Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Review Exercises - Page 1036: 57

Answer

$\frac{x^2}{9}-\frac{y^2}{9}=1$; see figure.
1584970476

Work Step by Step

Step 1. From the given equations, we have $sec(t)=\frac{x}{3}$ and $tan(t)=\frac{y}{3}$ Step 2. Using the identify $1+tan^2t=sec^2t$, we have $1+(\frac{y}{3})^2=(\frac{x}{3})^2$ or $\frac{x^2}{9}-\frac{y^2}{9}=1$ indicating a hyperbola centered at $(0,0)$ with $a=b=3$ Step 3. Given $0\leq t\leq \frac{\pi}{4}$ we have $0\leq y\leq 3$ and $3\leq x(=\frac{3}{cos(t)})\leq 3\sqrt 2$. We can graph the above equation as shown in the figure.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.