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Find the foci of the ellipse \(\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1\) ?
1. (0, - 4√2)
2. (0, 4√2)
3. None of these
4. Both (0, - 4√2) and (0, 4√2)
5. (0, - 4)

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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.

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