Question 1135980
given:

a focus at ({{{7}}},{{{ 0}}}) and a directrix at {{{x = -7}}}

since the vertex, exactly between the focus and directrix, must be at ({{{h}}}, {{{k}}}) = ({{{0}}},{{{ 0}}})

the absolute value of {{{p }}}is the distance between the vertex and the focus and the distance between the vertex and the directrix. (The sign on {{{p}}} tells me which way the parabola faces.) Since the focus and directrix are {{{7}}} units apart, then this distance has to be one unit, so {{{abs(p ) =7}}}

Since this is a "sideway" parabola, then the {{{y}}} part gets squared, rather than the {{{x}}} part. So the conics form of the equation must be:

{{{(y- y[1])^2 = 4p(x -x[1])}}}...plug in {{{p =7}}} and coordinates of the vertex  ({{{0}}},{{{ 0}}})

{{{(y- 0)^2 = 4(7)(x -0)}}}

 {{{y^2 = 28x }}}

{{{x=(1/28)y^2 }}}=> your answer


{{{drawing( 600,600, -10, 10, -10, 10,
circle(7,0,.12),locate(7,0.5,F(7,0)),line(-7,-10,-7,10),
 graph( 600,600, -10, 10, -10, 10,sqrt(28x),-sqrt(28x))) }}}