<s>[INST] 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.
 [/INST] 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.]
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