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

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

$x^2=8y-1$
$(x-0)^2=8 (y -\frac{1}{8})$
$=>\bigg(0,\frac{1}{8}\bigg)$
Hence 3 is the correct answer.