई राइट एंगल ट्रायंगल बना रहूंगा क्योंकि ये मुझे एक कार्ड लग रही है सिवान और सी के आपको क्वेश्चंस समझ आए तो आपके सामने है डाटा क्या-क्या निकलना है आपको जैज डब्लू और एक्सी ज़रा मुझे जल्दी से निकल के बताओ की आप कैसे सोचोगे आपको एक्स ए की इक्वेशन भी लिख कर दे दी है स्टूडेंट्स क्योंकि ये एक कॉमन चोर्ड है जो हम ऑलरेडी निकल चुके हैं तो स व को सोचने का थॉट क्या होगा सर जो को निकलने के लिए तरीका क्या हो सकता है आगे अन्य वैन अन्य थॉट्स ऑन डेट अच्छा चलो छोड़ो अब तो ये बताओ की अगर मैं सर्कल सी थ्री का सेंटर निकल लूं और वहां से अगर एक परपेंडिकुलर ड्रॉप कर डन तो क्या मैं चीज सोच सकता हूं देखो दोनों के लिए थॉट से है suniyega ध्यान से सर्कल C3 का जो सेंटर है 9/5 एप्रोक्सीमेटली कितना है 2.3 समथिंग है ना और ये मैन लो 1.1 में मैन लेता हूं अभी बस समझने के लिए आपको की जैसे मैन लो अब बस समझने की कोशिश करना ये आपका जो जिसका मैं अलग प्लॉट करता हूं है ना ये आपका जो ये आपकी स्ट्रेट लाइन है स व स्ट्रेट लाइन व्हिच इस थिस जो सुनेगा ध्यान से और आपके पास सर्कल का सेंटर है विच इस C3 में थोड़ा सा इसको ब्रोडें अप करके बना रहा हूं ताकि मैं आपको चीज दिखा समझा और बता पाऊं की देखो चीज आसान है ये आपने C3 से क्या-क्या अपने परपेंडिकुलर ड्रॉप किया स व पर है ना सो दिस वाज योर लाइन स व है ना दिस लेट्स से हेड अन मिड पॉइंट जैसे हम मैन लेते हैं कुछ भी लेट्स मी कॉल इट लेट मी कॉल इट के तो हमने नहीं लिया लेट्स कॉल C3 के cardinates पता है सर क्या 9/5 कमा 12/5 है ना तो ये है 9/5 दिस इस स्ट्रेट लाइन जो का इक्वेशन यस आई डू हैव ए और हो सो स व नथिंग बट आपके स्ट्रेट लाइन एक्स ए स्ट्रेट लाइन एक्स ए की इक्वेशन तो हमने निकल थी ये आपकी स्ट्रेट लाइन एक्स ए ही है याद ए रहा है क्या तो एक्सी की क्वेश्चंस हम निकल चुके हैं तो स व की इक्वेशन क्या थी सर याद करिए स व की इक्वेशन थी जो की एक्स ए की भी है यही इक्वेशन किसकी है यही इक्वेशन जो के भी जो है 3X + 4y - 9 सो ट्राईंग तू मेक यू अंडरस्टैंड यस आप प्लीज इस बात को समझो की सर अगर आप C3 के सेंटर से स व के स या व को milaoge तो shailise है वो C3 की रेडियस हो जाएगी बिल्कुल हो जाएगी सर तो C3 के सेंटर्स है मैंने इदर स दिया व को मिलाया है ना तो लिसन स को मिला है तो shailise ये जो है ये इसकी रेडियस है जो की कितनी है सिक्स तो सर आपके पास ये लेंथ पता है सिक्स ऑफ कोर्स ये कॉर्ड पर ड्रॉप किया गया परपेंडिकुलर है तो ये लेंथ परपेंडिकुलर डिस्टेंस किस से इसकी स्ट्रेट लाइन से निकल के जा सकती है हाइपोटेन्यूज और परपेंडिकुलर पता है तो आई थिंक मैं बेस पता कर लूंगा उसका डबल कर दूंगा तो स व ए जाएगा डू यू ऑल गेट डेट चलो सर सोचते हैं देखो अब 9 / 5 कमा 12/5 से इस लाइन की परपेंडिकुलर डिस्टेंस निकलना है सोचो कैसे करेंगे ध्यान से देखना सर आई होप आपको फॉर्मूला पता है क्या सर ए X1 सो 9 / 5 3 सो दिस इस गोइंग तू बी थ्री टाइम्स नाइन बाय फाइव प्लस 12 / 5 4 सो दिस इस गोइंग तू बी फोर टाइम्स 12/5 लगा लीजिए क्योंकि ये डिस्टेंस आप निकल रहे हो डिवाइडेड बाय अंडर रूट ओवर वही बात वापस है थ्री का स्क्वायर प्लस फोर का स्क्वायर जो के लिए कितना हो जाएगा √ में 25 यानी 5 कोई दिक्कत तो नहीं है कोई आपत्ति किसी भी स्टूडेंट को है तो पूछिए सर आप क्या करना है आपको मेरा कहना है सर एक छोटा सा कम कर लेते हैं क्या इस लाइन को क्या इस माइंस नाइन को मैं ऐसा लिख सकता हूं मेरी सहूलियत के लिए माइंस 9 5 / 5 है ना तो माइंस नाइन इन फाइव कितना 45 तो ये हो जाएगा 45 / 5 ऐसा क्यों किया आई होप आपको नजर ए रहा है ऐसा क्यों किया तो ऐसा करने का रीजन आय होप आप देख का रहे हो अब क्या हो रहा है देखो 9 3 27 9 3 27 प्लस 12 कितना 48 और यह कितना है 25 अब suniyega ध्यान से अब कम की बात बहुत कम की बात अब कार्स यहां मोड लगाना मत bhuliyega क्योंकि डिस्टेंस है 48 - 45 ये कितना ए रहा है आई थिंक ये थ्री ए रहा है क्या बस एक बार क्रॉस चेक कर लो 3 ए रहा है क्या हान थ्री ही ए रहा है है ना 27 + 3 कितना हो जाएगा सही हो जाएगा 30 तो 30 / 25 शल आई सही हो जाएगा 655 सो ये जो डिस्टेंस है डिस्टेंस आउट तू बी सिक्स बाई फाइव तो ये जो डिस्टेंस है दिस इस 6 / 5 और ये डिस्टेंस है तो क्या इस 6465 को उसे करके मैं स क्यों निकल सकता हूं अब जैसे ही आप स क्यों निकल लोग मैं अगली हिंट आपको देने वाला हूं अगली हिंट क्या है सर एक्स से C3 को कनेक्ट किया जब एक्स से C3 को कनेक्ट किया तो वापस आपको एक ट्रायंगल मिला कौन सा ट्रायंगल है C3 से आप एक्स को कनेक्ट करो जब आपने C3 से एक्स को कनेक्ट किया तो ये कौन सा सर्कल है अच्छा छोड़ो C3 से मत करो C3 से मत करो अब C2 से कर लो आई रिपीट माय स्टेटमेंट इस केस में C3 से मत करो उसे केस में आप C2 से कर लो है ना इसको मिटाता हूं क्योंकि उसे केस में हमारे पास रेडियस नहीं है तो छोड़ो अभी पहले इस वाले पार्ट को देखो तो अगर ये आपको सेक्स और 6 / 5 मिल गया है तो की हम सीधे सोच सकते हैं सर बहुत बड़ी बात नहीं है स के लेस निकल लिए ना तो सिक्स का ये सिक्स है स्क्वायर में से 655 का स्क्वायर करेंगे तो देखो भाई मैं स क्यों निकल रहा हूं एन तो स के जब निकलेंगे तो स कब क्यों निकलने से जो भी पूछा है ना तो स व क्या होगा जो होगा डबल ऑफ डबल ऑफ लेंथ का स्क्वायर माइंस 10 परपेंडिकुलर का स्क्वायर इनसाइड एंड अंडर रूट तो अंडर रूट में हाइपोटेन्यूज क्या सेक्स का स्क्वायर कितना 36 माइंस इसका स्क्वायर वो कितना हो जाएगा सर वो हो जाएगा 36 / 25 एंड दिस इसे गेटिंग अन लिटिल थोड़ा सा लेंडी हो रहा है टोटली एक्सेप्ट था अब देखो हम क्या करेंगे 25 36 या छोड़ो अब 36 कॉमन ले लो तो 36 बाहर आए तो कितना हो गया 6 तो ये हो जाएगा कितना 2 6 आप मेरी बात सन रही हो क्या अब ये कितना है सर ये है वैन माइंस वैन बाय 25 तो 25 - 1 / 25 यानी 24 / 25 / 25 को लेकर मेरा कहना है की देखो 25 का फाइव तो ये ए गया अंडर रूट में कितना बचा 24 24 कितना होता है 6 4 = 6 4 मतलब क्या 6 4 मतलब 2 बाहर ले आए तो फोर और अंदर क्या बचा ये तो दिस इसे योर नथिंग बट व्हाट स व और 6 4 कितना होता है 24 सो दिस इसे गोइंग तू बी दिस इस गोइंग तू बी 24/5 टाइम्स √6 जो की आपकी वैल्यू है स व की क्या ये सारी कहानियां इसी तरीके से जैसे आपने स व निकाला है जैसे आपने स व निकाला है क्या उसी तरीके से एक्स ए होता जा सकता है कहानी वही है स्टूडेंट्स सोच के देखो मेरा आपसे ये कहना है आपके पास अगेन नेक्स्ट लाइन के पास अगेन क्या है एक्स वे स्ट्रेट लाइन के साथ-साथ आपके पास C1 को ले लो क्योंकि ओरिजिनल है 0 से हमारा जीवन थोड़ा आसान हो जाएगा तो सोच के देखना मैं इस एक्सी के नहीं चाहिए आपका पॉइंट क्या है एक्स और ऑफ कोर्स आपका पॉइंट है ए और यहां कहीं है लेट से पॉइंट आपका क्या सी वैन मैं कहना चाह रहा हूं आपसे की सर C1 से लेट से एक परपेंडिकुलर ड्रॉ करते हैं सो C1 से जैसे ही आपने परपेंडिकुलर ड्रॉप किया तो क्या हो जाएगा बेस सोच के बताओ C1 से जब आपने से प्रक्रिया वापस रिप्लाई करो और बताओ क्या हो जाएगा C1 के कार्ड से C1 आपका पहला वाला सर्कल है जिसकी क्वाड्रेंट से जीरो कमा जीरो अब क्या सर सर C1 की रेडियस कितनी थी याद करोगे C1 की रेडियस की अगर मैं आपसे बात करूं तो C1 की रेडियस आपकी थी थ्री तो आप अगर C1 से अगर आप सी वैन से लेट से ए एक्स किसी भी एक को milaoge तो वो डिस्टेंस कितनी हो जाएगी इसे और डिस्टेंस आपकी क्लीयरली थ्री यूनिट्स हैं अच्छा सर मुझे एक बात बता सकते हो क्या लेट से मैंने यहां से एक पार्टिकुलर ड्रॉप किया जिसका एम एम पॉइंट लेट्स से मैन लेता हूं के तो सी के क्या आएगा सर सीके आएगा जो आपका वो वही होगा वही से बात वापस से सीकन है 0 से एक्सी एक्सी आपकी स्ट्रेट लाइन है जो की यही है 3X + 4y - 9 है ना तो ये आपकी स्क्रीन लाइन है जिसकी इक्वेशन क्या है 3X + 4y - 9 = 0 अब आप बस इतना बता दो मुझे की सर 0 से जब इस पर परपेंडिकुलर ड्रा किया तो वो लेंथ क्या होगी तो सीके को लेकर आपका क्या ओपिनियन होगा सर सीके होगा मोड ऑफ ध्यान से देखो 0 0 - 9 तो न्यूमैरेटर में -9 का मोड जिसे आप नाइन लिख लेना बिना सोचे समझे डिनॉमिनेटर में थ्री और फोर के स्क्वायर के अंडर रूट सैम का अंडर रूट तो हो जाएगा फाइव तो ये कितना ए जाता है सर ये ए जाता है 9 / 5 सो आपका सीके कितना आता है सर ये आता है 9/5 सीके आता है 9 / 5 और ये है थ्री तो क्या ऐसे y³ या एक्सी निकल सकता हूं सर जो एक्सी आएगा आपका वो डबल होगा y³ का और y³ क्या होगा माइंस इसका स्क्वायर कितना हो जाएगा सर हो जाएगा 81 / 25 डू यू ऑल एग्री विद डी कोई डाउट तो नहीं है सर एक छोटा सा कम कर लो आप यहां से और यहां से 9 कॉमन ले सकते हो तो जैसे ही नाइन कॉमन लिया तो क्या बन जा रहा है देखो एक्स ए जो ए रहा है वो है तू टाइम्स यहां से यहां से नाइन कॉमन लिया तो बाहर कितना आएगा थ्री अब अंडर रूट में क्या बचेगा वो देखो √me बचेगा 1 - कितना 9/25 अब प्लीज इस बात को ध्यान से देखो ये हो जाएगा 25 - 9 25 - 9 कितना होता है सर वो हो जाता है 16 16 25 मैंने 4/5 तो दिस इसे गोइंग तू बी फोर बाय फाइव ऐसे क्या लिखूंगा भाई मैं ऐसे लिखूंगा 4/5 है ना तो यहां पर मैं लिख रहा हूं अगर आप सिंपलीफाई करें तो कुछ दिख रहा है क्या देखो भाई ध्यान से 4 8 3 24/5 आपका X5 का वैल्यू ए रहा है एक्स वे कितना ए रहा है 24/5 सो ये आपकी कुछ वैल्यूज हैं किस-किस की एक्सी और स व की अब अगर मैं आपसे पूछूं जो आपसे क्वेश्चन में पूछा जा रहा है क्या लेंथ ऑफ जो अपॉन लेंथ ऑफ एक्सी तो स व कितना है सर ये और एक्सी कितना है वो भी इतना ही है बस रूट सिक्स का फर्क है अगर आप जो अपॉन एक्सी करेंगे तो 24/5 से 24 / कैंसिल जो न्यूमैरेटर में है तो बचेगा कितना अंडर रूट सिक्स बचेगा कितना √6 और टेक्निकल ही आपका सेक्शन का भी ऑप्शन क्यों है वो सामने ही रखा हुआ है भैया कल मत लगाना की सर इसका आंसर फिर यही होगा है ना जीवन इतना आसान भी नहीं है तो सेकंड के लिए हम माफ करेंगे सर सेकंड के लिए आप मार्क करोगे कब दो ये वाला ग्रिल विद डेट हम चीज कैसे सोच रहे हैं अब सर एरिया ऑफ मन और स म है ना एरिया ऑफ मन मन और ज़म अच्छा एक बात तो आप समझ रहे हो ना की सर्कल्स के सेंटर्स को ज्वाइन करने वाली लाइन उनके रेडिकल एक्स्पोज़र परपेंडिकुलर होती है ये बात तो मुझे कहने में बताने की जरूरत नहीं है ये बात आई होप अंडरस्टूड है तो अगर आपसे कोई पूछे जम एंड ट्रायंगल का एरिया आईएफ सोमबॉडी आस्क यू स में ट्रायंगल का एरिया तो मैं क्या कहूंगा सर ध्यान से देखना अगर मुझसे कभी भी कोई पूछे ट्रायंगल कौन सा जमाने का एरिया तो मैं कहूंगा सर वह में मैन सकता हूं सर हाइट हो जाएगी आपकी स और ये इसका पॉइंट ऑफ इंटरसेक्शन में मैन लेता हूं ना मैं के ही माना था जैसे क्या मैन लेता हूं मैं स क्यों तो ये कितना हो जाएगा सर ये हो जाएगा जेके डू यू ऑल एग्री विद डेट एंड मी वैन हेविंग अन्य क्वेरीज कंसर्न्स विद डेट कोई कन्फ्यूजन तो नहीं है अच्छा सर जैम का एरिया है स में के अलावा हम एक और एरिया में इंटरेस्ट रेट में के अलावा हम इंटरेस्टेड हैं किस्म ज़म में अब ज़म क्या होगा सर ध्यान से देखो ये आपका है स एम और व सर ज़म में भी ये 90 डिग्री है आई होप आपको दिख रहा है तो इसमें क्या एरिया होगा zeddw/m³ एम है ना तो ये कितना हो जाएगा ये हो जाएगा हाफ किस ट्रायंगल के लिए हम ढूंढ रहे हैं सर ज़म का है ना सो दिस इस दी एरिया ऑफ डी ट्रायंगल ज़म हाफ है ना क्या ले\| हमने बेस लिया है स व और m³ एस हाइट तो जो और क्या m³ अब सर आपको इसका अपॉन दिस यही चाहिए तो हम अगर दिस अपॉन दिस निकलना चाह रहे हैं तो दिस अपॉन दिस कर देंगे कोई कन्फ्यूजन डाउट तो नहीं भाई आई होप ये बातें आप समझ का रहे हो अब क्या अब मेरा ओपिनियन बड़ा शॉर्ट ट्रिक्स है सर क्या की अगर मैं पहले तो हाफ से हाफ कैंसिल कर दो बिल्कुल कर दो अब बच क्या रहा है वो देखो सर में यहां दिख रहा है है ना ये बाई डी वे कंफ्यूज मत हो जाना ये आपका तू नहीं सेट है ना यहां में दिख रहा है देखो में का रिलेशन कुछ है क्या देखो मेरी बात समझने की कोशिश करना सर आप बात समझो अब बहुत बेसिक सी बात समझो आप ये क्यों नहीं देख का रहे हो ये जो के पॉइंट है ये जो के पॉइंट है ये जो के पॉइंट है ये क्लीयरली बायसेक्स कर रहा होगा मैं में को लेकर तो नहीं जानता बट shailise ये जो पॉइंट है जो को तो बायसेक्स कर ही रहा है की बात हम शुरुआत से कहते हैं के पॉइंट जो है वो सर्कल का सेंटर से ड्रॉप किया गया है परपेंडिकुलर है तो ये जो आपकी जो चोर्ड है तो स व अगर कॉर्ड है तो बायसेक्स हो रही है के पर स डब्लू कोड है बायसेक्स हो रही है के पर रिपीट माय स्टेटमेंट जो जो इस डबल ऑफ जेके हमारे पास zqv है स डब है मैं आपसे ये कहना चाह रहा हूं की हम क्लीयरली देख का रहे हैं जो जो है डेट इस डबल ऑफ जेके है ना तो जो को मैं सर प्लस करूंगा देखो हाथ से हाफ कैंसिल यहां देखिएगा में यहां क्या दिखेगा जेके है ना ये दिख रहा है स के अपॉन नौ दिस इसे डेट व आई एम रिप्लेसिंग दिस इस आते डीएम रिप्लेसिंग भाई तू टाइम्स जेके और यहां पे मुझे क्या दिख रहा है यहां पर मुझे दिख ही रहा है ऑलरेडी एम आई होप आप सब इस बात को लेकर मेरे साथ हैं और सर अगर आप यहां से ध्यान से देखो तो आप एक और पार्ट कैंसिल कर का रहे हो अपने हाथ से तो हाफ कैंसिल डेट यू बी कैंसिल करो सो दिस थिंग गेट्स बोलिंग डाउन इट्स हाफ है ना में / m³ नौ कैन सी फिगर आउट नमू सर कोशिश की जा सकती है अब में और एम है क्या जरा देखो में और एम के अगर मैं आपसे पूछूं तो देखो भाई यह में सर मुझे जितना याद ए रहा है मैंने निकाला था मुझे जितना याद ए रहा है मैंने शायद में निकाला था आपको याद ए रहा है तो म तो मुझे मिल गया सर कितना 12 लेकिन क्या कोई भी स्टूडेंट मेरी एम क्यों निकले मैं हेल्प कर सकता है अन्य वैन हेविंग अन्य इतिहास अन्य थॉट्स की एम क्या होगा मेरा यकीन करिए बहुत टू नहीं है आप छोड़ो आपको C1 के पता है क्या क्योंकि सर mc1 मुझे पता है mc1 मुझे पता है अगर C1 के पता हो तो एम पता करना क्यों अभी निकाला था भाई जीरो कमा जीरो से साइन की परपेंडिकुलर डिस्टेंस याद करो मैं फिर से दिखाता हूं आपको जीरो कमा जीरो से इस लाइन की परपेंडिकुलर डिस्टेंस थी 9 / 5 आई होप यू हैव बिन फॉरगॉटेन डेट ये आपको याद ए रहा है ये क्या डिस्टेंस थी सर ये थी 9/5 और अगर ये डिस्टेंस के कितना है सर सर आपने जो निकाला तो आपका C1 के ए रहा है 9/5 और mc1 आई नो S3 तो एम होगा एम जो होगा वो कितना होगा सर mc13 प्लस 24 तो ये ए जाता है 24/5 तो एम कितना आता है सर एम के आता है आपका 24/5 सो दिस इसे गोइंग तू बी 24/5 कोई तकलीफ आपत्ति परेशानी कोई भी बात भाई किसी भी स्टूडेंट को सर सिंपलीफाई करके सॉल्व किया तो क्या मिल रहा है देखो भाई बड़ी सिंपल सी बात है बड़ी सिंपल सी बात है है ना अब देख का रहे हो क्या ये फाइव ऊपर चला जाएगा 12 का 24 डबल हो जाएगा तो ये फाइव अपॉन 2 2 यानी 5/4 हो जाएगा सो दिस इसे गोइंग तू बी फाइव बाय फोर तो ये जो रेश्यो आया है इन दोनों ट्रायंगल के एरिया का वो कितना है सर वो है 5 / 4 और 5 / 4 इंसीडेंटली मुझे वापस उसके सामने मिल गया तो इस बार इन्होंने जो क्वेश्चन सेट करें वो कुछ ऐसे ही कर दिए की सारे आंसर्स आपके सामने नहीं लिख दिया उन्होंने तो आई थिंक इस बात से तो किसी स्टूडेंट को कोई आपत्ति नहीं की अगर इसका एरिया जो ए रहा है इसका और इसका वो जो रेश्यो ए रहा है वो 5 / 4 यानी आर ही ए रहा है है ना अब हम में से कई सारे स्टूडेंट्स सोचेंगे सर पीके और आर तो मिल ही गए मतलब वैन का पीस है और थर्ड का तो बेशक सी बात है अल्फा की वैल्यू ऐसी होनी चाहिए है ना हमारा तो दिमाग ऐसे ही सोचता है पर आईआईटी जी एडवांस्ड है उसने आपके साथ खेल खेल के लिए तीनों दे दिए अब आपको गैस करना है और इसी से आपका क्वेश्चन पूरा कंप्लीट होगा की अगर आप इस फोर्थ पार्ट को सॉल्व करके मुझे बता दें की सेंट और यू में से कौन सी वैल्यू होगी और मैं फिर से रिपीट करूंगा की मैं इस पार्ट को यहीं पर होल्ड करूंगा इस पार्ट को मैं इसलिए होल्ड करूंगा क्योंकि अभी आपने नहीं पढ़ा है पैराबोला अभी आपने नहीं पढ़ा है पेरा बोला और अगर मैं इस पैराबोला वाले पार्ट को पटना भी चाहूं और मैं उसे कॉमन टांगें की इक्वेशन निकलोगे तो जिस तरीके से मैं निकलूंगा जैसे मैं निकलूंगा जो सारी बातें मैं करूंगा वो आपको अभी फिलहाल स्टेज पर नहीं समझ आई तो इस वाले पार्ट को जो इसका आखिरी पार्ट है इस वाले पार्ट को होल्ड पर रखते हैं और इस वाले पार्ट को जिस क्वेश्चन का ये वाला हिस्सा है इसे हम नेक्स्ट लेक्चर में सॉरी नेक्स्ट चैप्टर में जब हम किस पर बात करेंगे जब हम बात करेंगे पैराबोला पर तब हम बहुत डीटेल्ड डिस्कशन करेंगे मेरा आप यकीन करिए मेरा आप सच में यकीन करिए आप बहुत डिटेल डिस्कशन करिए अगर आपको हिंट चाहिए फिर स्टील वेरी क्यूरियस वेरी इनक्विजिटिव की नहीं सर अमिताभ बता दो की कैसे सोचना होगा तो जिन स्टूडेंट्स को अगर पैराबोला पता है जिन स्टूडेंट पैराबोला पढ़ रखा है मैं उनको थोड़ी सी ह टी इतिहास देने की कोशिश करता हूं की इस क्वेश्चन को कैसे एप्रोच करेंगे पहले तो ऐसे सोचो सर ये जो ये जो लाइन है ये जो लाइन है ये किसकी कमेंट टैसेंट है सर ये C1 और C3 की कमेंट्री अगर ये C1 और C3 की कॉमन टांगें तो कहना है की इक्वेशन होगी C1 - c3=0 एक तरीका तो ए जाएगा एक मेरा तरीका ये हमारे अकॉर्डिंग ए जाएगी बहुत आसानी से क्योंकि C1 भी पता है C3 भी पता है तो कॉमन टांगें की इक्वेशन ऐसे निकल जाएगी अब कॉमन टांगें की इक्वेशन अगर निकल गई है तो आप बताओ आपके पास पैराबोला होगा x² = 8 अल्फा / आपके पास पैराबोला क्या है आपके पास पैराबोला है x² = व्हाट 8 अल्फा ए तो सर मैं इस पैराबोला की टेंशन की इक्वेशन लिखूंगा मैं इस पैराबोला की टेंशन को जैसा भी आपको इतना लगे जैसा भी आपको ठीक लगे लिख लीजिए अब कोई सा भी तरीका ले लीजिए आप पैरामीटर फॉर्म ले लीजिए जो ठीक लगे लिख लीजिए ये तब समझ आएगा जब पैराबोला पढ़ चुके होंगे इसलिए अब मैं इसके नीचे नहीं जाना चाह रहे की इसके बाद क्या करना है आप क्या करेंगे इसकी टेंशन की इक्वेशन निकलेंगे और इस पैराबोला की टैसेंट और इसकी जो टैसेंट की इक्वेशन है ये से है वो यही का रहा है ना की वो जो कॉमन टांगें है वो इसकी जो कमेंट टैसेंट है वो वही उसे पैराबोला की भी tensionant है तो मेरा एक कहना है की दो लाइन से है तो उनके एक्स ए और कांस्टेंट हम जो रेश्योस हैं koficious के एक्स और ए के coefficiency के कांस्टेंट का रेशों वो से होगा तो उससे मैं उसे लाइन की स्लोप उसे लाइन की स्लो और उसके वो जो कांस्टेंट है उनमें रिलेशन निकल पाऊंगा उसके थ्रू में अल्फा तक ए जाऊंगा उसके थ्रू मैं अल्फा तक ए जाऊंगा और अल्फा के थ्रू फिर मैं फाइनल पार्ट भी मार्क कर दूंगा बट क्योंकि अभी आपने पेरा बोला नहीं पढ़ा है इसलिए मैं ये पार्ट अभी डिस्कस आपसे बिल्कुल भी नहीं कर रहा हूं क्या अभी तक कोई विदाउट क्या अभी तक कोई भी परेशानी और इस बेसिस पर जब हम से आंसर पूछे जाएंगे तो आई थिंक मैं बहुत आसानी से माफ कर दूंगा की आप कौन सा ऑप्शन मार्क करेंगे है ना तो पूछ रहा है विच ऑफ डी फॉलोइंग इसे डी ओनली करेक्ट कांबिनेशन तो इनमें से कौन सा इकलौता करेक्ट कांबिनेशन है जो है तो मैं देख का रहा हूं सर वैन के लिए पी था वैन के लिए पी नहीं दिख रहा है तू के लिए क्यों था तू के लिए क्यों यानी shailise है सर आपका जो पहला क्वेश्चन आएगा उसका सीधे-सीधे मार्क करेंगे ऑप्शन सी डू यू ऑल एग्री विद डेट आई थिंक इसको लेकर तो किसी भी स्टूडेंट को कोई डाउट या कन्फ्यूजन नहीं है आई होप ये बात आपको समझ आई इसी के साथ-साथ जब हम आगे बढ़े तो हमारे पास फिर से वही बात पूछे जा रही है विच ऑफ डी फॉलोइंग इसे दी ओनली करेक्ट कांबिनेशन है ना तो अभी फोर्थ हमने सॉल्व किया नहीं है तो फोर्थ को लेकर मेरा ओपिनियन बहुत सॉर्टेड है है ना सॉरी इस बार वो पूछ रहा है ओनली इनकरेक्ट है ना पिछले बारे में पिछली बार हमसे पूछ सकता हूं ओनली करेक्ट है ना तो सर 1p तो सही था तो ये गलत आंसर है मतलब ये आंसर नहीं हो सकता क्योंकि वो इनकरेक्ट पूछ रहा है थर्ड का आर भी हमने निकाला था तो ये सही था तो ये आपका आंसर नहीं होगा क्योंकि इनका एक पूछ रहा है और अब बस बात वही है अटक रही है की फोर्थ का एस होगा या यू तो ये पार्ट हमने सॉल्व नहीं किया है फोर्थ को हमने सॉल्व नहीं किया है इसलिए मैं इस पार्ट को स्किप कर रहा हूं और ये पार्ट हम नहीं डिस्कस कर रहे हैं आई होप यहां तक कोई डाउट नहीं है सर मुझे कहा जा रहा है डी सेंटर ऑफ तू सर्कल सी वैन एंड सी तू एच ऑफ यूनिट रेडियस आर ए डिस्टेंस ऑफ सिक्स यूनिट्स तो जीवन और C2 दो सर्कल से ठीक है सो आई हैव तू सर्कल्स सी वैन एंड सी तू और सर दोनों जो सर्कल्स हैं आपके जो दोनों सर्कल्स हैं आपके यह दोनों जो है आपके सर्कल्स इनकी रेडियस से है कितनी वैन यूनिट है ना एंड दे आर सेपरेटेड बाय अन डिस्टेंस ऑफ कितना सेपरेटेड बाय अन डिस्टेंस ऑफ कितना सिक्स यूनिट्स आई होप आपको बात समझ ए रही है और दोनों के वीडियो हैं मतलब दोनों आईडेंटिकल सर्कल है बस सेंटर अलग-अलग जगह सिचुएटेड है और ये दोनों सेंटर्स के बीच की अगर मैं डिस्टेंस निकलूं अगर दोनों सर्कल्स के बीच की अगर मैं डिस्टेंस निकलूं तो वो जो होगी सर वो होगी आपकी क्लीयरली कितनी भाई वो हो गई सर आपकी 6 यूनिट्स ड्यूल गेट डेट तू दिस इस हो थिंग्स आर बेसिकली प्लीज एवरीवन विथ मी सो फार कंफरटेबल कोई डाउट तो नहीं सो ये आपके दो सर्कल से इनकी रेडियस है वैन यूनिट एच एंड देयर सेपरेटेड बाय अन डिस्टेंस ऑफ कितनी बाय अन डिस्टेंस ऑफ 16 मिनिट्स है ना कोई तकलीफ आई थिंक नहीं होनी चाहिए अब वो क्या का रहा है लेट पी बी डी मिड पॉइंट ऑफ डी लाइन सेगमेंट जॉइनिंग डी सेंटर ऑफ C1 एंड C2 ऑल राइट तो अब आपके पास ये दो सर्कस जो हैं इसमें ये आपका सर्कल है C1 और ये सर्कल है C2 है ना इनको ज्वाइन कर रही है लाइन उसका मिड पॉइंट उसका मिड पॉइंट कौन है सर उसका मिड पॉइंट है पी तो शैल आय से मिड पॉइंट पी पॉसिबल आई यहां लाइक करना चाहिए मेरे अकॉर्डिंग तो यही लाइक करना चाहिए आपको अगर कहीं और दिख रहा हो तो जरूर बताइएगा अब क्या सर पी मिढ्वाइंट ऑफ डी लाइन सेगमेंट जॉइनिंग डी सेंटर C1 एंड C2 एंड सीबीआर सर्कल सो एंड आदर सर्कल हज बिन इंट्रोड्यूस्ड एक तीसरा सलकर पिक्चर में आता है टचिंग सर्कल्स C1 एंड C2 externalli तो सर एक और सर्कल पिक्चर में आया जिसने इन दोनों सर्कस को externalli टच किया ओके तो मैं एक और सर्कल बनाने की कोशिश करता हूं जो इन्हें externalli टच करें है ना सो आई एम गोइंग तू ड्रा अनादर सर्कल बेसिकली जो की थोड़ा सा बड़ा बन जाने वाला है एंड आय होप दिस डेजन बदर यू ऑल इट ऑल है ना तो ये आपका एक सर्कल बन रहा है आई होप इन सारी बातों से आपको कोई आपत्ति नहीं हो और ये इन दोनों सर्कल को इस तरीके से externalli टच करेगा तो सर एक और सर्कल है आपका जो की क्या है सी जो की क्लीयरली आपके सेंड सर्कल्स को इस तरीके से टच करेगा और ये इसका सेंटर अब आप एक बात खुद सोच के बताओ अगर मैंने एक लाइन ड्रॉ की होती है ड्रॉन है स्ट्रेट लाइन इस मिडिल पॉइंट से इस सर्कल के सेंटर तक आई होप आई नीड नॉट टेल यू ये क्लीयरली परपेंडिकुलर होगी के सर क्योंकि आप देखो ना ये जो दोनों सर्कल्स यू नो टच कर रहे हैं externalli यहां से 90° यहां से 90 डिग्री और इससे भी ये 90° इससे भी ये 90 डिग्री आप मेरी बात समझ का रहे हो मतलब क्लीयरली इसकी ज्वाइन करने वाली लाइन है इसकी ज्वाइन करने वाली लाइन 90 डिग्री पर जा रही है अब बात समझना जैसे की इनको जब आप ज्वाइन करेंगे इनको जब आप ज्वाइन करेंगे तो ये क्लीयरली इनके सेंटर्स के बीच के डिस्टेंस के रेडियस के इक्वल होगी और जब इनको भी आप जब ज्वाइन करेंगे सॉरी जब आप इसके सेंटर और इसके सेंटर को भी ज्वाइन करेंगे तो ये आपके लिए इस तरीके से दिखेगी है ना मतलब मैं आपसे ये कहना चाह रहा हूं बड़ी बेसिक सी सुलझी हुई इसी डायरेक्ट सी बात है सर ये एक आपका क्लीयरली आइसोसेलस ट्रायंगल बन रहा है क्यों आइसोसेलेस आप समझो इससे इसकी जो डिस्टेंस है वो क्या है वो इसकी रेडियस प्लस इसकी रेडियस इससे इसकी जो डिस्टेंस है वो क्या है सर इसकी रेडियस प्लस इसकी रेडियस ये रेडियस तो से है क्योंकि से सर्कल की और ये रेडियस भी तो से है आप समझ रहे हो तो ये आगे और ये आगे इक्वल लेंथ की होती है और पी जो है वो इस आगे का मिड पॉइंट है तो मैं क्या ये का सकता हूं एक आइसोसेलस एक आइसोसेलस ट्रायंगल में जब आप दो इक्वल साइड के कन्वर्जिंग पॉइंट से जो थर्ड साइड है वहां पर परपेंडिकुलर या एल्टीट्यूड ड्रॉप करते हैं तो वो उसे बायसेक्स करता है या अगर ये मेडियन है तो वो एल्टीट्यूड भी होगा राइट एंगल ट्रायंगल में इस वाली इस वाली लाइन के लिए बस आई होप ये बातें आपको समझ ए रही है मैं बड़ा बेसिक सा कंक्लुजन आपको देना चाह रहा हूं की सर ये जो लेंथ है ये जो लेंथ है इक्वल है और ये जो एंगल है ये 90 डिग्री एंगल है बिना किसी तकलीफ के आई होप एवरीवन ऑन डेट इस एवरीवन कोई डाउट यहां तक के पिक्चराइजेशन को लेकर आई होप आप समझ का रहे हो ये जो डिस्टेंस है ये वैन है और ये जो डिस्टेंस है ये भी है ये भी वैन है और ये भी वैन है और अगर C1 C2 के बीच की लेंथ अगर मुझसे किसी ने पूछी होती तो मैं कहता की वो डिस्टेंस कितनी है सर वो डिस्टेंस है क्लीयरली सिक्स यूनिट्स डू यू ऑल एग्री विद डेट आई होप नो वैन इस बौद्ध बौद्ध बाय दिस मच इनफार्मेशन कन्वेट सो फार नौ व्हाट डू सी हैव नेक्स्ट इफ अन कॉमन टांगें तू सी वैन एंड सी पासिंग थ्रू पी इस अलसो अन कॉमन टांगें इन C2 अब बात सुनना है स्टूडेंट्स वो ये का रहा है की अगर आपने अगर आपने एक कॉमन टांगें ड्रॉ की किस-किस की अगर आपने कमेंट ड्रॉ की किसके लिए सर C1 और सी के लिए वह को इंसीडेंटली कमेंट इंजन बन गई C1 और C2 की भी अगर मैं एक ऐसा कोई पॉइंट बनाने की कोशिश करूं तो मुझे वो लाइन कुछ ऐसी दिखेगी मुझे वो लाइन कुछ ऐसी जाती हुई देखनी चाहिए जो की मैं कुछ थोड़ा सा टिल्ट करके बनाना चाहूंगा इससे पहली बात तो वो लाइन आपकी कमेंट टेंट यहां से नहीं जा रही होगी वो कुछ ऐसी जा रही होगी है ना अब बस यही प्रॉब्लम है यहां पर थोड़ा ड्रॉ करने में दिक्कत आएगी तो मैं इसे कहूंगा की सर इसे थोड़ा सा आपको सहूलियत से ड्रॉ करना होगा और इसे थोड़ा तो ट्रीट करना पड़ेगा वर्ण बात नहीं बनेगी पर उसे प्रॉब्लम में यह प्रॉब्लम है आई होप आप समझ का रहे हो मुझे एक्चुअली ये सर्कस इस तरीके से लेने चाहिए द की वो कॉमन टैसेंट ऐसी बन जाती है ना अब अगर नहीं बन का रही है तो क्या तो बस मैं थोड़ा सा रफ्ता स्केच बना रहा हूं एक कॉमन सी टेंशन ऐसी की वो इन दोनों की टांगें है और वो ले लिया आपके इस सर्कल की भी टैसेंट है क्या ये पूरा सिनेरियो आपको समझ ए रहा है अच्छे से देखो इस बात को यहां पर वो कहना चाह रहा है की जो आपने कमेंट टैसेंट बनाई है सर जो आपके सी वैन और सी की कॉमन टांगें है वो क्लीयरली आपके C2 की भी कॉमन टेंसेंट है आप अगर मेरी बात से एग्री करें तो बिल्कुल सर मैन लिया आपकी बात इस बात से तो कोई प्रॉब्लम नहीं है तो अगर मैं एक tagent ड्रॉ करूं अगर मैं इस तरफ से करूं तो बन जाएगी क्या बस एक बार ड्रॉ करने की कोशिश करते हैं बन गई तो बहुत खुशी की बात होगी वर्ण लाइफ में जो है वो है चीज है ना अगर मैं थोड़ी सी इसको चीजों को शिफ्ट करूं तो सर वो कॉमन टांगें तो हम नहीं बना पाएंगे क्योंकि उसे चक्कर में प्रॉब्लम हो जाएगी की फिर वो इसकी टैसेंट नहीं बन पाएगी अगर मैंने इसकी बनाई तो है ना तो दिक्कत यही है वो फिर कहीं और निकल जा रही है तो इससे नहीं बनाते हैं और हम हमारे अकॉर्डिंग अपनी सवालिया से कमेंट इंजन बना लेते हैं मैंने कोशिश की मतलब एक्चुअली पता है सारा फेर इसमें ड्राइंग का है अगर आपने वो सर्कस इस तरीके से बनाए होते हैं की पहले अपने टैसेंट बना ली होती और फिर सर्कल बनाया होता तो और आसानी से मिल जाती है बट चूंकि अब बना चुके हैं तो मैं वो कमेंट इंजन बनाने की कोशिश कर रहा हूं की सर मैन लेते हैं वो कॉमन टांगें है ये है ना ये आपकी वो कमेंट टैसेंट है जो की आपके इस तरीके से ये जा रही है दिस इसे डेट कॉमन टेंसेंट आय होप यू आर नॉट बाय डेट सर हमारा आपसे सवाल है मैन लिया आपकी बात की ये जो कमेंट्री है वो C1 और सी की कमेंट्री आप समझ का रहे हो C1 और सी को यहां पर और सीटों यहां पर टच कर रही है एक्सेप्टेड डेट यू थिंक डेट इट विल गो और पास थ्रू दिस पॉइंट पी सी शल फिगर डेट आउट लेटर मैं पता कर लूंगा बहुत सारे मेरे पास तरीके होंगे ये निकलने के की वो पी से पास होगी या नहीं बट शैल आय से की सर यहां तक तो चीज मुझे समझ ए रही है की हान ये जो C1 और सी की जो कॉमन टांगें है वो क्लीयरली आपके सीट और सी की भी कमेंट इस तरीके से बनेगी क्या इस बात से कोई तकलीफ अब अगर मैं आगे बाद बधाई अब घर में चीज बेसिक सिमेट्री से सोचो तो क्वेश्चन में क्या दिया गया है उसे पे बात करते हैं आई थिंक अब जब हम से पूछा जा रहा है तुमसे फाइनली पूछा जा रहा है फाइंड डी रेडियस ऑफ डी सर्कल सी हमसे वो जो बड़ा वाला सर्कल है जहां से हमने कॉमन टांगें ड्रॉ की है उसे पॉइंट पर ये बात की जा रही है की सर वहां से जो आपने ड्रॉ किया है ये जो चीज है वो रेडियस क्या होगी वो कैसे आप निकलेंगे इस पर थोड़ी बात कर ली जाए तो बहुत खुशी होगी अब मैं बेसिक सेनेटरी अप्लाई करना चाह रहा हूं बेसिक सा ऑब्जर्वेशन आपको देना चाह रहा हूं की आप प्लीज इस को समझो यह सर्कल और यह सर्कल सिमिट्रिकल है एक्सेप्टेड अब सर जो दूसरी commentangent होगी वह बेसिकली यहीं से जाएगी वो कुछ इस तरीके से ही जा रही होगी की वो भी आपकी यहां से यू नो टच करते हुए जारी होती और आपको इस तरीके से ये निकल रही होती क्या आप मेरी बात समझ का रहे हैं क्योंकि चीज सिमिट्रिकल है सर आप ऐसा कैसे-केयर हो सिमेट्री करें चीज सिमिट्रिकल इसलिए है स्टूडेंट्स इस बात को समझो क्योंकि आप बात समझो इसकी रेडियस मैंने इसकी रेडियस 1 है इनके बीच में इनके बीच में जो सेंटर्स के बीच में डिस्टेंस है वो 6 यूनिट्स हैं और क्लीयरली ये जो सर्कल टच कर रहा है तो चीज perfectuli सिमिट्रिकल होंगी मतलब इसका रिलेशन कुछ और हो भी नहीं सकता अगर ये जो टैसेंट है जो इसे यहां टच कर रही है जो की ऑफ कोर्स उसे पे क्या है 90° एंगल बना रही है अगर आप इसे देखना चाहो तो ये जो होगा एंगल 90 डिग्री होगा तो सर अगर आपने इसके सेंटर से इस पर परपेंडिकुलर ड्रॉप किया होता तो ये भी क्या होता है 90 डिग्री और एक सिर्फ डेट अगर ये 90 डिग्री और ये 90 डिग्री है तो बड़ी ही बेसिक सी बात जो मैं देख का रहा हूं की सर इससे ध्यान से सुनना बहुत कम की बात है बहुत कम की बात है हम प्रूफ कर देंगे की ये से क्यों पास हो रही है आप खुद सोच के बताओ सर बहुत बेसिक सी बात ये लेंथ वैन है तो हम जान का रहे हैं कितनी है वैन बिल्कुल एक्सेप्ट किया सर आपकी बात किसी भी पॉइंट से पास कर लेते हैं मैन ले हमें नहीं पता अभी वो है देख लेंगे वो पी हो गया नहीं तो कैन आई से सर ये एंगल वर्टिकली अपोजिट होते हैं तो अगर ये मैन लो थोड़ी देर के लिए अल्फा है तो क्या ऐसे ही एंगल बी अल्फा होगा क्या इस बात से आप एग्री करो तो सर आप खुद सोच के बताओ ये क्या एंगल अल्फा उसके सामने की लेंथ ये वैन एक एंगल अल्फा सामने की लेंथ ये वैन है ना ये भी वही सर्कल ये भी वही सर्कल तो कहना इसे अगर ये पॉइंट मिड पॉइंट है अगर ये पॉइंट मिड पॉइंट है छोड़ो एक पॉइंट है जहां से आपने तेनजेन ड्रा किया तो इस tagent के लिए तो इस इंजन के लिए से होने वाली है या फिर आप बेसिकली ये देख का रहे हो क्या की ये वाला जो एंगल है इसको अगर मैं का डन सर्कल के सेंटर को मैं का देता हूं अभी फिलहाल लेट से ए और सर्कल के सेंटर को का देता हूं अभी तो क्या मैं आपसे का सकता हूं ये जो आप देख का रहे हो लेट्स से मैं इस पॉइंट ऑफ कॉन्टैक्ट को का देता हूं एम और इस पॉइंट ऑफ कॉन्टैक्ट को का देता हूं एन तो अब देख का रहे हो एपीएम और क्या और क्या बीपी यह दोनों ट्रायंगल सिमिलर है दे आर परफेक्टली बाकी दो साइज भी उसी प्रोपोर्शन में होंगी और इससे मुझे ज़ाहिर सी बात नज़र ए रही है की ये ट्रायंगल ना सिर्फ सिमिलर है बल्कि कांग्रेस हैं आप समझ का रहे हो ये ट्रायंगल कांग्रेस है मतलब एक्जेक्टली एपीएम वाला ही ट्रायंगल आपका बीपी बन रहा है यानी की आप कहोगे ये लेंथ और ये लेंथ दी इक्वल होगी और ऑफ कोर्स आपकी ए पी और पी बी इक्वल हो गया यानी पी इस डेफिनेटली गोइंग तू बी डी मिड पॉइंट डिड यू ऑल गेट डेट आई होप यहां तक कोई परेशानी आपत्ति नहीं है अब एक बात गौर से suniyega स्टूडेंट्स हम ये प्रूफ कर चुके हैं हम एक बड़ा ही बेसिक सा ऑब्जर्वेशन देख चुके हैं की सर ये जो एंगल है 90° है डेट सर ये जो एंगल है 90° है अगर ये एंगल 90 डिग्री था और अगर ये एंगल अल्फा है तो शैल आय से ये जो एंगल होगा ये होगा 90 - अल्फा आई होप आप इस बात को एग्री करते हो एक्सेप्ट करते हो सर इसके साथ-साथ मुझे पता है की ये भी 90° एंगल है अब suniyega ध्यान से यह 90 - अल्फा है तो दिस एंगल इसे अलसो वांट तू बी अल्फा आई हैव गिवन यू ए लॉट ऑफ इनफॉरमेशन लॉट ऑफ क्लूज अन लॉट ऑफ हिंट्स बेसिकली एंड नौ यू आर इन ए सूटेबल पोजीशन तू फिगर आउट दी रेडियस ऑफ दिस सर्कल सी ट्रस्ट मी आपके पास वो सारी इनफॉरमेशन और डाटा है जिससे आप इस बड़े वाले सर्कल की रेडियस निकल सकते हो आप बहुत आसानी से निकल सकते हो अगर आपको पॉइंट्स चाहिए सो लेट मी कॉल दिस पॉइंट लेट से अब और इससे मैं का लेता हूं सी तो सेंटर ही है लेट्स कॉल इट दी फॉर अवॉइड आप बहुत आसानी से सर्कल की रेडियस निकल सकते हो मेरा यकीन करिए है ना अगर आपने इसे कहा था तो लेट्स से हम का लेते हैं के बस कुछ रैंडम से पॉइंट्स में दे दे रहा हूं हमारे पास हर वो जरूरी चीज है जो निकल सकते हो एक बार बस इस वाले पॉइंट पर सोच के देखो की आप यहां पर इस सर्कल की रेडियस कैसे निकलोगे मैं आपसे दो ट्रायंगल में देखने को कहता हूं आप बड़े बेसिक से उन दोनों ट्रायंगल को देखो जिन पर हम बात करते ए रहे हैं इन बेसिक से सिंपल से दो ट्रायंगल सोच के देखो स्टूडेंट्स अगर मैं आपसे बोलूं की सर पहले तो आप ये देखो डीएमपी ट्रायंगल तो मेरा यकीन करिए आप बहुत आराम से इसकी रेडियस निकल पाएंगे सर क्या बात कर रहे हो ऐसा कैसे निकल लेंगे मेरा यकीन तो करिए आप आपको पता है चीजें ऐसे कैसे पता है बात समझो स्टूडेंट्स पहले तो बात करो इस पीएनबी ट्रायंगल में क्यों सर आप बस एक बात बताओ ये कितना है वैन अच्छा पेमेंट पॉइंट था अब का अब कितना था 6 तो कैन आई से सर ब जो होगा वो होगा 3 आप सन रहे हो ध्यान से सर मुझे पता है पीएनबी एक राइट एंगल ट्रायंगल है विच इस राइट एंगेल्ड आते डी वर्टेक्स एन तो अगर यह एंगल अल्फा देखना है ना बहुत ध्यान से देखना मेरी बात को ये एंगल अल्फा के सामने वाली साइड परपेंडिकुलर और ये आपका हाइपोटेन्यूज है क्योंकि यहां पे 90° है उसके अपोजिट तो shailise है की अगर मैं एंगल ट्रायंगल पीएनबी में साइन अल्फा निकलूं ट्रायंगल पी एंड वी में साइन α निकालो तो वो हो जाएगा 1 / 3 डू यू ऑल एग्री विद डेट इस से क्या हो रहा है सर आप इससे बस इस बात पर गौर फरमाइए की अगर मैं आपसे पूछता अगर मैं आपसे पूछता की आप देखो ट्रायंगल कौन सा डीएमपी ट्रायंगल दी एमपी में अगर मैं आपसे पूछूं क्या बहुत ध्यान से देखना स्टूडेंट्स मुझे डीएम निकलना है जो की क्या है आपके सर्कल की रेडियस क्या मुझे इस ट्रायंगल में ट्रायंगल डीएमपी में मुझे इस अल्फा के लिए इस आर से रिलेटेड कोई और इनफॉरमेशन पता है क्या रिपीट माय स्टेटमेंट अगर आप पता करना चाहें तो आपको एक बहुत जरूरी बात बताएं है ना आई रिपीट माय स्टेटमेंट अगर आप पता करना चाहें तो ये जो आपका ट्रायंगल है दी एमपी बहुत अच्छी से सिंपल से इंटरेस्ट क्वेश्चन हो जाएगा क्योंकि हम साइन α ऑलरेडी निकल चुके हैं तो आपके पास अल्फा है आप इसका साइन ले लीजिएगा आपको ये साइड आर निकालनी है और आपको सोचना है ट्रायंगल दी एमपी में जो की अब बहुत ही अपेरेंट बहुत ही एविडेंस ब्लू है जिससे आप सीधे सीधे सोच सकते हो अच्छा सर ऐसा कैसे सोच सकते हैं ऐसे कैसे ए सकते हैं बस ये बात बताओ मेरा कहना है एक सिंपल सी बात हो तो स्टूडेंट्स आप एक बात बताओ आप एमपी को लेकर के ओपिनियन रखते हो आप एमपी कैसे निकलोगे सर एमपी कैसे निकलेंगे एमपी के लिए बड़ा सिंपल सा थॉट है ये वैन है और आप कितना है सर आप वही बात है अगर मुझे पता है ap3 है और ये वैन है और आई होप आप देख का रहे हो ये क्लियर 90 डिग्री एंगल है क्योंकि ये इसकी कॉमन टांगें थी राइट तो इसकी रेडियस से भी है परपेंडिकुलर ड्रॉप होगा तो अगर मैं ट्रायंगल आपसे पीएम में पूछो अगर मैं आपसे पूछूं ट्रायंगल पता है परपेंडिकुलर पता है बेस पता करना है तो एमपी क्या होगा अगर मुझे बेस पता करना है तो एमपी की वैल्यू मैं लिखूंगा एमपी की वैल्यू क्या लिखूंगा 3 का स्क्वायर वैन का स्क्वायर 1 9 - 1 8 8 का अंडर रूट 2√2 तो एमपी कितना आता है सर एमपी आता है 2√2 तो ये जो एमपी की लेंथ आती है डेट्स हो तू बी तू रूट तू मुझे एमडी निकलना है अब सुनना ध्यान से बहुत कम की बात है एक एंगल अल्फा अल्फा के सामने वाली साइड और उसके नीचे वाली साइड मतलब मुझे परपेंडिकुलर और बेस से डील करना है तो मुझे टेक्निकल डील करना है कैसे 10 अल्फा से किस वाले ट्रायंगल में दी एमपी में ध्यान से 10 अल्फा क्या होगा आप बताओ सर 10 अल्फा जो होगा कुछ नहीं होगा बड़ी बेसिक सी बात है 10 अल्फा क्या होगा सर ध्यान से देखो 10 अल्फा विल बी परपेंडिकुलर अपॉन बेस यानी एमपी अपॉन एमडी ये कितना होगा सर एमपी अपॉन एमडी सर इन दोनों में कुछ जानता हूं क्या मैं टेन अल्फा निकल सकता हूं सिन अल्फा की हेल्प से मैं एमपी जानता हूं और शायद निकल सकता हूं क्या मेरा स्टेटमेंट सन समझ और सोच का रहे हैं तो सर पहले तो 10 अल्फा निकलने की कोशिश करते हैं तो फिर वही बात सिन अल्फा कितना आया था 1 / 3 आप सन रहे हो क्या सिन अल्फा वैन बाय थ्री आया था तो क्या मैं हाइपोटेन्यूज निकल सकता था बिल्कुल निकल सकता था सर यहां से देखना था क्या बिल्कुल निकल सकते द सर ध्यान से देखो वही बात अगर ये 2√2 है तो पीएनबी तो तू रूट तू होगा तो टाइम कितना हो जाएगा सर ये हो जाएगा 2√2 ड्यूल गेट माय पॉइंट अरे सर सीधी बात है एमपी और पी एन इक्वल लेंथ के ही तो होंगे आई होप आप ये बातें एग्री करते हो तो पीएम की वैल्यू होती है 2√2 अब अगर मैं आपसे पूछूं टेन अल्फा तो आप कहोगे 1 / 2√2 तो 10 अल्फा कितना ए जाएगा सर 10 अल्फा होगा 1 / 2√2 कोई दिक्कत तो नहीं इसे इक्वल तू एमपी एमपी कितना है सर एमपी है आपका कितना 2√2 और एमडी कितना है वही आप निकलने चले हो सर आई होप यहां तक कोई डाउट कोई कन्फ्यूजन कोई दिक्कत नहीं है क्या यहां से अब मैं फाइनली इस सर्कल की रेडियस का आंसर दे सकता हूं सर कोशिश करो और वहां पहुंचाओ ये यहां पहुंचाओ 2 2 4 √2 √2 4 2 कितना आते तो जो आपके सर्कल की रेडियस ए रही है वो कितनी ए रही है 8 एंड दिस 8 इस नथिंग बट योर करेक्ट आंसर फाइनली क्वेश्चन आपको समझ आया ही है इसमें क्या ठीक से प्लॉट करते हैं बेसिक ऑफ ज्यामिति अप्लाई करते तो आसानी से सोच पाते हमें किस किस से डील करना पड़ा एंगल से ट्रिगो से रेडियस ए कॉमन टांगें से सर्कल से डिस्टेंस पर यह सब करने में हमारी हेल्प किसने की इस विजुलाइजेशन ने इस प्लाटिंग ने इस ड्राइंग में अगर आप चीज अच्छे से सोच पाओ ना तो चीज आसान है मेरा यकीन करिए और बस कुछ ऐसा ही सोचना है पता है इस क्वेश्चन का ट्रिकी पार्ट किया था इस क्वेश्चन का ट्रिकी पार्टी था जहां पे सब स्टूडेंट शायद इस बात को लेकर घबरा रहे होंगे की सर ये कैसे बनाएंगे एक कॉमन टांगें जो C1 की भी कॉमन टेंट है C1 और सी की भी है और वही कमेंट ट और सी की भी है अगर ये पार्ट हम क्रैक कर लें तो मेरा यकीन करिए क्वेश्चन टू नहीं था बस यही पार्ट थोड़ा सा मुश्किल था जिससे शायद स्टूडेंट्स थोड़ा सा दिक्कत में करते हैं बाकी चीज इस क्वेश्चन की आसान थी जो की duabil थी आई होप ये सारी बातें आप याद रखेंगे और इन्हीं बातों के साथ मूव करते हैं अब आज के अगले क्वेश्चन पर नेक्स्ट क्वेश्चन क्या दिया है ध्यान से पढ़िए ये कब पूछा गया है सर ये आईआईटी जी एडवांस्ड ने 2011 में पूछा है अच्छा क्वेश्चन है ध्यान से suniyega वो का रहा है सर एक स्ट्रेट लाइन है जो आपके सर्कुलर रीजन यह कैसी सर्कुलर रीजन होप आप जानते हो x² + y² = 6 एक कल होगा जिसकी रेडियस होगी अंडर रूट 6 सेंटर होगा 0 पर उसके अंदर के अंदर वाले रीजन में लाइक करने वाले पॉइंट्स हैं अब वो ये का रहा है उसे सर्कल को ये लाइन दो हसन में डिवाइड कर रही है ठीक है सर फिर वो ये का रहा है की इनमें से इनमें से ये चार आपके पॉइंट्स हैं इनमें से कितने पॉइंट्स ऐसे हैं जो वो जो लाइन जब सर्कल को डिवाइड कर रही है ना दो हसन में तो उनके स्मॉलर पार्ट में वो लाइक कर रहे हैं ऐसे कितने पॉइंट्स आपके होंगे क्या आपको सबसे पहले तो ये क्वेश्चंस समझ आए आई होप इस क्वेश्चन को लेकर कोई डाउट या कन्फ्यूजन या कोई नहीं है पर सर सोचेंगे कैसे ये बात कैसे फिगर आउट करेंगे की कौन सा पार्ट्स स्मॉलर है और कौन सा पार्ट बिगड़ है ये थॉट आप कैसे लेकर आएंगे इस बारे में हमें थोड़ी कन्फ्यूजन है एक बात बताओ एक बड़ी बेसिक सी बात बताओ आप थोड़ा चीज में विजुलाइज तो करो कैसे करें सर विजुलाइज पहले तो ये सर्कल बनाओ X2 + y² = √6 है ना सो आई वांट तू प्लॉट दिस सर्कल मैं क्या करूंगा ये आपका ए एक्सेस हो जाएगा है ना और इसी तरीके से ही आपका क्या हो जाएगा सर एक्स में क्या करना है सर इसमें आप लेकर ए जाओ सर एक सर्कल सर्कल की खास बात क्या है सर्कल इस बेसिकली सेंटर सर्कल का सर अगर ओरिजिन के साथ-साथ में बात करूं आपसे की सर्कल के सेंटर की तो वो है ओरिजिन और उसकी रेडियस कितनी है अंडर रूट सिक्स तो ये आपका जीरो कमा जीरो और इसकी रेडियस है अंडर रूट सिक्स तो ये पॉइंट हो जाएगा अंडर रूट सिक्स कमा जीरो है ना ये पॉइंट क्या हो जाएगा सर दिस पॉइंट विल बी माइंस अंडर रूट 6 - √6 ड्यूल एग्री विद डेट आई होप डेट इस इन बदर यू ऑल इट ऑल नौ व्हाट मी गोइंग तू डू नेक्स्ट अब हम बात करेंगे सर इस स्ट्रेट लाइन के बारे में जिसको लेकर आपका क्या ओपिनियन है डी स्ट्रेट लाइन इस 2X - 3y = 1 है ना ये कितना लिखा है 2X - 3y = 1 कैन आई सिंपलीफाई दिस फादर अगर कोशिश की जाए तो बिल्कुल सर ध्यान से देखना आप ऐसे ऐसे दिख सकते हो आप लिख सकते हो कैसे एक्स / 2 + ए / 3 = 1 क्या इसे देखकर आपको कुछ स्ट्राइक कर रहा है सर कुछ तो कर रहा है क्या याद ए रहा है भाई आप याद कर रहे हो एक्स अपॉन ए + ए / बी = 1 जो की ए कमा जीरो कमा बी पर