What is the equation in standard form of the parabola with a focus at (3,0) and a directrix of x= -3?

1 Answer
Mar 29, 2017

Because the directrix is a vertical line, #x = -3# we know that the parabola is the type that opens to the left or right and the corresponding standard form is:

#x = ay^2 + by + c#

We know that the y coordinate of the focus and the y coordinate of the vertex, k, are the same:

#k =0#

We can use the equation, #k = -b/(2a)# to find the value of b:

#0 = -b/(2a)#

#b = 0#

The equation becomes:

#x = ay^2+c#

We know that the x coordinate of the vertex is halfway between the x coordinate of the directrix and the x coordinate of the focus:

#h = (3 + -3)/2#

#h = 0#

We can find the value of c by evaluating the equation at the vertex #(0,0)#:

#0 = a0^2+c#

#c = 0#

The equation becomes:

#x = ay^2#

Because the focus is to the right of the directrix, we know that the parabola opens to the right and, therefore, "a" is positive. To find the value of "a", we can use the equation:

a = 1/(4f)

where f is the horizontal distance from the vertex to focus:

#f = 3 - 0#

#f = 3#

#a = 1/(4(3))#

#a = 1/12#

The equation of the specified parabola is:

#x = 1/12y^2#