Question

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 $4\pi$.
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 $RS$ is

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

Answer 1

Answer: (A)

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