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The vertex of the parabola $ x^{2}=8y-1$ is

\[\begin{array}{1 1}(1)\bigg(-\frac{1}{8},0\bigg)&(2)\bigg(\frac{1}{8},0\bigg)\\(3)\bigg(0,\frac{1}{8}\bigg)&(4)\bigg(0,-\frac{1}{8}\bigg)\end{array}\]

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1 Answer

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$x^2=8y-1$
$(x-0)^2=8 (y -\frac{1}{8})$
$=>\bigg(0,\frac{1}{8}\bigg)$
Hence 3 is the correct answer.
answered May 15, 2014 by meena.p
 
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