Mathematics · Hyperbola

JEE Main 2026 — 5 April, Evening Shift — Question 37

Let the eccentricity $e$ of a hyperbola satisfy the equation $6e² - 11e + 3 = 0.$ If the foci of the hyperbola are $(3,5)$ and $(3,-4)$ , then the length of its latus rectum is:

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

Answer: C

Step-by-step solution

Given foci: $S_1(3,5)$ and $S_2(3,-4)$ . Distance between foci: $2ae = \sqrt{(3-3)^2 + (5+4)^2} = 9$ . Solve $6e^2 - 11e + 3 = 0$ : $(3e-1)(2e-3)=0$ , so $e = \frac{1}{3}$ (rejected, $e>1$ ) or $e = \frac{3}{2}$ . From $2ae = 9$ , $2a \cdot \frac{3}{2} = 9 \Rightarrow a = 3$ . Latus rectum: $\frac{2b^2}{a} = 2a(e^2-1) = 2 \cdot 3 \left(\frac{9}{4} - 1\right) = 6 \cdot \frac{5}{4} = \frac{15}{2}$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola