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10-2. Parabola, Ellipse, Hyperbola
easy
The eccentricity of the conjugate hyperbola of the hyperbola ${x^2} - 3{y^2} = 1$, is
A
$2$
B
$\frac{2}{{\sqrt 3 }}$
C
$4$
D
$\frac{4}{3}$
Solution
(a) Eccentricity of $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ is $e = \sqrt {\frac{{{a^2} + {b^2}}}{{{a^2}}}} $
Eccentricity of conjugate hyperbola, $e' = \sqrt {\frac{{{a^2} + {b^2}}}{{{b^2}}}} $
Write the given equation in standard form,
$\frac{{{x^2}}}{1} – \frac{{{y^2}}}{{1/3}} = 1$
==> ${a^2} = 1,\,\,{b^2} = \frac{1}{3}$
$e' = \sqrt {\frac{{1 + 1/3}}{{1/3}}} = \sqrt 4 = 2$.
Standard 11
Mathematics
