Correct Answer - Option 4 :

\(\dfrac{\sqrt{13}}{2}\)
**CONCEPT**:

The eccentricity of a hyperbola of the form \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) is given by: \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} \)

**CALCULATION**:

Given: The point (3tan θ, 2sec θ) lies on the hyperbola.

As we can see that, if we substitute x = 3tan θ and y = 2sec θ in the expression \(\frac{y^2}{4}-\frac{x^2}{9} \) we get

⇒ \(\frac{y^2}{4}-\frac{x^2}{9} = sec^2 \ \theta - tan^2 \ \theta = 1\)

So, we can say that, the point (3tan θ, 2sec θ) lies on the hyperbola \(\frac{y^2}{4}-\frac{x^2}{9} = 1\)

As we know that, eccentricity of the hyperbola of the form \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) is given by: \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}} \)

Here, a^{2} = 4 and b^{2} = 9

⇒ \(e = \sqrt {1 + \frac{{{9}}}{{{4}}}} = \frac{\sqrt {13}}{2} \)

Hence, correct option is 4.