## Precalculus (10th Edition)

$x^2=5(y-5)$
We are given the parametric equations: $\begin{cases} x=5\tan t\\ y=5\sec^2 t \end{cases}$ with $-\dfrac{\pi}{2}\lt t\lt \dfrac{\pi}{2}$ Rewrite the equations: $\begin{cases} \tan t=\dfrac{x}{5}\\ \sec^2 t=\dfrac{y}{5} \end{cases}$ Use the identity: $\tan^2 t+1=\sec^2 t$ $\left(\dfrac{x}{5}\right)^2+1=\dfrac{y}{5}$ $\dfrac{x^2}{25}+1=\dfrac{y}{5}$ $x^2+25=5y$ $x^2=5y-25$ $x^2=5(y-5)$ The curve represents a parabola.