Let line y=m x and ellipse 2 x^{2}+y^{2}=1 intersect at a ponit mathrm{P} in first quadrant. If norm

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10-2. Parabola, Ellipse, Hyperbola

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

A

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

B

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

C

$\frac{2 }{3}$

D

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

(JEE MAIN-2020)

Solution

Any normal to the ellipse is

$\frac{\mathrm{x} \sec \theta}{\sqrt{2}}-\mathrm{y} \cos \mathrm{ec} \theta=-\frac{1}{2}$

$\Rightarrow \frac{\mathrm{x}}{\left(\frac{-\cos \theta}{\sqrt{2}}\right)}+\frac{\mathrm{y}}{\left(\frac{\sin \theta}{2}\right)}=1$

$\Rightarrow \frac{\cos \theta}{\sqrt{2}}=\frac{1}{3 \sqrt{2}}$ and $\frac{\sin \theta}{2}=\beta$

$\Rightarrow \beta=\frac{\sqrt{2}}{3}$

Standard 11

Mathematics

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