Hello, magentarita!
A searchlight is shaped like a paraboloid of revolution.
The light source is located 2 feet from the base along the axis of symmetry
and the depth of the searchlight is 4 feet.
What should the width of the opening be? Code:
4
*      +      *

*  *
* 2* *
*  *
*  *
       *       

We are expected to know this equation: .$\displaystyle x^2 \:=\:4py$
. . where $\displaystyle p$ is the distance from the focus to the vertex.
Since $\displaystyle p = 2$, the equation is: .$\displaystyle x^2 \:=\:8y$
When $\displaystyle y = 4\!:\;\;x^2 \:=\:32\quad\Rightarrow\quad x \:=\:\pm4\sqrt{2}$
Therefore, the width of the opening is $\displaystyle 8\sqrt{2}$ feet.