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The eccentricity of the hyperbola with asymptotes $x+2y-5=0 ,2x-y+5=0 $ is

\[\begin{array}{1 1}(1)3&(2)\sqrt{2}\\(3)\sqrt{3}&(4)2\end{array}\]

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The equation of the asymptotes are $x+2y-5=0$ and $2x-y+5=0$
The asymptotes are perpendicular to each other .
$\therefore $ The hyperbola is rectangular hyperbola.
$\therefore e= \sqrt 2$
Hence 2 is the correct answer.
answered May 16, 2014 by meena.p
 

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