Gujarati
10-2. Parabola, Ellipse, Hyperbola
easy

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

A

$2x = \pm 25$

B

$5x = \pm 9$

C

$3x = \pm 10$

D

None of these

Solution

(d) $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1$

$e = \sqrt {1 – \frac{{16}}{{25}}} = \frac{3}{5}$

Therefore, directrices are $x \pm \frac{5}{{3/5}} = 0$ or $3x \pm 25 = 0$.

Standard 11
Mathematics

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