Parabola with Horizontal or Vertical AxisDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/