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= ⇒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
−=1 Hence, option
A is the correct answer.