Mathematics · Hyperbola

JEE Main 2026 — 6 April, Morning Shift — Question 31

If the eccentricity e of the hyperbola $\frac{x²}{a² }- \frac{y²}{b²} = 1$ passing through $(6,4\sqrt3)$ satisfies $15(e²+1)=34e,$ then the length of the latus rectum of the hyperbola $\frac{x²}{b²} - \frac{y²}{(2(a²+1)} = 1$ is:

Option A:
$10$
Correct
Option B:
$20$
Option C:
$25$
Option D:
$30$

Answer: A

Step-by-step solution

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ It passs through $(6,4 \sqrt{3})$

\Rightarrow \frac{36}{\mathrm{a}^{2}}-\frac{48}{\mathrm{~b}^{2}}=1 \end{gathered}$$ also $15 \mathrm{e}^{2}-34 \mathrm{e}+15=0$ $\Rightarrow 15 \mathrm{e}^{2}-25 \mathrm{e}-9 \mathrm{e}+15=0$ $\mathrm{e}=\frac{5}{3}$ or $\frac{3}{5} \Rightarrow \mathrm{e}=\frac{5}{3}$ $$\begin{gathered} 1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{25}{9} \Rightarrow \frac{\mathrm{~b}^{2}}{\mathrm{a}^{2}}=\frac{16}{9} \end{gathered}$$ using (1) and (2) $\Rightarrow \frac{36}{\mathrm{a}^{2}}-\frac{48}{16 \mathrm{a}^{2}} \times 9=1 \Rightarrow \mathrm{a}=3, \mathrm{~b}=4$ Length of L.R.

\frac{4\left(\mathrm{a}^{2}+1\right)}{\mathrm{b}}=\frac{4(10)}{4}=10

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Exam: JEE Main 2026
Paper: 6 April, Morning Shift
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola

Let e₁ and e₂ be two distinct roots of the equation x² - ax + 2 = 0. Let the sets { a∈R: e 1 and e 2 are the eccentricities of hyperbolas}…

Answer: A

Step-by-step solution

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