Area of a triangle formed by tangent and normal to the curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at \(p\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\) with the \(x\)-axis is

Given curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and point \(P\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\)
Slope of tangent at \(P\) is \(-b / a\).
Slope of normal at \(P\) is \(a / b\).
Equation of tangent on the curve
\[
b x+a y=\sqrt{2} a b
\]
Equation of normal on the tangent of the curve
\[
-a x+b y=\frac{\left(b^{2}-a^{2}\right)}{\sqrt{2}}
\]
Equation of \(x\)-axis \(y=0\)
Solving these equations in successive manner, we get three points of triangle
\(\left(\frac{a^{2}-b^{2}}{a \sqrt{2}}, 0\right),(\sqrt{2} \cdot a, 0)\) and \(\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\)
Now, area of triangle \(=\frac{1}{2}\left|\begin{array}{ccc}\frac{a^{2}-b^{2}}{a \sqrt{2}} & 0 & 1 \\ a / \sqrt{2} \cdot a & 0 / \sqrt{2} & 1 \\ 1\end{array}\right|\)
\(=\frac{-b}{2 \sqrt{2}}\left\{\frac{a^{2}-b^{2}}{a \sqrt{2}}-\sqrt{2} \cdot a\right\}=\frac{-b}{4 a}\left(-a^{2}-b^{2}\right)\)
\(=\frac{b\left(a^{2}+b^{2}\right)}{4 a}\)