Eccentricity of hyperbola 4{x^2} - 9{y^2} = 16 , is

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10-2. Parabola, Ellipse, Hyperbola

easy

The eccentricity of the hyperbola $4{x^2} - 9{y^2} = 16$, is

A

$\frac{8}{3}$

B

$\frac{5}{4}$

C

$\frac{{\sqrt {13} }}{3}$

D

$\frac{4}{3}$

Solution

(c) Given equation of hyperbola, $\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{(16/9)}} = 1$,

$a = 2,\,\,b = \frac{4}{3}$.

As we know, ${b^2} = {a^2}({e^2} – 1)$

==>$\frac{{16}}{9} = 4({e^2} – 1)$

==> ${e^2} = \frac{{13}}{9}$,

$\therefore e = \frac{{\sqrt {13} }}{3}$.

Standard 11

Mathematics

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