<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] 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.
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