Show that the tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at points whose eccentric angles differ by $90^\circ$ meet on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$
My attempt: Let the eccentric angles be $\phi$ and $\theta$
$\therefore$ $\phi -\theta=\pi/2$
I tried to combine the two equations but nothing came up I keep revolving in circles. Any hint will be greatly appreciated. Thank you.
Eq. of tangent at t point $(a\cos t, b \sin t)$ on the ellipse $x^2/a^2+y^2/b^2=1$ is $x c/a+y s/b=1 ~~~~(1)$ and at eccentic angle $t+\pi/2$ is $-x s/a+yc/b=1~~~~(2)$ Here $c$ is short for \cos t and $s$ is short for $\sin t$. The point of intersection of (1) and (2) comes out to be $x/a=(c-s),y/b=(c+s)].$ Squaring and adding we get the required locus as $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=[(c-s)^2+(c+s)^2]=2$$