Gujarati
10-2. Parabola, Ellipse, Hyperbola
medium

The eccentricity of the ellipse $\frac{{{{(x - 1)}^2}}}{9} + \frac{{{{(y + 1)}^2}}}{{25}} = 1$ is

A

$4\over5$

B

$3\over5$

C

$5\over4$

D

Imaginary

Solution

(a) ${a^2} = {b^2}(1 – {e^2})$ ,  $\{ \because \,\,a < b\} $

$9 = 25(1 – {e^2})$

==> $\frac{9}{{25}} = 1 – {e^2}$

==> ${e^2} = \frac{{16}}{{25}}$

==> $e = \frac{4}{5}$.

Standard 11
Mathematics

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