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οΏ½ Q & Aβ¦ Can we use the same method to find the determinant of a larger matrix? No, this method only works for 2 Γ 2 and 3 Γ 3 matrices. For larger matrices it is best to use a graphing utility or computer software. Using Cramerβs Rule to Solve a System of Three equations in Three Variables Now that we can find the det... |
2) Solution We begin by finding the determinants D, D x, and D y. 6x β 4y = 0 3x β 2y = 4 (1) D = β£ 3 β2 6 β4 β£ = 3(β4) β 6(β2) = 0 848 CHAPTER 9 systems oF eQuations and ineQualities We know that a determinant of zero means that either the system has no solution or it has an infinite number of solutions. To see which ... |
the planes are the same and they both intersect the third plane on a line. See Figure 2. SECTION 9.8 solving systems with cramer's rule 849 x β 2y + 3z = 0 2x β 4y + 6z = 0 3x + y + 2z = 0 Figure 2 Understanding Properties of Determinants There are many properties of determinants. Listed here are some properties that ... |
(2)(2) β 2(2)(1) β 2(2)(2 Property 4 states that if a row or column equals zero, the determinant equals zero. Thus, 1 2 0 0 ξ², det(A) = 1(0) β 2(0) = 0 A = ξ° Property 5 states that the determinant of an inverse matrix A β1 is the reciprocal of the determinant A. Thus, A = ξ° 1 2 3 4 ξ², det(A) = 1(4) β 3(2) = β2 A β1 = ξ°... |
ramer) β’ Solve a Systems of Three equations using Cramer's Rule (http://openstaxcollege.org/l/system3cramer) SECTION 9.8 section exercises 851 9.8 SeCTIOn exeRCISeS VeRBAl 1. Explain why we can always evaluate the determinant 2. Examining Cramerβs Rule, explain why there is no of a square matrix. 3. Explain what it mea... |
29. 4x + 3y = 23 2x β y = β1 26. 5x β 4y = 2 β4x + 7y = 6 27. 6x β 3y = 2 β8x + 9y = β1 28. 2x + 6y = 12 5x β 2y = 13 30. 10x β 6y = 2 β5x + 8y = β1 31. 4x β 3y = β3 2x + 6y = β4 32. 4x β 5y = 7 β3x + 9y = 0 33. 4x + 10y = 180 β3x β 5y = β105 34. 8x β 2y = β3 β4x + 6y = 4 For the following exercises, solve the system ... |
+ 18y β 24z = β 30 TeCHnOlOGY For the following exercises, use the determinant function on a graphing utility. 45 β£ β£ 469 1 3 β1 β2 0 β2 1 1 β£ 47. 1 _ 2 100,000 0 0 0 2 48 β£ β£ β£ β£ ReAl-WORlD APPlICATIOnS For the following exercises, create a system of linear equations to describe the behavior. Then, calculate the dete... |
At the end of one year, assuming simple interest, you have earned $2,470 in interest. The second account receives half a percent less than twice the interest on the first account. What are the interest rates for your accounts? 56. A concert venue sells single tickets for $40 each and coupleβs tickets for $65. If the t... |
were sold? of the population at a prison last year. This year, the same age groups made up 82.08% of the population. The 20β29 age group increased by 20%, the 30β39 age group increased by 2%, and the 40β49 age group decreased to 3 __ of their previous population. 4 Originally, the 30β39 age group had 2% more prisoners... |
W Key Terms addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable augmented matrix a coeff... |
x), revenue minus cost revenue function the function that is used to calculate revenue, simply written as R = xp, where x = quantity and p = price row a set of numbers aligned horizontally in a matrix row operations adding one row to another row, multiplying a row by a constant, interchanging rows, and so on, with the ... |
β’ A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables. See Example 4. β’ It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two eq... |
