## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.1 Parametric Equations - Exercises - Page 603: 14

#### Answer

$$y = \frac{\sqrt{1-x^2}}{x}$$

#### Work Step by Step

Given $$x=\cos t,\ \ \ \ y= \tan t$$ Since $x=\cos t$, then \begin{align*} y&= \tan t\\ &= \frac{\sin t}{\cos t} \\ &= \frac{\sqrt{1-\cos^2t}}{\cos t}\\ &= \frac{\sqrt{1-x^2}}{x} \end{align*} i.e. $$y = \frac{\sqrt{1-x^2}}{x}$$

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