JEE Main 2022 · 29 Jun · Shift 1 · Mathematics Q18

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MathematicsSingle correct+4 / -1Question 18 of 90

Let PQ be a focal chord of the parabola $y^2=4x$ such that it subtends an angle of $\dfrac\pi2$ at the point $(3,0)$ . Let the line segment PQ be also a focal chord of the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ , $a^2>b^2$ . If $e$ is the eccentricity of the ellipse E, then the value of $\dfrac1{e^2}$ is equal to :

1+\sqrt2

3+2\sqrt2

1+2\sqrt3

4+5\sqrt3

Answer and worked solution

Official answer: B

Focal chord through $(1,0)$ : $t_1t_2=-1$ . Right angle at $(3,0)$ : $(t_1^2-3)(t_2^2-3)+4t_1t_2=0$ , so $t_1^2+t_2^2=2$ and $t^2=1$ . PQ is the latus rectum $(1,\pm2)$ .

Ellipse. Focus $(1,0)$ : $ae=1$ , and $\frac{b^2}a=2$ with $b^2=a^2-1$ : $a=1+\sqrt2$ .

$\frac1{e^2}=a^2=3+2\sqrt2$ .

(B)

Topics:Parabola Ellipse

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JEE Main 2022 · 29 Jun · Shift 1, Mathematics question 18