JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 15)
Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of $${\pi \over 2}$$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse $$E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $${a^2} > {b^2}$$. If e is the eccentricity of the ellipse E, then the value of $${1 \over {{e^2}}}$$ is equal to :
$$1 + \sqrt 2 $$
$$3 + 2\sqrt 2 $$
$$1 + 2\sqrt 3 $$
$$4 + 5\sqrt 3 $$
Explanation
_29th_June_Morning_Shift_en_15_1.png)
As $$\angle PRQ = {\pi \over 2}$$
$$\left( {{{{2 \over t}} \over {3 - {1 \over {{t^2}}}}}} \right)\,.\,\left( {{{ - 2t} \over {3 - {t^2}}}} \right) = - 1$$
$$ \Rightarrow t = \pm \,1$$
$$\therefore$$ $$P \equiv (1,2)$$ & $$Q(1, - 2)$$
$$\therefore$$ for ellipse $${1 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$ and $$ae = 1$$
$$ \Rightarrow {1 \over {{a^2}}} + {4 \over {{a^2}(1 - {e^2})}} = 1$$
$$ \Rightarrow 1 + {4 \over {(1 - {e^2})}} = {1 \over {{e^2}}}$$
$$ \Rightarrow (5 - {e^2}){e^2} = 1 - {e^2}$$
$$ \Rightarrow {e^4} - 6{e^2} + 1 = 0$$
$$ \Rightarrow {e^2} = {1 \over {3 - 2\sqrt 2 }} \Rightarrow {1 \over {{e^2}}} = 3 + 2\sqrt 2 $$
Comments (0)
