Question 1185514
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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 answer to 2 decimal places.
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<pre>
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 = {{{ax^2}}}.      (1)


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


    4 = {{{a*5^2}}},   or   a = {{{4/25}}},


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


    y = {{{(4/25)*x^2}}}.     (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 = {{{(1/(4p))*x^2}}},  then the focal distance is p.


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


    y = {{{(4/25)x^2}}} = {{{(1/(25/4))*x^2}}} = {{{(1/(4*(25/16)))*x^2}}},


and we see that the parameter p is equal to  p = {{{25/16}}} = 1.5625 ft = 1.56 ft (rounded to two decimal places).


<U>ANSWER</U>.  The receiver should be 1.56 ft from the vertex  (the precise distance is  1.5625 ft).
</pre>

Solved.