आपके एक्सेस पर इंटरसेप्ट्स बनाती है तो एक सेंटर से क्या होगा 1 / 2 और ए इंटरसेप्ट क्या होगा जीरो कमा माइंस वैन बैटरी वैन बाय तू कितना होता है सर 1 / 2 होता है 0.53 मुझे आप थोड़ी सी हेल्प करिए अंडर रूट सिक्स को फिगर आउट करने में बड़ी सिंपल सी बात है ध्यान से देखो √4 कितना होता है सर अनरूट फोर होता है √93 तो टेक्निकल 2 और 3 के बीच में होगा आप मेरी इस बात से एग्री कर रहे हो तो shailise है मेरे बस अपनी सहूलियत के लिए अभी फिलहाल अभी फिलहाल बस ये मैन सकता हूं क्या की सर अभी अपने इस क्वेश्चन को सॉल्व करने के लिए अंडररूट सिक्स जो होगा मोटा-मोटा कितना होगा सर वो होगा 2.3 के आसपास बस ऐसे ही एक exomption बेस्ड बात में कर रहा हूं इस बात से तो कोई दिक्कत नहीं तो एक ले लिया आपका 2.3 कमा जीरो जैसा कोई पॉइंट होगा ये माइंस तू पॉइंट थ्री कमा जीरो होगा ये जीरो कमा 2.3 जैसा कोई पॉइंट होगा और ये जीरो कमा - 2.3 होगा अब इस बीच में आप क्या ढूंढना चाह रहे हो इस बीच में आप ढूंढ रहे हो क्या एक तो 1 / 2 तो सर ये है 2.3 के आसपास कुछ पॉइंट लेट्स से मैं आपको लिख देता हूं ये है एप्रोक्सीमेटली 2.3 कमा जीरो मोटा मोटा है 0.5 अगर यह 2.3 तो यहां कहीं वो आपका पॉइंट होगा जो आपका एक सेंटर सेट होगा जो की क्या होगा 1 / 2 कमा जीरो सिमिलरली अगर इस लाइन का हमें ए इंटरसेप्ट खोजना चाहूं तो डेट इसे जीरो पॉइंट थ्री थ्री डेट इस 0.33 आई होप आप ये बात समझ का रहे हो और वो भी नेगेटिव है ना तो नेगेटिव मतलब क्या नेगेटिव मतलब यहां कहीं और जब वो यहां कहीं होगा तो बड़ा बेसिक सा थॉट ही ए रहा है सर देखो ध्यान से 0.33 का प्रॉब्लम सर इतना क्यों बना रहे हो कोई खास जरूरत नहीं है बट आपको बस में समझा पाऊं की कौन सा लार्जेस्ट सेक्शन है कौन सा स्मॉलर सेक्शन है बहुत बहुत अच्छे से देख पाओ बस मेरा ये आइडिया है है ना तो सर 0.33 इसी के यहां कहीं होगा एप्रोक्सीमेटली यहां के तो धीस इसे गोइंग तू बी व्हाट दिस इस गोइंग तू बी जीरो कमा - 1 / 3 तो ये आपके दो पॉइंट्स होंगे और अगर मैं एक स्ट्रेट लाइन ड्रॉ करूं जो की इन दोनों पॉइंट से गुजरती है तो आप क्या कहोगे सर एक स्ट्रेट लाइन जो की इन दोनों पॉइंट से गुजरती है वो यही होगी क्या आप इस बात को देख का रहे हो एक स्ट्रेट लाइन जो इन दोनों पॉइंट से गुजरती है वो यही होगी एंड एफ आय जस्ट पुल्ड डाउन तो ये वो आपकी स्ट्रेट लाइन है कोई तकलीफ तो नहीं है इस बात से नहीं है सर तो ये वो आपकी स्ट्रेट लाइन कौन सी जो की उन्होंने कही थी जो की हमने इंटरसेक्ट फॉर्म में देखी हमने इंटरसेक्ट फॉर्म में देखिए लेकिन आपने इस तरीके से कही कौन सी 2X - 3y 2X - 3y = 1 अब ऑफ कोर्स जब कोई का रहा है x² + y² इस लेस दें इक्वल तू अंडर रूट से अंडर रूट सिक्स मतलब क्या मतलब इस सर्कल के अंदर लाइक करने वाले सारे पॉइंट्स बट मैं किस्म इंटरेस्ट दे डन मैं उन पॉइंट्स में इंटरेस्ट हूं सर मैं उन पॉइंट्स में इंटरेस्ट हूं जो की इस लाइन के इस सर्कल को डिवाइड करने पर स्मॉलर पार्ट के अंदर है जो की इस लाइन को इस सर्कल को डिवाइड करने पर जब यह लाइन है सर्कल को डिवाइड कर रही है तो जो स्मॉलर पार्ट है अब मेरा ऑब्जर्वेशन यही है सर अगर ये लाइन जीरो कमा जीरो से पास होती है तो इसका डायमीटर होती आप समझ का रहे हो जीरो कमा जीरो से वो नीचे है इस तरह से सर ये वाला जो जोन है ये वाला जो जोन है ये वो स्मॉलर पार्ट है डू यू ऑल रिलाइज डेट या आप मेरी इस बात से गई करते हो मैं बस इस तरीके से इसे देखना चाह रहा हूं आप ये बताओ ओरिजिनल किस तरफ है सर ओरिजिन इस लाइन के ऊपर है रिपीट माय स्टेटमेंट इस लाइन के ऊपर है ओरिजिन और इस लाइन के नीचे है वह सारे पॉइंट जो हमारे डिजायरेबल पॉइंट अच्छा बताओ ओरिजिनल की इक्वेशन को अगर आप ठीक से लिखोगे ना तो यह टेक्निकल है 2X - 3y है ना - 1 = 0 आप ऐसे ही बस ओरिजिन पास करो ये आपकी इक्वेशन है कौन सी L1 है ये आपकी लाइन है कौन सी एल वैन जैसे हमने का लेता हूं एल्बम अब अगर ओरिजिन पस्त क्या जीरो जो की इस लाइन के ऊपर की तरफ फ्लाई करता है इस तरफ तो जीरो जीरो माइंस वैन तो वो क्या देता है L1 की वैल्यू क्या ए जाती है नेगेटिव मतलब मुझे क्या चाहिए अगर इस तरफ लाइक करने वाले पॉइंट्स इस लाइन में रखने पर नेगेटिव वैल्यू देते हैं तो इस तरफ लाइक करने वाले सारे पॉइंट्स क्या देंगे इस तरफ इस लाइन के जो भी पॉइंट्स लाइक करेंगे वो सब मुझे देंगे पॉजिटिव वैल्यू तो पहली कंडीशन तो ये की मुझे वो सारे पॉइंट्स चाहिए इस पैसे में चेक करूंगा वो सारे पॉइंट्स जो इस लाइन में अगर मैं पास करूं तो वो क्या दें पॉजिटिव वैल्यू लेकिन आप bhuliyega मत की शायद आप ये गलती कर दोगे सर ने कहा की ये चारों पॉइंट से सर्कल के अंदर लाइक करते हैं आपको कैसे पता तो मुझे ये भी तो इंश्योर करना पड़ेगा की ये जो चारों पॉइंट्स दिए गए हैं वो इस सर्कल के अंदर लाइक करें मुझे नहीं पता लेकिन कहीं-कहीं लाइक कर रहे होंगे और अगर वो सर्कल के अंदर लाइक कर रहे होंगे तो shailai से वो एक और कंडीशन फुलफिल कर रहे होंगे सर की वो सर्कल की कंडीशन को भी तो inquality को भी तो सेटिस्फाई करेंगे की सर्कल की अगर इक्वेशन ये है तो वो इस रेडियस के अंदर ही लाइक करेंगे इस सर्कल पर या सर्कल के अंदर तो दोनों कंडीशंस फुलफिल होनी चाहिए तो आप बस ये देखो इनमें से कौन-कौन से पैर ऑफ पॉइंट्स हैं जो इसमें रखने पर पॉजिटिव वैल्यू और उसमें रखने पर सेटिस्फाई कर देते हैं इस कंडीशन को की वो लेस दें 6 है अगर यह दोनों कंडीशन सेटिस्फाई हो गई तो यही आपका आंसर होगा की इसमें से कौन-कौन से वो पॉइंट्स हैं एक बार ट्राई करिए स्टूडेंट्स और जल्दी बताइए आंसर क्या ए रहा है चलो तो पहले ट्राई कर लेते हैं इस ऑप्शन को 2 3 / 4 सो 2 3 / 4 पास क्या तो पहले यहां पास करते हैं है ना हमारे पास कर्व्स क्या-क्या है सर हमारे पास एक कप तो क्या है 2X - 3y -1 एक ये है और ये हमें चाहिए क्या पॉजिटिव और दूसरा क्या है आपका x² + y² है ना माइंस 6 और ये चाहिए मुझे लेस दें इक्वल तू जीरो तो मुझे ये दो कंडीशंस पर आपसे बात करनी है जो भी पॉइंट इन दो कंडीशंस को सेटिस्फाई कर रहा होगा वही हमारा आंसर होगा तो चलो सोच लेते हैं फटाफट पहले ये ट्राई किया तू कमा थ्री बाय फोर तो देखो तू यहां रखा एक्स की जगह तू रखा तो इन तू कितना फोर माइंस कितना 3 / 4 तो ये हो जाएगा कितना 9 / 4 है ना और ये कितना -1 ये हमें चाहिए पॉजिटिव अब सीधी सीधी सी बात सर एक छोटा सा कम ही कर लेते हैं -9/4 है ना एप्रोक्सीमेटली कितना होता है सर फोर का डबल होता है 8 अब सुनना ध्यान से तो हो जाएगा -2 के आसपास कुछ -2 -1 -3 के आसपास कुछ और फोर में से -3 क्या दो वैन कुछ थोड़ा भी वह पॉजिटिव होता है क्योंकि यह पार्ट फोर से तो छोटा है तो हान सर यहां पर ये तो सेटिस्फाई हुआ पर क्या ये भी सेटिस्फाई हो रहा है देख लेते हैं ध्यान से देखना सर 2 का स्क्वायर कितना फोर है ना 3/4 का स्क्वायर कितना 9 / 16 क्या आप सब इस बात से एग्री करते हो है ना 9/16 और कितना माइंस सिक्स अब अगर मैं इसे सॉल्व करना चाहूं तो अगेन थोड़ा सा एप्रोक्सीमेशन पर कम कर लेते हैं देखो बड़ी बेसिक सी बात है सोचना ये चार है ये चार है अगर ये कभी होता तो ये होता 4 + 1 5 और 5 में से भी छह को अगर हम सब्सट्रैक्ट करते हैं अगर पंच में से भी छह को 17 कहते हैं तो माइंस वैन मिलता वेल एनीवे ये एक नहीं है एक से कमी है तो ये नेगेटिव ही है तो ये भी सेटिस्फाइड है तो सर पहली बात तो यही की आपका 2 3/4 उसे रीजन में लाइक करेगा की वो सेंटर में तो होगा ही होगा सर सॉरी सर्कल में तो हो ही होगा और इस लाइन के उसे छोटे वाले स्मॉलर रीजन में भी लाइक करेगा व्हाट अबाउट 5/2 फिर वही बात 5/2 यहां रखा तो तू से तू कैंसिल तो कितना बचा फाइव है ना यहां पर 3/4 रखा तो कितना हो जाएगा माइंस वैन है ना सर देखो ये पॉजिटिव है या नहीं सोच के देखो भाई सिंपल सी बात दिमाग लगाओ चार का डबल होता है आते तो एप्रोक्सीमेटली तू है माइंस तू माइंस वैन माइंस थ्री फाइव माइंस थ्री तू एप्रोक्सीमेटली तू ही आएगा तू से थोड़ा भी वो छोटा होगा तो 1.8 1.9 आएगा जो की पॉजिटिव होगा तो सर ये तो सेटिस्फाई हो रहा है 5/2 क्या इससे सेटिस्फाई करें देख लेते हैं 5/2 का स्क्वायर कितना 25 / 4 कोई दिक्कत तो नहीं 3/4 मतलब क्या सर मतलब हो जाएगा क्या आपका 9 / 16 सही कर रहा हूं क्या और कितना सर माइंस सिक्स यहां पे थोड़ा ट्रिकी हो जा रहा है बट देख लेते हैं कितना ए रहा है देखो ध्यान से सुनना ध्यान से सुनना स्टूडेंट्स मैं बड़ी सिंपल सी बात आपसे कहना चाह रहा हूं की यहां पे थोड़ा सा ध्यान से कर लो क्योंकि बहुत क्लोज जा रहा है है ना तो मैं क्या करता हूं मैं पहले तो एलसीएम ले लेता हूं जो की थोड़ा सा इस क्वेश्चन को बहुत कैलकुलेशन वाइस बहुत परिसर हो रहा है मैं बस एक थॉट दे रहा हूं आपको सुनना ध्यान से कम की बात है फोर का सिक्स टाइम्स होता है 24 आप सन रहे हो फोरकास्ट सिक्स टाइम्स होता है 24 तो ये एप्रोक्सीमेटली सिक्स है सिक्स में कुछ ऐड किया तो 6 से बड़ा ही हुआ और सबसे बड़ी वैल्यू में से पॉजिटिव हो जाएगा मुझे क्या चाहिए था नेगेटिव तो ये सेटिस्फाई नहीं हुआ मतलब आपका ये एक एक्सेप्टेबल ऑप्शन नहीं है हम प्रॉपर्ली पूरा सॉल्व नहीं कर रहे हैं थोड़ा सा एप्रोक्सीमेशन पे कम करें जो की बेसिक कैलकुलेशन एक इंटेलिजेंस स्टूडेंट इस स्टेज पर कर सकता है सर बात की जाए बात कहूं ये सारे पॉइंट्स नहीं यहां पर भी प्लॉट करके देख सकते हो क्योंकि आपने बहुत प्रॉपर प्लानिंग किए हैं यहां पर आप ऐसे भी सोच सकते हो बट मेरा कहना है सर अगर मैथ्स हमारा एक बहुत प्रेसीजन के साथ आंसर एक्युरेटली दे पता है तो यहीं पर सोच लेते हैं तो अगर 1 / 4 - 1 / 4 रहा है क्या तो ये हो जाएगा 2 / 4 2 / 4 यानी 1 / 2 है ना ये हो जाएगा प्लस 3/4 आप सन रहे हो ना मेरी बात को कोई दिक्कत तो नहीं भाई और यहां पर क्या दिख रहा है -1 अब ध्यान से सुनो ये होता है आधा ये होता है 50.75 ये चला गया आपका 1.25 के आसपास सन रहे हो क्या और 1.25 में से वैन सब्सट्रैक्ट किया तो क्लीयरली ये पॉइंट 25 ए रहे हैं पॉजिटिव ए रहा है तो ये तो सेटिस्फाई हुआ कोई दिक्कत तो नहीं अब इसी बात से इसी बात से इसको भी चेक कर लो सर तो देखो ध्यान से वैन बाय फोर का स्क्वायर 1/4 का स्क्वायर कितना हो जाएगा सर 1 / 16 है ना ये भी कितना हो जाएगा 1 / 16 और इसमें से अपराध के बहुत छोटी वैल्यूज हैं सर ये बहुत छोटी वैल्यू है इनको सिक्स में से सब्सट्रैक्ट करोगे नेगेटिव आना तय है तो ये भी सेटिस्फाई हो गया और जो ये दोनों सेटिस्फाई हो गए तो आपका ये वाला ऑप्शन क्लियर ही इसे सेटिस्फाइड करता है ये भी आपका पैर है जो आपके डिजायर रीजन में लाइक करेगा इस पर बात कर लेते हैं तो 1/8 रखा तो 2 / 8 यानी कितना 1/4 यह 1/4 -3/4 और ये है माइंस वैन अब ध्यान से देखो भाई बहुत ध्यान से देखो सर मुझे रिजेक्शन दिख रहा है क्योंकि देखना 1 - 3 कितना हो जाएगा -2/4 -2/4 यानी -1 / 2 - 1 / 2 - 1 नेगेटिव लेकिन मुझे तो क्या चाहिए पॉजिटिव तो इसीलिए सेटिस्फाई नहीं किया तो आगे चेक नहीं करूंगा तो मुझे कितनी वैल्यूज मिल रही है सर मुझे क्लीयरली दो ऑप्शंस हैं मिल रहे हैं दो पॉइंट्स ऐसे मिल रहे हैं जो इसे क्लीयरली सेटिस्फाई कर रहे हैं कौन-कौन से सर 2 3 / 4 और 1 / 4 - 1 / 4 और अगर ये दोनों ने इसे सेटिस्फाई किया तो मैं कहूंगा सर इन चारों गिवन में से दो पॉइंट्स ऐसे हैं जो हमारे इस स्मॉल रीजन में लाइक करते हैं कोई दिक्कत कोई परेशानी आई थिंक ये अच्छे से क्लियर हो आपको इसी वीडियो के डिस्क्रिप्शन में जब आप जाएंगे ना तो दिखाई देगा फुल प्लेलिस्ट लिंक इंगेज मैथमेटिक्स फॉर आईआईटी जी मांस एंड एडवांस्ड उससे ये प्लेलिस्ट ओपन हो जाएगा इसमें सारे वीडियो आपको दिखाई देंगे वहां पर सेफ प्ले लिस्ट या फिर एक प्लस साइन दिखा देगा वहां क्लिक करके आप इसे अपने अकाउंट में से करके रख लीजिए ताकि आपके पास हमेशा ये प्ले लिस्ट और ये सारे वीडियो सुरक्षित और ऑर्गेनाइज्ड रहे हैं आज पहले क्वेश्चन में क्या लिख रहा है वो ये ज एडवांस्ड का 2017 का क्वेश्चन है एकदम ताजा ताजा ही कहूंगा मैं बहुत ज्यादा पुराना नहीं कुछ चार-पांच साल पुराना ही है ताजा तो नहीं पर हान एक अच्छा फ्रेश सही क्वेश्चंस है मानेंगे फॉर हो में वैल्यूज ऑफ पी किस कैटिगरी का क्वेश्चन है सर ये न्यूमेरिकल वैल्यू टाइप है ना पर हो में वैल्यूज ऑफ पी तो वो पी की वैल्यू उसमें इंटरेस्ट है ठीक है सर किस तरीके से की ये जो सर्कल है ये जो सर्कल है आपका सर ये सर्कल और cardinate एक्सिस के एक्जेक्टली तीन कॉमन पॉइंट्स हैं आप एक बात बताओ सारी बातें छोड़ो मेरे पास एक सर्कल है एक सर्कल क्वाड्रेटिक एक्सेस के साथ एक्जेक्टली तीन कॉमन पॉइंट्स किन-किन हालातो में बन जाता है की आप मुझे सवाल का जवाब देंगे सर एक तरीका तो यह होता एक तरीका आप सोच के देखो की वो सर्कल कुछ इस तरीके से बन रहा होता है कुछ इस तरीके से बन रहा होता क्या आपको दिख रहा है की ये सर्कल इस कोऑर्डिनेट्स एक्सिस ए एक्सिस और एक्स एक्सिस को कितने पॉइंट्स पर इंटरसेक्ट कर रहा है सर ध्यान से देखो आप एक दो और तीन एक्स एक्सिस ए एक्सिस दोनों को एक्स-एक्सिस ए-एक्सिस दोनों को विंटर सेट कर रहा है लेकिन ओरिजिन से पास हो रहा है और अगर ऐसा हो रहा है तो आपके क्वाड्रेंट एक्सेस को ये सर्कल एक्जेक्टली तीन पॉइंट्स पर इंटरसेक्ट कर रहा है मतलब सर्कल में और क्वाड्रेंट एक्सेस में तीन पॉइंट्स कॉमन है क्या आपको ये बात रिलाइज हुई अच्छा इसके अलावा और क्या हो सकता है इसके अलावा आप सोच के देखो ना ये आपका क्या रहा ए-एक्सिस और ये रहा मैन लो एक्स एक्सिस है ना अब सर अगर ऐसा हो रहा होता की आपका जो सर्कल है आपका जो सर्कल है वो एक्स एक्सिस को मैन लो टच कर रहा है लेकिन ए एक्सिस पे इंटरसेक्ट ले रहा है आप देख का रहे हो सर इसमें भी देखो आप एक पॉइंट दूसरा पॉइंट और तीसरा पॉइंट यहां पर भी तो तीन पॉइंट्स हैं और क्या बन सकता एक सिनेरियो और बन सकता है जनाब जरा गौर से देखिए की सर अच्छा बाय डी वे मैं सिर्फ फर्स्ट क्वाड्रेंट में ले रहा हूं ऐसा किसी भी क्वाड्रेंट में हो सकता था ये कोई बहुत बड़ी विशेष मुश्किल बात नहीं है मैन लो अगर यही बात कहां हो रही होती सर यही बात कुछ इस तरीके से हो रही है तो की ये रहा आपका सर्कल और इस बार वो ए एक्सिस को टच कर रहा है लेकिन एक्स एक्सिस पर इंटरसेप्ट रहे हैं तो क्लीयरली सर यहां भी आप देखो पहला दूसरा कॉमन पॉइंट है जो की सर्कल और एक्सेस में कॉमन है तो तीन कॉमन पॉइंट्स किसी सर्कल और एक्सेस में इन तीन हालातो में निकल कर आते हैं कोई तकलीफ कोई परेशानी अब अगर मैं सर्कल की इक्वेशन देखूं तो लिखा हुआ है x² + y² + 2X + 4y -p=0 क्या लिखा है इसे ये लिखा हुआ है रिपीट माय स्टेटमेंट एक्स स्क्वायर प्लस ए स्क्वायर की इक्वेशन है जो की ओरिजिन से पास होता है तो क्या कंडीशन मैच या फुलफिल हो रही होगी क्या आप बड़ा बेसिक सा थॉट लगा सकते हो की सर अगर ये ओरिजिन से पास होता है तो मुझे क्यों ज्यादा सोचना है लाइफ में अगर ये ओरिजिन से पास होता है तो ये आपकी पहली कंडीशन है की सर ये ओरिजिन अगर पास हो रहा है इफेक्ट पासेस थ्रू वो जो ओरिजिन इन डेट केस इट मस्ट बी सेटिस्फाइंग जीरो कमा जीरो कमा जीरो उसको सेट है स्वीकार कर रहे होंगे एंड इट मस्ट क्रिएट सेटिस्फाइड बाय डेट तो सर मैं कहूंगा जीरो जीरो तो पी की वैल्यू कितनी जीरो तो पहली कंडीशन तो यही की अगर पी की वैल्यू जीरो हो पहला आंसर यही की सर अगर आपके पी की वैल्यू जीरो हो तो नहीं सन्डे आप कहेंगे की सर आपका जो सर्कल होगा वो इंटरसेप्ट्स मतलब जो एक्सेस के साथ कॉमन पॉइंट्स रहेगा वो ही लेगा क्या बड़ी बेसिक सी डायरेक्ट सी बात समझ आई कोई दिक्कत तो नहीं अब अगर मैं आपसे पूछूं सर दूसरी बात यह कौन सा सिनेरियो है जब आपका सर्कल एक्स-एक्सिस को टच करता है और ए एक्सिस पर इंटरसेक्ट लेता है अब याद करिए हमने बहुत डिटेल में ये बात पढ़ी है की अगर कोई सर्कल एक्स-एक्सिस पर इंटरसेप्ट लेता है मतलब एक्स-एक्सिस को अगर वो टच करता है और ए एक्सिस पर इंटरसेक्ट लेता है तो उसे केस में क्या-क्या चीज ट्रू होनी चाहिए सर पहली बात तो अगर मैं एक्स इंटरसेक्ट की लेंथ की बात करूंगा तो वो जीरो हो जानी चाहिए क्योंकि एक्स एक्सेस पर टच कर रहा है इंटरसेक्ट है ही नहीं और ए एक्सिस पर इंटरसेप्ट की बात करूंगा तो वो आपका ग्रेटर दें जीरो होना चाहिए आपको याद है ना हमने इंटरसेप्ट्स के लेंथ की बात की थी जो की हमने पढ़ रखा है तू टाइम्स क्या जी स्क्वायर माइंस सी और क्या तू टाइम्स स्क्वायर माइंस ही आई होप आप ये बातें भूले नहीं हो अब ऑफ कोर्स x² - ये आपका ए इंटरसेक्ट होता है G2 - सी आपका एक्स इंटरसेप्ट होता है सर यहां पर सीधा-सीधा सो ओपिनियन बड़ा यही है की अगर आपका एक्स इंटरसेक्ट नहीं बन रहा है अब देखो ना एक सक्सेस को टच कर रहा है तो एक्स सेंटर से बन ही नहीं रहा है तो उसके लेंथ कितनी होगी जीरो और वही इंटरसेक्ट बन रहा है एंड इट हज सैम सिग्निफिकेंट लेंथ तो shailise भी पॉजिटिव बिल्कुल सही बात है सर अब इन दोनों कंडीशंस पर बात की जाए तो देखो क्या निकल कर ए रहा है पहले तो अगर मैं यहां से सॉल्व करूं तो मैं कनक्लूड करूंगा इस वाले पार्ट को की क्या सर जी स्क्वायर माइंस सी क्या होना चाहिए जीरो है ना तो g² क्या होना चाहिए आपका सी के इक्वल कोई तकलीफ तो नहीं है इस बात से बड़ी बेसिक्स सी बात है स्टूडेंट जी स्क्वायर क्या होना चाहिए आपका सी के इक्वल सिमिलरली क्या यहां से कोई कंक्लुजन किया जा सकता है सर यहां से अगर मैं कंक्लुजन डेरिव करूं तो यहां से क्या कंट्रोल अच्छा बाय डी वे जी और सी आप देख का रहे हो जी और सी क्या है जी और सी आपकी इस इक्वेशन से बने हुए आपके x² + y² + 2X + 4y माइंस पे है ना अल अगर मैं डायरेक्टली ही आपसे बात करूं तो इतने घुमा फिर के बात करने की वजह मैं डायरेक्ट कंक्लुजन पर आता हूं आपका सर्कल है x² + y² + आई थिंक 2X + 4y बिल्कुल सही बात तो ये है 2X + 4y - पी = 0 अब बताओ G2 G2 मतलब क्या G2 मतलब -1 का स्क्वायर यानी कितना वैन आप बात समझ का रहे हो है तो 1 - बहुत ध्यान से suniyega स्टूडेंट्स वैन माइंस वैन माइंस कितना सी है माइंस पी तो वैन माइंस माइंस पी यानी कितना 1 + पी एंड डेट शुड बी जीरो तो यहां से कंक्लुजन ए रहा है पी पी की वैल्यू माइंस होना चाहिए बट बाय डी वे आप ये बात सोच का रहे हो ना ये और ये दोनों honhana चाहिए सिर्फ ये नहीं होना चाहिए की एक्स-एक्सिस को टच करें ये भी दिस आगे वेल शुड हैव एन की वो एक्स एक्सिस के बजाय टच करने के साथ-साथ वो x² दो जगह इंटरसेप्ट केट तब ही जाकर वो आपके तीन पॉइंट्स पर कॉमन रहेगा सर्कल इन एक्सेस के साथ है ना अब सर अगर आप इस कंडीशन को देखो तो एफ की वैल्यू कितनी है सर एफ है -2 आई होप आप देख का रहे हो तो ये कितना हो जाएगा -2 का स्क्वायर यानी 4 आप समझ का रहे हो क्या 4 - -पी यानी कितना सर फोर प्लस पी एंड दिस शुड बी पॉजिटिव क्योंकि तू उधर गया अंडर रूट तू उधर गया तो यहां से आप क्या कनक्लूड कर का रहे हो सर जो आपका पी है जो आपका पी है सर वो माइंस फोर से बड़ा होना चाहिए आपका पी जो है वो माइंस फोर से बड़ा होना चाहिए और यहां पर अगर मैं देख का रहा हूं क्योंकि आप मुझे बता का रहे हो सर बड़ा सिंपल सा ऑब्जर्वेशन है की पी जो है वो माइंस वैन है और पी जो माइंस वैन है वो क्लीयरली आपका -4 से बड़ा है तो क्या -1 एक एक्सेप्टेबल वैल्यू है मतलब यह का रहा था की सर पी की वैल्यू आप लीजिएगा माइंस वैन ठीक है सर आपकी बात मैन ली ये का रहा है पी जो होना चाहिए वो आपको माइंस और से बड़ा होना चाहिए तो इन दोनों में इंटरसेक्शन क्या है कॉमन केयर माइंस फोर से बड़ी कोई भी वैल्यू इसने कहा माइंस वैन ले लो तो क्या माइंस फोर से माइंस वैन बड़ा है बिल्कुल बड़े नंबर स्केल पे माइंस वैन राइट हैंड साइड पर आता है तो हमें पी की एक और वैल्यू मिली जो की क्या थी माइंस वैन तो हमसे यहां पे पी की वैल्यू मिली जीरो और वहां पे मिली माइंस वैन तो दो वैल्यू चुका हूं पी की कौन-कौन सी सर पी की एक वैल्यू जीरो हो और एक वैल्यू -1 आगे बढ़ते हैं अब मैं थर्ड कंडीशन पर जाता हूं जहां पे फिर वही बात होगी की आपका सर्कल क्या है सर वो है x² + y² + कितना 2X + 4y - पी = 0 फिर वही बात की सर अगर ये सर्कल जो है आपका ये सर्कल अगर एक्स एक्सिस पर दो इंटरसेप्ट्स ले रहे हैं तो आपका जो अंडर रूट ओवर तू टाइम्स कितना G2 - सी है ना ये क्या होना चाहिए सर ये आपका पॉजिटिव होना चाहिए और ए इंटरसेप्ट कौन सा तू टाइम्स अंडर रूट ओवर x² - सी ये जीरो होना चाहिए समझ और देख समझ के सिख का रहे हैं सर यहां से कनक्लूड करें g² - सी तो क्या वही बात तू तो गया ये अंडर रूट भी गया तो वैन जी कितना है -1 तो 1 - -पी यानी कितना 1+p शुड बी पॉजिटिव वैन प्लस तू शुड बी पॉजिटिव मतलब पी शुड बी ग्रेटर दैन -1 टी जो है आपका माइंस में बड़ा होना चाहिए से बात यहां पे अप्लाई करिए जो लास्ट क्वेश्चन में भी हमने सोचिए वही बेसिक सी बात स्टूडेंट्स ये हो जाएगा तू सन रहे हो ना तू का स्क्वायर कितना फोर फोर माइंस माइंस पी यानी क्या फोर प्लस पी यानी पी शुड बी माइंस फोर यहां पे आपकी बात है ध्यान से देखो पी हो पी होगा माइंस हम से बड़ा यानी जीरो प्लस वैन तू थ्री इस तरफ है ना और ये का रहा है पी जो है आपका -4 है आपको गड़बड़ समझ ए रही है बेशक सर माइंस फोर तो माइंस में से छोटा होता है तो यहां पर कोई कॉमन इंटरसेक्शन नहीं ए रहा है क्योंकि ये दोनों बातें एक साथ रो होनी चाहिए थी तो यहां कोई कॉमन इंटरसेक्शन नहीं ए रहा है मतलब मुझे पी की सिर्फ दो ही एक्सेप्टेबल वैल्यूज में ही कौन-कौन सी सर वो मिली मुझे जीरो और माइंस वैन और जीरो और माइंस मिलती हैं तो कितनी वैल्यू आई से दो तो वो क्या पूछ रहा था और हो में वैल्यूज ऑफ पी तो मैं कहूंगा तू और यही आपका इस क्वेश्चन का आंसर होगा मेरा यकीन करिए एक इतना आसान इतना बेसिक इतना सिंपल क्वेश्चन था और आप देख रहे हो किस एग्जाम में पूछा है ये पूछा गया है जी एडवांस्ड ने 2017 में क्यों है ट्रिकी क्वेश्चन क्योंकि आप ये सिनेरियो अगर ठीक से इमेजिन कर लें और सारे इमेजिन कर लें तो आप आसानी से कर सकते हो कुछ ज्यादा इसमें सोचने की जरूरत ही नहीं है और अगर ये क्वेश्चन आप सभी को समझ आए तो एक और क्वेश्चन करेंगे यह कब पूछा गया है सर ये ज एडवांस ने 2019 में पूछा है तो एक और बेहतरीन मजेदार और आई गो अगेंस्ट है सिंपल है टू नहीं है एक अच्छा क्वेश्चन है आप सोच के देखिए फिर से पढ़ते हैं इस क्वेश्चन को देखते हैं सर इसमें क्या लिख रहा है वो का रहा है लेट डी पॉइंट बी बी डी रिफ्लेक्शन ऑफ डी पॉइंट ए तू कॉमन थ्री तो बी जो पॉइंट है आपका रिफ्लेक्शन है किसका एक पॉइंट ए का किसके रिस्पेक्ट में इस लाइन के रिस्पेक्ट में तो पहले इसी पर बात कर लेते हैं सर एक पॉइंट है मुझे बी मिल रहा है जो की इसका रिफ्लेक्शन है तो अब अगर मैं ए जो है आपका क्या ये जो आपका पॉइंट है वो क्या है सर तू कमा 3 है ना इसका अगर रिफ्लेक्शन खोज रहा हूं किसके रिस्पेक्ट में सर आपकी स्ट्रेट लाइन कौन सी है टैक्स - 6y - 23 तो ये है 8x - 6y - 26 तो मैं बड़ी बेसिक सी बात जानता हूं अगर इसका रिफ्लेक्शन है जैसे हम अभी भी फिलहाल का रहे हैं अगर कॉम के तो हमने एक सिंपल सा फॉर्मूला पढ़ा है ये पूरा फटाफट करेंगे ना ये आपको समझने के लिए मैंने लिखा है तो मैं कहूंगा इसका जो रिफ्लेक्शन है जैसे आप का रहे हो सर अभी बी उससे क्या कहेंगे सर हम निकलेंगे कैसे हम लिखेंगे ह - 2 / एक्स का कॉएफिशिएंट आते है ना इसे इक्वल तू के-3 के-3 विच इस -6 और ये दोनों बातें किसके इक्वल रखते हो सर आप रखते हो ध्यान से देखो माइंस तू टाइम्स क्योंकि आप मिरर इमेज ले रहे हो तो क्या हो जाएगा सर माइंस तू टाइम्स माइंस तू टाइम्स व्हाट यहां से देखो स्टूडेंट्स 2 8 कितना आप सन रहे हो 3 -6 डेट्स - 18 आई होप ऑल आर गेटिंग डेट और क्या सर - 23 ये आपकी स्ट्रेट लाइन आय होप यू ऑल आर गेटिंग डेट सो दिस इस - 23 हर डिनॉमिनेटर की अगर मैं बात करूं तो क्या सर अंडर रूट नहीं लगाना है फीट के और इमेज में अंडर रूट नहीं लेगा तो डिस्टेंस फॉर्मूला में लगता है है ना तो यहां पर क्या हो जाएगा 8² + 6² 64 + 36 डेट इसे स्केल इन नथिंग बट 100 डिड ऑल गेट डेट सर अगर आप इसे सिंपलीफाई करके चीज निकलने की कोशिश करें तो क्या मिलता है देखो भाई सर सबसे पहले तो आप ध्यान से देखो क्या देखें भाई अन इससे और इससे बाद में बात करेंगे मैं इसे थोड़ा देखना चाह रहा हूं एक बात बताओ ऑटो से आई थिंक 14 है ना कोई दिक्कत तो नहीं भाई आर्ट और 14 मतलब ऐसे जोड़ लो 18 में 6 ऐड किया तो 24 2410 34 तो कितना हो रहा है 34 34 एम रियली सॉरी मैं गड़बड़ कर रहा हूं मैंने ऐड किया माइंस 18 प्लस 16 कितना होगा -2 - 23 - 2 - 25 सो रहे हो क्या - 25 तू -2 डेट 50 और 50 / 100 कितना होता है 1/2 सो दिस इस गोइंग तू बी वैन बाय तू डिड ऑल अंडरस्टैंड दिस बस इसी तरीके से कैलकुलेशन करिएगा बस गलत मत करिएगा कोई इसके लिए मत करना सो दिस इस गोइंग तू बी वैन बाय तू नौ इफ यू सॉल्व दिस फॉर दिस अगर आप इस वाले हिस्से को सॉल्व करें तो क्या मिलेगा सर तू का 84 टाइम्स तू का 84 टाइम्स 4 उधर गया आप देख का रहे हो क्या तो ह-2=4 तो ह = 6 आई थिंक सिमिलरली सोच के देखो 2 का -6 - 3 टाइम्स -3 यहां आया -3 से -3 कैंसिल तो के की वैल्यू कितनी के की वैल्यू जीरो सो इस तरीके से आपके ये कोऑर्डिनेट्स हैं किसके हैं ये उसके इमेज के कोऑर्डिनेट्स हैं जो की आपकी है इस ए पॉइंट की इस लाइन के रिस्पेक्ट में मिलेगी जो की क्लीयरली है सिक्स कमा जीरो यहां तक तो कोई आपत्ति नहीं तो बी के कोऑर्डिनेट्स में क्या कहूंगा सर सिक्स कमा जीरो बहुत बढ़िया अब इस क्वेश्चन में क्या कहा जा रहा है वो जरा समझते हैं है तो यह जो बी पॉइंट है जिसकी मेरा इमेजेस के रिस्पेक्ट में जो की है पॉइंट भी अब वो का रहा है ताऊ ए और ताऊ बी है ना बी डी सर्कल ऑफ तू एंड वैन विद सेंटर्स ए बी तो सर आपके ए और बी जो पॉइंट्स हैं उन्हें अगर मैं सेंटर्स मैन लूं और उन पर तू और वैन रेडियस के सर्कल्स ड्रॉ करूं तो वो कहलाएंगे तो यही बात पहले प्लॉट कर लेते हैं है ना फिर वो कहना क्या चाह रहा है चलो बात आगे देखेंगे पहले इतना पार्ट प्लॉट करते हैं तो एक तो है तू कमा थ्री है ना और एक है सिक्स कमा जीरो बिल्कुल सर तो इसी पे बात कर लेते हैं एक तो है आपका तू कमा थ्री और एक है सिक्स कमा जीरो बहुत ध्यान से सुनना कम की बातें हम करेंगे पहले 6 स्ट्रेट लाइन लेता हूं इसी पर लाइक करेंगे आपके दोनों पॉइंट्स तू कमा थ्री और सिक्स कमा जीरो तो मैन लेता हूं कहीं ना कहीं तू कमा थ्री होंगे है ना तू कमा थ्री इधर मैन लेता हूं कोई बहुत बड़ी विशेष बात नहीं है तो ये है तू कमा थ्री और कहीं आगे जाकर लेते हैं आपके पास होता है सिक्स कमा जीरो और ज्यादा दूर ले लिया क्या थोड़ा पास ले ले कोई दिक्कत तो नहीं मतलब वो तो आपके ऊपर है ना की आप किस तरीके से उसने मानना चाह रहे हो अब सर आप बात समझना 2 वाला जो आपका सर्कल है ताऊ ए है ना उसकी रेडियस कितनी है सर उसकी रेडियस है तू बहुत बढ़िया और जो आपका ताऊ भी है उसकी रेडियस कितनी है वैन तो ठीक है सर ये भी बना लेते हैं तो हमने 2 3 वाले सर्कल की रेडियस कितनी ले ली तू यूनिट्स तो एक सर्कल बन गया आपका ऐसा कुछ है ना और इस केस में अगर मैं इसे प्लॉट करूंगा तो ये आपका सर्कल यहां पे ए जाएगा इसी तरीके से आपने वैन यूनिट का भी एक रेडियस का सर्कल लिया है सर तो 1 यूनिट का भी रेडियस का एक सर्कल ले लेते हैं और ये सर्कल आपका यहां मिल जाएगा इस तरीके से आई थिंक ये चीज इस तरीके से निकल कर ए रही है अब क्या टावर बी कोई दिक्कत तो नहीं भाई ये मैं ना भी लिखूं तो इन पर बात कर लेंगे हम जो भी वो कहना चाह रहे हैं इस बारे में अब वो आगे क्या पूछता है उसे पे बात कर रहे हैं लेट टी आर कॉमन टांगें तू डी सर्कल सच डेट बोथ डी सर्कल्स आर ऑन दी से साइड ऑफ दी अब सीधी बात है सर की मुझसे कभी भी कोई जीवन में कोई कॉमन टांगें की बात करता हूं मैं बहुत तरीके से बना सकता हूं मैं ऐसे बना सकता हूं मैं ऐसे बना सकता हूं आप समझ रहे हो मैं ये वाली बना सकता हूं मैं इसको घुमा के कैसे कर लेता हूं इसको घुमा के इस तरह से बना लेता है ऐसे बना सकता है ऐसे बना सकता है पर मैं कौन सी वाली लूंगा मैं वो कॉमेंटेटर लूंगा जिसमें देखो सम बनाऊंगा तो एक इस तरफ होगा तो मुझे वो डायरेक्ट कॉमन टांगें चाहिए ट्रांसफर कमेंट्री नहीं चाहिए तो जब आपको डायरेक्ट कॉमन टांगें चाहिए है ना डायरेक्ट कमेंट इंजन चाहिए तो ये कुछ ऐसे ही बन रही होगी राइट हम रियली सॉरी फॉर दिस हम इसे एक स्ट्रेट लाइन के तौर पर ही बना लेते हैं तो ये जो कमेंट्री आपकी बन रही होगी इस तरीके से बनने वाली होगी जो की कहीं आगे जाकर आपकी मिल रही होगी है ना इस तरीके से आपकी कमेंट टैसेंट होगी की आप मेरी बात समझ का रहे हो स्टूडेंट्स अरे वो बहुत ज्यादा बदर नहीं कर रही है बातें आपको बहुत अच्छे से आप चीज समझ का रहे हो इसलिए लेकर राइट ऐसे थोड़ा सा तिल कर डन तो चलेगा क्या आई थिंक किया जा सकता है बस ज्यादा आपको बॉर्डर ना करें ये बात इसको टिल्ट करने का स्कोप हमने खत्म कर लिया क्योंकि हमने सेंटर थोड़ा दूर बना लिया बट खैर कोई बहुत फर्क नहीं पड़ता बहुत बड़ी विशेष बात नहीं है अब क्या आप एक बहुत जरूरी बात ये जानते हो की सर अगर आपने इस लाइन को थोड़ा स्ट्रेच किया है यहां पर इफ यू वांट तू स्ट्रेच 10 लाइन फॉर आदर है ना अगर आप इस लाइन को स्ट्रेच करते हैं तो कैसे जा रही होगी बात समझना सर मैं इस लाइन को स्ट्रेच कर रहा हूं थोड़ा सा आगे है ना कैसे देखना मैं बड़ी बेसिक सी बात यह जानता हूं की जो डायरेक्ट कॉमन टांगें है यह जब इसे यहां कहीं इंटरसेक्ट कर रही होगी यह जहां कहीं इंटरसेक्ट कर रही होगी उसे पॉइंट पर हम बात करने वाले हैं आई होप ये बात आप जान का रहे हो समझ का रहे हो देख का रहे हो तो ये जो लाइन होगी आपकी किस तरीके से जा रही हो किस पे बात करते हैं सुनना ध्यान से यह जो आपके डायरेक्टर है इफ सी इसे डी पॉइंट ऑफ इंटरसेक्शन ऑफ टी है ना तो आपकी जो डायरेक्ट कमेंट टैसेंट है और जो लाइन है जो ए और बी से पास हो रही है ए और बी मतलब क्या इन दोनों सर्कस के सेंटर से जो लाइन पास हो रही है और जो आपकी डायरेक्ट कॉमन अटेंडेंट है वो दोनों चीज कहां इंटरसेक्ट करती हैं सर सी पॉइंट पे कोई दिक्कत तो नहीं तो ये आपका पॉइंट हो गया सी अच्छी बात है सर अब एक बात बताओ क्या हमने लाइफ में सिखा है की सर ये जो सी पॉइंट है जो डायरेक्ट कमेंट के सेंटर को ज्वाइन करने वाली स्ट्रेट लाइन से जो पॉइंट ऑफ विंटर विंटर सेशन है दूसरी भी जो आएगी वो सी पर ही जाएगी डायरेक्ट कमेंट टेंट है ना हम जानते हैं सर ये जो सी है ये एक्सटर्नल इनको आपके अकाउंट को ए और बी को इनकी रेडियस के रेश्यो में डिवाइड करती है यानी मैं एक कनक्लूड कर सकता हूं क्या की का है ना जो आपके डिस्टेंस होगी क्या सी ए और आपकी क्या डिस्टेंस व यानी आप इसे लिख लिए ई है ना ये आप बात समझ का रहे हो इनकी रेडियस के रेश्यो में डिवाइड करेंगे एक्सटर्नल मतलब क्या है externalli मतलब आपका ये जो रा होगा ए अगर यह सर्कल है मैन लो आर ए और इसके लीडर्स कितनी है रब तो ये इस रेश्यो में प्रोपोर्शन में डिवाइड करेगी और उनकी रेडियस में जानता हूं कितनी है 2 / 1 क्या पहले तो आप ये ऑब्जर्वेशन समझ का रहे हो अच्छे से इस पार्ट को देखो और मुझे बताओ क्या इस पार्ट में तो कोई दिक्कत नहीं है जब मैं का रहा हूं लेट्स सी ए या एक/ई विल बी इक्वल तू तू अपॉन वैन डिड यू ऑल गेट डेट बड़ा ही बेसिक सा फंड है ये हम पढ़ते हुए आए और इसी बारे में आज हम बात करेंगे यहां तक तो कोई दिक्कत नहीं है अब वो पूछ क्या रहा है जरा वो देखते हैं हमसे वो पूछ रहा है दें डी लेंथ ऑफ डी लाइन सेगमेंट एक तो हमें एक की लेंथ निकालनी है हमें किसकी लेंथ निकालनी है सर हमें एक की लाइन निकालनी है सी नीड तू फिगर आउट एक सर मैं एक निकल सकता हूं बट एक निकलने से पहले या तो मुझे ई चाहिए होगा पर फिर ई तक भी पहुंचना है तो ई तक पहुंचने का तरीका क्या होगा ई तक पहुंचने का तरीका या आइडिया क्या होगा मेरा कहना है आपके पास एक तरीका है अगर आप सोच पाए तो क्या तरीका है सर याद करो भाई उसने हिंट दी थी आप बस रखना भूल गए की तू कमा थ्री और सिक्स कमा जीरो कुछ ज्वाइन करने वाली लाइन है अगर उसे पर मैं एक परपेंडिकुलर लाइन ड्रॉ करूं बीच में कहीं एग्जैक्ट बीच में कहीं तो इस लाइन के बारे में मेरा एक ओपिनियन है सर आपको क्या ओपिनियन है सर आप भूल रहे हो शायद की तू कमा 3 की मिरर इमेज 360 किस लाइन के रिस्पेक्ट में माइंस 23 आई रिपीट माय स्टेटमेंट इस लाइन को हम क्या कहेंगे इसे हम कहेंगे 8x - 6y - 23 = 0 तो अगर मैं कुछ चीज अब शायद सोचता हूं तो बात बन सकती है क्या जैसे मैन लो ऑफ कोर्स ये तो हम जान का रहे हैं सर ये परपेंडिकुलर इस पॉइंट को मैं थोड़ी देर के लिए कहता हूं एल इससे हम बात करेंगे इस बारे में हम सोचेंगे बट अब जो हिंट आपको दी जा रही है वह यह है की क्या आप कुछ थोड़ा बहुत सोच के चीज निकल सकते हो क्या क्योंकि हमें फाइनली कहां तक पहुंचना है हमें फाइनली पहुंचना है किस लें तक एक तक तो लाइन सेगमेंट जो एक की लेंथ है वो निकल सकता हूं मैं बट अभी जो आपको हिंट दी है मैंने स्ट्रेट लाइन के थ्रू सोचने की क्या इससे कुछ चीज निकल जा सकती हैं क्या कोई थॉट क्या कोई एप्रोच क्या कोई बात जो आपके दिमाग में ए रही है की सर कैसे निकलेंगे आपको जो सलाह में देना चाह रहा हूं वो यही की पहले तो आप निकालो अल सर अल से क्या हो जाएगा ए एल से शायद मैं कहीं तक पहुंच जाऊं क्योंकि मैं ई को फिर एक के कॉन्टैक्ट में सोच पाऊंगा कैसे बात करते हैं बात सारी करते हैं पहले तो ये निकलती हैं की अल्की ए रहा होगा निकल लेते हैं पहले मैं आपको विज़न दे देता हूं की मैं कहां चाहना कहां सोचना क्या करना चाह रहा हूं फिर हम बात करेंगे एक और ई आपको क्या चाहिए आपको एक चाहिए तो मैं ई को एलिमिनेट करने की कोशिश कर रहा हूं अब सर ई कैसे निकलोगे आप तो एक तरीका है तरीका क्या कहता है गौर से suniyega तरीका ये कहता है प्लीज ध्यान से ऑफ कोर्स इसकी रेडियस कितनी है सर इसकी जो रेडियस आपने निकल है वो रेडियस कितनी इनके ऐसे फ्यूल एवर एस्क्ड के सर इस सर्कल की रेडियस कितनी है वो कितनी है भाई वो है आपकी 2 यूनिट्स आई होप यू हैवेंट फॉरगॉटेन डेट ये जो लेंथ है ये आपकी 2 यूनिट सिमिलरली हद यू मिन एस्क्ड अबाउट डी लेंथ ऑफ दिस पार्टिकुलर यू नो सर्कल सो दिस व्हाट हैपन के लिए कितना वैन ये हमसे कहा गया था ना ये वैन रेडियस और ये 2 रेडियस का सर्कल है अब अगर मैं यहां से चीजे सोचना चाहूं तो क्या मैं कुछ चीज वर्कआउट कर सकता हूं पहले तो मैं कोशिश करता हूं सर ये ए एल निकलने की है ना इससे क्या फायदा होगा इससे शायद मैं ई तक पहुंच पाऊं कैसे देखेंगे एक बार बताओ एल जो है वो बायसेक्स कर रहा है ए और बी को ज्वाइन करने वाली लाइन को क्यों क्योंकि एक ही मिरर इमेज स्टेटमेंट एक ही मिरर इमेज होती है यानी कैन आई कनक्लूड की सर जो आपकी ए एल होगी वही आपकी बैल होगी आई होप इस बात में तो कोई आपत्ति नहीं है रिपीट करता हूं वैसे की सर जो आपकी ए एल लेंथ होगी वही आपकी लेंथ क्या होगी बेल क्या ये बात आप सब ऑब्जर्व कर का रहे हो ये तो तय बात है ना एल जो है वो मिड पॉइंट है और अगर एल जो है वो मिड पॉइंट है तो बड़े आराम से आप का सकते हो की हान सर अल जो होगा वो क्लीयरली आपका क्या होगा इक्वल तू बी एल इस बात से तो कोई दिक्कत नहीं है ये बड़ी बेसिक सी बात है तो अब दोनों तरीके सोच सकते हो कैसे दोनों तरीके से या तो आप चाहो तो ए और बी के बीच में डिस्टेंस फॉर्मूला लगा लो और डिस्टेंस निकल लो उससे ए निकल लो या ए बी निकल लो अब बात समझ का रहे हो या फिर सर या फिर एक बहुत कॉम्प्लिकेटेड डिस्टेंस फॉर्मूला में उसे करना चाह रहा हूं तो 6 - 2 4 का स्क्वायर 16 - 0 3 3 का स्क्वायर 16 प्लस नाइन होगा ये 25 तो ये कितना हो जाएगा 25 का अंडर रूट 5 तो शैल आय से ए और बी के बीच की जो डिस्टेंस है ए और बी के बीच की जो डिस्टेंस है वो कितनी है 5 आप ऐसे ही सोचो बस तो ये डिस्टेंस कितनी निकल कर ए रही है फाइव क्या अब मैं ऐसे निकल सकता हूं मेरे ख्याल से सर अब तो आप सोच सकते हो ऐसे कैसे सो सकते हैं सर आप प्लीज ध्यान से सुनो इसको रिवाइट करता हूं क्या मैं लिख सकता हूं यहां से की सर ये जो एक होगा ये जो एक होगा डेट विल बी डबल ऑफ ई जो एक होगा सर डेट विल बी डबल ऑफ एसिड डू यू विल गेट माय पॉइंट एक जो होगा सर डेट विल बी डबल ऑफ ई नौ कैन आई कैन आई गेट रेड ऑफ समथिंग लेट्स से ई से मैं थोड़ी सी चीज सिंपलीफाई कर सकता हूं अगर मुझे ई निकलना है अगर मुझे ई निकलना है तो क्या मैं ई को ध्यान से देखना एक -अब लिख सकता हूं आई होप आप बात समझ का रहे हो ऐसा करने से क्या फायदा होगा अगर मैं ई को लिखना चाहता हूं तो देखना मैं कैसे लिखूंगा ये आपका एक है लेट्स कीप एक एक इट इस तू टाइम्स दिस ई दिस ई वुडन इट बी एक माइंस बी सी बी सी कितना हो जाएगा सर ये हो जाएगा आपका एक - अब आई थिंक बहुत ध्यान से देखना हो जाएगा 2 टाइम्स अब क्या आप मेरी बात समझ का रहे हो क्या आपको कुछ दिख रहा है जो मैं कहना चाह रहा हूं सर एक जो ए रहा है वो आपका ए रहा है 29 अब कितने भी है 5 5 का डबल कितना 10 तो एक की वैल्यू कितनी ए रही है रस्सी की वैल्यू ए रही है 10 आई थिंक सर ये बहुत डिफिकल्ट बहुत टू क्वेश्चन थोड़ी था तो बेसिक्स ऑब्जर्वेशन था बट ट्रिकी पार्ट ये था की आप ऐसा सोच पाओ की ये भी निकलने से मैं इसी तक पहुंच जाऊंगा अगर मैं ये वाला प्रॉपर्टी जानता हूं की सर आपके जो सर्कल्स होते हैं उनकी डायरेक्ट कमेंट टैसेंट जिस पॉइंट पर मिलती है वो पॉइंट इन दोनों के सेंटर्स को ज्वाइन करने वाली लाइन को एक्सटर्नल इनकी रेडियस के रेश्यो में डिवाइड करता है और इसी तरीके से मैं यहां पर लिखूंगा की जो लेंथ होगी सर ये लेंथ कितनी होगी 10 आई दोनों पार्ट्स समझ ए चुके हैं आपको आई होप आप एंजॉय कर रहे हो इन क्वेश्चंस को सिख का रहे हो चीजें कितनी आसान अच्छी बेहतर और सिंपल है और अगर ये क्वेश्चन समझ आए तो ट्राई करते हैं अगला क्वेश्चन यहां क्या लिखा हुआ है सुनेगा ध्यान से ये भी आपके जी एडवांस्ड मतलब बेसिकली एडवांस की कोशिश कर रहे हैं तो ये आपके 2020 ये तो आई थिंक बहुत फ्रेश है बिल्कुल ताजा सर दो तीन साल पुराना ही अभी हम 2022 में अगर बात करें तो ये 2020 में पूछा गया है 2 साल पुराना सवाल है न्यूमेरिकल आंसर टाइप क्वेश्चन सर का चैप्टर का फाइनल क्वेश्चन बिल्कुल एक अच्छी कमांड आप डिवेलप कर चुके हैं स्टेज पर आकर और एक बेहतरीन मजबूत सी कमांड आप डिवेलप कर चुके हो अगर आपने हर एक लेक्चर अच्छे से देखा है प्रॉपर्ली फॉलो किया है हर एक चीज को समझता अगर मैं बेसिक सी बात आपको समझाना चाहूं तो हम बेसिकली क्या कर रहे हैं हम हम हर चैप्टर को उठा रहे हैं मतलब आपका जो सिंह आगे का पूरा प्लेलिस्ट है मैं बहुत अच्छे से आपको समझाना चाह रहा हूं इस बात को प्लीज अच्छे से समझना यह सिंह आगे का जो आपका प्लेलिस्ट है हम इसमें सारी चीज कर रहे हैं हम इसमें क्या कर रहे हैं हम इसमें cardinate कर रहे हैं हम इसमें आपका कैलकुलस करेंगे हम इसमें आपका ऑफ कोर्स ट्रिगो करेंगे हम वेक्टर करेंगे हम अलजेब्रा करेंगे हर एक चीज करेंगे और यह आपका पूरा कंप्लीट सिलेबस है मतलब ये ऐसा नहीं है की कुछ चीज ये 11 12थ का पूरा कंप्लीट सिलेबस है मतलब अगर आप बोर्ड के प्रिंट है आईआईटी जी मैंस और एडवांस के experid है तो बेशक ये वीडियो ये लेक्चरर्स मैथमेटिक्स के आपके लिए ही हैं आपके लिए बहुत डिटेल में अंदर में हर एक चीज करवाई जा रही है स्टूडेंट्स मेरा यकीन करिए है ना अब आपको क्या करना है बस हम हम क्या करवा रहे हैं पहले इस बारे में बात कर लेते हैं हम जो चीज सोच रहे हैं वो क्या हम एक-एक करके हर बुक को उठेंगे पंच किताबें हैं और इसके थ्रू हम हर चैप्टर के हर चैप्टर के हर कॉन्सेप्ट को आपको पढ़ाएंगे उसके कॉन्सेप्ट को कैसे पढ़ाएंगे पहले थ्योरी अच्छे से समझ लेंगे फिर उसके क्या कर लेंगे सर उसके एग्जांपल्स कर लेंगे जो उसने दिए होंगे ताकि वो और ये एग्जांपल्स भी जी एडवांस्ड लेवल के ही हम करेंगे राइट फिर क्या करेंगे सर उसी चैप्टर के एंड में एक्सरसाइज दी गई है तो जब सारे कॉन्सेप्ट्स हो जाएंगे तो एक्सरसाइज करेंगे एक्सरसाइज से वो पूरा चैप्टर अच्छा कम सलूट हो जाएगा हर कोई चीज जो रिवाइज होनी है हो जाएगी कुछ पार्ट अगर मैं सो गया होगा या ठीक से नहीं समझ आएगा तो यहां से क्लियर हो जाएगा और फिर यही कॉन्फिडेंस आपके अंदर इन्फ्यूज करने की कोशिश करेंगे की देखो अब हम फाइनली उसे चैप्टर करेंगे आखिरी में जी एडवांस्ड के प्रीवियस इयर्स क्वेश्चन और उसकी थ्रू यही आपको दिखाना सीखना या यू नो भरोसा दिलाना चाह रहे हैं की आप आसानी से कर सकते हैं एग्जाम टू नहीं है अगर आपने हर एक चीज बहुत सिस्टमैटिकली ऑर्गेनाइज्ड मैनेजमेंट में पड़ी है तो यही प्रूफ यही यकीन मिलने की कोशिश करेंगे एग्जाम हवा बना दिया गया है ये एक आसान एग्जाम है बेशक हर एक चीज आसान है अगर उसे ठीक से तैयार स्टार्टिंग के लिए प्रिपेयर किया जाए और बहुत स्मार्टली बहुत meticulars के हम प्लान कर रहे हैं हर एक चीज की कब कौन सी चीज कैसे करनी है कैसे पठानी है क्या कैसे सीखना है बस आप शांतिपूर्ण तरीके से लेक्चर फॉलो करिए और यही बात आज की आपको इस क्वेश्चन में भी दिखाई देगी ये आपका सर्कस का फाइनल क्वेश्चन मतलब मैं ये बस बात कहना चाह रहा हूं की अब अगर आपको सर्कल का कोई भी क्वेश्चन आता है तो होना चाहिए अगर आपने सारे लेक्चरर्स अच्छे से देखे हैं हर एक चीज को अच्छे से पढ़ाया यार कॉन्सेप्ट को समझे नोट्स बनाए हैं रिवाइज कर रखा चीज को तो सर्कल्स का एडवांस लेवल के सारे क्वेश्चंस कर चुके हैं अब तो आपसे होना चाहिए साहब अब तो बहुत अच्छे से क्वेश्चंस होने चाहिए अगर आपने इस पूरा पैकेज सर्कस का देखा और ऐसे ही हर एक चैप्टर को एकदम चुन चुन के हम कवर करेंगे हर चैप्टर आईआईटी जी मांस और एडवांस 12th के मैथ्स का पूरा सिलेबस कवर करते हुए एक थारो कंप्रिहेंसिव वो डीटेल्ड तैयारी हम कर रहे हैं बहुत थारो बहुत तारों मतलब बहुत मेटिकुलोस एकदम बिल्कुल बारीक एकदम बिल्कुल गहन एकदम बिल्कुल यू नो रूट तो चलो इस क्वेश्चन को देख लें क्या ये देख लेते हैं सर पहले क्वेश्चन पढ़ते हैं लिखा क्या है बहुत सिंपल सी बात है सर पहले तो उसने बोला है सर वो आपका जो है वो एक सेंटर है एक सर्कल का सर्कल तो बड़ा मजेदार सा है क्योंकि उसको देख के हमारी आंखें खुश हो जाती हैं हमारी बिल्कुल आंखें चमचम उठाती हैं सर की ओर थी क्योंकि ये बहुत मजेदार कलर्स आप ये कैसा सर्कल है जो सेंटर रखता