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 $R=\left(2 cos \theta , 2 sin ⁡ \theta \right)$
Equation of the chord of contact $PQ$ is $hx+2yk=6\ldots ..\left(i\right)$
Again, the equation of the tangent $PQ$ is $\frac{x cos \theta }{2}+\frac{y sin ⁡ \theta }{1}=1\ldots ..\left(i i\right)$
From $\left(i\right)$ and $\left(i i\right)$ , we get,
$\frac{h 2}{cos \theta }=\frac{2 k}{s i n \theta }=6$
$cos \theta =\frac{h}{3},sin⁡\theta =\frac{k}{3}\Rightarrow x^{2}+y^{2}=9$
Hence, the perimeter of the circle is $6\pi $ units