# Find the equation for the parabola...

• April 8th 2010, 05:43 PM
uselessjack
Find the equation for the parabola...
Find the equation for the parabola that has its focus on the positive x-axis, 4 units away from the directrix.
• April 8th 2010, 06:52 PM
Soroban
Hello, uselessjack!

There are four forms of the parabola, each with its own equation.

I'll derive one of them . . .

Quote:

Find the equation for the parabola that has its focus on the positive x-axis,
4 units away from the directrix.

This is a "vertical" parabola, opening upward.

The focus is at $F(f,0)$
The directrix is: $y = -4$
Code:

            |             |         *  |                  *             |      F     - - -*- + - - - o - - - - -* - -           * |    (f,0)      *             *      :      *             |  *    :    *             |      o V             |    (f,-2)             |      :             |      :       - - - + - - - + - - - - -           -4|             |

The equation has the form: . $(x-h)^2 \:=\:4p(y-k)$

The vertex is at: . $V(f,-2)$

The value of $p$ is: . $p \,=\,2$

The equation is: . $(x-f)^2 \:=\:4(2)\left(y-[-2]\right) \quad\Rightarrow\quad (x-f)^2 \:=\:8(y+2)$

• April 8th 2010, 07:01 PM
uselessjack
Thank you for your help and time. I am left with confusion still. I thought there would be a specific answer. I am given five choices and still can't figure out which one is correct.

4y^2 = 4x
y^2 = 4x
y^2 = 8x
x^2 = 16y
4y^2 = x