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JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1
Let an ellipse E⇒x2a2+y2b2=1E \Rightarrow \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1, a2>b2{a^2} > {b^2}, passes through (32,  1)\left( {\sqrt {\frac{3}{2}} ,\,\,1} \right) and has eccentricity 13.\frac{1}{{\sqrt 3 }}. If a circle, centered
at focus F(α,0),  α>0F\left( {\alpha ,0} \right),\,\,\alpha > 0, of EE and radius 23\frac{2}{{\sqrt 3 }}, intersects EE at two points P  and  Q ,P\,\,\mathrm{and}\,\,Q\,, then PQ2P{Q^2} is equal to
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Correct option: C
163\frac{{16}}{3}
Option C (163\frac{{16}}{3}) is correct. By substituting the given point and eccentricity into the ellipse equation, we find the values of a2a^2 and b2b^2. Then, using the focus and radius of the circle, we find the intersection points and calculate the square of the distance between them.

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