Question

write an equation for a prabola with a vertex at the origin, passing through(5,6), and symmetric with respect to the x-axis

Answer 1

a parabola with symmetry or mirror image across the x-axis, will be a horizontal parabola, namely one that opens either to the right or left.

a horizontal parabola will be a parabola whose independent variable is the "y", so it'll be an x(y) type of equation.

$~~~~~~\textit{horizontal parabola vertex form} \\\\ x=a(y- k)^2+ h\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\supset}\qquad \stackrel{"a"~is~positive}{op ens~\subset} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=0\\ k=0 \end{cases}\implies x=a(y-0)^2+0\qquad \textit{we also know that} \begin{cases} x=5\\ y=6 \end{cases} \\\\\\ 5=a(6-0)^2 +0\implies 5=36a\implies \cfrac{5}{36}=a~\hfill \boxed{x=\cfrac{5}{36}y^2}$