Intersection of ellipse and hyperbola at a right angle

Question

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.

Answer 1

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... $${\beta^2+b^2 \over a^2-\alpha^2} =1$$

Answer 2

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.

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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.]

Answer 3

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.

Answer 4

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 $a^2-b^2=\alpha^2+\beta^2$.