Find an equation of the parabola with focus (4,0) and directrix x = 0

Using the standrard form for a parabola:


y-y_1=\frac1{4p} (x-x_1)^2


the vertex (x_1,y_1) is located midway between the focus and directrix, and the value 'p' is the distance from vertex to either the focus or directix. However - this gives a parabola that is "concave up" and whose directrix is al line of the form y = a. But since you have been told that the directrix is x =0, this tells you that the parabola is lying on its side, and the equation is of the form:


x-x_1=\frac1{4p} (y-y_1)^2


With this you have information to finish the problem.

Using the standard form for a parabola:


y-y_1=\frac1{4p} (x-x_1)^2


the vertex (x_1,y_1) is located midway between the focus and directrix, and the value 'p' is the distance from vertex to either the focus or directix. However - this gives a parabola that is "concave up" and whose directrix is al line of teh form y = a. But since yuo have been told that the directrix is x =0, this tells you that the parabola is lieing on its side, and the equation is of the form:


x-x_1=\frac1{4p} (y-y_1)^2


With this you have information to finish the problem.

How am I going to do it? I really don't know how. Please help me. Thanks!

I'll show you a slightly different example, then you can apply this techique to your homework problem.

Suppose you were told that the focus is at (3,2) and the directrix is the line x=1. First step is to find the vertex, which is midway between the point (3,2) and the line x=1; that point is (2,2). I got this by noting that the y_1 component is equal to the y coordinate of the focus, and the x_1 component is halfway between the x value of the directrix and the x component of the focus (to understand this better It helps to draw a graph of the focus and directrix). Next for the value 'p': the distance from the focus to the vertex is the difference between their x-coordinates, which is 3-2 = 1. Hence 'p'=1, and the equation of this parabola is:


x-2=\frac1{4} (y-2)^2


Now try to apply this same technique to your problem.