Answer
$\frac{(y-2)^2}{25}-\frac{(x-3)^2}{25}=1$
Work Step by Step
We know that $\tan^2 A+1=\sec^2 A$. $\tan t=\frac{x-3}{5},\sec t=\frac{y-2}{5}$
Hence here $(\frac{x-3}{5})^2+1=(\frac{y-2}{5})^2$, thus $\frac{(y-2)^2}{25}-\frac{(x-3)^2}{25}=1$
