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Step-by-Step Solution
Step 1: Write the general form of the ellipse
We have an ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$$
with the condition \(a^2 > b^2\). It passes through the point
\(\bigl(\sqrt{\tfrac{3}{2}},\,1\bigr)\) and has eccentricity
\(e = \tfrac{1}{\sqrt{3}}\).
Step 2: Form the equation using the point on the ellipse
Since \(\bigl(\sqrt{\tfrac{3}{2}},\,1\bigr)\) lies on the ellipse,
we substitute \(x^2 = \tfrac{3}{2}\) and \(y^2 = 1\) into the ellipse equation:
\[
\frac{\tfrac{3}{2}}{a^2} + \frac{1}{b^2} = 1
\quad\Longrightarrow\quad
\frac{3}{2a^2} + \frac{1}{b^2} = 1.
\]
Step 3: Use the eccentricity condition
For an ellipse, the eccentricity \(e\) is given by
\[
e^2 = 1 - \frac{b^2}{a^2}.
\]
Given \(e = \tfrac{1}{\sqrt{3}}\), we have \(e^2 = \tfrac{1}{3}\). Therefore:
\[
\frac{1}{3} = 1 - \frac{b^2}{a^2}
\quad\Longrightarrow\quad
\frac{b^2}{a^2} = 1 - \frac{1}{3} = \frac{2}{3}.
\]
Hence \(b^2 = \tfrac{2}{3}\,a^2.\)
Step 4: Solve for \(a^2\) and \(b^2\)
From
\(\tfrac{3}{2a^2} + \tfrac{1}{b^2} = 1\) and \(b^2 = \tfrac{2}{3}\,a^2,\) substitute \(b^2\) into the first equation:
\[
\frac{3}{2a^2} + \frac{1}{\tfrac{2}{3} a^2} = 1,
\]
\[
\frac{3}{2a^2} + \frac{3}{2\,a^2} = 1,
\]
\[
\frac{6}{2\,a^2} = 1
\quad\Longrightarrow\quad
\frac{3}{a^2} = 1
\quad\Longrightarrow\quad
a^2 = 3.
\]
Then
\(
b^2 = \frac{2}{3}\times 3 = 2.
\)
Step 5: Write the explicit equation of the ellipse
Thus, the ellipse becomes
\[
\frac{x^2}{3} + \frac{y^2}{2} = 1.
\]
Step 6: Identify the focus of the ellipse
For an ellipse \(\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1\),
the distance of each focus from the center is
\(c = \sqrt{a^2 - b^2}\). Here:
\[
c = \sqrt{3 - 2} = \sqrt{1} = 1.
\]
Since \(\alpha > 0\), the focus is \((1,\,0)\).
Step 7: Equation of the circle with center at the focus
The circle is centered at \(\bigl(1,\,0\bigr)\) and has radius \(\tfrac{2}{\sqrt{3}}\). Therefore, its equation is:
\[
(x - 1)^2 + y^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}.
\]
Step 8: Solve the system of ellipse and circle
We solve simultaneously:
\[
\begin{cases}
\frac{x^2}{3} + \frac{y^2}{2} = 1, \\
(x - 1)^2 + y^2 = \frac{4}{3}.
\end{cases}
\]
A key observation is to guess (and then verify) that \(x = 1\) might satisfy both equations. Substituting \(x=1\) into the ellipse gives
\[
\frac{1^2}{3} + \frac{y^2}{2} = 1
\quad\Longrightarrow\quad
\frac{1}{3} + \frac{y^2}{2} = 1
\quad\Longrightarrow\quad
\frac{y^2}{2} = \frac{2}{3}
\quad\Longrightarrow\quad
y^2 = \frac{4}{3}
\quad\Longrightarrow\quad
y = \pm \frac{2}{\sqrt{3}},
\]
which matches the circle equation \((1-1)^2 + y^2 = 4/3\). Therefore, the points of intersection are
\[
P\Bigl(1, \,\frac{2}{\sqrt{3}}\Bigr)
\quad\text{and}\quad
Q\Bigl(1, \,-\frac{2}{\sqrt{3}}\Bigr).
\]
Step 9: Compute the distance PQ
The points differ only in the \(y\)-coordinate by
\(\frac{4}{\sqrt{3}}\), so the distance PQ is
\[
PQ = \left|\frac{2}{\sqrt{3}} - \left(-\frac{2}{\sqrt{3}}\right)\right|
= \frac{4}{\sqrt{3}}.
\]
Hence,
\[
PQ^2 = \left(\frac{4}{\sqrt{3}}\right)^2 = \frac{16}{3}.
\]
Final Answer
\(\displaystyle \frac{16}{3}\)
