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QJEEM23Ma5143: JEE Main 2023 Mathematics - Class 12 JEEM23Ma5143 Solution: The area enclosed by the closed curve given by the differential …

The area enclosed by the closed curve C\mathrm{C} given by the differential equation

dydx+x+ay−2=0,y(1)=0\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0 is 4π4 \pi .

Let PP and QQ be the points of intersection of the curve C\mathrm{C} and the yy -axis. If normals at PP and QQ on the curve C\mathrm{C} intersect xx -axis at points RR and SS respectively, then the length of the line segment RSR S is :

Multiple Choice Options

  • A 433\frac{4 \sqrt{3}}{3}
  • B 232 \sqrt{3}
  • C 2
  • D 233\frac{2 \sqrt{3}}{3}

Step-by-Step Verified Solution

amp;dydx+x+ay−2=0amp;dydx=x+a2−yamp;(2−y)dy=(x+a)dxamp;2y−y2=x22+ax+camp;a+c=−12 as y(1)=0amp;X2+y2+2ax−4y−1−2a=0amp;πr2=4πamp;r2=4amp;4=a2+4+1+2aamp;(a+1)2=0\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±3)P, Q=(0,2 \pm \sqrt{3})

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

amp;y−2=−3(x−1)amp;R=(1−23,0)amp;S=(1+23,0)amp;RS=43=433\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}

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