Eccentricity of hyperbola conjugate to {x^2} - 3{y^2} = 2x + 8 is

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10-2. Parabola, Ellipse, Hyperbola

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The eccentricity of the hyperbola conjugate to ${x^2} - 3{y^2} = 2x + 8$ is

A

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

B

$\sqrt 3 $

C

$2$

D

None of these

Solution

(c) Given, equation of hyperbola is ${x^2} – 3{y^2} = 2x + 8$

==> ${x^2} – 2x – 3{y^2} = 8$

==> ${(x – 1)^2} – 3{y^2} = 9$

==> $\frac{{{{(x – 1)}^2}}}{9} – \frac{{{y^2}}}{3} = 1$

Conjugate of this hyperbola is $ – \frac{{{{(x – 1)}^2}}}{9} + \frac{{{y^2}}}{3} = 1$

and its eccentricity $(e) = \sqrt {\left( {\frac{{{a^2} + {b^2}}}{{{b^2}}}} \right)} $

Here, ${a^2} = 9$, ${b^2} = 3$;

$\therefore $ $e = \sqrt {\frac{{9 + 3}}{3}} = 2$.

Standard 11

Mathematics

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