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: $\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?
By conformal mapping
\begin{align*}
  x+yi &amp;= \lambda\cosh(u+vi) \\
  &amp;= \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} &amp;= 1 \\
  \frac{x^2}{\lambda^2 \cos^2 v}-\frac{y^2}{\lambda^2 \sin^2 v} &amp;= 1 \\
  a^2-b^2 &amp;= \lambda^2 (\cosh^2 u-\sinh^2 u) \\
  &amp;= \lambda^2 \\
  c^2+d^2 &amp;= \lambda^2 (\cos^2 v+\sin^2 v) \\
  &amp;= \lambda^2 \\
  a^2-b^2 &amp;= c^2+d^2
\end{align*}
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.&amp;text=In%20the%20mixture%20of%20confocal,orthogonally%20(at%20right%20angles).