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 r

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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 :

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Answer: A. \frac{4 \sqrt{3}}{3} \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} 	ext { 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} P, Q=(0,2 \pm \sqrt{3}) Equation of normal at \mathrm{P}, \mathrm{Q} are \mathrm{y}-2=\sqrt{3}(\mathrm{x}-1) \begin{aligned} & \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\ & \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0ight) \\ & \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0ight) \\ & \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3} \end{aligned}

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