A normal inclined at an angle of $45°$ to $x$-axis of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is drawn. It meets the major and minor axes in $P$ and $Q$ respectively. If $C$ is the centre of the ellipse, then area of $∆CPQ$ is

 Let $R(a\cos \varphi ,b\sin \varphi )$ be any point on the ellipse, then equation of normal at $R$ is$ax\sec \varphi -bycosec\varphi =a^2-b^2$

Putting $x=0 , y=0$ respectively in the equation we get points as  $P\left(\frac{\left(a^2-b^2\right)}{a}\cos \varphi ,0\right)$ and $Q\left(0,-\frac{\left(a^2-b^2\right)}{b}\sin \varphi \right)$ are respectively

$\Rightarrow CP=\left(\frac{a^2-b^2}{a}\right)\left|\cos \varphi \right|$ and $CQ=\left(\frac{a^2-b^2}{b}\right)\left|\sin \varphi \right|$

Since Area of $∆CPQ=\frac{1}{2}\times CP\times CQ$$=\frac{{\left(a^2-b^2\right)}^2\left|\sin \varphi \cos \varphi \right|}{2ab}$

As we know that $\sin 2\varphi =2\sin \varphi \cos \varphi$

But slope of normal $=\frac{a}{b}\tan \varphi =\tan 45°$ (given)

As we know that $\sin 2\varphi =\frac{2\tan \varphi }{1+{\tan }^2\varphi }$

$\Rightarrow$Area of $\Delta CPQ$$=\frac{{\left(a^2-b^2\right)}^2\left|\frac{\sin 2\varphi }{2}\right|}{2ab}$$=\frac{{\left(a^2-b^2\right)}^2\frac{ab}{\left(a^2+b^2\right)}}{2ab}$

Thus, area of triangle $APB$ is $\frac{{\left(a^2-b^2\right)}^2}{2\left(a^2+b^2\right)}\text{ sq units. }$

Hence option-$1$ is the correct answer.