The eccentricity of a hyperbola passing throu | vedclass

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Question

(b) $\frac{9}{{{a^2}}} = 1$

==> $a = 3$ and $\frac{{18}}{{{a^2}}} - \frac{4}{{{b^2}}} = 1$

==> ${b^2} = 4$

Therefore, $e = \sqrt {1 + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) $\frac{9}{{{a^2}}} = 1$

==> $a = 3$ and $\frac{{18}}{{{a^2}}} - \frac{4}{{{b^2}}} = 1$

==> ${b^2} = 4$

Therefore, $e = \sqrt {1 + \frac{4}{9}} = \frac{{\sqrt {13} }}{3}$.

The eccentricity of a hyperbola passing through the points $(3, 0)$, $(3\sqrt 2 ,\;2)$ will be

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