Question

Find coordinates of the foci, of the ellipse 4x2 +9y2 = 36.​

Answer 1

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Answer 2

$4 {x}^{2} + 9 {y}^{2} = 36 \\ Divide \: \: equations \: \: by \: \: 36 \\ \frac{4 {x}^{2} }{36} + \frac{9 {y}^{2} }{36} = \frac{36}{36} \\ \frac{ {x}^{2} }{9} + \frac{ {y}^{2} }{4} = 1⟹⓵ \\ \frac{ {x}^{2} }{ {3}^{2} } + \frac{ {y}^{2} }{ {2}^{2} } = 1 \\ \\ Since, \: \: 9>4 \\ Hence \: \: the \: \: above \: \: equation \\ is \: \: of \: \: the \: \: form : \\ \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1⟹ ⓶\\ \\ Comparing \: \: ⓵ \: \: and \: \: ⓶ \\ {a}^{2} = 9 \\ a = \sqrt{9} \\ a = 3 \\ \\ {b}^{2} = 4 \\ b = \sqrt{4} \\ b = 2 \\ \\ We \: \: know \: \: that \\ c = \sqrt{ {a}^{2} - {b}^{2} } \\ = \sqrt{9 - 4} \\ = \sqrt{5} \\ \\ Coordinates \: \: of \: \: foci \: \: are : \\ ( \sqrt{5} , \: \: 0) \: \: and \: \: ( - \sqrt{5} , \: \: 0)$