## Precalculus (6th Edition) Blitzer

$\frac{x^2}{16}+\frac{y^2}{9}=1$; see figure.
Step 1. From the given equations, we have $sin(t)=\frac{x}{4}$ and $cos(t)=\frac{y}{3}$ Step 2. Use the identify $sin^2t+cos^2t=1$. We have $(\frac{x}{4})^2+(\frac{y}{3})^2=1$ or $\frac{x^2}{16}+\frac{y^2}{9}=1$, indicating an ellipse with $a=4, b=3$ centered at $(0,0)$ Step 3. Given $0\leq t\lt \pi$, we have $0\leq x\leq 4$ and $-3\lt y\leq 3$. We can graph the above equation as shown in the figure.