है अपने जीरो कमा जीरो पर और इसकी रेडियस हम देखे रहे हैं यार और उसने एक और पॉइंट कहा है की जो आपकी रेडियस है वैन रूट फाइव बाय तू से बड़ी है बात करेंगे इस बारे में सपोज पीके इसे अन पीके जो है कार्ड है सर किसकी सर्कल की है ना एंड इक्वेशन ऑफ डी लाइन पीके पासिंग थ्रू पी एंड कस तो वही इक्वेशन ऑफ कट जो दे रहा है वो ये दे रहा है फिर वो का रहा है स्थापना समझना बंद हो रहा है तो क्या विजुलाइज कर लो प्लॉट कर लो प्लॉट करने का मतलब पहले तो सर्कल बनाया था सर्कल क्या हो रहा है सर्कल आपका दिख रहा है मुझे x² + y² तो ये आपका क्या हो गया ए एक्सिस है ना सिमिलरली ये क्या हो जाएगा सर आपका एक्स एक्सिस कोई दिक्कत तो नहीं अब अगर मेरे पास एक सर्कल है विच इस सेंटर डेट originaf आई हैव अन सर्कल विच इस सेंटर आते ओरिजिन है ना तो इसको क्या कहूंगा मैं सर ये सर्कल का सेंटर अगर ओरिजिन पर है तो ये आपका सर्कल हो गया बिल्कुल सही बात अब क्या करेंगे इस बारे में बात करते हैं देखो जरा ध्यान से अब क्या करेंगे भाई अब एक सिंपल सा थॉट निकल कर ए रहा है की सर ये जो सर्कल आपने बनाया है ये सर्कल है आपका ये वाला तो इसकी रेडियस कुछ ऐसा कोई रिलेशन मुझे दे रही है ठीक है अब आपके पास पीके एक कार्ड है जो की इस स्ट्रेट लाइन को फॉलो करती है 2X + 4y = 5 सर इस सेकंड में सोचना चाहो तो कैसी होगी देखो जरा एक्स की जगह जीरो रखो तो ए कितना आता है 5 / 4 आप सन रहे हो क्या और अगर ए की जगह जीरो रखा तो एक्स कितना आता है 5 / 2 तो एक्स और yordinate क्या ए जाएंगे 5 / 2 और 5 / 4 मतलब 5 / 2 कमा जीरो और जीरो कमा 5/4 तो दोनों पॉजिटिव है मतलब ये जो लाइन होगी आपकी टेक्निकल ये कुछ ऐसी कुछ पैसे की रैंडम सी एक लाइन बना ले रहा हूं ये कुछ ऐसी सी एक लाइन होगी है ना तो दिस इसे डेट कट बेसिकली ये आपकी वो गोद होने वाली है आई होप यू ऑल हैव अंडरस्टूड दिस मच इनफॉरमेशन सो फार और ये आपकी quadcos की है दिस इस योर कोड पीके तो ये आपके सर्कल की क्वाड है पीके अब बहुत कृष्ण और कम की बात निकल कर आने वाली है इसी पर हम बात करेंगे ये आपका ए जाता है पीके अब क्या सर अब suniyega ध्यान से वो ये कहना चाह रहा है डी सेंटर ऑफ डी सरकम सर्कल ऑफ डी ट्रायंगल ओपीक्यू डिड यू ऑल अंडरस्टैंड ओपीक्यू मतलब क्या सर ओपीक्यू मतलब ये तीन बातें आप के मतलब ये तीन साइड्स आई होप यू आर रिलीजिंग ये आपका ओरिजिन है तो ओ पी के मतलब क्या है सर ओ को आप पी से मिला दो मैं बस अभी एक रैंडम ये आपका ओ है और वहां पर आपका क्या है क्यों तो ये आपका हो गया के सिर्फ मेरा कहना है की आपको दिख नहीं रहा है की आपको समझ नहीं ए रही चीज अगर आपको नहीं दिख रही है तो पता नहीं क्या लेकिन सर आपको चीज देखनी चाहिए आपको कैसे चीज देखनी चाहिए सर ध्यान से देखो ये बड़ी बेसिक सी बात है की सर ये जो आपका ओपीक्यू है ट्रायंगल इसका circumsar का अगर मैं बनाऊंगा तो क्लीयरली एक ऐसा सर्कल जो इससे और इससे पास होगा बेसिक सी बात है इस सर्कल को लेकर आपका क्या खास ओपिनियन या पॉइंट है ट्रायंगल ओपीक्यू जो है आपका जो circumsar है सरकमसर्किल जो है ऑफ डी ट्रायंगल ओपीक्यू उसका जो सेंटर है मतलब सर कम सेंटर वो इस स्ट्रेट लाइन पर ले रहता है आई रिपीट माय स्टेटमेंट आपका जो सरकम सेंटर है किसका इस ट्रायंगल ओ पी के का वो किस टाइम पर लाइक करता है एक्स + 2y = 4 इस इनफॉरमेशन क्या करना है तो सोचेंगे इनफॉरमेशन से क्या डिटेल निकल जा सकती है बट फिलहाल अभी मेरे लिए जो कम की बात है वो ये सोचना की सर इससे और इससे आप क्या डेड कर सकते हो बड़ी बेसिक सी बात बड़ी ही बेसिक और बड़ी ही जरूरी बात की सर अगर एक स्ट्रेट लाइन और एक सर्कल इंटरसेक्ट करें तो उनके पॉइंट ऑफ इंटरसेक्शन से पास होने वाले सारे जो भी सर्कल्स बन सकते हैं उनकी इक्वेशन हम क्या लिखेंगे सर आपके पास एक सर्कल है लेट्स कॉल इट फॉर ए विले S1 आपके पास एक स्ट्रेट लाइन है लेट्स कॉल इट L1 है ना तो इसके और इसके इंटरसेक्शन से मतलब S1 और L1 ने जहां इंटरसेक्ट किया है इन दोनों से infainight सर्कल्स के पास है ना मल्टीपल सर्कल्स के पास सो व्हाट आई एम ट्राईंग तू मेक यू अंडरस्टैंड इसकी अगर इनसे बनने वाले वो सारे इंफिनिटी सर्कल्स की कंबाइंड की इक्वेशन लिखना चाहूं तो मैं लिखूंगा S1 + लामबीडीए l1=0 आई होप ये बात आप समझ का रहे हो उससे क्या 1 मिनट आप लिखिए तो जैसे ही आप लिखते हो S1 प्लस वैन एस वैन क्या है सर S1 आपका x² + ए स्क्वायर क्या अपनी बात समझ का रहे हो प्लस लामबीडीए टाइम से L1 का ये वाली स्ट्रेट लाइन ये वाली स्ट्रेट लाइन मतलब कौन सी ये वाली 2X + 4y -5 तो ये कितना हो जाएगा 2X + 4y - 5 = 0 सो दिस इसे गोइंग तू बी डी सर्कल यू आर लुकिंग फॉर बट ये एक सर्कल नहीं है डीज आर मल्टीपल सर्कल्स की एक से ज्यादा सर्कस है बहुत सारे सर्कल्स हैं तू बी वेरी ऑनेस्ट तू बी वेरी प्रेसिस अब आप बात समझने की कोशिश करना बहुत बहुत क्रोशिया है इस बात को ध्यान से सुनो सर लामबीडीए की अलग-अलग वैल्यूज के लिए मुझे अलग अलग सर्कल्स मिलेंगे मतलब आपको कैसे कैसे सेट करवाएंगे आपको बहुत सारे सर्कल मिल सकते हैं और वो infainight सर्कल जो आपके किस जीरो पी और के से पास हो रहे होंगे मतलब आपके किस किस से पास हो रहे होंगे वो ओरिजिनल है के से और आपके जीरो से यानी ओ से अब मेरी बात समझ का रहे हो जैसे बहुत सारे सर्कल होंगे अभी ड्रा करने बैठूंगा तो बहुत सारे बन जाएंगे जो की इन तीनों से पास हो रहे होंगे टेक्निकल और ऐसे बहुत सर्कल्स कैन क्रिएट अन डिपेंडिंग उसका कोई एक पार्ट थोड़ा छोटा होगा कोई बड़ा होगा बट आप और के को पास होते हुए मतलब ऐसे सर्कल जो ओपीक्यू से गुजर रहे हो जैसे infyanite सर्कल्स के पास और उन सारे अलग-अलग सर्कल लामबीडीए की अलग-अलग वैल्यूज के लिए वेरी करेगी सारी बातें छोड़ो मुझे तो ये बताओ मैं इसकी इक्वेशन देखू तो इस सर्कल की इक्वेशन में मुझे इस इस सर्कल के बीच में सेंटर के कोऑर्डिनेट्स क्या मिलते हैं सो जब आप लामबीडीए अंदर मल्टीप्लाई कर दोगे ना तो कुछ मिलेगा क्या तुलुम दाई एक्स + 4y -5 λ है ना अब देखो आप एक सर्कल है x² + y² अगर इसमें मैं सेंटर के अकॉर्डिंग इस सर्कल के सेंटर के अकॉर्डिंग सर्कल का सेंटर क्या होगा सर ये सर्कल के सेंटर के जब आप कोऑर्डिनेट्स निकलोगे ना तो वो क्या होंगे देखो 2 लामबीडीए एक्स का नेगेटिव बात समझ का रहे हो सिमिलरली फोर लामबीडीए वही का नेगेटिव हाफ आई थिंक - रूल सो दिस इस गोइंग तू बी माइंस तू टेक्निकल - लामबीडीए - 2 लामबीडीए आपके उसे सर्कल के सेंटर के cardinate होंगे मैं किस खास सर्कल को ढूंढना चाह रहा हूं सर वह खास सर्कल ढूंढना चाह रहा हूं जिसका सेंटर इस पर लाइक करता है तो सर जो आपने ढूंढा है क्या जो आपने सेंटर ढूंढ रहा है माइंस लामबीडीए कमा - 2 लामबीडीए वैसे सेटिस्फाई करेगा क्योंकि इस पर लाइक करता है तो माइंस लें दा + 2 टाइम्स -2 λ विल बी इक्वल तू फोर डिड ऑल गेट डी डेट आई होप ये बात आपको समझ आई सर यहां से अगर आप सॉल्व करें तो ये हो जाता है माइंस लाइन डाउन ये हो जाता है कितना -4 सही लिख रहा हूं क्या इस इक्वल तू व्हाट फोर कोई दिक्कत तो नहीं तो ये हो जाएगा -5 लामबीडीए है ना तो लामबीडीए की वैल्यू कितनी ए जाएगी सर वो हो जाएगी -4 / 5 अब जब लामबीडीए की वैल्यू आई है -4 / 5 तो आपको क्या चाहिए आपको तो ये आर की वैल्यू चाहिए अरे हान या ना तो आर की वैल्यू कैसे निकलेंगे आई थिंक सर आर की वैल्यू निकलना आसान है क्या थॉट रहेगा आई थिंक अगर सर आप यहां से चीज सोचना जाए तो आप बहुत आराम से यहां से आर की वैल्यू निकल सकते हैं ऐसे कैसे हैं और क्या थॉट या क्या प्रक्रिया रहेगी जो आप आर की वैल्यू निकल पाएंगे अच्छा एक सारी बातें छोड़ो मुझे ये बात पता है क्या की ये वाला जो सर्कल आपने बनाया है यह वाला जो सरकार आपने बनाया है सर यह सर्कल एक बहुत खास पॉइंट ओरिजिन से पास होता है होता है क्या सर होता है तो यहां पर ओरिजिन यानी जीरो कमा जीरो पास करेंगे तो सेटिस्फाई होगा बिल्कुल सर आपके पास एक और कंडीशन थी बस आपने ध्यान नहीं दिया तो इस कंडीशन को देखो जीरो तो यहां क्या बचा फ्री बचा -r² 00 ये क्या बचा सर -5 लामबीडीए आप मेरी बात समझ का रहे हो इस इक्वल तू जीरो यानी अगर मैं आपसे पूछूं r² तो r² किसके इक्वल होगा सर r² अगर मैं पूछूं आपसे तो वो होगा -5 लामबीडीए के इक्वल और लामबीडीए कितना है सर सो लामबीडीए जस्ट अभी हमने निकाला - 4 / 5 एंड डी जस्ट हमने निकाला -4/5 तो r² कितना ए जाएगा सर वो होगा -5 -4/5 आई थिंक आर स्क्वायर की वैल्यू कितनी आती है 4 हमसे क्या पूछा गया था स्टूडेंट्स आई थिंक हमसे कुछ तो पूछा गया था हमसे पूछा गई थी आर की वैल्यू अब चुकी है रेडियस है तो ये +2 भी हो सकती है और -2 भी हो सकती है बट प्रिफरेबली मैं क्या लूंगा क्योंकि ये रेडियस एक डिस्टेंस होगी तो -2 नहीं लूंगा तो इसका आंसर क्या हो जाएगा प्लस तू और इस तरीके से आप इस क्वेश्चन का आंसर मार्क करेंगे सीधे सीधे डायरेक्टली इस एप्रोच इस मेथड थॉट से आई थिंक चीजें आसान है मुश्किल और डिफिकल्ट नहीं है और इसी के साथ आपका ये पार्ट हो रहा है खत्म और ये चैप्टर हो रहा है का नेक्स्ट लेक्चर से हम शुरुआत करेंगे पैराबोला आज अन न्यू चैप्टर वही कहानी फिर से दोहराएंगे पहले हर कॉन्सेप्ट को पढ़ेंगे उसे कॉन्सेप्ट को कंसोलिडेटेड करेंगे उसी के आईआईटी जी मांस और एडवांस लेवल के इलस्ट्रेशंस और एग्जांपल से फिर एक्सरसाइज करेंगे उसे चैप्टर की पैराबोला की जिसमें हर कैटिगरी का हम क्वेश्चन करेंगे सिंगल करेक्ट मल्टीपल करेक्ट यू नो न्यूमेरिकल आंसर टाइप हर तरह का हर कैटिगरी का हम क्वेश्चन करेंगे और उसे देखेंगे और इसी कहानी को कंटिन्यू करते हुए हम आगे बढ़ाएंगे बात पैराबोला के अंडरस्टैंडिंग को लेते हुए आगे हम उसमें करेंगे जी एडवांस्ड के प्रीवियस इयर्स क्वेश्चंस को और बहुत अच्छे तरीके से हम चीज समझ रहे होंगे और देख रहे होंगे की चीज बहुत टू और मुश्किल नहीं है ये सब कुछ हम करवा रहे हैं स्टूडेंट्स आपको बिल्कुल फ्री ऑफ कॉस्ट कोई हिडन कॉस्ट कोई चार्जेस नहीं है कोई सब्सक्रिप्शन आपको नहीं खरीदना है बिल्कुल मुफ्त बिल्कुल एकदम फ्री ऑफ कॉस्ट इन डी गार्डन से 12 स्टैंडर्ड तक अलग-अलग वोट्स अलग-अलग मीडियम और कई सारे कॉम्पिटेटिव एग्जाम्स के हम आपको तैयारी करते हैं ये प्लेलिस्ट से करके रख लीजिए प्लीज इंगेज विथ डी मजिक्स फॉर आईआईटी जी मांस एंड एडवांस्ड फुल प्लेलिस्ट लिंक आपको दिखाई देगी इसी वीडियो के डिस्क्रिप्शन में आप हमारे नोट्स इबुक्स और पीडीएफ के लिए भी टीम से कॉन्टैक्ट कर सकते हैं बड़ा ही आसान सा तरीका है आपको कहां जाना है इसी वीडियो के डिस्क्रिप्शन में वहां आपको गेट नोट्स लिंक दिखाई देगा उसे पर क्लिक करके चीजे आई थिंक आसान है सॉर्टेड है सुलझी हुई है बस अच्छे से पढ़िए नोट्स बनाया रिवाइज करिए क्वेश्चंस प्रैक्टिस करिए बस यही एक तरीका है एग्जाम को क्रैक करने का सिंह आगे बुक जो की पंच हसन में डिवाइड है कोऑर्डिनेट्स कैलकुलस वेक्टर एंड 3D अलजेब्रा को ये हर एक चीज हर एक चैप्टर 11 12 का पूरा मैथ्स का सिलेबस कवर करते हुए हम पढ़ेंगे बहुत डिटेल कंपैरिजन में या तो आप हमारे यूट्यूब ब्रिंग्स या फिर वेबसाइट www.com या फिर हमारे एंड्रॉयड एप्लीकेशन पर सारी चीज हैं सिस्टमैटिकली ऑर्गेनाइज्ड मैनर में एक्सेस कर सकते हैं डेली दो वीडियो रिलीज हो रहे हैं ताकि आपका टाइमली कोर्स कंप्लीट कर रहा है और आपको कॉन्फिडेंट एग्जामिनेशन हॉल में भेजे और आप अच्छे से अपने पेपर को दे और एक अच्छा स्कोर एक अच्छी रैंक एक अच्छा प्रेसिडेंट आई बस यही एक सिंपल सा थॉट रहेगा हमारा हमेशा आय थिंक चीजें लग चुकी हैं चीज सॉर्टेड है और कोई परेशानी डिफिकल्टी दिक्कत नहीं है इस चैप्टर में कोई भी परेशानी हो कमेंट में likhiyega कोई रियल 11th होगी तो जरूर रेस्पॉन्ड करेंगे विद डेट स्टूडेंट्स मिलते हैं नेक्स्ट लेक्चर में पैराबोलिक चैप्टर के साथ मिलेंगे अच्छे से बने रहिए खूब मेहनत करिए खूब पढ़ करिए थैंक यू सो मच स्टूडेंट्स टेक अन गुड केयर ऑफ योर सेल्फ सी आर कॉलिंग टुडे हर
17079	https://www.jeffreythompson.org/collision-detection/line-circle.php	Collision Detection ←Collision Detection→ [ your browser does not support the canvas tag ] LINE/CIRCLE To check if a circle is hitting a line, we use code from previous examples — a practice that we'll use through the rest of the book. The resulting math behind this gets a little hairy, but we'll simplify the harder parts. First, let's test if either of the ends of the line are inside the circle. This is likely to happen if the line is much smaller than the circle. To do this, we can use Point/Circle from the beginning of the book. If either end is inside, return true immediately and skip the rest. boolean inside1 = pointCircle(x1,y1, cx,cy,r); boolean inside2 = pointCircle(x2,y2, cx,cy,r); if (inside1 \|\| inside2) return true; Next, we need to get closest point on the line. To start, let's get the length of the line using the Pythagorean Theorem: float distX = x1 - x2; float distY = y1 - y2; float len = sqrt( (distXdistX) + (distYdistY) ); Then, we get a value we're calling dot. If you've done vector math before, this is the same as doing the dot product of two vectors. If this isn't familiar, no worry! Consider this step a lot of math you can be glad not to have to solve by hand: float dot = ( ((cx-x1)(x2-x1)) + ((cy-y1)(y2-y1)) ) / pow(len,2); Finally, we can use this equation to find the closest point on the line: float closestX = x1 + (dot (x2-x1)); float closestY = y1 + (dot (y2-y1)); However, this returns a point anywhere on the line as it extends to infinity in both directions. In other words, it could give us a point off the end of the line! So let's check if that closest point is actually on the line using the Line/Point algorithm we just made. This is the first of many times we'll nest previous functions when working on more complex collisions. If the point is on the line, we can keep going. If not, we can immediately return false, since that means the closest point is off one of the ends: boolean onSegment = linePoint(x1,y1,x2,y2, closestX,closestY); if (!onSegment) return false; Finally, we get the distance from the circle to the closest point on the line, once again using the Pythagorean Theorem: distX = closestX - cx; distY = closestY - cy; float distance = sqrt( (distXdistX) + (distYdistY) ); If that distance is less than the radius, we have a collision (same as Point/Circle). if (distance <= r) { return true; } return false; Here's a full example putting everything together. Notice that we have three functions at the bottom: the one we just built and two previous functions. ``` float cx = 0; // circle position (set by mouse) float cy = 0; float r = 30; // circle radius float x1 = 100; // coordinates of line float y1 = 300; float x2 = 500; float y2 = 100; void setup() { size(600,400); strokeWeight(5); // make it a little easier to see } void draw() { background(255); // update circle to mouse position cx = mouseX; cy = mouseY; // check for collision // if hit, change line's stroke color boolean hit = lineCircle(x1,y1, x2,y2, cx,cy,r); if (hit) stroke(255,150,0, 150); else stroke(0,150,255, 150); line(x1,y1, x2,y2); // draw the circle fill(0,150,255, 150); noStroke(); ellipse(cx,cy, r2,r2); } // LINE/CIRCLE boolean lineCircle(float x1, float y1, float x2, float y2, float cx, float cy, float r) { // is either end INSIDE the circle? // if so, return true immediately boolean inside1 = pointCircle(x1,y1, cx,cy,r); boolean inside2 = pointCircle(x2,y2, cx,cy,r); if (inside1 \|\| inside2) return true; // get length of the line float distX = x1 - x2; float distY = y1 - y2; float len = sqrt( (distXdistX) + (distYdistY) ); // get dot product of the line and circle float dot = ( ((cx-x1)(x2-x1)) + ((cy-y1)(y2-y1)) ) / pow(len,2); // find the closest point on the line float closestX = x1 + (dot (x2-x1)); float closestY = y1 + (dot (y2-y1)); // is this point actually on the line segment? // if so keep going, but if not, return false boolean onSegment = linePoint(x1,y1,x2,y2, closestX,closestY); if (!onSegment) return false; // optionally, draw a circle at the closest // point on the line fill(255,0,0); noStroke(); ellipse(closestX, closestY, 20, 20); // get distance to closest point distX = closestX - cx; distY = closestY - cy; float distance = sqrt( (distXdistX) + (distYdistY) ); if (distance <= r) { return true; } return false; } // POINT/CIRCLE boolean pointCircle(float px, float py, float cx, float cy, float r) { // get distance between the point and circle's center // using the Pythagorean Theorem float distX = px - cx; float distY = py - cy; float distance = sqrt( (distXdistX) + (distYdistY) ); // if the distance is less than the circle's // radius the point is inside! if (distance <= r) { return true; } return false; } // LINE/POINT boolean linePoint(float x1, float y1, float x2, float y2, float px, float py) { // get distance from the point to the two ends of the line float d1 = dist(px,py, x1,y1); float d2 = dist(px,py, x2,y2); // get the length of the line float lineLen = dist(x1,y1, x2,y2); // since floats are so minutely accurate, add // a little buffer zone that will give collision float buffer = 0.1; // higher # = less accurate // if the two distances are equal to the line's // length, the point is on the line! // note we use the buffer here to give a range, // rather than one # if (d1+d2 >= lineLen-buffer && d1+d2 <= lineLen+buffer) { return true; } return false; } ``` Math using lines can benefit from some of the built-in functionality of the PVector class. If you haven't used PVectors before, it may be worth some time to get familiar with them. The Processing website has a good tutorial. Daniel Shiffman's excellent Nature of Code book deals with vectors quite a bit and is a very friendly introduction. We'll cover PVectors a little bit when we start working with polygons, if you want a very short introduction. This example was based on code by Philip Nicoletti. This CodeGuru post inclues a lot more discussion of how this algorithm works and the math behind it, if you're so inclined. NEXT: Line/Line [ intro, source, issues ]
17080	https://www.geeksforgeeks.org/videos/automorphic-number//	Automorphic number - GeeksforGeeks \| Videos Data Structure Java Python HTML Interview Preparation Sign In Tutorials Courses Tracks DSA Practice Problems C C++ Java Python JavaScript Data Science Machine Learning Courses Linux DevOps SQL Web Development System Design Aptitude Videos Video Player is loading. Play Video Play Mute Current Time 0:00 / Duration 0:00 Loaded: 0% 0:00 Stream Type LIVE Seek to live, currently behind live LIVE Remaining Time-0:00 1x Playback Rate 2x 1.5x 1x, selected 0.5x Chapters Chapters Descriptions descriptions off, selected Captions captions settings, opens captions settings dialog captions off, selected English Captions auto Quality Audio Track Fullscreen This is a modal window. Beginning of dialog window. Escape will cancel and close the window. Text Color Transparency Background Color Transparency Window Color Transparency Font Size Text Edge Style Font Family Reset restore all settings to the default values Done Close Modal Dialog End of dialog window. June 07, 2022 \|3.4K Views Automorphic number CPP,Maths,C++,Magic Numbers,CPP Programs Save Share Like Description Discussion Given a number N, the task is to check whether the number is Automorphic number or not. A number is called Automorphic number if and only if its square ends in the same digits as the number itself. Examples : Input : N = 76 Output : Automorphic Explanation: As 7676 = 5776 Input : N = 25 Output : Automorphic As 2525 = 625 Input : N = 7 Output : Not Automorphic As 77 = 49 Automorphic number: Read More Recommendations 10:14 67.4K Views \| 11/01/2025... Program to Find the Sum of All Digits of a Number 10:37 22.4K Views \| 09/01/2025... Graham Scan Algorithm 03:53 16.7K Views \| 08/01/2025... Primality Test (Introduction and School Method) 04:41 9.7K Views \| 08/01/2025... C++ Program to Find Roots of a Quadratic Equation 08:23 5.9K Views \| 03/01/2025... C++ Program on Multiplication of Two Matrices 08:00 3.8K Views \| 03/01/2025... JavaScript Program to Check if Two Strings are Anagrams 03:36 3.0K Views \| 16/08/2024... C++ Program to Return Maximum occurring Character in an Input String 11:51 850 Views \| 18/07/2024... Palindrome Number 03:12 109.9K Views \| 18/07/2024... C++ Program to Find GCD 10:49 43.4K Views \| 17/07/2024... C++ Program to Calculate Power of a Number Corporate & Communications Address: A-143, 7th Floor, Sovereign Corporate Tower, Sector- 136, Noida, Uttar Pradesh (201305) Registered Address: K 061, Tower K, Gulshan Vivante Apartment, Sector 137, Noida, Gautam Buddh Nagar, Uttar Pradesh, 201305 Advertise with us Company About Us Legal Privacy Policy Contact Us Advertise with us GFG Corporate Solution Campus Training Program Explore POTD Job-A-Thon Community Blogs Nation Skill Up Tutorials Programming Languages DSA Web Technology AI, ML & Data Science DevOps CS Core Subjects Interview Preparation GATE Software and Tools Courses IBM Certification DSA and Placements Web Development Programming Languages DevOps & Cloud GATE Trending Technologies Videos DSA Python Java C++ Web Development Data Science CS Subjects Preparation Corner Aptitude Puzzles GfG 160 DSA 360 System Design @GeeksforGeeks, Sanchhaya Education Private Limited, All rights reserved
17081	https://discrete.openmathbooks.org/dmoi2/sec_polyfit.html	(\def\d{\displaystyle} \def\course{Math 228} \newcommand{\f}{\mathfrak #1} \newcommand{\s}{\mathscr #1} \def\N{\mathbb N} \def\B{\mathbf{B}} \def\circleA{(-.5,0) circle (1)} \def\Z{\mathbb Z} \def\circleAlabel{(-1.5,.6) node[above]{$A$}} \def\Q{\mathbb Q} \def\circleB{(.5,0) circle (1)} \def\R{\mathbb R} \def\circleBlabel{(1.5,.6) node[above]{$B$}} \def\C{\mathbb C} \def\circleC{(0,-1) circle (1)} \def\F{\mathbb F} \def\circleClabel{(.5,-2) node[right]{$C$}} \def\A{\mathbb A} \def\twosetbox{(-2,-1.5) rectangle (2,1.5)} \def\X{\mathbb X} \def\threesetbox{(-2,-2.5) rectangle (2,1.5)} \def\E{\mathbb E} \def\O{\mathbb O} \def\U{\mathcal U} \def\pow{\mathcal P} \def\inv{^{-1}} \def\nrml{\triangleleft} \def\st{:} \def\~{\widetilde} \def\rem{\mathcal R} \def\sigalg{$\sigma$-algebra } \def\Gal{\mbox{Gal}} \def\iff{\leftrightarrow} \def\Iff{\Leftrightarrow} \def\land{\wedge} \def\And{\bigwedge} \def\entry{\entry} \def\AAnd{\d\bigwedge\mkern-18mu\bigwedge} \def\Vee{\bigvee} \def\VVee{\d\Vee\mkern-18mu\Vee} \def\imp{\rightarrow} \def\Imp{\Rightarrow} \def\Fi{\Leftarrow} \def\var{\mbox{var}} \def\Th{\mbox{Th}} \def\entry{\entry} \def\sat{\mbox{Sat}} \def\con{\mbox{Con}} \def\iffmodels{\bmodels\models} \def\dbland{\bigwedge !!\bigwedge} \def\dom{\mbox{dom}} \def\rng{\mbox{range}} \def\isom{\cong} \DeclareMathOperator{\wgt}{wgt} \newcommand{\vtx}{node[fill,circle,inner sep=0pt, minimum size=4pt,label=#1:#2]{}} \newcommand{\va}{\vtx{above}{#1}} \newcommand{\vb}{\vtx{below}{#1}} \newcommand{\vr}{\vtx{right}{#1}} \newcommand{\vl}{\vtx{left}{#1}} \renewcommand{\v}{\vtx{above}{}} \def\circleA{(-.5,0) circle (1)} \def\circleAlabel{(-1.5,.6) node[above]{$A$}} \def\circleB{(.5,0) circle (1)} \def\circleBlabel{(1.5,.6) node[above]{$B$}} \def\circleC{(0,-1) circle (1)} \def\circleClabel{(.5,-2) node[right]{$C$}} \def\twosetbox{(-2,-1.4) rectangle (2,1.4)} \def\threesetbox{(-2.5,-2.4) rectangle (2.5,1.4)} \def\ansfilename{practice-answers} \def\shadowprops{{fill=black!50,shadow xshift=0.5ex,shadow yshift=0.5ex,path fading={circle with fuzzy edge 10 percent}}} \newcommand{\hexbox}{ \def\x{-cos{30}\r#1+cos{30}#2\r2} \def\y{-\r#1-sin{30}\r#1} \draw (\x,\y) +(90:\r) -- +(30:\r) -- +(-30:\r) -- +(-90:\r) -- +(-150:\r) -- +(150:\r) -- cycle; \draw (\x,\y) node{#3}; } \renewcommand{\bar}{\overline} \newcommand{\card}{\left\| #1 \right\|} \newcommand{\twoline}{\begin{pmatrix}#1 \ #2 \end{pmatrix}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} ) Discrete MathematicsAn Open Introduction Oscar Levin Section2.3Polynomial Fitting ¶ Investigate!20 A standard (8 \times 8) chessboard contains 64 squares. Actually, this is just the number of unit squares. How many squares of all sizes are there on a chessboard? Start with smaller boards: (1\times 1\text{,}) (2 \times 2\text{,}) (3\times 3\text{,}) etc. Find a formula for the total number of squares in an (n\times n) board. So far we have seen methods for finding the closed formulas for arithmetic and geometric sequences. Since we know how to compute the sum of the first (n) terms of arithmetic and geometric sequences, we can compute the closed formulas for sequences which have an arithmetic (or geometric) sequence of differences between terms. But what if we consider a sequence which is the sum of the first (n) terms of a sequence which is itself the sum of an arithmetic sequence? Before we get too carried away, let's consider an example: How many squares (of all sizes) are there on a chessboard? A chessboard consists of (64) squares, but we also want to consider squares of longer side length. Even though we are only considering an (8 \times 8) board, there is already a lot to count. So instead, let us build a sequence: the first term will be the number of squares on a (1 \times 1) board, the second term will be the number of squares on a (2 \times 2) board, and so on. After a little thought, we arrive at the sequence \begin{equation} 1,5,14,30, 55,\ldots \end{equation} This sequence is not arithmetic (or geometric for that matter), but perhaps it's sequence of differences is. For differences we get \begin{equation} 4, 9, 16, 25, \ldots \end{equation} Not a huge surprise: one way to count the number of squares in a (4 \times 4) chessboard is to notice that there are (16) squares with side length 1, 9 with side length 2, 4 with side length 3 and 1 with side length 4. So the original sequence is just the sum of squares. Now this sequence of differences is not arithmetic since it's sequence of differences (the differences of the differences of the original sequence) is not constant. In fact, this sequence of second differences is \begin{equation} 5, 7, 9, \ldots \end{equation} which is an arithmetic sequence (with constant difference 2). Notice that our original sequence had third differences (that is, differences of differences of differences of the original) constant. We will call such a sequence (\Delta^3)-constant. The sequence (1, 4, 9, 16, \ldots) has second differences constant, so it will be a (\Delta^2)-constant sequence. In general, we will say a sequence is a (\Delta^k)-constant sequence if the (k)th differences are constant. Example2.3.1 Which of the following sequences are (\Delta^k)-constant for some value of (k\text{?}) (2, 3, 7, 14, 24, 37,\ldots\text{.}) (1, 8, 27, 64, 125, 216, \ldots\text{.}) (1,2,4,8,16,64,128,\ldots\text{.}) Solution This is the sequence from Example 2.2.6, in which we found a closed formula by recognizing the sequence as the sequence of partial sums of an arithmetic sequence. Indeed, the sequence of first differences is (1,4,7, 10, 13,\ldots\text{,}) which itself has differences (3,3,3,3,\ldots\text{.}) Thus (2, 3, 7, 14, 24, 37,\ldots) is a (\Delta^2)-constant sequence. These are the perfect cubes. The sequence of first differences is (7, 19, 37, 61, 91, \ldots\text{;}) the sequence of second differences is (12, 18, 24, 30,\ldots\text{;}) the sequence of third differences is constant: (6,6,6,\ldots\text{.}) Thus the perfect cubes are a (\Delta^3)-constant sequence. If we take first differences we get (1,2,4,8,16,\ldots\text{.}) Wait, what? That's the sequence we started with. So taking second differences will give us the same sequence again. No matter how many times we repeat this we will always have the same sequence, which in particular means no finite number of differences will be constant. Thus this sequence is not (\Delta^k)-constant for any (k\text{.}) The (\Delta^0)-constant sequences are themselves constant, so a closed formula for them is easy to compute (it's just the constant). The (\Delta^1)-constant sequences are arithmetic and we have a method for finding closed formulas for them as well. Every (\Delta^2)-constant sequence is the sum of an arithmetic sequence so we can find formulas for these as well. But notice that the format of the closed formula for a (\Delta^2)-constant sequence is always quadratic. For example, the square numbers are (\Delta^2)-constant with closed formula (a_n= n^2\text{.}) The triangular numbers (also (\Delta^2)-constant) have closed formula (a_n = \frac{n(n+1)}{2}\text{,}) which when multiplied out gives you an (n^2) term as well. It appears that every time we increase the complexity of the sequence, that is, increase the number of differences before we get constants, we also increase the degree of the polynomial used for the closed formula. We go from constant to linear to quadratic. The sequence of differences between terms tells us something about the rate of growth of the sequence. If a sequence is growing at a constant rate, then the formula for the sequence will be linear. If the sequence is growing at a rate which itself is growing at a constant rate, then the formula is quadratic. You have seen this elsewhere: if a function has a constant second derivative (rate of change) then the function must be quadratic. This works in general: Finite Differences The closed formula for a sequence will be a degree (k) polynomial if and only if the sequence is (\Delta^k)-constant (i.e., the (k)th sequence of differences is constant). This tells us that the sequence of numbers of squares on a chessboard, (1, 5, 14, 30, 55, \ldots\text{,}) which we saw to be (\Delta^3)-constant, will have a cubic (degree 3 polynomial) for its closed formula. Now once we know what format the closed formula for a sequence will take, it is much easier to actually find the closed formula. In the case that the closed formula is a degree (k) polynomial, we just need (k+1) data points to “fit” the polynomial to the data. Example2.3.2 Find a formula for the sequence (3, 7, 14, 24,\ldots\text{.}) Assume (a_1 = 3\text{.}) Solution First, check to see if the formula has constant differences at some level. The sequence of first differences is (4, 7, 10, \ldots) which is arithmetic, so the sequence of second differences is constant. The sequence is (\Delta^2)-constant, so the formula for (a_n) will be a degree 2 polynomial. That is, we know that for some constants (a\text{,}) (b\text{,}) and (c\text{,}) \begin{equation} a_n = an^2 + bn + c. \end{equation} Now to find (a\text{,}) (b\text{,}) and (c\text{.}) First, it would be nice to know what (a_0) is, since plugging in (n = 0) simplifies the above formula greatly. In this case, (a_0 = 2) (work backwards from the sequence of constant differences). Thus \begin{equation} a_0 = 2 = a\cdot 0^2 + b \cdot 0 + c, \end{equation} so (c = 2\text{.}) Now plug in (n =1) and (n = 2\text{.}) We get \begin{equation} a_1 = 3 = a + b + 2 \end{equation} \begin{equation} a_2 = 7 = a4 + b 2 + 2. \end{equation} At this point we have two (linear) equations and two unknowns, so we can solve the system for (a) and (b) (using substitution or elimination or even matrices). We find (a = \frac{3}{2}) and (b = \frac{-1}{2}\text{,}) so (a_n = \frac{3}{2} n^2 - \frac{1}{2}n + 2\text{.}) Example2.3.3 Find a closed formula for the number of squares on an (n \times n) chessboard. Solution We have seen that the sequence (1, 5, 14, 30, 55, \ldots) is (\Delta^3)-constant, so we are looking for a degree 3 polynomial. That is, \begin{equation} a_n = an^3 + bn^2 + cn + d. \end{equation} We can find (d) if we know what (a_0) is. Working backwards from the third differences, we find (a_0 = 0) (unsurprisingly, since there are no squares on a (0\times 0) chessboard). Thus (d = 0\text{.}) Now plug in (n = 1\text{,}) (n =2\text{,}) and (n =3\text{:}) \begin{align} 1 = \amp a + b + c\ 5 = \amp 8a + 4b + 2c\ 14 = \amp 27a + 9b + 3c. \end{align} If we solve this system of equations we get (a = \frac{1}{3}\text{,}) (b = \frac{1}{2}) and (c = \frac{1}{6}\text{.}) Therefore the number of squares on an (n \times n) chessboard is (a_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n\text{.}) Note: Since the squares-on-a-chessboard problem is really asking for the sum of squares, we now have a nice formula for (\d\sum_{k=1}^n k^2\text{.}) Not all sequences will have polynomials as their closed formula. We can use the theory of finite differences to identify these. Example2.3.4 Determine whether the following sequences can be described by a polynomial, and if so, of what degree. (1, 2, 4, 8, 16, \ldots) (0, 7, 50, 183, 484, 1055, \ldots) (1,1,2,3,5,8,13,\ldots) Solution As we saw in Example 2.3.1, this sequence is not (\Delta^k)-constant for any (k\text{.}) Therefore the closed formula for the sequence is not a polynomial. In fact, we know the closed formula is (a_n = 2^n\text{,}) which grows faster than any polynomial (so is not a polynomial). The sequence of first differences is (7, 43, 133, 301, 571,\ldots\text{.}) The second differences are: (36, 90, 168, 270,\ldots\text{.}) Third difference: (54, 78, 102,\ldots\text{.}) Fourth differences: (24, 24, \ldots\text{.}) As far as we can tell, this sequence of differences is constant so the sequence is (\Delta^4)-constant and as such the closed formula is a degree 4 polynomial. This is the Fibonacci sequence. The sequence of first differences is (0, 1, 1, 2, 3, 5, 8, \ldots\text{,}) the second differences are (1, 0, 1, 1, 2, 3, 5\ldots\text{.}) We notice that after the first few terms, we get the original sequence back. So there will never be constant differences, so the closed formula for the Fibonacci sequence is not a polynomial. SubsectionExercises ¶ 1 Use polynomial fitting to find the formula for the (n)th term of the sequences ((a_n)_{n \ge 0}) below. 2, 5, 11, 21, 36,… 0, 2, 6, 12, 20,… 1, 2, 4, 8, 15, 26 … 3, 6, 12, 22, 37, …. After finding a formula here, compare to part (a). Solution Notice that the third differences are constant, so (a_n = an^3 + bn^2 + cn + d\text{.}) Use the terms of the sequence to solve for (a, b, c,) and (d) to get (a_n = \frac{1}{6} (12+11 n+6 n^2+n^3)\text{.}) (a_n = n^2 + n\text{.}) Here we know that we are looking for a quadratic because the second differences are constant. So (a_n = an^2 + bn + c\text{.}) Since (a_0 = 0\text{,}) we know (c= 0\text{.}) So just solve the system \begin{align} 2 \amp = a + b \ 6 \amp = 4a + 2b \end{align} 2 Make up a sequences that have 3, 3, 3, 3, … as its second differences. 1, 2, 3, 4, 5, … as its third differences. 1, 2, 4, 8, 16, … as its 100th differences. 3 Consider the sequence (1, 3, 7, 13, 21, \ldots\text{.}) Explain how you know the closed formula for the sequence will be quadratic. Then “guess” the correct formula by comparing this sequence to the squares (1, 4, 9, 16, \ldots) (do not use polynomial fitting). Solution The first differences are (2, 4, 6, 8, \ldots\text{,}) and the second differences are (2, 2, 2, \ldots\text{.}) Thus the original sequence is (\Delta^2)-constant, so can be fit to a quadratic. Call the original sequence (a_n\text{.}) Consider (a_n - n^2\text{.}) This gives (0, -1, -2, -3, \ldots\text{.}) That sequence has closed formula (1-n) (starting at (n = 1)) so we have (a_n - n^2 = 1-n) or equivalently (a_n = n^2 - n + 1\text{.}) 4 Use a similar technique as in the previous exercise to find a closed formula for the sequence (2, 11, 34, 77, 146, 247,\ldots\text{.}) 5 In their down time, ghost pirates enjoy stacking cannonballs in triangular based pyramids (aka, tetrahedrons), like those pictured here: Note, in the picture on the right, there are some cannonballs (actually just one) you cannot see. The next picture would have 4 cannonballs you cannot see. The stacks are not hollow. The pirates wonder how many cannonballs would be required to build a pyramid 15 layers high (thus breaking the world cannonball stacking record). Can you help? Let (P(n)) denote the number of cannonballs needed to create a pyramid (n) layers high. So (P(1) = 1\text{,}) (P(2) = 4\text{,}) and so on. Calculate (P(3)\text{,}) (P(4)) and (P(5)\text{.}) Use polynomial fitting to find a closed formula for (P(n)\text{.}) Show your work. Answer the pirate's question: how many cannonballs do they need to make a pyramid 15 layers high? 6 Suppose (a_n = n^2 + 3n + 4\text{.}) Find a closed formula for the sequence of differences by computing (a_n - a_{n-1}\text{.}) Solution (a_{n-1} = (n-1)^2 + 3(n-1) + 4 = n^2 + n + 2\text{.}) Thus (a_n - a_{n-1} = 2n+2\text{.}) Note that this is linear (arithmetic). We can check that we are correct. The sequence (a_n) is (4, 8, 14, 22, 32, \ldots) and the sequence of differences is thus (4, 6, 8, 10,\ldots) which agrees with (2n+2) (if we start at (n = 1)). 7 Repeat the above assuming this time (a_n = an^2 + bn + c\text{.}) That is, prove that every quadratic sequence has arithmetic differences. 8 Can you use polynomial fitting to find the formula for the (n)th term of the sequence 4, 7, 11, 18, 29, 47, …? Explain why or why not. 9 Will the (n)th sequence of differences of (2, 6, 18, 54, 162, \ldots) ever be constant? Explain. 10 Consider the sequences (2, 5, 12, 29, 70, 169, 408,\ldots) (with (a_0 = 2)). Describe the rate of growth of this sequence. Find a recursive definition for the sequence. Find a closed formula for the sequence. If you look at the sequence of differences between terms, and then the sequence of second differences, the sequence of third differences, and so on, will you ever get a constant sequence? Explain how you know.
