Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.5 - Implicit Differentiation - 3.5 Exercises - Page 216: 44

Answer

At the point $(x_0,y_0),$ the equation of a tangent line is: $~~\frac{x_0~x}{a^2}+\frac{y_0~y}{b^2} = 1$

Work Step by Step

$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ We can find an expression for $y'$: $\frac{2x}{a^2}+\frac{2y}{b^2}~y' = 0$ $\frac{2y}{b^2}~y' = -\frac{2x}{a^2}$ $y' = -\frac{xb^2}{ya^2}$ At the point $(x_0, y_0)$, the slope of a tangent line is $-\frac{x_0b^2}{y_0a^2}$ We can find the equation of a tangent line at the point $(x_0,y_0)$: $y-y_0 = -\frac{x_0b^2}{y_0a^2}~(x-x_0)$ $\frac{y_0~y}{b^2}-\frac{y_0^2}{b^2} = \frac{x_0^2}{a^2}-\frac{x_0~x}{a^2}$ $\frac{x_0~x}{a^2}+\frac{y_0~y}{b^2} = \frac{x_0^2}{a^2}+\frac{y_0^2}{b^2}$ $\frac{x_0~x}{a^2}+\frac{y_0~y}{b^2} = 1$ At the point $(x_0,y_0),$ the equation of a tangent line is: $~~\frac{x_0~x}{a^2}+\frac{y_0~y}{b^2} = 1$
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