Question

The locus of the point of intersection of the tangents at the extremities of the chord of the ellipse $x^2+2y^2=6$ which also touches the ellipse $x^2+4y^2=4,$ is

• A. $x^2+y^2=4$
• B. $x^2+y^2=6$
• C. $x^2+y^2=9$
• D. $x^2+y^2=12$

Answer 1

The correct option is C $x^2+y^2=9$

Given,
$S:x^2+2y^2=6\Rightarrow \frac{x^2}{6}+\frac{y^2}{3}=1$
$S^{\prime }:x^2+4y^2=4\Rightarrow \frac{x^2}{4}+\frac{y^2}{1}=1$


Equation of chord of contact $AB$ with respect to point $P(h,k)$ is $T=0$
$\Rightarrow \frac{hx}{6}+\frac{ky}{3}=1\dots \left(1\right)$

Equation of tangent to $S^{\prime }=0$
$\Rightarrow \frac{x\cos \theta }{2}+\frac{y\sin \theta }{1}=1\dots \left(2\right)$

As both $\left(1\right)$ and $\left(2\right)$ represent the same line $AB.$
$\therefore$ On comparing, we get
$\Rightarrow \frac{\frac{h}{6}}{\frac{\cos \theta }{2}}=\frac{\frac{k}{3}}{\frac{\sin \theta }{1}}=1$
$\Rightarrow \cos \theta =\frac{h}{3},\sin \theta =\frac{k}{3}$

We know that,
${\sin }^2\theta +{\cos }^2\theta =1$
$\Rightarrow {\left(\frac{h}{3}\right)}^2+{\left(\frac{k}{3}\right)}^2=1$
Hence, required locus is
$\Rightarrow x^2+y^2=9$