1. Understanding the Problem and Key Concepts

We are given an ellipse $E$ and information about its properties (passing point, eccentricity). We need to determine its equation. Then, we are given a circle centered at one of the ellipse's foci with a specified radius. This circle intersects the ellipse at two points, P and Q. Our ultimate goal is to find the square of the distance between P and Q, i.e., $PQ^2$.

To solve this problem, we will utilize the following key concepts and formulas related to ellipses and coordinate geometry:

• Standard Equation of an Ellipse: For an ellipse centered at the origin with foci on the x-axis, the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the semi-major axis, $b$ is the semi-minor axis, and $a^2 > b^2$.
• Eccentricity: The eccentricity $e$ of an ellipse is a measure of its "roundness." It is related to $a$ and $b$ by the formula $b^2 = a^2(1-e^2)$.
• Foci of an Ellipse: For an ellipse with foci on the x-axis, the coordinates of the foci are $(\pm ae, 0)$.
• Focal Distance Property: For any point $P(x,y)$ on the ellipse, its distance from the focus $F(ae,0)$ is $PF = a - ex$ (when considering the right focus). Similarly, for the left focus $F'(-ae,0)$, $PF' = a + ex$. The sum of focal distances is constant: $PF + PF' = 2a$.
• Equation of a Circle: A circle with center $(h,k)$ and radius $R$ has the equation $(x-h)^2 + (y-k)^2 = R^2$.
• Distance Formula: The square of the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $D^2 = (x_2-x_1)^2 + (y_2-y_1)^2$.

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2. Determining the Ellipse Parameters ($a^2$ and $b^2$)

Our first step is to find the specific equation of the ellipse $E$. We are given two pieces of information: its eccentricity and a point it passes through.

Step 2.1: Using Eccentricity to Relate $a^2$ and $b^2$ The given eccentricity is $e = \frac{1}{\sqrt{3}}$. The fundamental relation between $a$, $b$, and $e$ for an ellipse is: $b^2 = a^2(1-e^2)$ Substitute the given value of $e$: $b^2 = a^2\left(1 - \left(\frac{1}{\sqrt{3}}\right)^2\right)$ $b^2 = a^2\left(1 - \frac{1}{3}\right)$ $b^2 = a^2\left(\frac{2}{3}\right) \quad \text{(Equation 1)}$ Explanation: This equation is crucial as it connects the semi-minor axis ($b$) to the semi-major axis ($a$) using the eccentricity. We now have one equation with two unknowns ($a^2$ and $b^2$).

Step 2.2: Using the Given Point to Form a Second Equation The ellipse $E$ passes through the point $\left(\sqrt{\frac{3}{2}}, 1\right)$. Since this point lies on the ellipse, it must satisfy the ellipse's equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ Substitute $x = \sqrt{\frac{3}{2}}$ and $y = 1$: $\frac{\left(\sqrt{\frac{3}{2}}\right)^2}{a^2} + \frac{1^2}{b^2} = 1$ $\frac{\frac{3}{2}}{a^2} + \frac{1}{b^2} = 1$ $\frac{3}{2a^2} + \frac{1}{b^2} = 1 \quad \text{(Equation 2)}$ Explanation: Any point on a curve must satisfy the curve's equation. This gives us a second independent equation involving $a^2$ and $b^2$.

Step 2.3: Solving for $a^2$ and $b^2$ Now we solve the system of Equation 1 and Equation 2. From Equation 1, we have $b^2 = \frac{2}{3}a^2$. Substitute this expression for $b^2$ into Equation 2: $\frac{3}{2a^2} + \frac{1}{\frac{2}{3}a^2} = 1$ $\frac{3}{2a^2} + \frac{3}{2a^2} = 1$ $\frac{6}{2a^2} = 1$ $\frac{3}{a^2} = 1$ $a^2 = 3$ Now, substitute $a^2 = 3$ back into Equation 1 to find $b^2$: $b^2 = \frac{2}{3}(3)$ $b^2 = 2$ Explanation: We have successfully determined the values of $a^2$ and $b^2$, which completely defines the ellipse's equation. The equation of the ellipse $E$ is therefore: $\frac{x^2}{3} + \frac{y^2}{2} = 1$

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3. Identifying the Focus and Circle Equation

Next, we need to find the center of the circle, which is one of the foci of the ellipse, and then write the circle's equation.

