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Question: $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1 $ meets $\frac{x^2}{c^2} - \frac{y^2}{d^2}=1$ in such a way that the tangent lines at the points of intersection are perpendicular to each other. Show that $a^2-b^2=c^2+d^2$


I've been stuck on a while now , first I tried finding the intersection between the ellipse and the hyperbola and then creating tangent lines based on those specific $x_1 , y_1$ but that didn't turn out pretty. Second of all I tried taking the derivative of both the ellipse and hyperbola and using $m_1 \cdot m_2 =-1 $ but that didn't get me far either. What should I do?

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    $\begingroup$ You need to do both off those things you mention above. And this will give you two sets of equations. Substitute from one to the other, and lots of things cancel. $\endgroup$ – Doug M Aug 4 '16 at 1:12
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Intersection points are given by $$\frac{x^2}{a^2} + \frac{y^2}{b^2}=\frac{x^2}{c^2} - \frac{y^2}{d^2}\iff(\frac xy)^2=-(\frac{ac}{bd})^2\frac{b^2+d^2}{c^2-a^2}\qquad (*)$$ Besides by implicit derivation we have for the ellipse $$-\frac{xb^2}{ya^2}$$ and for the hyperbola $$\frac{xd^2}{yc^2}$$ Now from the equality for perpendicularity $$(-\frac{xb^2}{ya^2})(\frac{xd^2}{yc^2})=-1\iff(\frac xy)^2\frac{(bd)^2}{(ac)^2}=1\qquad (**)$$ Thus from $(*)$ and $(**)$ $$a^2-b^2=c^2+d^2$$

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By conformal mapping \begin{align*} x+yi &= \lambda\cosh(u+vi) \\ &= \lambda \cosh u \cos v+\lambda i\sinh u \sin v \\ \frac{x^2}{\lambda^2 \cosh^2 u}+\frac{y^2}{\lambda^2 \sinh^2 u} &= 1 \\ \frac{x^2}{\lambda^2 \cos^2 v}-\frac{y^2}{\lambda^2 \sin^2 v} &= 1 \\ a^2-b^2 &= \lambda^2 (\cosh^2 u-\sinh^2 u) \\ &= \lambda^2 \\ c^2+d^2 &= \lambda^2 (\cos^2 v+\sin^2 v) \\ &= \lambda^2 \\ a^2-b^2 &= c^2+d^2 \end{align*}

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Hint:

Solve for $x^2,y^2$

Find $\dfrac{dy}{dx}$ for both the curve and multiply. Replace the values of $x^2,y^2$ in terms of $a,b,c,d$?

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If the tangents to the two curves are orthogonal at their intersection, then so are their normals. A normal to the ellipse at a point is given by $$\nabla\left({x^2\over a^2}+{y^2\over b^2}\right)=2\left\langle{x\over a^2},{y\over b^2}\right\rangle$$ and a normal to the hyperbola by $$\nabla\left({x^2\over c^2}-{y^2\over d^2}\right)=2\left\langle{x\over c^2},-{y\over d^2}\right\rangle.$$ For these to be orthogonal, their dot product must vanish, which yields the condition $${x^2\over a^2c^2}={y^2\over b^2d^2}.\tag{*}$$ At the intersection of the two curves we have $${x^2\over a^2}+{y^2\over b^2}={x^2\over c^2}-{y^2\over d^2}$$ and substituting the values of $y^2/b^2$ and $y^2/d^2$ from the previous equation gives $$(c^2+d^2){x^2\over a^2c^2}=(a^2-b^2){x^2\over a^2c^2},$$ from which the required result follows. The condition (*) can also be derived from the slopes of the tangents to the two curves.

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