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  • The area enclosed by the closed curve given by the differential equation <img alt="\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0"

The area enclosed by the closed curve 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 C and the y-axis. If normals at P and Q on the curve Cintersect x-axis at points R and S respectively, then the length of the line segment R S is

Option: 1

2


Option: 2

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


Option: 3

2 \sqrt{3}


Option: 4

\frac{2 \sqrt{3}}{3}


Answers (1)

best_answer

\begin{aligned} & \frac{d y}{d x}+\frac{x+\alpha}{y-2}=0, y(1)=0 \\ & \frac{d y}{d x}=\frac{-(x+\alpha)}{y-2} \\ & \int(y-2) d y=-\int(x+\alpha) d x \\ & \frac{y^2}{2}-2 y=-\left[\frac{x^2}{2}+\alpha x\right]+\lambda \end{aligned}

\begin{aligned} & y(1)=0 \\ & x=1 \Rightarrow y=0 \\ & 0-0=-\left[\frac{1}{2}+\alpha\right]+\lambda \\ & \frac{y^2}{2}-2 y=-\left[\frac{x^2}{2}+\alpha x\right]+\frac{1}{2}+\alpha \\ & \frac{x^2+y^2}{2}=2 y-\alpha x+\frac{1}{2}+\alpha \end{aligned}

\begin{aligned} & x^2+y^2+2 \alpha x-4 y-1-2 \alpha=0 \\ & \text { Area }=4 \pi \\ & \pi r^2=4 \pi \\ & r^2=4 \\ & \alpha^2+4+1+2 \alpha=4 \\ & \alpha^2+2 \alpha+1=0 \\ & (\alpha+1)^2=0 \Rightarrow[\alpha=-1] \\ & x^2+y^2-2 x-4 y+1=0 \end{aligned}

\begin{array}{ll} y-2=\sqrt{3}(x-1) & \, \, \, \, \, \, \, y-2=-\sqrt{3}(x-1) \\ y=0 & \, \, \, \, \, \, \, y=0 \end{array}

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

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