Question

Consider two concentric circles $C_1:x^2+y^2-4=0$ and $C_2:x^2+y^2-9=0.$ A parabola is drawn through the points where $C_1$ meets $y-$axis and having an arbitrary tangent of $C_2$ as its directrix. If $C$ is the curve of locus of focus of drawn parabola and $e$ is the eccentricity of the curve $C,$ then the value of $12e$ is

Answer 1

Let $P(h,k)$ be the focus of parabola.
Circle $C_1$ meets $y-$axis at $(0,\pm 2)$
An arbitrary tangent to $C_2$ is $x\cos \theta +y\sin \theta =3$
Parabola passes through $(0,\pm 2)$ and its directrix is
$x\cos \theta +y\sin \theta -3=0$

We know that in a parabola, distance between a point and the directrix and the distance between focus and the point are equal.
$\Rightarrow h^2+(k-2{)}^2=(2\sin \theta -3{)}^2\cdots \left(1\right)$
and $h^2+(k+2{)}^2=(2\sin \theta +3{)}^2\cdots \left(2\right)$
From equations $\left(1\right)$ and $\left(2\right),$
$\frac{k}{3}=\sin \theta ,\frac{h}{\sqrt{5}}=\cos \theta$
Since ${\sin }^2\theta +{\cos }^2\theta =1$
$\frac{k^2}{(3{)}^2}+\frac{h^2}{(\sqrt{5}{)}^2}=1$
$\therefore$ Required locus is $\frac{x^2}{5}+\frac{y^2}{9}=1$
Eccentricity, $e=\sqrt{1-\frac{5}{9}}$
$\Rightarrow e=\frac{2}{3}$
$\Rightarrow 12e=8$