17082	https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence?srsltid=AfmBOoo6JmOifh3nSYtZeB36V0FqP2hWXhpgPM7tFTN52nmmJTNDY5SR	Art of Problem Solving Arithmetic sequence - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Arithmetic sequence Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Arithmetic sequence In algebra, an arithmetic sequence, sometimes called an arithmetic progression, is a sequence of numbers such that the difference between any two consecutive terms is constant. This constant is called the common difference of the sequence. For example, is an arithmetic sequence with common difference and is an arithmetic sequence with common difference ; however, and are not arithmetic sequences, as the difference between consecutive terms varies. More formally, the sequence is an arithmetic progression if and only if . A similar definition holds for infinite arithmetic sequences. It appears most frequently in its three-term form: namely, that constants , , and are in arithmetic progression if and only if . Contents [hide] 1 Properties 2 Sum 3 Problems 3.1 Introductory problems 3.2 Intermediate problems 4 See Also Properties Because each term is a common distance from the one before it, every term of an arithmetic sequence can be expressed as the sum of the first term and a multiple of the common difference. Let be the first term, be the th term, and be the common difference of any arithmetic sequence; then, . A common lemma is that given the th term and th term of an arithmetic sequence, the common difference is equal to . Proof: Let the sequence have first term and common difference . Then using the above result, as desired. Another common lemma is that a sequence is in arithmetic progression if and only if is the arithmetic mean of and for any consecutive terms . In symbols, . This is mostly used to perform substitutions, though it occasionally serves as a definition of arithmetic sequences. Sum An arithmetic series is the sum of all the terms of an arithmetic sequence. All infinite arithmetic series diverge. As for finite series, there are two primary formulas used to compute their value. The first is that if an arithmetic series has first term , last term , and total terms, then its value is equal to . Proof: Let the series be equal to , and let its common difference be . Then, we can write in two ways: Adding these two equations cancels all terms involving ; and so , as required. The second is that if an arithmetic series has first term , common difference , and terms, it has value . Proof: The final term has value . Then by the above formula, the series has value This completes the proof. Problems Here are some problems with solutions that utilize arithmetic sequences and series. Introductory problems 2005 AMC 10A Problem 17 2006 AMC 10A Problem 19 2012 AIME I Problems/Problem 2 2004 AMC 10B Problems/Problem 10 2006 AMC 10A, Problem 9 2006 AMC 12A, Problem 12 Intermediate problems 2003 AIME I, Problem 2 Find the roots of the polynomial , given that the roots form an arithmetic progression. See Also Geometric sequence Harmonic sequence Sequence Series Retrieved from " Categories: Algebra Sequences and series Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
17083	https://t5k.org/prove/prove1.html	\| \| \| \| \| --- --- \| \| \| \| \| --- \| \| Finding primes & proving primality 1: Introduction \| \| \| \| \| Home > Primality Proving > Chapter One: Introduction \| \| Introduction How do we go about finding primes? And once we have found them, how do we prove they are truly prime? The answer depends on the size of the primes and how sure we need to be of their primality. In these pages we present the appropriate answers in several sections. Let us preview these chapters one at a time. Chapter Two: The quick tests for small numbers and probable primes For very small primes we can use the Sieve of Eratosthenes or trial division. These methods are sure, and are the best methods for small numbers, but become far too time consuming before our numbers reach thirty digits. If we are going to use our primes for "industrial" uses (e.g., for RSA encryption) we often do not need to prove they are prime. It may be enough to know that the probability they are composite is less than 0.000000000000000000000001%. In this case we can use (strong) probable primality tests. These probable primality tests can be combined to create a very quick algorithm for proving primality for integers less than 340,000,000,000,000. Chapter Three: The classical tests A quick look at the list of largest known primes shows numbers with hundreds of thousands (even millions) of digits--and these are all proven primes (not probable primes)! So how can we know they are prime? Look at a portion of this list and decide what the numbers all have in common. ``` 111 1892^34233-1 10308 Z 89 112 152^34224+1 10304 D 93 113 (5452545+10^5153)10^5147+1 10301 D 90 Palindrome 114 23801#+1 10273 C 93 primorial plus one 115 632^34074+1 10260 Y 95 116 2138192^33869+1 10201 Y 93 ``` They are all trivial to factor if we either add, or subtract, one! This is no accident. It is possible to turn the probable-primality tests of chapter two for an integer n into primality proofs, if we know enough factors of either n+1 and/or n-1. These proofs are called the classical tests and we survey them in our third chapter. These tests have been used for over 99.99% of the largest known primes. They include special cases such as the Lucas-Lehmer test for Mersenne primes and Pepin's Test for Fermat primes. Chapter Four: The General Purpose Tests Finally, the obvious problem with the classical tests is that they depend on factorization--and it appears factoring is much harder than primality proving for the "average" integer. In fact this is the key assumption behind the popular RSA encryption method! Using complicated modern techniques, the classical tests have been improved into tests for general numbers that require no factoring such as the APR, APRT-CL and the ECPP algorithms. In chapter four we say a few words about these methods, discuss which of these test to use (classical, general purpose...), and then leave you with a few references with which to pursue these tests. In 2002 a long standing question was answered: can integers be proven prime in "polynomial time" (that is, with time bounded by a polynomial evaluated at the number of digits). Some of the previous algorithms come close (ECPP is almost always polynomial, and is conjectured to always be polynomial bounded). Agrawal, Kayal and Saxena answered this question in the affirmative by giving a "simple" polynomial time algorithm. We present this algorithm in chapter four. \| \| \| \| --- \| \| \| \| [ next page \| contents ] \| \| Copyright © 2023 PrimePages (Originally written by Chris Caldwell) \| \|
17084	https://math.stackexchange.com/questions/2904333/prove-equality-involving-binomial-coefficients	probability - Prove equality involving binomial coefficients - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Prove equality involving binomial coefficients Ask Question Asked 7 years ago Modified7 years ago Viewed 156 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I was solving a probability problem and I got a different answer than the one given in the book. Seems the authors were using a different way of counting/arguing. For the two answers to be equal, the following equality should hold true. b−1∑k=0(a+k−1 a−1)p a(1−p)k=a+b−1∑k=a(a+b−1 k)p k(1−p)a+b−k−1 How can this be proved? And the problem itself was: in a series of Bernoulli trials with probability for success in a single trial equal to p, what is the probability to get a successes before getting b failures? I think you guys will reverse engineer how I counted and how the authors counted. probability probability-theory Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 3, 2018 at 21:23 peter.petrovpeter.petrov asked Sep 3, 2018 at 21:06 peter.petrovpeter.petrov 13k 2 2 gold badges 23 23 silver badges 40 40 bronze badges 9 @callculus I think the address function works even if there are no comments from the person in question. Not sure about that...lulu –lulu 2018-09-03 21:13:47 +00:00 Commented Sep 3, 2018 at 21:13 @callculus it works! That is, I got notified of your comment.lulu –lulu 2018-09-03 21:19:02 +00:00 Commented Sep 3, 2018 at 21:19 @callculus Ah, true. I missed one. But, on reflection, I sometimes cite other users responses to other posts and I use the address system to refer to them so they know they are being quoted. That appears to work.lulu –lulu 2018-09-03 21:24:06 +00:00 Commented Sep 3, 2018 at 21:24 @lulu It sounds tricky.callculus42 –callculus42 2018-09-03 21:27:26 +00:00 Commented Sep 3, 2018 at 21:27 1 @lulu Hmm, interesting. OK, I'll also try a few corner cases. By the way my answer is on the left. No wait, it is not 6/8, it is 4/8 on the right in that corner case you described.peter.petrov –peter.petrov 2018-09-03 21:44:31 +00:00 Commented Sep 3, 2018 at 21:44 \|Show 4 more comments 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. This is a nice question! The short answer is: You've already proved it! By showing that the two sides are different counts of the same thing, you've proved that they're equal. On the assumption that you're looking for an algebraic proof, here's a longer answer. Since on the left you only consider as many trials as you need until the result is decided, whereas on the right the authors consider all a+b−1 trials until the result is decided no matter what it is, we can transform your sum into their sum by adding the missing irrelevant b−k−1 trials: b−1∑k=0(a+k−1 a−1)p a(1−p)k=b−1∑k=0(a+k−1 a−1)p a(1−p)k 1 b−k−1=b−1∑k=0(a+k−1 a−1)p a(1−p)k(p+(1−p))b−k−1=b−1∑k=0(a+k−1 a−1)p a(1−p)k b−k−1∑j=0(b−k−1 j)p j(1−p)b−k−1−j=b−1∑k=0 b−k−1∑j=0(a+k−1 a−1)p a(1−p)k(b−k−1 j)p j(1−p)b−k−1−j=b−1∑j=0 b−j−1∑k=0(a+k−1 a−1)p a(1−p)k(b−k−1 j)p j(1−p)b−k−1−j=b−1∑j=0 p a p j(1−p)b−1−j b−j−1∑k=0(a+k−1 a−1)(b−k−1 j)=b−1∑j=0 p a p j(1−p)b−1−j b−j−1∑k=0(b−k−1 b−j−1−k)(a+k−1 k)=b−1∑j=0 p a p j(1−p)b−1−j(a+b−1 b−j−1)=b−1∑j=0 p a p j(1−p)b−1−j(a+b−1 a+j)=a+b−1∑k=a(a+b−1 k)p k(1−p)a+b−k−1, where the inner sum is evaluated using the relationship n∑k=0(x+n−k−1 n−k)(y+k−1 k)=(x+y+n−1 n), of which several proofs are given at Proving that (x+y+n−1 n)=∑n k=0(x+n−k−1 n−k)(y+k−1 k). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 4, 2018 at 13:19 answered Sep 3, 2018 at 22:51 jorikijoriki 243k 15 15 gold badges 311 311 silver badges 548 548 bronze badges 3 1 Thank you for this detailed explanation peter.petrov –peter.petrov 2018-09-04 11:24:14 +00:00 Commented Sep 4, 2018 at 11:24 (+1) I hope you don't mind that I posted my answer, which is essentially the same as yours, but with each step annotated.robjohn –robjohn♦ 2018-09-04 14:38:46 +00:00 Commented Sep 4, 2018 at 14:38 @robjohn: No, sure, I upvoted it, the more the merrier :-)joriki –joriki 2018-09-04 14:52:45 +00:00 Commented Sep 4, 2018 at 14:52 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. I didn't post this because joriki posted first and his answer is essentially the same (+1). However, since some explanation of the steps might be useful, I will post mine to provide an explanation for his steps. b−1∑k=0(a+k−1 a−1)p a(1−p)k=b−1∑k=0(a+k−1 a−1)p a(1−p)k(p+(1−p))b−k−1=b−1∑k=0 b−k−1∑j=0(a+k−1 a−1)(b−k−1 j)p a(1−p)k p j(1−p)b−j−k−1=b−1∑j=0 b−j−1∑k=0(a+k−1 a−1)(b−k−1 j)p a+j(1−p)b−j−1=b−1∑j=0(a+b−1 a+j)p a+j(1−p)b−j−1=a+b−1∑j=a(a+b−1 j)p j(1−p)a+b−j−1 Explanation: (2): multiply by 1=(p+(1−p))b−k−1 (3): apply the Binomial Theorem to (p+(1−p))b−k−1 (4): switch order of summation and combine terms (5): sum in k using ∑b−c k=0(a+k a)(b−k c)=(a+b+1 a+c+1) (6): substitute j↦j−a Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Sep 4, 2018 at 14:36 robjohn♦robjohn 355k 38 38 gold badges 499 499 silver badges 892 892 bronze badges Add a comment\| This answer is useful 1 Save this answer. Show activity on this post. Starting from b−1∑k=0(a+k−1 a−1)p a(1−p)k=a+b−1∑k=a(a+b−1 k)p k(1−p)a+b−k−1 we simplify to b−1∑k=0(a+k−1 a−1)p a(1−p)k=b−1∑k=0(a+b−1 a+k)p a+k(1−p)b−k−1 or b−1∑k=0(a+k−1 a−1)(1−p)k=b−1∑k=0(a+b−1 a+k)p k(1−p)b−k−1. We get for the LHS ∑k≥0(a+k−1 a−1)(1−p)k=∑k≥0(a+k−1 a−1)(1−p)k[z b−1]z k 1−z=[z b−1]1 1−z∑k≥0(a+k−1 a−1)(1−p)k z k=[z b−1]1 1−z 1(1−(1−p)z)a. The RHS is b−1∑k=0 p k(1−p)b−k−1[z b−1−k]1(1−z)a+k+1=[z b−1]1(1−z)a+1 b−1∑k=0 p k(1−p)b−k−1 z k(1−z)k. There is no contribution to the coefficient extractor in front when k>b−1 and we may extend k to infinity, getting (1−p)b−1[z b−1]1(1−z)a+1∑k≥0 p k(1−p)−k z k(1−z)k=(1−p)b−1[z b−1]1(1−z)a+1 1 1−p z/(1−p)/(1−z)=(1−p)b−1[z b−1]1(1−z)a 1 1−z−p z/(1−p)=[z b−1]1(1−(1−p)z)a 1 1−(1−p)z−p z=[z b−1]1 1−z 1(1−(1−p)z)a. The LHS and the RHS are seen to be the same and we may conclude. Remark. The first one is the easy one and follows by inspection. The Iverson bracket may be of interest here as an example of the method. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 6, 2018 at 14:16 answered Sep 4, 2018 at 13:43 Marko RiedelMarko Riedel 64.9k 5 5 gold badges 67 67 silver badges 147 147 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability probability-theory See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 10Proving that (x+y+n−1 n)=∑n k=0(x+n−k−1 n−k)(y+k−1 k) Related 11Probability of n successes in a row at the k-th Bernoulli trial... geometric? 0Random variable related to binomial 2Interesting variant of binomial distribution. 1Probability of binomial n success before m failures? 3Bernoulli trials - proving that a sequence will occur infinitely many times 0There are 10 trials with a probability of success of .3 1Independent Bernoulli trials and the gambler's ruin 2A negative binomial distribution for two random variables? 0Comparison of different solutions to the problem of points Hot Network Questions The rule of necessitation seems utterly unreasonable How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? Passengers on a flight vote on the destination, "It's democracy!" Can a GeoTIFF have 2 separate NoData values? What is a "non-reversible filter"? Is it ok to place components "inside" the PCB Xubuntu 24.04 - Libreoffice Who is the target audience of Netanyahu's speech at the United Nations? Lingering odor presumably from bad chicken Is encrypting the login keyring necessary if you have full disk encryption? What meal can come next? What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left? How can the problem of a warlock with two spell slots be solved? Making sense of perturbation theory in many-body physics Why are LDS temple garments secret? Origin of Australian slang exclamation "struth" meaning greatly surprised Does the curvature engine's wake really last forever? What's the expectation around asking to be invited to invitation-only workshops? What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? My dissertation is wrong, but I already defended. How to remedy? I have a lot of PTO to take, which will make the deadline impossible How to home-make rubber feet stoppers for table legs? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
17085	https://www.youtube.com/watch?v=Hhm1TjLnb9o	How to Find Domain of a Function (Radicals, Rational Functions, and Interval Notation) Wrath of Math 290000 subscribers 20 likes Description 621 views Posted: 27 Jun 2023 We show how to find the domain of a function that contains radicals, fractions, and square roots, and even square roots in the denominator. We will use a number line and interval notation to describe the domain. This video contains plenty of examples and practice problems, try them out yourself before watching the solutions! Let me know what you want to see in future precalculus video tutorials! 0:00 Polynomials 1:14 Problem Functions 1:46 Overview of Examples 1:57 Functions with x in the Denominator 7:17 Functions with Square Roots 13:21 More Complicated Examples 19:26 Outro Precalculus playlist: Precalculus Exercises playlist: ◉Textbooks I Like◉ Graph Theory: Real Analysis: Abstract Algebra: Linear Algebra: Calculus: Proofs and Set Theory: (available for free online) Statistics: Discrete Math: Number Theory: ★DONATE★ ◆ Support Wrath of Math on Patreon for early access to new videos and other exclusive benefits: ◆ Donate on PayPal: Thanks to Loke Tan, Matt Venia, Micheline, Doug Walker, Odd Hultberg, Marc, Roslyn Goddard, Shlome Ashkenazi, Barbora Sharrock, Mohamad Nossier, Rolf Waefler, Shadow Master, and James Mead for their generous support on Patreon! Outro music is mine. You cannot find it anywhere, for now. Follow Wrath of Math on... ● Instagram: ● Facebook: ● Twitter: My Math Rap channel: 2 comments Transcript: Polynomials the domain of a function is the set of all X values that we can plug into the function it's where the function is defined for example what's the domain of 3x plus 1. this is a linear function it looks like this it's just a line that goes on forever as you might expect then the domain is all real numbers which in interval notation looks like this everything from negative Infinity to positive Infinity there's no x value you can plug into 3x plus 1 that causes a problem similarly for a quadratic function like this one 2x squared minus X plus 5 which looks like this you can plug any real number you want in here so the domain is everything from negative Infinity to positive Infinity more generally every polynomial function here's one with degree seven has a domain of all real numbers every polynomial function has a domain of all real numbers anything from negative Infinity to positive Infinity you can plug in and by the way a function is a polynomial if it's created by adding and or subtracting powers of X where those powers are non-negative integers like 0 1 2 3 4 Etc so finding the domain of Problem Functions polynomials is really easy it's just all real numbers the task becomes more difficult when we start to consider some problem functions these are functions that do have domain restrictions values of X that we can't plug into them for example Division and the square root if we've got a function with division we need the denominator to not equal zero and if we have a square root we need the input of the square root to be non-negative it needs to be greater than or equal to zero I'm going to show you Overview of Examples how to find the domains of these types of functions you can see on screen now all the examples we're going to go over I'll also show you how to write the domains in interval notation for a first Functions with x in the Denominator example consider f of x equals 2x to the negative 1. what's the domain of this function it kind of looks like a polynomial so you might be tempted to say it's all real numbers but as we can see from the graph that's not true there's very clearly an issue here and that's because this is not a polynomial remember in a polynomial every Power of X has to be non-negative and it has to be an integer this power is negative one two x to the negative 1 is the same as 2 over X these are just two different ways for writing the same thing so clearly when you have a negative power what you're dealing with is division so we do have a domain restriction the Restriction is that X can't equal zero we cannot divide by zero so if we want to draw our domain on a number line we have an open circle at zero because the domain doesn't include zero but anything else is fair game the division by zero is the only potential issue here to write that in interval notation we have an interval going from negative Infinity to zero we have parentheses on both ends because we're not including negative Infinity since it's not a number and we're not including 0 because we don't want a division by zero and then we Union this that's a big U with the interval that goes from zero to positive Infinity this is just all real numbers except zero next example G of x equals four over x minus five notice we're calling this function G you can call your functions whatever you want f and g are both common names if we look at this function what are the potential problems well we have Division and we can't divide by zero so whatever we're dividing by must not equal zero x minus five must not equal zero and that means that X must not equal five you can just add five to both sides and get that that's what we see in the graph there's this asymptote at x equals five if we want to represent that on our number line we can say 5 is maybe right about there and we need an open circle because 5 is not included in the domain everything else though is perfectly fine so we'll shade in the rest of the line in interval notation that looks just like our previous domain where we had to cut out zero the only difference of course is that now we have to cut out five so we'll go from negative Infinity to five Union that with the interval from 5 to Infinity these are called open intervals by the way because they're not including the endpoints next h of x equals two X plus one divided by x squared minus eight X plus 15. there's no issues in the numerator that's just a linear function you could plug anything you want in there but we have a denominator we have division we can't divide by zero so that denominator x squared minus ax plus 15 must not equal zero now we treat this just like solving an equation in order to find what values of x we need to prohibit so we'll Factor the left side of this equation we need two numbers that multiply to 15 and add to negative eight three and five multiply to 15 but they add to positive eight but if we make them negative that'll work because negative 3 times negative 5 is positive 15 and they add to negative 8. so we can Factor the left side as x minus 3 times x minus five and we know that this must not equal zero but the zero product property then that means that X must not equal 3 because if it did then this would be zero and also X must not equal 5 because if it did then that would be zero thus we have two numbers we need to exclude from the domain which you could see from the graph going to our number line three is right there so we'll put an open circle there and 5 is there everything else is totally fine so we'll draw an arrow going off to positive Infinity an arrow going off to negative infinity and connect the space in between those gaps in interval notation this is similar to our previous examples but slightly more complicated we need to go from negative Infinity to 3 and we also need to go from 5 to positive Infinity but then we also need to get the space in between which is from 3 to 5. next example s of x equals 3x plus 1 divided by x squared plus nine again the only problem function here is the division we can't divide by zero x squared plus nine must not equal zero and that means that x squared must not equal negative nine however x squared is never equal to negative nine if you square zero you get zero if you square anything else you get a positive number you cannot Square something and get a negative number so actually any value of x here is fine which the graph suggests we can see there's no gaps there's no asymptotes the domain here is actually all real numbers on the number line that would just mean that we're shading the entire thing in interval notation again this is just everything from negative Infinity to positive Infinity next f of x equals Functions with Square Roots x to the one-half what's the domain of this there's no division to worry about kind of looks like a polynomial so maybe the domain is all reals but it's not a polynomial remember for a polynomial the powers of X need to be non-negative integers one-half is not an integer it's a fraction it's not whole this is actually the same as the square root of x these are just two ways of writing the same thing and we know you cannot take the square root of a negative number the only restriction here is that X has to be non-negative so for our domain we can include zero zero is totally allowed so we have a solid dot there and then everything in positive direction we can have as well we just can't have negatives what that looks like in interval notation is a bracket with zero because again we can include zero we can plug zero in this and we go up to positive Infinity you can see that from the graph x starts at zero and just goes off to positive Infinity next example f of x equals the square root of x minus two very similar idea we cannot take the square root of a negative number so whatever we're taking the square root of in this case it's x minus 2 has to be non-negative has to be greater than or equal to zero adding two to both sides of that inequality tells us that X has to be greater than or equal to two so we'll put a solid dot at 2 because 2 is allowed and then we'll shade everything towards positive Infinity The Domain here is a closed bracket at 2 because we can include two and then go up to positive Infinity again that's reflected in the graph next h of T equals the square root of t squared plus 10 T minus 24. just like we can change the name of our function we can also change the independent variable X is the most common choice but T is another common Choice doesn't really make a big difference again we're taking a square root here so what we take the square root of which in this case is t squared plus 10 T minus 24 must be greater than or equal to zero we then want to solve for t to see what restrictions are placed on T and we can do that by factoring the left side of the inequality we need two numbers that multiply to negative 24 and add to 10. 2 and 12 multiply to 24 2 and negative 12 multiply to negative 24 however they add to negative 10. we need positive 12 and negative 2. those add to 10 and multiply to negative 24. so we can fact after the left side as t plus 12 and T minus two that's the left side and it must be greater than or equal to zero now this is a little tricky it's not as simple as just using the zero product property we need this thing to be greater than or equal to zero we know that it equals zero based on our factorization it equals zero at T equals negative twelve and at T equals positive two so if this is zero we know our function hits zero at negative twelve and then it hits it again at two but we don't know what it's doing besides that we got to figure out where is it positive and where is it negative what parts do we need to exclude because wherever it's negative well we cannot have that for this type of situation a sine chart can be really useful we just draw a line and put our two points on there where the function hits zero negative 12 and 2 in this case it doesn't matter how they're spaced yeah we really just want to focus on the fact that there are these three intervals everything to the left of negative 12 everything to the right of positive 2 and everything in between we need to see where is the function positive and where is it negative to do that just take a point from each interval and plug it into the domain and see what you get for example 0 is between negative 12 and 2. so let's take that point and plug it in to the function if we plug 0 into the function it's clear that we will just get Negative 24 which means all the values between negative 12 and 2 are negative now what about this interval to the right of 2 we could plug in 3 for example if we plug 3 in we get 9 plus 30 minus 24 and that's positive 30 minus 24 is 6 Plus 9 is 15. so that means to the right of 2 the function is positive now to the left of negative 12 we could choose negative 20 for example that's a multiple of 10 so it will work nicely and it will probably be easier if we plug it into the factored form of the function which works just as well if we plug negative 20 into this we get negative 20 plus 12 which is negative 8 times negative 20 minus 2 which is negative 22. it's two negatives they multiply to a positive so all the values of our function to the left of negative 12 have to be positive so by using sample points from each interval we're able to fill out our sign chart to see where our function is positive and negative in this case we see it's negative between negative 12 and positive 2. that's the interval we need to restrict from our domain so what is our domain well it's everything except for that part which means that we're going from negative Infinity to negative 12 and we can include negative 12 because that makes the input Z 0 and the square root of 0 is 0 that's no problem and then we Union that with everything from positive two again we're including positive 2 because the square root of 0 is just zero and that goes up to positive Infinity notice we've completely excluded everything between negative 12 and 2. on our number line it looks like that next X of T equals three More Complicated Examples t plus four divided by the square root of t plus two notice our Strange naming here instead of naming the function f we've named it X and instead of our independent variable being X it's t this could be like the horizontal position of something it's X position as a function of time just for an example regarding the domain this has two potential problems not only do we have Division and so the square root of t plus 2 must not equal zero but also we have the square root the thing that we're taking the square root of t plus 2 must be greater than or equal to zero focusing on the first restriction the square root of t plus two is not equal to zero we can square both sides to find that t plus 2 must not equal zero and that means that t must not equal negative two so that's our first restriction the Restriction from the square root is that t plus 2 must be at least zero which means that t must be at least negative two we then need to combine the two restrictions to make sure that we're satisfying everything T can't equal negative two and T needs to be at least negative two in total that means that t has to be greater than negative two on the number line that looks like this an open circle at negative two because we can't include it but anything bigger than that we can include so we just shade to the right the domain is a parenthesis at negative two and we're going up to positive Infinity were back to normal names for this next example f of x equals the square root of x minus 9 divided by x squared minus 16. in this case we have a square root in the numerator so that forces the Restriction that x minus 9 must be at least zero we can't take the square root of a negative quickly dealing with this that means that X has to be at least nine on the other hand we also have division we have division by x squared minus 16. that means x squared minus 16 must not equal zero which means adding 16 to both sides the x squared must not equal 16 then taking the square root and remembering it could be positive or negative means that X can't equal positive or negative 4. remember that's because not only does positive 4 squared equal 16 but also negative 4 squared equals 16. we then have to combine these two restrictions X is at least 9 and X is not equal to Plus or negative 4. in this case the first restriction actually already forces the second restriction so our total amount of restrictions is just that X has to be at least nine if x is at least nine it's certainly not plus or minus four on our number line that looks like this a solid dot at nine because we can include it and then everything greater than that we can also include in interval notation we have a bracket for nine because we can include it and we're going up to positive Infinity finally our last example G of a equals the square root of a plus 2 divided by the square root of a squared minus nine we've got a square root in the numerator which forces the Restriction a plus two has to be at least zero because we can't take the square root of a negative and that means that a has to be at least negative two then we're dividing by the square root of a squared minus nine so that thing we're dividing by the square root of a squared minus 9 must not equal zero squaring both sides that means that a squared minus 9 must not equal zero and then adding 9 to both sides we have the a squared must not equal nine taking the square root and remembering the plus or minus this means that a must not equal plus or minus 3 because both positive three and negative 3 Square to give positive 9. but then there is one more restriction because remember this thing in the denominator is a square root so the thing inside of it has to be non-negative a squared minus 9 has to be greater than or equal to zero which means that a squared has to be greater than or equal to nine and this leads to the somewhat strange restriction that a needs to be greater than or equal to 3 because if you square 3 or something great greater you'll get something that's at least nine or a needs to be less than or equal to negative three because if you square negative 3 or anything less you will again get something that is 9 or bigger for example 5 squared is 25 that is at least nine negative 7 squared is positive 49 which is at least nine now we need to add up all the domain restrictions to see what the final domain is that satisfies all requirements we figured out that it's possible a is less than or equal to negative three but one of the other restrictions was that a has to be at least negative two so that completely eliminates the less than or equal to negative three that can't be allowed because we also need to satisfy that but then this restriction that a is at least negative 2 doesn't really matter either because we also have that a has to be at least positive three which is a stronger requirement than a being at least negative two we then also also have that a can't be positive or negative three of course it can't be negative 3 because it has to be at least three the fact that it can't equal positive three though is an additional restriction our final restriction is that a is greater than three on the number line of course that's an open circle at three with a line going off to positive Infinity the domain is a parenthesis at three because we're not including three going off to Outro positive infinity and now you know how to find the domains of polynomials radical functions and rational functions those are the division ones remember when you have division you cannot divide by zero whatever you're dividing by it can't equal zero whatever you take the square root of that thing needs to be non-negative you need to take these restrictions figure out how they affect your independent variable and if you have multiple requirements by the end of this process make sure you carefully add them all up to see what your final domain is which satisfies all restrictions
17086	https://stackoverflow.com/questions/7365562/de-bruijn-like-sequence-for-2n-1-how-is-it-constructed	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog De Bruijn-like sequence for 2^n - 1: how is it constructed? Ask Question Asked Modified 3 years, 5 months ago Viewed 6k times This question shows research effort; it is useful and clear 34 Save this question. Show activity on this post. I'm looking at the entry Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup from Bit Twiddling hacks. I can easily see how the second algorithm in that entry works ``` static const int MultiplyDeBruijnBitPosition2 = { 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9 }; r = MultiplyDeBruijnBitPosition2[(uint32_t)(v 0x077CB531U) >> 27]; ``` which calculates n = log2 v where v is known to be a power of 2. In this case 0x077CB531 is an ordinary De Bruijn sequence, and the rest is obvious. However, the first algorithm in that entry ``` static const int MultiplyDeBruijnBitPosition = { 0, 9, 1, 10, 13, 21, 2, 29, 11, 14, 16, 18, 22, 25, 3, 30, 8, 12, 20, 28, 15, 17, 24, 7, 19, 27, 23, 6, 26, 5, 4, 31 }; v \|= v >> 1; v \|= v >> 2; v \|= v >> 4; v \|= v >> 8; v \|= v >> 16; r = MultiplyDeBruijnBitPosition[(uint32_t)(v 0x07C4ACDDU) >> 27]; ``` looks a bit more tricky to me. We begin by snapping v to a nearest greater 2^n - 1 value. This 2^n - 1 value is then multiplied by 0x07C4ACDD, which in this case acts the same way as the DeBruijn sequence in the previous algorithm did. My question is: how do we construct this magic 0x07C4ACDD sequence? I.e. how do we construct a sequence that can be used to generate unique indices when multiplied by a 2^n - 1 value? For 2^n multiplier it is just an ordinary De Bruijn sequence, as we can see above, so it is clear where 0x077CB531 came from. But what about 2^n - 1 multiplier 0x07C4ACDD? I feel like I'm missing something obvious here. P.S. To clarify my question: I'm not really looking for an algorithm to generate these sequences. I'm more interested in some more-or-less trivial property (if one exists) that makes 0x07C4ACDD work as we want it to work. For 0x077CB531 the property that makes it work is pretty obvious: it contains all 5-bit combinations "stored" in the sequence with 1-bit stepping (which is basically what De Bruijn sequence is). The 0x07C4ACDD, on the other hand, is not a De Bruijn sequence by itself. So, what property were they aiming for when constructing 0x07C4ACDD (besides the non-constructive "it should make the above algorithm work")? Someone did come up with the above algorithm somehow. So they probably knew that the approach is viable, and that the appropriate sequence exists. How did they know that? For example, if I were to construct the algorithm for an arbitrary v, I'd do ``` v \|= v >> 1; v \|= v >> 2; ... ``` first. Then I'd just do ++v to turn v into a power of 2 (let's assume it doesn't overflow). Then I'd apply the first algorithm. And finally I'd do --r to obtain the final answer. However, these people managed to optimize it: they eliminated the leading ++v and the trailing --r steps simply by changing the multiplier and rearranging the table. How did they know it was possible? What is the math behind this optimization? algorithm bit-manipulation discrete-mathematics logarithm Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications edited Nov 22, 2017 at 19:56 Alex Fainshtein 1,15311 gold badge99 silver badges1717 bronze badges asked Sep 9, 2011 at 17:57 AnT stands with RussiaAnT stands with Russia 322k4444 gold badges547547 silver badges790790 bronze badges 8 to compute log 2, why not just repeat right bit shift until you reach zero and count how many shifts had you done? This had always been a usual practice! – Tomas Commented Sep 9, 2011 at 19:12 1 @Tomas T: Well, shifting is a viable approach. But in some cases this can be faster. Moreover, this is pretty elegant. The first one at least. Maybe the second one is too, if I knew how to make that constant. – AnT stands with Russia Commented Sep 9, 2011 at 19:30 Since 0x00010000 and 0x0000FFFF are different, they need different DeBruijn sequences. How does one generate 0x077CB531U? You'll need to use an equivalent theory to find the other number. Unless you can solve a2^n=(2^n-1)b, a=0x077CB531U, b=0x07C4ACDD for all n Another aproach may have been to solve: l2[(v+1)a] l2[va+a] l1[vb]+t l1[vb] – nulvinge Commented Sep 9, 2011 at 22:00 1 @nulvinge: Well, I don't immediately see why the 0x07C4ACDD sequence should be DeBruijn at all. Why? In the first case it is obvious: by multiplying by v we are simply shifting the sequence, so all we need is a sequence that can represent all 5-bit numbers in a 32-bit word. This is obviously classic DeBruijn. But in the second case the multiplication by v can be seen as a shift followed by a subtraction. So, I'd say that 0x07C4ACDD should be derived from DeBruijn, but is not DeBruijn by itself. If fact, it isn't if you look at it. – AnT stands with Russia Commented Sep 9, 2011 at 22:14 1 Very nice question and very interesting topic I never heared about it before. – Saeed Amiri Commented Sep 10, 2011 at 5:57 \| Show 3 more comments 4 Answers 4 Reset to default This answer is useful 18 Save this answer. Show activity on this post. A De Bruijn sequence of order n over k symbols (and of k^n length) have a property that every possible n-length word appears as consecutive characters in it, some of them with cyclic wrapping. For example, in the case of k=2, n=2, the possible words are 00, 01, 10, 11, and a De Bruijn sequence is 0011. 00, 01, 11 appears in it, 10 with wrapping. This property naturally means that left shifting a De Bruijn sequence (multiplying with power of two) and taking its upper n bits results in a unique number for each power of two multiplier. Then you only need a lookup table to determine which one it is. It works on a similar principle to numbers which are one less than power of two, but the magic number in this case is not a De Bruijn sequence, but an analogue. The defining property simply changes to "every possible n-length word appears as the sum of the first m subsequences of length n, mod 2^n". This property is all that is needed for the algorithm to work. They simply used this different class of magic numbers to speed up the algorithm. I did as well. One possible method of construction of De Bruijn numbers is the generation of a Hamiltonian path of the De Bruijn graph, Wikipedia provides an example of such a graph. In this case, the nodes are 2^5=32-bit integers, the directed edges are transitions between them, where a transition is a left shift and a binary or operation according to the label of the edge, 0 or 1. There might be a direct analogue to 2^n-1 type magic numbers, it might be worth exploring, but this isn't a way people usually construct such algorithms. In practice you might try to construct it differently, especially if you want it to behave in a tad different manner. For example, the implementation of leading/trailing number of zeros algorithms on the bit twiddling hacks page can only return values in [0..31]. It needs additional checking for the case of 0, which has 32 zeros. This check requires a branching and can be way too slow on some CPUs. The way I did it, I used a 64 element lookup table instead of 32, generated random magic numbers, and for each of them I built up a lookup table with power of two inputs, checked its correctness (injectivity), then verified it for all 32-bit numbers. I continued till I encountered a correct magic number. The resulting numbers do not fulfill a property of "every possible n-length word appears", since only 33 numbers appear, which are unique for all 33 possible input. Exhaustive brute force search sounds slow, especially if good magic numbers are rare, but if we first test known power of two values as inputs, the table is filled quickly, rejection is fast, and the rejection rate is very high. We only need to clear the table after each magic number. In essence I exploited a high rejection rate algorithm to construct magic numbers. The resulting algorithms are ``` int32 Integer::numberOfLeadingZeros (int32 x) { static int32 v = { 32, -1, 1, 19, -1, -1, -1, 27, -1, 24, 3, -1, 29, -1, 9, -1, 12, 7, -1, 20, -1, -1, 4, 30, 10, -1, 21, -1, 5, 31, -1, -1, -1, -1, 0, 18, 17, 16, -1, -1, 15, -1, -1, -1, 26, -1, 14, -1, 23, -1, 2, -1, -1, 28, 25, -1, -1, 13, 8, -1, -1, 11, 22, 6}; x \|= x >> 1; x \|= x >> 2; x \|= x >> 4; x \|= x >> 8; x \|= x >> 16; x = 0x749c0b5d; return v[cast(x) >> 26]; } int32 Integer::numberOfTrailingZeros (int32 x) { static int32 v = { 32, -1, 2, -1, 3, -1, -1, -1, -1, 4, -1, 17, 13, -1, -1, 7, 0, -1, -1, 5, -1, -1, 27, 18, 29, 14, 24, -1, -1, 20, 8, -1, 31, 1, -1, -1, -1, 16, 12, 6, -1, -1, -1, 26, 28, 23, 19, -1, 30, -1, 15, 11, -1, 25, 22, -1, -1, 10, -1, 21, 9, -1, -1, -1}; x &= -x; x = 0x4279976b; return v[cast(x) >> 26]; } ``` As for your question of how did they know, they probably didn't. They experimented, tried to change things, just like me. After all, it isn't a big stretch of imagination that 2^n-1 inputs might work instead of 2^n inputs with different magic number and lookup table. Here, I made a simplified version of my magic number generator code. It checks all possible magic numbers in 5 minutes if we check only for power of two inputs, finding 1024 magic numbers. Checking against other inputs are pointless since they are reduced to 2^n-1 form anyway. Does not construct the table, but it is trivial once you know the magic number. ``` include include using namespace Frigo::Lang; using namespace std; class MagicNumberGenerator { public: static const int32 log2n = 5; static const int32 n = 1 << log2n; static const bool tryZero = false; MagicNumberGenerator () {} void tryAllMagic () { for( int32 magic = 0; magic < Integer::MAX_VALUE; magic++ ){ tryMagic(magic); } tryMagic(Integer::MAX_VALUE); for( int32 magic = Integer::MIN_VALUE; magic < 0; magic++ ){ tryMagic(magic); } } bool tryMagic (int32 magic) { // clear table for( int32 i = 0; i < n; i++ ){ table[i] = -1; } // try for zero if( tryZero and not tryInput(magic, 0) ){ return false; } // try for all power of two inputs, filling table quickly in the process for( int32 i = 0; i < 32; i++ ){ if( not tryInput(magic, 1 << i) ){ return false; } } // here we would test all possible 32-bit inputs except zero, but it is pointless due to the reduction to 2^n-1 form // we found a magic number cout << "Magic number found: 0x" << Integer::toHexString(magic) << endl; return true; } bool tryInput (int32 magic, int32 x) { // calculate good answer int32 leadingZeros = goodNumberOfLeadingZeros(x); // calculate scrambled but hopefully injective answer x \|= x >> 1; x \|= x >> 2; x \|= x >> 4; x \|= x >> 8; x \|= x >> 16; x = magic; x = Integer::unsignedRightShift(x, 32 - log2n); // reject if answer is not injective if( table[x] != -1 ){ return table[x] == leadingZeros; } // store result for further injectivity checks table[x] = leadingZeros; return true; } static int32 goodNumberOfLeadingZeros (int32 x) { int32 r = 32; if( cast<uint32>(x) & 0xffff0000 ){ x >>= 16; r -= 16; } if( x & 0xff00 ){ x >>= 8; r -= 8; } if( x & 0xf0 ){ x >>= 4; r -= 4; } if( x & 0xc ){ x >>= 2; r -= 2; } if( x & 0x2 ){ x >>= 1; r--; } if( x & 0x1 ){ r--; } return r; } int32 table[n]; }; int32 main (int32 argc, char argv[]) { if(argc\|\|argv){} measure{ MagicNumberGenerator gen; gen.tryAllMagic(); } } ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Sep 12, 2011 at 2:24 community wiki 15 revsFrigo 7 tl;dr: The magic number produces a unique number when multiplied by 2^m-1 and shifted right by 2^n-n, for all m in [1..2^n], n=5 in this case. In other words, it produces a unique number when summing its first m subsequence of length n, mod 2^n. 0x07C4ACDD is the smallest such magic number for n=5. Generation is straightforward with random magic number candidates and injectivity (i.e. unique) checking. – Frigo Commented Sep 12, 2011 at 1:17 It's a wrong criterion that the magic number produces a unique number when you sums up its first subsequences of length n. Let me give you an example. 0x87c4acdd works as the magic for n = 5. For m = 1, 2, ...32, it produces the following sequence after the multiplication and shift: {16, 18, 22, 30, 14, 13, 11, 7, 0,...}. This magic's first 5-bit subsequences are: {16, 1, 3, 7, 15, 31, 30, 28, 24,...}. Now try to sum up these subsequences by mod 32, you will get: {16, 17, 20, 27, 10, 9, 7, 3, 27,...}. Oops... the fourth and the ninth numbers are the same. – facetus Commented Oct 24, 2014 at 0:43 You misunderstood the process, the multiplication by 2^m-1 (and right shift) IS the summing of the subsequences. – Frigo Commented Oct 27, 2014 at 19:45 It IS NOT. You don't take into account overflow in bits after the subsequences. Please, consider the example. It is obvious from it that the sums of the subsequences ARE NOT unique. – facetus Commented Oct 29, 2014 at 13:13 "how did they know, they probably didn't" -- actually, an "ordinary" de Bruijn sequence can be generated very efficiently (linear time) by finding any Eulerian cycle in the (n-1)-dimensional de Bruijn graph, rather than a Hamiltonian path in the n-dimension graph. (They mention this on the Wikipedia page now, though maybe not when you wrote this.) Using this, you could, e.g., find a magic constant for 64-bit (or 1024-bit, etc.) integers in less than 1ms, while it could take centuries to find via exhaustive testing. Please mention this at the top, and I'll +1. – j_random_hacker Commented Nov 26, 2017 at 15:22 \| Show 2 more comments This answer is useful 3 Save this answer. Show activity on this post. It's based on the paper Using de Bruijn Sequences to Index a 1 in a Computer Word. I'm going to guess that they did a search for a perfect hash function to map 2^n-1 to [0..31]. They describe a method for searching for counting zeroes of integers with up to two bits set which involves incrementally building up the multiplier. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Sep 10, 2011 at 2:18 MSNMSN 54.8k77 gold badges7878 silver badges108108 bronze badges 3 Constructing perfect hash functions are an even larger problem, and [0..2^n-1] -> [0..31] is not a perfect hash function. The problem is simply searching for an injective function that maps numbers with the same properties to the same bucket. Multiplication and an n-bit part of the answer is sufficient. – Frigo Commented Sep 10, 2011 at 2:57 @Frigo, I wasn't saying it maps [0..2^n-1], I'm saying it maps 2^n-1 for n= 0 to 31. – MSN Commented Sep 12, 2011 at 16:47 Ah sorry, misunderstood that. In that case yeah, that's pretty similar to what I did. Except I was more interested in making an injective hash function rather than a perfect one. Unused table entries are not a problem, the algorithm not accepting 0 as input is a big one. – Frigo Commented Sep 12, 2011 at 17:05 Add a comment \| This answer is useful 3 Save this answer. Show activity on this post. From: ``` 130329821 0x07C4ACDD 00000111110001001010110011011101B bit 31 - bit 27 00000 0 bit 30 - bit 26 00001 1 bit 29 - bit 25 00011 3 bit 28 - bit 24 00111 7 bit 27 - bit 23 01111 15 bit 26 - bit 22 11111 31 bit 25 - bit 21 11110 30 bit 24 - bit 20 11100 28 bit 23 - bit 19 11000 24 bit 22 - bit 18 10001 17 bit 21 - bit 17 00010 2 bit 20 - bit 16 00100 4 bit 19 - bit 15 01001 9 bit 18 - bit 14 10010 18 bit 17 - bit 13 00101 5 bit 16 - bit 12 01010 10 bit 15 - bit 11 10101 21 bit 14 - bit 10 01011 11 bit 13 - bit 9 10110 22 bit 12 - bit 8 01100 12 bit 11 - bit 7 11001 25 bit 10 - bit 6 10011 19 bit 9 - bit 5 00110 6 bit 8 - bit 4 01101 13 bit 7 - bit 3 11011 27 bit 6 - bit 2 10111 23 bit 5 - bit 1 01110 14 bit 4 - bit 0 11101 29 bit 3 - bit 31 11010 26 bit 2 - bit 30 10100 20 bit 1 - bit 29 01000 8 bit 0 - bit 28 10000 16 ``` It seems to me that 0x07C4ACDD is a 5-bit de Bruijn sequence. