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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