Answer
$7X^2-7Y^2+48XY-40X-30Y=0$
Work Step by Step
Here, we have $x=X \cos \phi+Y\sin \phi$ and $y=-X \sin \phi+Y \cos \phi$
$\cos \phi =\dfrac{3}{5}$ and $\sin \phi =\dfrac{4}{5}$
$x=X (\dfrac{3}{5}) -Y (\dfrac{4}{5}) \implies x=\dfrac{3X-4Y}{5}$
and $y=X (\dfrac{4}{5})+Y (\dfrac{3}{5}) \implies y=\dfrac{4X+3Y}{5}$
$x^2-y^2=2y$
This gives: $(\dfrac{3X-4Y}{5})^2-(\dfrac{4X+3Y}{5})^2=2(\dfrac{4X+3Y}{5})$
Thus, $7X^2-7Y^2+48XY-40X-30Y=0$
