Write the standard form equation of the parabola that passes through (48,−12), if the vertex at the origin and the axis of symmetry is horizontal.

Answer 1

This will be a parabola sitting on its side opening to the right. Draw a general sketch of the parabola with the information given to see this so equation is x= a(y-k)^2 +h

hand k are the coordinates of the vertex and are both zero. Solving for a we have

48=a(_12)^2 + 0

48= 144a a=1/3 so equation is

x=(1/3)y^2