Answer
$x^2 =-\dfrac{25}{4} (y-48)$
Work Step by Step
We need to write the Standard form of a parabola with a vertical axis as follows:
$(x-h)^2 =4p (y-k)$
Now, $(x-0)^2 =4p (y-48)$
$h=0$ and $k=58$
Vertex: $(0, 48)$
Points are: $(10 \sqrt 3,0)$
$(10\sqrt 3-0)^2 =4p (0-48) \implies p=-\dfrac{25}{16}$
Back substitution:
$(x-0)^2 =4(-\dfrac{25}{16}) (y-48)$
Our answer is: $x^2 =-\dfrac{25}{4} (y-48)$