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Nov 9, 2015 at 20:55 answered Nov 9, 2015 at 20:46 Rudy VelthuisRudy Velthuis 28.9k55 gold badges5050 silver badges9797 bronze badges Add a comment \| This answer is useful 1 Save this answer. Show activity on this post. The answer to your question is perhaps a bit simpler than expected. First, both 0x077CB531 and 0x07C4ACDD are de Bruijn sequences. Let's call them B and C. Now let's look at finding the top bit of n - 1 where n is a power of 2 (thus non-zero). Note that: We calculate (n - 1) C which is the same as n C - C. We only look at the top 5 bits of n C - C. n C is a left-shifted version of C. The top 5 bits of C are zero: subtracting C might decrement the top 5 bits of n C by 1 if C is larger than the bottom 27 bits of n C. Here's what's special about C: it was chosen such that it is always larger than those bottom 27 bits. C was chosen such that after its top 5 bits which are zero, the next 5 bits are all 1s. Every k-bit de Bruijn sequence has exactly one set of k consecutive 0s and one set of k consecutive 1s and none longer; however, these two subsequences are not adjacent in all de Bruijn sequences. They are adjacent in C. Assuming n is non-zero, the rotated n C never has 5 consecutive 1s in those next 5 bits: it always has at least a zero there, so the subtraction by C always results in decrementing the top 5 bits of n C. In other words, using n - 1 instead of n doesn't really change anything, except rotate the lookup table by one entry. Generating such de Bruijn sequences One approach is to simply apply the above constraint to any method of generating de Bruijn sequences. Here's a simple example using LFSRs. While LFSRs generate de Bruijn sequences of essentially any bit length, they only find a few with the above constraint, so this is illustrative only. Linear feedback shift registers (LFSRs) behave much like de Bruijn sequences: a maximal-period n-bit LFSR produces a periodic sequence of (2^n)-1 bits, for which the last n bits at any point cycles through all n-bit numbers (except one, typically zero). This zero is trivially added back by adding a zero bit in the LFSR output sequence at the point where it outputs n-1 consecutive zeroes (which it must do, covering all n-bit numbers save zero). Here's example C code to find all maximal-period Galois LFSRs of a given range of bit widths, and display the corresponding de Bruijn sequence by adding the missing zero. It turns out trivial to do by starting the LFSR with the top bit set when doing right shifts (starting with the value 1 ends up with zeroes at the end instead of at the beginning). It also displays FOUND if the sequence meets the constraints stated earlier. ``` include int lfsr_period(int width, int taps, int show) { int max = 1 << (width - 1), n = max; int period = 0, lastbit = 0, adjacent = 1; do { / Compute LFSR / int bit = n & 1; n >>= 1; if (bit) n ^= taps; period++; if (show) printf("%d", bit); if (lastbit && !bit && period < width 2) adjacent = 0; lastbit = bit; } while (n != max); if (show && adjacent) printf(" FOUND"); return period; } int main() { for (int width = 2; width <= 12; width++) { printf("%d bits:\n", width); int max = 1 << width; for (int taps = max / 2; taps < max; taps++) { int period = lfsr_period(width, taps, 0); if (period == max - 1) { printf("0x%X: 0", taps); lfsr_period(width, taps, 1); printf("\n"); } } } return 0; } ``` For example, it finds the following sequences: ``` 2 bits: 0x3: 0011 FOUND 3 bits: 0x5: 00011101 FOUND 0x6: 00010111 4 bits: 0x9: 0000111101011001 FOUND 0xC: 0000100110101111 5 bits: 0x12: 00000101011101100011111001101001 0x14: 00000100101100111110001101110101 0x17: 00000110010011111011100010101101 0x1B: 00000110101001000101111101100111 0x1D: 00000111001101111101000100101011 0x1E: 00000101101010001110111110010011 6 bits: 0x21: 0000001111110101011001101110110100100111000101111001010001100001 FOUND 0x2D: 0000001110000100100011011001011010111011110011000101010011111101 0x30: 0000001000011000101001111010001110010010110111011001101010111111 0x33: 0000001101110011000111010111111011010001000010110010101001001111 0x36: 0000001011111100101010001100111101110101101001101100010010000111 0x39: 0000001111001001010100110100001000101101111110101110001100111011 7 bits: 0x41: 00000001111111010101001100111011101001011000110111101101011011001001000111000010111110010101110011010001001111000101000011000001 FOUND 0x44: 00000001001001101001111011100001111111000111011000101001011111010101000010110111100111001010110011000001101101011101000110010001 [...] ``` Looking carefully, you might notice it provides symmetrical (reversed) pairs of sequences. Also, none match the other de Bruijn sequences already mentioned, whether shifted, inverted or reversed: while LFSRs find several de Bruijn sequences, they certainly do not find all of them. They also don't find constrained sequences for 5 bits and very few above 7 bits; the ones it does find seem to always consist of two taps, at the top and bottom bits. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Mar 16, 2022 at 11:57 answered Mar 16, 2022 at 9:27 uptimeuptime 8355 bronze badges Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm bit-manipulation discrete-mathematics logarithm See similar questions with these tags. The Overflow Blog Robots in the skies (and they use Transformer models) Research roadmap update, August 2025 Featured on Meta Policy: Generative AI (e.g., ChatGPT) is banned Updated design for the new live activity panel experiment Further Experimentation with Comment Reputation Requirements Community activity Last 1 hr Users online activity 7461 users online 19 questions 12 answers 75 comments 164 upvotes Popular tags pythonc#c++performancemultiple-inheritancejavascript Popular unanswered question Nuxt 3: Meta Tags Not Updating in "View Page Source" but Visible in DevTools vuejs3nuxt3.jsvue-composition-apii18nextvue-i18n Sheharzad Salahuddin 23 Linked 47 Find most significant bit (left-most) that is set in a bit array Trying to compute Armstrong numbers in C: program prints 000..... and hangs 1 Number of byte of my int How to generate the "count trailing zeroes" table composed of a de Bruijn sequence? Hot Network Questions Issue with \integral command from intexgral package in math mode Find an explicit real number that is not in the set produced by the convergent rational series Why do we introduce the continuous functional calculus for self-adjoint operators? Non-committing? Have we been using deniable authenticated encryption all along? Spectral sequences every mathematician should know Having trouble identifying a font Is kernel memory mapped once or repeatedly for each spawned process How did Louis XIV raise an army half the size of Aurangzeb's with only a tenth of the revenue? How NOT to get hyper-inflation in a vassal-state that is printing/minting money at exorbitant speed? How is the term "average" used in classical physics? Can I use super glue or hot glue as an insulator/solder mask for a small circuit? Film about alien spaceships ominously hovering over Earth, only to form a shield to protect it Inconsistent volume numbers of old journal article in contemporary and modern citations? Incoming water pipe has no apparent ground Are trills only for high-pitched notes? Gift of showing mercy Do strong winds also cause a hovering creature to fall? Resonance in hexa-2,4-dienyl cation Is it possible to identify this residual pattern as heteroscedastic or homoscedastic? Using adverbs of frequency in the Future Simple and Present Continuous tenses How are the word-searches made for mass-produced puzzle books? How can I get my security door to stay shut in place as to not obstruct the deadbolt? How are the crosswords made for mass-produced puzzle books? Why wrap fruitcake in grease-proof paper? Question feed By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
17087	https://www.bbc.co.uk/bitesize/articles/zqxgvwx	KS2 Converting time intervals Part of MathsTime Units of time The passing of time is a big part of our life. You're born on a particular day of a month, in a year. We celebrate birthdays every year we get older. At school, there are set times for registration, lunch, breaks and home time. There are lots of different units of time: seconds, minutes, hours, days, weeks, months and years. We use units of time to say when an event happens, or to measure time between events. Back to top Activity: Converting time intervals Complete this interactive activity to learn more about this topic and then put your knowledge to the test. Back to top Equivalent time units It's important to know the value of these different measures of time and be able to convert them to find equivalent (equal in value) time intervals. Here is a handy conversion guide: Every four years is a leap year. Instead of having 365 days, leap years have 366 days. Back to top Weeks and years Let’s use this table to work out how many weeks there are in 3 years. First, look in the 'Years' column to find 3 years. Then, read across to see how many weeks are in the ‘Weeks’ column. It shows there are 154 weeks in 3 years. Back to top Converting time intervals When you know the equivalent time facts, you can use these to convert times. Bar models are very useful to help with conversions. Let's look at a question. How many minutes are there in (\frac{1}{3}) of an hour? There are 60 minutes in an hour, so let’s break 60 minutes into thirds using a bar model. You can now see that (\frac{1}{3}) of an hour is  20 minutes. Let's try another question. How many months are there in 3 years and 4 months? There are 12 months in one year. To find the number of months in 3 years, multiply 12 by 3. 12 × 3 = 36 Now add the remaining 4 months to 36 to find the total. 36 + 4 = 40 So, 3 years and 4 months is the same as 40 months. Back to top Example 1 Many time problems need you to convert the units so they are all the same. The Henderson family have lived in their house for 140 days. The Johnson family have lived in their house for 22 weeks. Which family has lived in their house for longer? To compare them, convert them to the same units, either days or weeks. ✓ There are 7 days in a week, so 70 days is 10 weeks and 140 days is 20 weeks. A table can help you with this. We know there are 7 days in a week. Look for numbers that multiply easily to get to the answer. Look back at the problem. The Henderson family have lived in their house for 20 weeks. The Johnson family have lived in their house for 22 weeks. So, the Johnson family have lived in their house for longer. Back to top Example 2 When you convert seconds to minutes, or minutes to hours, you can calculate with the number 60. How many seconds are there in 5 and a half minutes? ✓ 5 and a half minutes is the same as 330 seconds. There are 60 seconds in a minute. First work out 5 lots of 60 seconds: 60 × 5 = 300 Half of 60 is 30, so there are 30 seconds in half a minute. Now add 300 and 30 together to find the answer. 300 + 30 = 330 Back to top More on Time Find out more by working through a topic Telling the time to the nearest minute count 1 of 6
17088	https://www.lehman.edu/faculty/anchordoqui/lab2.pdf	Physics-131: Lab 2 Prof. Anchordoqui Ohm’s law I. OBJECTIVES • Become familiar with the use of a digital voltmeter and a digital ammeter to measure DC voltage and current. • Construct a series circuit using resistors, wires, and a breadboard. • Test the validity of Ohm’s law. II. BACKGROUND In the U.S., most of us use electricity every day. That electricity is handled in circuits: a closed loop of con-ductors travelling from power plants to neighborhoods to households and back again. That closed loop, with all of its many parts, forms one huge electrical circuit. Today we will use the three essential parts of a circuit: power supply (or battery), wires, and resistors. We will learn how resistors aﬀect the current of electrons that ﬂows through them, and how connecting resistors in diﬀerent ways changes their behavior. A. Key concepts One of the fundamental laws describing how electri-cal circuits behave is Ohm’s law. According to Ohm’s law, there is a linear relationship between the voltage drop across a circuit element and the current ﬂowing through it. Therefore the resistance is viewed as a con-stant independent of the voltage and the current. More explicitely, V = iR , (1) where V is the voltage applied across the circuit in volts (V), i is the current ﬂowing through the circuit in units of amperes (A), and R is the resistance of the circuit with units of ohms (Ω). (1) implies that, for a resistor with constant resistance, the current ﬂowing through it is proportional to the voltage across it. If the voltage is held constant, then the current is inversely proportional to the resistance. If the voltage polarity is reversed (that is, if the applied voltage is negative instead of positive), the same current ﬂows but in the opposite direction. If Ohm’s law is valid, it can be used to deﬁne resistance as: R = V i , (2) where R is a constant, independent of V and i. 2. Ohm’s Law (a) Series Figure 2.1: Schematics of circuits illustrating r and in parallel. Series Parall VS = V1 + V2 VP = V1 = IS = I1 = I2 IP = I1 + RS = R1 + R2 1 RP = 1 R1 + 1 R2 or Table 2.1: Equations for two resistors in Using these relationships, a complex circuit with a single resistor. You may wish to review equivalent resistance of circuits in your physics 2.5 In today’s lab Today we’ll become accustomed to some stan we’ll ﬁgure out how to use the circuit boards, re series and parallel circuits. Then we’ll learn h through a wire, as well as the voltage betwee Finally, we’ll verify that the equations that ar section are actually correct. 22 Last updated January 24, 20 FIG. 1: Schematics of circuits illustrating resistors connected in series. 2. Ohm’s Law (a) Series (b) Parallel Figure 2.1: Schematics of circuits illustrating resistors connected in series and in parallel. Series Parallel VS = V1 + V2 VP = V1 = V2 IS = I1 = I2 IP = I1 + I2 RS = R1 + R2 1 RP = 1 R1 + 1 R2 or RP = R1 R2 R1+R2 Table 2.1: Equations for two resistors in series and parallel. Using these relationships, a complex circuit can be redrawn as a circuit ith i l i t Y i h t i th f ﬁdi th FIG. 2: Schematics of circuits illustrating resistors connected in parallel. It is important to understand just what is meant by these quantities. The current (i) is a measure of how many electrons are ﬂowing past a given point during a set amount of time. The current ﬂows because of the elec-tric potential (V), sometimes referred to as the voltage, applied to a circuit. In much the same way that a grav-itational potential will cause mass to move, the electric potential will cause electrons to move. If you lift a book and release it from a height (high gravitational potential) it will fall downward (to a lower potential). The electric potential works in a similar way; if one point of the cir-2 cuit has a high electric potential, it means that it has a net positive charge and another point of the circuit with a low potential will have a net negative charge. Electrons in a wire will ﬂow from low electric potential with its net negative charge to high electric potential with its net positive charge because unlike charges attract and like charges repel.1 As these electrons ﬂow through the wire, they are scat-tered by atoms in the wire. The resistance of the circuit is just that; it is a measure of how diﬃcult it is for the electrons to ﬂow in the presence of such scattering. This resistance is a property of the circuit itself, and just about any material has a resistance. Materials that have a low resistance are called conductors and materials that have a very high resistance are called insulators. Some ma-terials have a moderate resistance and still allow some current to ﬂow. These are the materials that we use to make resistors like the ones we will use in this experi-ment. In short, the electric potential causes the current to ﬂow and the resistance impedes that ﬂow. Two or more resistors can be connected in series (that is connected one after another as shown in Fig. 1), or in parallel (that is the current can split: there is more than one way for the current to ﬂow as shown in Fig. 2). When two resistors R1 and R2 are connected in series, the equivalent resistance Req is given by Req = R1 + R2 . (3) Hence, the circuit behaves as if it contained a single resis-tor with resistance Req and it draws current from a given applied voltage like such a resistor. When those same resistors are connected in parallel instead, the equivalent resistance is related to R1 and R2 according to 1 Req = 1 R1 + 1 R2 . (4) B. Standard electronic symbols and resistor color codes Standard electronic symbols are shown in Fig. 3. The positive side of a battery or power supply is indicated with the longer vertical line. Most resistors are coded with color bands around one end of the resistor body. Using the resistor color code system is similar to using scientiﬁc notation. Scientiﬁc notation uses a number between 0 and 9.9 multiplied by some power of ten. The resistor color code system uses 1 Note that we say the current ﬂows from high poential to low poten-tial, but electrons ﬂow from low to high. This is because current is deﬁned as the ﬂow of positive charges, and electrons are negatively charged. A negative charge ﬂowing in one direction is like a positive charge ﬂowing in the other. Yes, it is confusing, but we cannot make the whole world start calling electrons positively charged, so we are stuck with it. 2. Ohm’s Law Figure 2.5: Standard symbols all (i.e. 0.00), you have a very small current and can then move the to the 400 mA plug. An ammeter must be connected in series with the circuit ment of interest, as shown in Fig. 2.4(b). This means that unlike me ing voltage, if you want to measure current you must break the circui wire the ammeter in. All of the current must ﬂow through the amme order for it to be measured. If you use your ﬁnger to trace the path charge in Fig. 2.4(b) after it leaves the power supply, you will see t must go through both the resistor and the ammeter. In contrast, tr the path of a charge in Fig. 2.4(a) you will see that it has two “par paths through which it can go (do not connect an ammeter in this ner). An ammeter typically has a very small internal resistance. Ther the current in the circuit is approximately the same before and afte ammeter is connected. Standard electronic symbols are shown in Fig. 2.5. The positive s a battery or power supply is indicated with the longer vertical line. The Breadboard The breadboard is designed for quick construction of simple electron cuits and is shown in Figure 2.6. Electronic elements (e.g. resistor 26 Last updated January 24, 2013 FIG. 3: Standard symbols. a number between 1 and 99 multiplied by some power of ten. These color bands tell the value of the resistance. Starting from the end, the ﬁrst band represents the ﬁrst digit of the resistance value and the second band the second digit. The third band represents the power of ten multiplying the ﬁrst two digits. The fourth band represents the tolerance. If the fourth band is absent, it means the tolerance is 20%. In Fig. 4 we show an example together with a color code chart, from which one can tell the resistance of a resistor. C. The DC power supply A DC power supply is used to provide varying volt-age to a circuit. The power supply used in this lab is shown in Fig. 5. The black and red connectors are the negative (−) and positive (+) output terminals, respec-tively. The voltage knob controls the power supply’s output voltage. The current knob sets a limiting current. Here, adjust the current control to its maximum setting (all the way clockwise) at all times. D. The digital multimeter The digital multimeter is shown in Fig. 5. As its name suggests, a multimeter has multiple functions. It can be used for several diﬀerent purposes: (i) a voltage measur-ing device (a voltmeter), (ii) a current measuring device (an ammeter), and (iii) a resistance measuring device (an ohmmeter). We will use all these functions in this experiment. To use the multimeter as a voltmeter, the dial selector is set to one of the positions labeled “V.” The probing ca-bles are then connected to the plugs labeled “VΩ′” and “COM.” There are two types of “V” settings. The setting 3 2. Ohm’s Law Table 2.2: Resistor color codes Color 1st digit 2nd digit Power of 10 Tolerance black 0 0 0 -brown 1 1 1 -red 2 2 2 -orange 3 3 3 -yellow 4 4 4 -green 5 5 5 -blue 6 6 6 -violet 7 7 7 -gray 8 8 8 -white 9 9 9 -gold ---5% silver ---10% none ---20% Figure 2.11: Example resistor. 32 Last updated January 24, 2013 FIG. 4: The upper panel shows the resistor color codes. The ﬁrst digit of the resistance in the lower panel is a 4, the second digit is a 7 and the multiplier is a 3, so the resistance is R = 47 × 103 Ω. The fourth color is used to calculate the uncertainty in the resistance. The tolerance of this resistor is 5% (corresponding to the gold band). The uncertainty in the resistance is given by: uncertainty = R× tolerance = 4.7×104Ω×5/100 = 2.35×103Ω. 2. Ohm’s Law Figure 2.2: DC power supply Figure 2.3: Digital multimeter To use the multimeter as a voltmeter, the dial selector is set to one of the positions labeled “V”. The probing cables are then connected to the plugs labeled “VΩ” and “COM”. There are two types of “V” settings. The setting with the tilde (⇠) over it is used for measuring AC voltage. The other type of “V” setting has two lines over the V, one line is solid and the second line is dashed, to indicate DC voltage. AC is an abbreviation for alternating current. An AC voltage is a voltage whose magnitude and polarity vary with time. DC is an abbreviation for direct current. A DC voltage is a constant voltage. During this experiment, only the DC setting is used. There are two DC voltage settings on the multimeter: “V” and “mV”. When using the “mV” setting, the output of the multimeter will be in millivolts. Whether the multimeter is used to measure voltage (as a 24 Last updated January 24, 2013 FIG. 5: The DC power supply (left) and the digital multimeter (right). with the tilde (−) over it is used for measuring AC volt-age. The other type of “V” setting has two lines over the V, one line is solid and the second line is dashed, to in-dicate DC voltage. AC is an abbreviation for alternating current. An AC voltage is a voltage whose magnitude and polarity vary with time. DC is an abbreviation for di-rect current. A DC voltage is a constant voltage. During this experiment, we will only use the DC setting. There are two DC voltage settings on the multimeter: “V” and “mV.” When using the “mV” setting, the output of the multimeter will be in millivolts. Whether the multime-ter is used to measure voltage (as a voltmeter) or current (as an ammeter), one cable is always connected to the COM plug. If the multimeter is used to measure current, the other lead is connected to either the 10 A plug or the 400 mA plug. A voltmeter must be connected in parallel (across) to the circuit element of interest, as shown in Fig. 6. Since the voltmeter measures potential diﬀerence between two points, it is easy to connect. To measure the potential dif-ference (voltage drop) across a resistor, use two cables to connect the plugs of the voltmeter to the circuit across the resistor (one cable before the resistor and a second cable after the resistor). A voltmeter typically has a very large internal resistance; therefore very little current will ﬂow through it. Consequently, the current in the cir-cuit will be approximately the same before and after the voltmeter is connected. To use the multimeter as an ammeter, the dial selector is set to one of the positions labeled “A.” Similar to the voltmeter settings there are AC and DC settings. Like the voltmeter, two cables must be connected to the ammeter. One of your cables must be connected to the plug labeled “COM.” The second cable can be connected to one of two possible plugs: either the 10 A plug or the 400 mA plug. If you have a large amount of current (anything above 400 mA), you must connect the cable to the terminal marked 10 A. If you put it in the 400 mA terminal you could damage the multimeter. If you are unsure if you have too much current for the 400 mA plug, start with the 10 A plug. If you do not get any reading at all (that is 0.00), you have a very small current and can then move the cable to the 400 mA plug. An ammeter must be connected in series with the cir-cuit element of interest, as shown in Fig. 6. This means that unlike measuring voltage, if you want to measure current you must break the circuit and wire the ammeter in. All of the current must ﬂow through the ammeter in order for it to be measured. If you use your ﬁnger to trace the path of a charge in Fig. 6 (right) after it leaves the power supply, you will see that it must go through both the resistor and the ammeter. In contrast, tracing the path of a charge in Fig. 6 (left) you will see that it has two “parallel” paths through which it can go (do not connect an ammeter in this manner). An ammeter typi-cally has a very small internal resistance. Therefore, the current in the circuit is approximately the same before and after the ammeter is connected. An ohmmeter does not function with a circuit con-nected to a power supply. If you want to measure the resistance of a particular component, you must take it out of the circuit altogether and test it separately. 4 2.6. Equipment (a) Voltmeter connected in parallel (b) Ammeter connected in series Figure 2.4: Schematics of meters being connected in a circuit. voltmeter) or current (as an ammeter), one cable is always connected to the COM plug. If the multimeter is used to measure current, the other lead is connected to either the 10A plug or the 400mA plug. A voltmeter must be connected in parallel (across) to the cir-cuit element of interest, as shown in Fig. 2.4(a). Since the voltmeter measures potential di⇥erence between two points, it is easy to connect. To measure the potential di⇥erence (voltage drop) across a resistor, use two cables to connect the plugs of the voltmeter to the circuit across the resis-tor (one cable before the resistor and a second cable after the resistor). A voltmeter typically has a very large internal resistance; therefore very little current will ﬂow through it. Consequently, the current in the circuit will be approximately the same before and after the voltmeter is connected. To use the multimeter as an ammeter, the dial selector is set to one of the positions labeled “A”. Similar to the voltmeter settings there are AC and DC settings. Like the voltmeter, two cables must be connected to the ammeter. One of your cables MUST be connected to the plug labeled “COM”. The second cable can be connected to one of two possible plugs —- either the “10A” plug or the “400mA” plug. If you have a large amount of current (anything above 400 mA), you must connect the cable to the terminal marked “10A”. If you put it in the “400mA” terminal you could damage the multimeter. If you are unsure if you have too much current for the 400 mA plug, start with the 10A plug. If you do not get any reading at Last updated January 24, 2013 25 FIG. 6: Schematics of meters being connected in a circuit. Voltmeter connected in parallel (left) and ammeter conected in series (right). 2.6. Equipment Figure 2.6: Breadboard easily attached using the metal spring clips in the middle of the bread-board. Each metal clip is electrically connected to a plug connector by a metallic strip. The resistance between the metal clip and the plug connec-tor is negligible; therefore, you can assume that these two points are at the same electrical potential (voltage) and the same point in a circuit. Circuits are constructed by connecting the electronic circuit elements and the power supply together using cables with banana plugs. The banana plugs ﬁt se-curely into the plug connectors on the breadboard, the multimeter and the power supply. Wires You will have to hook up wires to make the circuits described in the circuit diagrams. Each line without any circuit element should correspond to a wire in your circuit. A wire (or line in the diagram) represents a path where current can ﬂow.4 All points on a wire/line have the same voltage. Because of this, a circuit may be realized by several di⇥erent arrangements of wire. For example, see Fig. 2.7. 4A wire is actually a resistor with very low resistance compared to the resistors we typically use in class. Therefore, we can usually neglect (ignore) any resistance that it has. On the other hand, this resistance is a big factor in long-distance electrical transmission lines, since there is so much wire involved. Last updated January 24, 2013 27 FIG. 7: The breadboard. E. The breadboard The breadboard is designed for quick construction of simple electronic circuits and is shown in Fig. 7. Elec-tronic elements (e.g. resistors) are easily attached using the metal spring clips in the middle of the breadboard. Each metal clip is electrically connected to a plug con-nector by a metallic strip. The resistance between the metal clip and the plug connector is negligible; there-fore, you can assume that these two points are at the same electrical potential (voltage) and the same point in a circuit. Circuits are constructed by connecting the elec-tronic circuit elements and the power supply together using cables with banana plugs. The banana plugs ﬁt securely into the plug connectors on the breadboard, the multimeter, and the power supply. F. Wires You will have to hook up wires to make the circuits described in the circuit diagrams. Each line without any circuit element should correspond to a wire in your circuit. A wire (or line in the diagram) represents a path where current can ﬂow.2 All points on a wire/line have the same voltage. Because of this, a circuit may be realized by several diﬀerent arrangements of wire. For example, see Fig. 8. G. Suggestions for building circuits The schematic representation of electronic circuits typ-ically shows wires as straight lines and changes in the directionof thewiresare indicatedbyabruptbends inthe wires. In practice, the ﬂexible wires are not straight and as you might expect changes in direction are not abrupt 90 degree bends in the wires. Adding measuring devices (e.g. ammeters, voltmeters) to the circuit increases the circuits complexity. The following steps will guide you through the construction of a simple circuit that includes an ammeter and a voltmeter. To avoid confusion, all of the wires used in the following example have diﬀerent colors. The ﬁgures in this guide show both the circuit represented schematically and how the circuit actually looks in practice. 1. Start by building the circuit without any meters. Where two lines meet, you will need two wires. Although it may seem eﬃcient to initially construct the circuit with the meters included, experience has shown that this method often leads to wiring errors. In Fig. 9 we show a simple circuit with a power supply and a single resistor. The green wire is connected to the positive terminal of the power supply and the white wire is connected to the negative terminal. 2 A wire is actually a resistor with very low resistance compared to the resistors we typically use in class. Therefore, we can usually neglect (ignore) any resistance that it has. On the other hand, this resistance is a big factor in long-distance electrical transmission lines, since there is so much wire involved. 5 Figure 2.7: These two circuits are equivalent — they have the same conﬁg-uration of elements and will act in exactly the same manner. Suggestions for building circuits The schematic representation of electronic circuits typically shows wires as straight lines and changes in the direction of the wires are indicated by abrupt bends in the wires. In practice, the ﬂexible wires are not straight and as you might expect changes in direction are not abrupt 90 degree bends in the wires. Adding measuring devices (e.g. ammeters, voltmeters) to the circuit increases the circuit’s complexity. The following steps will guide you through the construction of a simple circuit that includes an ammeter and a voltmeter. To avoid confusion, all of the wires used in the following example have di⇥erent colors. The ﬁgures in this guide show both the circuit represented schematically and how the circuit actually looks in practice. 1. Start by building the circuit without any meters. Where two lines meet, you will need two wires. Although it may seem e⇧cient to initially construct the circuit with the meters included, experience has shown that this method often leads to wiring errors. Figs. 2.8(a) (schematically) and 2.8(b) (in practice) show a simple circuit with a power supply and a single resistor. The green wire is connected to the positive terminal of the power supply and the white wire is connected to the negative terminal. 28 Last updated January 24, 2013 FIG. 8: These two circuits are equivalent. They have the same conﬁguration of elements and will act in exactly the same manner. 2.6. Equipment (a) Schematic (b) In practice Figure 2.8: Building a circuit that includes a power supply and resistor. 2. Include an ammeter in the circuit to measure current. Attach a single wire to the “COM” input of your ammeter; in this example, this is the purple wire. Identify the element in your circuit through which the desired current is ﬂowing (in this case the resistor). Unplug the wire (or wires) leading into one end of that element and plug all of them into either the 400mA or 10A input of your ammeter, depending on the size of the current you are measuring. In this example, there is only one wire leading to the resistor (the green wire) and we are using the 400mA setting of the ammeter. Plug the free end of the purple wire into the plug on the breadboard where you removed the circuit’s wire (or wires) — i.e. the place where the green wire was connected in Fig. 2.8(b). You have now forced all of the current carried by the wire (or wires) to go through the ammeter in addition to the circuit element of interest. The ammeter is now properly connected in series with the resistor. Figs. 2.9(a) (schematically) and 2.9(b) (in practice) show our simple circuit with a power supply, a single resistor, and an ammeter. Turn the dial to read mA or A. By default, it is set to read AC current. We have DC current, so press the yellow button to change the mode to DC. You’ll have to do this again if the multimeter turns o⇥automatically. Note that the ammeter should FIG. 9: Building a circuit that includes a power supply and resistor; schematic (left) and in practice (right). 2. Include an ammeter in the circuit to measure cur-rent. Attach a single wire to the “COM” input of your ammeter; in this example, this is the purple wire. Identify the element in your circuit through which the desired current is ﬂowing (in this case the resistor). Unplug the wire (or wires) leading into one end of that element and plug all of them into either the 400 mA or 10 A input of your am-meter, depending on the size of the current you are measuring. In this example, there is only one wire leading to the resistor (the green wire) and we are using the 400 mA setting of the ammeter. Plug the free end of the purple wire into the plug on the breadboard where you removed the circuit’s wire (or wires), that is the place where the green wire was connected in Fig. 9. You have now forced all of the current carried by the wire (or wires) to go through the ammeter in addition to the circuit el-ement of interest. The ammeter is now properly connected in series with the resistor. In Fig. 10 we show our simple circuit with a power supply, a sin-gle resistor, and an ammeter. Turn the dial to read mA or A. By default, it is set to read AC current. We have DC current, so press the yellow button to change the mode to DC. You will have to do this again if the multimeter turns oﬀautomatically. Note that the ammeter should display “DC” just to the right of the numbers and an “A” for ampers (the unit for current). 3. Include a voltmeter in the circuit to measure volt-age. Attach two wires to the voltmeter inputs. In the example below the red wire is connected to the V input and the black wire is connected to the COM input of the voltmeter. Attach the free end of each wire across the circuit element whose voltage you would like to measure; in this case the red wire is connected to the right of the resistor and the black wire is connected to the left of the resistor. The voltmeter is now properly connected in parallel with the resistor, as seen in Fig. 11. Never connect an ammeter in this fashion as it can damager the 6 2. Ohm’s Law (a) Schematic (b) In practice Figure 2.9: Building a circuit that includes a power supply, ammeter, and resistor. display “DC” just to the right of the numbers and an “A” for Amps (the unit for current). 3. Include a voltmeter in the circuit to measure voltage. Attach two wires to the voltmeter inputs. In the example below the red wire is connected to the “VΩ” input and the black wire is connected to the “COM” input of the voltmeter. Attach the free end of each wire across the circuit element whose voltage you would like to measure – in this case the red wire is connected to the right of the resistor and the black wire is connected to the left of the resistor. The voltmeter is now properly connected in parallel with the resistor, as seen in Fig. 2.10. Never connect an ammeter in this fashion as it can damager the meter. Once you have constructed a circuit, no matter how complicated, you can use steps two and three to measure the current ﬂowing through a given element in the circuit and the voltage across that circuit element. Resistor color codes Most resistors are coded with color bands around one end of the resistor body. Using the resistor color code system is similar to using scientiﬁc 30 Last updated January 24, 2013 FIG. 10: Building a circuit that includes a power supply, ammeter, and resistor; schematic (left) and in practice (right). 2.6. Equipment (a) Schematic (b) In practice Figure 2.10: Building a circuit that includes a power supply, voltmeter, resistor, and ammeter. notation. Scientiﬁc notation uses a number between 0 and 9.9 multiplied by some power of ten. The resistor color code system uses a number between 01 and 99 multiplied by some power of ten. These color bands tell the value of the resistance. Starting from the end, the ﬁrst band represents the ﬁrst digit of the resistance value and the second band the second digit. The third band represents the power of ten multiplying the ﬁrst two digits. The fourth band represents the tolerance. If the fourth band is absent, it means the tolerance is 20%. Table 2.2 is a color code chart, from which one can tell the resistance of a resistor. Example Suppose the color code on a resistor is yellow, violet, orange and gold like the resistor depicted in Fig. 2.11. What is its resistance and what is the uncertainty of this resistance? The value of the resistance can be found from the ﬁrst three colors. From the table above, the ﬁrst digit is 4 (corresponding to the yellow band), the FIG. 11: Building a circuit that includes a power supply, voltmeter, resistor, and ammeter; schematic (left) and in practice (right). meter. Once you have constructed a circuit, no matter how complicated, you can use steps two and three to measure the current ﬂowing through a given element in the circuit and the voltage across that circuit element. H. Safety tips • When plugging or unplugging wires, ﬁrst turn oﬀ all electronics that are connected or will become connected to the circuit. • Prior to making any change in the circuit, always turn the voltage knob to its minimum setting (all the way counterclockwise) and turn oﬀthe power supply! So the next time you turn on the power supply its output will be zero volts. III. MATERIALS • DC power supply; see Fig. 5. • Two digital multimeters; see Fig. 5. • Breadboard; see Fig. 7. • Several banana-tobanana wires. 7 Ohm’s Law Figure 2.13: Schematic for 2 resistors connected in series. Graphical test of Ohm’s law for a constant resistor. Use the same circuit as in Step 1. Begin with a very small positive voltage and gradually increase the voltage. For ﬁve settings throughout the range, record both the voltage reading from the voltmeter and the current reading from the ammeter in the top half of Data Table 2. Decrease the supply voltage to its minimum value and change the po-larity of the voltage (make the electricity ﬂow in the opposite direction through the circuit). You do this by switching the wires connecting your circuit to the power supply. Again, gradually increase the supply voltage. For ﬁve settings throughout the range, record voltage and current measurements in the bottom half of Data Table 2. Using all of the data in Data Table 2, plot V (vertical axis) vs. I (horizontal axis). Have Kaleidagraph ﬁt your data with a best ﬁt line, display the equation of the best ﬁt line and the uncertainties in the slope and intercept (don’t forget to brieﬂy comment on your graph). Record the slope and its uncertainty in your spreadsheet. Compare this to your value for the resistance determined by Ohm’s Law (see Question 2). Two resistors connected in series Construct the circuit shown in Fig. 2.13. Use two di⇥erent resistors having resistances of approximately 1 kΩand 2 kΩ. Set the power supply voltage to the middle of its range and record your measured voltage and current in Data Table 3. Last updated January 24, 2013 FIG. 12: Schematic for two resistors connected in series. TABLE I: Measured voltage, current, and resistance. Voltage (V) Current (A) R = V/i (Ω) 0.20 0.78 1.36 1.94 2.52 3.10 3.68 4.26 4.84 5.42 6.00 IV. ASSEMBLY AND OPERATION 1. Wire up the circuit shown in Fig. 11. Have your instructor check the connections before you plug in the power supply. If it is ok, turn it on, and vary the voltage by turning the “DC oﬀset” knob. Note the changes in the voltmeter and ammeter readings. Vary the voltage in ten steps, from a low of about 0.2 V to the maximum possible (about 6 V). Record the voltage and current in Table I, and compute the resistance R from the ratio. Determine the average value of the resistance. 2. Disconnect the circuit. Use the ohmmeter function of the multimeter to make a direct reding of the resistance used in part 1, R1 =−−−−−Ω. Compare this with the resistance measured in part 1. Com-pare the results with the value from the color code in the resistor and verify your results are within errors. Choose another resistor on the board and measure its resistance, R2 =−−−−−Ω. 3. Consider the series circuit shown in Fig. 1. Use Ohm’s law in the form i = V/R to predict the cur-rent in this circuit, if the power supply is set to 5 V. (Remember, resistors in series add.) 4. Wire up the circuit shown in Fig. 12 and have your instructor check the connections. Adjust the power supply so that the voltmeter reads 5 V. How does the current compare with what you predicted in part 3? 5. Remove the voltmeter connections. Connect the voltmeter across R1 and record the reading, V1 =−−−−−V. Now, connect the voltmeter across R2 and record the reading, V2 =−−−−−V. What is the relation between V1, V2, and the power supply voltage of 5 V? Use Ohm’s law in the form V = iR to explain this result.
17089	https://scienceready.com.au/pages/acidic-basic-and-neutral-salts?srsltid=AfmBOooIkfmEmrE7ddkC8SVaayhsMeqsB-VtoiQAj5RNfkyu7gGXMZa8	Acidic, Basic and Neutral Salts – HSC Chemistry – Science Ready Skip to content Submit Close search Home Free Resources Year 11 Physics Module 1: Kinematics Module 2: Dynamics Module 3: Waves and Thermodynamics Module 4: Electricity and Magnetism Year 12 Physics Module 5: Advanced Mechanics Module 6: Electromagnetism Module 7: The Nature of Light Module 8: From the Universe to the Atom Year 11 Chemistry Module 1: Properties and Structure of Matter Module 2: Introduction to Quantitative Chemistry Module 3: Reactive Chemistry Module 4: Drivers of Reaction Year 12 Chemistry Module 5: Equilibrium and Acid Reactions Module 6: Acid/base Reactions Module 7: Organic Chemistry Module 8: Applying Chemical Ideas Study Skills & Advice Scientific Skills Premium Resources HSC Chemistry Topic Tests HSC Chemistry Practice Exams HSC Physics Topic Tests HSC Physics Practice Exams About About Us Contact Us 1-on-1 Mentoring YouTube Membership Member-only Discord Server Log inCartCurrency Home Free Resources Free Resources Menu Free Resources Year 11 Physics Year 11 Physics Menu Year 11 Physics Module 1: Kinematics Module 2: Dynamics Module 3: Waves and Thermodynamics Module 4: Electricity and Magnetism Year 12 Physics Year 12 Physics Menu Year 12 Physics Module 5: Advanced Mechanics Module 6: Electromagnetism Module 7: The Nature of Light Module 8: From the Universe to the Atom Year 11 Chemistry Year 11 Chemistry Menu Year 11 Chemistry Module 1: Properties and Structure of Matter Module 2: Introduction to Quantitative Chemistry Module 3: Reactive Chemistry Module 4: Drivers of Reaction Year 12 Chemistry Year 12 Chemistry Menu Year 12 Chemistry Module 5: Equilibrium and Acid Reactions Module 6: Acid/base Reactions Module 7: Organic Chemistry Module 8: Applying Chemical Ideas Study Skills & Advice Scientific Skills Premium Resources Premium Resources Menu Premium Resources HSC Chemistry Topic Tests HSC Chemistry Practice Exams HSC Physics Topic Tests HSC Physics Practice Exams About About Menu About About Us Contact Us 1-on-1 Mentoring YouTube Membership Member-only Discord Server Currency Acidic, Basic and Neutral Salts This is part of the HSC Chemistry course under the topic Quantitative Analysis. Chemistry textbooks HSC Chemistry Syllabus write ionic equations to represent the dissociation of acids and bases in water, conjugate acid/base pairs in solution and amphiprotic nature of some salts, for example: – sodium hydrogen carbonate – potassium dihydrogen phosphate What are Acidic, Basic and Neutral Salts? This video explores the acidic, basic and neutral nature of various salts. You will learn how to identify these different types of salts. Chemistry textbooks What are Salts? Salts are ionic compounds produced from an acid-base reaction. Many salts, despite being neutrally charged, produce acidic and basic solutions when dissolved in water. The properties of acidic and basic salts can be understood using the Brønsted-Lowry acid/base theory. Acidic Salts Acidic salts are acidic because they contain conjugate acids of weak bases. An example of an acidic salt is ammonium chloride, which is formed from the neutralisation between hydrochloric acid and ammonia: H C l(a q)+N H 3(a q)→N H 4 C l(a q)H C l(a q)+N H 3(a q)→N H 4 C l(a q) Ammonium is the conjugate acid of ammonia (weak base). Ammonium ions are Brønsted-Lowry acids as they donate protons to water: N H+4(a q)+H 2 O(l)⇌N H 3(a q)+H 3 O+(a q)N H 4+(a q)+H 2 O(l)⇌N H 3(a q)+H 3 O+(a q) The production of hydronium ions causes a solution of ammonium chloride to become acidic. Chloride ions are spectator ions and thus do not contribute to the pH of the solution. Acidic salts such as ammonium chloride are not Arrhenius acids as they do not contain and dissociate to produce hydrogen ions. Other examples of acidic salts include: Ammonium nitrate Ammonium sulfate Basic Salts Basic salts are basic because they contain conjugate bases of weak acids. An example of a basic salt is sodium acetate, which is formed from the neutralisation between acetic acid and sodium hydroxide: C H 3 C O O H(a q)+N a O H(a q)→N a C H 3 C O O(a q)+H 2 O(l)C H 3 C O O H(a q)+N a O H(a q)→N a C H 3 C O O(a q)+H 2 O(l) Acetate is the conjugate base of acetic acid (weak acid). Acetate ions are Brønsted-Lowry bases as they accept proton from water: C H 3 C O O−(a q)+H 2 O(l)⇌C H 3 C O O H(a q)+O H−(a q)C H 3 C O O−(a q)+H 2 O(l)⇌C H 3 C O O H(a q)+O H−(a q) The production of hydroxide ions causes a solution of sodium acetate to become basic. Sodium ions are spectator ions and thus do not contribute to the pH of the solution. Basic salts such as sodium acetate arenotArrhenius bases as they do not contain and dissociate to produce hydroxide ions. Other examples of basic salts include: Potassium acetate Sodium fluoride Sodium hypochlorite Sodium citrate Neutral Salts Neutral salts are neutral because they contain neither of conjugate acids of weak bases or conjugate bases of weak acids. Neutral salts are typically formed from neutralisation between strong acids and strong bases. An example of a neutral salt is sodium chloride, which is formed from the neutralisation between hydrochloric acid and sodium hydroxide: H C l(a q)+N a O H(a q)→N a C l(a q)+H 2 O(l)H C l(a q)+N a O H(a q)→N a C l(a q)+H 2 O(l) Sodium and chloride ions are both spectator ions. Chloride ions are unable to accept protons from water because they are the conjugate base of hydrochloric acid which is a strong acid. Strong acids like HCl completely dissociate in water: H C l(a q)→H+(a q)+C l−(a q)H C l(a q)→H+(a q)+C l−(a q) Sometimes a neutral salt can be formed from the neutralisation between a strong acid and weak base. For example, the reaction between hydrochloric acid and sodium carbonate (weak base) produces sodium chloride: H C l(a q)+N a 2 C O 3(a q)→N a C l(a q)+H 2 O(l)+C O 2(g)H C l(a q)+N a 2 C O 3(a q)→N a C l(a q)+H 2 O(l)+C O 2(g) Other examples of neutral salts include: Potassium chloride Sodium nitrate Potassium nitrate RETURN TO MODULE 6: ACID-BASE REACTIONS Resources HSC Chemistry Topic Tests HSC Chemistry Practice Exams HSC Physics Topic Tests HSC Physics Practice Exams Community Youtube Instagram Tiktok Facebook Patreon Quick links Contact Us About Us Privacy Terms of Service Refund policy Patreon Membership YouTube Membership Facebook Instagram YouTube Payment methods © 2025 Science Ready ABN: 95665321837 Use left/right arrows to navigate the slideshow or swipe left/right if using a mobile device choosing a selection results in a full page refresh press the space key then arrow keys to make a selection Opens in a new window. Opens external website. Opens external website in a new window.
