Let the line y=mx and the ellipse 2x^{2}+y^{2 | vedclass

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(D) The equation of the ellipse is $2x^{2}+y^{2}=1$,which can be written as $\frac{x^{2}}{(1/\sqrt{2})^{2}} + \frac{y^{2}}{1^{2}} = 1$. Let the point

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

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(D) The equation of the ellipse is $2x^{2}+y^{2}=1$,which can be written as $\frac{x^{2}}{(1/\sqrt{2})^{2}} + \frac{y^{2}}{1^{2}} = 1$. Let the point $P$ be $(\frac{1}{\sqrt{2}}\cos	heta, \sin	heta)$. The equation of the normal at $P(\frac{1}{\sqrt{2}}\cos	heta, \sin	heta)$ is given by $\frac{ax}{\cos	heta} - \frac{by}{\sin	heta} = a^{2}-b^{2}$,where $a=\frac{1}{\sqrt{2}}$ and $b=1$. Substituting the values,we get $\frac{x}{\sqrt{2}\cos	heta} - \frac{y}{\sin	heta} = \frac{1}{2} - 1 = -\frac{1}{2}$. This simplifies to $\frac{x}{(-\frac{1}{\sqrt{2}}\cos	heta)} + \frac{y}{(\frac{1}{2}\sin	heta)} = 1$. Comparing this with the given intercepts $(-\frac{1}{3\sqrt{2}}, 0)$ and $(0, \beta)$,we have $-\frac{1}{\sqrt{2}}\cos	heta = -\frac{1}{3\sqrt{2}} \Rightarrow \cos	heta = \frac{1}{3}$. Since $\sin^{2}	heta = 1 - \cos^{2}	heta = 1 - \frac{1}{9} = \frac{8}{9}$,we have $\sin	heta = \frac{2\sqrt{2}}{3}$ (as $P$ is in the first quadrant). Thus,$\beta = \frac{1}{2}\sin	heta = \frac{1}{2} 	imes \frac{2\sqrt{2}}{3} = \frac{\sqrt{2}}{3}$.

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

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