abola; (2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See Example 1. β’ There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one so... |
a repeated irreducible quadratic factor, when the irreducible quadratic factors are repeated, powers of the denominator factors must be represented in increasing powers as Ax + B __ + (a x 2 + bx + c) A 2 x + B 2 __ (a x 2 + bx + c __ n (a x 2 + bx + c). See Example 4. 9.5 Matrices and Matrix Operations β’ A matrix is ... |
Back-substitute to find the solutions. See Example 7 and Example 8. β’ A calculator can be used to solve systems of equations using matrices. See Example 9. β’ Many real-world problems can be solved using augmented matrices. See Example 10 and Example 11. 9.7 Solving Systems with Inverses β’ An identity matrix has the pr... |
columns are identical, the determinant equals zero. β If a matrix contains either a row of zeros or a column of zeros, the determinant equals zero. β The determinant of an inverse matrix A β1 is the reciprocal of the determinant of the matrix A. β If any row or column is multiplied by a constant, the determinant is mu... |
. 11. 0.5x β 0.5y = 10 β 0.2y + 0.2x = 4 0.1x + 0.1z = 2 14. 2x β 3y + z = β1 x + y + z = β4 4x + 2y β 3z = 33 17. 2x β 3y + z = 0 2x + 4y β 3z = 0 6x β 2y β z = 0 12. 5x + 3y β z = 5 3x β 2y + 4z = 13 4x + 3y + 5z = 22 15. 3x + 2y β z = β10 x β y + 2z = 7 βx + 3y + z = β2 18. 6x β 4y β 2z = 2 3x + 2y β 5z = 4 6y β 7z ... |
+ y 2 + 2x < 3 y > β x 2 β 3 29. x 2 β 2x + y 2 β 4x < 4 y < β x + 4 30 PARTIAl FRACTIOnS For the following exercises, decompose into partial fractions. 31. β2x + 6 __________ x 2 + 3x + 2 34. x β 18 ____________ x 2 β 12x + 36 32. 10x + 2 ___________ 4 x 2 + 4x + 1 35. β x 2 + 36x + 70 _____________ x 3 β 125 33. 7x ... |
= 31 55. x + 3z = 12 βx + 4y = 0 y + 2z = β 7 860 CHAPTER 9 systems oF eQuations and ineQualities For the following exercises, solve the system of linear equations using Gaussian elimination. 56. 3x β 4y = β 7 β6x + 8y = 14 57. 3x β 4y = 1 β6x + 8y = 6 58. β1.1x β 2.3y = 6.2 β5.2x β 4.1y = 4.3 59. 2x + 3y + 2z = 1 β4x... |
brownies and chocolate chip cookies. They priced the brownies at $2 and the chocolate chip cookies at $1. They raised $250 and sold 175 items. How many brownies and how many cookies were sold? SOlVInG SYSTeMS WITH CRAMeR'S RUle For the following exercises, find the determinant. 71. ξ° 100 0 ξ² 0 0 72. ξ° 0.2 β0.6 0.7 β1.... |
5x β 4y β 3z = 0 2x + y + 2z = 0 x β 6y β 7z = 0 8. y = x 2 + 2x β 3 y = x β 1 For the following exercises, graph the following inequalities. 10. y < x 2 + 9 11. 4x β 6y β 2z = 1 ___ 10 x β 7y + 5z = β 1 __ 4 3x + 6y β 9z = 6 __ 5 9. y 2 + x 2 = 25 y 2 β 2 x 2 = 1 For the following exercises, write the partial fractio... |
z = 6 For the following exercises, use the inverse of a matrix to solve the systems of equations. 25. 4x β 5y = β50 βx + 2y = 80 z = β49 26. 1 ___ 100 3 ___ 100 9 ___ 100 x β 3 ___ 100 x β 7 ___ 100 x β 9 ___ 100 y + 1 ___ 20 y β 1 ___ 100 y β 9 ___ 100 z = 13 z = 99 For the following exercises, use Cramerβs Rule to s... |
circular. He presumed that the planets moved in circular orbits around Earth, and for nearly 2,000 years this was the commonly held belief. It was not until the Renaissance movement that Johannes Kepler noticed that the orbits of the planet were not circular in nature. His published law of planetary motion in the 1600... |
of the 33. Architect of the Capitol. http://www.aoc.gov. Accessed April 15, 2014. SECTION 10.1 the ellipse 865 equation for the ellipse. 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). We ... |
