The eccentricity of the hyperbola x^2 16 - y^2 25 = 1 is - Vidyadip

MCQ

The eccentricity of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{25}} = 1$ is

A
$3\over4$
B
$3\over5$
✓
$\sqrt {41} /4$
D
$\sqrt {41/5} $

✓

Answer

Correct option: C.

$\sqrt {41} /4$

c
(c) Equation of hyperbola is $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{{25}} = 1$

Eccentricity is ${e^2} = \frac{{{b^2}}}{{{a^2}}} + 1$

$i.e.$, ${e^2} = \frac{{25}}{{16}} + 1$

==> ${e^2} = \frac{{41}}{{16}}$

==> $e = \frac{{\sqrt {41} }}{4}$.

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