Question:

Step 1: Solve the given differential equation. The given differential equation is $\dfrac{dy}{dx}+\dfrac{x+a}{y-2}=0$. We can separate the variables: $$(y-2)dy = -(x+a)dx$$ Integrate both sides: $$\int (y-2)dy = -\int (x+a)dx$$ $$\frac{y^2}{2} - 2y = -\left(\frac{x^2}{2} + ax\right) + C'$$ Multiply by 2 and rearrange the terms: $$x^2 + 2ax + y^2 - 4y = C$$ where $C = 2C'$. Step 2: Identify the type of curve by completing the square. To identify the curve, we complete the square for $x$ and $y$ terms: $$(x^2 + 2ax + a^2) + (y^2 - 4y + 4) = C + a^2 + 4$$ $$(x+a)^2 + (y-2)^2 = C + a^2 + 4$$ This is the equation of a circle centered at $(-a, 2)$. Let $R^2 = C + a^2 + 4$. So, the equation of the curve C is $(x+a)^2 + (y-2)^2 = R^2$. Step 3: Use the initial condition to find the constant term. The initial condition is $y(1)=0$. Substitute $x=1$ and $y=0$ into the equation of the curve: $$(1+a)^2 + (0-2)^2 = R^2$$ $$(1+a)^2 + 4 = R^2$$ $$1 + 2a + a^2 + 4 = R^2$$ $$a^2 + 2a + 5 = R^2$$ So, the equation of the curve C is $(x+a)^2 + (y-2)^2 = a^2+2a+5$. Step 4: Use the given area to find the value of $a$. The area enclosed by the curve C is $4\pi$. Since C is a circle, its area is $\pi R^2$. Therefore, we have: $$\pi (a^2+2a+5) = 4\pi$$ $$a^2+2a+5 = 4$$ $$a^2+2a+1 = 0$$ $$(a+1)^2 = 0$$ This implies $a = -1$. Step 5: Write the final equation of curve C. Substitute $a=-1$ into the equation of C: $$(x+(-1))^2 + (y-2)^2 = (-1)^2+2(-1)+5$$ $$(x-1)^2 + (y-2)^2 = 1-2+5$$ $$(x-1)^2 + (y-2)^2 = 4$$ This is a circle centered at $(1, 2)$ with a radius $R=2$. Step 6: Find the points P and Q where C intersects the y-axis. The y-axis is defined by $x=0$. Substitute $x=0$ into the equation of C: $$(0-1)^2 + (y-2)^2 = 4$$ $$1 + (y-2)^2 = 4$$ $$(y-2)^2 = 3$$ $$y-2 = \pm\sqrt{3}$$ $$y = 2 \pm \sqrt{3}$$ So, the points of intersection are $P(0, 2+\sqrt{3})$ and $Q(0, 2-\sqrt{3})$. Step 7: Determine the slopes of the normals at P and Q. From the differential equation, the slope of the tangent is $m_T = \dfrac{dy}{dx} = -\dfrac{x+a}{y-2}$. Since $a=-1$, $m_T = -\dfrac{x-1}{y-2}$. The slope of the normal, $m_N$, is the negative reciprocal of the tangent's slope: $$m_N = -\frac{1}{m_T} = -\left(\frac{y-2}{-(x-1)}\right) = \frac{y-2}{x-1}$$ For point $P(0, 2+\sqrt{3})$: $$m_{NP} = \frac{(2+\sqrt{3})-2}{0-1} = \frac{\sqrt{3}}{-1} = -\sqrt{3}$$ For point $Q(0, 2-\sqrt{3})$: $$m_{NQ} = \frac{(2-\sqrt{3})-2}{0-1} = \frac{-\sqrt{3}}{-1} = \sqrt{3}$$ Step 8: Find the equations of the normals at P and Q. Equation of normal at P, $L_P$, using point-slope form $y-y_1 = m(x-x_1)$: $$y - (2+\sqrt{3}) = -\sqrt{3}(x-0)$$ $$y - 2 - \sqrt{3} = -\sqrt{3}x$$ $$\sqrt{3}x + y - (2+\sqrt{3}) = 0$$ Equation of normal at Q, $L_Q$: $$y - (2-\sqrt{3}) = \sqrt{3}(x-0)$$ $$y - 2 + \sqrt{3} = \sqrt{3}x$$ $$\sqrt{3}x - y - (\sqrt{3}-2) = 0$$ Step 9: Determine the points R and S where the normals intersect the x-axis. The x-axis is defined by $y=0$. For the normal $L_P$ (point R): $$\sqrt{3}x_R + 0 - (2+\sqrt{3}) = 0$$ $$\sqrt{3}x_R = 2+\sqrt{3}$$ $$x_R = \frac{2+\sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3}} + 1$$ So, $R\left(1+\frac{2}{\sqrt{3}}, 0\right)$. For the normal $L_Q$ (point S): $$\sqrt{3}x_S - 0 - (\sqrt{3}-2) = 0$$ $$\sqrt{3}x_S = \sqrt{3}-2$$ $$x_S = \frac{\sqrt{3}-2}{\sqrt{3}} = 1 - \frac{2}{\sqrt{3}}$$ So, $S\left(1-\frac{2}{\sqrt{3}}, 0\right)$. Step 10: Calculate the length of RS. Since R and S lie on the x-axis, the distance between them is the absolute difference of their x-coordinates: $$RS = |x_R - x_S|$$ $$RS = \left|\left(1+\frac{2}{\sqrt{3}}\right) - \left(1-\frac{2}{\sqrt{3}}\right)\right|$$ $$RS = \left|1+\frac{2}{\sqrt{3}} - 1+\frac{2}{\sqrt{3}}\right|$$ $$RS = \left|\frac{4}{\sqrt{3}}\right|$$ $$RS = \frac{4}{\sqrt{3}}$$ Rationalizing the denominator: $$RS = \frac{4\sqrt{3}}{3}$$ The final answer is $\boxed{\dfrac{4\sqrt{3}}{3}}$.