Question

Let from a point $A\left(h,k\right)$ chords of contact are drawn to the ellipse $x^2+2y^2=6$ where all these chords touch the ellipse $x^2+4y^2=4$ . Then, the perimeter (in units) of the locus of point $A$ is

• A. $2\pi$
• B. $3\pi$
• C. $4\pi$
• D. $6\pi$

Answer 1

Answer: (D)

Let the chord of contact is $PQ$ which touches $x^2+4y^2=4$ at $R$
Now, assume $R=\left(2cos\theta ,2sin\theta \right)$
Equation of the chord of contact $PQ$ is $hx+2yk=6\dots ..\left(i\right)$
Again, the equation of the tangent $PQ$ is $\frac{xcos\theta }{2}+\frac{ysin\theta }{1}=1\dots ..\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$ , we get,
$\frac{h2}{cos\theta }=\frac{2k}{sin\theta }=6$
$cos\theta =\frac{h}{3},sin\theta =\frac{k}{3}\Rightarrow x^2+y^2=9$
Hence, the perimeter of the circle is $6\pi$ units