Answer
$x^2-2\sqrt 3 xy-y^2=-4$
Work Step by Step
Here, we have $x'=x \cos \phi+y \sin \phi$ and $y'=-x \sin \phi+y \cos \phi$ and $\phi =30^{\circ}$
The polar equation for x-component along X-axis is given as:
$x'=x \cos \phi+y\sin \phi=x \cos 30^{\circ} +y\sin 30^{\circ}\\=\dfrac{x\sqrt 3}{2}+\dfrac{y}{2}$
The polar equation for y-component along Y-axis is given as:
$y'=-x \sin \phi+y\cos \phi=-x \sin 30^{\circ}+y\cos 30^{\circ}\\=\dfrac{x}{2}+\dfrac{y\sqrt 3}{2}$
Consider $x'^2+2\sqrt 3x'y'-y'^2=4$
$(\dfrac{x\sqrt 3}{2}+\dfrac{y}{2})^2+2\sqrt 3(\dfrac{x\sqrt 3}{2}+\dfrac{y}{2}) (\dfrac{x}{2}+\dfrac{y\sqrt 3}{2}) -(\dfrac{x}{2}+\dfrac{y\sqrt 3}{2})^2=4$
This gives: $3x^2+y^2+2\sqrt xy-6x^2+4\sqrt 3 xy+6y^2-x^2-3y^2+2\sqrt 3 xy=16$
Thus, $x^2-2\sqrt 3 xy-y^2=-4$