866 CHAPTER 10 analytic geometry Deriving the Equation of an Ellipse Centered at the Origin To derive the equation of an ellipse centered at the origin, we begin with the foci (βc, 0) and (c, 0). The ellipse is the set of all points (x, y) such that the sum of the distances from (x, y) to the foci is constant, as show... |
+ y 2 ] β 2 x 2 + 2cx + c2 + y 2 = 4a2 β 4a β (x β c)2 + y 2 + (x β c)2 + y 2 (x β c)2 + y 2 + x 2 β 2cx + c2 + y 2 β 2cx = 4a2 β 4a β (x β c)2 + y 2 β 2cx β 4cx β 4a2 = β4a β β β (x β c)2 + y 2 (x β c)2 + y 2 2 (x β c)2 + y 2 ξ² cx β a2 = β a β = a2 ξ° β c2 x 2 β 2a2 cx + a4 = a2 ξ’ x 2 β 2cx + c2 + y 2 ξͺ 2 ξ° cx β a2 ξ² ... |
in Standard Form Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations weβve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the rela... |
β b2. When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form. y (0, b) Minor Axis (βa, 0) (βc, 0) Major Axis (0, 0) (c, 0) (a, 0) x (βb, 0) Minor Axis (0, βb) (a) Figure 6 (a) Horizontal ellipse with center (0, 0) (b) ... |
so c = 5 and c2 = 25. We know that the vertices and foci are related by the equation c 2 = a 2 β b 2. Solving for b 2, we have: c 2 = a 2 β b2 25 = 64 β b2 b 2 = 39 Substitute for c 2 and a 2. Solve for b2. Now we need only substitute a2 = 64 and b2 = 39 into the standard form of the equation. The equation of the elli... |
of the foci are (h Β± c, k), where c2 = a2 β b2. See Figure 7a. The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the y-axis is where (x β h)2 _ b2 + (y β k)2 _ a2 = 1 β’ a > b β’ the length of the major axis is 2a β’ the coordinates of the vertices are (h, k Β± a) β’ the length o... |
vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form (x β h)2 _ b2 + (y β k)2 _ a2 = 1. 870 CHAPTER 10 analytic geometry 2. Identify the center of the ellipse (h, k) using the midpoint formula and the given coordinates for the vertices. 3. Find a2 by solving for the leng... |
of these points to solve for c. k + c = 1 β3 + c = 1 c = 4 So c2 = 16. Next, we solve for b2 using the equation c2 = a2 β b2. c2 = a2 β b2 16 = 25 β b2 b2 = 9 Finally, we substitute the values found for h, k, a2, and b2 into the standard form equation for an ellipse: (y + 3)2 _ 25 (x + 2)2 _ 9 = 1 + Try It #2 What is ... |
center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 3 Graphing an Ellipse Centered at the Origin x 2 _ 9 + y 2 _ 25 Graph the ellipse given by the equation, = 1. Identify and label the center, vertices, co-vertices, and foci. Solution First, we determin... |
_ 4 x 2 _ 25 = 1 = + Next, we determine the position of the major axis. Because 25 > 4, the major axis is on the x-axis. Therefore, the equation is in the form y 2 x 2 _ _ a2 + b2 = 1, where a2 = 25 and b2 = 4. It follows that: β’ the center of the ellipse is (0, 0) β’ the coordinates of the vertices are (Β±a, 0) = (Β± β ... |
β¦ Given the standard form of an equation for an ellipse centered at (h, k), sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci. a. If the equation is in the form = 1, where a > b, then (x β h)2 _ a2 + (y β k)2 _... |
ices are (h Β± b, k) = (β2 Β± β β’ the coordinates of the foci are (h, k Β± c), where c 2 = a 2 β b 2. Solving for c, we have: 9 ) = (β2, 5 Β± 3), or (β2, 2) and (β2, 8) 4, 5) = (β2 Β± 2, 5), or (β4, 5) and (0, 5) β β c = Β± β = Β± β = Β± β β β a2 β b2 9 β 4 5 β Therefore, the coordinates of the foci are (β2, 5 β β β 5 ) and (β... |
Form Graph the ellipse given by the equation 4x 2 + 9y 2 β 40x + 36y + 100 = 0. Identify and label the center, vertices, co-vertices, and foci. Solution We must begin by rewriting the equation in standard form. 4x 2 + 9y 2 β 40x + 36y + 100 = 0 Group terms that contain the same variable, and move the constant to the o... |
Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure 11. β β c = Β± β = Β± β β β a2 β b2 9 β 4 5 5, β2). β SECTION 10.1 the ellipse 875 β1 0 β1 β2 β3 β4 1 (5 ββ5, β2) 1 2 3 4 (5 +β5, β2) 6 7 8 9 x (5, 0) 5 (2, β2) (5, β2) (8, β2) (5, β4) Figur... |
two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot. 46 feet 96 feet Figure 13 876 Solution CHAPTER 10 analytic geometry a. We are assuming a horizontal ellipse with center (0, 0), so we need to find an equation of the form y 2 x 2 _ _ b... |
axcollege.org/l/grphellorigin) β’ Graph an ellipse with Center not at the Origin (http://openstaxcollege.org/l/grphellnot) SECTION 10.1 section exercises 877 10.1 SeCTIOn exeRCISeS VeRBAl 1. Define an ellipse in terms of its foci. 3. What special case of the ellipse do we have when the 2. Where must the foci of an ellip... |
β 360y + 864 = 0 23. 4x 2 + 40x + 25y 2 β 100y + 100 = 0 25. 4x 2 + 24x + 25y 2 + 200y + 336 = 0 19. 4x 2 β 8x + 9y 2 β 72y + 112 = 0 For the following exercises, find the foci for the given ellipses. 27. (x + 3)2 _ 25 + (y + 1)2 _ 36 = 1 28. (x + 1)2 _ 100 + (y β 2)2 _ 4 = 1 29. x 2 + y 2 = 1 30. x 2 + 4y 2 + 4x + 8y... |
878 CHAPTER 10 analytic geometry 48. Center at the origin, symmetric with respect to the x- and y-axes, focus at (3, 0), and major axis is twice as long as minor axis. 49. Center (4, 2); vertex (9, 2); one focus: (4 + 2 β 6, 2). β 50. Center (3, 5); vertex (3, 11); one focus: (3, 5 + 4 β For the following exercises, g... |
an equation for the ellipse, and use that to find the height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center. 66. A bridge is to be built in the shape of a semi- elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 fe... |
1 A shock wave intersecting the ground forms a portion of a conic and results in a sonic boom. Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the spe... |
diagonals of the central rectangle. See Figure 3. y Conjugate axis Transverse axis Co-vertex Focus Vertex Vertex x Focus Co-vertex Center Asymptote Asymptote Figure 3 Key features of the hyperbola In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordina... |
x β ( β c))2 + (y β 0)2 β β (x β c)2 + (y β 0)2 = 2a Distance formula β (x + c)2 + y 2 β β β d2 β d1 = β (x β c)2 + y 2 = 2a β β (x + c)2 + y 2 = 2a + β β (x β c)2 + y 2 β (x + c)2 + y 2 = (2a + β x 2 + 2cx + c2 + y 2 = 4a2 + 4a β β (x β c)2 + y 2 )2 β (x β c)2 + y 2 + (x β c)2 + y 2 x 2 + 2cx + c2 + y 2 = 4a2 + 4a β β... |
the squares. Expand remaining square. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2. Combine like terms. Rearrange terms. Factor common terms Set b2 = c2 β a2. Divide both sides by a2b2. This equation defines a hyperbola centered at the origin with vertices ... |
) (0, βb) (a) y = a x b x (b, 0) y (0, c ) (0, a) (0, 0) (0, βa) (0, βc) (b) Figure 5 (a) Horizontal hyperbola with center (0, 0) (b) Vertical hyperbola with center (0, 0) How Toβ¦ Given the equation of a hyperbola in standard form, locate its vertices and foci. 1. Determine whether the transverse axis lies on the x- or... |
, y = Β± β β 49 = Β± 7 c = β β a2 + b2 = β β 49 + 32 = β β 81 = 9 Therefore, the vertices are located at (0, Β± 7), and the foci are located at (0, 9). Try It #1 Identify the vertices and foci of the hyperbola with equation x 2 _ 9 β y 2 _ 25 = 1. Writing equations of Hyperbolas in Standard Form Just as with ellipses, wri... |
the standard form of the equation determined in Step 1. 884 CHAPTER 10 analytic geometry Example 2 What is the standard form equation of the hyperbola that has vertices (Β±6, 0) and foci (Β±2 β Finding the Equation of a Hyperbola Centered at (0, 0) Given its Foci and Vertices 10, 0)? β Solution The vertices and foci are... |
