Question

A tangent to the ellipse $x^2+4y^2=4$ meets the ellipse $x^2+2y^2=6$ at $P$ and $Q$. The angle between the tangents at $P$ and $Q$ of the ellipse $x^2+2y^2=6$ is

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

Answer 1

The correct option is D $\frac{\pi }{2}$

Given ellipse are $x^2+4y^2=4\Rightarrow \frac{x^2}{2^2}+\frac{y^2}{1^2}=1$ ...(1)

and $x^2+2y^2=6\Rightarrow \frac{x^2}{{\left(\sqrt{6}\right)}^2}+\frac{y^2}{{\left(\sqrt{3}\right)}^2}=1$ ....(2)

Let $R(\alpha ,\beta )$ be the point of intersection of tangents to ellipse (2) at $P$ and $Q$, then $PQ$ will be chord of contacdt of $R$.

$\therefore$ its equation is

$\frac{\alpha x}{6}+\frac{\beta y}{3}=1\Rightarrow \alpha x+2y\beta =6\Rightarrow y=-\frac{\alpha }{2\beta }x+\frac{3}{\beta }$ ...(3)

Since (3) touches (1)

$\therefore {\left(\frac{3}{\beta }\right)}^2=2^2.\frac{{\alpha }^2}{4{\beta }^2}+1^2[c^2=a^2{m}^2+b^2]$

$\Rightarrow \frac{9}{{\beta }^2}=\frac{{\alpha }^2}{{\beta }^2}+1=\frac{{\alpha }^2+{\beta }^2}{{\beta }^2}\Rightarrow {\alpha }^2+{\beta }^2=9$

$\therefore$ Locus of $(\alpha ,\beta )$ is

$x^2+y^2=9={\left(\sqrt{6}\right)}^2+{\left(\sqrt{3}\right)}^2$

i.e. director circle

$\therefore$ tangent at $P,Q$ meets at right angles.