Question

Find the locus of the point of intersection of tangents of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if the difference of the eccentric angles of their points of contact is ${90}^o$.

Answer 1

$h=\frac{a\cos \left((\alpha +\beta )/2\right)}{\cos \left((\alpha -\beta )/2\right)}$

$⟹h=\frac{a\cos \left((\alpha +\beta )/2\right)}{1/\sqrt{2}}$

$⟹h=\sqrt{2}a\cos \left(\frac{\alpha +\beta }{2}\right)$

$k=\sqrt{2}b\sin \left(\frac{\alpha +\beta }{2}\right)$

${\sin }^2\left(\frac{\alpha +\beta }{2}\right)+{\cos }^2\left(\frac{\alpha +\beta }{2}\right)=1$

$⟹\frac{k^2}{2b^2}+\frac{h^2}{2a^2}=1$

Thus,

$⟹\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$