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Step-by-Step Solution
Step 1: Write down the given differential equation
We have the differential equation
$ \frac{dy}{dx} + \frac{x+a}{y-2} = 0, $
along with the initial condition
$ y(1) = 0, $
and we are told that the closed curve described by this equation encloses an area of
$ 4\pi. $
Step 2: Rearrange the differential equation for separation of variables
Rewrite the equation:
$ \frac{dy}{dx} = -\frac{x+a}{y-2}. $
Multiply both sides by
$ (y-2)\,dx $
to separate variables:
$ (y-2)\,dy = -\,(x+a)\,dx. $
Step 3: Integrate both sides
Integrate left side with respect to
$ y $
and right side with respect to
$ x $:
$ \displaystyle \int (y-2)\,dy = \int -(x+a)\,dx. $
Compute each integral:
$ \int (y-2)\,dy = \frac{y^2}{2} - 2y, $
$ \int -(x+a)\,dx = -\left(\frac{x^2}{2} + ax\right) = -\frac{x^2}{2} - ax. $
Hence, combining both:
$ \frac{y^2}{2} - 2y = -\frac{x^2}{2} - ax + C, $
where
$ C $
is the constant of integration.
Step 4: Apply the initial condition
The initial condition given is
$ y(1) = 0. $
Substitute
$ x = 1 $
and
$ y = 0 $
into
$ \frac{y^2}{2} - 2y = -\frac{x^2}{2} - ax + C. $
This becomes:
$ 0 = -\frac{1^2}{2} - a(1) + C \ \Longrightarrow \ -\frac{1}{2} - a + C = 0. $
Therefore,
$ C = \frac{1}{2} + a. $
Step 5: Write the implicit solution for the curve
Substituting
$ C = \tfrac{1}{2} + a $
gives
$ \frac{y^2}{2} - 2y = -\frac{x^2}{2} - ax + \frac{1}{2} + a. $
Multiply through by 2 to simplify:
$ y^2 - 4y = -x^2 - 2ax + 1 + 2a. $
Rearrange:
$ x^2 + y^2 + 2ax - 4y - (1 + 2a) = 0. $
This is the implicit form of the closed curve
$ C. $
Step 6: Determine the value of a using the enclosed area of the curve
We are told that the area enclosed by this curve is
$ 4\pi. $
Notice that completing the square in
$ x $
and
$ y $
suggests this curve might form a circle. Let us complete the square:
$ x^2 + 2ax + a^2 - a^2 + y^2 - 4y + 4 - 4 - (1 + 2a) = 0. $
Rewrite and group terms:
$ (x+a)^2 - a^2 + (y-2)^2 - 4 - (1 + 2a) = 0. $
The left-hand portion
$ (x+a)^2 + (y-2)^2 $
suggests the circle form. We want the resulting circle to have area
$ 4\pi. $
A circle of radius
$ r $
has area
$ \pi r^2. $
If that area is
$ 4\pi, $
then
$ r^2 = 4. $
Through consistency checks and the initial condition, we find
$ a = -1. $
One may verify that this value yields the correct enclosed area
$ 4\pi. $
Step 7: Substitute a = -1 to get the circle equation
Plug
$ a=-1 $
into
$ x^2 + y^2 + 2ax - 4y - (1 + 2a) = 0. $
This becomes:
$ x^2 + y^2 - 2x - 4y - [1 + 2(-1)] = 0, $
i.e.
$ x^2 + y^2 - 2x - 4y - (1 - 2) = 0 \ \Longrightarrow \ x^2 + y^2 - 2x - 4y + 1 = 0. $
Complete the square in
$ x $
and
$ y $:
$ (x^2 - 2x + 1) + (y^2 - 4y + 4) = 4, $
which simplifies to
$ (x - 1)^2 + (y - 2)^2 = 4. $
This is a circle of center
$ (1,2) $
and radius
$ 2, $
hence enclosing an area
$ \pi \cdot 2^2 = 4\pi, $
confirming the given information.
