text stringlengths 235 3.08k |
|---|
solutions by 578. 0.3x β 0.1y = β 10 β0.1x + 0.3y = 14 579. 0.4x β 0.2y = β 0.6 β0.1x + 0.05y = 0.3 580. 4x + 3y β 3z = β 4.3 5x β 4y β z = β 6.1 x + z = β 0.7 This content is available for free at https://cnx.org/content/col11758/1.5 581. β2x β 3y + 2z = 3 βx + 2y + 4z = β 5 β2y + 5z = β 3 For the following exercises... |
β 9y β 3z = 3 2 x β 5y β 5z = 5 2 3 10 1 10 593. x β 1 5 x β 1 10 x β 1 2 y β 3 10 50 z = β 9 50 z = β 1 5 2 5 CHAPTER 11 PRACTICE TEST Is the following ordered pair a solution to the system of equations? 602. y2 + x2 = 25 y2 β 2x2 = 1 594. β5x β y = 12 x + 4y = 9 with ( β 3, 3) For the following exercises, solve the ... |
6, what would be the determinant if you switched rows 1 and 3, multiplied the second row by 12, and took the inverse? 621. 0.1x + 0.1y β 0.1z = β 1.2 0.1x β 0.2y + 0.4z = β 1.2 0.5x β 0.3y + 0.8z = β 5.9 For the following exercises, solve using a system of linear equations. 622. A factory producing cell phones has the ... |
at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1341 12 | ANALYTIC GEOMETRY Figure 12.1 (a) Greek philosopher Aristotle (384β322 BCE) (b) German mathematician and astronomer Johannes Kepler (1571β1630) Chapter Outline 12.1 The Ellipse 12.2 The Hyperbola 12.3 The Parabola 12.4 Rotation of Axes 12.5... |
C., shown in Figure 12.2, is such a room.[1] It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can... |
of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See Figure 12.5. Figure 12.5 In this ... |
x, y) on the ellipse. We know that the sum of these distances is 2a for the vertex (a, 0). It follows that d1 + d2 = 2a for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic. d1 + d2 = (x β ( β c))2 + (y β 0)2 + (x β c)2 + (y β 0)2 = 2a Dist... |
+ a2 y2 = a2 b2 a2 b2 x2 b2 a2 b2 + a2 b2 x2 a2 + a2 y2 a2 b2 = y2 b2 = 1 β + a2 y2 = a2 β 2 Expand the squares. Expand remaining squares. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2. Rewrite. Factor common terms. Set b2 = a2 β c2. Divide both sides by a2 ... |
coordinates of the vertices are (Β±a, 0) the length of the minor axis is 2b the coordinates of the co-vertices are (0, Β± b) the coordinates of the foci are (Β±c, 0), where c2 = a2 β b2. See Figure 12.7a The standard form of the equation of an ellipse with center (0, 0) and major axis on the y-axis is x2 b2 + y2 a2 = 1 w... |
oci, to solve for b2. 3. Substitute the values for a2 and b2 into the standard form of the equation determined in Step 1. Example 12.1 Writing the Equation of an Ellipse Centered at the Origin in Standard Form What is the standard form equation of the ellipse that has vertices (Β±8, 0) and foci (Β±5, 0)? Solution The foc... |
form of the equation we saw previously, with x replaced by (x β h) and y replaced by β βy β kβ β . Standard Forms of the Equation of an Ellipse with Center (h, k) The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the x-axis is (x β h)2 a2 + where β’ a > b 2 (12.3) β βy β kβ β ... |
tic Geometry Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form. 1. Determine whether the major axis is parallel to the x- or y-axis. If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard f... |
+ 2 2 β β = (β2, β3) Next, we find a2. The length of the major axis, 2a, distance between the y-coordinates of the vertices. is bounded by the vertices. We solve for a by finding the 2a = 2 β (β8) 2a = 10 a = 5 So a2 = 25. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic G... |
