Question

The area enclosed by the closed curve $C$ 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 $C$ 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

• A. $\frac{4 \sqrt{3}}{3}$
• B. $\frac{2 \sqrt{3}}{3}$
• C. 2
• D. $\quad 2 \sqrt{3}$

Answer 1

Answer: (A)

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