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

Answer: (D)

Given ellipses are $x^2+4y^2=4$
i.e.,$\frac{x^2}{2^2}+\frac{y^2}{1^2}=\mathrm{1...}$(1)
and $x^2+2y^2=6$
i.e., $\frac{x^2}{(\sqrt{6}{)}^2}+\frac{y^2}{(\sqrt{3}{)}^2}=\mathrm{1...}$(2)
Let $R(\alpha ,\beta )$ be the point of intersection of the tangents to ellipse (2) at $P$ and $Q$. then $PQ$ will be chord of contact of $R$.
$\therefore$ its equation is
$\frac{\alpha x}{6}+\frac{\beta y}{3}=1$
i.e., $\alpha x+2y\beta =6$
or $y=-\frac{\alpha }{2\beta }x+\frac{3}{\beta }...$(3)
Since (3) touches (1)
$\therefore {\left(\frac{3}{\beta }\right)}^2=2^2\cdot \frac{{\alpha }^2}{4{\beta }^2}+1^2\left(c^2=a^2{m}^2+b^2\right)$
$\Rightarrow \frac{9}{{\beta }^2}=\frac{{\alpha }^2}{{\beta }^2}+1=\frac{{\alpha }^2+{\beta }^2}{{\beta }^2}$
$\Rightarrow {\alpha }^2+{\beta }^2=9$
$\therefore$ locus of $(\alpha ,\beta )$ is
$x^2+y^2=9=(\sqrt{6}{)}^2+(\sqrt{3}{)}^2$
i.e., director circle.
$\therefore$ tangent at $P,Q$ meet at right angles.