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Finding Focus and Directrix


Date: 04/12/2001 at 09:23:35
From: Brent 
Subject: Parabola

I'm having a very hard time understanding how to find the focus and 
directrix when given a formula for a parabola. I understand how to 
identify the vertex and axis of symmetry.

Can you help, please?


Date: 04/14/2001 at 14:06:28
From: Doctor Jaffee
Subject: Re: Parabola

Hi Brent,

The key to unlocking the mystery is getting the equation into the 
appropriate form, namely (x - h)^2 = 4p(y - k), where (h, k) is the 
vertex of the parabola and, consequently, x = h will be the line of 
symmetry.

If p is positive, then the focus will be p units above the vertex, and 
the directrix will be the line perpendicular to the axis of symmetry 
and passing through a point p units below the vertex.

If p is negative, then just have the words "above" and "below" trade 
places in the previous paragraph.

For example, suppose you have the equation:

     x^2 - 6x - 8y + 1 = 0

Keep x^2 - 6x on the left side, but add 8y - 1 to both sides. You now 
have:

              x^2 - 6x = 8y - 1

Add 9 to both sides (to complete the square) and get:

          x^2 - 6x + 9 = 8y + 8

Factor on each side:

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

and you should see that you have a parabola whose vertex is at the 
point (3, -1), whose axis of symmetry is the line x = 3, and whose 
focus is at the point 2 units above the vertex, (3, 1). The directrix 
is the line that is perpendicular to the axis of symmetry and passes 
through a point 2 units below the vertex. The equation of the line is 
y = -3.

I hope this explanation is helpful. Write back if it needs 
clarification or if you have a specific problem that needs attention.

- Doctor Jaffee, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations

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