Let the line y=mx and the ellipse 2x^2+y^2=1 intersect at a point P in the first quadrant.

Question

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 $\left(-\dfrac1{3\sqrt2},0\right)$ and $(0,\beta)$, then $\beta$ is equal to:

Accepted Answer

For $2x^2+y^2=1$, the normal at $(x_1,y_1)$ runs along the gradient direction $(2x_1,y_1)$: $\dfrac{x-x_1}{2x_1}=\dfrac{y-y_1}{y_1}$.

$x$-intercept: $x=-x_1=-\dfrac1{3\sqrt2}\Rightarrow x_1=\dfrac1{3\sqrt2}$. $y$-intercept: $\beta=\dfrac{y_1}2$.

From the ellipse: $2x_1^2+y_1^2=1\Rightarrow\dfrac19+y_1^2=1\Rightarrow y_1^2=\dfrac89\Rightarrow y_1=\dfrac{2\sqrt2}3$ (first quadrant).

$$\beta=\frac{y_1}2=\frac{\sqrt2}3.$$

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