### Area Under Curves Question 12

#### Question 12 - 01 February - Shift 1

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 $RS$ is

(1) $2 \sqrt{3}$

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

(3) 2

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

## Show Answer

#### Answer: (4)

#### Solution:

#### Formula: Method of Separable variable, Equation of Normal

$\frac{d y}{d x}+\frac{x+a}{y-2}=0$

$\frac{d y}{d x}=\frac{x+a}{2-y}$

$(2-y) d y=(x+a) d x$

$2 y - \frac{y^2}{2}=\frac{x^{2}}{2}+ax+c$

$a+c=-\frac{1}{2}$ as $y(1)=0$

$x^{2}+y^{2}+2 ax-4 y-1-2 a=0$

$\pi r^{2}=4 \pi$

$r^{2}=4$

$4=\sqrt{a^{2}+4+1+2 a}$

$(a+1)^{2}=0$

$P, Q=(0,2 \pm \sqrt{3})$

Equation of normal at $P, Q$ are $y-2=\sqrt{3}(x-1)$

$y-2=-\sqrt{3}(x-1)$

$R=(1-\frac{2}{\sqrt{3}}, 0)$

$S=(1+\frac{2}{\sqrt{3}}, 0)$

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