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 :

The equation of a hyperbola with center at the origin and transverse axis along the x-axis is given by x²/a² - y²/b² = 1. The length of the transverse axis is 2a, so 2a = 4, which means a = 2. The equation becomes x²/4 - y²/b² = 1.

Since the hyperbola passes through the point (4,2), we can substitute x = 4 and y = 2 into the equation:

(4)²/4 - (2)²/b² = 1

16/4 - 4/b² = 1

4 - 4/b² = 1

3 = 4/b²

b² = 4/3

The eccentricity (e) of a hyperbola is given by e = √(a² + b²)/a. Substituting a = 2 and b² = 4/3, we get:

e = √(4 + 4/3)/2

e = √(16/3)/2

e = (4/√3)/2

e = 2/√3

Rationalizing the denominator, we get:

e = 2√3/3

However, none of the given options match this exactly. Let's re-examine the calculation. The eccentricity is given by e = √(1 + b²/a²) . Substituting a=2 and b² = 4/3, we get:

e = √(1 + (4/3)/4)

e = √(1 + 1/3)

e = √(4/3)

e = 2/√3 = 2√3/3

The closest option is 2√3. There might be a slight discrepancy in the problem statement or options.