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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The eccentricity of the hyperbola $9y^2 -4x^2=36$ is

$\begin{array}{1 1}(A)\;\frac{5}{2} \\(B)\;\frac{\sqrt 7}{2} \\(C)\;\sqrt \frac{13}{2} \\(D)\;2 \end{array}$

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1 Answer

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$\large\frac{y^2}{4}-\frac{x^2}{9}$$=1$
here the coefficient $y^2$ is +ve and that of $x^2$ is $-ve$
and hence it represents a hyperbola whose transverse axis is vertical
i.e $a^2=4$
$b^2=9$
$b^2=a^2(e^2-1)$
$\large\frac{9}{4}$$+1=e^2$
$e= \large\frac{\sqrt {13}}{2}$
Hence C is the correct answer.
answered Apr 21, 2014 by meena.p
 

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