Eccentricity of ellipse 25{x^2} + 16{y^2} = 100 , is

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10-2. Parabola, Ellipse, Hyperbola

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The eccentricity of the ellipse $25{x^2} + 16{y^2} = 100$, is

A

$\frac{5}{{14}}$

B

$\frac{4}{5}$

C

$\frac{3}{5}$

D

$\frac{2}{5}$

Solution

(c) $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{(25/4)}} = 1$.

Here, $a = 2,\,\,b = 5/2$

$\therefore b > a$, therefore ${a^2} = {b^2}(1 – {e^2})$

==> $4 = \frac{{25}}{4}(1 – {e^2})$

==> $\frac{{16}}{{25}} = 1 – {e^2}$ ==> ${e^2} = 1 – \frac{{16}}{{25}} = \frac{9}{{25}}$,

$e = \frac{3}{5}$ .

Standard 11

Mathematics

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