Finding Focus and DirectrixDate: 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/ |
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