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The curve $y^{2}(x-2)=x^{2}(1+x)$ has

\[\](1)an asymptote parallel to $x$-axis\[\](2)an asymptote parallel to $y$-axis\[\](3)asymptotes parallel to both axes\[\](4)no asymptotes

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$y^2 (x-2) =x^2(1+x)$
$y^2 =\large\frac{x^2(1+x)}{x-2}$
When $x=2\;y = \infty$
$x=2$ is a asymptote to the curve which is parallel to y axis
Hence 2 is the correct answer
answered May 20, 2014 by meena.p
 

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