JEE MAIN - Mathematics (2021 - 25th July Morning Shift - No. 14)
Let an ellipse $$E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $${a^2} > {b^2}$$, passes through $$\left( {\sqrt {{3 \over 2}} ,1} \right)$$ and has eccentricity $${1 \over {\sqrt 3 }}$$. If a circle, centered at focus F($$\alpha$$, 0), $$\alpha$$ > 0, of E and radius $${2 \over {\sqrt 3 }}$$, intersects E at two points P and Q, then PQ2 is equal to :
$${8 \over 3}$$
$${4 \over 3}$$
$${{16} \over 3}$$
3
Explanation
$${3 \over {2{a^2}}} + {1 \over {{b^2}}} = 1$$ and $$1 - {{{b^2}} \over {{a^2}}} = {1 \over 3}$$
$$ \Rightarrow {a^2} = 3{b^2} = 3$$
$$ \Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1$$ ...... (i)
Its focus is (1, 0)
Now, equation of circle is
$${(x - 1)^2} + {y^2} = {4 \over 3}$$ ..... (ii)
Solving (i) and (ii) we get
$$y = \pm {2 \over {\sqrt 3 }},x = 1$$
$$ \Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}$$
$$ \Rightarrow {a^2} = 3{b^2} = 3$$
$$ \Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1$$ ...... (i)
Its focus is (1, 0)
Now, equation of circle is
$${(x - 1)^2} + {y^2} = {4 \over 3}$$ ..... (ii)
Solving (i) and (ii) we get
$$y = \pm {2 \over {\sqrt 3 }},x = 1$$
$$ \Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}$$
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