Find conditions on a1,a2,b1,b2 so that the intersections of two second-order curves are perpendicular $a_1x^2+b_1y^2=1;$
$a_2 x^2+b_2y^2=1;$
I'm tried to use derivatives in points of crossing (that it crossed at right angle too), 
but nothing has helped. Thanks!
 A: Note that $y^2 =  \frac{1}{b_1} - \frac{a_1}{b_1}x^2$
and
$y^2 =  \frac{1}{b_2} - \frac{a_2}{b_2}x^2$
so to find the points of intersection, set the two equal:
$$\frac{1}{b_1} - \frac{a_1}{b_1}x^2 =  \frac{1}{b_2} - \frac{a_2}{b_2}x^2$$
and solve the resulting quadratic for $x$, and then use one of the original equations to solve for the corresponding $y$ (and note that because these are quadratics, you'll likely find more than one solution, corresponding to more than one point of intersection. Be thorough in your analysis through this step, there can be as many as $4$ points of intersection).
To check whether the intersection is at a right angle, the derivative will come into play, as you suggest.
Use implicit differentiation to find
$$
2a_1 x + 2b_1 y \frac{dy}{dx} = 0 \\
\frac{dy}{dx} = -\frac{a_1 x}{b_1 y}
$$
on the first curve, and
$$
\frac{dy}{dx} = -\frac{a_2 x}{b_2 y}
$$
on the second. Now, if you plug in the points of intersection $(x,y)$ you found in the first part, you can compare the two derivatives to test for a right-angle intersection. This should finally allow you to come up with conditions on $a_1,a_2,b_1,b_2$ so that this happens.
