Answer
$$x^2+\sqrt 3 xy=16$
Work Step by Step
Given: $\phi =60^{\circ}$
The polar equation for x-component along X-axis is given as:
$X=x \cos \phi+y\sin \phi=x \cos 60^{\circ} +y \sin 60^{\circ}\\=x\dfrac{1}{2}+y\dfrac{\sqrt 3}{2}\\=\dfrac{x-\sqrt 3y}{2}$
The polar equation for y-component along Y-axis is given as:
$Y=-x \sin \phi+y\cos \phi=x(\dfrac{\sqrt 3}{2})+y(\dfrac{1}{2})\\=\dfrac{x\sqrt 3+y}{2}$
Now, $x^2-3y^2=4$
This gives: $(\dfrac{x-\sqrt 3y}{2})^2-3(\dfrac{x\sqrt 3+y}{2})^2=4$
This implies that
$-8x^2-8\sqrt 3xy=16$
Thus, we have $x^2+\sqrt 3 xy=16$
