Explanation Given: The given equation is, 16 x 2 − 9 y 2 − 96 x + 288 = 0 Approach: The basic equation for an hyperbola with a vertical transverse axis which is used is as follows, y 2 a 2 − x 2 b 2 = 1 Where a and b are constant values. Expression to find vertices is, V = ( 0 , ± a ) Expression to find Foci is, F = ( 0 , ± c ) , Where, c 2 = a 2 − b 2 Expression to find Asymptotes is, y = ± a b ( x − b ) Calculation: Consider the given equation, 16 x 2 − 9 y 2 − 96 x + 288 = 0 Now complete the square as shown below, 16 x 2 − 96 x + 144 − 144 − 9 y 2 = − 288 [ ( 4 x ) 2 − 2 ( 4 x ) ( 12 ) + ( 12 ) 2 ] − 9 y 2 = − 288 + 144 ( 4 x − 12 ) 2 − 9 y 2 = − 144 12 ( x − 3 ) 2 − 9 y 2 = − 144 Further solve the above equation. − 16 ( x − 3 ) 2 144 + 9 y 2 144 = 1 − ( x − 3 ) 2 9 + y 2 16 = 1 y 2 16 − ( x − 3 ) 2 9 = 1 This is an equation of a hyperbola with center ( 3 , 0 ) and a vertical transverse axis. Its graph will have the same shape as of un-shifted hyperbola. y 2 16 − x 9 = 1 Hyperbola with center at origin. Here, a 2 = 16 And, b 2 = 9 Then, c 2 = a 2 + b 2 c 2 = 16 + 9 c 2 = 25 c = ± 5 Thus the un-shifted ellipse foci are,

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Chapter 12.4, Problem 53E

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The graph of the equation 16x2−9y2−96x+288=0 and its characteristics.

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Chapter 12.4, Problem 52E

Chapter 12.4, Problem 54E

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Ch. 12.1 - A parabola is the set of all points in the plane...Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E Ch. 12.1 - Prob. 4E Ch. 12.1 - Prob. 5E Ch. 12.1 - Prob. 6E Ch. 12.1 - 5-10 Graphs of Parabolas Match the equation with...Ch. 12.1 - Prob. 8E Ch. 12.1 - 5-10 Graphs of Parabolas Match the equation with...Ch. 12.1 - Prob. 10E

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The graph of the equation 16 x 2 − 9 y 2 − 96 x + 288 = 0 and its characteristics.

Solution Summary: The author explains the basic equation for an hyperbola with a vertical transverse axis which is used.

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The graph of the equation 16x2−9y2−96x+288=0 and its characteristics.

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