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Question 825266: Find the equation of the parabola
a) that is symmetric about the line y=4 and passing through the points (4,4) & (10,0)
b) whose focus and vertex are the vertex and focus, respectively, of the parabola with equation x+4y^2-y=0
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
This post contains 2 questions
I will answer the first:
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The general equation of a horizontal parabola with vertex (h,k) is:
x = a(y – k)^2 + h
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given:
(4,4) and
the parabola is symmetric about the line y=4
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(4,4) must be the vertex
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x = a(y – k)^2 + h
x = a(y – 4)^2 + 4
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given:
(10,0)
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x = a(y – 4)^2 + 4
10 = a(0 – 4)^2 + 4
10 = a(-4)^2 + 4
10 = 16a + 4
16a = 14
a = 14/16
a = 7/8
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answer:
x = (7/8)(y – 4)^2 + 4
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in standard form:
x = (7/8)(y – 4)(y – 4) + 4
x = (7/8)(yy – 8y + 16) + 4
x = (7/8)yy – 7y + 14 + 4
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x = (7/8)y^2 – 7y + 18
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