Let an ellipse E : x^2/a^2 + y^2 /b^2 = 1, a^2>b^2

Question

Let an ellipse E : \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) , a² > b² , 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 PQ² is equal to :

(1) \(\frac{8}{3}\)

(2) \(\frac{4}{3}\)

(3) \(\frac{16}{3}\)

(4) 3

Answer 1

Correct option (3) \(\frac{16}{3}\)

\(\frac{3}{2a^2} + \frac{1}{b^2} = 1\) and 1 - \(\frac{b^2}{a}\) = \(\frac{1}{3}\)

\(\Rightarrow\) a² = 3b² = 3

\(\Rightarrow\) \(\frac{x^3}{3} + \frac{y^2}{2} = 1\) .......... (1)

Its focus is (1,0) 

Now, eqn of circle is

(x-1)² + y² = \(\frac{4}{3}\) = ......... (ii)

Solving (i) and (ii) we get