$$\begin{gathered} \text{ Let the line }y=mx\text{ and the ellipse }2x^2+y^2=1\text{ intersect at a point }\mathrm{P}\text{ in the first quadrant. } \\ \text{the area of the traingle formed by the normal to this ellipse at P with axes is } \\ \left(Normal is pas\sin g through the point \left(-\frac{1}{3\sqrt{2}},0\right)\right) \end{gathered}$$

$\begin{array}{l}\text{ Let }\mathrm{P}\text{ be }\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)\\ \text{ Equation of normal at }\mathrm{P}\text{ is }\frac{x}{2x_1}-\frac{y}{y_1}=-\frac{1}{2}\end{array}$

$\begin{array}{l}\text{ It passes through }\left(-\frac{1}{3\sqrt{2}},0\right)\Rightarrow \frac{-1}{6\sqrt{2}{x}_1}=-\frac{1}{2}\Rightarrow x_1=\frac{1}{3\sqrt{2}}\\ \text{ So }\left.y_1=\frac{2\sqrt{2}}{3}\text{ (as }\mathrm{P}\text{ lies in }{1}^{\text{st }}\text{ quadrant }\right)\\ \text{ The area of the triangle formed by this normal with axes is }\frac{1}{2}\left(\frac{\sqrt{2}}{3}\right)\left(\frac{1}{3\sqrt{2}}\right)\\ Therefore the area of the triangle is \frac{1}{18}square units\end{array}$