Question 80216
<pre><font size = 3><b>
A field Microphone has a parabolic cross section and is 18 inches deep.
The focus is 4 inches from the vertex.  Find the width of the microphone.


Let's consider its cross-section to be a parabola with vertex at the 
origin and opening to the right.  Then its equation is

                   y² = 4px

where p is the distance from the vertex to the focus.  This is given
to be p = 4. So the equation is

                   y² = 4(4)x

or

                   y² = 16x

which has this graph:

{{{drawing(371,600,-2,19,-17,17, 
               
                locate(3.85,.45,O),
                locate(3.85,.45,X),
                locate(3.1,1.5,F(4,0)),
                line (18,-17,18,17),

graph(371,600,-2, 19, -17,17,4*sqrt(x)*sqrt(18-x)/sqrt(18-x), -4*sqrt(x))*sqrt(18-x)/sqrt(18-x) )}}}

To find the top and bottom points of the microphone,
we substitute 18 for x in the equation:

                   y² = 16x

                   y² = 16(18)
                          ___
                    y = ±<font face = "symbol">Ö</font>288
                          _____
                    y = ±<font face = "symbol">Ö</font>2·144
                            _
                    y = ±12<font face = "symbol">Ö</font>2
                           _
So the top point is (18,12<font face = "symbol">Ö</font>2) and the bottom point
           _  
is (18,-12<font face = "symbol">Ö</font>2)
                                     _
So the width of the microphone is 24<font face = "symbol">Ö</font>2 or about 33.94 inches.

Edwin</pre>