17090	https://www.semanticscholar.org/paper/Electrocardiographic-changes-in-patients-with-Krenke-Nasi%C5%82owski/90e6fb2affee6fc7a6f1424714bfc95e0c75d9ed/figure/1	Figure 1 from Electrocardiographic changes in patients with spontaneous pneumothorax. \| Semantic Scholar Skip to search formSkip to main contentSkip to account menu Search 229,291,922 papers from all fields of science Search Sign In Create Free Account Corpus ID: 261269513 Electrocardiographic changes in patients with spontaneous pneumothorax. @article{Krenke2008ElectrocardiographicCI, title={Electrocardiographic changes in patients with spontaneous pneumothorax.}, author={Rafał Krenke and Jacek Nasiłowski and Tadeusz Przybyłowski and Ryszarda Chazan}, journal={Journal of physiology and pharmacology : an official journal of the Polish Physiological Society}, year={2008}, volume={59 Suppl 6}, pages={ 361-73 }, url={ } R. Krenke, J. Nasiłowski, +1 authorR. Chazan Published in Journal of Physiology and…1 December 2008 Medicine TLDR It is concluded that ECG in subjects with pneumothorax often reveals significant abnormalities, and the most significant abnormalities were seen in patients with massive right-sided pneumothsorax.Expand View on PubMed Save to Library Save Create Alert Alert Cite Share 12 Citations Highly Influential Citations 2 Background Citations 6 Results Citations 1 View All Figures and Tables 12 Citations 28 References Related Papers Figures and Tables from this paper table 1 figure 1 table 2 figure 2 table 3 figure 3 View All 6 Figures & Tables 12 Citations Citation Type Has PDF Author More Filters More Filters Filters ### A retrospective study B. KlinItai GuetaH. BibiS. BaramI. Abu-Kishk Medicine 2021 TLDR Electrocardiography changes that accompanied PSP in relation to side and size of pneumothorax were examined to rule out pneumothorax among children presented with chest pain, dyspnea and ECG changes.Expand 10 Highly Influenced 6 Excerpts Save ### Changes in electrocardiographic findings after closed thoracostomy in patients with spontaneous pneumothorax W. LeeYoonje Lee+6 authorsJ. Shin Medicine Clinical and experimental emergency medicine 2017 TLDR Electrocardiographic (ECG) findings in spontaneous pneumothorax patients before and after closed thoracostomy are described to find only minor changes in ECG after closed Thoracstomy in spontaneous pneumonia patients.Expand 5 Highly Influenced ( 6 Excerpts Save ### Electrocardiographic changes in young men with left-sided spontaneous pneumothorax. Gen-Min LinShih-Chung HuangYi-Hwei LiChih-lu Han Medicine International journal of cardiology 2014 2 1 Excerpt Save ### Acute right bundle branch block due to pneumothorax Manish RuhelaG. KhandelwalSamiksha GuptaA. Bansal Medicine Journal of family medicine and primary care 2018 TLDR The first case of acute-onset complete RBBB due to left-sided pneumothorax in a young patient after blunt chest trauma from a motor vehicle accident is reported.Expand 2 2 Excerpts Save ### Left spontaneous pneumothorax presenting with ST-segment elevations: a case report and review of the literature. A. ShiyovichZeldetz VladimirLior Nesher Medicine Heart & lung : the journal of critical care 2011 23 Save ### Electrocardiographic changes in pneumothorax: an updated review Hina ArshNandhini Iyer+12 authorsJahanzeb Malik Medicine Annals of medicine and surgery 2024 TLDR A narrative review examines the existing literature to provide insights into the various ECG abnormalities observed in patients with pneumothorax, their underlying mechanisms, and clinical implications, highlighting the commonly reported changes.Expand 1 PDF 1 Excerpt Save ### Uncomplicated circulatory shock: a narrative review Mauro Dirlando Conte de OliveiraO. F. D. dos SantosGiancarlo ColomboThiago Domingos CorrêaMiguel Cendoroglo Medicine Einstein 2024 TLDR Although lactate-guided management has proven to be effective, the use of capillary refill time, other biochemical markers of perfusion, and preload-directed resuscitation have the potential to avoid volume overload and improve outcomes.Expand PDF Save ### Accidental Hypothermia-Induced J Wave Coupled With Giant R Wave Augmented by Premature Atrial Contraction: A Case Report Koji TakahashiH. MoriokaShigeki UemuraTakafumi OkuraKatsuji Inoue Medicine Cureus 2024 TLDR Observations indicate that the underlying mechanism for the genesis of J waves is indeed conduction delay and not transient outward currents, and this case of hypothermia-induced J waves together with giant R waves, which have not been previously reported during hypothermia, is presented.Expand ( 1 Excerpt Save ### Mechanical ventilation-associated pneumothorax presenting with paroxysmal supraventricular tachycardia in patients with acute respiratory failure J. EomM. Lee+6 authorsS. Park Medicine 2015 TLDR The results show clear improvements in the quality and efficiency of the care provided to patients with chronic obstructive lung disease and the quality of their care has improved significantly.Expand PDF 1 Excerpt Save ### Diagnosis and Management of Spontaneous Pneumothorax in the Emergency Department: A Review of the Most Current Clinical Evidence for Diagnosis and Treatment Ian ChongA. Chao+4 authorsP. Perera Medicine 2016 TLDR This research presents a novel, scalable, scalable and scalable approaches that can be used to improve the quality and quantity of emergency care in patients with complex medical conditions.Expand 4 PDF Save ... 1 2 ... 28 References Citation Type Has PDF Author More Filters More Filters Filters ### Electrocardiographic patterns in pneumothorax. R. N. ArmenT. V. Frank Medicine Diseases of the chest 1949 TLDR Characteristic ECG patterns both in right and left sided pneumothorax have been shown and described, and available evidence points to rotation of theheart and the presence of air between the heart and the chest wall as very probable causative factors.Expand 31 Highly Influential 8 Excerpts Save ### Electrocardiographic changes in pulmonary collapse; artificial and spontaneous left-sided pneumothorax studied by conventional and unipolar methods. C. SilverbergR. Kingsland+4 authorsHos-Pital Medicine Diseases of the chest 1950 TLDR The use of the CP precordial leads in the presence of left-sided pneumothorax might lead to a false electrocardiographic impression of coronary artery disease, or even of old myocardial infarction.Expand 4 PDF Save ### The electrocardiographic changes in pneumothorax in which the heart has been rotated A. Master Medicine 1928 43 Save ### Severe transmyocardial ischemia in a patient with tension pneumothorax U. JanssensK. KochJ. GrafP. Hanrath Medicine Critical care medicine 2000 TLDR The specific condition of TP may lead to impaired systolic and diastolic coronary artery blood flow affecting ventricular repolarization and T‐wave configuration in ECG indicative of transmyocardial ischemia.Expand 34 2 Excerpts Save ### Right pneumothorax with the S1Q3T3 electrocardiogram pattern usually associated with pulmonary embolus. Roy GoddardRoy GoddardR. ScofieldR. Scofield Medicine The American journal of emergency medicine 1997 19 Highly Influential 5 Excerpts Save ### Are electrocardiogram changes the first sign of impending peri‐operative pneumothorax? G. BotzJ. Brock‐Utne Medicine Anaesthesia 1992 TLDR A patient in the right lateral position underwent left nephrectomy, after which he was placed supine for insertion of an arteriovenous fistula, and following the position change it was noted that the amplitude of the electrocardiogram complexes were dramatically reduced.Expand 21 PDF Save ### Electrocardiographic changes with right-sided pneumothorax. M. AlikhanJ. Biddison Medicine Southern medical journal 1998 TLDR Five cases of right-sided pneumothorax in which electrocardiographic findings resolved with reexpansion of the lung are presented, and a new finding is reported in two cases.Expand 14 2 Excerpts Save ### The electrocardiogram as a diagnostic aid in pneumothorax. Roland S. Summers Medicine Chest 1973 TLDR This study shows that during some common activities, there appears to be shallow, irregular breathing involving both chest and diaphragmatic movement, and this is followed by a period of increased ventilation, which is appreciated as dyspnea in patients with obstructive lung disease.Expand 8 PDF 1 Excerpt Save ### ECG changes in pneumothorax. A unique finding and proposed mechanism. T. FeldmanC. January Medicine Chest 1984 TLDR A patient with anterior myocardial infarction had bilateral pneumothoraces, which suggests the importance of air insulating the chest wall, rather than cardiac rotation, dilatation, or displacement as a mechanism of the ECG change.Expand 29 2 Excerpts Save ### Unusual ECG variations in left-sided pneumothorax. M. KoželjP. RakovecM. Sok Medicine Journal of electrocardiology 1997 19 1 Excerpt Save ... 1 2 3 ... Related Papers Showing 1 through 2 of 0 Related Papers Fig. 1. A - Right bundle-branch block in a 30-year-old male with left-sided pneumothorax, ECG on admission; B - ECG after lung re-expansion. Published in Journal of Physiology and Pharmacology 2008 Electrocardiographic changes in patients with spontaneous pneumothorax. R. KrenkeJ. NasiłowskiT. PrzybyłowskiR. Chazan Figure 2 of 6 Stay Connected With Semantic Scholar Sign Up What Is Semantic Scholar? Semantic Scholar is a free, AI-powered research tool for scientific literature, based at Ai2. Learn More About About UsPublishersBlog (opens in a new tab)Ai2 Careers (opens in a new tab) Product Product OverviewSemantic ReaderScholar's HubBeta ProgramRelease Notes API API OverviewAPI TutorialsAPI Documentation (opens in a new tab)API Gallery Research Publications (opens in a new tab)Research Careers (opens in a new tab)Resources (opens in a new tab) Help FAQLibrariansTutorialsContact Proudly built by Ai2 (opens in a new tab) Collaborators & Attributions •Terms of Service (opens in a new tab)•Privacy Policy (opens in a new tab)•API License Agreement The Allen Institute for AI (opens in a new tab) By clicking accept or continuing to use the site, you agree to the terms outlined in ourPrivacy Policy (opens in a new tab), Terms of Service (opens in a new tab), and Dataset License (opens in a new tab) ACCEPT & CONTINUE or Only Accept Required
17091	https://www.youtube.com/watch?v=edrKK20wqQI	Using Series to Evaluate Limits \| 3-Step Easy Process! \| Calculus 2 \| Math with Professor V Math with Professor V 50500 subscribers 120 likes Description 4514 views Posted: 12 Jul 2023 Need help using series to evaluate limits? Look no further, I got you covered! I break down the process into 3 simple steps, and walk you through several examples so you can master this topic in no time. Enjoy! Need more help with sequences and series? Check out my other video lectures already on my YouTube Channel: Power Series: Representation of Functions as Power Series: Finding Taylor and Maclaurin Series for Functions: Calculus 2 Video Lectures: Don't forget to LIKE, Comment, & Subscribe! xoxo, Professor V Calculus 2 Lecture Videos on Integration: Integration by parts: Tabular Integration: Trigonometric integrals: Trigonometric Substitution: Partial Fraction Decomposition: Strategy for Integration: Improper Integrals: Trig Review: Unit Circle: Trig Identities: Sum and Difference Formulas: Double Angle & Half-Angle Formulas: Calculus 3 Video Lectures: mathwithprofessorv #integration #partialfractions #partialfractionmethod #partialfractiondecomposition #trigonometricsubstitution #trigintegrals #trigsubstitution #integrals #integralcalculus #calculus2 #calculusvideos #calculus2videos #integralvideos #trigonometricintegrals #math #youtubemath #mathvideos #mathtutor #mathprofessor #calculusvideos #integrationbyparts #integralcalculus Join this channel to get access to perks: Socials: IG: @mathwithprofessorv TikTok: @mathwithprofessorv I'm also an Amazon influencer, so feel free to support and shop: EXCITING NEWS: You can now sign up for my Patreon at the link below! My Patreon is a place for students to have access to exclusive ad-free content not available on my YouTube channel! Access to the library of additional videos, worksheets, and more is available with the "Star Pupil Package" tier for just $9.99/month (USD). This video is copy protected and cannot be downloaded or used in any capacity without my permission. 14 comments Transcript: welcome to math TV with Professor V in this video we're going to learn how to use series to evaluate limits and when I say series I mean infinite Series in particular when this kind of problem pops up you usually have to use a mclen series so try to break it down into three basic steps that you can follow anytime you have to tackle one of these problems the first thing is you want to identify the term or terms in the limit that you can replace with a known McLaren series and I just pasted in some known important mclen series that typically will come up in these kinds of problems that you should have studied already in this topic in this lesson then the second thing you're going to do is write out the first few terms of the mclen series typically when you're working with the limits you don't want to um use the sigma notation version you want to use the expanded version and then you can put plus dot dot dot in the limit I'll show you exactly how it's all going to work and then usually these problems have been constructed in such a way that terms will cancel and simplify really nicely you'll see while we work through some examples so just simplify your little heart out and find the limit all right this is just a you know there's a small set of exercises that have to do with this topic and it always kind of stumbles students but it's not that difficult it just takes a little bit of practice so that's what we will do right now get out your pencil and paper and join me okay let's have a little math party so the directions will usually say use series to evaluate the limit obviously um like you know if I had it my way and the directions didn't say use series to evaluate the limit I'm looking here I have an indeterminant form Z over zero I would probably just use lyal's rule you know and take the limit directly but part of being a success uccessful math student entails following directions all right so we're going to use series to evaluate the limit so first step was find the term or terms that can be replaced using a mclen series so I'm noticing ooh ln1 + x go back and look yes that is a known mclen series now what if you don't have these basic muren series memorized O then you're stuck okay so make sure you put them all to memory or at least know how to derive them like quite a few you know sometimes in a pinch I'll derive this one but I you know when I'm working I memorize it okay so off to the side watch what I'm going to do I'm just going to list out the first few terms of natural log of 1 plus X like I said in the limit we usually don't want to work with the sigma notation version I'm going to use the expanded version and just put some dot dot dots so we've got ellipses I know what they're called I just like saying dot dot dot okay so x - x^2 over 2 + x X Cub over 3 - x 4/ 4 + dot dot dot I love it okay so we're going to go through take that replace it where we have natural log 1+ X in the limit you see yes we see Professor V oh good job okay limit X approaches zero X I didn't mean to change colors there The Perils of being left-handed okay x minus and then all of that goes in parentheses so x - x^2 / 2 + x Cub over 3 - x 4 over 4 plus yep keep them dot dot dots going over X2 beautiful and then look ah I told you something would always cancel in like a cutesy kind of way see how this x cancels with this x here and then be careful because this minus sign will distribute and change the signs of all of the terms that remain so I'm going to list out some more so here we go we've got limit X approaches zero now this is going to be positive x^2 / 2 - x Cub over 3+ x 4 over 4 let me just write out a couple more what would be next - x to the 5th over five right and then plus da da da over x^2 now what to do here with this over x^2 well basically you're going to split it up and write x squared under each of the terms here and simplify you can do it in your head but basically what we're doing is we're like going to divide this guy by x s this guy by X SAR everybody gets an X squ Etc all right so then let's see what's going to result we'll have the limit as X approaches zero now in this first term these X will just cancel out completely so all I have left is 12 minus now when you cancel here this x squ is gone but this is going to be to the first so - X over3 Plus again this is gone now this is going to be X2 so that's going to be X2 over 4 minus do you see what's going on this will be X cubed yes okay X x cubed over 5 plus blah blah blah now notice all of the other terms after - x over 3 everybody from here on out has an X in the numerator and just a constant in the denominator so if I use direct substitution and plug in that X is going to zero well this term is going to go to zero this term's going to go to zero right you just have zero over some number all the rest they go to Zer to so who's remaining in this limit just this positive 1/2 here bomom and there we go okay how was that not too bad right I hope it wasn't too terrible okay let's try another like I said you just got to practice these and then they get super repetitive they really do so example number two we have the limit as X approaches zero of sin x - x + 16 x Cub over X 5th so first things first identify which term or terms can be replaced with a McLaren Series so you just have to know that table you're on the lookout for typically e to the something s of x cosine X tan inverse of X natural log 1 plus X or 1 plus X raised to some power okay it's going to be sinx right do you remember the mclen series representation for sinx so I just think to myself s we know from trig s is an odd function so it involves X raised to odd powers divided by the corresponding odd number factorials and it's alternating all right so sin of X is x - x Cub over 3 factorial it's alternating so then it's going to be plus x 5 over 5 factorial - x 7th over 7 factorial Etc so that should be enough to get the ball rolling we are going to replace sin x in the limit with the first few terms from its mclen series all right good so here we go we've got the limit as X approaches zero let's write it out we have x - x cubed over 3 factorial + x 5 over 5 factorial - x 7th over 7 factorial plus dot dot dot - x + 16 x cubed oop I need a little more there we go and then over x to the 5th okay anything you notice we can cancel out yes look this is X we have a X beautiful um 3 factorial 3 factorial is 3 2 1 so that's 6 so I have Nega X Cub over 6 here's positive X Cub over 6 those cancel as well Oh Me Oh My and that looks like all the cancellation I can do in the numerator so now we've got limit X goes to zero let's see what's left up top x to the 5th over 5 factorial - x 7th over 7 factorial should we write a few more plus x to the you don't have to I find it therapeutic at times that's all over 11 factorial over X to 5th and then now you should be getting memories of the problem that we just did a couple minutes ago it looks pretty much identical you're obviously just going to simplify and cancel out x to the 5th from each of the terms in the numerator at this point we're Advanced so I am not rewriting x to the 5th under every little term you're in calc 2 suck it up and let's just do it mentally okay yes we can do it hold yourself to a high standard of Excellence all right so now we're going to have one over 5 factorial minus this will be x^2 over 7 factorial I'm just canc out the X to 5th from the numerator mentally oo X 4th over 9 factorial minus x to the 6 that's enough that is certainly enough same thing as before all of these terms right here forever onward will have x's in the numerators constants in the denominator so as X goes to zero yes this goes to zero this goes to zero everybody goes to zero do one for the dot dot dots they go to zero too all I'm left with now is 1 over five factorial 5 factorial is 5 4 3 2 1 which is 120 so this is 1 over 120 oh are you getting the hang of it they can get a wee bit spicier okay but uh I'll save those for like the end I think I think you guys can try this one on your own I really do why don't you pause the video and see if you can knock this one out okay did you pause the video I certainly hope so otherwise you're just cheating yourself okay so what term can be replaced with mclen series it's the tan inverse of X so let me write out the first few terms tan inverse of X we don't need the sigma notation version so it's just x minus So Tan inverse this is how I remember it looks so much like sign it just doesn't have the factorials in the denominator and why does it look like s instead of cosine I just go oh CU tangent is an odd function I mean that's not really why but that's just how I remember okay now there is a three in the front that's fine don't have a panic attack you're going to put parentheses around all of this when you replace it for Tan inverse of X in the limit and you're going to be fine fine okay it's like too aggressively dark of a purple okay so here we go limit as X goes to zero here's the numerator we have X Cub - 3x + 3 x - x Cub over 3 + x 5 over 5 - x to the 7th over 7 plus yada yada yada over x to the 5th okay now from here let's see shall we start Distributing and then cancelling it's up to you it's just how good is your mental math I could probably start cancelling now but I don't want to lose anybody so let's just take it one step further so this is X Cub - 3x plus I'm going to distribute are you ready yes we're ready Professor V we're so thrilled to do this good good good good so we have Plus 3x minus now those threes are going to cancel so we just have X cubed + 35s x to 5th - 37 X to the 7th Oh Me Oh my I have run out of space What would be next you know I just like one more term plus 3 9 x to 9th minus blah blah blah good over x to the 5th good okay anything wild happening um X Cub cancels 3x cancels and at this point you should be like oh this old scenario limit X goes to zero 35s x 5 - 37 x 7 + 3 9 x to the 99th minus blah blah blah dot dot dot over x to the 5th we've been here twice already so now it's time to just simplify divide everything in the numerator by x to the 5th will it always work out this way yes because someone sits there and artificially manufactures these limit problems okay I know I know minus this is going to be 37th x^2 + 39 x to 4th minus blah blah blah and then as the limit as X goes to zero we can see that these terms onward and all of the rest encapsulated in the ellipses will also go to zero so all we're left with is 3 fths you know this is why I always believe in doing homework a good amount of homework because perfect practice makes perfect and by now we've done three of these and I'm sure you're feeling a lot better than after we did the first one just cuz you see the process over and over and over you know know okay now I told you it was going to get spicier so here we go use series to evaluate the limit which terms can be replaced with a mclen series in this case we have two wow cosine of x as well as e to the X okay so let's list them out off to the side before we get rolling all right so cosine of x cosine is an even trig function so I just think of it it's all even um powers on the variables and even factorials corresponding in the denominator it's also alternating so it's 1 - x^2 over 2 factorial + x 4th over 4 factorial also if you take the derivative of the mclen series representation for sign you get cosine isn't math just beautiful okay and then e to the x is it's basically absolute value so no alternating s plus cosine so everybody odds and evens all together living in Perfect Harmony 1 + x + x^2 over 2 factorial + x Cub over 3 factorial + x okay that's enough right yeah we could go on forever literally okay so we're going to replace um both cosine X and E to x with this expanded version of their McLaren series okay in the limit so now we have the limit X approaches zero let's see what's going on in the numerator 1 minus let me switch colors for your learning pleasure 1 - x^2 over 2 factorial + x 4 over 4 factorial - x 6 over 6 factorial plus blah blah blah over 1 + x - and then now e to X is 1 + x + x^2 / 2 factorial + x Cub over 3 factorial plus x 4 over 4 factorial excuse me etc etc okay by now you should be Old Pros it's just this one might make you nervous CU you know you've got stuff going on infinite number of terms in the top and bottom but don't fret we got this the ones cancel out yeah okay just make sure we distribute the negative now so here we go in the numerator I've got x^2 over 2 factorial - x 4th over 4 factorial + x 6 over 6 factorial Etc and then what about in the denominator okay look the one cancels the X cancels and then now I'm also going to distribute this negative so what do we have here ax 2 over 2 factorial - x Cub over 3 factorial everybody's negative now in the denominator blah blah blah okay now in this case we don't just have one term to cancel out from the denominator right and typically when you have this kind of setup where you have um variables in the numerator variables in the denominator you just want want to divide out or cancel the greatest common factor so I can't take out any more than x² from these two terms so that's what I'm going to divide Everybody by not the highest power of X in the denominator you only do that for limits that go to Infinity in this case I'm just going to cancel out the GCF I can divide out an X squ from everyone so it's kind of like the lowest exponent even though it is the greatest common factor that's the biggest I can take out from everybody which happens to be the smallest power on X I know math terminology can be confusing okay so let's see if I divide Everybody by x^2 in the top I'm going to have 1 over 2 factorial - x^2 over 4 factorial + x 4 over 6 factorial minus blah blah blah and then in the denominator let's see I'm dividing Everyone by x^2 so -1 /2 factorial - x 3 factorial - x^2 over 4 factorial minus a bunch more terms and now here we're pretty much done so as X approaches zero this is just a constant this is just a constant all of the other terms have X in the numerator and a constant in the denominator so as X goes to zero all of the other terms will be going to zero and I don't have an indeterminant form no I don't have 0 over 0 I have 1 over 2 factorial over- 1 over 2 factorial remaining I don't have to write Lim anymore I have no more variables here 1 over 2 factorial / - 1 over 2 factorial why that's just one okay now I did mention you know sometimes or most of the time if they didn't tell me to use series to evaluate the limit we could do it much more easily it's 1 - cosine X over 1 + x - e x so if you want to check that you did it correctly let's see let's see if using our old methods it would come up quicker 1 - cosine X over 1 + x - E X I think this one will be nice to use ly T's rule so if you try direct substitution you have 1 - cosine of 0 is 1 over 1 + 0 - e 0 is 1 we have 0 over 0 so we can use lal's rule so this is the limit X approaches zero derivative of 1 is zero derivative of negative cosine X is positive sinx derivative of 1 is zero derivative of x is 1 minus this is e to the X can I plug in 0 for X I'm going to have 0 over 1 - 1 nope no good still so one more ly rule iteration and I think we should be good so now in the numerator we have cosine X over derivative of 1 is 0 e to X and now if I direct substitute I plug in 0 cosine of 0 is 1 over e to the 0 is 1 so this is - 1 oh cool so why aren't we just using lal's rule because did you read with the direction said the direction said use series to evaluate the limit so that's what we're going to do okay don't question it just do it um sometimes it does make it easier some of the earlier problems ly tall's role would have been nightmarish don't believe me go try it I did I waited until this problem to show you ly tall's rle worked for a reason because it wouldn't have been so cute on this one it wouldn't have been so fab on some of the others do you get what I'm saying it would have taken like so many iterations it could have gotten gnarly I waited to show you that this was a an alternative method on this guy cuz it came out nice okay so one more and then we'll call it a day this one is on my practice test for my class I'm going to do Shameless plug if you want more worksheets practice exams and stuff it's all on my patreon so if you want to sign up the link is in the description um there's different tiers but the mainly two so the basic one will give you pretty much everything you need okay use series to evaluate the limit ooh so which term or terms are we going to replace with the known mclen series it's the e to the 5x and you might be like wait a minute Professor B we do not have a mcloren series for e to the 5x well you do sort of so you know e to the x or you should is the sum n = 0 To infinity x to the N Over N factorial and then expanded we know what is looks like so if you have e to the -5x you're just going to come through and replace this x here with a 5x to the N okay over n factorial and then you do want to keep in mind that negative means it's going to be an alternating series so you pretty much have -1 to the N 5 to the n x the N Over N factorial and I'm going to list out the first few terms right now so we're ready to go so if n is zero it's going to be one 1 if n is 1 it's -5x over 1 factorial plus then 5^2 x^2 over 2 factorial - 5 Cub X Cub over 3 factorial Etc and that should be good for our purposes all right so then let's see what we have we have now the limit I think this I put this as extra credit on the on the final or on one of the exams seems familiar okay instead of e to the 5x 1 - 5x + let's write this as 25 x^2 over 2 factorial - 125 x cubed over 3 factorial o plus dot dot dot and then these guys just come along for the ride so Min -1 + 5x most likely they'll cancel that's how this has been going over X2 Oh indeed they do look minus one + one - 5x + 5x goodbye and then who's left oh this is just like all the other problems we did WOW 25x over 2 factorial - 125x cubed over 3 factorial plus blah blah blah on honestly that's enough over X2 you're going to divide out X2 from everybody who's left and then we'll be done so limit X goes to zero this is 25 over 2 factorial - 125x over 3 factorial + Well the next one you don't have to do a next one I'm just filling inspir 5 4 X2 over 4 factorial minus blah blah blah and then again as X goes to zero this term goes to zero this term goes to zero all of these guys would go to zero only lone survivors 25 over 2 factorial what's 2 factorial 2 1 yes so it's just 25 over 2 boom and we're done and that concludes the video I hope you enjoyed it give it a thumbs up if you found it helpful share it with your friends your classmates give me a subscribe and hit that thumbs up button it really helps support my channel and if you would like to see more content check out the rest of my YouTube channel I have everything organized into playlists I have full length video lectures for all of Cal 1 2 and three if you want to see what I'm up to on the daily you can follow me on Instagram and Tik Tok at mathtv with Professor V and I hope you have a great day you guys see you soon
17092	https://www.knowledgedoor.com/2/elements_handbook/speed_of_sound_part_2.html	\| \| \| \| \| --- --- \| \| \| \| \| \| \| \| \| \| Speed of Sound Navigation \| \| Aluminum to CopperDysprosium to NickelNiobium to PlutoniumPotassium to Zirconium \| \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| \| By Name \| \| \| By Symbol \| \| \| By Number \| \| \| \| \| Element \| \| Speed of Sound Click to see citations \| \| Notes \| --- --- \| \| \| \| Dysprosium \| \| \| \| solid, 293 K \| \| \| \| --- \| \| 2710 m/s \| \| \| calculated value \| \| Erbium \| \| \| \| solid, 293 K \| \| \| \| --- \| \| 2830 m/s \| \| \| calculated value \| \| Gadolinium \| \| \| \| solid, 293 K \| \| \| \| --- \| \| 2680 m/s \| \| \| calculated value \| \| Gallium \| \| \| \| liquid, 50 °C, longitudinal wave \| \| \| \| --- \| \| 2740 m/s \| \| \| \| \| Germanium \| \| \| \| solid \| \| \| \| --- \| \| 5400 m/s \| \| \| \| \| Gold \| \| \| \| solid \| \| \| \| room temperature, hard-drawn, extensional wave \| \| \| \| --- \| \| 2030 m/s \| \| \| \| \| 20 °C \| \| \| \| longitudinal wave \| \| \| \| --- \| \| 3240 m/s \| \| \| \| \| shear wave \| \| \| \| --- \| \| 1200 m/s \| \| \| \| \| 288 K to 293 K, annealed \| \| \| \| --- \| \| 1740 m/s \| \| \| \| \| 283 K \| \| \| \| --- \| \| 2110 m/s \| \| \| \| \| Hafnium \| \| \| \| α-hafnium, solid, 293 K \| \| \| \| --- \| \| 2740 m/s to 3280 m/s \| \| \| \| \| Helium \| \| \| \| gas \| \| \| \| 273.2 K \| \| \| \| --- \| \| 971.9 m/s \| \| \| \| \| 0 °C, 1 atm \| \| \| \| --- \| \| 965 m/s \| \| \| \| \| liquid \| \| \| \| -268.9 °C \| \| \| \| --- \| \| 180 m/s \| \| \| \| \| -269 °C, longitudinal wave \| \| \| \| --- \| \| 211 m/s \| \| \| \| \| Holmium \| \| \| \| solid, 293 K \| \| \| \| --- \| \| 2760 m/s \| \| \| calculated value \| \| Hydrogen \| \| \| \| gas \| \| \| \| 400.0 K, 1 atm \| \| \| \| --- \| \| 1518.0 m/s \| \| \| parahydrogen \| \| 380.0 K, 1 atm \| \| \| \| --- \| \| 1479.0 m/s \| \| \| parahydrogen \| \| 360.0 K, 1 atm \| \| \| \| --- \| \| 1439.0 m/s \| \| \| parahydrogen \| \| 340.0 K, 1 atm \| \| \| \| --- \| \| 1398.0 m/s \| \| \| parahydrogen \| \| 320.0 K, 1 atm \| \| \| \| --- \| \| 1355.0 m/s \| \| \| parahydrogen \| \| 300.0 K, 1 atm \| \| \| \| --- \| \| 1310.0 m/s \| \| \| parahydrogen \| \| 27 °C, 1 atm \| \| \| \| --- \| \| 1310 m/s \| \| \| \| \| 280.0 K, 1 atm \| \| \| \| --- \| \| 1263.0 m/s \| \| \| parahydrogen \| \| 270.0 K, 1 atm \| \| \| \| --- \| \| 1239.0 m/s \| \| \| parahydrogen \| \| 260.0 K, 1 atm \| \| \| \| --- \| \| 1214.0 m/s \| \| \| parahydrogen \| \| 250.0 K, 1 atm \| \| \| \| --- \| \| 1189.0 m/s \| \| \| parahydrogen \| \| 240.0 K, 1 atm \| \| \| \| --- \| \| 1163.0 m/s \| \| \| parahydrogen \| \| 230.0 K, 1 atm \| \| \| \| --- \| \| 1136.0 m/s \| \| \| parahydrogen \| \| 220.0 K, 1 atm \| \| \| \| --- \| \| 1110.0 m/s \| \| \| parahydrogen \| \| 210.0 K, 1 atm \| \| \| \| --- \| \| 1082.0 m/s \| \| \| parahydrogen \| \| 200.0 K, 1 atm \| \| \| \| --- \| \| 1054.0 m/s \| \| \| parahydrogen \| \| 190.0 K, 1 atm \| \| \| \| --- \| \| 1026.0 m/s \| \| \| parahydrogen \| \| 180.0 K, 1 atm \| \| \| \| --- \| \| 997.7 m/s \| \| \| parahydrogen \| \| 170.0 K, 1 atm \| \| \| \| --- \| \| 969.1 m/s \| \| \| parahydrogen \| \| 160.0 K, 1 atm \| \| \| \| --- \| \| 940.5 m/s \| \| \| parahydrogen \| \| 150.0 K, 1 atm \| \| \| \| --- \| \| 911.9 m/s \| \| \| parahydrogen \| \| 140.0 K, 1 atm \| \| \| \| --- \| \| 883.4 m/s \| \| \| parahydrogen \| \| 130.0 K, 1 atm \| \| \| \| --- \| \| 855.3 m/s \| \| \| parahydrogen \| \| 125.0 K, 1 atm \| \| \| \| --- \| \| 841.3 m/s \| \| \| parahydrogen \| \| 120.0 K, 1 atm \| \| \| \| --- \| \| 827.5 m/s \| \| \| parahydrogen \| \| 115.0 K, 1 atm \| \| \| \| --- \| \| 813.7 m/s \| \| \| parahydrogen \| \| 110.0 K, 1 atm \| \| \| \| --- \| \| 800.1 m/s \| \| \| parahydrogen \| \| 105.0 K, 1 atm \| \| \| \| --- \| \| 786.4 m/s \| \| \| parahydrogen \| \| 100.0 K, 1 atm \| \| \| \| --- \| \| 772.7 m/s \| \| \| parahydrogen \| \| 95.0 K, 1 atm \| \| \| \| --- \| \| 758.7 m/s \| \| \| parahydrogen \| \| 90.0 K, 1 atm \| \| \| \| --- \| \| 744.4 m/s \| \| \| parahydrogen \| \| 85.0 K, 1 atm \| \| \| \| --- \| \| 729.6 m/s \| \| \| parahydrogen \| \| 80.0 K, 1 atm \| \| \| \| --- \| \| 713.9 m/s \| \| \| parahydrogen \| \| 75.0 K, 1 atm \| \| \| \| --- \| \| 697.1 m/s \| \| \| parahydrogen \| \| 70.0 K, 1 atm \| \| \| \| --- \| \| 678.7 m/s \| \| \| parahydrogen \| \| 65.0 K, 1 atm \| \| \| \| --- \| \| 658.5 m/s \| \| \| parahydrogen \| \| 62.0 K, 1 atm \| \| \| \| --- \| \| 645.3 m/s \| \| \| parahydrogen \| \| 60.0 K, 1 atm \| \| \| \| --- \| \| 636.0 m/s \| \| \| parahydrogen \| \| 58.0 K, 1 atm \| \| \| \| --- \| \| 626.3 m/s \| \| \| parahydrogen \| \| 56.0 K, 1 atm \| \| \| \| --- \| \| 616.2 m/s \| \| \| parahydrogen \| \| 54.0 K, 1 atm \| \| \| \| --- \| \| 605.7 m/s \| \| \| parahydrogen \| \| 52.0 K, 1 atm \| \| \| \| --- \| \| 594.8 m/s \| \| \| parahydrogen \| \| 50.0 K, 1 atm \| \| \| \| --- \| \| 583.6 m/s \| \| \| parahydrogen \| \| 48.0 K, 1 atm \| \| \| \| --- \| \| 571.9 m/s \| \| \| parahydrogen \| \| 46.0 K, 1 atm \| \| \| \| --- \| \| 559.8 m/s \| \| \| parahydrogen \| \| 44.0 K, 1 atm \| \| \| \| --- \| \| 547.4 m/s \| \| \| parahydrogen \| \| 42.0 K, 1 atm \| \| \| \| --- \| \| 534.6 m/s \| \| \| parahydrogen \| \| 40.0 K, 1 atm \| \| \| \| --- \| \| 521.4 m/s \| \| \| parahydrogen \| \| 38.0 K, 1 atm \| \| \| \| --- \| \| 507.7 m/s \| \| \| parahydrogen \| \| 36.0 K, 1 atm \| \| \| \| --- \| \| 493.4 m/s \| \| \| parahydrogen \| \| 34.0 K, 1 atm \| \| \| \| --- \| \| 478.7 m/s \| \| \| parahydrogen \| \| 32.0 K, 1 atm \| \| \| \| --- \| \| 463.3 m/s \| \| \| parahydrogen \| \| 30.0 K, 1 atm \| \| \| \| --- \| \| 447.3 m/s \| \| \| parahydrogen \| \| 28.0 K, 1 atm \| \| \| \| --- \| \| 430.5 m/s \| \| \| parahydrogen \| \| 273.2 K, D2 \| \| \| \| --- \| \| 890 m/s \| \| \| \| \| liquid \| \| \| \| -252.9 °C \| \| \| \| --- \| \| 1101 m/s \| \| \| \| \| -258 °C, longitudinal wave \| \| \| \| --- \| \| 1242 m/s \| \| \| \| \| Indium \| \| \| \| solid, 293 K \| \| \| \| --- \| \| 1200 m/s to 1230 m/s \| \| \| calculated value \| \| Iridium \| \| \| \| solid, 293 K \| \| \| \| --- \| \| 4800 m/s to 4850 m/s \| \| \| calculated value \| \| Iron \| \| \| \| solid \| \| \| \| room temperature, cast, longitudinal wave \| \| \| \| --- \| \| 4994 m/s \| \| \| \| \| room temperature, cast, shear wave \| \| \| \| --- \| \| 2809 m/s \| \| \| \| \| room temperature, cast, extensional wave \| \| \| \| --- \| \| 4480 m/s \| \| \| \| \| room temperature, electrolytic, longitudinal wave \| \| \| \| --- \| \| 5950 m/s \| \| \| \| \| room temperature, electrolytic, shear wave \| \| \| \| --- \| \| 3240 m/s \| \| \| \| \| room temperature, extensional wave \| \| \| \| --- \| \| 5120 m/s \| \| \| \| \| room temperature, Armco, longitudinal wave \| \| \| \| --- \| \| 5960 m/s \| \| \| \| \| room temperature, Armco, shear wave \| \| \| \| --- \| \| 3240 m/s \| \| \| \| \| room temperature, Armco, extensional wave \| \| \| \| --- \| \| 5200 m/s \| \| \| \| \| 473 K \| \| \| \| --- \| \| 4720 m/s \| \| \| \| \| 373 K \| \| \| \| --- \| \| 5300 m/s \| \| \| \| \| 20 °C \| \| \| \| longitudinal wave \| \| \| \| --- \| \| 5957 m/s \| \| \| \| \| shear wave \| \| \| \| --- \| \| 3224 m/s \| \| \| \| \| 288 K to 293 K, wrought \| \| \| \| --- \| \| 5120 m/s \| \| \| \| \| 283 K to 293 K, wire \| \| \| \| --- \| \| 4910 m/s \| \| \| \| \| Krypton \| \| \| \| gas, 273.2 \| \| \| \| --- \| \| 213 m/s \| \| \| \| \| liquid \| \| \| \| --- \| \| 1120 m/s \| \| \| calculated value \| \| Lanthanum \| \| \| \| α-lanthanum, solid, 293 K \| \| \| \| --- \| \| 2460 m/s to 2490 m/s \| \| \| calculated value \| \| Lead \| \| \| \| solid \| \| \| \| room temperature, rolled, longitudinal wave \| \| \| \| --- \| \| 1960 m/s \| \| \| \| \| room temperature, rolled, shear wave \| \| \| \| --- \| \| 690 m/s \| \| \| \| \| room temperature, rolled, extensional wave \| \| \| \| --- \| \| 1210 m/s \| \| \| \| \| room temperature, annealed, extensional wave \| \| \| \| --- \| \| 1190 m/s \| \| \| \| \| 20 °C \| \| \| \| longitudinal wave \| \| \| \| --- \| \| 2160 m/s \| \| \| \| \| shear wave \| \| \| \| --- \| \| 700 m/s \| \| \| \| \| 288 K to 293 K \| \| \| \| --- \| \| 1260 m/s \| \| \| \| \| Lithium \| \| \| \| liquid \| \| \| \| --- \| \| 5000 m/s \| \| \| calculated value \| \| solid \| \| \| \| --- \| \| 6000 m/s \| \| \| \| \| Magnesium \| \| \| \| solid \| \| \| \| room temperature, annealed, longitudinal wave \| \| \| \| --- \| \| 5770 m/s \| \| \| \| \| room temperature, annealed, shear wave \| \| \| \| --- \| \| 3050 m/s \| \| \| \| \| room temperature, annealed, extensional wave \| \| \| \| --- \| \| 4940 m/s \| \| \| \| \| 20 °C \| \| \| \| longitudinal wave \| \| \| \| --- \| \| 5823 m/s \| \| \| \| \| shear wave \| \| \| \| --- \| \| 3163 m/s \| \| \| \| \| Manganese \| \| \| \| solid, 20 °C, longitudinal wave \| \| \| \| --- \| \| 4600 m/s \| \| \| \| \| Mercury \| \| \| \| liquid, 20 °C, longitudinal wave \| \| \| \| --- \| \| 1454 m/s \| \| \| \| \| solid, 203 K \| \| \| \| --- \| \| 2670 m/s \| \| \| \| \| Molybdenum \| \| \| \| solid \| \| \| \| room temperature, longitudinal wave \| \| \| \| --- \| \| 6250 m/s \| \| \| \| \| room temperature, shear wave \| \| \| \| --- \| \| 3350 m/s \| \| \| \| \| room temperature, extensional wave \| \| \| \| --- \| \| 5400 m/s \| \| \| \| \| 20 °C \| \| \| \| longitudinal wave \| \| \| \| --- \| \| 6475 m/s \| \| \| \| \| shear wave \| \| \| \| --- \| \| 3505 m/s \| \| \| \| \| Neodymium \| \| \| \| α-neodymium, solid, 293 K \| \| \| \| --- \| \| 2330 m/s \| \| \| calculated value \| \| Neon \| \| \| \| gas, 0 °C, 1 atm \| \| \| \| --- \| \| 435 m/s \| \| \| \| \| liquid \| \| \| \| --- \| \| 936 m/s \| \| \| calculated value \| \| Nickel \| \| \| \| solid \| \| \| \| room temperature, longitudinal wave \| \| \| \| --- \| \| 6040 m/s \| \| \| \| \| room temperature, shear wave \| \| \| \| --- \| \| 3000 m/s \| \| \| \| \| room temperature, extensional wave \| \| \| \| --- \| \| 4900 m/s \| \| \| \| \| 20 °C \| \| \| \| longitudinal wave \| \| \| \| --- \| \| 5700 m/s \| \| \| \| \| shear wave \| \| \| \| --- \| \| 3000 m/s \| \| \| \| \| 291 K \| \| \| \| --- \| \| 4970 m/s \| \| \| \| References (Click the next to a value above to see complete citation information for that entry) de Podesta, Michael. Understanding the Properties of Matter, 2nd edition. London: Taylor & Francis, 2002. Lide, David R., editor. CRC Handbook of Chemistry and Physics, 88th edition. Boca Raton, Florida: Taylor & Francis Group, 2008. Samsonov, G. V., editor. Handbook of the Physicochemical Properties of the Elements. New York: Plenum Publishing Corporation, 1968. Younglove, B. A. "Thermophysical Properties of Fluids. I. Argon, Ethylene, Parahydrogen, Nitrogen, Nitrogen Trifluoride, and Oxygen." Journal of Physical and Chemical Reference Data, volume 11, supplement 1, 1982, pp. 1–1 to 1–353. © KnowledgeDoor LLC\|Home\|Facebook\|Contact Us\|Terms of Use\|Privacy Policy\|Disclaimer
17093	https://www.ixl.com/math/lessons/customary-units-of-measurement	Skill previewSearch shortcut: SKIP TO CONTENT Customary units of measurement Print this lesson Share lesson: Share this lesson: Copy the link to this lesson Copy link Google Classroom Facebook Twitter What is the customary system? The customary system of measurement is used in the United States. Customary units are used to measure length, weight, and volume. Customary units of length You can use the customary units inch, foot, yard, and mile to measure length. A U.S. quarter is about 1 inch across. A ruler is 1 foot long. A baseball bat is about 1 yard long. Four laps around a running track is about 1 mile. You can also convert between these units. Use the table to see how. \| \| \| Length \| \| 1 foot (ft.) = 12 inches (in.) 1 yard (yd.) = 3 ft. 1 mile (mi.) = 1,760 yd. 1 mi. = 5,280 ft. \| Try some practice problems! Which customary unit of length is appropriate? Customary units of weight You can use the customary units ounce, pound, and ton to measure weight. A slice of bread weighs about 1 ounce. A football weighs about 1 pound. A small car weighs about 1 ton. You can convert between these units. Use the table to see how. \| \| \| Weight \| \| 1 pound (lb.) = 16 ounces (oz.) 1 ton = 2,000 lb. \| Try some practice problems! Which customary unit of weight is appropriate? Customary units of capacity (volume) You can use the customary units fluid ounce, cup, pint, quart, and gallon to measure capacity. Two bottles of nail polish hold about 1 fluid ounce in all. A teacup holds about 1 cup. A large drinking glass holds about 1 pint. A spray bottle of cleaner contains about 1 quart. A jug of milk contains 1 gallon. You can convert between these units. Use the table to see how. \| \| \| Volume \| \| 1 cup (c.) = 8 fluid ounces (fl. oz.) 1 pint (pt.) = 2 c. 1 quart (qt.) = 2 pt. 1 gallon (gal.) = 4 qt. \| Try some practice problems! Which customary unit of volume is appropriate? Related
17094	https://math.stackexchange.com/questions/4061181/probability-question-from-brilliant	dice - Probability question from Brilliant - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Probability question from Brilliant Ask Question Asked 4 years, 6 months ago Modified4 years, 6 months ago Viewed 366 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Which is more likely? You roll two dice 5 times and, every time, one of the two comes up as 1 and the other as 6. You roll 10 dice all at once. 5 come up as 1s and the other 5 come up as 6s. The answer is 2 and I'm confused by it. Here's my thinking. When rolling two dice the probability of getting 1 and 6 or 6 and 1 is 2⋅(1/6⋅1/6)=2/36=1/18.2⋅(1/6⋅1/6)=2/36=1/18. This means that probability of getting this 5 times in a row is (2/36)5(2/36)5. In the second case there are 2 5 2 5 dice combinations of 1s and 6s where we get 5 of each. The probability of every combination is (1/6)10(1/6)10 and so the probability of the second case is 2 5⋅(1/6)10=(2/36)5 2 5⋅(1/6)10=(2/36)5 which is exactly the same as the first case. What's wrong with my thinking? probability dice Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Mar 14, 2021 at 12:11 amWhy 211k 198 198 gold badges 283 283 silver badges 505 505 bronze badges asked Mar 14, 2021 at 7:13 Deniss SudakDeniss Sudak 21 2 2 bronze badges 3 2 The answer to the first one is correct . But for the second one number of combinations is not 2 5 2 5 rather it (10 5)(10 5)Vinanth S Bharadwaj –Vinanth S Bharadwaj 2021-03-14 07:19:35 +00:00 Commented Mar 14, 2021 at 7:19 1 Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.José Carlos Santos –José Carlos Santos 2021-03-14 07:22:33 +00:00 Commented Mar 14, 2021 at 7:22 The first is more likely, since I'm much more likely to be rolling two dice at a time than ten.Carmeister –Carmeister 2021-03-14 15:45:19 +00:00 Commented Mar 14, 2021 at 15:45 Add a comment\| 6 Answers 6 Sorted by: Reset to default This answer is useful 8 Save this answer. Show activity on this post. Here's some easy intuition as to why 2 2 has to be more probable than 1 1. Suppose you roll the 10 10 dice two at a time. If the first pair comes up as a pair of 1 1 s, you're out of the running for scenario 1 1, but you're still alive for scenario 2 2. But there is no sequence of rolls you can make that knocks you out of the running for scenario 2 2 but leaves it possible to achieve scenario 1 1. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 14, 2021 at 7:32 Robert ShoreRobert Shore 26.4k 3 3 gold badges 21 21 silver badges 49 49 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Look at the type of patterns of two pictures, and it will be quite clear: Two dice rolled 5 times each 1 6 6 1 1 6 6 1 6 1 1 6 6 1 1 6 6 1 6 1 Ten dice rolled together 1 1 6 6 6 1 1 6 6 1 1 1 6 6 6 1 1 6 6 1 The second case can generate all of the types the first one can, and then more... Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 14, 2021 at 9:58 answered Mar 14, 2021 at 9:52 true blue aniltrue blue anil 49.2k 4 4 gold badges 31 31 silver badges 64 64 bronze badges Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. Which has more ways of occurring? Obtaining five ‘1’s and five ‘6’s among ten concurrent die rolls in a row, with the ‘1’s spaced apart. Obtaining five ‘1’s and five ‘6’s among ten concurrent die rolls in a row, with no restriction on the numbers’ order of appearance. (Whether the die rolls are successive or concurrent is immaterial to the answer.) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 14, 2021 at 14:01 answered Mar 14, 2021 at 9:32 ryangryang 1 Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. the second is (10 5,5)×(1 6)5×(1 6)5=10!5!5!×(1 6)5×(1 6)5(10 5,5)×(1 6)5×(1 6)5=10!5!5!×(1 6)5×(1 6)5 thus it is most likely being 10!5!5!>2 5 10!5!5!>2 5 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 14, 2021 at 7:24 tommiktommik 33.2k 4 4 gold badges 18 18 silver badges 36 36 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. The mistake is subtle: When rolling the dice 5 times in 1), you are required for each roll to include a 1 and a 6. However, if you roll ten dice, you allow for the two dice to be 6, or 1, or both. So the correct probability for the first five dice in 2) is in fact higher, than the first five dice in 1). So while your calculation in 1) is correct, yours in 2) is incorrect. The probability of 2) will range between (2 6)9∗(1 6)1(2 6)9∗(1 6)1 at most, and (2 6)5∗(1 6)5(2 6)5∗(1 6)5 at worst, depending on what you roll. Whatever the actual answer, 2) will always have a higher probability. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Mar 14, 2021 at 7:39 veryverdeveryverde 252 1 1 silver badge 8 8 bronze badges Add a comment\| This answer is useful 0 Save this answer. Show activity on this post. Alternative approach. Label the dice D 1,D 2,⋯,D 10 D 1,D 2,⋯,D 10. Let E 1 E 1 denote the event represented by problem 1. Let E 2 E 2 denote the event represented by problem 2. Assume that for the purposes of the first problem, when the dice are rolled, D 1,D 2 D 1,D 2 are rolled together, D 3,D 4,D 3,D 4, are rolled together, and so forth. Assume that each pair of dice, re above paragraph are rolled. E 2 E 2 will be deemed a success if exactly 5 5 of the 10 10 dice are a 1 1. Note that ifE 2 E 2 does not occur, then E 1 E 1 can not occur. Assume that E 2 E 2, succeeded. Given that, what is the probability that E 1 E 1 succeeded? Without loss of generality (WLOG) D 1 D 1 was a 1 1. Then the chance that D 2 D 2 was a 6 6 is (5/9)(5/9). Note that if D 1 D 1 had been a 6 6, the chance that D 2 D 2 was a 1 1 would similarly have been (5/9)(5/9). Assuming that D 1 D 1 and D 2 D 2 have been navigated successfully, then WLOG, D 3 D 3 was a 1 1 and the chance that D 4 D 4 was a 6 6 is (4/7)(4/7). Continuing to the end, you see that under the assumption that E 2 E 2 succeeded, the chance that E 1 E 1then succeeded is (5/9)×(4/7)×(3/5)×(2/3).(5/9)×(4/7)×(3/5)×(2/3). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 14, 2021 at 8:48 answered Mar 14, 2021 at 8:33 user2661923user2661923 42.9k 3 3 gold badges 22 22 silver badges 47 47 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions probability dice See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 0Game Dice Probability 22Probability: 6 Dice are rolled. Which is more likely, that you get exactly one 6, or that you get 6 different numbers? 4Why are the probability of rolling the same number twice and the probability of rolling pairs different? 1Dice probability with limited reroll 6Probability of rolling dice twice 16Rolling 2 2 dice: NOT using 36 36 as the base? 0How to predict the likelihood of multiple future occurrences based on probability 1Binomial probability doesn't give the correct number? 5Keep rolling two dice until the cumulative sum hits 1000 2Probability of at least k same specific digits in an n-digit sequence? Hot Network Questions Interpret G-code Another way to draw RegionDifference of a cylinder and Cuboid Is direct sum of finite spectra cancellative? Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Bypassing C64's PETSCII to screen code mapping Xubuntu 24.04 - Libreoffice Exchange a file in a zip file quickly Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward" Countable and uncountable "flavour": chocolate-flavoured protein is protein with chocolate flavour or protein has chocolate flavour Discussing strategy reduces winning chances of everyone! Drawing the structure of a matrix how do I remove a item from the applications menu Why are LDS temple garments secret? Do we need the author's permission for reference What's the expectation around asking to be invited to invitation-only workshops? How do you emphasize the verb "to be" with do/does? ConTeXt: Unnecessary space in \setupheadertext Is there a way to defend from Spot kick? A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man Who is the target audience of Netanyahu's speech at the United Nations? Gluteus medius inactivity while riding Making sense of perturbation theory in many-body physics Why does LaTeX convert inline Python code (range(N-2)) into -NoValue-? What is the meaning and import of this highlighted phrase in Selichos? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
17095	https://helpyourmath.com/immersion/elementary-algebra-chapters/chapter3	Immersion The online open educational resources(OER). Course Information Open Stax Elementary Algebra Homework Exam Elementary Algebra Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Subscribe Subscribe Our YouTube Channel for more math Video Chapter 3 Linear Equations and Inequalities in Two Variables Detail for Chapter 3 3.1 Paired Data and Graphing Ordered Pairs 3.2 Solution to Linear Equations in Two Variables 3.3 Graphing Linear EQUATIONS In Two Variables 3.4 More on Graphing: Intercepts 3.5 The Slope of a Line 3.6 Finding the Equation of a Line OER Book Page# ( Click Below Numbers) 404 - 409 & 419 - 420 410 - 422 424 - 443 444 - 459 459 - 486 512 - 529 Introduction to the Coordinate Plane Linear Equations in Two Variables - Different Forms Find the Slope and Y-intercept from an Equation of a Line CEAFE Problem Example More Examples and Practice Problems Slope and Y-intercept Find the Equation of a Line Given That You Know Two Points - Slope Intercept Form CEAFE Problem Example Calculate the Slope from Two Points Find the Slope of a Line Given That You Know Two Points More Examples and Practice Problems Find the Equation of a Line Given That You Know Two Points Video - Another Method Point Slope Form - Find the Equation of a Line Given That You Know Two Points Graphing Linear Equations CEAFE Problem Example How to find X, Y Intercepts? X and Y Intercepts Slopes, Horizontal Line and Vertical Line CEAFE Problem Example More about Slope Different Types of Slopes Finding the Slope of a Line from its Graph CEAFE Problem Example Icon Copyright by HelpYourMath©2019 All rights reserved Quick Menu Help Your Math Course Information Textbook Elementary Algebra Homework Exam Subscribe
