JEE Main 2026 Apr 5 Shift 2, Mathematics Q13: Introduction to Hyperbola

JEE Main2026Apr 5, Shift 2Mathematics

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

A
$\frac{11}{3}$
B
$\frac{17}{3}$
C
$\frac{15}{2}$
D
$\frac{17}{2}$

Solution

Distance between foci: $2 a e = (3 - 3)^{2} + (5 - (- 4))^{2} = 9$ .

Solve $6 e^{2} - 11 e + 3 = 0$ : roots $e = 3/2$ and $e = 1/3$ . A hyperbola requires $e > 1$ , so $e = 3/2$ .

Then $2 a \cdot \frac{3}{2} = 9 \Rightarrow a = 3$ .

Length of latus rectum $= \frac{2 b ^{2}}{a} = 2 a (e^{2} - 1) = 2 \cdot 3 \cdot (\frac{9}{4} - 1) = 6 \cdot \frac{5}{4} = \frac{15}{2}$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q13: Introduction to Hyperbola

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