Question

How do i find the focus of the parabola from the equation x^2-12y=0?

Answer 1

`\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \boxed{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\ x^2-12y=0\implies x^2=12y\implies (x-0)^2=12(y-0) \\\\\\ (x-\stackrel{h}{0})^2=4(\stackrel{p}{3})(y-\stackrel{k}{0})`

so, first off, notice, the squared variable is the "x", meaning is a vertical parabola.

the coefficient of the x² is a positive value, thus the vertical parabola is opening upwards.

the vertex is at 0,0, namely the origin.

the "p" distance is 3.

so the focus point is 3 units above the vertex, and you surely know where that is.