SOLUTION: The owner of an animal farm who happens to be an architect is trying to design a parabolic arch of the entrance to his farm. He wants the highest point of the arch to be 10ft from

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Question 1167613: The owner of an animal farm who happens to be an architect is trying to design a parabolic arch of the entrance to his farm. He wants the highest point of the arch to be 10ft from the ground. Also included in his design is that at the height of 8ft the width of the arc is to be 6ft. How wide does he want the arch at ground level?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The owner of an animal farm who happens to be an architect is trying to design a parabolic arch of the entrance to his farm.
He wants the highest point of the arch to be 10ft from the ground.
Also included in his design is that at the height of 8ft the width of the arc is to be 6ft.
How wide does he want the arch at ground level?
:
Write an equation that the axis of symmetry goes thru the origin
ax^2 + 10 = y
x=-3, y=8
9a + 10 = 8
9a = -2
a = -2/9
a = -.2222
The equation for this parabola
-.2222x^2 + 10
graphically, green line is 8 ft, intersects + and -3 ft
+graph%28+300%2C+200%2C+-10%2C+10%2C+-4%2C+15%2C+-.2222x%5E2+%2B+10%2C+8%29+
:
Find arch at ground level, (x when y=0)
-.2222x^2 + 10 = 0
x%5E2+=+%28-10%29%2F%28-.2222%29
x = +/-sqrt%2845.0045%29
x = +/- 6.71ft
Ground level width: 2 * 6.71 = 13.42 ft