Question 1198212

Since focus is ({{{8.6}}},{{{0}}}) and vertex at the origin, an equation for the parabola of the form:

{{{y^2 = 4ax}}} where {{{a =8.6}}}

{{{y^2 = 4*8.6x}}}

{{{y^2 = 34.4x}}}


{{{ drawing( 600, 600, -10, 15, -10, 10,
circle(8.6,0,.12),locate(8.6,0.5,F(8.6,0)),

graph( 600, 600, -10, 15, -10, 10, sqrt(34.4x), -sqrt(34.4x))) }}}