## Calculus (3rd Edition)

Published by W. H. Freeman

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

#### Answer

$$y =(1-x)^{-2}$$

#### Work Step by Step

Given $$x=1+t^{-1},\ \ \ \ y= t^2$$ Since $t=\frac{1}{x-1}$, then \begin{align*} y&= t^2\\ &= \frac{1}{(x-1)^2}\\ &=(1-x)^{-2} \end{align*} i.e. $$y =(1-x)^{-2}$$

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