Question

Which is the directrix of a parabola with equation x^2=y

Answer 1

`\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})} \\ \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ x^2=y=(x-0)^2=1(y-0)\implies (x-0)^2=4\stackrel{p}{\left( (1)/(4) \right)}(y-0)`

now, notice, the vertex is at h = 0, k = 0, or 0,0, at the origin, the squared variable is the "x", and thus is a vertical parabola. The leading term's coefficient is positive, so is going upwards, and the "p" distance is that much.

So the directrix is from the vertex, down "p" distance, check the picture below.