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JEE Main 20265 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2026 · 5 April · Evening Shift · Mathematics · Hyperbola

Let the eccentricity e of a hyperbola satisfy the equation 6e^2 -...

Let the eccentricity ee of a hyperbola satisfy the equation 6e2−11e+3=06e^2 - 11e + 3 = 0. If the foci of the hyperbola are (3,5)(3, 5) and (3,−4)(3, -4), then the length of its latus rectum is :

Options

  1. A113\dfrac{11}{3}
  2. B173\dfrac{17}{3}
  3. C152\dfrac{15}{2}
  4. D172\dfrac{17}{2}

Correct answer

C. 152\dfrac{15}{2}

Step-by-step solution

Given the equation for eccentricity: 6e2−11e+3=06e^2 - 11e + 3 = 0

6e2−9e−2e+3=06e^2 - 9e - 2e + 3 = 0

(3e−1)(2e−3)=0(3e - 1)(2e - 3) = 0

e=13e = \dfrac{1}{3} or e=32e = \dfrac{3}{2}

For a hyperbola, e>1e > 1, so e=32e = \dfrac{3}{2}.

The distance between the foci (3,5)(3, 5) and (3,−4)(3, -4) is 2ae2ae.

2ae=(3−3)2+(5−(−4))2=92ae = \sqrt{(3-3)^2 + (5 - (-4))^2} = 9

2a(32)=9⇒3a=9⇒a=32a \left(\dfrac{3}{2}\right) = 9 \Rightarrow 3a = 9 \Rightarrow a = 3

Using the relation b2=a2(e2−1)b^2 = a^2(e^2 - 1):

b2=9(94−1)=9(54)=454b^2 = 9 \left(\dfrac{9}{4} - 1\right) = 9 \left(\dfrac{5}{4}\right) = \dfrac{45}{4}

The length of the latus rectum is 2b2a\dfrac{2b^2}{a}:

2(454)3=456=152\dfrac{2 \left(\dfrac{45}{4}\right)}{3} = \dfrac{45}{6} = \dfrac{15}{2}

Answer: 152\dfrac{15}{2}
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