17096	https://www.hajim.rochester.edu/optics/undergraduate/senior-design/pdf/spr18/coherence-team_coherence_final_ddd.pdf	Coherence Length Measurement System Design Description Document 00007 Rev G page 1 Coherence Length Measurement System Design Description Document ASML / Tao Chen Faculty Advisor: Professor Thomas Brown Pellegrino Conte (Scribe & Document Handler) Lei Ding (Customer Liaison) Maxwell Wolfson (Project Coordinator) Document Number 00007 Revisions Level Date G 5-06-2018 This is a computer-generated document. The electronic master is the official revision. Paper copies are for reference only. Paper copies may be authenticated for specifically stated purposes in the authentication block. Authentication Block Coherence Length Measurement System Design Description Document 00007 Rev G page 2 Rev Description Date Authorization A Initial DDD 1-22-2018 All B Edited System Overview 2-05-2018 All Added Lab Results C Edited System Overview 2-19-2018 All Added New Results D Added New Lab Results 2-26-2018 All Updated FRED Progress E Edited System overview 4-02-2018 All Finalized cost analysis Added new results F Edited Cost Analysis 4-20-2018 All Added new results Added to code analysis G Added Final Results 5-06-2018 All Added Customer Instructions Added to Code Coherence Length Measurement System Design Description Document 00007 Rev G page 3 Table of Contents Revision History 2 Table of Contents 3 Vision Statement 4 Project Scope 4 Theoretical Background 5-7 System Overview 8-10 Cost Analysis 11-13 Spring Semester Timeline 14-15 Lab Results 16-22 FRED Analysis 23-26 Visibility Analysis 27 Design Day Description 28 Conclusions and Future Work 29 Appendix A: Table of All Lab Results 30-40 Appendix B: Visibility Processing Code 41-43 Appendix C: Customer Instructions 44-49 References 50 Coherence Length Measurement System Design Description Document 00007 Rev G page 4 Vision Statement: This projects’ goal is to design and assemble an interferometer capable of measuring and reporting information regarding the coherence length of a laser. The device will be used to characterize the lasers used in the semiconductor industry to improve the performance of lithography systems. Project Scope: Interferometer Design: Our responsibilities include the design and development of a working prototype interferometer capable of measuring and reporting visibility measurements of a laser over a path length difference of 500 mm. This system should be capable of calculating visibility of the interference pattern every 10 um of path length difference and be able to calculate a visibility of at least 0.01. The device must be able to analyze multispectral lasers. Additionally, the system is to be housed in a maximum enclosed area of 1 m x 1 m x 0.5 m and is to operate in a lab setting. The laser will be introduced to the system from an optical fiber and the gathered data will be exported to a connected computer. We are not responsible for vibration isolation. The budget for our system is $5,000. Delivery of Device In addition to the building the prototype interferometer, our team is responsible for the delivery of the system to our customer. This will be done by mailing: 1) The breadboard with all the mechanical components still mounted on the breadboard, but with the optics removed 2) All optics in their original cases 3) The detector ASML will pay for the shipping of all the components. Coherence Length Measurement System Design Description Document 00007 Rev G page 5 Theoretical Background: The “coherence” of a source, describes the degree to which there exists “a fixed phase relationship between the electric field values at different locations or at different times” . Characterizing the coherence of a source is important, as it is indicative of the lights ability to interfere. When two coherent waves are combined, the result is an interference pattern, where the relative phase relationship of the waves at different locations, results in fringes (areas of maximal and minimal intensity). On the other hand, when two incoherent waves are combined, the lack of a relative phase relationship, results in no distinguishable fringes and rather a uniform intensity pattern. It should be noted, that in reality no source exists that is entirely coherent or incoherent; all physical sources have varying degrees of coherence that depend upon how long a relative phase relationship can be maintained . Figure 1: Top left is a representation of the phase relationship of polychromatic light. Top right is a representation of the phase relationship of low coherence monochromatic light. On the bottom is a representation of the phase relationship of high coherence monochromatic light . The term used to describe the longevity of the phase relationship is temporal coherence. Temporal coherence can be quantified in terms of coherence time which relays the maximum delay in which a wave can be combined with a copy of itself and still produce an interference pattern. The coherence time, can be expressed in terms of coherence length, where Coherence Length equals the Coherence Time multiplied by the Speed of Light. In words, it can be expressed that the “coherence length is a measure of the largest optical path length diﬀerence two waves can sustain before they can no longer interfere” . Coherence Length Measurement System Design Description Document 00007 Rev G page 6 One way to determine the coherence length of a source is by using a Michelson Interferometer. A Michelson Interferometer is an amplitude splitting interferometer, that takes a collimated beam and divides it into two paths. Part of the light goes towards and is reflected back by the mirror in the measurement arm and the other part of the light goes toward and is reflected back by the mirror in the reference arm. The two beams are then recombined to create an interference pattern. To determine the coherence length, the optical path difference (OPD) between the two arms is increased until interference is no longer observed. Instead of relying on a subjective approach to estimate when the source is no longer coherent, the strength of the interference pattern can be quantified using the metric of visibility. The visibility of a source is the difference in the maximum and minimum intensity, divide by the sum of the maximum and minimum intensity. A visibility of 1 indicates complete coherence, while a visibility of 0 indicates complete incoherence. While different values of decay can be used to quantify the coherence length, the most common value used is when the fringe visibility is 1/e or approximately 37% . It should be noted, that the Fourier Transform of the source irradiance, can be used to determine the interferograms visibility . Figure 2: On the left is an image of the longitudinal modes for a HeNe laser with a cavity length of 20 cm. On the right is an image of the longitudinal modes for a HeNe laser with a cavity length of 80 cm . With the concept of coherence length established, the source properties that influence this metric can now be discussed. Directly addressing laser sources, there exists a strong degree of coherence, on account of stimulated emission creating photons that have a fixed phase relationship. The coherence length of a laser, depends upon the number of longitudinal modes (which are modes determined by the axial dimensions of the resonant cavity) and therefore the shape of the spectrum curve . A narrow bandwidth results in a longer coherence length and a broad bandwidth results in a shorter coherence length. Additionally, lasers that sustain multi-longitudinal modes have resurgence peaks of visibility. Coherence Length Measurement System Design Description Document 00007 Rev G page 7 In a multimode helium-neon (HeNe) laser, the typical coherence length is about 20cm. However, in a singlemode HeNE lasers, the typical coherence length exceeds 100 m . A standard laser diode usually has a shorter coherence length of less than a millimeter. A standard light emitting diode (LED), would have a very short coherence length on the order of microns. Shape of Spectrum Visibility vs. OPL Figure 3: Depiction of how the shape of the spectrum influences the visibilty as the optical path length is increased. Going left to right is a narrow spectrum source, a broad spectrum source, and a multi-spectral source . Coherence Length Measurement System Design Description Document 00007 Rev G page 8 System Overview: Design and Performance Constraints: 1. Path difference range: 500 mm 2. Path difference incrementation: 10 µm 3. Minimum visibility measurable: 0.01 4. Wavelength range: 500-900 nm 5. Interface: FC/PC connector 6. Data output: raw data of visibility over entire measurement range 7. Packaging size: 1 x 1 x 0.5 m Final layout of the Device: Figure 4: Set-up of current interferometer design. Looking at Figure 4, the final dimensions of our interferometer can be seen to be 450 mm x 450 mm. A total of 7 mirrors are used in the system and are labeled M0 through M6. In Figure 4, BS indicates a beamsplitter and PL indicates a polarizer. Coherence Length Measurement System Design Description Document 00007 Rev G page 9 In one arm of the interferometer, hereafter called the grating arm, is a 25 mm x 50 mm grating with 31.6 grooves/mm and a blaze angle of 63°. This grating is tilted by 63° to be in the Littrow configuration, such that light is made perpendicular to each line of the grating and, therefore, reflected back with fine steps of OPL information (See Figure 5). Additionally, the grating is also given a very small vertical tilt which allows for a continuous measurement of OPD through each grating step. For the final design, on account of the grating being tilted, the maximum OPL that could be achieved from the furthest beam reflected back by the grating, in comparison to the closest beam reflected by the grating, was 44.6 mm. Of this information, only 40 mm was considered since the light from the edge of the grating proved to not be as useful. Figure 5: Drawing of how grating act as a staircase reflector. In the other arm of the interferometer, hereafter called the measurement arm, is a seven mirror configuration. By rotating M0 towards the other mirrors, it is possible to measure the visibility over the entire measurement range. This is accomplished by placing M0 such that it and the top of the grating are at an equal path length from the BS. Since the interference of M0 and the grating gives visibility information over a ΔOPL range of 0 mm to 80 mm, M1 is placed 40 mm from M0. When M0 is oriented towards M1, the interreference pattern created between M1 and the grating will therefore give visibility information over the ΔOPL range of 80 mm to 160 mm. This process is continued with all subsequent mirrors to give a maximum ΔOPL measurement of 560 mm (See Figure 6). Coherence Length Measurement System Design Description Document 00007 Rev G page 10 Figure 6: Drawing mathematically depicting how measurement arm extends measurement capabilities over desired range. In order to resolve the discrepancy between the light intensity from each arm, on account of the fact that the mirrors in the measurement arm are more reflective than the grating in the grating arm, our design uses a two polarizer approach. A stationary polarizer is placed in the measurement arm and a rotatable polarizer is placed in front of the detector. By rotating the polarizer in front of the detector it is possible to make the detector receive approximately equal strength beams from the two arms of the interferometer. Enclosure and Mounting: The device will be mounted upon its own breadboard and enclosed by a carboard box to prevent stray light from entering the outside environment. This entire system will be mounted onto a vibration isolation table by the customer. Coherence Length Measurement System Design Description Document 00007 Rev G page 11 Cost Analysis: Optomechanical Components: Part Company Product Number Qty. Cost per Unit Total Cost Collimating Lens Mount Thorlabs FMP1 1 $ 16.12 $ 16.12 Imaging Lens Mount Thorlabs FMP1 1 $ 16.12 $ 16.12 Beam Splitter Mount Thorlabs KM100S 1 $ 80.58 $ 80.58 Grating Mount Part 1 Thorlabs FP90 1 $ 67.83 $ 67.83 Grating Mount Part 2 Thorlabs KGM40 1 $ 134.64 $ 134.64 Grating Mount Part 3 Thorlabs KM100 1 $ 38.70 $ 38.70 Mirrors 0, 1-5 Mounts Thorlabs KM100S 6 $ 80.58 $ 483.48 Mirror 6 Mount Thorlabs KM100 1 $ 38.70 $ 38.70 Rotation Stage Thorlabs ELL8K/M 1 $ 391.68 $ 391.68 M6 Cap Screws (Pack of 25) Thorlabs SH6MS12 1 $ 8.11 $ 8.11 20 mm Posts (Packs of 5) Thorlabs TR20/M-P5 2 $ 21.33 $ 42.66 20 mm Posts (Single) Thorlabs TR20/M 3 $ 4.74 $ 14.22 100 mm Post (Single) Thorlabs TR100/M 1 $ 5.87 $ 5.87 Clamping Forks (Packs of 5) Thorlabs CF125-P5 2 $ 42.45 $ 84.90 Clamping Forks (Single) Thorlabs CF125 4 $ 8.95 $ 35.80 20 mm Post Holders (Pack of 5) Thorlabs PH20/M-P5 2 $ 35.15 $ 70.30 20 mm Post Holders (Single) Thorlabs PH20/M 3 $ 7.03 $ 21.09 75 mm Post Holder (Single) Thorlabs PH75/M 1 $ 8.27 $ 8.27 Post Holder Base (Pack of 5) Thorlabs BE1/M-P5 2 $ 47.43 $ 94.86 Post Holder Base (Single) Thorlabs BE1/M 4 $ 9.49 $ 37.96 Coherence Length Measurement System Design Description Document 00007 Rev G page 12 Breadboard Thorlabs MB4545/M 1 $ 273.36 $ 273.36 Fiber Plug-in Mount Thorlabs FMP1 1 $ 16.12 $ 16.12 Fiber Plug-in Port Thorlabs S1FC 1 $ 29.58 $ 29.58 Glass Windows for Polarizer Edmund Optics 84-455 4 $ 125.00 $ 500.00 Fixed Polarizer Mount Thorlabs FMP1 1 $ 16.12 $ 16.12 Rotating Polarizer Mount Thorlabs RSP1 1 $ 86.19 $ 86.19 Detector Mount Thorlabs XT34TR3/ M 1 $ 42.84 $ 42.84 $ 2,656.10 Table 1: Cost breakdown of optomechanical components. Interferometer Components: Part Company Product Number Qty. Cost per Unit Total Cost Collimating Lens Edmund Optics 49-361 1 $ 96.50 $ 96.50 Beam Splitter Thorlabs BSW26R 1 $ 294.78 $ 294.78 Blazed Grating Thorlabs GE2550-0363 1 $ 223.38 $ 223.38 Mirrors 1-5 Thorlabs PFSQ10-03-P01 5 $ 53.30 $ 266.50 Mirror 6 Thorlabs PF10-03-P01 1 $ 52.02 $ 52.02 Rotating Flat Mirror Thorlabs PFR10-P01 1 $ 83.39 $ 83.39 Wire Grid Polarizing Film Edmund Optics 34-254 2 $ 55.00 $ 110.00 Detector High Point Scientific Inc. ASI183MM 1 $ 629.00 $ 629.00 Imaging Lens Edmund Optics 49-361 1 $ 96.50 $ 96.50 $ 1,852.07 Table 2: Cost breakdown of interferometer components. Coherence Length Measurement System Design Description Document 00007 Rev G page 13 Complete Cost Analysis: Component Final Cost Optomechanical $ 2,656.10 Interferometer $ 1,852.07 Total $ 4,508.17 Table 3: Cost breakdown of combined components. Coherence Length Measurement System Design Description Document 00007 Rev G page 14 Spring Semester Timeline January 1/21-1/27 • (1/25) Met with customer to discuss direction of project. • (1/26) Met with advisor to discuss current design concerns and questions posed by customer. • Decision made to pursue interferometer design utilizing a blazed grating. 1/28-2/3 • Assembled a simplified version of current design to gain insight into design practicality issues. • Investigated possibility of a custom made grating via diamond turning. February 2/4-2/10 • Investigated algorithm to optimize visibility measurements. • Investigated best mounting method for grating. • Learned basics of and began modelling with FRED. • Began testing the use of multiple mirrors in reference arm. 2/11-2/17 • (2/14) Met with customer to provide update on project. • (2/16) Met with advisor to discuss design and lab set-up. • Investigated use of OAP mirror for large beam collimation. • Investigation of alternate way to divide up reference arm. • Modeled current set-up with FRED. 2/18-2/24 • Updated lab set-up and began testing the rotation mirror method. • Updated FRED analysis by creating a new grating and implementing a more detailed/realistic source. 2/25-2/28 • Used FRED to determine test if rotation stage possessed adequate specifications. • Found suitable achromatic doublet to replace OAP mirror. March 3/1-3/3 3/4-3/10 • (3/5) Met with Advisor to discuss ways to improve system design and methods of data analysis. • Investigated using a tarp as a “soft” enclosure. • Started writing code for data analysis. 3/11-3/17 • Spring Break. 3/18-3/24 • (3/23) Met with customer to provide an update on progress and ask a few questions. • Updated lab set-up to test if manipulation of polarization characteristics could create equal output beam intensity from both arms. 3/25-3/31 • (3/30) Met with Advisor to discuss a finalization of components. • (3/31) Met with Professor Eastman to discuss detector choice. Coherence Length Measurement System Design Description Document 00007 Rev G page 15 • Decided on the two polarizer method to balance the two arms. • Decided on a detector. • Performed final mathematical calculations for spacing of components on breadboard • Finalized BOM. • Continued refining code to analyze visibility. April 4/1-4/7 • (4/1) Sent completed BOM to customer for ordering. • Continued refining computer analysis method to determine visibility 4/8-4/14 • Measured coherence length of a short coherence length source • Continue refining computer analysis method to determine visibility 4/15-4/21 • (4/17) Received confirmation from customer that parts were delivered to him and that they are currently in transit to the U of R. • (4/19) Received all parts except for detector • Roughly assembled prototype components 4/22-4/28 • Calibrated and determined angle for rotation stage • Performed fine adjustments of prototype set-up • Began tolerancing prototype • Began writing customer instructions for operation 4/29-4/30 • Printed poster for Senior Design Day • Continued testing prototype • Finished writing customer instructions • (5/4) Senior Design Day May 5/1-5/5 5/6-5/12 • Package and ship materials to customer Table 4: Spring Semester Timeline Coherence Length Measurement System Design Description Document 00007 Rev G page 16 Lab Results: Final Testing Set-up: Figure 7: Image of testing set-up. The system tested was the final prototype, with all of the components specified in the above listed cost analysis. Also, used was an aperture that was borrowed from the teaching lab. The source used to run the experiment was a fiber-coupled red laser with a wavelength of λ=650 nm. The purpose of this final experiment was to produce the best images possible, in order to determine the limits of our system. One major factor in producing these best possible images, was the fact that our ordered detector, which had a significantly higher resolution, had arrived. Another factor in yielding the best images possible, was that in a previous experiment our group noticed that part of an additional unwanted diffraction order from the grating was making it to the detector. To eliminate this light, our group used an aperture to block out this light. Additionally, our team re-checked and improved the degree of collimation of the light, by using a shearing interferometer. The procedure of our experiment followed how the interferometer is intended to be used. First, the rotation stage was put into the M0 orientation and a power meter was used to measure the power from each arm. The rotatable polarizer was adjusted until the two arms were balanced. Next, the mirror was rotated into the various path length configurations and images were captured. Finally, the images were processed using the visibility code. Coherence Length Measurement System Design Description Document 00007 Rev G page 17 Final Results: Mirror Image M0 M1 M2 Coherence Length Measurement System Design Description Document 00007 Rev G page 18 M3 M4 M5 M6 Table 5: Images of interference patterns created by interferometer in the various mirror configurations. Coherence Length Measurement System Design Description Document 00007 Rev G page 19 Figure 8: Plot of Visibility vs. ΔOPL. Image was created by applying the analysis code described in Appendix B, to the images in Table 5. From looking at the images contained in Table 5 and the plot presented in Figure 8, their exist a couple of points to be discussed. First, it is important to observe global decay in fringe visibility that can be seen as the OPD is increased. In regard to subjectively viewing the fringe contrast shown in Table 5 by eye, M0 clearly has the best contrast, while M2, M3, and M4 show a decay in fringe visibility, and, finally, M5 and M6 display no fringes at all. This observation is mostly confirmed by Figure 8, in that the visibility decays from M0-M4; however, M5 and M6 show a spiked visibility that is clearly not indicated by our images. We attribute this strange result to be a byproduct of our visibility code picking up on artifacts within the M5 and M6 images, which represents a new problem that will need to be addressed. Second, it is important to note the high level of noise that exists over each mirror measurement region. We attribute a significant portion of this noise be a result of our fringes having a strange bend. Despite all of our attempts at adjusting the mirrors so that the fringes would be very straight, we could not eliminate this deformation of the fringes. As soon as any adjustment to move past the point of inflection was made, the large bend would appear and distort the fringes. This in turn, negatively impacted our visibility plot, as a uniform frequency of the fringes is a necessary condition for achieving good results with our visibility code. Coherence Length Measurement System Design Description Document 00007 Rev G page 20 Summary of Significant Results from Earlier Lab Sessions: The following table is a condensed summary of our most important results from our lab sessions. If desired for reference, all relevant images captured in the lab are attached in appendix A. Date Testing Conditions Significant Images and Discussion 1/28-2/3 Light Source 532 nm Laser Pointer The purpose of this initial lab session was to test the possibility of producing interference fringes using a reflective grating. In the above image it can be seen that our initial tests were promising in that they did yield the creation of prominent interference fringes. Grating Arm Reflector Blazed Grating (20 grooves/ mm, 26° 45’ blaze) Measurement Arm Reflector Single Flat Mirror 2/11-2/17 Light Source 633 nm HeNe Laser The purpose of this lab session was to test the possibility of using a tiered reference mirror structure. Our results indicate that this design will not work for our system, as the interference patterns from each reference mirror were not observable at the same time. Additionally, the portion of the detector where the two reference beams overlapped became unusable. Grating Arm Reflector Blazed Grating (20 grooves/ mm, 26° 45’ blaze) Measurement Arm Reflector Two Flat Mirrors Coherence Length Measurement System Design Description Document 00007 Rev G page 21 3/18-3/24 Light Source 633 nm HeNe Laser The purpose of this lab session was to further confirm our initial positive results seen when using the rotation mirror method and also to produce high-contrast fringes that could be used for image analysis. In the lab, we were able to create very straight and vertical fringes that became the first images from our system that we were actually able to analyze using our code. Grating Arm Reflector Blazed Grating (20 grooves/ mm, 26° 45’ blaze) Measurement Arm Reflector Three Flat Mirrors in Rotation Method Configuration 4/8-4/14 Light Source Red Laser Pointer M0 M1 The purpose of this lab session was to test a version of our final design with a short coherence length source. In a global sense, our results were promising in that our system did capture how the visibility was decreasing with greater OPD. On a more local scale, the visibility analysis of each image by itself, was found to be hindered by excessive noise that we attributed to vibration and the non-ideal quality of the lab equipment. Grating Arm Reflector Blazed Grating (20 grooves/ mm, 26° 45’ blaze) Measurement Arm Reflector Three Flat Mirrors in Rotation Method Configuration Coherence Length Measurement System Design Description Document 00007 Rev G page 22 4/22-4/28 Light Source Fiber-Coupled Red Laser The purpose of this lab session was to test the capabilities of our completed prototype except for the detector, which had not yet arrived. Similar to the results achieved using a short coherence length source, the global trend of visibility decreasing as OPD increased was observed, however on a local level our visibility for each individual mirror was still filled with excessive noise. Our group attributes this noise to unwanted diffraction in our system and also to the low resolution of the camera as our ordered detector had not yet arrived. Grating Arm Reflector Blazed Grating (31.6 grooves/ mm, 63° blaze) Measurement Arm Reflector Seven Flat Mirrors in Rotation Method Configuration Table 6: Summary of key lab results. Coherence Length Measurement System Design Description Document 00007 Rev G page 23 FRED Analysis: Early concerns regarding both the propagation of light and the impact of diffraction in our system pushed us to model our system using FRED. Our first FRED model was a very simplified version of our system consisting of a collimated light source, a beamsplitter, one mirror in the measurement arm, and a blazed grating with approximately 20 grooves/mm and a 26° 45’ blaze, that was tilted at 26° 45’. The purpose of this initial model, was to continue getting familiar with modeling in FRED and as a basic proof of concept. We were unsure if we could see fringes or if diffraction from the grating would dominate the system. The initial FRED model gave us confidence that our design would produce results. Figure 9: Shows a simplified model of our system in FRED. The top figures show the physical system. And the figures below show the irradiance pattern observed by the detector (the yellow object in the upper figure) Coherence Length Measurement System Design Description Document 00007 Rev G page 24 The next model that was created was a more realistic representation of our system. This model utilized a non-trivial light source which was collimated. The component used for collimation was a parabolic mirror, that we had intended to use at that time. Two mirrors were used in the measurement arm to simulate the rotation stage mechanism and a blazed grating with 31.6 grooves/mm and a 63° blaze was placed in the grating arm and tilted 63° to be in the Littrow configuration. This model proved useful as an aid to developing our system and also as a means to corroborate our lab results. It allowed for tolerancing and the ability to quickly make theoretical changes to our system and see immediate results. Figure 10: Model of current set-up. This model includes the grating of 31.6 grooves/mm and a 63° blaze angle that we intend to use, a realistic source collimated with a parabolic mirror, and a CCD detector with a glass cover plate. Figure 11: Results from a plane wave and a gaussian source modeling a HeNe laser. . Coherence Length Measurement System Design Description Document 00007 Rev G page 25 The next model that was created was a model to test how the error of our selected rotation stage would impact results. After applying the maximum possible error of the rotation stage, no noticeable change occurred to the imaged interference pattern. This led to the conclusion that our selected rotation stage was accurate enough to be purchased. Figure 12: A visualization of our model used to check if the repeatability of our rotation stage would cause any problems. Coherence Length Measurement System Design Description Document 00007 Rev G page 26 The final FRED model created was a model of all the optical components used in the prototype. Compared to previous models, this includes the addition of a collimating and imaging lens and the additional mirrors used in the measurement arm. Figure 13: Our final system modeled in FRED. The top figures show the physical system. The bottom image shows the modeled interference pattern observed from using a Gaussian laser beam source with a spectrum modeled after that of a HeNe laser. Coherence Length Measurement System Design Description Document 00007 Rev G page 27 Visibility Analysis: This section illustrates the software analysis method that was used to create a plot of the visibility as a function of optical path length difference through calculating the image along each line of the grating and though calculating the grating line. For the full code see Appendix B. Once an image is collected, the visibility is calculated by taking the Fourier transform of each grating line. The visibility of a fringe pattern can be determined by taking twice the amplitude of the fringe frequency and dividing it by the zero order frequency. The FFT of the system was obtained in python with the use of the numpy toolbox command FFT: FringFft = np.abs(fft.fftshift(fft.fft(fft.fftshift(FringeArray)))) Figure 14: The fast Fourier transform of a 1-dimensional array along the line of the grating. Circled in orange is the shifted zero order frequency peak. Circled in red are the secondary peaks that correspond to the fringes. Coherence Length Measurement System Design Description Document 00007 Rev G page 28 Design Day Description: Our design day presentation consisted of two parts: 1. Our Prototype Interferometer o This interferometer was attached to a red-fiber coupled laser that could be turned on to help demonstrate how light propagated through the system. We also demonstrated how the rotation stage was programed to quickly rotate to the different angles necessary for extending the measurement range of our device. 2. Poster Coherence Length Measurement System Design Description Document 00007 Rev G page 29 Conclusions and Future Work: Our final prototype interferometer is promising in its design, but currently lacking in practical implementation. Theoretically our design meets and exceeds the major design restraints that made the customer unable to measure the coherence length of his various sources. Our system is able to outperform a traditional Michelson interferometer, in that the grating allows for a measurement value of the visibility in increments less than 10 µm. Also, our system can outperform a spectral analysis method of the coherence length, in that the mechanical nature of our device does not obfuscate lower visibility values. However, when trying to physically test the prototype our team encountered issues that led to excessive noise in our results. Problems with our system set-up resulted in warped fringes that yielded undesirable results. A number of factors, probably working together, may have caused the distortion of our fringes. The thinness of our beamsplitter and rotating mirror, along with the type of mount being used upon them, may have resulted in these components being bent and therefore impacting fringe quality. One way to eliminate this concern regarding the mirror, would be to fix it to a more solid base and then mount that component. In regard to the beamsplitter, it may be necessary to use a different type of mount. Another factor impacting our results, could be unwanted reflections caused by the beamsplitter and additional orders reflected by the grating. A new design that takes into account this factor of unwanted light when positioning optics may be needed to improve fringe quality. An additional factor that impacted our results, was the quality of our lenses. A future design could be made with better quality optics in order to avoid issues with aberrations, nonuniform brightness, and vignetting that negatively impacted our image quality. Another possible avenue to explore in improving fringe quality, is to perhaps initially use a smaller collimated beam and then a beam expander in the grating arm to cover the grating. By doing this issues with the large beam being clipped and diffracted by mounts and other optics could be avoided. Also, a future avenue for improvement of image quality, could also be to create a custom made grating with a larger step spacing and a higher accuracy in step-size uniformity. Another possible future design change, could be to use another spatial filter set-up in addition to the fiber coupler, to improve the quality of the collimated beam. In support of this design change, is that our best lab results occurred during the 3/18-3/24 lab week, when a spatial filter was being used. In addition to these mechanical concerns, our visibility code is also in need of improvement. While the visibility code can report the visibility when fringes exist, in the presence of artifacts and the absence of fringes our code produces results that do not follow what is experimentally observed. Ultimately, our prototype is a unique type of interferometer that shows promise in achieving the capabilities desired by the customer. Hopefully, with some modifications of this prototype design, the issues that impact fringe quality can be eliminated and the device can, subsequently, be implemented to ensure the efficacy of the systems used to measure wafer quality. Coherence Length Measurement System Design Description Document 00007 Rev G page 30 Appendix A: Table of All Lab Results Date Results 1/28-2/3 Equipment Specifications • 532 nm Laser Pointer • Blazed Grating (20 grooves/mm, 26° 45’ blaze) System Layout Images Grating Arm Only Reference Arm Only Interference of Two Arms Coherence Length Measurement System Design Description Document 00007 Rev G page 31 2/4-2/10 Equipment Specifications • 633 nm HeNe Laser • Note: Used three flat mirrors. Two in the reference arm and one in the measurement arm. System Layout Images Measurement Arm Only Reference Arm Only Interference of Two Arms Coherence Length Measurement System Design Description Document 00007 Rev G page 32 2/11-2/17 Equipment Specifications • 633 nm HeNe Laser • Blazed Grating (20 grooves/ mm, 26° 45’ blaze) System Layout Images Grating Arm Only First Reference Mirror Only Second Reference Mirror Only Reference Arm Only Coherence Length Measurement System Design Description Document 00007 Rev G page 33 Interference of Two Arms 2/18-2/24 Equipment Specifications • 633 nm HeNe Laser • Rotation stage method. • OPL values: Grating Top=158mm, Grating Bottom=140mm, Mirror 0=130mm, Mirror 1=284mm, Mirror 2=750mm, Mirror 3=1540mm System Layout Images Measurement Arm Only Flat Mirror 0 Only Interference of Flat and Mirror 0 Coherence Length Measurement System Design Description Document 00007 Rev G page 34 Grating Arm Only Mirror 0 Only Interference of Grating and Mirror 0 Grating Arm Only Mirror 1 Only Interference of Grating and Mirror 1 Grating Arm Only Mirror 2 Only Interference of Grating and Mirror 2 Grating Arm Only Mirror 3 Only Interference of Grating and Mirror 3 Coherence Length Measurement System Design Description Document 00007 Rev G page 35 3/18-3/24 Equipment Specifications • 633 nm HeNe Laser • Rotation stage method System Layout Images Interference between Grating and Rotating Mirror (Narrow Fringes) Interference between Grating and Rotating Mirror (Wide Fringes) Interference between Grating and Extended Mirror (Narrow Fringes) Interference between Grating and Extended Mirror (Wide Fringes) Coherence Length Measurement System Design Description Document 00007 Rev G page 36 Visibility Plots Visibility determined Once per Grating Line Visibility determined Multiple Times per Grating Line 4/8-4/14 Equipment Specifications • Red Laser Pointer • Rotation stage method • Base-to-Base OPD values: Mirror 0 = 0 cm , Mirror 1 = 50 cm, Mirror 2 = 130 cm System Layout Coherence Length Measurement System Design Description Document 00007 Rev G page 37 Images Interference between Grating and M0 Interference between Grating and M1 Interference between Grating and M2 Visibility Plots Visibility of M0 and M1 Configurations Visibility of M2 Configuration 4/22-4/28 Equipment Specifications • Fiber-Coupled Red Laser Pointer (λ=650nm) • Complete prototype except for detector • OPD Ranges Mirror 0 = 0-40 mm , Mirror 1 = 40-80 mm, Mirror 2 = 80-120 mm, Mirror 3 = 120-160 mm, Mirror 4 = 160-200 mm, Mirror 5 = 200-240 mm, Mirror 6 = 240-280 mm System Layout Coherence Length Measurement System Design Description Document 00007 Rev G page 38 Images Grating Arm Only M0 Only Interference between Grating and M0 Interference between Grating and M1 Interference between Grating and M2 Interference between Grating and M3 Interference between Grating and M4 Interference between Grating and M5 Interference between Grating and M6 Visibility Plot Coherence Length Measurement System Design Description Document 00007 Rev G page 39 4/29-5/5 Equipment Specifications • Fiber-Coupled Red Laser Pointer (λ=650nm) • Complete prototype with detector and a borrowed aperture • OPD Ranges Mirror 0 = 0-40 mm , Mirror 1 = 40-80 mm, Mirror 2 = 80-120 mm, Mirror 3 = 120-160 mm, Mirror 4 = 160-200 mm, Mirror 5 = 200-240 mm, Mirror 6 = 240-280 mm System Layout Images Interference between Grating and M0 Interference between Grating and M1 Interference between Grating and M2 Interference between Grating and M3 Interference between Grating and M4 Interference between Grating and M5 Interference between Grating and M6 Coherence Length Measurement System Design Description Document 00007 Rev G page 40 Visibility Plot Table 7: Record of all lab results. Coherence Length Measurement System Design Description Document 00007 Rev G page 41 Appendix B: Visibility Processing Code #!/usr/bin/env python3 # -- coding: utf-8 -- """ Created on Fri Mar 30 17:05:40 2018 @author: pellegrinoconte """ import PIL import os import numpy as np import matplotlib.pyplot as plt from numpy import array from numpy import fft from PIL import Image from scipy import ndimage print(os.path.realpath(file)) os.chdir("/Users/pellegrinoconte/Desktop/310/53") def getv(name): img = PIL.Image.open(name +".png") #.convert('LA') arr = array(img) ar = arr #[:,:,0] v = np.empty(0, dtype = float) j = 0 for each in np.transpose(ar)[::-1]: # j+=1 # print(each) segm = each # [int(inp.shape(each)/roll):int((i+1)np.shape(each)/roll)] altimg = np.abs(fft.fftshift(fft.fft(fft.fftshift(segm)))) # if j == 100: # plt.plot(altimg) m1 = (np.amax(np.abs(altimg[:int((altimg.shape/2 - altimg.shape/50 ))]))) m2 = (np.amax(np.abs(altimg))) vis = 2m1/m2 Coherence Length Measurement System Design Description Document 00007 Rev G page 42 v = np.append(v, vis) return v fig1 = plt.figure() ax = fig1.add_subplot(111) ax.invert_xaxis() nums =0 v = getv("M0") x = np.linspace(nums, nums+80, np.shape(v)) fig1 = plt.figure() ax = fig1.add_subplot(111) p7, = ax.plot(x, getv("M0") , label = "M0") v = getv("M1") nums = nums+80 x = np.linspace(nums, nums+80, np.shape(v)) p1, = ax.plot(x, getv("M1") , label = "M1") v = getv("M2") nums = nums+80 x = np.linspace(nums, nums+80, np.shape(v)) p2, = ax.plot(x, getv("M2") , label = "M2") v = getv("M3") nums = nums+80 x = np.linspace(nums, nums+80, np.shape(v)) p3, = ax.plot(x, getv("M3" ), label = "M3") v = getv("M4") nums = nums+80 x = np.linspace(nums, nums+80, np.shape(v)) p4, = ax.plot(x, getv("M4") , label = "M4") v = getv("M5") nums = nums+80 x = np.linspace(nums, nums+80, np.shape(v)) p5, = ax.plot(x, getv("M5") , label = "M5") Coherence Length Measurement System Design Description Document 00007 Rev G page 43 v = getv("M6") nums = nums+80 x = np.linspace(nums, nums+80, np.shape(v)) p6, = ax.plot(x, getv("M6") , label = "M6") ax.legend(handles=[p7, p1, p2, p3, p4, p5, p6], loc = 'best') ax.set_xlabel("$\Delta$ OPL") ax.set_ylabel("Visibility") ax.set_title("Initial System Results") Coherence Length Measurement System Design Description Document 00007 Rev G page 44 Appendix C: Customer Instructions The purpose of our customer instructions manual was to provide the customer with a document that could explain how to set-up and operate the interferometer in case the system was ever in need of reassembly, as well as to clearly detail the procedure that our group followed when making the measurements provided earlier in this document. Depicted are screenshots of the manual that will be sent electronically to the customer along with the interferometer. Coherence Length Measurement System Design Description Document 00007 Rev G page 45 Coherence Length Measurement System Design Description Document 00007 Rev G page 46 Coherence Length Measurement System Design Description Document 00007 Rev G page 47 Coherence Length Measurement System Design Description Document 00007 Rev G page 48 Coherence Length Measurement System Design Description Document 00007 Rev G page 49 Coherence Length Measurement System Design Description Document 00007 Rev G page 50 References: Paschotta, Rüdiger. “Coherence.” Encyclopedia of Laser Physics and Technology - Coherence, Coherent, Light, Spatial and Temporal Coherence, Monochromaticity, 20 Feb. 2017, www.rp-photonics.com/coherence.html. “Properties of Lasers.” Worldoflasers.com, 6 Apr. 2015, worldoflasers.com/laserproperties.htm. “7 (D.g.). Coherent Light.” Andres Robotics and Science, www.goodrichscience.com/7-dg-coherent-light.html. Vamivakas, Nick. Introduction to Wave Optics. “Coherence Length.” Wikipedia, Wikimedia Foundation, 14 Mar. 2018, en.wikipedia.org/wiki/Coherence_length. “Helium-Neon Lasers.” Sam's Laser FAQ - Helium-Neon Lasers, donklipstein.com/laserhen.htm#henmtmhl. Customer PowerPoint Presentation Vladimyros Devrelis, Martin O'Connor, and Jesper Munch, "Coherence length of single laser pulses as measured by CCD interferometry," Appl. Opt. 34, 5386-5389 (1995)
17097	https://emedicine.medscape.com/article/1071854-overview	close Please confirm that you would like to log out of Medscape. If you log out, you will be required to enter your username and password the next time you visit. Log out Cancel processing.... Tools & Reference>Dermatology Trichotillomania Updated: Oct 30, 2024 Author: Dirk M Elston, MD; Chief Editor: William D James, MD more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Trichotillomania Sections Trichotillomania Overview Background Pathophysiology Etiology Epidemiology Prognosis Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Histologic Findings Show All Treatment Approach Considerations Behavioral Interventions Pharmacologic Therapy Diet and Activity Consultations Show All Medication Antidepressants, SSRIs Antidepressants, TCAs Pulmonary, Other Serotonin-Dopamine Activity Modulators (SDAM) Show All Media Gallery;) References;) Overview Background Trichotillomania (hair-pulling disorder) is characterized by the persistent and excessive pulling of one’s own hair, resulting in noticeable hair loss. [1, 2, 3] Hair pulling can occur in any area of the body where hair grows. The scalp is the area most commonly affected, followed by the eyelashes and eyebrows. The hair loss that results from hair pulling can range from small undetectable areas of hair thinning to complete alopecia. Trichotillomania most commonly presents in early adolescence, with the peak prevalence between ages 4 and 17 years. The disorder has both physical and psychosocial implications. Affected patients may experience distress, moderate impairment of social or academic functioning, and adverse impacts on family relationships. Although trichotillomania is more often a focus of behavioral and psychiatric publications than of dermatologic publications, patients are more likely to present to dermatologists than to mental health professionals. Accordingly, it is important for dermatologists to be familiar with the clinical features and treatment options for these patients. Trichotillomania must be differentiated clinically from other alopecias (eg, alopecia areata, traction alopecia, androgenetic alopecia, pseudopelade, alopecia mucinosa) through careful history-taking and physical examination. Dermatologists, psychologists, and psychiatrists should be familiar with the key features of the disorder because earlier treatment yields a better prognosis and can prevent complications such as trichobezoar and scarring. [6, 7] Diagnostic criteria (DSM-5) The American Psychiatric Association’s Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition (DSM-5) placed trichotillomania in the category of obsessive-compulsive and related disorders and noted that it is characterized by recurrent body-focused repetitive behavior (hair pulling) and repeated attempts to decrease or stop the behavior. The behavior can occur during both relaxed and stressful times, but there is often a mounting sense of tension before hair pulling occurs or when attempts are made to resist the behavior. Some authors have advocated for the distinction between “automatic” pulling occurring during sedentary activities with little awareness and “focused” pulling in response to negative or stressful emotions, in that these different styles may respond to different treatment strategies. The specific DSM-5 criteria for trichotillomania (hair-pulling disorder) are as follows : Recurrent pulling out of one’s hair, resulting in hair loss Repeated attempts to decrease or stop the hair-pulling behavior The hair pulling causes clinically significant distress or impairment in social, occupational, or other important areas of functioning The hair pulling or hair loss cannot be attributed to another medical condition (eg, a dermatologic condition) The hair pulling cannot be better explained by the symptoms of another mental disorder (eg, attempts to improve a perceived defect or flaw in appearance, such as may be observed in body dysmorphic disorder) Next: Pathophysiology Pathophysiology From a dermatologic standpoint, trichotillomania is a form of traumatic alopecia. The trauma to the follicle occurs as a result of the patient’s repetitive hair-pulling behavior. The hair pulling may present in conjunction with other repetitive grooming behaviors, such as nail biting and skin picking. Trichotillomania results in highly variable patterns of hair loss. The scalp is the most common area of hair pulling, followed by the eyebrows, eyelashes, pubic and perirectal areas, axillae, limbs, torso, and face. The resulting alopecia can range from thin unnoticeable areas of hair loss to total baldness in the area(s) being plucked. In addition, trichophagia (ingestion of the hair) is common in persons who pull out their hair. This chewing or mouthing behavior can frequently lead to the formation of trichobezoars (ie, hair casts) in the stomach or small intestines. Trichobezoars can result in anemia, abdominal pain, hematemesis, nausea or vomiting, bowel obstruction, perforation, gastrointestinal (GI) bleeding, acute pancreatitis, and obstructive jaundice. Previous Next: Pathophysiology Etiology The etiology of trichotillomania is largely unknown, though both environmental and genetic causes have been suspected. Explanations that have been proposed for the onset and maintenance of the hair-pulling behavior include the following: Coping mechanism for anxiety or stressful events Benign habit that developed from a sensory event (eg, itchy eyelash) or another event and resulted in trichotillomania Occurrence in conjunction with another habitual behavior (ie, thumbsucking) in young children Serotonin deficiency - A link may exist between a deficiency of the neurotransmitter serotonin (5-hydroxytryptamine [5-HT]) and trichotillomania; the hypothesized connection is based on the success of selective serotonin reuptake inhibitors (SSRIs) in treating some people with trichotillomania Structural brain abnormalities - On magnetic resonance imaging (MRI), some individuals with trichotillomania have abnormalities of subcortical regions involved in habit generation, inhibitory control, and regulation of affect Abnormal brain metabolism - On positron emission tomography (PET), some individuals with trichotillomania have a high metabolic glucose rate in the global, bilateral, cerebellar, and right superior parietal areas Genetic susceptibility - DSM-5 cited some evidence that genetic vulnerability plays a role ; trichotillomania occurs more frequently in people with obsessive-compulsive disorder (OCD) and their first-degree relatives Psychological factors - Several psychological theories (eg, psychodynamic, behavioral, and ethologic) have attempted to explain trichotillomania in children; such theories have included stress reduction, emotional regulation, and sensory stimulation [11, 12] Disordered reward processing - Preliminary data have suggested that trichotillomania may represent a disorder of altered reward processing within the central nervous system (CNS); a study of reward processing in trichotillomania patients demonstrated altered nucleus accumbens activations and a decreased functional connection between the dorsal anterior cingulate and nucleus accumbens and basolateral amygdala and reward network ; input was through glutamatergic projections, identifying a possible intervention point with agents that modulate glutamate Neurodegenerating disease associations - Reports have also suggested a possible association between neurodegenerating diseases (eg, Parkinson disease and dementia ) and trichotillomania in older populations Previous Next: Pathophysiology Epidemiology United States and international statistics Although US epidemiologic studies on the prevalence of trichotillomania are rare, it has been estimated that approximately 8 million people have trichotillomania. The overall frequency is probably underestimated, because only persons who present for treatment are counted; denial of the disorder is frequent, and many individuals with the disorder do not seek professional intervention. Further epidemiologic studies are needed. In a study of college students, approximately 1-2% had past or current symptoms of trichotillomania. The rate fell to 0.6% when patients were restricted to the group having related mental tension and relief; without such restrictions, the rate of hair pulling resulting in visible hair loss was 1.5% for males and 3.4% for females. A survey at a historically Black university (N = 248) found that 6.3% of those surveyed had a history of pulling out their hair. In the authors’ experience, the number of patients with trichotillomania is approximately 5% of the number of patients with alopecia areata. The incidence of alopecia areata is approximately 50% of all patients presenting with alopecia, and the total number of hair-loss patients is approximately 2% of all dermatologic patients. Age-, sex-, and race-related demographics Trichotillomania is frequently a chronic disorder that lasts weeks to decades, with a variable age of onset. Hair-pulling sites may vary with the age of onset: Patients with a very early onset of trichotillomania are more likely to pull eyelashes, whereas those with a later onset are more likely to pull pubic hair. In a study by Walther et al, it was reported that the 27 children in the preschool age group (0-5 y) pulled only from the scalp and that more than half of those in the 5- to 10-year age group children pulled from other body areas in addition to the scalp. Although empiric data are not available, this condition appears to be substantially more common in children than in adults. In general, prognosis is related to patient age. Children typically have a time-limited disorder, with an excellent prognosis. Adolescents have more severe disease, with a guarded prognosis. Adults, many of whom were diagnosed before reaching adulthood, have a poor prognosis. With regard to sex-related differences, the younger the patient, the more equal the sex distribution. However, a cross-sectional study of 110 young children (age range, 0-10 y) demonstrated that a female predominance still exists, even among younger patients. In adult groups, most patients are women. In adolescents, girls are affected more often than boys. DSM-5 cited an overall female-to-male ratio of 10:1. No racial differences in prevalence have been reported. Trichotillomania appears to be equally common in Whites, Blacks, and Asians. Previous Next: Pathophysiology Prognosis In very young children, the prognosis is excellent; hair pulling that occurs in young children may be described more accurately as a short-term habit disorder. In late childhood and adolescence, the prognosis is usually good but should be considered guarded; the alopecia quite often continues for months or a couple of years and then recurs after a variable time. In adult patients, the prognosis is poor, and permanent recovery is uncommon. Trichotillomania results in highly variable patterns of hair loss, ranging from small undetectable patches of hair loss to total baldness. Ingestion of the pulled hair can result in trichobezoar formation and subsequent anemia, abdominal pain, hematemesis, nausea or vomiting, bowel obstruction, perforation, GI bleeding, pancreatitis, and obstructive jaundice. Trichotillomania can become a chronic and persistent condition. Specifically, symptoms of trichotillomania can persist for weeks to decades. Therefore, comprehensive treatment planning is critical and may require consultations with mental health professionals. Treating trichotillomania in children may be difficult because of the low reliability and validity of self-reporting. Mortality is not reported with trichotillomania. Most patients with trichotillomania in dermatologic clinics are children and early adolescents. Patients may try to conceal the alopecic area and may have some restrictions in their school activities. In adults, trichotillomania may cause distress and impairment in occupational and social or marital relations. Previous Clinical Presentation References Chamberlain SR, Odlaug BL, Boulougouris V, Fineberg NA, Grant JE. Trichotillomania: neurobiology and treatment. Neurosci Biobehav Rev. 2009 Jun. 33 (6):831-42. [QxMD MEDLINE Link]. Harrison JP, Franklin ME. Pediatric trichotillomania. Curr Psychiatry Rep. 2012 Jun. 14 (3):188-96. [QxMD MEDLINE Link].[Full Text]. Franklin ME, Zagrabbe K, Benavides KL. Trichotillomania and its treatment: a review and recommendations. Expert Rev Neurother. 2011 Aug. 11 (8):1165-74. [QxMD MEDLINE Link].[Full Text]. Franklin ME, Flessner CA, Woods DW, Keuthen NJ, Piacentini JC, Moore P, et al. The child and adolescent trichotillomania impact project: descriptive psychopathology, comorbidity, functional impairment, and treatment utilization. J Dev Behav Pediatr. 2008 Dec. 29 (6):493-500. [QxMD MEDLINE Link]. Keren M, Ron-Miara A, Feldman R, Tyano S. Some reflections on infancy-onset trichotillomania. Psychoanal Study Child. 2006. 61:254-72. [QxMD MEDLINE Link]. Nadipelli AR, Duggineni D. A Rare Case of Gastric Trichobezoar Managed by Laparotomy. Cureus. 2024 Aug. 16 (8):e67357. [QxMD MEDLINE Link].[Full Text]. Salaam K, Carr J, Grewal H, Sholevar E, Baron D. Untreated trichotillomania and trichophagia: surgical emergency in a teenage girl. Psychosomatics. 2005 Jul-Aug. 46 (4):362-6. [QxMD MEDLINE Link]. Keuthen NJ, Tung ES, Woods DW, Franklin ME, Altenburger EM, Pauls DL, et al. Replication study of the milwaukee inventory for subtypes of trichotillomania-adult version in a clinically characterized sample. Behav Modif. 2015 Jul. 39 (4):580-99. [QxMD MEDLINE Link]. American Psychiatric Association. Diagnostic and Statistical Manual of Mental Disorders, Fifth Edition. 5th ed. Washington, DC: American Psychiatric Association; 2013. 251-4. Isobe M, Redden SA, Keuthen NJ, Stein DJ, Lochner C, Grant JE, et al. Striatal abnormalities in trichotillomania: a multi-site MRI analysis. Neuroimage Clin. 2018. 17:893-898. [QxMD MEDLINE Link]. Diefenbach GJ, Tolin DF, Meunier S, Worhunsky P. Emotion regulation and trichotillomania: a comparison of clinical and nonclinical hair pulling. J Behav Ther Exp Psychiatry. 2008 Mar. 39 (1):32-41. [QxMD MEDLINE Link]. Meunier SA, Tolin DF, Franklin M. Affective and sensory correlates of hair pulling in pediatric trichotillomania. Behav Modif. 2009 May. 33 (3):396-407. [QxMD MEDLINE Link]. White MP, Shirer WR, Molfino MJ, Tenison C, Damoiseaux JS, Greicius MD. Disordered reward processing and functional connectivity in trichotillomania: a pilot study. J Psychiatr Res. 2013 Sep. 47 (9):1264-72. [QxMD MEDLINE Link]. De Sousa A, Mehta J. Trichotillomania in a case of vascular dementia. Int J Trichology. 2013 Jan. 5 (1):38-9. [QxMD MEDLINE Link].[Full Text]. Christenson GA, Pyle RL, Mitchell JE. Estimated lifetime prevalence of trichotillomania in college students. J Clin Psychiatry. 1991 Oct. 52 (10):415-7. [QxMD MEDLINE Link]. Mansueto CS, Thomas AM, Brice AL. Hair pulling and its affective correlates in an African-American university sample. J Anxiety Disord. 2007. 21 (4):590-9. [QxMD MEDLINE Link]. Flessner CA, Lochner C, Stein DJ, Woods DW, Franklin ME, Keuthen NJ. Age of onset of trichotillomania symptoms: investigating clinical correlates. J Nerv Ment Dis. 2010 Dec. 198 (12):896-900. [QxMD MEDLINE Link]. Walther MR, Snorrason I, Flessner CA, Franklin ME, Burkel R, Woods DW. The trichotillomania impact project in young children (TIP-YC): clinical characteristics, comorbidity, functional impairment and treatment utilization. Child Psychiatry Hum Dev. 2014 Feb. 45 (1):24-31. [QxMD MEDLINE Link]. Diefenbach GJ, Tolin DF, Hannan S, Crocetto J, Worhunsky P. Trichotillomania: impact on psychosocial functioning and quality of life. Behav Res Ther. 2005 Jul. 43 (7):869-84. [QxMD MEDLINE Link]. Falkenstein MJ, Haaga DA. Symptom accommodation, trichotillomania-by-proxy, and interpersonal functioning in trichotillomania (hair-pulling disorder). Compr Psychiatry. 2016 Feb. 65:88-97. [QxMD MEDLINE Link]. Murphy C, Redenius R, O'Neill E, Zallek S. Sleep-isolated trichotillomania: a survey of dermatologists. J Clin Sleep Med. 2007 Dec 15. 3 (7):719-21. [QxMD MEDLINE Link].[Full Text]. Chen KL, Chiu HY, Chan CC, Chan JY, Lin SJ. Extensive cicatricial alopecia in a patient with long-term trichotillomania. J Dermatol. 2016 Feb. 43 (2):226-8. [QxMD MEDLINE Link]. Radmanesh M, Shafiei S, Naderi AH. Isolated eyebrow and eyelash trichotillomania mimicking alopecia areata. Int J Dermatol. 2006 May. 45 (5):557-60. [QxMD MEDLINE Link]. Lacarrubba F, Micali G, Tosti A. Scalp dermoscopy or trichoscopy. Curr Probl Dermatol. 2015. 47:21-32. [QxMD MEDLINE Link]. Tolin DF, Diefenbach GJ, Flessner CA, Franklin ME, Keuthen NJ, Moore P, et al. The trichotillomania scale for children: development and validation. Child Psychiatry Hum Dev. 2008 Sep. 39 (3):331-49. [QxMD MEDLINE Link]. Ihm CW, Han JH. Diagnostic value of exclamation mark hairs. Dermatology. 