JEE Main 2026 · 5 Apr · Shift 2 · Mathematics Q13

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MathematicsSingle correct+4 / -1Question 13 of 75

Let the eccentricity e of a hyperbola satisfy the equation $6e^2-11e+3=0$ . If the foci of the hyperbola are $(3,5)$ and $(3,-4)$ , then the length of its latus rectum is :

\dfrac{11}3

\dfrac{17}3

\dfrac{15}2

\dfrac{17}2

Answer and worked solution

Official answer: C

Eccentricity. $6e^2-11e+3=0$ gives $e=\tfrac32$ or $\tfrac13$ . A hyperbola needs $e>1$ , so $e=\tfrac32$ .

a and b. The foci are 9 apart, so $2ae=9$ and $a=3$ . Then $b^2=a^2(e^2-1)=\tfrac{45}4$ .

Latus rectum. $\dfrac{2b^2}a=\dfrac{15}2$ .

(C)

Topics:Hyperbola Nature of roots

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JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 13