JEE Main 2022 · 24 Jun · Shift 2 · Mathematics Q29

← JEE Main 2022 · 24 Jun · Shift 2

MathematicsInteger+4 / -1Question 29 of 90

Let the hyperbola $H:\dfrac{x^2}{a^2}-y^2=1$ and the ellipse $E:3x^2+4y^2=12$ be such that the length of latus rectum of $H$ is equal to the length of latus rectum of $E$ . If $e_H$ and $e_E$ are the eccentricities of $H$ and $E$ respectively, then the value of $12\left(e_H^2+e_E^2\right)$ is equal to ______.

Your answer

Answer and worked solution

Official answer: 42

Ellipse $\dfrac{x^2}4+\dfrac{y^2}3=1$ : latus rectum $\dfrac{2\cdot3}2=3$ , $e_E^2=1-\frac34=\frac14$ .

Hyperbola $\dfrac{x^2}{a^2}-y^2=1$ : latus rectum $\dfrac2a=3$ , so $a=\frac23$ and $e_H^2=1+\dfrac1{a^2}=\dfrac{13}4$ .

Value. $12\left(\frac{13}4+\frac14\right)=42$ .

Answer: 42

Topics:Hyperbola Ellipse

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JEE Main 2022 · 24 Jun · Shift 2, Mathematics question 29