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 :
Solution
${3 \over {2{a^2}}} + {1 \over {{b^2}}} = 1$ and $1 - {{{b^2}} \over {{a^2}}} = {1 \over 3}$<br><br>$\Rightarrow {a^2} = 3{b^2} = 3$ <br><br>$\Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1$ ...... (i)<br><br>Its focus is (1, 0)<br><br>Now, equation of circle is <br><br>${(x - 1)^2} + {y^2} = {4 \over 3}$ ..... (ii)<br><br>Solving (i) and (ii) we get<br><br>$y = \pm {2 \over {\sqrt 3 }},x = 1$<br><br>$\Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}$
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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