Question 1091792
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A satellite dish has a shape of paraboloid. The signals that it receives is reflected to the receiver that is located 
at the focus of the paraboloid. If dish is 8 feet across at its opening and 1 foot deep at its vertex, determine the location 
(distance from the vertex of the dish) of its focus.
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<pre>
Let us represent the parabola by the equation y = {{{ax^2}}}  (1).

Then this phrase " . . . dish is 8 feet across at its opening and 1 foot deep at its vertex" means that
the point (x,y) = (4,1) lies on the parabola.

In turn, it then  means that  y = 1 at  x = 4.

It implies that in the equation (1)   1 = {{{a*4^2}}},  or  a = {{{1/16}}}.


So, the equation of the parabola is y = {{{(1/16)*x^2}}}.


It is the <U>canonical equation</U> of the parabola

    (see the lesson <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Parabola-definition--canonical-equation--characteristic-points-and-elements.lesson>Parabola definition, canonical equation, characteristic points and elements</A>  in this site).


The vertex of this parabola is the point  (0,0)  (the origin of the coordinate system).

The focus of this parabola is the point  ( 0, {{{(1/2)*(16/2)}}} ) = (0,4).

    (See again the lesson referred above).


Thus the distance from the vertex to the focus is 4 feet.
</pre>

<U>Answer</U>.  The distance from the vertex to the focus of the parabola / paraboloid is 4 feet.



Solved.



The solution by @josgarithmetic and his answer "1 feet" is {{{highlight(WRONG)}}}.



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You have this free of charge online textbook in ALGEBRA-II in this site

&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic 
"<U>Conic sections: Parabolas. Definition, major elements and properties. Solved problems</U>".