SOLUTION: A satellite dish in the shape of a paraboloid is 10 ft across, and 4 ft deep at its vertex. How far is the receiver from the vertex, if it is placed at the focus? Round off your an

In plane section of the paraboloid, we have a parabola.

We consider this parabola as opened upward with the vertex at the origin of the coordinate system.


About this parabola, we know that the y-value is 4 ft, when the x-value is 5 ft. 


We want to restore the parabola equation, so we write


    y = .      (1)


To find the coefficient "a", we substitute y= 4 and x= 5 into this equation. We get then


    4 = ,   or   a = ,


so the equation (1)  of the parabola takes the form  


    y = .     (2)


     +----------------------------------------------------------------------------------+
     |                     Half of the problem is just solved.                          |
     |                                                                                  |
     |   Now we should find the focal distance of the parabola, given by equation (2).  |
     +----------------------------------------------------------------------------------+


Generally, if the parabola has equation  y = ,  then the focal distance is p.


So, we should write equation (2) in this form.  For it, we write


    y =  =  = ,


and we see that the parameter p is equal to  p =  = 1.5625 ft = 1.56 ft (rounded to two decimal places).


ANSWER.  The receiver should be 1.56 ft from the vertex  (the precise distance is  1.5625 ft).