Thank you for reporting, we will resolve it shortly
Q.
Area of a triangle formed by tangent and normal to the curve $\frac {x^2}{a^2}+ \frac {y^2}{b^2}=1$ at $P (\frac {a}{\sqrt2}, \frac {b}{\sqrt2})$ 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
$bx+a y=\sqrt{2} a b$...(i)
Equation of normal on the tangent of the curve
$-ax+b y=\frac{\left(b^{2}-a^{2}\right)}{\sqrt{2}}$...(ii)
Equation of $x$ -axis $y=0$...(iii)
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 $=\frac{1}{2}\left|\begin{array}{ccc}\frac{a^{2}-b^{2}}{a \sqrt{2}} & 0 & 1 \\ \sqrt{2} \cdot a & 0 & 1 \\ a / \sqrt{2} & b / \sqrt{2} & 1\end{array}\right|$
$=\frac{-b}{2 \sqrt{2}}\left\{\frac{a^{2}-b^{2}}{a \sqrt{2}}-\sqrt{2} \cdot a\right\}=\frac{-b}{4 a}\left(-a^{2}-b^{2}\right)$
$= \frac{b\left(a^{2}+b^{2}\right)}{4 a}$