UPSC NDA Math Model Paper 3

Given: The equation of ellipse is 4x2+9y2+16x+18y −11=0
The given equation can be re-written as

(x+2)2

9

+

(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

(x−h)2

a2

+

(y−k)2

b2

=1 we get,
⇒ h = - 2, k = - 1, a = 3 and b = 2
As we know that, eccentricity of an ellipse is given by e=

√a2−b2

a

⇒e=

√32−22

3

=

√5

3

So, a⋅e=√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.