Question

Find the equation of a parabola with vertex at the origin and focus (0, –7). Group of answer choices y equals 1 over 28 x squaredx2 = 128y y equals negative 1 over 28 x squared x equals 1 over 28 y squared x equals negative 1 over 28 y squared

Answer 1

Check the picture below, so the parabola looks more or less like that one, opening downwards and thus with a negative "p" distance.

`\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{p~is~negative}{op ens~\cap}\qquad \stackrel{p~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill`

`\begin{cases} h=0\\ k=0\\ p=-7 \end{cases}\implies 4(-7)(~~y-0~~) = (~~x-0~~)^2 \\\\\\ -28y=x^2\implies {\Large \begin{array}{llll} y=-\cfrac{1}{28}x^2 \end{array}}`