JEE Math Practice Question
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 :
- A.$$\frac{4 \sqrt{3}}{3}$$
- B.$$2 \sqrt{3}$$
- C.2
- D.$$\frac{2 \sqrt{3}}{3}$$
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