Intersection of ellipse and hyperbola at a right angle Need to show that two functions intersect at a right angle. 
Show that the ellipse
$$ \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1
$$
and the hyperbola
$$
\frac{x^2}{α^2} −\frac{y^2}{β^2} = 1
$$
will intersect at a right angle if
$$α^2 ≤ a^2 \quad \text{and}\quad a^2 − b^2 = α^2 + β^2$$
Not sure how to tackle this question, graphing didn't help.

Apparently, if $\alpha^2>a^2$, the two curves never intersect in the first place. Now let's just find the derivatives at the point of intersection and show that their product is $-1$.
Ellipse:
$$y=b\sqrt{1-\left({x\over a}\right)^2}\\
y'=b\cdot{-2x/a^2\over2\sqrt{1-\left({x\over a}\right)^2}}=-{b^2x\over a^2y}\tag{1}$$
Hyperbola:
$$y=\beta\sqrt{\left({x\over\alpha}\right)^2-1}\\
y'=\beta\cdot{2x/\alpha^2\over2\sqrt{\left({x\over\alpha}\right)^2-1}}={\beta^2x\over\alpha^2y}\tag{2}$$
Point of intersection:
$${x^2\over\alpha^2} − {y^2\over\beta^2} = 1\\
{x^2\over a^2} + {y^2\over b^2} = 1$$
Let's multiply the first equation by $\beta^2$, the second by $b^2$, and add them together.
$${\beta^2x^2\over\alpha^2}+{b^2x^2\over a^2} = \beta^2+b^2\\
x^2={\beta^2+b^2\over{\beta^2\over\alpha^2}+{b^2\over a^2}}=a^2\alpha^2{\beta^2+b^2\over a^2\beta^2+\alpha^2b^2}\\
y^2=b^2\left(1-{x^2\over a^2}\right)=b^2\cdot{a^2\beta^2+\alpha^2b^2-\alpha^2\beta^2-\alpha^2b^2\over a^2\beta^2+\alpha^2b^2}=b^2\beta^2\cdot{a^2-\alpha^2\over a^2\beta^2+\alpha^2b^2}\tag{3}$$
Now back to the derivatives:
$$-{b^2x\over a^2y}\cdot{\beta^2x\over\alpha^2y}=-1\\
{b^2\beta^2x^2\over a^2\alpha^2y^2}=1
$$
Plug in $x^2$ and $y^2$ from (3). Then nearly everything magically cancels out, and we're left with...
$${\beta^2+b^2
\over
a^2-\alpha^2}
=1$$
