Question

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

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

Answer 1

Answer: (D)

Given ellipses are $x^{2}+4 y^{2}=4$
i.e.,$ \frac{x^{2}}{2^{2}}+\frac{y^{2}}{1^{2}}=1 ...$(1)
and $x^{2}+2 y^{2}=6 $
i.e., $\frac{x^{2}}{(\sqrt{6})^{2}}+\frac{y^{2}}{(\sqrt{3})^{2}}=1 ...$(2)
Let $R(\alpha, \beta)$ be the point of intersection of the tangents to ellipse (2) at $P$ and $Q$. then $P Q$ will be chord of contact of $R$.
$\therefore $ its equation is
$ \frac{\alpha x}{6}+\frac{\beta y}{3}=1 $
i.e., $ \alpha x+2 y \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.