ices, and foci. a. b. If the equation is in the form x2 a2 + the major axis is the x-axis y2 b2 = 1, where a > b, then the coordinates of the vertices are (Β±a, 0) the coordinates of the co-vertices are (0, Β± b) the coordinates of the foci are (Β±c, 0) If the equation is in the form x2 b2 + the major axis is the y-axis y... |
4). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 12.9. Figure 12.9 12.3 Graph the ellipse given by the equation x2 36 + y2 4 and foci. = 1. Identify and label the center, vertices, co-vertices, Example 12.4 Graphing an Ellipse Centered at ... |
, co-vertices, and foci. Graphing Ellipses Not Centered at the Origin When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, (h, k), we use the standard forms (x β h)2 + 2 β β βy β kβ b2 = 1, a > b for horizon... |
c using the equation c2 = a2 β b2. 3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 12.5 Graphing an Ellipse Centered at (h, k) Graph the ellipse given by the equation, (x + 2)2 4 + vertices, and foci. β β βy β 5β 9 2 = 1. Identify and la... |
, and foci. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1357 Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form. 1. Recognize that an ellipse described by an equation in the form ax2 + by2 + cx + dy +... |
β100 + 100 + 36 4 Rewrite as perfect squares. 4(x β 5)2 + 9β βy + 2β β 2 = 36 Divide both sides by the constant term to place the equation in standard form. (x β 5)2 9 + β β βy + 2β 4 2 = 1 Now that the equation is in standard form, we can determine the position of the major axis. Because 9 > 4, the major axis is para... |
Involving Ellipses Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Som... |
, we have 2b = 46, so b = 23, and b2 = 529. 1360 Chapter 12 Analytic Geometry Therefore, the equation of the ellipse is x2 2304 + y2 529 = 1. b. To find the distance between the senators, we must find the distance between the foci, (Β±c, 0), where c2 = a2 β b2. Solving for c, we have: c2 = a2 β b2 c2 = 2304 β 529 c = Β± ... |
the origin and foci along the y-axis? Algebraic For the following exercises, determine whether the given equations represent ellipses. If yes, write in standard form. 6. 7. 8. 9. 2x2 + y = 4 4x2 + 9y2 = 36 4x2 β y2 = 4 4x2 + 9y2 = 1 10. 4x2 β 8x + 9y2 β 72y + 112 = 0 For the following exercises, write the equation of ... |
β 16 β β βy β 7β 9 11. 12. 13. 14. 15. 16. 17. 18. Graphical For the following exercises, graph the given ellipses, noting center, vertices, and foci x2 25 x2 16 + + y2 36 y2 9 = 1 = 1 4x2 + 9y2 = 1 81x2 + 49y2 = 1 32. 33. 34. 35. 36. 1362 (x β 2)2 64 + β β βy β 4β 16 2 = 1 Chapter 12 Analytic Geometry 37. 38. 39. 40. ... |
3, 11) ; one focus: β β3, 5+4 2β β 51. Center (β3, 4) ; vertex (1, 4) ; one focus: ββ3 + 2 3, 4β β β For the following exercises, given the graph of the ellipse, determine its equation. 52. This content is available for free at https://cnx.org/content/col11758/1.5 54. 55. Chapter 12 Analytic Geometry 1363 height to the... |
2 4x2 β 8x + 9y2 β 72y + 112 = 0 9x2 β 54x + 9y2 β 54y + 81 = 0 Real-World Applications Find the equation of the ellipse that will just fit inside a 62. box that is 8 units wide and 4 units high. Find the equation of the ellipse that will just fit inside a 63. box that is four times as wide as it is high. Express in t... |
separate unbounded curves that are mirror images of each other. See Figure 12.16. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1365 Figure 12.16 A hyperbola Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyper... |
