UPSC NDA Math Model Paper 5

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=

√a2+b2

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

x2

36

−

y2

28

=1
Hence, option A is the correct answer.