Step 8: Find the intersection points of the curve with the y-axis
For the intersection with the
$ y $-axis, set
$ x = 0 $
in the circle equation
$ (x - 1)^2 + (y - 2)^2 = 4. $
That is:
$ (0 - 1)^2 + (y - 2)^2 = 4 \ \Longrightarrow \ 1 + (y - 2)^2 = 4. $
So
$ (y - 2)^2 = 3 \ \Longrightarrow \ y - 2 = \pm \sqrt{3} \ \Longrightarrow \ y = 2 \pm \sqrt{3}. $
Therefore, the intersection points are
$ P = (0,\,2 + \sqrt{3}) \quad\text{and}\quad Q = (0,\,2 - \sqrt{3}). $
Step 9: Determine the slopes of the tangents and hence normals at P and Q
Recall the differential equation (substituting
$ a = -1 $):
$ \frac{dy}{dx} = -\frac{x-1}{y-2}. $
• At
$ P(0,\,2 + \sqrt{3}): $
$ \left.\frac{dy}{dx}\right|_P = -\frac{0-1}{(2 + \sqrt{3}) - 2}
= -\frac{-1}{\sqrt{3}}
= \frac{1}{\sqrt{3}}. $
Hence the slope of the normal at
$ P $
is the negative reciprocal:
$ m_{\text{normal at }P} = -\sqrt{3}. $
• At
$ Q(0,\,2 - \sqrt{3}): $
$ \left.\frac{dy}{dx}\right|_Q = -\frac{0-1}{(2 - \sqrt{3}) - 2}
= -\frac{-1}{-\sqrt{3}}
= -\frac{1}{\sqrt{3}}. $
The slope of the normal at
$ Q $
is the negative reciprocal of
$ -\tfrac{1}{\sqrt{3}}, $
which is
$ m_{\text{normal at }Q} = \sqrt{3}. $
Step 10: Find equations of the normals and their intersection with the x-axis
• Normal at
$ P(0,\,2 + \sqrt{3}) $
has slope
$ -\sqrt{3}. $
The equation of this normal line is
$ y - (2 + \sqrt{3}) = -\sqrt{3}\,(x - 0), $
or
$ y - 2 - \sqrt{3} = -\sqrt{3}\,x. $
To find the intersection with the
$ x $-axis, set
$ y = 0 $:
$ 0 - (2 + \sqrt{3}) = -\sqrt{3}\,x
\ \Longrightarrow \ x = \frac{2 + \sqrt{3}}{\sqrt{3}}. $
So
$ R = \left(\frac{2 + \sqrt{3}}{\sqrt{3}},\,0\right). $
• Normal at
$ Q(0,\,2 - \sqrt{3}) $
has slope
$ \sqrt{3}. $
The equation of this normal is
$ y - (2 - \sqrt{3}) = \sqrt{3}\,(x - 0), $
or
$ y - 2 + \sqrt{3} = \sqrt{3}\,x. $
Set
$ y = 0 $
to find the intersection with the
$ x $-axis:
$ 0 - (2 - \sqrt{3}) = \sqrt{3}\,x
\ \Longrightarrow \ x = -\frac{2 - \sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3} - 2}{\sqrt{3}}. $
Thus
$ S = \left(\frac{\sqrt{3} - 2}{\sqrt{3}},\,0\right). $
Step 11: Calculate the length of RS
Points
$ R $
and
$ S $
both lie on the
$ x $-axis, so the distance
$ RS $
is the absolute difference of their
$ x $-coordinates:
$ x_R = \frac{2 + \sqrt{3}}{\sqrt{3}}, \quad x_S = \frac{\sqrt{3} - 2}{\sqrt{3}}. $
Hence,
$ RS = \bigl|\,x_R - x_S\bigr|
= \left|\frac{2 + \sqrt{3}}{\sqrt{3}} - \frac{\sqrt{3} - 2}{\sqrt{3}}\right|
= \frac{1}{\sqrt{3}}\Bigl[\,(2 + \sqrt{3}) - (\sqrt{3} - 2)\Bigr]. $
Simplify within the bracket:
$ (2 + \sqrt{3}) - (\sqrt{3} - 2) = 2 + \sqrt{3} - \sqrt{3} + 2 = 4. $
So,
$ RS = \frac{1}{\sqrt{3}} \times 4 = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}. $
Final Answer
The length of the line segment
$ RS $
is
$ \boxed{\frac{4\sqrt{3}}{3}}. $
