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

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

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