Question

Find the equation of the hyperbola whose foci are (± 5, 0) and the conjugate axis is of length 8 ?
1. \(\frac{{{x^2}}}{{{9}}} - \frac{{{y^2}}}{{{16}}} = 1\)
2. \(\frac{{{x^2}}}{{{9}}} - \frac{{{y^2}}}{{{7}}} = 1\)
3. \(\frac{{{x^2}}}{{{25}}} - \frac{{{y^2}}}{{{16}}} = 1\)
4. None of these

Answer 1

Correct Answer - Option 1 : \(\frac{{{x^2}}}{{{9}}} - \frac{{{y^2}}}{{{16}}} = 1\)

CONCEPT:

The properties of a rectangular hyperbola \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) are:

• Its centre is given by: (0, 0)
• Its foci are given by: (- ae, 0) and (ae, 0)
• Its vertices are given by: (- a, 0)  and (a, 0)
• Its eccentricity is given by: \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)
• Length of transverse axis = 2a and its equation is y = 0.
• Length of conjugate axis = 2b and its equation is x = 0.
• Length of its latus rectum is given by: \(\frac{2b^2}{a}\)

CALCULATION:

Here, we have to find the equation of the hyperbola whose foci are (± 5, 0) and the conjugate axis is of length 8.

By comparing the foci (± 5, 0) with (± ae, 0)

⇒ ae = 5

∵ Length of the conjugate axis is given by 2b

⇒ 2b = 8

⇒ b = 4

As we know that, eccentricity of a hyperbola is given by \(e = \frac{{\sqrt {{a^2} + {b^2}} }}{a}\)

⇒ a²e² = a² + b²

⇒ 25 = a² + 16

⇒ a² = 9

So, the equation of the required hyperbola is \(\frac{{{x^2}}}{{{9}}} - \frac{{{y^2}}}{{{16}}} = 1\)

Hence, option A is the correct answer.