Question

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. \(\dfrac{\sqrt{13}}{2}\)

Answer 1

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.