Explanation Given: Hyperbola is x 2 a 2 − y 2 b 2 = 1 and the point is ( x 0 , y 0 ) . Concept used: Equation of a line passing through the point (a, b) with slope m is, y − b = m ( x − a ) Calculation: Slope of the tangent is the derivative of the curve at the given point. So, differentiating given curve with respect to x, we get 2 x a 2 − 2 y b 2 ⋅ d y d x = 0 ⇒ 2 y b 2 ⋅ d y d x = 2 x a 2 ⇒ d y d x = b 2 x a 2 y Now slope of the tangent at ( x 0 , y 0 ) is, d y d x | ( x 0 , y 0 ) = b 2 x 0 a 2 y 0 Hence equation of the tangent line at ( x 0 , y 0 ) is, y − y 0 = b 2 x 0 a 2 y 0 ( x − x 0 ) ⇒ a 2 y 0 y − a 2 y 0 ⋅ y 0 = b 2 x 0 x − b 2 x 0 ⋅ x</

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 3.6, Problem 56PS

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To find the equation of the tangent line to the hyperbola x2a2−y2b2=1 at the point (x0 , y0) .

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Chapter 3.6, Problem 55PS

Chapter 3.6, Problem 57PS

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Chapter 3 Solutions

Calculus

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To find the equation of the tangent line to the hyperbola x 2 a 2 − y 2 b 2 = 1 at the point ( x 0 , y 0 ) .

Solution Summary: The author calculates the slope of the tangent line to the hyperbola, based on the equation of a line passing through the point.

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To find the equation of the tangent line to the hyperbola x2a2−y2b2=1 at the point (x0 , y0) .

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