Question

If the point of intersection of the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$ be at the extremities of the conjugate diameters of the former, then $\frac{a^2}{{\alpha }^2}+\frac{b^2}{{\beta }^2}=$

Answer 1

Point A and B are extremities of conjugate diameters

Let point B$=(a\cos \theta ,b\sin \theta )$

So point A is $=(a\cos (90+\theta ),b\sin (90+\theta ))$

Since eccentric angles of ends of conjugate diameters differ by 90

Substituting B in $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$

$\frac{a^2{\cos }^2\theta }{{\alpha }^2}+\frac{b^2{\sin }^2\theta }{{\beta }^2}=1$

Substituting A in $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1...\left(1\right)$

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

Adding the above equations

$\frac{a^2({\sin }^2\theta +{\cos }^2\theta )}{{\alpha }^2}+\frac{b^2({\sin }^2\theta +{\cos }^2\theta )}{{\beta }^2}=2$

$\frac{a^2}{{\alpha }^2}+\frac{b^2}{{\beta }^2}=2$