Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.1 - Curves Defined by Parametric Equations - 10.1 Exercises - Page 647: 41


The parametric form of the answer is: $$x=a\,cos\theta\\y=b\,sin\theta$$ Removing the parameter, we get the ellipse: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

Work Step by Step

Draw a line such that the angle between the line and the $x$-axis is $\theta$. Draw two line segments perpendicular to the $x$-axis with each going through points where the line we drew earlier intersects with each circle. Looking at the drawing, we can see that the distance point $P$ is from the $x$-axis is equal to the height of the smaller triangle and the distance point $P$ is from the $y$-axis is equal to the width of the larger triangle. Since the hypotenuse of the smaller triangle is $b$ and the hypotenuse of the larger triangle is $a$, we can represent all possible points $P$ with the equations: $$x=a\,cos\theta\\y=b\,sin\theta$$ To remove the parameter, we can square each equation, getting: $$x^2=a^2cos^2\theta\\y^2=b^2sin^2\theta$$ We can divide each equation by $a^2$ and $b^2$ respectively getting: $$\frac{x^2}{a^2}=cos^2\theta\\\frac{y^2}{b^2}=sin^2\theta$$ Adding the two equations together, we get: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=cos^2\theta+sin^2\theta$$ Using the Pythagorean trig identity, we get the ellipse: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
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