HyperbolamediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Hyperbola: Let Eccentricity Hyperbola Satisfy 11e Foci Length Latus (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

Let the eccentricity

e

of a hyperbola satisfy

6e^2-11e+3=0

. If the foci are

(3,5)

and

(3,-4)

, then the length of its latus rectum is:

\dfrac{11}{3}

\dfrac{17}{3}

\dfrac{15}{2}

correct

\dfrac{17}{2}

Solution

Step 1: Solve for

e

(3e-1)(2e-3)=0 \implies e=\frac{1}{3} \text{ (rejected, need }e>1) \text{ or } e=\frac{3}{2}

Step 2: Determine the semi-major axis

a

The distance between the foci equals

2ae

2ae = \sqrt{0^2+9^2} = 9 \implies 2a\cdot\frac{3}{2} = 9 \implies a = 3

Step 3: Compute

b^2

b^2 = a^2(e^2-1) = 9\!\left(\frac{9}{4}-1\right) = 9\cdot\frac{5}{4} = \frac{45}{4}

Step 4: Evaluate the latus rectum

\text{LR} = \frac{2b^2}{a} = \frac{2\cdot\frac{45}{4}}{3} = \frac{45}{6} = \frac{15}{2}

Answer: (3)

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