the conjugate axis is 2b β’ the coordinates of the co-vertices are (h, k Β± b) β’ the distance between the foci is 2c, where c 2 = a 2 + b 2 β’ the coordinates of the foci are (h Β± c, k) SECTION 10.2 the hyperBola 885 The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the r... |
x β h) + k b x Like hyperbolas centered at the origin, hyperbolas centered at a point (h, k) have vertices, co-vertices, and foci that are related by the equation c2 = a2 + b2. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and f... |
β h)2 _ a2 β (y β k)2 _ b2 = 1 First, we identify the center, (h, k). The center is halfway between the vertices (0, β2) and (6, β2). Applying the midpoint formula, we have Next, we find a2. The length of the transverse axis, 2a, is bounded by the vertices. So, we can find a2 by finding the distance between the x-coor... |
hyperBola 887 How To⦠Given a standard form equation for a hyperbola centered at (0, 0), sketch the graph. 1. Determine which of the standard forms applies to the given equation. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the a. If the equation is in... |
), where c = Β± β β a2 + b2. Solving for c, we have c = Β± β β a2 + b2 = Β± β β 64 + 36 = Β± β β 100 = Β± 10 Therefore, the coordinates of the foci are (0, Β± 10) a 8 4 __ __ __ The equations of the asymptotes are Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are p... |
. How To⦠Given a general form for a hyperbola centered at (h, k), sketch the graph. 1. Determine which of the standard forms applies to the given equation. Convert the general form to that standard form. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the... |
constant to the opposite side of the equation. Factor the leading coefficient of each expression. (9x 2 β 36x) β (4y 2 + 40y) = 388 9(x 2 β 4x) β 4(y 2 + 10y) = 388 Complete the square twice. Remember to balance the equation by adding the same constants to each side. 9(x 2 β 4x + 4) β4(y 2 + 10y + 25) = 388 + 36 β 100... |
, β5) (8, β5) (2, β14) (2, β5) Figure 9 Try It #5 Graph the hyperbola given by the standard form of an equation vertices, co-vertices, foci, and asymptotes. (y + 4)2 _ β 100 (x β 3)2 _ 64 = 1. Identify and label the center, Solving Applied Problems Involving Hyperbolas As we discussed at the beginning of this section, ... |
: We must find the values of a2 and b2 to complete the model. First, we find a2. Recall that the length of the transverse axis of a hyperbola is 2a. This length is represented by the distance where the sides are closest, which is given as 65.3 meters. So, 2a = 60. Therefore, a = 30 and a2 = 900. To solve for b2, we nee... |
staxcollege.org/l/hyperbolaorigin) β’ Graph a Hyperbola with Center not at Origin (http://openstaxcollege.org/l/hbnotorigin) SECTION 10.2 section exercises 893 10.2 SeCTIOn exeRCISeS VeRBAl 1. Define a hyperbola in terms of its foci. 2. What can we conclude about a hyperbola if its asymptotes intersect at the origin? 3.... |
360y + 864 = 0 22. β4x 2 + 40x + 25y 2 β 100y + 100 = 0 24. β9x 2 + 72x + 16y 2 + 16y + 4 = 0 For the following exercises, find the equations of the asymptotes for each hyperbola. (y + 4)2 _______ 22 (x β 3)2 _______ 52 β 26. 27. = 1 x 2 y 2 __ __ 32 = 1 32 β (y β 3)2 _______ 32 28. β (x + 5)2 _______ 62 30. 16y 2 + 9... |
) and one focus at (0, β8). 47. Vertices at (1, 1) and (11, 1) and one focus at (12, 1). 48. Center: (0, 0); vertex: (0, β13); one focus: (0, β β 313 ). 49. Center: (4, 2); vertex: (9, 2); one focus: (4 + β β 26, 2). 50. Center: (3, 5); vertex: (3, 11); one focus: (3, 5 + 2 β β 10 ). For the following exercises, given ... |