, 0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x, y) such that the difference of the distances from (x, y) to the foci is constant. See Figure 12.18. Figure 12.18 If (a, 0) is a vertex of the hyperbola, the distance from (βc, 0) to (a, 0) is a β (βc) = a + c. The distanc... |
y2 + (x β c)2 + y2 x2 + 2cx + c2 + y2 = 4a2 + 4a (x β c)2 + y2 + x2 β 2cx + c2 + y2 2cx = 4a2 + 4a (x β c)2 + y2 β 2cx 4cx β 4a2 = 4a (x β c)2 + y2 cx β a2 = a (x β c)2 + y2 2 2 βcx β a2β β β β£ (x β c)2 + y2β€ β¦ βx2 β 2cx + c2 + y2β β = a2 β‘ c2 x2 β 2a2 cx + a4 = a2 β c2 x2 β 2a2 cx + a4 = a2 x2 β 2a2 cx + a2 c2 + a2 y... |
2 = 1 (12.5) where β’ β’ β’ β’ β’ β’ β’ the length of the transverse axis is 2a the coordinates of the vertices are (Β±a, 0) the length of the conjugate axis is 2b the coordinates of the co-vertices are (0, Β± b) the distance between the foci is 2c, where c2 = a2 + b2 the coordinates of the foci are (Β±c, 0) the equations of the... |
located at ( Β± a, 0), and the foci are located at (Β±c, 0). y2 b2 = 1, then the transverse axis lies on the x-axis. The vertices If the equation has the form y2 a2 β are located at (0, Β± a), and the foci are located at (0, Β± c). x2 b2 = 1, then the transverse axis lies on the y-axis. The vertices 2. Solve for a using t... |
hyperbolas centered at (0, 0), we see that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2. Note that this equation can also be rewritten as b2 = c2 β a2. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices. Given the vertices... |
tic Geometry 1371 Finally, we substitute a2 = 36 and b2 = 4 into the standard form of the equation, x2 a2 β y2 b2 = 1. The equation of the hyperbola is x2 36 β y2 4 = 1, as shown in Figure 12.20. Figure 12.20 12.9 What is the standard form equation of the hyperbola that has vertices (0, Β± 2) and foci β β0, Β± 2 5β β ? Hy... |
) and transverse axis parallel to the y-axis is β βy β kβ β 2 a2 β (x β h)2 b2 = 1 (12.8) where β’ β’ β’ β’ β’ β’ the length of the transverse axis is 2a the coordinates of the vertices are (h, k Β± a) the length of the conjugate axis is 2b the coordinates of the co-vertices are (h Β± b, k) the distance between the foci is 2c,... |
οΏ½οΏ½ β 2 (x β h)2 b2 2. 3. Find a2 by solving for the length of the transverse axis, 2a, which is the distance between the given vertices. 4. Find c2 using h and k found in Step 2 along with the given coordinates for the foci. 5. Solve for b2 using the equation b2 = c2 β a2. 6. Substitute the values for h, k, a2, and b2 ... |
b2 = c2 β a2 = 25 β 9 = 16 Finally, substitute the values found for h, k, a2, and b2 into the standard form of the equation. (x β 3)2 9 β (y + 2)2 16 = 1 What is the standard form equation of the hyperbola that has vertices (1, β2) and (1, 8) and foci 12.10 (1, β10) and (1, 16)? Graphing Hyperbolas Centered at the Ori... |
of the foci are (0, Β± c) the equations of the asymptotes are y = Β± x a b 3. Solve for the coordinates of the foci using the equation c = Β± a2 + b2. 4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola. Example 12.11 Graphing a Hyperbola Centered... |
olas Not Centered at the Origin Graphing hyperbolas centered at a point (h, k) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms (x β h)2 a2 β 2 β β βy β kβ b2 = 1 for horizontal hyperbolas, and β βy β kβ β 2 a2 β (x β h)2 b2 = 1 for vertical hype... |
h, k Β± a) the coordinates of the co-vertices are (h Β± b, k) the coordinates of the foci are (h, k Β± c) the equations of the asymptotes are y = Β± a b(x β h) + k 3. Solve for the coordinates of the foci using the equation c = Β± a2 + b2. 4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate pla... |
