UPSC NDA Math Model Paper 6

Given: Equation of hyperbola is

y2

9

−

x2

27

=1
As we can see that, the given hyperbola is a vertical hyperbola.
So, by comparing the given equation of hyperbola with

y2

a2

−

x2

b2

=1 we get,
⇒a2=9 and b2=27
As we know that, foci of a horizontal hyperbola is given by: (0, - ae) and (0, ae)
First we have to find the value of e
As we know that, eccentricity of a hyperbola is given by: e=

√a2+b2

a

⇒e=

√9+27

3

=2
So, the foci of the given hyperbola are : (0, ± 6)
Hence, option D is the correct answer.