Calculus: Early Transcendentals 8th Edition

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

Chapter 3 - Review - Exercises - Page 268: 66

Answer

The points on the ellipse are $~~(-\sqrt{\frac{2}{3}}, \frac{\sqrt{6}}{6})~~$ and $~~(\sqrt{\frac{2}{3}}, -\frac{\sqrt{6}}{6})~~$

Work Step by Step

$x^2+2y^2 = 1$ We can find $\frac{dy}{dx}$: $x^2+2y^2 = 1$ $2x+4y\frac{dy}{dx} = 0$ $4y\frac{dy}{dx} = -2x$ $\frac{dy}{dx} = -\frac{2x}{4y}$ $\frac{dy}{dx} = -\frac{x}{2y}$ The tangent line has a slope of $1$ at points on the ellipse where $\frac{dy}{dx} = 1$ We can find an expression for $y$ when $\frac{dy}{dx} = 1$: $\frac{dy}{dx} = -\frac{x}{2y} = 1$ $2y = -x$ $y = -\frac{x}{2}$ We can find the x-coordinates of the points on the ellipse that satisfy this condition: $x^2+2y^2 = 1$ $x^2+2(-\frac{x}{2})^2 = 1$ $x^2+\frac{x^2}{2} = 1$ $\frac{3x^2}{2} = 1$ $x^2 = \frac{2}{3}$ $x = \pm \sqrt{\frac{2}{3}}$ When $x = -\sqrt{\frac{2}{3}}$: $y = -\frac{x}{2}$ $y = -\frac{-\sqrt{\frac{2}{3}}}{2}$ $y = \sqrt{\frac{1}{4}}\cdot \sqrt{\frac{2}{3}}$ $y = \sqrt{\frac{2}{12}}$ $y = \sqrt{\frac{1}{6}}$ $y = \frac{\sqrt{6}}{6}$ When $x = \sqrt{\frac{2}{3}}$: $y = -\frac{x}{2}$ $y = -\frac{\sqrt{\frac{2}{3}}}{2}$ $y = -\sqrt{\frac{1}{4}}\cdot \sqrt{\frac{2}{3}}$ $y = -\sqrt{\frac{2}{12}}$ $y = -\sqrt{\frac{1}{6}}$ $y = -\frac{\sqrt{6}}{6}$ The points on the ellipse are $~~(-\sqrt{\frac{2}{3}}, \frac{\sqrt{6}}{6})~~$ and $~~(\sqrt{\frac{2}{3}}, -\frac{\sqrt{6}}{6})~~$
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