# If any point on a hyperbola is (3tan θ, 2sec θ), then what is the eccentricity of the hyperbola?

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If any point on a hyperbola is (3tan θ, 2sec θ), then what is the eccentricity of the hyperbola?
1. $\dfrac{3}{2}$
2. $\dfrac{5}{2}$
3. $\dfrac{\sqrt{11}}{2}$
4. $\dfrac{\sqrt{13}}{2}$

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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, a2 = 4 and b2 = 9

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

Hence, correct option is 4.