JEE Mains · Maths · STD 12 - 9. differential equations
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 \(C\) intersect \(x\)-axis at points \(R\) and \(S\) respectively, then the length of the line segment \(RS\) is
- A \(2 \sqrt{3}\)
- B \(\frac{2 \sqrt{3}}{3}\)
- C \(2\)
- D \(\frac{4 \sqrt{3}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{4 \sqrt{3}}{3}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}+\frac{x+a}{y-2}=0\) \(\frac{d y}{d x}=\frac{x+a}{2-y}\) \((2-y) d y=( x + a ) dx\) \(2 y \frac{-y}{2}=\frac{x^2}{2}+ ax + c\) \(a+c=-\frac{1}{2} \text { as } y(1)=0\) \(X^2+y^2+2 a x-4 y-1-2 a=0\) \(\pi r^2=4 \pi\) \(r^2=4\) \(4=\sqrt{a^2+4+1+2 a}\)…
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