1993. 186 (2):99-102. [QxMD MEDLINE Link]. Miteva M, Romanelli P, Tosti A. Pigmented casts. Am J Dermatopathol. 2014 Jan. 36 (1):58-63. [QxMD MEDLINE Link]. Franklin ME, Edson AL, Ledley DA, Cahill SP. Behavior therapy for pediatric trichotillomania: a randomized controlled trial. J Am Acad Child Adolesc Psychiatry. 2011 Aug. 50 (8):763-71. [QxMD MEDLINE Link].[Full Text]. Weidt S, Zai G, Drabe N, Delsignore A, Bruehl AB, Klaghofer R, et al. Affective regulation in trichotillomania before and after self-help interventions. J Psychiatr Res. 2016 Apr. 75:7-13. [QxMD MEDLINE Link]. Grant JE, Odlaug BL, Kim SW. N-acetylcysteine, a glutamate modulator, in the treatment of trichotillomania: a double-blind, placebo-controlled study. Arch Gen Psychiatry. 2009 Jul. 66 (7):756-63. [QxMD MEDLINE Link]. Grant JE, Odlaug BL, Chamberlain SR, Kim SW. Dronabinol, a cannabinoid agonist, reduces hair pulling in trichotillomania: a pilot study. Psychopharmacology (Berl). 2011 Dec. 218 (3):493-502. [QxMD MEDLINE Link]. Van Ameringen M, Mancini C, Patterson B, Bennett M, Oakman J. A randomized, double-blind, placebo-controlled trial of olanzapine in the treatment of trichotillomania. J Clin Psychiatry. 2010 Oct. 71 (10):1336-43. [QxMD MEDLINE Link]. Lee MT, Mpavaenda DN, Fineberg NA. Habit Reversal Therapy in Obsessive Compulsive Related Disorders: A Systematic Review of the Evidence and CONSORT Evaluation of Randomized Controlled Trials. Front Behav Neurosci. 2019. 13:79. [QxMD MEDLINE Link]. Sani G, Gualtieri I, Paolini M, Bonanni L, Spinazzola E, Maggiora M, et al. Drug Treatment of Trichotillomania (Hair-Pulling Disorder), Excoriation (Skin-picking) Disorder, and Nail-biting (Onychophagia). Curr Neuropharmacol. 2019. 17 (8):775-786. [QxMD MEDLINE Link]. Khan S, Vij K, Lopez E. Off-Label Use of Naltrexone in Pica and Other Compulsive Behaviors: A Report of Two Cases. Cureus. 2024 Jul. 16 (7):e65845. [QxMD MEDLINE Link].[Full Text]. Bloch MH, Landeros-Weisenberger A, Dombrowski P, Kelmendi B, Wegner R, Nudel J, et al. Systematic review: pharmacological and behavioral treatment for trichotillomania. Biol Psychiatry. 2007 Oct 15. 62 (8):839-46. [QxMD MEDLINE Link]. Morris SH, Zickgraf HF, Dingfelder HE, Franklin ME. Habit reversal training in trichotillomania: guide for the clinician. Expert Rev Neurother. 2013 Sep. 13 (9):1069-77. [QxMD MEDLINE Link]. Papadopoulos AJ, Janniger CK, Chodynicki MP, Schwartz RA. Trichotillomania. Int J Dermatol. 2003 May. 42 (5):330-4. [QxMD MEDLINE Link]. Rothbart R, Amos T, Siegfried N, Ipser JC, Fineberg N, Chamberlain SR, et al. Pharmacotherapy for trichotillomania. Cochrane Database Syst Rev. 2013 Nov 8. 321 (8):CD007662. [QxMD MEDLINE Link]. Jones G, Keuthen N, Greenberg E. Assessment and treatment of trichotillomania (hair pulling disorder) and excoriation (skin picking) disorder. Clin Dermatol. 2018 Nov - Dec. 36 (6):728-736. [QxMD MEDLINE Link]. Kiliç F, Keleş S. Repetitive Behaviors Treated With N-Acetylcysteine: Case Series. Clin Neuropharmacol. 2019 Jul/Aug. 42 (4):139-141. [QxMD MEDLINE Link]. Bloch MH, Panza KE, Grant JE, Pittenger C, Leckman JF. N-Acetylcysteine in the treatment of pediatric trichotillomania: a randomized, double-blind, placebo-controlled add-on trial. J Am Acad Child Adolesc Psychiatry. 2013 Mar. 52 (3):231-40. [QxMD MEDLINE Link].[Full Text]. Golubchik P, Sever J, Weizman A, Zalsman G. Methylphenidate treatment in pediatric patients with attention-deficit/hyperactivity disorder and comorbid trichotillomania: a preliminary report. Clin Neuropharmacol. 2011 May-Jun. 34 (3):108-10. [QxMD MEDLINE Link]. Leombruni P, Gastaldi F. Oxcarbazepine for the treatment of trichotillomania. Clin Neuropharmacol. 2010 Mar-Apr. 33 (2):107-8. [QxMD MEDLINE Link]. Virit O, Selek S, Savas HA, Kokaçya H. Improvement of restless legs syndrome and trichotillomania with aripiprazole. J Clin Pharm Ther. 2009 Dec. 34 (6):723-5. [QxMD MEDLINE Link]. Peabody T, Reitz S, Smith J, Teti B. Clinical management of trichotillomania with bimatoprost. Optom Vis Sci. 2013 Jun. 90 (6):e167-71. [QxMD MEDLINE Link]. Media Gallery Geometric patch of incomplete alopecia in teenage boy. Bizarre-patterned lesion covered with short hairs in 11-year-old girl. Typical geometric shape trichotillomania in 7-year-old boy. Smooth baldness of scalp surface at this age is rare. In eyebrow involvement, characteristic geometric shape is not made. Sometimes, alopecia is not circumscribed but simply shows deficient hair volume, as in this 9-year-old girl. When entire scalp is involved, trichotillomania resembles keratinization disorder of hairs (eg, monilethrix). Tonsure trichotillomania (so named because of similarity to medieval monks' tonsures). In this patient, hair is preserved only in posterior margin of scalp. Close-up picture of lesion of usual trichotillomania shows combination of newly growing young hair, broken shafts, comedolike black dots, empty orifices, and vellus or intermediate hairs. Contrast card examination helps demonstrate nature of alopecia to parents of children with trichotillomania. It shows broken hairs and newly growing hairs with slender tips among long intact hairs. Woman with severe long-standing lesions from trichotillomania. Close-up picture of severe long-standing lesion in which hairs are regressed to vellus- or intermediate-type hairs and scalp is rather smooth. Histopathologically, trichomalacia (twisted pigmented soft cortex) with catagen follicles is characteristic of trichotillomania with empty follicles. of 12 Tables Back to List Contributor Information and Disclosures Author Dirk M Elston, MD Professor and Chairman, Department of Dermatology and Dermatologic Surgery, Medical University of South Carolina College of Medicine Dirk M Elston, MD is a member of the following medical societies: American Academy of Dermatology Disclosure: Nothing to disclose. Coauthor(s) Carly A Elston The Commonwealth Medical College Disclosure: Nothing to disclose. Chief Editor William D James, MD Emeritus Professor, Department of Dermatology, University of Pennsylvania School of Medicine William D James, MD is a member of the following medical societies: American Academy of Dermatology, American Contact Dermatitis Society, Association of Military Dermatologists, Association of Professors of Dermatology, American Dermatological Association, Women's Dermatologic Society, Medical Dermatology Society, Dermatology Foundation, Society for Investigative Dermatology, Pennsylvania Academy of Dermatology Disclosure: Received income in an amount equal to or greater than $250 from: ElsevierServed as a speaker for various universities, dermatology societies, and dermatology departments. Additional Contributors Cynthia R Ellis, MD Director of Developmental Medicine, Associate Professor, Department of Pediatrics and Psychiatry, Munroe Meyer Institute for Genetics and Rehabilitation, University of Nebraska Medical Center Cynthia R Ellis, MD is a member of the following medical societies: Nebraska Medical Association Disclosure: Nothing to disclose. Connie J Schnoes, MA, PhD Director, National Behavioral Health Dissemination, Supervising Practitioner, Boys Town Center for Behavioral Health, Father Flanagan’s Boys’ Home, Boys Town Disclosure: Nothing to disclose. Holly Jean Roberts, PhD Assistant Professor, Department of Pediatrics, Munroe-Meyer Institute, University of Nebraska Medical Center Holly Jean Roberts, PhD is a member of the following medical societies: Autism Society, National Association of School Psychologists, Psi Chi Disclosure: Nothing to disclose. Megha Patel The Commonwealth Medical College Disclosure: Nothing to disclose. Acknowledgements David F Butler, MD Professor of Dermatology, Texas A&M University College of Medicine; Chair, Department of Dermatology, Director, Dermatology Residency Training Program, Scott and White Clinic, Northside Clinic David F Butler, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Dermatology, American Medical Association, American Society for Dermatologic Surgery, American Society for MOHS Surgery, Association of Military Dermatologists, and Phi Beta Kappa Disclosure: Nothing to disclose. Edward F Chan, MD Clinical Assistant Professor, Department of Dermatology, University of Pennsylvania School of Medicine Edward F Chan, MD is a member of the following medical societies: American Academy of Dermatology, American Society of Dermatopathology, and Society for Investigative Dermatology Disclosure: Nothing to disclose. Cynthia R Ellis, MD Director of Developmental Medicine, Associate Professor, Department of Pediatrics and Psychiatry, Munroe Meyer Institute for Genetics and Rehabilitation, University of Nebraska Medical Center Cynthia R Ellis, MD is a member of the following medical societies: Nebraska Medical Association Disclosure: Nothing to disclose. Chull-Wan Ihm, MD Professor, Department of Dermatology, Chonbuk National University, Korea Chull-Wan Ihm is a member of the following medical societies: American Academy of Dermatology and American Society of Dermatopathology Disclosure: Nothing to disclose. Chet Johnson, MD Professor and Chair of Pediatrics, Associate Director, Developmental Pediatrician, Center for Child Health and Development, Shiefelbusch Institute for Life Span Studies, University of Kansas School of Medicine; LEND Director, University of Kansas Medical Center Chet Johnson, MD is a member of the following medical societies: American Academy of Pediatrics Disclosure: Nothing to disclose. Caroly Pataki, MD Clinical Professor of Psychiatry and Pediatrics, Keck School of Medicine of the University of Southern California Caroly Pataki, MD is a member of the following medical societies: American Academy of Child and Adolescent Psychiatry, New York Academy of Sciences, and Physicians for Social Responsibility Disclosure: Nothing to disclose. CH Rhee, MD Disclosure: Nothing to disclose. Holly Jean Roberts, PhD Assistant Professor, Department of Pediatrics, Munroe-Meyer Institute, University of Nebraska Medical Center Holly Jean Roberts, PhD is a member of the following medical societies: Autism Society of America, National Association of School Psychologists, and Psi Chi Disclosure: Nothing to disclose. Connie J Schnoes, MA, PhD Psychologist, Director of Training, Supervising Practitioner, Father Flanagan's Boys' Home, Boys Town Disclosure: Nothing to disclose. Leonard Sperling, MD Chair, Professor, Department of Dermatology, Uniformed Services University of the Health Sciences Leonard Sperling, MD is a member of the following medical societies: American Academy of Dermatology Disclosure: Nothing to disclose. Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Nothing to disclose. Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article encoded search term (Trichotillomania) and Trichotillomania What to Read Next on Medscape Related Conditions and Diseases Obsessive-Compulsive Disorder Scarring Alopecia Excoriation Disorder Fast Five Quiz: Embarrassing Health Conditions Trichorrhexis Nodosa Psychiatry Fast Five Quiz: How Much Do You Know About Obsessive-Compulsive Disorder? News & Perspective Hair Loss in Children: How to Spot and Treat Different Causes Alzheimer's Drug May Ease Hair Pulling, Skin-Picking Disorders Treatment Options Limited for Skin-Picking, Hair-Pulling Disorders Trichotillomania: What You Need to Know About This Common Hair-Pulling Disorder Drug Interaction Checker Pill Identifier Calculators Formulary Hair Disorders: Finding the Root of the Problem Log in or register for free to unlock more Medscape content Unlimited access to our entire network of sites and services Log in or Register Log in or register for free to unlock more Medscape content Unlimited access to industry-leading news, research, resources, and more
17098	https://www.wowt.com/2024/08/28/explainer-relative-humidity-vs-dew-point-whats-difference/	EXPLAINER: Relative humidity vs. dew point... what’s the difference? Skip to content We are Local Omaha Everyday Advertise With Us Heartland Heroes Sketch the Sky Becka's Talking 62° Omaha, NE News Streaming First Alert 6 Weather 24/7 Weather We the People Sports on 6 Contact Us News Tips Home News Get the First Alert News App Crime & Public Safety Education Forecast Health International National Regional Sports State We the People 24/7 Weather First Alert 6 Traffic First Alert 6 Investigations First Alert 6 Weather Get The First Alert Weather App Interactive Radar Severe Weather Dashboard Weather Maps Union Bank & Trust CityCam Network Traffic Tornado Closings Sketch the Sky Sports on 6 Olympics High School College World Series Scoreboard Nebraska Huskers UNO Mavericks Creighton Bluejays Iowa Sports Omaha Everyday Community Community Calendar We Are Local Heartland Heroes Partnerships Contact Us News Tips NextGen TV Meet the Team Meet the Sales Team Cozi TV Heroes & Icons ION Television Start TV Circle TV Contests Employment Opportunities Politics Election Results Interactive Results Nebraska Interactive Results Iowa Interactive Results Newsletter TV Listings Submit Photos and Videos NextGen TV Zeam - News Streams Take Our Poll Circle Country Gray DC Bureau InvestigateTV Watching Your Wallet PowerNation Digital Marketing Sketch the Sky EXPLAINER: Relative humidity vs. dew point... what’s the difference? Close What's the difference between relative humidity and dew point? 6 News Meteorologist Jade Steffens explains. By Jade Steffens Published: Aug. 28, 2024 at 9:28 PM UTC OMAHA, Neb. (WOWT) - The moisture content in the air can be measured in both the relative humidity and the dew point, however, the way they are calculated is very different. The dew point is the temperature air must be cooled to in order to reach a relative humidity of 100%, also known as saturation. In other words, when the dew point rises, there is simply more moisture in the air. This will directly impact how comfortable it will feel outside. On the muggy meter, a dew point below 60° will feel pleasant, while a dew point above 70° will feel steamy. Dew Point Scale(WOWT) On the other hand, relative humidity measures water vapor relative to the temperature of the air. It’s a measure of the current water vapor in the air compared to the maximum amount of water it can hold at its current temperature. The higher the temperature, the more water it can hold. Relative humidity can be misleading on a day with very warm temperatures. For example, a 100° day with a dew point of 75° will feel very humid, but only have a relative humidity of 45%. A day that is 50° with a dew point of 50° has a relative humidity of 100%, but will feel comfortable and dry. Dew Point VS Relative Humidity(WOWT) If the 100° day and the 50° day both have 100% humidity, one would have a feels like temperature of 195° and the other would have a feels like temperature of 49°! For a true measure of just how comfortable it will feel outside, the dew point is generally a better tool to use, especially on hot and humid summer days. Copyright 2024 WOWT. All rights reserved. Bo’s First Alert 6 Forecast: Warm & muggy on Monday, a warm start to October Discover Bo’s First Alert 6 Forecast: Sunny & warm weather continues, gusty winds at times Discover Junie Boutique A Lifetime of Bags, One Final Sale — Up to 80% Off Read More Skip Bo’s First Alert 6 Forecast: Sunny & breezy today, muggy Sunday then more sunshine next week Discover Top 6 on 6: This week’s most-watched videos - September 26 Discover South Carolina New Policy for Senior Drivers The Quote General \| SponsoredSponsored Undo Founders Are Using This Card to Get Paid Back for Daily Expenses Ramp \| SponsoredSponsored Undo This Business Card Tracks Spending and Gives 1.5% Back Ramp \| SponsoredSponsored Undo Little Known Brain Hack To Manifest Money bravethinkinginstitute \| SponsoredSponsored Learn More Undo Flight Attendant Reveals How Seniors Can Fly Business Class For The Price Of Economy Americans, you'll want to check this out ASAP!Online Shopping Tools \| SponsoredSponsored Undo Discover Effortless Glucose Monitoring: Request a Free Trial Dexcom \| SponsoredSponsored Undo She Manifested 7 Figures & Her Dream Home bravethinkinginstitute \| SponsoredSponsored Undo by Taboolaby Taboola You might be interested in the content above- \| SponsoredSponsored Undo Flight Attendant Reveals How Seniors Can Fly Business Class For the Price Of Economy Online Shopping Tools \| SponsoredSponsored Undo Explore new and used vehicles across the board.Carxplore \| SponsoredSponsored Shop Now Undo What Neck Pain Could Mean (And When to Act)NeuroMD \| SponsoredSponsored Learn More Undo Most Read Omaha Police Department, Mayor Ewing issue statements after Bud Crawford traffic stop Four Starbucks locations in Omaha to close on Saturday Omaha celebrates boxing champ Bud Crawford with parade, festivities Nebraska Department of Transportation planning new I-80 interchange Nebraska mourns former Cornhusker and beloved business owner Gunman in truck smashes into Michigan church and opens fire, killing at least 4 and injuring 8 Bud Crawford parade: What to know before heading out on Saturday Listeria found in Walmart meatball meals may be linked to deadly outbreak Latest News 6-HOUR LATE FORECAST 9.28.25 10-DAY PM FORECAST 9.28.25 3-DAY PM FORECAST 9.28.25 6-HOUR PM FORECAST 9.28.25 Bo’s First Alert 6 Forecast: Warm & muggy on Monday, a warm start to October 10-DAY AM FORECAST 9.28.25 6-HOUR AM FORECAST 9.28.25 Bo’s First Alert 6 Forecast: Sunny & warm weather continues, gusty winds at times News First Alert 6 Investigations WOWT 6 News Crime Stoppers First Alert 6 Weather Omaha Everyday Contact Us Election Results Get the First Alert 6 Weather app WOWT 3555 Farnam St. Omaha, NE 68131 402-346-6666 Public Inspection File sixonline@wowt.com - 402-346-6666 Terms of Service Privacy Policy FCC Applications EEO Statement Advertising Digital Marketing Closed Captioning/Audio Description Customize Click here to learn more about our approach to artificial intelligence. A Gray Local Media Station © 2002-2025 [x] search by querylyAdvanced Search This website uses cookies and other tracking technologies to enhance user experience, to display personalized advertisements and to analyze performance and traffic on our website. We also share information about your use of our site with our social media, advertising and analytics partners. You can exercise your rights to opt-out of sale of processing personal data for targeted advertising by clicking the "Customize" link. For more details see our Privacy Policy Customize Reject All Accept All Customize Options Your Privacy Strictly Necessary Cookies Marketing Cookies Your Privacy When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link. More information Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Marketing Cookies [x] Marketing Cookies You have the right to opt-out from the sale or sharing of your personal information at any time across business platform, services, businesses and devices. You can opt-out of the sale and sharing of your personal information by using this toggle switch. You have the right to opt-out from the sale of your personal data and the processing of your personal data for targeted advertising. You can opt-out of the sale of your personal data and targeted advertising by using this toggle switch. For more information on your rights, see our privacy notice. Targeting Cookies [x] Switch Label label These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Cookie List Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Clear [x] checkbox label label Apply Cancel Confirm My Choices Allow All We are Local Omaha Everyday Advertise With Us Heartland Heroes Sketch the Sky Becka's Talking 62° Omaha, NE Keep on reading Nebraska Department of Transportation planning new I-80 interchange NDOT will now start with environmental review and early design.6 News Undo Pumpkin patch owners face child abuse charges Aaron and Kara Smith, who own Cedar Creek Pumpkin Patch, are facing child abuse charges.6 News Undo Parents say middle schooler was attacked by another student in cafeteria, had to have parts of skull removed A middle school student in Texas is recovering from serious injuries after his family says he was attacked by another student at school and had his head slammed into a metal pole.6 News Undo Costco’s new shopping policy officially starts Monday Costco says a grace period is ending and the policy will take effect on Labor Day.6 News Undo Man sent to prison for raping 6-month-old baby, authorities say A Kentucky man has been sentenced to prison in what authorities are calling a horrific abuse case.6 News Undo South Carolina New Policy for Senior Drivers Don't pay your next auto insurance bill until you read this The Quote General \| SponsoredSponsored Learn More Undo Founders Are Using This Card to Get Paid Back for Daily Expenses Ramp \| SponsoredSponsored Undo High school cheerleader accused of instigating ‘love triangle’ fight that led to her boyfriend’s death A high school cheerleader in South Carolina is accused of instigating a fight that led to her boyfriend’s death, according to multiple reports.6 News Undo Longtime real estate agent found dead in basement prompts homicide investigation Authorities in Indiana are investigating the death of a longtime real estate agent.6 News Undo High school football coach resigns, hoping to force the cancellation of the season An Iowa high school football coach has resigned, saying he hopes the move forces the school to cancel the football season.6 News Undo by Taboolaby Taboola Brought to you by- \| SponsoredSponsored Undo EMS team under fire for treating man with antivenom after he was bitten by a mamba snake An EMS team in Kentucky is in hot water after they treated a man who had been bitten by a Mamba snake with antivenom.6 News Undo Big Ten announces network details for Huskers vs. Michigan game Big Ten Network announced the network details for the Nebraska vs. Michigan game on Sept. 20.6 News Undo This Business Card Tracks Spending and Gives 1.5% Back Ramp \| SponsoredSponsored Undo Little Known Brain Hack To Manifest Money bravethinkinginstitute \| SponsoredSponsored Learn More Undo Hikers injured while fighting off bear, troopers say A pair of hikers in Alaska were injured while fighting off a bear, according to Alaska State Troopers.6 News Undo 82-year-old killed her husband at assisted living facility, sheriff’s office says An 82-year-old woman in South Carolina is facing charges after officials said she shot and killed her husband last week.6 News Undo
17099	https://www.sciencedirect.com/topics/medicine-and-dentistry/uhthoffs-phenomenon	Uhthoff's Phenomenon - an overview \| ScienceDirect Topics Skip to Main content Journals & Books Uhthoff's Phenomenon In subject area:Medicine and Dentistry Uhthoff's phenomenon is defined as a transient worsening of visual function in patients with demyelinating optic neuropathy, such as multiple sclerosis, that occurs with increases in body temperature, likely due to heat-induced conduction block in the optic nerve. Vision typically returns to baseline when body temperature normalizes. AI generated definition based on:Encyclopedia of the Neurological Sciences (Second Edition), 2014 How useful is this definition? Press Enter to select rating, 1 out of 3 stars Press Enter to select rating, 2 out of 3 stars Press Enter to select rating, 3 out of 3 stars About this page Add to MendeleySet alert Discover other topics 1. On this page On this page Definition Chapters and Articles Related Terms Recommended Publications On this page Definition Chapters and Articles Related Terms Recommended Publications Chapters and Articles You might find these chapters and articles relevant to this topic. Review article Common Neurologic Disorders 2009, Medical Clinics of North AmericaArdith M. Courtney DO, ... Elliot Frohman MD, PhD Uhthoff's phenomenon Exposure to heat, prolonged exercise, infection, perimenstrual and even psychologic stress may cause transient appearance or worsening of neurologic symptoms or dysfunction in chronically demyelinated pathways (Uhthoff's phenomenon) because of conduction slowing or block.32–34 It is important to differentiate transient worsening (pseudoexacerbation) secondary to inciting factors (heat, fatigue, infection) from symptoms that represent a true acute relapse that may warrant therapy modification. Transient symptoms secondary to a pseudoexacerbation usually last less than 24 hours, fluctuate during the day, and resolve once the factor is removed. It is sometimes difficult to differentiate a pseudoexacerbation from a bona fide exacerbation, and clinical judgment must be exercised in deciding whether to pursue further investigation or modification of treatment. View article Read full article URL: Journal2009, Medical Clinics of North AmericaArdith M. Courtney DO, ... Elliot Frohman MD, PhD Chapter The pathophysiology of multiple sclerosis 2006, McAlpine's Multiple Sclerosis (Fourth Edition)Kenneth Smith, ... Hans Lassmann Uhthoff's phenomenon Many patients with multiple sclerosis exhibit a worsening of symptoms with a rise in body temperature – an effect known as Uhthoff's phenomenon, after the German neuro-ophthalmologist Wilhelm Uhthoff (1890) (Selhorst and Saul 1995). In fact, Uhthoff's original description was restricted to the worsening of vision upon exercise, and the role of temperature was missed, even though one of his patients reported that she experienced the same deterioration of vision when standing in front of a hot stove. Interest in Uhthoff's phenomenon was resurrected after the First World War when induced pyrexia became a fashionable therapy for multiple sclerosis (see Chapter 1), and many authors reported benefits even though it was appreciated that warming could result in neurological deterioration, and even death. Indeed, the worsening was sufficiently robust and reproducible that the phenomenon later underpinned the diagnostic ‘hot bath test’ (Guthrie 1951; Malhotra and Goren 1981). This procedure was shown to be associated with clinical deterioration in >80% of patients, and 60% seemingly developed new signs. In fact, the Uhthoff phenomenon can be provoked by a hot shower (Waxman and Geschwind 1983), sunbathing (Avis and Pryse-Phillips 1995; J.R. Berger and Sheremata 1985; Harbison et al 1989), use of a hair dryer (Brickner 1950), exercise (van Diemen et al 1992) as slight as ascending stairs (Edmund and Fog 1955), or even by the normal circadian change in body temperature (D.G. Baker 2002; F.A. Davis et al 1973; Namerow 1968a; Romani et al 2000). For example, a pilot with multiple sclerosis experienced blurred vision each afternoon (changing from an acuity of 20/30 in the morning to 20/50 by the afternoon), associated with corresponding changes in the pattern visual evoked potential (Figure 13.33; Scherokman et al 1985). The exacerbations due to warming occur very promptly. In extreme instances, support to prevent drowning in a hot bath has been described (Guthrie 1951). Deaths due to scalding and hyperthermia have been reported from immersion in hot water and sunbathing, respectively (Harbison et al 1989; Waxman and Geschwind 1983) indicating that patients can be overcome by weakness so rapidly that they are unable to summon help (Avis and Pryse-Phillips 1995). Cooling can not only reverse the deficits caused by warming, but also prove inherently beneficial (Schwid et al 2003). Thus an improvement in clinical function is often reported by patients following a cold bath, or even after drinking cold water. The pilot referred to above found that several minutes after drinking iced water his normally blurred vision was improved sufficiently to read newsprint for 30–40 minutes. This improvement was accompanied by an increase in amplitude of the pattern evoked response (Figure 13.33). On a different day, a similar improvement in acuity was accompanied by reduction in temperature of the tympanic membrane of 0.25ºC. In another case, improvements in vision and amplitude of the visual evoked potential were documented over time after drinking iced water (Figure 13.34; Hopper et al 1972; W.I. McDonald 1986). These effects of temperature are typically reversible, although persistent deficits are reported (J.R. Berger and Sheremata 1983; 1985; F.A. Davis 1985). (A website regarding the effects of temperature, organized by educated patients and without the contents necessarily having been professionally scrutinized, has been established at The clinical observations suggest that warming and cooling can cause transient block and restoration of conduction, respectively, and these effects are easily reproduced in experimentally demyelinated axons both in central axons (Figure 13.35; K.J. Smith et al 2000) and peripheral axons (Bostock et al 1981; F.A. Davis and Jacobson 1971; F.A. Davis et al 1975; Pencek et al 1980; Pender and Sears 1984; Rasminsky 1973; Sears and Bostock 1981; Sears et al 1978). Evidence for conduction failure in patients has been provided by Persson and Sachs (1981), who showed that impairment of visual acuity induced by exercise is accompanied by a reduction in amplitude of the visual evoked potential without a change in latency, indicating that conduction block has developed. Recovery of acuity was accompanied by restored amplitude of the evoked potential. An earlier study showed a prominent (40%) reduction in flicker fusion frequency upon exposure to heat, even though the patient underwent only a 0.2°C rise in temperature (Figure 13.36; Namerow 1972). Because heating one arm caused weak muscles in other limbs, an effect that could be prevented by applying a tourniquet to the arm, Uhthoff's phenomenon was first attributed to humoral effects or to a reflex reduction in cerebral blood flow (Guthrie 1951). However, this hypothesis was challenged when a large quantity of blood, sampled at a time when heat-induced visual loss was at its maximum, failed to induce the same symptoms after transfusion back into the cooled patient once normal vision had been restored. It is now believed, on the basis of experimental studies, that the exacerbations upon warming are primarily due to shortened duration of the action potential at the node just before the demyelinated region, namely the node responsible for driving the current to depolarize the demyelinated axolemma to its firing threshold (Paintal 1966). Duration of the action potential changes because the temperature coefficient for sodium inactivation is larger than that for activation (F.A. Davis and Schauf 1981). We have described above how demyelinated axons have a reduced safety factor for conduction, such that many hover between conduction block and successful transmission of the nerve impulse. In this situation, the minute shortening in duration of the action current can be the deciding factor in causing conduction failure. Although the generally accepted view is that temperature-mediated conduction block arises from effects on duration of the action potential, a different mechanism is suggested by a study that examined the consequences, both on body temperature and leucocyte nitric oxide production, of wearing a cooling garment (Beenakker et al 2001). Cooling did not reduce core temperature as measured from the tympanic membrane but did decrease mean leucocyte nitric oxide production by 41%, when this was assessed from blood samples taken 3 hours later. The authors argue that if core temperature was unchanged, the traditional explanation involving prolongation of action potential duration cannot be offered for supposed improvement in the success of axonal conduction. Thus, although temperature has profound effects on axonal conduction in demyelinated axons, these effects may not account for the measured improvement in function associated with wearing a cooling garment. Rather, the authors attribute the beneficial effects of cooling to the reduction in nitric oxide production, as nitric oxide can promote conduction block, especially in demyelinated axons (Redford et al 1997). The reduction in nitric oxide production has been attributed to elevated plasma norepinephrine (noradrenaline) caused by cooling the sympathetic nervous system, which in turn leads to reduced release of proinflammatory cytokines by mononuclear cells and lowered NOS activity in leucocytes (Beenakker et al 2001). These effects may offer a partial explanation for the decreased sense of fatigue reported by patients during a month of daily cooling (Schwid et al 2003). If these effects are confirmed, it seems to us that cooling may be beneficial for two reasons. First, in circumstances where core cooling is reliably achieved, it will restore conduction in demyelinated axons. This is a very robust observation in the laboratory, where our unpublished observations show that conduction can be blocked and restored many times per minute if the temperature of the lesion is rapidly modulated. Secondly, cooling may have prolonged beneficial effects arising from the modulation of nitric oxide production. This latter mechanism would explain why patients can experience a sustained benefit from wearing a cooling garment for 45–60 minutes, two to three times each day (Beenakker et al 2001; Schwid et al 2003). It is a curious fact that Uhthoff's phenomenon is rarely encountered in demyelinating peripheral neuropathy. Why this should be is unclear. One of us (KJS) has formed the casual impression that whereas conduction in all demyelinated axons is very insecure when first restored, peripheral axons achieve a higher security than central fibres over time. In patients, any such benefit will be augmented by the fact that the temperature of peripheral nerves is often substantially cooler than central nerve fibres. This difference will tend to persist even in a hot patient due to evaporative cooling of the periphery, although it is acknowledged that this argument would not apply if lesions were present in the intrathecal portions of spinal roots. Another consideration is that peripheral axons are more likely to be repaired by remyelination than their central counterparts, and remyelinated axons do not exhibit the conduction changes responsible for Uhthoff's phenomenon. Curiously, some patients with peripheral neuropathy report a negative Uhthoff's finding, such that their symptoms increase with body cooling. We suspect that this finding is due to the markedly different magnitude of temperature change that can occur in peripheral compared with central nervous system axons. On a cold day, temperature of the extremities may reduce to the point where other factors come into play, including the depression of metabolic activity and compromise in blood supply due to vasoconstriction. The beneficial effects of cooling (Figure 13.35) have encouraged the view that a useful symptomatic therapy for multiple sclerosis may result from drugs that, at normal body temperature, mimic the effects of body cooling by prolonging duration of the action potential (Davis and Schauf 1981; Sears and Bostock 1981; Waxman 1992; Waxman et al 1994b). This can be achieved by potassium channel blockade using 4-aminopyridine or by delaying sodium channel inactivation with scorpion venom or pyrethroids. Both strategies can be shown to restore conduction in the laboratory (Bostock et al 1978; 1981; Bowe et al 1987; Lees 1998; Sherratt et al 1980; Targ and Kocsis 1985). As discussed in Chapter 17, 4-aminopyridine is an effective symptomatic therapy in multiple sclerosis (reviewed in Bever 1994; Schwid et al 1997b), although whether its efficacy depends on prolonging action potentials at the driving node, or effects on the demyelinated axolemma remains uncertain. Indeed, it is not even clear that 4-aminopyridine acts at the site of the lesion (K.J. Smith et al 2000). Laboratory examination has revealed that the drug has no discernible influence on the success of conduction through central demyelinating lesions at therapeutic concentrations (which are orders of magnitude below the concentrations applied directly to demyelinated axons in the initial experimental studies), but it strongly potentiates synaptic transmission (K.J. Smith et al 2000). However, if 4-aminopyridine does not restore conduction, what underlies its beneficial effects? It seems possible that, in pathways where conduction in many axons fails due to demyelination or degeneration, potentiating the synaptic efficacy of surviving axons may help to compensate for conduction failure. Whatever its mechanism of action, the adoption of 4-aminopyridine as a therapy has been limited, largely because most patients find its modest advantages are offset by the inherent risk of convulsions (Bever et al 1994; Blight et al 1991). It is even less surprising that scorpion toxin has also failed to achieve clinical use, although inadvertent ‘field trials’ in desert regions have apparently resulted in some favourable chance consequences (Breland and Currier 1983). Show more View chapterExplore book Read full chapter URL: Book 2006, McAlpine's Multiple Sclerosis (Fourth Edition)Kenneth Smith, ... Hans Lassmann Review article Multiple Sclerosis Rehabilitation 2013, Physical Medicine and Rehabilitation Clinics of North AmericaAlexius E.G. Sandoval MD Heat Intolerance A frequent concern with exercise in MS is potentially triggering Uhthoff phenomenon. Uhthoff phenomenon was originally described as a transient amblyopia due to overheating from exercise.27,61 The term has since been expanded to include other symptoms triggered by overheating.27,62 The exact mechanism of Uhthoff phenomenon is unclear. It may be due to heat-worsened conduction across partially demyelinated axons, fatigue of damaged neuronal pathways with repetitive nerve transmission,27,63 or a hormonal factor produced with cooling.4,64 Exercise-induced Uhthoff phenomenon should not be regarded as a contraindication to exercise.27 It is usually reversible and often resolves within an hour or even sooner with rapid cooling.27 Furthermore, it is still more common for heating to produce just general fatigue than an Uhthoff phenomenon with focal neurologic deficits.27,49 How much does the body temperature change with exercise? Studies have shown that routine exercise does not significantly increase core body temperature. Ponichtera-Mulcare and colleagues 55 noted a mean rectal temperature change of 0.1°C during land-based exercise and −0.1°C during water-based exercise.27 Alternatively, normal thermoregulatory reflexes (eg, sweating and vasodilatation) that maintain a steady core temperature during routine exercise may be impaired in persons with MS. In such cases, a rise in core temperature of even less than 1°C may be enough to trigger heat-related symptoms.27,65 The use of cooling devices and strategies seems to provide some modest benefits for persons with MS. One such device, used by Capello and colleagues 66 and Kraft and Alquist,67 was a head-vest liquid cooling garment. The former found a slight improvement in pyramidal and cerebellar function 27 whereas the latter demonstrated a treatment effect for strength, dynamic coordination, and endurance capacity, with greater heat loss associated with greater motor function gain.27,67 Syndulko and colleagues 68 saw reduced fatigue and improved ambulation for up to 3 hours postcooling with the use of either the liquid cooling system or an icepack suit.27 When engaging in pool-based exercises, the ideal pool temperature for heat-sensitive MS individuals seems between 27°C and 29°C (80°F–84°F).27,54,69 Temperatures below 27°C can paradoxically increase spasticity.27,70 View article Read full article URL: Journal2013, Physical Medicine and Rehabilitation Clinics of North AmericaAlexius E.G. Sandoval MD Review article Multiple sclerosis in ophthalmology: Beyond optic neuritis 2020, Medicina Clínica (English Edition)Rafel Alcubierre, ... Silvia Muñoz Visual phenomena Commonly associated with demyelinating optic neuritis, and by extension with MS, the Uhthoff phenomenon consists of a visual loss due to heat or exercise. A saturation effect may also appear with exposure to high levels of light. Another typical anomaly, the Pulfrich phenomenon, is the abnormal perception of a moving object’s path. These phenomena are not exclusive to MS and may appear in other patients who have suffered from an asymmetric optic nerve abnormality. View article Read full article URL: Journal2020, Medicina Clínica (English Edition)Rafel Alcubierre, ... Silvia Muñoz Review article Visual dysfunction in retinal and optic nerve disease 2003, Neurologic ClinicsTimothy Murtha MD, Steven F. Stasheff MD, PhD Exercise- or heat-induced transient visual loss (Uhthoff's phenomenon) frequently is seen with optic neuritis and rarely with compressive optic neuropathy, sarcoidosis, or Leber's hereditary optic neuropathy (LHON) . Transient visual loss precipitated by eye movements may herald compression of the optic nerve by an intraconal mass or a kink in the optic nerve (especially with pseudotumor cerebri or demyelinating disease) [17,22]. Other visual disturbances that occur with head movement may indicate involvement of the vestibular system, such as a malfunctioning vestibulo-ocular reflex (VOR), tested most easily by headshake acuity. View article Read full article URL: Journal2003, Neurologic ClinicsTimothy Murtha MD, Steven F. Stasheff MD, PhD Chapter NEURO-OPHTHALMOLOGY 2008, Moorfields Manual of Ophthalmology Other ocular features of MS ▪ Uhthoff's phenomenon: transiently decreased vision with exertion or in hot conditions. ▪ Intermediate uveitis: sheathing of retinal venules may be seen. ▪ Cranial nerve palsy: commonly 6th, but any may be affected. ▪ Internuclear ophthalmoplegia: the ipsilateral eye fails to adduct or adducts slowly, whilst the contralateral eye has horizontal nystagmus on abduction. ▪ Wall-eyed bilateral internuclear ophthalmoplegia (WEBINO): both eyes have no adduction. Prisms may reduce diplopia. ▪ One-and-a-half syndrome: one eye has no voluntary horizontal movements and the other can only abduct. If supranuclear, the palsy is overcome by dolls-head movements. Other supranuclear gaze palsies may occur. ▪ Cerebellar disease: dysmetric saccades and gaze-evoked nystagmus. ▪ Acquired pendular nystagmus: associated with reduced acuity; the amplitude may differ greatly in the two eyes. View chapterExplore book Read full chapter URL: Book 2008, Moorfields Manual of Ophthalmology Chapter The Conduction Properties of Demyelinated and Remyelinated Axons 2005, Multiple Sclerosis As A Neuronal DiseaseKenneth J. Smith Ph.D., Stephen G. Waxman M.D., Ph.D. 4.Temporary Exacerbation Caused by Warming: Uhthoff's Phenomenon The security of conduction in demyelinated axons is markedly affected by temperature, and this effect can readily be demonstrated in the laboratory in both central (Fig. 7) (Smith et al., 2000) and peripheral experimentally demyelinated axons (Bostock et al., 1981; Davis et al., 1975; Rasminsky, 1973). The effects can be sufficiently strong that even subtle changes in body temperature can have a profound effect on the expression of symptoms by patients with MS. Indeed, many patients with MS exhibit a worsening of some symptoms upon body warming, and the worsening is sufficiently robust and reproducible that at one time the phenomenon was used in the diagnosis of MS in the form of the “hot bath test” (Guthrie, 1951; Malhotra and Goren, 1981). The effect was first described in 1890 by Uhthoff, and the phenomenon now bears his name. Uhthoff-like phenomena can be provoked by a hot shower (Waxman and Geschwind, 1983), sunbathing (Berger and Sheremata, 1985; Harbison et al., 1989; Avis and Pryse-Phillips, 1995), exercise (van Diemen et al., 1992; Edmund and Fog, 1955), or even by the normal circadian change in body temperature (Fig. 8) (Scherokman et al., 1985). Just as warming can be deleterious to function, so cooling can be beneficial and can sometimes be realized after a cool bath or simply after drinking cold water. Indeed, improvement in vision has been documented after drinking iced water, and these functional improvements are accompanied by an increase in the amplitude of the visual-evoked potential (McDonald, 1986; Hopper et al., 1972). Sign in to download full-size image Figure 7. Two families of superimposed records obtained from excised dorsal columns examined in vitro, showing the effects of temperature on axonal conduction. The records on the left are from normal tissue, while those on the right include an experimental demyelinating lesion in the conduction pathway, induced 21 days previously. The records were obtained as the temperature in the central recording lane (containing the lesion, where present) was raised from 25°C to 37°C in 1°C intervals. (A) The temperature changes had little effect on conduction along normal axons. (B) In contrast, temperature changes had prominent effects on conduction in demyelinated axons, distinguished by their longer latency. At cooler temperatures many more demyelinated axons were able to conduct than at normal body temperature, when nearly all the demyelinated axons failed to conduct. (Reproduced from Smith et al., 2000.)Copyright © 2000 Sign in to download full-size image Figure 8. Visual-evoked potentials from a pilot with MS who experienced blurred vision each afternoon (visual acuity indicated). Visual acuity improved substantially in the afternoon within 3 minutes of drinking iced water, and this improvement was accompanied by an increase in the visual-evoked response. On a different day a similar improvement in acuity was accompanied by a reduction in the temperature of the tympanic membrane of 0.25°C. (Reproduced from Scherokman et al., 1985.)Copyright © 1985 It is generally accepted that temperature affects the success of conduction in demyelinated axons mainly by affecting the duration of the action potential (Paintal, 1966) at the driving node, the node just before the demyelinated region that is responsible for driving the current to depolarize the demyelinated axolemma to its firing threshold. The change in duration arises from the fact that the temperature coefficient for sodium inactivation is larger than that for sodium activation (Davis and Schauf, 1981). In axons balanced on the “knife edge” between conduction block and successful conduction (see Safety Factor) this change can easily be sufficient to decide whether conduction succeeds or fails. Thus although warming will increase the conduction velocity (Fig. 9) (Swadlow et al., 1981), thereby offsetting the reduction in conduction velocity produced by demyelination, the net effect of warming is typically deleterious to the patient, at least with regard to the expression of symptoms. It is worth mentioning that although the mechanism described (changes in action potential duration) may be paramount in most lesions in patients, the expression of Uhthoff's phenomenon may involve multiple mechanisms, perhaps including the modulation of nitric oxide production (see Chapter 18). Sign in to download full-size image Figure 9. Axonal conduction velocity increases, but safety factor is reduced, as temperature increases. (A) Changes in conduction latency for callosal axons, measured in rabbits, as body temperature is altered. Note the decrease in latency, reflecting increased conduction velocity, as temperature rises. (B, C) Extracellular recordings of action potentials elicited in a callosal efferent neuron by stimulation of its axon in the corpus callosum, at 36.5°C (B) and at 38.7°C (C), display more rapid conduction (resulting in earlier onset of the action potential) at higher temperature. For the second action potential evoked at high frequency, the safety factor is reduced, resulting in an inflection (arrow in B) between the initial segment spike and somatodendritic spike at 36.5°C. At 38.7C° the action potential fails to invade the soma (arrow in C) because of a superimposed temperature-induced decrease in safety factor. (Reproduced from Swadlow et al., 1981.)Copyright © 1981 Show more View chapterExplore book Read full chapter URL: Book 2005, Multiple Sclerosis As A Neuronal DiseaseKenneth J. Smith Ph.D., Stephen G. Waxman M.D., Ph.D. Chapter Thermoregulation: From Basic Neuroscience to Clinical Neurology, Part II 2018, Handbook of Clinical NeurologyScott L. Davis, ... Thad E. Wilson Reversibility of heat sensitivity One redeeming feature of impaired heat sensitivity in MS is that effects are transient and temporary. Typically, deficits caused by increases in temperature are reversible by removing heat stressors and allowing subsequent cooling. Davis et al. (2008) quantified Uhthoff's phenomenon during passive whole-body heating and its reversibility with the subsequent application of active cooling by quantifying horizontal eye movement velocities in MS patients with internuclear ophthalmoparesis (INO), an abnormality characterized by the slowing of horizontal eye movements moving toward the nose (adduction). INO is caused by an MS lesion localized in the medial longitudinal fasciculus, a central periventricular tract system within the brainstemtegmentum. The speed of horizontal eye movements already slowed in MS patients with INO was further slowed from pre-heating baseline when core body temperature was raised by ~0.8°C using passive whole-body heating via water-perfused suits. Eye movements in patients with INO returned to pre-heating baseline following the perfusion of cool water through the water-perfused suit for a period of 1 hour (Fig. 42.2) (Davis et al., 2008). Interestingly, MS patients without INO and likely no lesion burden in the medial longitudinal fasciculus showed no decrements in horizontal eye movements, indicating that Uhthoff's phenomenon is localized to areas of the CNS where MS damage is present or has occurred (Fig. 42.2) (Davis et al., 2008). Sign in to download hi-res image Fig. 42.2. Data from healthy controls (control), multiple sclerosis (MS) patients without internuclear ophthalmoparesis (MS-control), and MS patients diagnosed with internuclear ophthalmoparesis (MS-INO) showing ocular function responses expressed as velocity versional dysconjugacy index (VDI) during passive whole-body heating and subsequent passive whole-body cooling. A significant slowing of horizontal eye movements (increased velocity VDI) was observed in the MS-INO group during heating (indicated by ). However, ocular function was restored to preheating baseline with subsequent cooling (indicated by #). (Modified from Davis SL, Frohman TC, Crandall CG, et al. (2008) Modeling Uhthoff's phenomenon in MS patients with internuclear ophthalmoparesis. Neurology 70: 1098–1106.) View chapterExplore book Read full chapter URL: Handbook2018, Handbook of Clinical NeurologyScott L. Davis, ... Thad E. Wilson Review article Neurology Case Studies 2006, Neurologic ClinicsJ.D. Bartleson MD A differential diagnosis of pseudoclaudication is shown in Table 4. Multiple sclerosis and arteriovenous malformations of the spinal cord can cause pseudoclaudication-like symptoms. In patients who have multiple sclerosis, the exertion involved with walking can raise their body temperature slightly and cause a Uhthoff's phenomenon with the development of temporary lower limb numbness or weakness that resolves with rest. Although uncommon, spinal cord vascular malformations also can cause pseudoclaudication-like symptoms. Dural arteriovenous fistulas (DAVF) are the most common spinal cord vascular malformation and often present with symmetric or asymmetric sensory loss, lower limb weakness, and sometimes pain in the low back and lower limbs that can worsen with standing, walking, and Valsalva's maneuver. DAVF affect middle-aged and older men more often than women, and it is believed that high-pressure arterial flow from a radicular artery causes venous hypertension, engorgement, and secondary spinal cord ischemia. The large majority of DAVF involve the thoracolumbar spinal cord. Subtle evidence of myelopathy on examination often is encountered; bowel and bladder dysfunction usually occurs relatively late . EMG can be normal or show polyradiculopathy or even anterior horn cell disease. MRI with and without gadolinium is the diagnostic test of choice, but the spinal cord also must be imaged. MRI in DAVF shows areas of intramedullary increased T2 signal in the spinal cord with patchy, diffuse spinal cord enhancement often with significant, enlarged pial blood vessels. Myelography with CT scanning sometimes is used to help confirm the presence of serpiginous blood vessels on the surface of the spinal cord. Spinal magnetic resonance angiography is used to confirm the presence of a DAVF and guide the performance of conventional catheter spinal angiography, which can define the abnormal vascular anatomy and help plan surgical intervention. Surgical disconnection of the DAVF reverses the pathophysiology and, especially if performed early in the course, can result in symptomatic improvement . Table 4. Differential diagnosis of lumbar stenosis Vascular claudication—usually resulting from atherosclerotic disease Osteoarthritis of hips or knees Lumbar disk protrusion Unrecognized neurologic disease Multiple sclerosis Intraspinal tumor Spinal cord arteriovenous malformations Peripheral neuropathy Rarely, communicating hydrocephalus Adapted from Bartleson JD, O'Duffy JD. Spinal stenosis. In: Koopman WJ, editor. Arthritis and allied conditions. 14th ed. Baltimore: Lippincott Williams & Wilkins; 2001. p. 2048; with permission. Show more View article Read full article URL: Journal2006, Neurologic ClinicsJ.D. Bartleson MD Review article Neurology (Part 1 of 2) 2016, MedicineAngus Kennedy, Rasheed Zakaria Fluctuating levels of severity Pathological processes, such as inflammatory or autoimmune disease (common examples being MS and myasthenia gravis), often present with episodic symptoms but with a longer time frame. Patients may be asymptomatic between exacerbations. The level of function at different time points, especially before and after treatment, should be assessed. Useful questions include the time taken to perform a certain task, how far the patient can walk unaided and how many activities of daily living, such as washing and dressing, they can perform alone. These factors are often formalized in scoring systems, such as the Expanded Disability Status Scale for MS 4 or the more general Barthel Index.5 With suspected myasthenia gravis, the patient should be quizzed about fatigability, whereas in MS, symptoms may be more pronounced in summer or after a hot bath (Uhthoff's phenomenon). The pattern of symptoms in MS is particularly important in determining the subtypes of disease (e.g. relapsing–remitting; Figure 3). Sign in to download hi-res image Figure 3. Line graph of disability against time for relapsing–remitting multiple sclerosis (MS). View article Read full article URL: Journal2016, MedicineAngus Kennedy, Rasheed Zakaria Related terms: Optic Neuritis Amaurosis Nerve Block Multiple Sclerosis Optic Nerve Cytokine Neurologic Disease Visual Acuity Visual Impairment Demyelination View all Topics Recommended publications Neurologic ClinicsJournal The Lancet NeurologyJournal Multiple Sclerosis and Related DisordersJournal Survey of OphthalmologyJournal Browse books and journals About ScienceDirect Remote access Advertise Contact and support Terms and conditions Privacy policy Cookies are used by this site. Cookie settings All content on this site: Copyright © 2025 or its licensors and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply. We use cookies that are necessary to make our site work. We may also use additional cookies to analyze, improve, and personalize our content and your digital experience. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy Cookie Settings Accept all cookies Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. You may also be able to exercise your privacy choices as described in our Privacy Policy Allow all Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. Cookie Details List‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Cookie Details List‎ Contextual Advertising Cookies [x] Contextual Advertising Cookies These cookies are used for properly showing banner advertisements on our site and associated functions such as limiting the number of times ads are shown to each user. Cookie Details List‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm my choices

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.