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Given Conic Equation 2 9x² - 44² +548 +164 +29

Given Conic Equation 2 9x² - 44² +548 +164 +29

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Given Conic Equation 9x² - 4y² +54x +16y +29 = 0 9x² - 4y² + 54x + 1by = -29 Complete the square for 9x²+54x foome 92²+bx+c to find. Use the a, b, and c a=9 6=54 Consider the vertex form of parabola a(x+d)² + e Find the value da b 29 2 54 2x9 b2 पव Value CHO of d using the = Find the value of e using the formula е e= c- 20-(54)2 value of e 9(2+3)²-81 = -81 4x9 Subtitute the values of a, dande in the vertex of d = 5/20₁ b 29 1/2 2 Subtitute 9(2+3)²-81 for 9x² +54x in the equation to find the value of 9x² - 4y² +54x + 16y=-29 9(2+3) ²2²-81-4y² + 16y = - 9 (x+3)² - 4y²+1by 9(x+3)² - 4y² +1by C=O = C- = 3 Complete the Square for _4y² +167 Use the form ax²+bx+c, to find the value of a, band c a = - 4 b=16 Consider the vertex form of parabola G (a+d)² + e = e formula 52 49 16 2X (-4) = -2 =-29+8) = +62 (16)2 = 16 4 X (-4) Subtitute the value of a, d ande into the vertex form -4(y-2)² +16 Subtitute -4 (9-2)² +16 for - 4y² +16y in the equation. Gx² - 4y² + 54x+16y=-29 9 (x+3)² - 4(y + 2)² + 16 = -29+8|
Transcribed Image Text:Given Conic Equation 9x² - 4y² +54x +16y +29 = 0 9x² - 4y² + 54x + 1by = -29 Complete the square for 9x²+54x foome 92²+bx+c to find. Use the a, b, and c a=9 6=54 Consider the vertex form of parabola a(x+d)² + e Find the value da b 29 2 54 2x9 b2 पव Value CHO of d using the = Find the value of e using the formula е e= c- 20-(54)2 value of e 9(2+3)²-81 = -81 4x9 Subtitute the values of a, dande in the vertex of d = 5/20₁ b 29 1/2 2 Subtitute 9(2+3)²-81 for 9x² +54x in the equation to find the value of 9x² - 4y² +54x + 16y=-29 9(2+3) ²2²-81-4y² + 16y = - 9 (x+3)² - 4y²+1by 9(x+3)² - 4y² +1by C=O = C- = 3 Complete the Square for _4y² +167 Use the form ax²+bx+c, to find the value of a, band c a = - 4 b=16 Consider the vertex form of parabola G (a+d)² + e = e formula 52 49 16 2X (-4) = -2 =-29+8) = +62 (16)2 = 16 4 X (-4) Subtitute the value of a, d ande into the vertex form -4(y-2)² +16 Subtitute -4 (9-2)² +16 for - 4y² +16y in the equation. Gx² - 4y² + 54x+16y=-29 9 (x+3)² - 4(y + 2)² + 16 = -29+8|
9 (x+3)² - 4(4+2)² 9(2+3)² -4(y+2)² = 36 divide each term by 36 to make the right side equal to one. 9 (x+3)² 36 (2+3)² 4 1 (x-4)² 92 4(५+2)2 36 (y-2)2 9 This is the form of a hyperbola. Standrad form of hyperbolo (Y-K 2² b2 = -29 +81-16 Here, a=2, b= 3, K=2, h=-3 The center of hyperbola (hik) (-3,2) Find the distance from the center to a focus <= √6²45² = √(²)² + (3)² = √√4+9 =√13 Find the eccentricity eccentricity = Find the focal parameter Focal Parameter 36 = 1 Find the vertices The first vertex of a hyperbola can be found by adding a to h (h+a, k) (-1,2) 19²+52 = 1 The second vestex of a hyperbola can be by found by Subtracting a from h (n-a, k) 36 (-5,2) The vertices of a hyperbola (-1,2), (-5,2) Find the foci The first focus of a hyperbola can be found by adding (toh (h+(₂ k) (-3+√√13,2) Seond second focus of a hyperbola found by Subtracting c from h (h-C, K) (-3-√√13, 2) The foci of a hyperbola (-3+√73, 2) (-3-√73,2) First Asymptotes y = 3x + 13 2 2 Second Asymptates y = ± 3 (x-(-3)+2 2 √√/2²+32 2 62 =32 √a²+5² 113 Asymptotes follow the form y: I bla-h) + because this hyperbola open left and right a y=-321-57/2 can be 2 =9/13 13 +K
Transcribed Image Text:9 (x+3)² - 4(4+2)² 9(2+3)² -4(y+2)² = 36 divide each term by 36 to make the right side equal to one. 9 (x+3)² 36 (2+3)² 4 1 (x-4)² 92 4(५+2)2 36 (y-2)2 9 This is the form of a hyperbola. Standrad form of hyperbolo (Y-K 2² b2 = -29 +81-16 Here, a=2, b= 3, K=2, h=-3 The center of hyperbola (hik) (-3,2) Find the distance from the center to a focus <= √6²45² = √(²)² + (3)² = √√4+9 =√13 Find the eccentricity eccentricity = Find the focal parameter Focal Parameter 36 = 1 Find the vertices The first vertex of a hyperbola can be found by adding a to h (h+a, k) (-1,2) 19²+52 = 1 The second vestex of a hyperbola can be by found by Subtracting a from h (n-a, k) 36 (-5,2) The vertices of a hyperbola (-1,2), (-5,2) Find the foci The first focus of a hyperbola can be found by adding (toh (h+(₂ k) (-3+√√13,2) Seond second focus of a hyperbola found by Subtracting c from h (h-C, K) (-3-√√13, 2) The foci of a hyperbola (-3+√73, 2) (-3-√73,2) First Asymptotes y = 3x + 13 2 2 Second Asymptates y = ± 3 (x-(-3)+2 2 √√/2²+32 2 62 =32 √a²+5² 113 Asymptotes follow the form y: I bla-h) + because this hyperbola open left and right a y=-321-57/2 can be 2 =9/13 13 +K
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