Find the condition so that the line $px +qy=r$ intersects the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in points whose eccentric angles differ by $\frac{\pi}4$.
Though I know how to solve it using parametric coordinates, I was wondering if there's an another approach which is less time consuming.
Let $P_1 = (a \cos t , b \sin t) $ and $P_2 = (a \cos(t + \phi), b \sin (t + \phi))$ where $\phi = \dfrac{\pi}{4} $, then we need to satisfy two conditions:
$$ p a \cos t + q b \sin t = r \hspace{20pt}(1) $$
$$ p a \cos(t + \phi) + q b \sin(t + \phi) = r \hspace{20pt} (2) $$
This last equation becomes
$$ \cos t ( p a \cos \phi + q b \sin \phi ) + \sin t ( - p a \sin \phi + q b \cos \phi ) = r \hspace{20pt} (3)$$
And after substituting $ \cos \phi = \sin \phi = \dfrac{1}{\sqrt{2}} $, it becomes,
$$ \cos t ( p a + q b ) + \sin t ( q b - p a ) = \sqrt{2} r \hspace{20pt} (4)$$
Now solving $(1)$ and $(4)$ for $\cos t$ and $\sin t $, we find that
$ \cos t = -r \dfrac{ (q b - p a) - \sqrt{2} q b }{ p^2 a^2 + q^2 b^2 } $
$\sin t = -r \dfrac{ \sqrt{2} p a - (p a + q b) }{ p^2 a^2 + q^2 b^2 }$
Since $\cos^2 t + \sin^2 t = 1$, then the condition is
$ \dfrac{ ( p^2 a^2 + q^2 b^2 )^2 }{ r^2 } = (q b - p a)^2 + (pa + q b)^2 + 2 ( q^2 b^2 + p^2 a^2 ) - 2 \sqrt{2} ( q b (q b - p a) + p a ( pa + q b) ) $
Simplifying the right hand side,
$ \dfrac{ ( p^2 a^2 + q^2 b^2 )^2 }{ r^2 } = 4 (p^2 a^2 + q^2 b^2) - 2 \sqrt{2} ( q^2 b^2 + p^2 a^2 ) $
Dividing through by $(p^2 a^2 + q^2 b^2)$, yields,
$ \dfrac{p^2 a^2 + q^2 b^2}{r^2} = 4 - 2 \sqrt{2} $
And this is the condition on the line.