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QJEEM22Ma6c70: JEE Main 2022 Mathematics - Class 11 JEEM22Ma6c70 Solution: Let PQ be a focal chord of the parabola y 2 = 4x such that it…

Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of π2{\pi \over 2} at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E:x2a2+y2b2=1E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1 , {a^2} > {b^2} . If e is the eccentricity of the ellipse E, then the value of 1e2{1 \over {{e^2}}} is equal to :

Multiple Choice Options

  • A 1+21 + \sqrt 2
  • B 3+223 + 2\sqrt 2
  • C 1+231 + 2\sqrt 3
  • D 4+534 + 5\sqrt 3

Step-by-Step Verified Solution

JEE Main 2022 (Online) 29th June Morning Shift Mathematics - Parabola Question 64 English Explanation

As ∠PRQ=π2\angle PRQ = {\pi \over 2}

(2t3−1t2) . (−2t3−t2)=−1\left( {{{{2 \over t}} \over {3 - {1 \over {{t^2}}}}}} \right)\,.\,\left( {{{ - 2t} \over {3 - {t^2}}}} \right) = - 1

⇒t=± 1\Rightarrow t = \pm \,1

∴\therefore P≡(1,2)P \equiv (1,2) & Q(1,−2)Q(1, - 2)

∴\therefore for ellipse 1a2+4b2=1{1 \over {{a^2}}} + {4 \over {{b^2}}} = 1 and ae=1ae = 1

⇒1a2+4a2(1−e2)=1\Rightarrow {1 \over {{a^2}}} + {4 \over {{a^2}(1 - {e^2})}} = 1

⇒1+4(1−e2)=1e2\Rightarrow 1 + {4 \over {(1 - {e^2})}} = {1 \over {{e^2}}}

⇒(5−e2)e2=1−e2\Rightarrow (5 - {e^2}){e^2} = 1 - {e^2}

⇒e4−6e2+1=0\Rightarrow {e^4} - 6{e^2} + 1 = 0

⇒e2=13−22⇒1e2=3+22\Rightarrow {e^2} = {1 \over {3 - 2\sqrt 2 }} \Rightarrow {1 \over {{e^2}}} = 3 + 2\sqrt 2

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