The area enclosed by the closed curve C given | vedclass

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(D) Given the differential equation $\frac{dy}{dx} = -\frac{x+a}{y-2}$.Integrating both sides,we get $\int (y-2) dy = -\int (x+a) dx$,which gives $\fr

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

Vedclass · Answer

(D) Given the differential equation $\frac{dy}{dx} = -\frac{x+a}{y-2}$.Integrating both sides,we get $\int (y-2) dy = -\int (x+a) dx$,which gives $\frac{(y-2)^2}{2} = -\frac{(x+a)^2}{2} + k$.This simplifies to $(x+a)^2 + (y-2)^2 = 2k$. Since the area is $4\pi$,the radius $r = 2$,so $2k = 4$,meaning $(x+a)^2 + (y-2)^2 = 4$.Using $y(1) = 0$,we have $(1+a)^2 + (0-2)^2 = 4$,so $(1+a)^2 = 0$,which gives $a = -1$.The equation of the circle is $(x-1)^2 + (y-2)^2 = 4$.For intersection with the $y$-axis,set $x=0$: $(0-1)^2 + (y-2)^2 = 4 \implies (y-2)^2 = 3 \implies y = 2 \pm \sqrt{3}$.So $P = (0, 2+\sqrt{3})$ and $Q = (0, 2-\sqrt{3})$.The center of the circle is $(1, 2)$. The slope of the radius to $P$ is $\frac{(2+\sqrt{3})-2}{0-1} = -\sqrt{3}$.The slope of the normal at $P$ is the same as the slope of the radius,which is $-\sqrt{3}$.The equation of the normal at $P$ is $y - (2+\sqrt{3}) = -\sqrt{3}(x-0) \implies y = -\sqrt{3}x + 2 + \sqrt{3}$.Setting $y=0$ for $R$,we get $0 = -\sqrt{3}x + 2 + \sqrt{3} \implies x_R = 1 + \frac{2}{\sqrt{3}}$.Similarly,for $Q$,the slope of the radius is $\frac{(2-\sqrt{3})-2}{0-1} = \sqrt{3}$.The equation of the normal at $Q$ is $y - (2-\sqrt{3}) = \sqrt{3}(x-0) \implies y = \sqrt{3}x + 2 - \sqrt{3}$.Setting $y=0$ for $S$,we get $0 = \sqrt{3}x + 2 - \sqrt{3} \implies x_S = 1 - \frac{2}{\sqrt{3}}$.The length $RS = |x_R - x_S| = |(1 + \frac{2}{\sqrt{3}}) - (1 - \frac{2}{\sqrt{3}})| = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$.

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