SOLUTION: Please answer this question, I tried but I don't understand it. Suppose a designer of a 10 ft. parabolic solar cooker wants to place the cooking pot 5 ft. above the vertex. For

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Please answer this question, I tried but I don't understand it. Suppose a designer of a 10 ft. parabolic solar cooker wants to place the cooking pot 5 ft. above the vertex. For       Log On


   



Question 1167899: Please answer this question, I tried but I don't understand it.
Suppose a designer of a 10 ft. parabolic solar cooker wants to place the cooking pot 5 ft. above the vertex. For reference, the first considers a parabolic string with a base 10 ft. and a focus at 5 ft.from the vertex. How deep is the parabolic solar cooker?
Thank you for your help!

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
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Please answer this question, I tried but I don't understand it.
Suppose a designer of a 10 ft. parabolic solar cooker wants to place the cooking pot 5 ft. above the vertex.
For reference, the first considers a parabolic string with a base 10 ft. and a focus at 5 ft. from the vertex.
How deep is the parabolic solar cooker?
Thank you for your help!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


In this problem, you are given the size 10 ft of the opening of the parabolic solar cooker,
and the distance 5 ft from the vertex to the focus of the paraboloid (same as the place for cooking).

They want you find the depth of the parabolic mirror.


                                 S O L U T I O N

To solve such problems,  use an equation of the parabolic cross-section (which is a parabola)
in the form

    y = %281%2F%284p%29%29%2Ax%5E2.    (1)


In this form, 'p' is the focal distance, i.e. the distance of 5 ft from the vertex 
to the cooking place;  x is half of the opening, i.e. 10/2 = 5 ft;  y is the depth of the paraboloid.


So, we substitute  p = 5 ft  and  x = 5 ft  into equation (1), and we get

    y = %281%2F%284%2A5%29%29%2A5%5E2 = 25%2F20 = 5%2F4 ft = 11%2F4 ft.


At this point, the problem is completely solved.


ANSWER.  The depth of the paraboloid is 11%2F4 ft.

Solved.