Let an ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a^2>b^2$, passes through $\left(\sqrt{\frac{3}{2}}, 1\right)$ and has eccentricity $\frac{1}{\sqrt{3}}$. If a circle, centered at focus $F(\alpha, 0), \alpha>0$, of $E$ and radius $\frac{2}{\sqrt{3}}$, intersects $E$ at two points $P$ and $Q$, then $PQ^2$ is equal to :
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When a circle is centered at a point on the x-axis, its intersections with a standard ellipse will always be symmetric about the x-axis. Thus $P$ and $Q$ have the same x-coordinate, and $PQ$ is simply the difference in their y-coordinates.
Step 1: Understanding the Concept:
The problem requires finding the specific equation of the ellipse using the given point and eccentricity.
Once the ellipse is defined, we find its focus, which acts as the center of a circle.
The intersection of the circle and ellipse provides the coordinates of $P$ and $Q$, leading to the distance $PQ$. Step 2: Key Formula or Approach:
1. Relation between $a, b,$ and $e$: \(b^2 = a^2(1 - e^2)\).
2. General equation of ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
3. Focus of the ellipse: \((\pm ae, 0)\).
4. Circle equation: \((x - h)^2 + (y - k)^2 = r^2\). Step 3: Detailed Explanation:
Given eccentricity \(e = \frac{1}{\sqrt{3}}\), so \(e^2 = \frac{1}{3}\).
Using the relation \(b^2 = a^2(1 - e^2) \Rightarrow b^2 = a^2(1 - \frac{1}{3}) = \frac{2}{3}a^2\).
The ellipse passes through \(\left(\sqrt{\frac{3}{2}}, 1\right)\):
\[ \frac{(\sqrt{3/2})^2}{a^2} + \frac{1^2}{b^2} = 1 \Rightarrow \frac{3}{2a^2} + \frac{1}{(2/3)a^2} = 1 \]
\[ \frac{3}{2a^2} + \frac{3}{2a^2} = 1 \Rightarrow \frac{6}{2a^2} = 1 \Rightarrow a^2 = 3 \]
Then, \(b^2 = \frac{2}{3}(3) = 2\).
The equation of the ellipse is \(\frac{x^2}{3} + \frac{y^2}{2} = 1\).
The focus \(F(\alpha, 0)\) with \(\alpha>0\) is \(F(ae, 0) = (\sqrt{3} \cdot \frac{1}{\sqrt{3}}, 0) = (1, 0)\). So \(\alpha = 1\).
The circle is centered at \((1, 0)\) with radius \(r = \frac{2}{\sqrt{3}}\):
\[ (x - 1)^2 + y^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3} \]
To find intersection points, substitute \(y^2 = 2\left(1 - \frac{x^2}{3}\right)\) into the circle equation:
\[ x^2 - 2x + 1 + 2 - \frac{2x^2}{3} = \frac{4}{3} \]
\[ \frac{x^2}{3} - 2x + 3 = \frac{4}{3} \Rightarrow x^2 - 6x + 9 = 4 \Rightarrow (x - 3)^2 = 4 \]
\[ x - 3 = 2 \Rightarrow x = 5 \text{ (Rejected as } x^2 \le 3) \]
\[ x - 3 = -2 \Rightarrow x = 1 \]
At \(x = 1\), \(y^2 = 2(1 - 1/3) = \frac{4}{3} \Rightarrow y = \pm \frac{2}{\sqrt{3}}\).
Points $P$ and $Q$ are \((1, 2/\sqrt{3})\) and \((1, -2/\sqrt{3})\).
\(PQ^2 = (1 - 1)^2 + (\frac{2}{\sqrt{3}} - (-\frac{2}{\sqrt{3}}))^2 = (\frac{4}{\sqrt{3}})^2 = \frac{16}{3}\). Step 4: Final Answer:
The value of $PQ^2$ is \(\frac{16}{3}\).