or a = 6 and b = 9. Thus, the transverse axis is parallel to the x-axis. It follows that: β’ β’ β’ β’ the center of the ellipse is (h, k) = (2, β5) the coordinates of the vertices are (h Β± a, k) = (2 Β± 6, β5), or (β4, β5) and (8, β5) the coordinates of the co-vertices are (h, k Β± b) = (2, β 5 Β± 9), or (2, β 14) and (2, 4)... |
.24. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide! Figure 12.24 Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr) 1380 Chapter 12 Analytic Geometry The first hyperbolic towers were designed in 1914 and were 35 met... |
meters. Therefore, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1381 x2 a2 β y2 b2 = 1 b2 = = y2 x2 a2 β 1 (79.6)2 (36)2 900 β 1 Standard form of horizontal hyperbola. Isolate b2 Substitute for a2, x, and y β 14400.3636 Round to four decimal places The sides o... |
β y2 9 = 1 5y2 + 4x2 = 6x 25x2 β 16y2 = 400 β9x2 + 18x + y2 + 4y β 14 = 0 For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. x2 25 β y2 36 = 1 x2 100 β y2 9 = 1 y2 4 β x2 81 = 1 9y2 β 4x2 = 1 (x... |
4x2 + 16x + 112 = 0 Graphical the following exercises, sketch a graph of For hyperbola, labeling vertices and foci. the 99. x2 49 β y2 16 = 1 100. 101. x2 64 y2 9 β y2 4 = 1 β x2 25 = 1 102. 81x2 β 9y2 = 1 Chapter 12 Analytic Geometry 1383 103. 104. 105. 106. 107. 108. 109. 110. 111. 112. β β βy + 5β 9 2 β (x β 4)2 25... |
οΏ½ β β . the following exercises, given the graph of For hyperbola, find its equation. the 122. 119. 1384 Chapter 12 Analytic Geometry will hedge The asymptotes y = x and y = β x, and its closest distance to the center fountain is 5 yards. follow the The hedge 130. y = 2x and y = β2x, center fountain is 6 yards. will fol... |
solar system along a path approximated by the line y = β x + 2. The object enters along a path approximated by the 135. line y = 2x β 2 and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = β2x +... |
paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror (see Figure 12.27), which focuses light rays from the sun to ignite the flame. Parabolic mirrors (or reflectors) are able to capture energy and focus it to ... |
P to the directrix. Figure 12.29 Key features of the parabola To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former. 1388 Chapter 12 Analytic Geometry Figure 12.30 Let (x, y) be a point... |
οΏ½p, 0β β β β0, pp, Β± 2pβ β β βΒ±2p, pβ β This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1389 Figure 12.31 (a) When p > 0 and the axis of symmetry is the x-axis, the parabola opens right. (b) When p < 0 and the axis of symmetry is the x-axis, the parabola opens lef... |
directrix, x = β p βͺ use p to find the endpoints of the latus rectum, β βp, Β± 2pβ β . Alternately, substitute x = p into the original equation. b. If the equation is in the form x2 = 4py, then βͺ βͺ the axis of symmetry is the y-axis, x = 0 set 4p equal to the coefficient of y in the given equation to solve for p. If p >... |
focus, directrix, and endpoints of the latus rectum. Example 12.15 Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry Graph x2 = β6y. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution 1392 Chapter 12 Analytic Geometry The standard form that applies to the g... |
the equation determined in Step 1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1393 Example 12.16 Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix What is the equation for the parabola with focus β ββ 1 2 β β and directrix x =... |
