Let the line y=m x and the ellipse
Question:

Let the line $y=m x$ and the ellipse $2 x^{2}+y^{2}=1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse

at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3 \sqrt{2}}, 0\right)$ and $(0, \beta)$, then $\beta$ is equal to:

1. (1) $\frac{2 \sqrt{2}}{3}$

2. (2) $\frac{2}{\sqrt{3}}$

3. (3) $\frac{2}{3}$

4. (4) $\frac{\sqrt{2}}{3}$

Correct Option: , 4

Solution:

Let $P$ be $\left(x_{1}, y_{1}\right)$

So, equation of normal at $P$ is

$\frac{x}{2 x_{1}}-\frac{y}{y_{1}}=-\frac{1}{2}$

It passes through $\left(-\frac{1}{3 \sqrt{2}}, 0\right)$

$\Rightarrow \quad \frac{-1}{6 \sqrt{2} x_{1}}=-\frac{1}{2} \Rightarrow \quad x_{1}=\frac{1}{3 \sqrt{2}}$

So, $y_{1}=\frac{2 \sqrt{2}}{3}$ (as $P$ lies in I $^{\text {st }}$ quadrant)

So, $\beta=\frac{y_{1}}{2}=\frac{\sqrt{2}}{3}$