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