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

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