Question

If a variable tangent of the circle $x^2+y^2=1$ intersects the ellipse $x^2+2y^2=4$ at points $P$ and $Q,$ then the locus of the point of intersection of tangent at $P$ and $Q$ is

Answer 1

Let the intersection of the tangent at $P$ and $Q$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$ be $(x_1,y_1)$
Then the equation of $PQ$ is $T=0$
$\frac{x{x}_1}{4}+\frac{y{y}_1}{2}=1$
i.e., $y=-\frac{x{x}_1}{2y_1}+\frac{2}{y_1}$
This is a tangent to the circle $x^2+y^2=1$
So, $c^2=a^2(1+m^2)$
$\Rightarrow \frac{4}{y_1^2}=1(1+\frac{x_1^2}{4y_1^2})$
$\Rightarrow 16=4y_1^2+x_1^2$
$\Rightarrow \frac{x_1^2}{16}+\frac{y_1^2}{4}=1$
which is the equation of an ellipse.
Length of latus rectum $=\frac{2b^2}{a}=\frac{8}{4}=2$