Gujarati
10-2. Parabola, Ellipse, Hyperbola
medium

The eccentricity of the ellipse $4{x^2} + 9{y^2} + 8x + 36y + 4 = 0$ is

A

$\frac{5}{6}$

B

$\frac{3}{5}$

C

$\frac{{\sqrt 2 }}{3}$

D

$\frac{{\sqrt 5 }}{3}$

Solution

(d) $4{x^2} + 8x + 4 + 9{y^2} + 36y + 36 = 36$

$ \Rightarrow \frac{{{{(x + 1)}^2}}}{9} + \frac{{{{(y + 2)}^2}}}{4} = 1$;

$e = \sqrt {1 – \frac{4}{9}} = \frac{{\sqrt 5 }}{3}$.

Standard 11
Mathematics

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