Let the eccentricity e of a hyperbola satisfy the equation 6e^2 - 11e + 3 = 0. If the foci of the hyperbola are (3, 5) and (3, -4), then the length…
JEE Main 2026 — Mathematics Coordinate Geometry
2026mcqmedium
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 :
Official previous-year question
Held on 5 Apr 2026 · Verified 6 Jul 2026.
Options
A
311
B
317
C
215
D
217
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Solution
Given the equation for eccentricity: 6e2−11e+3=0
6e2−9e−2e+3=0
(3e−1)(2e−3)=0
e=31 or e=23
For a hyperbola, e>1, so e=23.
The distance between the foci (3,5) and (3,−4) is 2ae.
2ae=(3−3)2+(5−(−4))2=9
2a(23)=9⇒3a=9⇒a=3
Using the relation b2=a2(e2−1):
b2=9(49−1)=9(45)=445
The length of the latus rectum is a2b2:
32(445)=645=215
Answer: 215
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