Solve this following


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

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

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

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

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

Correct Option: , 4


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