x = h Table 12.2 Equation Focus Directrix Endpoints of Latus Rectum β βy β kβ β 2 = 4p(x β h) (x β h)2 = 4pβ βy β kβ β β βh + p, kβ β β βh, k + ph + p, k Β± 2pβ β β βh Β± 2p, k + pβ β 1394 Chapter 12 Analytic Geometry Figure 12.35 (a) When p > 0, p > 0, the parabola opens up. (d) When p < 0, the parabola opens right. (b... |
h and p to find the equation of the directrix, x = h β p βͺ use h, k, and p to find the endpoints of the latus rectum, β βh + p, k Β± 2pβ β b. If the equation is in the form (x β h)2 = 4pβ βy β kβ β , then: βͺ use the given equation to identify h and k for the vertex, (h, k) βͺ use the value of h to determine the axis of s... |
β’ β’ β’ the coordinates of the focus are β βh + p, kβ β = (β3 + (β4), 1) = (β7, 1) the equation of the directrix is x = h β p = β3 β (β4) = 1 the endpoints of the latus rectum are β (β7, 9) βh + p, k Β± 2pβ β = (β3 + (β4), 1 Β± 2(β4)), or (β7, β7) and Next we plot the vertex, axis of symmetry, focus, directrix, and latus ... |
x = h = 4 since p = 7, p > 0 and so the parabola opens up the coordinates of the focus are β βh, k + pβ β = (4, β8 + 7) = (4, β1) the equation of the directrix is y = k β p = β8 β 7 = β15 the endpoints of (18, β1) the latus rectum are β βh Β± 2p, k + pβ β = (4 Β± 2(7), β8 + 7), or (β10, β1) and Next we plot the vertex, ... |
is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds. a. Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane. b. Use the equation found in part (a) to find the depth of the f... |
openstaxcollege.org/l/parabola1) β’ Conic Sections: The Parabola Part 2 of 2 (http://openstaxcollege.org/l/parabola2) β’ Parabola with Vertical Axis (http://openstaxcollege.org/l/parabolavertcal) β’ Parabola with Horizontal Axis (http://openstaxcollege.org/l/parabolahoriz) 1400 Chapter 12 Analytic Geometry 12.3 EXERCISES ... |
οΏ½y β 2β β β βy β 4β β 2 = 4 5 (x + 4) 2 = 2(x + 3) (x + 1)2 = 2β βy + 4β β (x + 4)2 = 24β βy + 1β β β βy + 4β β 2 = 16(x + 4) y2 + 12x β 6y + 21 = 0 x2 β 4x β 24y + 28 = 0 5x2 β 50x β 4y + 113 = 0 y2 β 24x + 4y β 68 = 0 x2 β 4x + 2y β 6 = 0 y2 β 6y + 12x β 3 = 0 3y2 β 4x β 6y + 23 = 0 x2 + 4x + 8y β 4 = 0 Graphical For... |
). Vertex 184. (β4, 0). 185. Vertex is (2, 2); directrix is x = 2 β 2, focus is β2 + 2, 2β β β . 186. Vertex is (β2, 3); directrix is x = β 7 2, focus is β ββ 1 2 β, 3 β . 187. Vertex is β β 2, β 3β β ; directrix is x = 2 2, focus is β0, β 3β β β . 188. Vertex is (1, 2); directrix is y = 11 3, focus is 191. β β1, 1 3 β β . ... |
across at its opening and 4 feet deep at its center, where should the receiver be placed? 202. Consider the satellite dish from the previous exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver? 193. This content is available for free at https://cnx.org/content/col1... |
cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices ... |
οΏ½ = 0 parallel lines (x β 4)(x β 9) = 0 a point 4x2 + 4y2 = 0 no graph 4x2 + 4y2 = β 1 Table 12.3 General Form of Conic Sections A nondegenerate conic section has the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (12.9) where A, B, and C are not all zero. Table 12.4 summarizes the different conic sections where B = 0,... |
6y β 4 = 0 d. β25x2 β 4y2 + 100x + 16y + 20 = 0 Solution a. Rewriting the general form, we have A = 4 and C = β9, hyperbola. so we observe that A and C have opposite signs. The graph of this equation is a 1408 Chapter 12 Analytic Geometry b. Rewriting the general form, we have A = 0 and C = 9. We can determine that th... |
y-axes and the resulting xβ²β and yβ²βaxes formed by a rotation by an angle ΞΈ. The original coordinate x- and y-axes have unit vectors i and j. The rotated coordinate axes have unit vectors iβ² and jβ². The angle ΞΈ is known as the angle of rotation. See Figure 12.44. We may write the new unit vectors in terms of the origi... |
