NDA Math Conic Section

CONCEPT: The eccentricity of a hyperbola of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \; \;$ is given by: $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$ CALCULATION: Given: The point $(3 \tan \theta, 2 \sec \theta$ ) lies on the hyperbola. As we can see that, if we substitute$x = 3 \tan \theta$ and $y = 2 \sec \theta$ in the expression $\frac{y^{2}}{4} - \frac{x^{2}}{9}$ we get $\Rightarrow \frac{y^{2}}{4} - \frac{x^{2}}{9} = \sec^{2} \theta - \tan^{2} \theta = 1$ So, we can say that, the point $(3 \tan \theta, 2 \sec \theta$ ) lies on the hyperbola $\frac{y^{2}}{4} - \frac{x^{2}}{9} = 1$ As we know that, eccentricity of the hyperbola of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$ is given by: $e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$ Here, $a^{2} = 4$ and $b^{2} = 9$ $\Rightarrow e = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}$ Hence, correctoption is 4 .