The hedge will follow the asymptotes y = x and 2 1 __ y = β x, and its closest distance to the center 2 2 __ 64. The hedge will follow the asymptotes y = x and 3 2 __ y = β x, and its closest distance to the center 3 fountain is 10 yards. fountain is 12 yards. 3 __ 65. The hedge will follow the asymptotes y = x and 4 ... |
the line y = β3x + 9. 896 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: β’ Graph parabolas with vertices at the origin. β’ Write equations of parabolas in standard form. β’ Graph parabolas with vertices not at the origin. β’ Solve applied problems involving parabolas. 10. 3 THe PARABOlA Figur... |
rix. In Quadratic Functions, we learned about a parabolaβs vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See Figure 3. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between... |
2 x 2 β 2py = 2py x 2 = 4py The equations of parabolas with vertex (0, 0) are y 2 = 4px when the x-axis is the axis of symmetry and x 2 = 4py when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features. 898 CHAPTER 10 analytic geometry standard forms o... |
features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in Figure 6. y y 2 = 24x (6, 12) (0, 0) (6, 0) x (6, β12) x =... |
as the Axis of Symmetry Graph y 2 = 24x. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is y 2 = 4px. Thus, the axis of symmetry is the x-axis. It follows that: β’ 24 = 4p, so p = 6. Since p > 0, the parabola opens right β’ the co... |
form the parabola. β3, β 3 2 y (0, 0) 0, β 3 2 Figure 8 y = 3 2 x 3, β 3 2 x2 = β6y Try It #2 Graph x 2 = 8y. Identify and label the focus, directrix, and endpoints of the latus rectum. Writing equations of Parabolas in Standard Form In the previous examples, we used the standard form equation of a parabola to calcula... |
translation results in the standard form of the equation we saw previously with x replaced by (x β h) and y replaced by (y β k). To graph parabolas with a vertex (h, k) other than the origin, we use the standard form (y β k)2 = 4p(x β h) for parabolas that have an axis of symmetry parallel to the x-axis, and (x β h)2 ... |
Figure 9 (a) When p > 0, the parabola opens right. (b) When p < 0, the parabola opens left. (c) When p > 0, the parabola opens up. (d) When p < 0, the parabola opens down. 902 CHAPTER 10 analytic geometry How To⦠Given a standard form equation for a parabola centered at (h, k), sketch the graph. 1. Determine which of ... |
find the endpoints of the latus rectum, (h Β± 2p, k + p) 3. Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Example 4 Graphing a Parabola with Vertex ( h, k) and Axis of Symmetry Parallel to the x-axis Graph (y β 1)2 = β16(x + 3). Identify and label t... |
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is (x β h)2 = 4p (y β k). Thus, the axis of symmetry is parallel to the y-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable x in order... |
of light parallel to the parabolaβs axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus. See Figure 12. This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror. 904 CHAPTER 10 analytic geometry Parallel rays of sunlight Focu... |
about 0.74 inches deep. SECTION 10.3 the paraBola 905 Try It #6 Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1,600 mm. The sunβs rays reflect off the parabolic mirror toward the βcooker,β which is placed 320 mm from the base. a. Find an equation that ... |
4x 2 10. y 2 + 12x β 6y β 51 = 0 8. 3x 2 β 6y 2 = 12 For the following exercises, rewrite the given equation in standard form, and then determine the vertex (V), focus (F), and directrix (d) of the parabola. 11. x = 8y 2 1 __ 12. y = x 2 4 13. y = β4x 2 1 __ 14. x = y 2 8 15. x = 36y 2 16. x = 1 ___ 36 y 2 17. (x β 1)... |
