UPSC NDA Math Model Paper 3

Given: The equation of ellipse is $4 x^{2}+9 y^{2}+16 x+18 y$ $-11=0$ The given equation can be re-written as $\frac{(x+2)^{2}}{9}+\frac{(y+1)^{2}}{4}=1$ ......(1) As we can see that, the given ellipse is a horizontal ellipse So, by comparing equation (1) with respect to $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$ we get, ⇒ h = - 2, k = - 1, a = 3 and b = 2 As we know that, eccentricity of an ellipse is given by $e=\frac{\sqrt{a^{2}-b^{2}}}{a}$ $\Rightarrow e=\frac{\sqrt{3^{2}-2^{2}}}{3}=\frac{\sqrt{5}}{3}$ So, $a \cdot e = \sqrt{5}$ As we know that, foci of a horizontal ellipse are given by: (h ± a, k) So, the required foci of the given ellipse are: (- 2 ± √5, - 1) Hence, option B is the correct answer.