Question

If a light 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?

Stuck on this, any insight into how I can go about solving would be excellent. Thank you.

Answer 1

`f = 2/3" ft"`

Explanation:

Let the vertex be the point `(0,0)`.

The vertex form of the equation is:

`y = a(x-0)^2 + 0`

Simplifying:

`y = ax^2`

We know that `a = 1/(4f)` where `f` is the signed distance from the vertex to the focus.

`y = 1/(4f)x^2`

We have chosen the vertex so that the line `x =0` divides the parabola in half, therefore, if the beam is 4 feet wide 1.5 feet above the vertex, then the point `(2" ft",1.5" ft")` must be on the parabola:

`1.5" ft" = 1/(4f)(2" ft")^2`

`6" ft"f = 4" ft"^2`

`f = 2/3" ft"`