# (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.

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

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 $$(0, 0)$$