The area enclosed by the closed curve C given by the differential...

Given:

$\dfrac{dy}{dx} + \dfrac{x + a}{y - 2} = 0$

$⇒ \dfrac{dy}{dx} = \dfrac{x + a}{2 - y}$

$⇒ \left(2 - y\right)dy = \left(x + a\right)dx$

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

Put $x = 1$, then we get $y\left(1\right) = 0$.

$a + c = - \dfrac{1}{2}$

Now,

$4y - {y}^{2} = {x}^{2} + 2ax + 2c$

$⇒ {x}^{2} + {y}^{2} + 2ax - 4y + 2c = 0$

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

This is a circle having radius

$r = \sqrt{{a}^{2} + 4 + 1 + 2a} = \sqrt{{\left(a + 1\right)}^{2} + 4}$

Now,

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

${r}^{2} = 4$

$⇒ \sqrt{{\left(a + 1\right)}^{2} + 4} = 4$

$⇒ {\left(a + 1\right)}^{2} = 0$

So, $a = - 1$

Hence, circle is

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

$⇒ {\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} = 4$

It cuts $y - axis$ at $x = 0$, so

$P ≡ \left(0, 2 + \sqrt{3}\right) \text{ and } Q ≡ \left(0, 2 - \sqrt{3}\right)$

Now,

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

$⇒ 2x + 2y\left(\dfrac{dy}{dx}\right) - 2 - 4\left(\dfrac{dy}{dx}\right) = 0$

$⇒ \left(\dfrac{dy}{dx}\right) = \dfrac{1 - x}{y - 2}$

$⇒ - \left(\dfrac{dx}{dy}\right) = \dfrac{y - 2}{x - 1}$

Slope of normal at $P$ is

${m}_{1} = \dfrac{2 + \sqrt{3} - 2}{0 - 1} = - \sqrt{3}$

Slope of normal at $Q$ is

${m}_{2} = \dfrac{2 - \sqrt{3} - 2}{0 - 1} = \sqrt{3}$

Equation of normal at $Q$ is

$y - \left(2 - \sqrt{3}\right) = \sqrt{3}\left(x - 0\right)$

$⇒ y - \left(2 - \sqrt{3}\right) = \sqrt{3}x$

Equation of normal at $P$ is

$y - 2 - \sqrt{3} = - \sqrt{3}\left(x - 0\right)$

$⇒ y = - \sqrt{3}x + 2 + \sqrt{3}$

So,

$R ≡ \left(1 + \dfrac{2}{\sqrt{3}}, 0\right)$ and $S ≡ \left(1 - \dfrac{2}{\sqrt{3}}, 0\right)$

Hence,

$RS = \dfrac{4}{\sqrt{3}} = \dfrac{4\sqrt{3}}{3}$ units