Question

A tangent to $E_1:x^2+4y^2=4$ meets $E_2:x^2+2y^2=6$ at $P$ and $Q$. Tangents at $P$ and $Q$ of $E_2$ make an angle $\frac{\pi }{n}(n\in N)$. Then $n^2=$

Answer 1

Let any point on $\frac{x^2}{4}+\frac{y^2}{1}=1$ be $(2\cos \theta ,\sin \theta )$.
tangent at that point is $\frac{x\cos \theta }{2}+y\sin \theta =1\dots \left(1\right)$

Let point of intersection of tangents at $P$
and $Q$ be $R(h,k)$
Chord of contact for $E_2:\frac{hx}{6}+\frac{ky}{3}=1\dots \left(2\right)$
$\left(1\right)$ and $\left(2\right)$ are indentical
$\Rightarrow \frac{\cos \theta }{2\cdot \frac{h}{6}}=\frac{\sin \theta }{\frac{k}{3}}=1$
$\cos \theta =\frac{h}{3},\sin \theta =\frac{k}{3}$
$\Rightarrow h^2+k^2=9$
$x^2+y^2=6+3$
Clearly, locus of point of intersection of tangents lie on director circle of $E_2$
$\therefore$ Angle between tangents $=\frac{\pi }{2}$
$\therefore n^2=2^2=4$