(Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola

Question

I searched but couldn't find the proof.

Isosceles hyperbola equation: $${H:x^{2}-y^{2} = a^{2}}$$

And let's take any point $P(x, y)$ on this hyperbola. Now, the product of the distances of this point $P(x, y)$ to the foci of the isosceles hyperbola is equal to the square of the distance from point $P$ to the center of the hyperbola.

Proof?

I took this question from my analytical geometry project assignment. I tried various ways (I found the foci $F(x,y)$ and $F^{'}(x,y)$ terms of $x$, $y$ and chose any point on the hyperbola...) but I couldn't prove. I request your help.

Answer 1

$$PS_1.PS_2=\sqrt{[(x-ae)^2+y^2][(x+ae)^2+y^2]}$$ $$=\sqrt{(x^2-a^2e^2)^2+y^2(x^2+a^2e^2+2aex)+y^2(x^2+a^2e^2-2aex)+y^4} $$ $$=\sqrt{x^4+a^4e^4-2a^2e^2x^2+y^2x^2+y^2a^2e^2+2aexy^2+y^2x^2+y^2a^2e^2-2aexy^2}$$ $$=\sqrt{x^2+2x^2y^2+y^4+a^4e^4-2a^2e^2(x^2-y^2)}$$ $$=\sqrt{(x^2+y^2)^2+a^4e^4-2a^4e^2}$$ Since $e=\sqrt{2}$ for this hyperbola, so finally we prove that $$PS_1.PS_2= x^2+y^2=OP^2.$$

Answer 2

For any point on the hyperbola, $x^2 - y^2 = a^2$

Foci of the hyperbola are $(\pm a\sqrt2,0)$ and the center is $(0, 0)$.

So product of distance of point $P(x,y)$ on the hyperbola to foci is

$\sqrt{(x-a\sqrt2)^2 + y^2} \times \sqrt{(x+a\sqrt2)^2 + y^2}$

$\sqrt{x^2 + 2a^2 + y^2 - 2 \sqrt2 a x} \times \sqrt{x^2 + 2a^2 + y^2 + 2 \sqrt2 a x}$

Using $y^2 = x^2 - a^2$,

$ = \sqrt{((2x^2+a^2) - 2\sqrt2 a x) ((2x^2+a^2) + 2\sqrt2 a x)}$

$ = \sqrt{ 4x^4 + a^4+4a^2x^2 - 8a^2x^2}$

$ = 2x^2 - a^2 = x^2 + y^2$

Which is the square of distance of point $P$ to the center of the hyperbola $(0, 0)$