Question

Let the line $y=mx$ and the ellipse $2x^2+y^2=1$ intersect at point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $(-\frac{1}{3\sqrt{2}},0)$ and $(0,\beta )$, then $\beta$ is equal to:

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{2}{3}$
• C. $\frac{2\sqrt{2}}{3}$
• D. $\frac{\sqrt{2}}{3}$

Answer 1

The correct option is D $\frac{\sqrt{2}}{3}$


Let $P\equiv (x_1,y_1)$
Given equation of ellipse is $2x^2+y^2=1$
$\Rightarrow 4x+2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}{|}_{(x_1,y_1)}=-\frac{2x_1}{y_1}$

Therefore, slope of normal at $P(x_1,y_1)$ is $\frac{y_1}{2x_1}$
Equation of normal at $P(x_1,y_1)$ is
$y-y_1=\frac{y_1}{2x_1}(x-x_1)$
It passes through $(-\frac{1}{3\sqrt{2}},0)$
$\Rightarrow -y_1=\frac{y_1}{2x_1}(-\frac{1}{3\sqrt{2}}-x_1)$
$\Rightarrow x_1=\frac{1}{3\sqrt{2}}$
$\Rightarrow y_1=\frac{2\sqrt{2}}{3}$ as $P$ lies in first quadrant.

Since $(0,\beta )$ lies on the normal of the ellipse at point $P$, hence we get
$\beta -y_1=\frac{-y_1}{2}$
$\Rightarrow \beta =\frac{y_1}{2}=\frac{\sqrt{2}}{3}$