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 $\mathrm{PQ}^{2}$ is equal to :

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Ellipse

Let an ellipse

E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a^{2} > b^{2}

, passes through

(\frac{3}{2}, 1)

and has eccentricity

\frac{1}{3}

. If a circle, centered at focus

F (α, 0), α > 0

, of

E

and radius

\frac{2}{3}

, intersects

E

at two points

P

and

Q

, then

PQ^{2}

is equal to :

Answer: (A)

From

e^{2} = \frac{1}{3} = 1 - \frac{b ^{2}}{a ^{2}}

comes

b^{2} = \frac{2 a ^{2}}{3}

. Putting

(\frac{3}{2}, 1)

into the equation,

\frac{3/2}{a ^{2}} + \frac{1}{b ^{2}} = 1 \Rightarrow \frac{3}{2 a ^{2}} + \frac{3}{2 a ^{2}} = 1 \Rightarrow a^{2} = 3, b^{2} = 2

The ellipse is

\frac{x ^{2}}{3} + \frac{y ^{2}}{2} = 1

and the focus with positive abscissa is

(a e, 0) = (1, 0)

, so the circle is

(x - 1)^{2} + y^{2} = \frac{4}{3}

. Substituting

y^{2} = 2 - \frac{2 x ^{2}}{3}

from the ellipse,

(x - 1)^{2} + 2 - \frac{2 x ^{2}}{3} = \frac{4}{3} \Rightarrow \frac{x ^{2}}{3} - 2 x + \frac{5}{3} = 0 \Rightarrow x^{2} - 6 x + 5 = 0

x = 1

x = 5

, and only

x = 1

lies on the ellipse. Then

y^{2} = 2 - \frac{2}{3} = \frac{4}{3}

, so the common points are

(1, \frac{2}{3})

and

(1, - \frac{2}{3})

, and

P Q^{2} = (\frac{4}{3})^{2} = \frac{16}{3}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option