Correct Answer - Option 4 : Both (0, - 4√2) and (0, 4√2)
CONCEPT:
The properties of a vertical ellipse \(\frac{{{x^2}}}{b^2} + \frac{{{y^2}}}{a^2} = 1\) where 0 < b < a are as follows:
- Centre of ellipse is (0, 0)
- Vertices of ellipse are: (0, - a) and (0, a)
- Foci of ellipse are: (0, - ae) and (0, ae)
- Length of major axis is 2a
- Length of minor axis is 2b
- Eccentricity of ellipse is given by: \(e = \frac{{\sqrt {{a^2} - {b^2}} }}{a}\)
CALCULATION:
Given: Equation of ellipse is \(\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1\)
As we can see that the given ellipse is a vertical ellipse.
By comparing the given equation of ellipse with \(\frac{{{x^2}}}{b^2} + \frac{{{y^2}}}{a^2} = 1\) where 0 < b < a we get,
⇒ a = 6, b = 2
As we know that, eccentricity of ellipse is given by: \(e = \frac{{\sqrt {{a^2} - {b^2}} }}{a}\)
⇒ \(e = \frac{{\sqrt {{6^2} - {2^2}} }}{6} = \frac{2\sqrt 2}{3}\)
⇒ ae = 4√2
As we know that, foci of a vertical ellipse are: (0, - ae) and (0, ae)
So, the foci of the given ellipse are: (0, - 4√2) and (0, 4√2)
Hence, option D is the correct answer.