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.
 A: Note that if $a^2=b^2+\alpha^2+\beta^2$ holds, then $a^2\ge \alpha^2$ holds since $$a^2=(b^2+\beta^2)+\alpha^2\ge 0+\alpha^2=\alpha^2.$$
There are four intersection points $(x,y)$ where
$$x^2=\frac{a^2\alpha^2 (b^2+\beta^2)}{\alpha^2 b^2+a^2\beta^2},\qquad y^2=\frac{\beta^2b^2(a^2-\alpha^2)}{\alpha^2b^2+a^2\beta^2}=\frac{\beta^2b^2(b^2+\beta^2)}{\alpha^2b^2+a^2\beta^2}\tag1$$
Now
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\implies \frac{yy_{\text{e}}'}{b^2}=-\frac{x}{a^2}$$
$$\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1\implies \frac{yy_{\text{h}}'}{\beta^2}=\frac{x}{\alpha^2}$$
giving
$$\frac{yy_{\text{e}}'}{b^2}\cdot \frac{yy_{\text{h}}'}{\beta^2}=-\frac{x^2}{a^2\alpha^2}\tag2$$
From $(1)(2)$, we have
$$y_{\text{e}}'y_{\text{h}}'=-\frac{x^2}{a^2\alpha^2}\cdot\frac{b^2\beta^2}{y^2}=-\frac{b^2\beta^2}{a^2\alpha^2}\cdot\frac{a^2\alpha^2}{b^2\beta^2}=-1$$
Therefore, the two curves intersect at right angle.
A: The two curves are an ellipse and a hyperbola, having coordinate axes as symmetry axes. It is well known (and easy to prove) that if $A$ is a point on an ellipse and $F$, $G$ are its foci, then the bisector of angle $FAG$ is the line normal to the ellipse at $A$; and if $A$ is a point on a hyperbola and $F'$, $G'$ are its foci, then the bisector of angle $F'AG'$ is the line tangent to the hyperbola at $A$.
For the two curves to intersect at right angles, it is necessary these two lines to be the same, and that happens only if they have the same foci. The squared distance of a focus from the center is given by $a^2-b^2$ for the ellipse and by $\alpha^2+\beta^2$ for the hyperbola, meaning that the curves have the same foci if $a^2-b^2=\alpha^2+\beta^2$.
A: The gradient vector $\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right)$ is normal to the curve $f(x,y)=0$.
Then
$$\left(\frac{2x}{a^2},\frac{2y}{b^2}\right)\cdot\left(\frac{2x}{\alpha^2},-\frac{2y}{\beta^2}\right)=\frac{4x^2}{a^2\alpha^2}-\frac{4y^2}{b^2\beta^2}=0.$$
We can eliminate $x^2$ and $y^2$ from the three equations by
$$\left|\begin{matrix}\dfrac1{a^2}&\dfrac1{b^2}&1\\\dfrac1{\alpha^2}&-\dfrac1{\beta^2}&1\\\dfrac1{a^2\alpha^2}&-\dfrac1{b^2\beta}&0\end{matrix}\right|=\frac{\beta^2+\alpha^2+b^2-a^2}{a^2b^2\alpha^2\beta^2}=0.$$
Hence the claim.

I have not investigated the inequality $\alpha^2<a^2$. Most probably you obtain it by expressing that $x^2,y^2>0.$ [Confirmed by the expressions provided by @mathlove.]
A: 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...
$${\beta^2+b^2
\over
a^2-\alpha^2}
=1$$
A: There is absolutely no need to indulge y' and tedious calculations. We can simply utilise the reflection properties of ellipse and hyperbola to note that confocal ellipse and hyperbola shall always intersect orthogonally. Here, ellipse focus is sqrt(a^2-b^2) and that of hyperbola is sqrt(alpha^2+beta^2) which are equal (given). Hence the result.
An important takeaway from this answer is that the best methods to solve conics questions is to involve BOTH Euclidean as well as analytical geometry. A video on YouTube by Mathologer on parabolas explains this in the best possible manner.
A: Wiki link explains the situation clearly in the paragraph Confocal ellipses and hyperbolas. 
Note that when ray is correctly traced,
Normal to ellipse is angle bisector between foci and enables internal reflection from one focus to the other as shown.$(\alpha=\alpha)$
Normal to hyperbola is angle bisector between foci and enables external reflection from one focus to the other as shown.$(\beta=\beta)$
