The area enclosed by the closed curve $\mathrm{C}$ given by the differential equation

$\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.

Let $P$ and $Q$ be the points of intersection of the curve $\mathrm{C}$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $\mathrm{C}$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is :

$$ \begin{aligned} & \frac{d y}{d x}+\frac{x+a}{y-2}=0 \\\\ & \frac{d y}{d x}=\frac{x+a}{2-y} \\\\ & (2-y) d y=(x+a) d x \\\\ & 2 y \frac{-y}{2}=\frac{x^2}{2}+\mathrm{ax}+\mathrm{c} \\\\ & \mathrm{a}+\mathrm{c}=-\frac{1}{2} \text { as } \mathrm{y}(1)=0 \\\\ & \mathrm{X}^2+\mathrm{y}^2+2 \mathrm{ax}-4 \mathrm{y}-1-2 \mathrm{a}=0 \\\\ & \pi \mathrm{r}^2=4 \pi \\\\ & \mathrm{r}^2=4 \\\\ & 4=\sqrt{a^2+4+1+2 a} \\\\ & (\mathrm{a}+1)^2=0 \end{aligned} $$ <br/><br/>$P, Q=(0,2 \pm \sqrt{3})$ <br/><br/>Equation of normal at $\mathrm{P}, \mathrm{Q}$ are $\mathrm{y}-2=\sqrt{3}(\mathrm{x}-1)$ <br/><br/>$$ \begin{aligned} & \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\\\ & \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3} \end{aligned} $$