= 0 44. β2x 2 + 8x β 4y β 24 = 0 For the following exercises, find the equation of the parabola given information about its graph. 45. Vertex is (0, 0); directrix is y = 4, focus is (0, β4). 46. Vertex is (0, 0); directrix is x = 4, focus is (β4, 0). 47. Vertex is (2, 2); directrix is x = 2 β β β (2 + β 2, 2). β 2, fo... |
a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as x 2 = 4y. At what coordinates should you place the light bulb? 62. If we want to construct the mirror from the previous exercise such that the focus is located at (0, 0.25), what should the equation of ... |
equations of rotated conics in standard form. β’ Identify conics without rotating axes. 10. 4 ROTATIOn OF AxIS As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which w... |
zero. Conic Sections Example ellipse circle hyperbola parabola one line intersecting lines parallel lines a point no graph 4x 2 + 9y 2 = 1 4x 2 + 4y 2 = 1 4x 2 β 9y 2 = 1 4x 2 = 9y or 4y 2 = 9x 4x + 9y = 1 (x β 4) (y + 4)= 0 (x β 4)(x β 9) = 0 4x 2 + 4y 2 = 0 4x 2 + 4y 2 = β 1 Table 1 general form of conic sections A ... |
as shown below, the conic section is rotated. Notice the phrase βmay beβ in the definitions. That is because the equation may not represent a conic section at all, depending on the values of A, B, C, D, E, and F. For example, the degenerate case of a circle or an ellipse is a point: Ax 2 +By 2=0, when A and B have the... |
, we have Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 (β25)x 2 + 0xy + (β4)y 2 + 100x + 16y + 20 = 0 A = β25 and C = β4. Because AC > 0 and A β C, the graph of this equation is an ellipse. Try It #1 Identify the graph of each of the following nondegenerate conic sections. a. 16y 2 β x 2 + x β 4y β 9 = 0 b. 16x 2 + 4y 2 + 16x +... |
οΏ½ βsin ΞΈ x' sin ΞΈ x i' i ΞΈ cos ΞΈ Figure 5 Relationship between the old and new coordinate planes. SECTION 10.4 rotation oF axis 913 Consider a vector u in the new coordinate plane. It may be represented in terms of its coordinate axes. u = xβ² iβ² + yβ² jβ² u = xβ²(i cos ΞΈ + j sin ΞΈ) + yβ²( β i sin ΞΈ + j cos ΞΈ) u = ix'cos ΞΈ ... |
angle of ΞΈ = 45Β°. Solution Find x and y, where x = xβ² cos ΞΈ β yβ² sin ΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ. Because ΞΈ = 45Β°, and x = xβ² cos(45Β°) β yβ² sin(45Β°) ξͺ β yβ² ξ’ 1 _ β 2 β xβ² β yβ² sin(45Β°) + yβ² cos(45Β°) ξͺ + yβ² ξ’ 1 _ β 2 β xβ² + yβ² _ 2 β β y = Substitute x = xβ² cosΞΈ β yβ² sinΞΈ and y = xβ² sin ΞΈ + yβ² cos ΞΈ into 2x 2 β xy + 2... |
. Figure 6 shows the graph. xβ² 2 _ 20 + yβ² 2 _ 12 = 1 xβ ΞΈ = 45Β° 2 3 1 4 x yβ β4 β3 β2 β1 4 3 2 1 0 β1 β2 β3 β4 Figure 6 Writing equations of Rotated Conics in Standard Form Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic g... |
in the given equation, and then simplify. 5. Write the equations with xβ² and yβ² in the standard form with respect to the rotated axes. Example 3 Rewriting an Equation with respect to the xΒ΄ and yΒ΄ axes without the xΒ΄yΒ΄ Term Rewrite the equation 8x 2 β 12xy + 17y 2 = 20 in the xβ² yβ² system without an xβ² yβ² term. Soluti... |
y into in the given equation, and then simplify. 2 1 _ _ ξͺ ξͺ + yβ² ξ’ y = xβ² ξ’ β β 5 5 β β xβ² + 2yβ² _ 5 y = β β 2 2xβ² β yβ² ξͺ 8 ξ’ _ 5 β β β 12 ξ’ 2xβ² β yβ² ξͺ ξ’ _ 5 β xβ² + 2yβ² _ 5 β β β ξͺ + 17 ξ’ 2 xβ² + 2yβ² ξͺ _ = 20 5 β β 8 ξ’ (2xβ² β yβ²)(2xβ² β yβ²) __ 5 ξͺ = 20 8 (4xβ² 2 β 4xβ² yβ² + yβ² 2) β 12(2xβ² 2 + 3xβ² yβ² β 2yβ² 2) + 17(xβ² 2 + ... |
