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 eccentricity of the conjugate hyperbola of hyperbola $S _2$, is

• A. $\frac{4}{3}$
• B. $\frac{5}{3}$
• C. $\frac{8}{3}$
• D. 3

Answer 1

Answer: (D)

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 $S : x ^2+ y ^2=\left(\frac{8}{3}\right)^2$
We have $e _{ h }^2=1+\frac{1}{8}=\frac{9}{8}$
Using, $\frac{1}{ e _{ c }^2}+\frac{1}{ e _{ h }^2}=1 \Rightarrow \frac{1}{ e _{ c }^2}=1-\frac{1}{9 / 8}=1-\frac{8}{9}=\frac{1}{9}$ $\therefore e _{ c }=3$.