JEE Main 2023 — Area Under Curves Question with Solution

Given:

$\frac{dy}{dx}+\frac{x+a}{y-2}=0$

$\Rightarrow \frac{dy}{dx}=\frac{x+a}{2-y}$

$\Rightarrow \left(2-y\right) dy=\left(x+a\right) dx$

$\Rightarrow 2y-\frac{y^2}{2}=\frac{x^2}{2}+ax+c$

Put $x=1$, then we get $y\left(1\right)=0$.

$a+c=-\frac{1}{2}$

Now,

$4y-y^2=x^2+2ax+2c$

$\Rightarrow x^2+y^2+2ax-4y+2c=0$

$\Rightarrow x^2+y^2+2ax-4y-1-2a=0$

This is a circle having radius

$r=\sqrt{a^2+4+1+2a}=\sqrt{{\left(a+1\right)}^2+4}$

Now,

$\pi r^2=4\pi$

$r^2=4$

$\Rightarrow \sqrt{{\left(a+1\right)}^2+4}=4$

$\Rightarrow {\left(a+1\right)}^2=0$

So, $a=-1$

Hence, circle is

$x^2+y^2-2x-4y+1=0$

$\Rightarrow {\left(x-1\right)}^2+{\left(y-2\right)}^2=4$

It cuts $y-\mathrm{axis}$ at $x=0$, so

$P\equiv \left(0,2+\sqrt{3}\right) \& Q\equiv \left(0,2-\sqrt{3}\right)$

Now,

$x^2+y^2-2x-4y+1=0$

$\Rightarrow 2x+2y\left(\frac{dy}{dx}\right)-2-4\left(\frac{dy}{dx}\right)=0$

$\Rightarrow \left(\frac{dy}{dx}\right)=\frac{1-x}{y-2}$

$\Rightarrow -\left(\frac{dx}{dy}\right)=\frac{y-2}{x-1}$

Slope of normal at $P$ is

$m_1=\frac{2+\sqrt{3}-2}{0-1}=-\sqrt{3}$

Slope of normal at $Q$ is

$m_2=\frac{2-\sqrt{3}-2}{0-1}=\sqrt{3}$

Equation of normal at $Q$ is

$y-\left(2-\sqrt{3}\right)=\sqrt{3}\left(x-0\right)$

$\Rightarrow y-\left(2-\sqrt{3}\right)=\sqrt{3}x$

Equation of normal at $P$ is

$y-2-\sqrt{3}=-\sqrt{3}\left(x-0\right)$

$\Rightarrow y=-\sqrt{3}x+2+\sqrt{3}$

So,

$R\equiv \left(1+\frac{2}{\sqrt{3}},0\right)$ and $S\equiv \left(1-\frac{2}{\sqrt{3}},0\right)$

Hence,

$RS=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$ units