= β4 cot(2ΞΈ) = A β C ______ B 1 β (β4) ________ 12 cot(2ΞΈ) = cot(2ΞΈ) = 5 ___ 12 SECTION 10.4 rotation oF axis 917 Because cot(2ΞΈ) =, we can draw a reference triangle as in Figure 9. 5 __ 12 y 12 cot(2ΞΈ) = 5 12 x 2ΞΈ 5 Figure 9 cot(2ΞΈ) = = 5 ___ 12 adjacent _ opposite Thus, the hypotenuse is 52 + 122 = h2 25 + 144 = h2 ... |
3yβ² _ 13 ξͺ β 4 ξ’ β β 2 2xβ² + 3yβ² _ = 30 13 ξͺ β β 918 CHAPTER 10 analytic geometry 1 ξ’ ___ 13 ξͺ [(3xβ² β 2yβ² )2 + 12(3xβ² β 2yβ² )(2xβ² + 3yβ² ) β 4 (2xβ² + 3yβ² )2] = 30 1 ξͺ [9xβ² 2 β 12xβ² yβ² + 4yβ² 2 + 12 (6xβ² 2 + 5xβ² yβ² β 6yβ² 2) β 4 (4xβ² 2 + 12xβ² yβ² + 9yβ² 2)] = 30 ξ’ ___ 13 1 ξͺ [9xβ² 2 β 12xβ² yβ² + 4yβ² 2 + 72xβ² 2 + 60xβ² yβ² β 72... |
, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section. using the discriminant to identify a conic If the equation Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 is transformed by rotating axes into the equation Aβ²xβ² 2 + ... |
2 β Try It #3 Identify the conic for each of the following without rotating axes. a. x 2 β 9xy + 3y 2 β 12 = 0 b. 10x 2 β 9xy + 4y 2 β 4 = 0 Access this online resource for additional instruction and practice with conic sections and rotation of axes. β’ Introduction to Conic Sections (http://openstaxcollege.org/l/intro... |
β β 3 xy + 6y 2 β 6x β 3 = 0 17. 8x 2 + 4 β β 2 xy + 4y 2 β 10x + 1 = 0 For the following exercises, find a new representation of the given equation after rotating through the given angle. 18. 3x 2 + xy + 3y 2 β 5 = 0, ΞΈ = 45Β° 20. 2x 2 + 8xy β 1 = 0, ΞΈ = 30Β° 22. 4x 2 + β 2 xy + 4y 2 + y + 2 = 0, ΞΈ = 45Β° β 19. 4x 2 β x... |
= 30Β° 2 SECTION 10.4 section exercises 921 For the following exercises, graph the equation relative to the xβ² yβ² system in which the equation has no xβ² yβ² term. 39. xy = 9 41. x 2 β 10xy + y 2 β 24 = 0 43. 6x 2 + 2 β β 3 xy + 4y 2 β 21 = 0 45. 21x 2 + 2 β 3 xy + 19y 2 β 18 = 0 β 47. 16x 2 + 24xy + 9y 2 β 60x + 80y = 0... |
128 = 0, 59. Given kx 2 + 8xy + 8y 2 β 12x + 16y + 18 = 0, find k for the graph to be a hyperbola. find k for the graph to be a parabola. 60. Given 6x 2 + 12xy + ky 2 + 16x + 10y + 4 = 0, find k for the graph to be an ellipse. 922 CHAPTER 10 analytic geometry leARnInG OBjeCTIVeS In this section, you will: β’ Identify a... |
= 2 + y 2 shown in Figure 2. P(r, ΞΈ) D r ΞΈ Polar axis Directrix F, Focus @ pole x = 2 + y2 Figure 2 SECTION 10.5 conic sections in polar coordinates 923 In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any ... |
the type of conic, the directrix, and the eccentricity. 1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form. 2. Identify the eccentricity e as the coefficient of the trigonometric function in the denominator. 3. Compare e with 1 to det... |
οΏ½οΏ½ 4 + 5 cos ΞΈ 1 ξͺ ξ’ _ 4 1 _ ξͺ 12 ξ’ 4 __ 1 1 _ _ ξͺ cos __ Because cos ΞΈ is in the denominator, the directrix is x = p. Comparing to standard form + cos ΞΈ 4 Therefore, from the numerator, 3 = ep 5 __ = ξ’ ξ’ __ __ __ p 4 5 5 12 ___ = p 5 5 __ and the directrix is x = Since e > 1, the conic is a hyperbola. The eccentricity... |
the conic in standard form by multiplying the numerator and denominator by the 1 __. reciprocal of 3, which is 3 1 ξͺ 5 ξ’ _ 3 __ 1 1 ξͺ cos cos ΞΈ 5 __ 3 _ 1 + 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 direc... |
in Polar Form Graph r = 10 __________ 5 β 4 cos ΞΈ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the 1 __. reciprocal of 5, which is 5 r = 10 _________ 5 β 4 cos ΞΈ = 1 _ ξͺ 10 ξ’ 5 __ 1 1 _ _ ξͺ cos __ 1 β sin ΞΈ 5 4 __, e < 1, so we will graph an ellipse with a focu... |
If p < 0, use subtraction. If p > 0, use 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 5 Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Di... |
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