UPSC NDA Math Model Paper 2

Here, we have to find the equation of the hyperbola whose vertices are (0, ± 3) and the eccentricity is 4/3
By comparing the vertices (0, ± 3) with (0, ± a) we get
⇒ a = 3
As we know that, e=

√a2+b2

a

⇒a2e2=a2+b2
⇒16=9+b2
⇒b2=7
so, the equation of the hyperbola is

y2

9

−

x2

7

=1
Hence, option C is the correct answer.