<s>[INST] (Proof) Equality of the distances of any point $P(x, y)$ on the isosceles hyperbola to the foci and center of the hyperbola 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.
 [/INST] 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 $(0, 0)$
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