Explanation Proof: Consider the equation A x 2 + C y 2 + D x + E y + F = 0 . Assume that the conic is not degenerate. (a) Consider the condition: A = C , A ≠ 0 Substitute A = C in the equation A x 2 + C y 2 + D x + E y + F = 0 . A x 2 + A y 2 + D x + E y + F = 0 A x 2 + y 2 + D A x + E A y + F A = 0 x 2 + D A x + y 2 + E A y = − F A Make the terms of x and y perfect squares. x 2 + D A x + D 2 4 A 2 + y 2 + E A y + E 2 4 A 2 = D 2 4 A 2 + E 2 4 A 2 − F A x + D 2 A 2 + y + E 2 A 2 = D 2 + E 2 − 4 A F 4 A 2 This equation is in the form of the standard equation of the circle x − h 2 + y − k 2 = r 2 , where center is h , k and radius is r . So, this represents a circle with center − D 2 A , − E 2 A and radius D 2 + E 2 − 4 A F 2 A . (b) Consider the condition: A = 0 or C = 0 , but not both So, take C = 0 . Substitute C = 0 in the equation A x 2 + C y 2 + D x + E y + F = 0 . A x 2 + 0 y 2 + D x + E y + F = 0 A x 2 + D x + E y + F = 0 A x 2 + D A x + E A y + F A = 0 x 2 + D A x = − E A y − F A Make the terms of x a perfect square. x 2 + D A x + D 2 4 A 2 = − E A y − F A + D 2 4 A 2 x + D 2 A 2 = − E A y + F E − D 2 4 A E x + D 2 A 2 = − 4 ⋅ E 4 A y + 4 A F − D 2 4 A E This equation is in the form of the standard equation of the parabola x − h 2 = − 4 a y − k , where vertex is h , k . So, this represents a parabola with vertex − D 2 A , D 2 − 4 A F 4 A E . (c) Consider the condition: A C > 0 , A ≠ C A x 2 + C y 2 + D x + E y + F = 0 A x 2 + D A x + C y 2 + E C y = − F Make the terms of x and y perfect squares. A x 2 + D A x + D 2 4 A 2 + C y 2 + E C y + E 2 4 C 2 = D 2 4 A + E 2 4 C − F A x + D 2 A 2 + C y + E 2 C 2 = C D 2 + A E 2 − 4 A C F 4 A C A x + D 2 A 2 C D 2 + A E 2 − 4 A C F 4 A C + C y + E 2 C 2 C D 2 + A E 2 − 4 A C F 4 A C = 1 x + D 2 A 2 C D 2 + A E 2 − 4 A C F 4 A 2 C 2 + y + E 2 C 2 C D 2 + A E 2 − 4 A C F 4 A C 2 2 = 1 If A C > 0 , then either both A and C are positive or both are negative

10th Edition

ISBN: 9781337282291

Author: Ron Larson

Publisher: Cengage Learning

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Question

Chapter 4, Problem 11PS

To determine

To prove: The graph of the equation Ax2+Cy2+Dx+Ey+F=0 is one of the given conic-sections (except in degenerate cases).

Conic Conditiona Circle A=C, A≠0b Parabola A=0 or C=0, but not bothc Ellipse AC>0, A≠Cd Hyperbola AC<0

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Chapter 4, Problem 10PS

Chapter 5.1, Problem 1ECP

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Chapter 4 Solutions

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To prove: The graph of the equation A x 2 + C y 2 + D x + E y + F = 0 is one of the given conic-sections (except in degenerate cases). Conic Condition a Circle A = C , A ≠ 0 b Parabola A = 0 or C = 0 , but not both c Ellipse A C > 0 , A ≠ C d Hyperbola A C < 0

Solution Summary: The author explains how to prove that the graph of the equation Ax2+Cy2,+Dx+Ey+F=0 is a conic-section.

To prove: The graph of the equation Ax2+Cy2+Dx+Ey+F=0 is one of the given conic-sections (except in degenerate cases).

Conic Conditiona Circle A=C, A≠0b Parabola A=0 or C=0, but not bothc Ellipse AC>0, A≠Cd Hyperbola AC<0

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Chapter 4 Solutions

To prove: The graph of the equation A x 2 + C y 2 + D x + E y + F = 0 is one of the given conic-sections (except in degenerate cases). Conic Condition a Circle A = C , A ≠ 0 b Parabola A = 0 or C = 0 , but not both c Ellipse A C &gt; 0 , A ≠ C d Hyperbola A C &lt; 0

Solution Summary: The author explains how to prove that the graph of the equation Ax2+Cy2,+Dx+Ey+F=0 is a conic-section.

To prove: The graph of the equation Ax2+Cy2+Dx+Ey+F=0 is one of the given conic-sections (except in degenerate cases). Conic Conditiona Circle A=C, A≠0b Parabola A=0 or C=0, but not bothc Ellipse AC>0, A≠Cd Hyperbola AC<0

Want to see the full answer?

Chapter 4 Solutions

To prove: The graph of the equation A x 2 + C y 2 + D x + E y + F = 0 is one of the given conic-sections (except in degenerate cases). Conic Condition a Circle A = C , A ≠ 0 b Parabola A = 0 or C = 0 , but not both c Ellipse A C > 0 , A ≠ C d Hyperbola A C < 0

To prove: The graph of the equation Ax2+Cy2+Dx+Ey+F=0 is one of the given conic-sections (except in degenerate cases).

Conic Conditiona Circle A=C, A≠0b Parabola A=0 or C=0, but not bothc Ellipse AC>0, A≠Cd Hyperbola AC<0