Question

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 $R S$ is 

• A. $\frac{4 \sqrt{3}}{3}$
• B. $\frac{2 \sqrt{3}}{3}$
• C. 2
• D. $\quad 2 \sqrt{3}$

Answer 1

The Correct Option is A

$\frac{dy}{dx}+\frac{x+a}{y-2}=0$
$\frac{dy}{dx}=\frac{x+a}{2-y}$
$(2-y)dy=(x+a)dx$
$2y\frac{-y}{2}=\frac{x^2}{2}+ax+c$
$a+c=-\frac{1}{2}\text{as}y(1)=0$
$X^2+y^2+2ax-4y-1-2a=0$
$\pi r^2=4\pi$
$r^2=4$
$4=\sqrt{a^2+4+1+2a}$
$(a+1{)}^2=0$
$P,Q=(0,2\pm \sqrt{3})$
Equation of normal at $P,Q$ are $y-2=\sqrt{3}(x-1)$
$y-2=-\sqrt{3}(x-1)$
$R=\left(1-\frac{2}{\sqrt{3}},0\right)$
$S=\left(1+\frac{2}{\sqrt{3}},0\right)$
$RS=\frac{4}{\sqrt{3}}=4\frac{\sqrt{3}}{3}$

Answer 2

Step 1: Given Differential Equation

We are given the following differential equation: 

\[ \frac{dy}{dx} = \frac{x + a}{y - 2} \] This is a separable equation that we can manipulate to solve for \( y \) in terms of \( x \). 

Step 2: Rearranging the Equation

By multiplying both sides by \( (2 - y) \) and integrating, we get:

\[ (2 - y) \, dy = (x + a) \, dx \] 

Step 3: Integrating \p>Now, we integrate both sides:

 

\[ 2y - y^2 = \frac{x^2}{2} + ax + c \] where \( a + c = -\frac{1}{2} \) as \( y(1) = 0 \). Thus, we get the equation:

 

\[ x^2 + y^2 + 2ax - 4y - 1 - 2a = 0 \] 

Step 4: Solving for \( a \) and \( c \)

We now solve for \( a \) and \( c \) using the equation:

\[ \pi r^2 = 4\pi \quad \Rightarrow \quad r^2 = 4 \] Thus, the equation simplifies to: \[ 4 = \sqrt{a^2 + 4 + 1 + 2a} \] Which leads to: \[ (a + 1)^2 = 0 \quad \Rightarrow \quad a = -1 \] 

Step 5: Finding the Points P and Q

We know that \( P, Q = (0, 2 \pm \sqrt{3}) \). The equation of the normal line at these points is:

\[ y - 2 = -\sqrt{3}(x - 1) \] 

Step 6: Calculating the Lengths R, S, and RS

The points \( R \) and \( S \) are given as:

\[ R = \left( 1 - \frac{2}{\sqrt{3}}, 0 \right), \quad S = \left( 1 + \frac{2}{\sqrt{3}}, 0 \right) \] The distance \( RS \) is calculated as: \[ RS = \frac{4}{\sqrt{3}} \quad \Rightarrow \quad RS = 4\sqrt{3} \] 

Conclusion \p>The final answer is found by calculating the length \( RS \), which is \( 4\sqrt{3} \).