JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 11)
The area enclosed by the closed curve $$\mathrm{C}$$ 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 $$\mathrm{C}$$ and the $$y$$-axis. If normals at $$P$$ and $$Q$$ on the curve $$\mathrm{C}$$ intersect $$x$$-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$R S$$ is :
$$\frac{4 \sqrt{3}}{3}$$
$$2 \sqrt{3}$$
2
$$\frac{2 \sqrt{3}}{3}$$
Explanation
$$
\begin{aligned}
& \frac{d y}{d x}+\frac{x+a}{y-2}=0 \\\\
& \frac{d y}{d x}=\frac{x+a}{2-y} \\\\
& (2-y) d y=(x+a) d x \\\\
& 2 y \frac{-y}{2}=\frac{x^2}{2}+\mathrm{ax}+\mathrm{c} \\\\
& \mathrm{a}+\mathrm{c}=-\frac{1}{2} \text { as } \mathrm{y}(1)=0 \\\\
& \mathrm{X}^2+\mathrm{y}^2+2 \mathrm{ax}-4 \mathrm{y}-1-2 \mathrm{a}=0 \\\\
& \pi \mathrm{r}^2=4 \pi \\\\
& \mathrm{r}^2=4 \\\\
& 4=\sqrt{a^2+4+1+2 a} \\\\
& (\mathrm{a}+1)^2=0
\end{aligned}
$$
$$ P, Q=(0,2 \pm \sqrt{3}) $$
Equation of normal at $\mathrm{P}, \mathrm{Q}$ are $\mathrm{y}-2=\sqrt{3}(\mathrm{x}-1)$
$$ \begin{aligned} & \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\\\ & \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3} \end{aligned} $$
$$ P, Q=(0,2 \pm \sqrt{3}) $$
Equation of normal at $\mathrm{P}, \mathrm{Q}$ are $\mathrm{y}-2=\sqrt{3}(\mathrm{x}-1)$
$$ \begin{aligned} & \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\\\ & \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\\\ & \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3} \end{aligned} $$
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