Explanation The given equation of the curve is x 2 16 − y 2 9 = 1 . Note that slope of a tangent line at a point on the curve is the value of the first derivative d y d x of the curve evaluated at that point. Differentiate both sides of the given equation. d d x ( x 2 16 ) − d d x ( y 2 9 ) = d d x ( 1 ) ( 1 16 ) 2 x − ( 1 9 ) 2 y d y d x = 0 d y d x = 9 x 16 y Therefore, the slope of the tangent line at point ( 20 3 , − 4 ) si calculated as follows. d y d x | ( 20 3 , − 4 ) = 9 x 16 y | ( 20 3 , − 4 ) = ( 9 ) ( 20 3 ) ( 16 ) ( − 4 ) = − 15 16 Therefore slope of the tangent line is − 15 16

Single Variable Calculus Format: Unbound (saleable)

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ISBN: 9780134765761

Author: Briggs, William L.^cochran, Lyle^gillett, Bernard^

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Chapter 12, Problem 58RE

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The equation of the tangent line to the curve x216−y29=1 at point (203,−4) and verify the answer with graphing utility.

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Chapter 12, Problem 57RE

Chapter 12, Problem 59RE

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The equation of the tangent line to the curve x 2 16 − y 2 9 = 1 at point ( 20 3 , − 4 ) and verify the answer with graphing utility.

Solution Summary: The author calculates the slope of the tangent line at a point on the curve x216-

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The equation of the tangent line to the curve x216−y29=1 at point (203,−4) and verify the answer with graphing utility.

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