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)

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

Answer 2

dxdy​+y−2x+a​=0
dxdy​=2−yx+a​
(2−y)dy=(x+a)dx
2y2−y​=2x2​+ax+c
a+c=−21​ as y(1)=0
X2+y2+2ax−4y−1−2a=0
πr2=4π
r2=4
4=a2+4+1+2a​
(a+1)2=0
P,Q=(0,2±3​)
Equation of normal at P,Q are y−2=3​(x−1)
y−2=−3​(x−1)
R=(1−3​2​,0)
S=(1+3​2​,0)
RS=3​4​=433​​