Area of the triangle formed by the tangent, normal to the curve $x^{2}/a^{2}+y^{2}/b^{2}=1$ at the point $(a/\sqrt{2} , b/\sqrt{2})$ and the $x-$axis is

Solution-

Area of $\Delta=\frac{1}{2}$ Base $\times$ Height $=\frac{1}{2} \times \mid$ Narmal $|\times|$ Targent $\mid$

$\begin{array}{l} \text { Length of Normal }=|y| \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} \\ =\frac{b}{\sqrt{2}} \sqrt{1+\left(\frac{-b x}{a \sqrt{a^{2}-x^{2}}}\right)_{x=a / \sqrt{2}}^{2}} \\ =\frac{b}{\sqrt{2}} \sqrt{\left(1+\frac{b^{2}}{a^{2}}\right)}=\frac{b}{\sqrt{2} a} \sqrt{a^{2}+b^{2}} \text {. } \\ \text { length of Targent }=|y| \sqrt{1+\frac{a^{2}}{b^{2}}}=\frac{b}{\sqrt{2} b} \sqrt{b^{2}+a^{2}}=\frac{\sqrt{a^{2}+b^{2}}}{\sqrt{2}} \\ \text { Area }=\frac{}{}\left({}+\right)^{} \frac{1}{2} \times \frac{b}{\sqrt{2} a} \sqrt{a^{2}+b^{2}} \times \frac{\sqrt{a^{2}+b^{2}}}{\sqrt{2}}=\frac{b}{4 a}\left(a^{2}+b^{2}\right) \text {. } \\ \end{array}$

$C$ is correct.