Expert: Steve Holleran - 4/12/2008

Question
OK trigonometric word problem for ya.  
A light source is to be placed on the axis of symmetry of the parabolic reflector shown (it looks exactly like the nose of an aircraft, with a length of 6" and width at the widest of 6", with the axis of symmetry traveling from the vertex at the nose straight back through the opening) How far to the right of the vertex point should the light source be located if the designer wishes the reflected light rays to form a beam of parallel rays?

Answer
Hi Rich,

I hope I've got the figure visualized correctly.

Let's orient the parabola with vertex at the origin, and opening into the positive x's, horizontally.

If its 6 in long and 6 in wide at its widest,then the points on the vertical line through the parabola are (6,3) and (6, -3).  These points have to fit the equation for the parabola, which in this case is

                             x = 1/4p * y^2

where p is the distance from the vertex to the  focus.

So, lets use (6,3):           6 = 1/4p * 3^2

                             6 = 1/4p * 9

                            4p = 1/6 * 9 = 3/2

                             p = 3/2 * 1/4 = 3/8.

That's where the light source belongs, at the focus of the  parabola.

Hope this helps.
Steve