# How to find p in a parabola?

Rewrite the equation for the parabola in standard form (x-h)^2=a(y-k) or (y-k)^2=a(x-h). Then, p=a/4.

Parabolas can be thought of as the graph of a quadratic polynomial. Geometrically, however, they are the locus of points equidistant from a fixed point (called the focus) and a fixed line (called the directrix). If you draw the segment from the focus perpendicular to the directrix, the length of...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Parabolas can be thought of as the graph of a quadratic polynomial. Geometrically, however, they are the locus of points equidistant from a fixed point (called the focus) and a fixed line (called the directrix). If you draw the segment from the focus perpendicular to the directrix, the length of the segment is 2p, and the point on the parabola intersecting this segment is p units from both the focus and directrix.

Consider the parabola whose directrix is parallel to one of the axes. (Rotations of the axes are handled with trigonometry.) Then the standard form for a parabola opening up or down is `(x-h)^2=4p(y-k)` and for a parabola opening left or right it is `(y-k)^2=4p(x-h).`

The key to finding p is to rewrite the equation in standard form. (If you are given the graph as opposed to an equation, find the midpoint of the segment joining the focus and the directrix; divide this length by two to get p.)

For example:

`y^2-6y-4x+1=0` Get the variables on opposite sides of the equation:

`y^2-6y=4x-1` Use completing the square on the left side:

`y^2-6y+9=4x-1+9`

`(y-3)^2=4x+8`

`(y-3)^2=4(x+2)`

This is a parabola opening to the right with vertex (-2,3), focus (-1,3), directrix x=-3, where p=1.