Question

Tangents are drawn from any point on the director circle of ellipse $S_1: x^2+\frac{y^2}{8}=1$ to auxiliary circle of hyperbola $S_2: \frac{x^2}{8}-y^2=1$. The locus of mid-point of the chord of contact is a circle $S$.
The locus of the point $( h , k )$ for which the line $hx + ky =1$ touches the ellipse $S _1$ is

• A. $8 x^2+y^2=1$
• B. $x ^2+8 y ^2=1$
• C. $8 x^2-y^2=1$
• D. $x^2-8 y^2=1$

Answer 1

Answer: (B)

We have,
$x(3 \cos \theta)+y(3 \sin \theta)=8$....(1)
Also, $hx + ky = h ^2+ k ^2$....(2)
As (1) and (2) are same, so
$\frac{3 \cos \theta}{ h }=\frac{2 \sin \theta}{ k }=\frac{8}{ h ^2+ k ^2} $
$\Rightarrow \cos \theta=\frac{8 h }{3\left( h ^2+ k ^2\right)} ; \sin \theta=\frac{8 k }{3\left( h ^2+ k ^2\right)}$
$\therefore$ The locus of midpoint of chord of contact is $S : x ^2+ y ^2=\left(\frac{8}{3}\right)^2$
$h x+k y=1 \Rightarrow y=\left(\frac{-h}{k}\right) x+\frac{1}{k}$
So, using condition of tangency, we get $\frac{1}{ k ^2}=\frac{ h ^2}{ k ^2}+8$
$\Rightarrow 1= h ^2+8 k ^2$
$\therefore$ The locus of $( h , k )$ is $x ^2+8 y ^2=1$