Equation of hyperbola having tranverse axis along x−axis is a2x2−b2y2=1
From the given information, equation of hyperbola can be taken as
4x2−b2y2=1, since a=2

Since it passes through (4,2),
Hence b2=34
Now e=1+a2b2=1+31=32
JEE Main 2019 — Mathematics Coordinate Geometry
A hyperbola has its centre at the origin, passes through the point (4,2) and has transverse axis of length 4 along the x−axis. Then the eccentricity of the hyperbola is:
Held on 9 Jan 2019 · Verified 6 Jul 2026.
3
23
32
2
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