The area (in sq. units) of triangle formed by the tangent and normal at a point a 2 , b 2 to the curve x 2 a…

We need to find the area of $∆PMN$

$ar\left(∆PMN\right)=\frac{1}{2}\times \left|\text{Length of normal}\right|\times \left|\text{Length of tangent}\right|$

Now,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{dy}{dx}=\frac{-bx}{a\sqrt{a^2-x^2}}$

Now, length of normal is given by

$=\left|y\right|\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$

$=\left|\frac{b}{\sqrt{2}}\right|\sqrt{1+{\left(\frac{-bx}{a\sqrt{a^2-x^2}}\right)}^2}$

At $x=\frac{a}{\sqrt{2}}$

$=\frac{b}{\sqrt{2}a}\sqrt{a^2+b^2}$

Now, length of tangent is given by

$\left|y\right|\sqrt{1+{\left(\frac{dx}{dy}\right)}^2}$

$=\sqrt{\frac{a^2+b^2}{2}}$

Therefore, area of triangle is 

$=\frac{1}{2}\left(\sqrt{\frac{a^2+b^2}{2}}\right)\left(\frac{b}{\sqrt{2}a}\sqrt{a^2+b^2}\right)$

$=\frac{b}{4a}\left(a^2+b^2\right)$