The area enclosed by the closed curve C given by the differential equation dydx+x+ay-2=0, …

Question

The area enclosed by the closed curve $C$ given by the differential equation $\dfrac{dy}{dx}+\dfrac{x+a}{y-2}=0$, $y(1)=0$, 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 the $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is:

Accepted Answer

$$
\begin{aligned}
& \frac{d y}{d x}+\frac{x+a}{y-2}=0 \\
& \frac{d y}{d x}=\frac{x+a}{2-y} \\
& (2-y) d y=(x+a) d x \\
& 2 y \frac{-y}{2}=\frac{x^2}{2}+\mathrm{ax}+\mathrm{c} \\
& \mathrm{a}+\mathrm{c}=-\frac{1}{2} \text { as } \mathrm{y}(1)=0 \\
& \mathrm{X}^2+\mathrm{y}^2+2 \mathrm{ax}-4 \mathrm{y}-1-2 \mathrm{a}=0 \\
& \pi \mathrm{r}^2=4 \pi \\
& \mathrm{r}^2=4 \\
& 4=\sqrt{a^2+4+1+2 a} \\
& (\mathrm{a}+1)^2=0
\end{aligned}
$$

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

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

$$
\begin{aligned}
& \mathrm{y}-2=-\sqrt{3}(\mathrm{x}-1) \\
& \mathrm{R}=\left(1-\frac{2}{\sqrt{3}}, 0\right) \\
& \mathrm{S}=\left(1+\frac{2}{\sqrt{3}}, 0\right) \\
& \mathrm{RS}=\frac{4}{\sqrt{3}}=4 \frac{\sqrt{3}}{3}
\end{aligned}
$$

Solutions · 1