The area enclosed by the closed curve C given by the differential equation dy dx+ x+a y-2=0,y(1)=0 is 4π. Let P and Q be the points of intersection…

Given:

$$\frac{dy}{dx}+\frac{x+a}{y-2}=0$$dxdy​+y−2x+a​=0

$$\Rightarrow \frac{dy}{dx}=\frac{x+a}{2-y}$$⇒dxdy​=2−yx+a​

$$\Rightarrow (2-y)dy=(x+a)dx$$⇒(2−y)dy=(x+a)dx

$$\Rightarrow 2y-\frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+ax+c$$⇒2y−2y2​=2x2​+ax+c

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

$$a+c=-\frac{1}{2}$$a+c=−21​

Now,

$$4y-{y}^{2}={x}^{2}+2ax+2c$$4y−y2=x2+2ax+2c

$$\Rightarrow {x}^{2}+{y}^{2}+2ax-4y+2c=0$$⇒x2+y2+2ax−4y+2c=0

$$\Rightarrow {x}^{2}+{y}^{2}+2ax-4y-1-2a=0$$⇒x2+y2+2ax−4y−1−2a=0

This is a circle having radius

$$r=\sqrt{{a}^{2}+4+1+2a}=\sqrt{{(a+1)}^{2}+4}$$r=a2+4+1+2a​=(a+1)2+4​

Now,

$$\pi {r}^{2}=4\pi$$πr2=4π

$${r}^{2}=4$$r2=4

$$\Rightarrow \sqrt{{(a+1)}^{2}+4}=4$$⇒(a+1)2+4​=4

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

So, $a=-1$a=−1

Hence, circle is

$${x}^{2}+{y}^{2}-2x-4y+1=0$$x2+y2−2x−4y+1=0

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

It cuts $y-\mathrm{axis}$y−axis at $x=0$x=0, so

P\equiv (0,2+\sqrt{3})&Q\equiv (0,2-\sqrt{3})

Now,

$${x}^{2}+{y}^{2}-2x-4y+1=0$$x2+y2−2x−4y+1=0

$$\Rightarrow 2x+2y(\frac{dy}{dx})-2-4(\frac{dy}{dx})=0$$⇒2x+2y(dxdy​)−2−4(dxdy​)=0

$$\Rightarrow (\frac{dy}{dx})=\frac{1-x}{y-2}$$⇒(dxdy​)=y−21−x​

$$\Rightarrow -(\frac{dx}{dy})=\frac{y-2}{x-1}$$⇒−(dydx​)=x−1y−2​

Slope of normal at $P$P is

$${m}_{1}=\frac{2+\sqrt{3}-2}{0-1}=-\sqrt{3}$$m1​=0−12+3​−2​=−3​

Slope of normal at $Q$Q is

$${m}_{2}=\frac{2-\sqrt{3}-2}{0-1}=\sqrt{3}$$m2​=0−12−3​−2​=3​

Equation of normal at $Q$Q is

$$y-(2-\sqrt{3})=\sqrt{3}(x-0)$$y−(2−3​)=3​(x−0)

$$\Rightarrow y-(2-\sqrt{3})=\sqrt{3}x$$⇒y−(2−3​)=3​x

Equation of normal at $P$P is

$$y-2-\sqrt{3}=-\sqrt{3}(x-0)$$y−2−3​=−3​(x−0)

$$\Rightarrow y=-\sqrt{3}x+2+\sqrt{3}$$⇒y=−3​x+2+3​

So,

$R\equiv (1+\frac{2}{\sqrt{3}},0)$R≡(1+3​2​,0) and $S\equiv (1-\frac{2}{\sqrt{3}},0)$S≡(1−3​2​,0)

Hence,

$RS=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}$RS=3​4​=343​​ units