Let a,b be positive numbers. Sketch the parabolas

$\displaystyle y^2=4a^2-4ax $

$\displaystyle y^2=4b^2+4bx $

and show that they have a common focus, and that they intersect at right angles.

I was able to show that the focus of each parabola was at the origin by reducing these equations to standard form:

$\displaystyle y^2=-4a(x-a) $

$\displaystyle y^2=4b(x+b) $

But I don't know how to prove that they intersect at right angles. I got the coordinates $\displaystyle \left( \frac{a^2-b^2}{a+b},\pm 2\sqrt{ab} \right)$ as the points of intersection. How do I show that the parabolas cross through these points at right angles to one another? Any help would be appreciated.