An ellipse meets a hyperbola that the tangent lines at the points of intersection are perpendicular to each other . Show that $a^2-b^2=c^2+d^2$

Question

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 $a^2-b^2=c^2+d^2$

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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?

Answer 1

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 $(**)$ $$a^2-b^2=c^2+d^2$$

Answer 2

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 $a,b,c,d$?

Answer 3

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.

Answer 4

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*}

Answer 5

Or just recognize that an ellipse and hyperbola that have a perpendicular intersection of tangent lines must be confocal (link below). Thus, the c value, or distance from center to focii for both ellipse and hyperbola must be equal. The focus of an ellipse is calculated by (a^2 - b^2)^0.5 while the focus of a hyperbola is (c^2 + d^2)^0.5. Set both these equal and you get a^2 - b^2 = c^2 + d^2.

https://en.wikipedia.org/wiki/Confocal_conic_sections#:~:text=In%20geometry%2C%20two%20conic%20sections,they%20have%20the%20same%20foci.&text=In%20the%20mixture%20of%20confocal,orthogonally%20(at%20right%20angles).