# Parabola, vertex-focus distance

• Aug 16th 2012, 04:21 PM
Greymalkin
Parabola, vertex-focus distance
A spotlight has a parabolic cross section that is 4 ft wide at the opening and 1.5 ft deep at the vertex, how far is the vertex from the focus?

I'm quite confused as my textbook gives no illustration, im assuming that the light is 4ft wide and 1.5ft deep from the opening/plexiglass.
Saying the light is on a cartesian coordinate system with vertex (0,0) gives me no clue as to how this question is solved.

Answer is 2/3, how did you get such an answer??
• Aug 16th 2012, 04:55 PM
Soroban
Re: Parabola, vertex-focus distance
Hello, Greymalkin!

You are expected to know this form of a parabola: .$\displaystyle x^2 \:=\:4py$
. . where $\displaystyle p$ is the distance from the vertex to the focus.

Quote:

A spotlight has a parabolic cross section that is 4 ft wide at the opening and 1.5 ft deep at the vertex.
How far is the vertex from the focus?

Code:

                |  (-2,1.5)      |        (2,1.5)     * - - - - - + - - - - - *                 |     *          |          *       *        |        *       *        |        *         *      |      *             *  |  *   --------------*--------------                 |
We have the equation: .$\displaystyle x^2 \:=\:4py$

We are told that when $\displaystyle x = 2,\:y =\tfrac{3}{2}$

Hence: .$\displaystyle 2^2 \:=\:4p\left(\tfrac{3}{2}\right) \quad\Rightarrow\quad 4 \:=\:6p$

Therefore: .$\displaystyle p \,=\,\tfrac{2}{3}$