Find the coordinates of focus of parabola $$\left(y-x\right)^{2}=16\left(x+y\right)$$
rewriting:
$(\frac{x-y}{\sqrt{2}})^2=8\sqrt2(\frac{x+y}{\sqrt{2}})$
comparing with $Y^2=4aX$
$4a=8\sqrt2,a=2\sqrt2 $
$\Rightarrow$ coordinates of focus=2,2
Is this the correct approach?
Using Rotation of axes,
let $$x=X\cos t- Y\sin t,y=X\sin t+Y\cos t$$
set $\cos2t=0$ to eliminate $XY$
If $t=\dfrac\pi4$
$$2Y^2=16(\sqrt2X)\iff Y^2=4(2\sqrt2)X$$
Rewrite $$(x-y)^2=16(x+y)$$ as $$(\frac{x-y}{\sqrt{2}})^2=8\sqrt{2}\frac{(x+y)}{\sqrt{2}}\implies Y^2=4AX,~~~ A=2\sqrt{2}$$ The co-0rdinates of focus are given by $(X=A,Y=0) \implies x+y=4,x=y \implies x=2, y=2$ So the focus is at $(2,2)$.
