## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Problems Plus - Problems - Page 271: 1

#### Answer

The point $P$ is $(-\frac{\sqrt{3}}{2},\frac{1}{4})$ The point $Q$ is $(\frac{\sqrt{3}}{2},\frac{1}{4})$

#### Work Step by Step

If the triangle $\triangle ABC$ is an equilateral triangle, then the angle $\angle ABC$ must be $60^{\circ}$. We can find the slope of the line $BA$: $m = tan ~60^{\circ}$ $m = \sqrt{3}$ The slope of the line $BA$ is $\sqrt{3}$ so the slope at point $P$ must also be $\sqrt{3}$ We can find the point $x$ on the parabola where the slope is $\sqrt{3}$: $y = 1-x^2$ $y' = -2x = \sqrt{3}$ $x = -\frac{\sqrt{3}}{2}$ When $x = -\frac{\sqrt{3}}{2},$ then $y = 1-(-\frac{\sqrt{3}}{2})^2 = \frac{1}{4}$ Therefore, the point $P$ is $(-\frac{\sqrt{3}}{2},\frac{1}{4})$ By symmetry, the point $Q$ is $(\frac{\sqrt{3}}{2},\frac{1}{4})$

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