The area enclosed by the closed curve C given by the differential equation $\frac{dy}{dx} + \frac{x + a}{y - 2} = 0, y (1) = 0 is 4 π .$
Let P and Q be the points of intersection of the curve C and the y-axis. If normals at P and Q on the curve C intersect x-axis at points R and S respectively, then the length of the line segment $R S$ is

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$\frac{4 \sqrt{3}}{3}$

$\frac{2 \sqrt{3}}{3}$

2 $\sqrt{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} (y - 2) dy + (x + a) dx = 0 \\ \frac{y^{2}}{2} - 2 y + \frac{x^{2}}{2} + ax + c = 0; y (1) = 0 \Rightarrow 2 c = - 1 - 2 a \\ (x + a)^{2} + (y - 2)^{2} = {(a + 1)}^{2} + 4; r = 2 \Rightarrow a = - 1 \\ C : {(x - 1)}^{2} + (y - 2)^{2} = 2^{2} \end{array}$
y - axis, x = 0, $P, Q = (0, 2 \pm \sqrt{3})$
Normal P ; $y - 2 = \frac{2 + \sqrt{3} - 2}{- 1} (x - 1)$
Normal Q ; $y - 2 = \frac{2 - \sqrt{3} - 2}{- 1} (x - 1)$
$\begin{array}{l} y - 2 = - \sqrt{3} (x - 1) \Rightarrow x - axis; y = 0 : 1 + \frac{2}{\sqrt{3}} \\ y - 2 = \sqrt{3} (x - 1); y = 0 : x = 1 - \frac{2}{\sqrt{3}} \\ R S = \frac{4}{\sqrt{3}} = \frac{4 \sqrt{3}}{3} \end{array}$