JEE Main 2026 · 6 Apr · Shift 1 · Mathematics Q11

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MathematicsSingle correct+4 / -1Question 11 of 75

If the eccentricity $e$ of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ , passing through $\left(6,4\sqrt3\right)$ , satisfies $15(e^2+1)=34e$ , then the length of the latus rectum of the hyperbola $\dfrac{x^2}{b^2}-\dfrac{y^2}{2(a^2+1)}=1$ is:

Answer and worked solution

Official answer: A

Eccentricity. $15e^2-34e+15=0$ gives $e=\tfrac53$ (taking $e>1$ ). Then $b^2=a^2(e^2-1)=\tfrac{16}9a^2$ .

Through $(6,4\sqrt3)$ .
$\frac{36}{a^2}-\frac{48\cdot9}{16a^2}=\frac9{a^2}=1,$
so $a^2=9$ and $b^2=16$ .

New hyperbola. $\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$ , with latus rectum $\dfrac{2\cdot20}4=10$ .

(A)

Topics:Hyperbola

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JEE Main 2026 · 6 Apr · Shift 1, Mathematics question 11