## Precalculus (6th Edition) Blitzer

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