1. ## Parabola directrix

Could someone help me with these two types of problems?

1) Find the directrix of the parabola (y-7)[squared]=4(x+3).

2) Find the equation of the parabola whose focus is (2,3) and the directrix is x=12.

Thanks,
Gretch

2. Originally Posted by gretchen
Could someone help me with these two types of problems?

1) Find the directrix of the parabola (y-7)[squared]=4(x+3).

2) Find the equation of the parabola whose focus is (2,3) and the directrix is x=12.

Thanks,
Gretch
Here is one way.

The directrix of a parabola is "p" away from the vertex along the axis of symmetry. [Also, the focus of a parabola is "p" away from the vertex.]

1) The directrix of parabola y^2 = 4(x+3).

The y is squared, so the parabola is horizontal, or the axis of symmetry is horizontal.
A standard form of the equation of a horizontal parabola, in terms of "p", is
(x- h) = [1/(4p)](y -k)^2 -------------(1)
where
(h,k) is the vertex
p is vertex to focus, or vertex to direcrix along the axis of symmetry.
If the 1/(4p) is negative, then the parabola opens to the left.
If the 1/(4p) is positive, then the parabola opens to the right.
So,
y^2 = 4(x+3) -----------------------(i)
Convert that into the form of (1),
(x+3) = [1/(4*1)](y -0)^2 ---------------(ii)
Therfore,
vertex is (-3,0)
p = 1 unit.

The 1/(4p) is 1/(4*1) is positive, so the directrix is to the left of the vertex.
The vertex (-3,0) is on the verical line x = -3. One unit to the left of that is x = -4.

Therefore, the directrix is the vertical line x = -4. -----------answer.

===========================================
2) Find the equation of the parabola whose focus is (2,3) and the directrix is x=12.

The vertex is halfway of the distance between the focus and the directrix along the axis of symmetry.

The directrix x = 12 is a vertical line, therefore the parabola is horizontal. The axis of symmetry is the horizontal line y = 3 because the focus is always on the axis of symmetry.

The focus (2,3) is on the vertical line x = 2.
So distance from focus to directrix along the axis of symmetry is 12 -2 = 10 units.
Hence the "p" is 10/2 = 5 units.
And so the vertex is (2+5,3) = (7,3)

The focus is at (2,3) and the directrix is the vertical line x=12. Therefore, the parabola opens to the left. Hence the 1/(4p) is negative.

Then, the parabola is:
(x -h) = -[1/(4p)](y -k)^2
(x -7) = -[1/(4*5)](y -3)^2
x -7 = -(1/20)(y -3)^2 --------------------answer.

3. Thanks for your help. I see how you did the first one and I will try the others on my own.

On the 2nd problem, I ended up with -20(x-7)=(y-3)[squared] which is what you came up with. Thanks.