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QJEEM21Macd56: JEE Main 2021 Mathematics - Class 11 JEEM21Macd56 Solution: Let an ellipse , , passes through and has eccentricity . If a…

Let an ellipse E:x2a2+y2b2=1E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1 , {a^2} > {b^2} , passes through (32,1)\left( {\sqrt {{3 \over 2}} ,1} \right) and has eccentricity 13{1 \over {\sqrt 3 }} . If a circle, centered at focus F( α\alpha , 0), α\alpha > 0, of E and radius 23{2 \over {\sqrt 3 }} , intersects E at two points P and Q, then PQ2 is equal to :

Multiple Choice Options

  • A 83{8 \over 3}
  • B 43{4 \over 3}
  • C 163{{16} \over 3}
  • D 3

Step-by-Step Verified Solution

32a2+1b2=1{3 \over {2{a^2}}} + {1 \over {{b^2}}} = 1 and 1−b2a2=131 - {{{b^2}} \over {{a^2}}} = {1 \over 3}

⇒a2=3b2=3\Rightarrow {a^2} = 3{b^2} = 3

⇒x23+y22=1\Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1 ...... (i)

Its focus is (1, 0)

Now, equation of circle is

(x−1)2+y2=43{(x - 1)^2} + {y^2} = {4 \over 3} ..... (ii)

Solving (i) and (ii) we get

y=±23,x=1y = \pm {2 \over {\sqrt 3 }},x = 1

⇒PQ2=(43)2=163\Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}

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