Question 756237
The way the focus and directrix are positioned, this parabola has a minimum.  The vertex is directly in the exact middle of focus and directrix, so vertex is (0,0).  Without going to derivation process for a parabola equation, you can use {{{x^2=4py}}}, where p is the focal length, same as distance from vertex to the closest point on the directrix.


In your question, p=5, and since the parabola has a minimum, the coefficient on {{{x^2}}} is positive (when written in the form y=something...).


{{{4py=x^2}}}
{{{y=(1/(4p))x^2}}}
and knowing p=5
{{{y=(1/(4*5))x^2}}}
{{{highlight(y=(1/20)x^2)}}}


{{{graph(300,300,-15,15,-15,15,(1/20)x^2)}}}