Find the area of the triangle formed by the x-axis and the tangent and the normal to the curve $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ at the point $\left(\frac{\mathrm{a}}{\sqrt{2}},\frac{\mathrm{b}}{\sqrt{2}}\right)$

$\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$
$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-{\mathrm{b}}^2}{{\mathrm{a}}^2} \frac{\mathrm{x}}{\mathrm{y}} \\ m={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(\frac{\mathrm{a}}{\sqrt{2}}, \frac{\mathrm{b}}{\left(\sqrt{2}\right)}\right)} =-\frac{b}{a} \end{gathered}$$

eq of Tangent $\mathrm{bx}+\mathrm{ay}=\sqrt{2} \mathrm{ab}$

eq of normal   $\mathrm{ax}-\mathrm{by}=\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{\sqrt{2}}$

$\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right), \left(\sqrt{2}a, 0\right), \left(\frac{a^2-b^2}{{\displaystyle a}\sqrt{2}}, 0\right)$   

$$\begin{gathered} area =\frac{1}{2}\left|\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{\sqrt{2}\mathrm{a}}-\sqrt{2} \mathrm{a}\right| \\ =\frac{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right) \mathrm{b}}{4\mathrm{a}} \end{gathered}$$