Find the equation of the hyperbola with vertices at (± 6, 0) and foci at (± 8, 0) ?

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 with vertices at (± 6, 0) and foci at (± 8, 0)

By comparing the given vertices  at (± 6, 0) with (± a, 0) we get

⇒ a = 6

Similarly, by comparing the given foci at (± 8, 0) with (± ae, 0) we get,

⇒ ae = 8

⇒ 6e = 8 ⇒ e = 4/3

As we know that, \(e = \frac{{\sqrt{{a^2} + {b^2}} }}{a}\)

⇒ a²e² = a² + b²

By substituting a² = 36 and e² = 16/9 in the above equation we get,

⇒ 36 × (16/9) = 36 + b²

⇒ b² = 64 - 36 = 28

So, the required equation of hyperbola is \(\frac{{{x^2}}}{{{36}}} - \frac{{{y^2}}}{{{28}}} = 1\)

Hence, option A is the correct answer.