conic, find a new representation after rotating through an angle. 1. Find x and y where x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ. 2. Substitute the expression for x and y into in the given equation, then simplify. 3. Write the equations with xβ² and yβ² in standard form. Example 12.21 This content is availabl... |
yβ² 2 β 30 = 0 Combine like terms. 2xβ² 2 + 2yβ² 2 β = 30 Combine like terms. 2 β β 2xβ² 2 + 2yβ² 2 β β 4xβ² 2 + 4yβ² 2 β (xβ² 2 β yβ² 2) = 60 4xβ² 2 + 4yβ² 2 β xβ² 2 + yβ² 2 = 60 3xβ² 2 60 5yβ² 2 60 = 60 60 β β = 2(30) β + Multiply both sides by 2. Simplify. Distribute. Set equal to 1. (xβ² 2 β yβ² 2) 2 (xβ² 2 β yβ² 2) 2 Write the equa... |
A = C, then ΞΈ = 45Β°. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1413 Given an equation for a conic in the xβ² yβ² system, rewrite the equation without the xβ² yβ² term in terms of xβ² and yβ², where the xβ² and yβ² axes are rotations of the standard axes by ΞΈ degree... |
οΏ½ β yβ² β x = xβ² β β 1 5 β 2 β 5 2xβ² β yβ² 5 and x = y = xβ² sin ΞΈ + yβ² cos ΞΈ β β β + yβ² β y = xβ² + 2yβ² 5 y = 2 Substitute the expressions for x and y into in the given equation, and then simplify. 8 β β 2xβ² β yβ² 5 xβ² + 2yβ² 2xβ² β yβ² β β β β β β β β + 17 β 12 β β β β β β 5 5 (xβ² + 2yβ²)(xβ² + 2yβ²) (2xβ² β yβ²)(xβ² + 2yβ²) β β β ... |
.21 Rewrite the 13x2 β 6 3xy + 7y2 = 16 in the xβ² yβ² system without the xβ² yβ² term. Example 12.23 Graphing an Equation That Has No xβ²yβ² Terms Graph the following equation relative to the xβ² yβ² system: x2 + 12xy β 4y2 = 30 Solution First, we find cot(2ΞΈ). x2 + 12xy β 4y2 = 20 β A = 1, B = 12, and C = β4 cot(2ΞΈ) = cot(2ΞΈ... |
2 β β β β + 12 3xβ² β 2yβ² 13 3xβ² β 2yβ² 13 β β β β β β β β β β β β β β‘ β£(3xβ² β 2yβ²)2 + 12(3xβ² β 2yβ²)(2xβ² + 3yβ²) β 4(2xβ² + 3yβ²)2β€ β β€ β‘ β4xβ² 2 + 12xβ² yβ² + 9yβ² 2β β β6xβ² 2 + 5xβ² yβ² β 6yβ² 2β β β β£9xβ² 2 β 12xβ² yβ² + 4yβ² 2 + 12 β¦ = 30 β β β β‘ β£9xβ² 2 β 12xβ² yβ² + 4yβ² 2 + 72xβ² 2 + 60xβ² yβ² β 72yβ² 2 β 16xβ² 2 β 48xβ² yβ² β 36yβ² 2β€ β ... |
2 + Dx + Ey + F = 0 1418 Chapter 12 Analytic Geometry Aβ² xβ²2 + Bβ² xβ² yβ² + Cβ² yβ²2 + Dβ² xβ² + Eβ² yβ² + Fβ² = 0 It may be shown that B2 β 4AC = Bβ²2 β 4Aβ² Cβ². The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2 β 4AC, is invariant and remains unchanged after rotation. Because... |
and C. x2 + 2 3 β B 5 β A xy + 12 β C y2 β 5 = 0 Now, we find the discriminant. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1419 B2 β 4AC = β = 4(3) β 240 = 12 β 240 = β 228 < 0 Therefore, 5x2 + 2 3xy + 12y2 β 5 = 0 represents an ellipse. β2 3β β 2 β 4(5)(12)... |
2 + Dx + Ey + F = 0, the equation the value of ΞΈ that satisfies cot(2ΞΈ) = A β C B gives us what information? Algebraic For the following exercises, determine which conic section is represented based on the given equation. 214. 215. 216. 217. 218. 219. 220. 221. 222. 223. 224. 225. 9x2 + 4y2 + 72x + 36y β 500 = 0 x2 β 1... |
, ΞΈ = β 45β 240. x = y2, ΞΈ = 45β x2 4 y2 16 + y2 1 + x2 9 = 1, ΞΈ = 45β = 1, ΞΈ = 45β 241. 242. 243. Chapter 12 Analytic Geometry 1421 y2 β x2 = 1, ΞΈ = 45β For the following exercises, determine the value of k based on the given equation. 264. Given 4x2 + kxy + 16y2 + 8x + 24y β 48 = 0, find k for the graph to be a parab... |
β 6 3xy + 7y2 β 16 = 0 4x2 β 4xy + y2 β 8 5x β 16 5y = 0 For the following exercises, determine the angle of rotation in order to eliminate the xy term. Then graph the new set of axes. 258. 259. 260. 261. 262. 263. 6x2 β 5 3xy + y2 + 10x β 12y = 0 6x2 β 5xy + 6y2 + 20x β y = 0 6x2 β 8 3xy + 14y2 + 10x β 3y = 0 4x2 + 6... |
