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 ee of a hyperbola satisfy 6e2−11e+3=06e^2-11e+3=0. If the foci are (3,5)(3,5) and (3,−4)(3,-4), then the length of its latus rectum is:
A113\dfrac{11}{3}
B173\dfrac{17}{3}
C152\dfrac{15}{2}correct
D172\dfrac{17}{2}
Solution
Step 1: Solve for ee
(3e−1)(2e−3)=0  ⟹  e=13 (rejected, need e>1) or e=32(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 aa The distance between the foci equals 2ae2ae:
2ae=02+92=9  ⟹  2a⋅32=9  ⟹  a=32ae = \sqrt{0^2+9^2} = 9 \implies 2a\cdot\frac{3}{2} = 9 \implies a = 3
 Step 3: Compute b2b^2
b2=a2(e2−1)=9 ⁣(94−1)=9⋅54=454b^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
LR=2b2a=2⋅4543=456=152\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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