Explanation Rewrite the equation x 2 − 4 y 2 + 4 x + 32 y − 64 = 0 in terms of y as follows: x 2 − 4 y 2 + 4 x + 32 y − 64 = 0 − 4 y 2 + 32 y + ( x 2 + 4 x − 64 ) = 0 4 y 2 − 32 y − ( x 2 + 4 x − 64 ) = 0 Note that, 4 y 2 − 32 y − ( x 2 + 4 x − 64 ) = 0 is of the form a y 2 + b y + c = 0 where a = 4 , b = − 32 , c = − ( x 2 + 4 x − 64 ) . Substitute a = 4 , b = − 32 , c = − ( x 2 + 4 x − 64 ) in y = − b ± b 2 − 4 a c 2 a to obtain the solution of the equation as follows: y = − ( − 32 ) ±

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

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Author: Allyn Washington, Richard Evans

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Chapter 21.6, Problem 37E

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To sketch: The graph of the hyperbola x2−4y2+4x+32y−64=0 using a calculator.

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Chapter 21.6, Problem 36E

Chapter 21.6, Problem 38E

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Chapter 21 Solutions

Pearson eText for Basic Technical Mathematics with Calculus -- Instant Access (Pearson+)

Ch. 21.1 - Find the distance between (−2, 6) and (5, −3). Ch. 21.1 - Prob. 2PE Ch. 21.1 - Prob. 1E Ch. 21.1 - Prob. 2E Ch. 21.1 - Prob. 5E Ch. 21.1 - In Exercises 5–14, find the distance between the...Ch. 21.1 - Prob. 7E Ch. 21.1 - Prob. 8E Ch. 21.1 - Prob. 9E Ch. 21.1 - Prob. 10E

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The graph of the hyperbola x 2 − 4 y 2 + 4 x + 32 y − 64 = 0 using a calculator.

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To sketch: The graph of the hyperbola x2−4y2+4x+32y−64=0 using a calculator.

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Chapter 21 Solutions