Question

A variable tangent to the circle $x^2+y^2=1$ intersects the ellipse $2x^2+y^2=4$ at $P$ and $Q$. The locus of the point of intersection of the tangents at $P$ and $Q$ is

• A. an ellipse of eccentricity $\frac{\sqrt{3}}{4}$
• B. a parabola with focus at $(2,3)$
• C. an ellipse of area $8\pi$ sq. units
• D. a parabola with latus rectrum $12$

Answer 1

Answer: (C)

an ellipse of area $8\pi$ sq. units

R.E.F image

Let $y=mx+c$ is tangent to $x^2+y^2=1$

$\therefore \frac{C}{\sqrt{m^2+1}}=1$

$\therefore C=\sqrt{m^2+1}$

$\therefore y=mx+\sqrt{m^2+1}$ ...(1)

Let $P(2\cos \alpha ,\sqrt{2}\sin \alpha ),Q(2\cos \beta ,\sqrt{2}\sin \beta )$

Since, $p$ lies on (1)

$\sqrt{2}\sin \alpha =m\times 2\cos \alpha +\sqrt{4{\cos }^2\alpha +1}$

$PQ$ as a

$T=0$

$\frac{y\times h}{4}+\frac{x\times k}{2}=1$

$\frac{kx}{2}-\frac{hy}{4}-120$

$mx-y+\sqrt{m^2+1}=0$

$\frac{k}{2m}=\frac{h}{4}=\frac{-1}{\sqrt{m^2+1}}$

$\therefore h=\frac{4}{\sqrt{m^2+1}},k=\frac{2m}{\sqrt{m^2+1}}$

${\left(\frac{h}{4}\right)}^2+{\left(\frac{k}{2}\right)}^2=\frac{1}{m^2+1}+\frac{m^2}{m^2+1}$

$\therefore \frac{h^2}{16}+\frac{k^2}{4}=1$

$\frac{x^2}{16}+\frac{y^2}{4}=1$

It represent ellipse $A=ab\pi =4\times 2\times \pi$

$\therefore A=8\pi$