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
$\frac{3\cos \theta }{h}=\frac{2\sin \theta }{k}=\frac{8}{h^2+k^2}$
$\Rightarrow \cos \theta =\frac{8h}{3\left(h^2+k^2\right)};\sin \theta =\frac{8k}{3\left(h^2+k^2\right)}$
$\therefore$ 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, $\frac{1}{e_c^2}+\frac{1}{e_h^2}=1\Rightarrow \frac{1}{e_c^2}=1-\frac{1}{9/8}=1-\frac{8}{9}=\frac{1}{9}$ $\therefore e_c=3$.