orbiting body breaks free of the celestial bodyβs gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system. Identifying a Conic in Polar Form Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix... |
ic has a polar equation r = ep 1 Β± e cos ΞΈ For a conic with a focus at the origin, if the directrix is y = Β± p, where p eccentricity is a positive real number e, the conic has a polar equation is a positive real number, and the r = ep 1 Β± e sin ΞΈ 1424 Chapter 12 Analytic Geometry Given the polar equation for a conic, i... |
1.5 Chapter 12 Analytic Geometry 1425 r = 12 4 + 5 cos = 12 β β 1 β β 4 β β β 1 β cos cos ΞΈ Because cos ΞΈ is in the denominator, the directrix is x = p. Comparing to standard form, e = 5 4. Therefore, from the numerator, p 3 = ep 3 = β β 5 5 12 = p 5 β β p 5 4 Since e > 1, the conic is a hyperbola. The eccentricity is e... |
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is cos ΞΈ r = 5 3 + 3 cos + cos ΞΈ Because e = 1, we will graph a parabola with a focus at the origin. The function has a cos ΞΈ, and there is an addition sign in the denominator, so the directrix is x ... |
ΞΈ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 1 5. 1430 Chapter 12 Analytic Geometry 10 β β 1 β β 5 β β β 1 β cos = 10 5 β 4cos ΞΈ = r = 2 1 β 4 5 cos ΞΈ Because e = 4 5, e < 1, so we will graph an ellipse with a focus at the origin.... |
addition. 3. Write the coefficient of the trigonometric function as the given eccentricity. 4. Write the absolute value of p in the numerator, and simplify the equation. Example 12.29 Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix Find the polar form of the co... |
use the identities r = x2 + y2, x = r cos ΞΈ, and y = r sin ΞΈ. 1 5 β 5 sin ΞΈ 1 5 β 5 sin ΞΈ β
(5 β 5 sin ΞΈ) Eliminate the fraction. r = r β
(5 β 5 sin ΞΈ) = 5r β 5r sin ΞΈ = 1 5r = 1 + 5r sin ΞΈ 25r 2 = (1 + 5r sin ΞΈ)2 25(x2 + y2) = (1 + 5y)2 25x2 + 25y2 = 1 + 10y + 25y2 25x2 β 10y = 1 Distribute. Isolate 5r. Square both s... |
equation? Algebraic For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. 274. 275. 276. 277. 278. 279. 280. 281. r = 6 1 β 2 cos ΞΈ r = 3 4 β 4 sin ΞΈ r = 8 4 β 3 cos ΞΈ r = 5 1 + 2 sin ΞΈ r = 16 4 + 3 cos ΞΈ r = 3 10 + 10 cos ΞΈ r = 2 1 β cos ΞΈ r = 4 7 + ... |
308. r(3 β 4sin ΞΈ) = 9 309. r(3 β 2sin ΞΈ) = 6 310. r(6 β 4cos ΞΈ) = 5 For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Chapter 12 Analytic Geometry 324. xy = 2 325. x2 + xy + y2 = 4 326. 327. 2x2 + 4xy + 2y2 = 9 16x2 + 24xy + 9y2 = 4 32... |
bola that is perpendicular to the transverse axis and has the co-vertices as its endpoints degenerate conic sections any of the possible shapes formed when a plane intersects a double cone through the apex. Types of degenerate conic sections include a point, a line, and intersecting lines. directrix a line perpendicula... |
at origin Horizontal ellipse, center (h, k) Vertical ellipse, center (h, k) x2 a2 + y2 b2 = 1, a > b x2 b2 + y2 a2 = 1, a > b (x β h)2 a2 + (x β h)2 b2 + β βy β kβ β 2 b2 = 1, a > b β βy β kβ β 2 a2 = 1, a > b Hyperbola, center at origin, transverse axis on x-axis Hyperbola, center at origin, transverse axis on y-axis... |
ipse β’ An ellipse is the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). β’ When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See Ex... |
