Parabola with Horizontal or Vertical Axis

Date: 05/03/2001 at 09:53:49
From: Jamie
Subject: Finding the focus and directrix of 2x^2 = -y

I know how to find the focus and directrix of regular problems, but I 
just have no clue on how to do this problem:

     2x^2 = -y

Please help!

Date: 05/03/2001 at 11:27:48
From: Doctor Rob
Subject: Re: Finding the focus and directrix of 2x^2 = -y

Thanks for writing to Ask Dr. Math, Jamie.

Here is the process to find the focus and directrix when the axis of 
the parabola is either horizontal or vertical.

Find the variable that is squared. Complete the square on it. Solve 
for that square. Absorb the constant term into the linear part. The 
result should have one of the two forms:

     (x-h)^2 = 4*p*(y-k)   (axis vertical)
     (y-k)^2 = 4*p*(x-h)   (axis horizontal)

Then the vertex is (h,k) in each case. In the first case the focus is 
(h,k+p), the axis is x = h, and the directrix is y = k - p. In the 
second case, the focus is (h+p,k), the axis is y = k, and the 
directrix is x = h - p.

Example:

                   y^2 - y - 11 = 3*x

     (y^2 - y + 1/4) - 1/4 - 11 = 3*x

               (y-1/2)^2 - 45/4 = 3*x

                      (y-1/2)^2 = 3*x + 45/4

                      (y-1/2)^2 = 3*(x+15/4)

Then we are in the second case, with horizontal axis. We read off 
(h,k) = (-15/4,1/2), the vertex, and 4*p = 3, so p = 3/4. That tells 
us that the focus is (-15/4+3/4,1/2) = (-3,1/2), the axis is y = 1/2, 
and the directrix is the line x = -15/4 - 3/4, or x = -9/2.

In your problem, complete the square on x (already done), and solve 
for the square:

     x^2 = (-1/2)*y

Absorb the constant term into the linear part (done). See that the 
first form is the right one, with vertical axis. Read off h = 0, 
k = 0, and 4*p = -1/2, so p = -1/8. Use these values to write down the 
focus and directrix.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/