UPSC NDA Math Model Paper 6

Given: Equation of hyperbola is $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$ As we can see that, the given hyperbola is a vertical hyperbola. So, by comparing the given equation of hyperbola with $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$ we get, $\Rightarrow a^{2}=9$ and $b^{2}=27$ 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: $e=\frac{\sqrt{a^{2}+b^{2}}}{a}$ $\Rightarrow e=\frac{\sqrt{9+27}}{3}=2$ So, the foci of the given hyperbola are : (0, ± 6) Hence, option D is the correct answer.