and a fixed point (the focus) not on the directrix. β’ The standard form of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry can be used to graph the parabola. If p > 0, the parabola opens right. If p < 0, the parabola opens left. See Example 12.14. β’ The standard form of a parabola with vertex (0, ... |
form can be transformed into an equation in the xβ² and yβ² coordinate system without the xβ² yβ² term. See Example 12.22 and Example 12.23. β’ An expression is described as invariant if it remains unchanged after rotating. Because the discriminant is invariant, observing it enables us to identify the conic section. See Ex... |
y2 + 16x + 4y β 44 = 0 336. 2x2 + 3y2 β 20x + 12y + 38 = 0 For the following exercises, use the given information to find the equation for the ellipse. 337. Center at (0, 0), focus at (3, 0), vertex at (β5, 0) Center at (2, β2), vertex at (7, β2), focus at 338. (4, β2) 339. A whispering gallery is to be constructed su... |
. 346. x2 β 4y2 + 6x + 32y β 91 = 0 Rotation of Axes 347. 2y2 β x2 β 12y β 6 = 0 For the following exercises, find the equation of the hyperbola. 348. Center at (0, 0), vertex at (0, 4), focus at (0, β6) 349. Foci at (3, 7) and (7, 7), vertex at (6, 7) The Parabola For the following exercises, write the equation of the... |
conic given in polar form. If it the vertex, focus, and directrix. If it is an ellipse or a hyperbola, label the vertices and foci. is a parabola, label 373. r = 3 1 β sin ΞΈ 374. r = 8 4 + 3 sin ΞΈ 375. r = 10 4 + 5 cos ΞΈ 376. r = 9 3 β 6 cos ΞΈ For the following exercises, given information about the graph of a conic w... |
+ 112 = 0 399. 16x2 + 24xy + 9y2 β 125x = 0 For the following exercises, graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes. For the following exercises, identify the conic with focus at the origin, and then give the directrix and eccentricity. 387. (x β 3)2 25 β β β βy +... |
y term. 1444 Chapter 12 Analytic Geometry This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1445 13 | SEQUENCES, PROBABILITY, AND COUNTING THEORY Figure 13.1 (credit: Robert S. Donovan, Flickr.) Chapter Outline 13.1 Sequences and Their Nota... |
13.1 If their model continues, how many hits will there be at the end of the month? To answer this question, weβll first need to know how to determine a list of numbers written in a specific order. In this section, we will explore these kinds of ordered lists. Writing the Terms of a Sequence Defined by an Explicit For... |
the last day of the month, we need to find the 31st term of the sequence. We will substitute 31 for n in the formula. a31 = 231 = 2,147,483,648 If the doubling trend continues, the company will get 2,147,483,648 hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unre... |
the term. A sequence that continues indefinitely is an infinite sequence. the second term of the sequence, a3 the third term of the sequence, and Does a sequence always have to begin with a1? No. In certain problems, it may be useful to define the initial term as a0 instead of a1. In these problems, the domain of the ... |
βs take a look at the following sequence. Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence. {2, β4, 6, β8} 1450 Cha... |
of the sequence: an = 4n ( β 2) n Investigating Piecewise Explicit Formulas Weβve learned that sequences are functions whose domain is over the positive integers. This is true for other types of functions, including some piecewise functions. Recall that a piecewise function is a function defined by multiple subsection... |
the sequence. Sometimes, the explicit formula for the nth term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.