The area enclosed by the closed curve C given by the differential equation dy/dx + (x+a)/(y-2) = 0,

Q: Wait, why do we assume the curve is a circle? Could it be an ellipse or something else?

Great observation! When we integrate the differential equation, we get $(x+a)^2 + (y-2)^2 = C'$. Since the coefficients of $x^2$ and $y^2$ are identical (both are 1), the equation describes a circle. If the coefficients were different, we would be looking at an ellipse!

Q: Why do the normals at $P$ and $Q$ pass through the center? Is this always true?

Yes, this is a beautiful geometric property of circles. By definition, the tangent at any point is perpendicular to the radius at that point. Therefore, the normal—which is perpendicular to the tangent—must lie along the line containing the radius, which by definition passes through the center.

Q: I got $a = -1$ but I'm confused about the intersection points. How do I find them?

To find the intersection with the $y$-axis, you must set $x=0$ in your circle equation $(x-1)^2 + (y-2)^2 = 4$. This leaves you with $(y-2)^2 = 3$, which gives you two values for $y$: $2+\sqrt{3}$ and $2-\sqrt{3}$. These are your $y$-coordinates for $P$ and $Q$.

Q: Is there a faster way to find the length $RS$ without finding the full equations of the lines?

You could use similar triangles or coordinate geometry properties, but finding the line equations is very robust. Since you know the center $(1, 2)$ and the points $P$ and $Q$, the slopes are easy to calculate. Don't rush the algebra; the symmetry of the circle will make the final subtraction very clean.

Q: Why did we use the $x$-axis to find $R$ and $S$?

The problem explicitly defines $R$ and $S$ as the points where the normals intersect the $x$-axis. In coordinate geometry, the $x$-axis is defined by the equation $y=0$. By setting $y=0$ in the equations of your normal lines, you are mathematically forcing the intersection with the $x$-axis.

Question

The area enclosed by the closed curve $C$ given by the differential equation $\frac{dy}{dx} + \frac{x+a}{y-2} = 0, y(1) = 0$ 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

Options:

(A) $2\sqrt{3}$

(B) $\frac{2\sqrt{3}}{3}$

(C) $2$

(D) $\frac{4\sqrt{3}}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:

$$\frac{dy}{dx} + \frac{x+a}{y-2} = 0$$

By rearranging the terms, we isolate the variables to perform integration:

$$(y-2) dy = -(x+a) dx$$

Integrating both sides yields:

$$\frac{(y-2)^2}{2} = -\frac{(x+a)^2}{2} + C'$$

This simplifies to the standard form of a circle equation:

$$(x+a)^2 + (y-2)^2 = r^2$$

We have identified that the curve $C$ is a circle with center $(-a, 2)$ and radius $r$.

### The Power of Constraints

We are given that the area enclosed by the circle is $4\pi$. Since the area is defined as $\pi r^2$, we find:

$$r^2 = 4 \Rightarrow r = 2$$

Using the initial condition $y(1) = 0$, we substitute these values into the circle equation:

$$(1+a)^2 + (0-2)^2 = 4$$

$$(1+a)^2 + 4 = 4 \Rightarrow (1+a)^2 = 0$$

This confirms that $a = -1$. Consequently, the equation of our circle is:

$$(x-1)^2 + (y-2)^2 = 4$$

### The Geometry of Intersections

To find the intersection points $P$ and $Q$ on the $y$-axis, we set $x=0$:

$$(0-1)^2 + (y-2)^2 = 4$$

$$(y-2)^2 = 3 \Rightarrow y = 2 \pm \sqrt{3}$$

Thus, the points are $P(0, 2+\sqrt{3})$ and $Q(0, 2-\sqrt{3})$.

### The Elegance of the Normal

A normal to a circle at any point on its circumference must pass through the center $(1, 2)$. We calculate the slopes of the normals at $P$ and $Q$ using the center $(1, 2)$.

For point $P(0, 2+\sqrt{3})$, the slope $m_P$ is:

$$m_P = \frac{2 - (2+\sqrt{3})}{1 - 0} = -\sqrt{3}$$

The equation of the normal at $P$ is $y - 2 = -\sqrt{3}(x - 1)$. Setting $y=0$ to find the $x$-intercept $R$:

$$-2 = -\sqrt{3}(x - 1) \Rightarrow x = 1 + \frac{2}{\sqrt{3}}$$

For point $Q(0, 2-\sqrt{3})$, the slope $m_Q$ is:

$$m_Q = \frac{2 - (2-\sqrt{3})}{1 - 0} = \sqrt{3}$$

The equation of the normal at $Q$ is $y - 2 = \sqrt{3}(x - 1)$. Setting $y=0$ to find the $x$-intercept $S$:

$$-2 = \sqrt{3}(x - 1) \Rightarrow x = 1 - \frac{2}{\sqrt{3}}$$

### Final Calculation

The distance $RS$ is the difference between the $x$-coordinates of $R$ and $S$:

$$RS = \left| \left(1 + \frac{2}{\sqrt{3}}\right) - \left(1 - \frac{2}{\sqrt{3}}\right) \right| = \frac{4}{\sqrt{3}}$$

Rationalizing the denominator, we obtain the final result:

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

Select Answer:

Visualized Solution

Analyze the Differential Equation

Variable Separation

Integration and Curve Identification

Finding the Radius

Finding the Constant $a$

The Final Equation of Curve $C$

Intersection with $y$ -axis

The Normal Property

Normal at $P$ and Point $R$

Normal at $Q$ and Point $S$

Calculate Length $R S$

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Constraints

The Geometry of Intersections

The Elegance of the Normal

Final Calculation

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