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

We have the ellipse $$2x^{2}+y^{2}=1$$ and the straight line $$y=mx$$. Putting $$y=mx$$ in the equation of the ellipse gives the co-ordinates of their point of intersection P:

$$2x^{2}+(mx)^{2}=1 \quad\Longrightarrow\quad (2+m^{2})x^{2}=1 \;\Longrightarrow\;x=\dfrac{1}{\sqrt{\,2+m^{2}}}.$$

Because the point lies in the first quadrant, $$x>0,\;y>0,$$ so

$$P\left(\dfrac{1}{\sqrt{\,2+m^{2}}},\;\dfrac{m}{\sqrt{\,2+m^{2}}}\right).$$

The slope of the tangent to the ellipse is obtained by implicit differentiation. Differentiating $$2x^{2}+y^{2}=1$$ gives

$$4x+2y\dfrac{dy}{dx}=0 \quad\Longrightarrow\quad \dfrac{dy}{dx}=-\dfrac{2x}{y}.$$

Therefore, at P

$$\text{slope of tangent}=m_{\text{tan}}=-\dfrac{2x_{P}}{y_{P}} =-\dfrac{2}{m}.$$

Since the normal is perpendicular to the tangent,

$$m_{\text{tan}}\;m_{\text{norm}}=-1 \;\Longrightarrow\; \left(-\dfrac{2}{m}\right)m_{\text{norm}}=-1 \;\Longrightarrow\; m_{\text{norm}}=\dfrac{m}{2}.$$

The normal at P is therefore

$$y-\dfrac{m}{\sqrt{\,2+m^{2}}} =\dfrac{m}{2}\!\left(x-\dfrac{1}{\sqrt{\,2+m^{2}}}\right).$$

This normal meets the x-axis at $$\!\left(-\dfrac{1}{3\sqrt{2}},\,0\right)\!.$$ Putting $$y=0$$ in the equation of the normal and letting $$x=x_{i}$$ gives

$$0-\dfrac{m}{\sqrt{\,2+m^{2}}} =\dfrac{m}{2}\!\left(x_{i}-\dfrac{1}{\sqrt{\,2+m^{2}}}\right).$$

Dividing by $$m\;(m\neq0)$$ and multiplying by 2, we get

$$-\dfrac{2}{\sqrt{\,2+m^{2}}}=x_{i}-\dfrac{1}{\sqrt{\,2+m^{2}}} \;\Longrightarrow\; x_{i}=-\dfrac{1}{\sqrt{\,2+m^{2}}}.$$

But the given x-intercept is $$-\dfrac{1}{3\sqrt{2}},$$ so

$$-\dfrac{1}{\sqrt{\,2+m^{2}}}=-\dfrac{1}{3\sqrt{2}} \;\Longrightarrow\; \sqrt{\,2+m^{2}}=3\sqrt{2} \;\Longrightarrow\; 2+m^{2}=18 \;\Longrightarrow\; m^{2}=16 \;\Longrightarrow\; m=4\;(\text{positive for first quadrant}).$$

Now we find the y-intercept. Put $$x=0$$ in the normal:

$$y-\dfrac{m}{\sqrt{\,2+m^{2}}} =\dfrac{m}{2}\!\left(0-\dfrac{1}{\sqrt{\,2+m^{2}}}\right) =-\dfrac{m}{2}\cdot\dfrac{1}{\sqrt{\,2+m^{2}}}.$$

Hence

$$y=\dfrac{m}{\sqrt{\,2+m^{2}}}-\dfrac{m}{2}\cdot\dfrac{1}{\sqrt{\,2+m^{2}}} =\dfrac{m}{\sqrt{\,2+m^{2}}}\left(1-\dfrac12\right) =\dfrac{m}{2\sqrt{\,2+m^{2}}}.$$

Substituting $$m=4$$ and $$\sqrt{\,2+m^{2}}=\sqrt{18}=3\sqrt{2},$$ we obtain

$$\beta=\dfrac{4}{2\cdot3\sqrt{2}} =\dfrac{2}{3\sqrt{2}} =\dfrac{\sqrt{2}}{3}.$$

Hence, the correct answer is Option D.