Step 3.1: Locating the Focus F($\alpha$, 0) The foci of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are at $(\pm ae, 0)$. We have $a^2 = 3 \Rightarrow a = \sqrt{3}$ (since $a>0$). We are given $e = \frac{1}{\sqrt{3}}$. So, $ae = \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1$. The foci are at $(\pm 1, 0)$. The problem states that the circle is centered at focus $F(\alpha, 0)$ with $\alpha > 0$. Therefore, the center of the circle is $F(1, 0)$. So, $\alpha = 1$. Explanation: The value $ae$ represents the distance of each focus from the center of the ellipse. Since $\alpha > 0$ is specified, we choose the positive focus.

Step 3.2: Writing the Equation of the Circle The circle is centered at $F(1, 0)$ and has a radius $R = \frac{2}{\sqrt{3}}$. Using the standard circle equation $(x-h)^2 + (y-k)^2 = R^2$: $(x-1)^2 + (y-0)^2 = \left(\frac{2}{\sqrt{3}}\right)^2$ $(x-1)^2 + y^2 = \frac{4}{3} \quad \text{(Equation 3)}$ Explanation: This is the equation of the circle whose intersection with the ellipse we need to find.

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4. Finding the Intersection Points P and Q

The points P and Q are the intersection points of the ellipse $E$ and the circle (Equation 3). Instead of directly solving the system of equations (which can be algebraically intensive), we can use a crucial property of the ellipse: the focal distance property.

Step 4.1: Utilizing the Focal Distance Property For any point $P(x,y)$ on the ellipse, its distance from the focus $F(ae,0)$ is given by $PF = a - ex$. We found $a = \sqrt{3}$ and $e = \frac{1}{\sqrt{3}}$. So, for points on the ellipse, the distance from the focus $F(1,0)$ is: $PF = \sqrt{3} - \frac{1}{\sqrt{3}}x$ Explanation: This formula is a direct consequence of the definition of an ellipse. It states that the distance from a point on the ellipse to a focus is linearly related to its x-coordinate. It's often easier to use than the standard distance formula involving square roots, especially when the focus is involved.

The points P and Q are on the circle centered at $F(1,0)$ with radius $R = \frac{2}{\sqrt{3}}$. This means that the distance from $F$ to P (and Q) is equal to the radius of the circle. So, $PF = R$. Therefore, for the intersection points P and Q: $PF = \frac{2}{\sqrt{3}}$ Equating the two expressions for $PF$: $\sqrt{3} - \frac{1}{\sqrt{3}}x = \frac{2}{\sqrt{3}}$ Multiply by $\sqrt{3}$ to clear denominators: $3 - x = 2$ $x = 1$ Explanation: This is a significant shortcut! Since the circle is centered at a focus, any point of intersection with the ellipse must satisfy both the circle's radius condition and the ellipse's focal distance property. This directly gives us the x-coordinate of the intersection points.

Step 4.2: Finding the y-coordinates of P and Q Now that we have the x-coordinate ($x=1$) for both P and Q, we can substitute it back into the ellipse equation to find the corresponding y-coordinates. The ellipse equation is $\frac{x^2}{3} + \frac{y^2}{2} = 1$. Substitute $x=1$: $\frac{1^2}{3} + \frac{y^2}{2} = 1$ $\frac{1}{3} + \frac{y^2}{2} = 1$ $\frac{y^2}{2} = 1 - \frac{1}{3}$ $\frac{y^2}{2} = \frac{2}{3}$ $y^2 = \frac{4}{3}$ $y = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}}$ Explanation: The ellipse is symmetric with respect to the x-axis. Since the focus is on the x-axis and the intersection points have the same x-coordinate, their y-coordinates must be opposites. This gives us two distinct points.

So, the two intersection points are: $P = \left(1, \frac{2}{\sqrt{3}}\right) \quad \text{and} \quad Q = \left(1, -\frac{2}{\sqrt{3}}\right)$ Common Mistake: Attempting to solve the system of equations by substituting $y^2 = \frac{4}{3} - (x-1)^2$ from the circle equation into the ellipse equation would lead to a more complex polynomial in $x$. Recognizing the focal distance property simplifies this significantly.

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5. Calculating PQ²