$25 x^{2} + 16 y^{2} = 400$ . Find the directrices.

y = + \frac{25}{3}

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y = - \frac{25}{3}

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y = \pm \frac{25}{3}

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None of these

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Solution

The correct option is B $y = \pm \frac{25}{3}$
$25 x^{2} + 16 y^{2} = 400 O r, \frac{x^{2}}{16} + \frac{y^{2}}{25} = 1 e = \sqrt{1 - \frac{16}{25}} O r, e = \sqrt{\frac{9}{25}} O r, e = \frac{3}{5} E q u a t i o n o f d i r c t r i x i s x = \pm \frac{b}{e} ∴ E q u a t i o n i s : x = \pm \frac{25}{3}$

Option [C]

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Similar questions

The equation of the directrices of the ellipse $16 x^{2} + 25 y^{2} = 400$ are

25 x^{2} - 16 y^{2} = 400

x = l o g_{3} 5, y = l o g_{17} 25

x = l o g_{3} 5, y = l o g_{17} 25

5 x y + 25 y + 3 x + 15

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