SOLUTION: I can not figure out how to write an equation for parabola focus (3,8),directrix y=4

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Question 81488This question is from textbook Algebra 2
: I can not figure out how to write an equation for parabola
focus (3,8),directrix y=4 This question is from textbook Algebra 2

Found 2 solutions by stanbon, scott8148:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
write an equation for parabola
focus (3,8),directrix y=4
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Draw the line y=4.
Plot the point (3,8)
The focus is above the directrix and is 2p away from it.
So 2p=4; p=2
The vertex is half way between the focus and the directrix.
So the vertex is (3,6)
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EQUATION:
4p(y-k)=(x-h)^2
8(y-6)=(x-3)^2
8y-48=x^2-6x+9
8y=x^2-6x+57
y=(1/8)x^2-(3/4)x+57/8
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Cheers,
Stan H.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
a parabola is made up of points that are equidistant from a given point (focus) and a given line (directrix)

the "distance formula" gives the distance between (x,y) and (3,8) as sqrt%28%28x-3%29%5E2%2B%28y-8%29%5E2%29

the distance between (x,y) and y=4 is y-4 ... x is not involved

setting these equal gives sqrt%28%28x-3%29%5E2%2B%28y-8%29%5E2%29=y-4 ... squaring both sides gives %28x-3%29%5E2%2B%28y-8%29%5E2=%28y-4%29%5E2

x%5E2-6x%2B9%2By%5E2-16y%2B64=y%5E2-8y%2B16 ... x%5E2-9x%2B64%2B9-16=y%5E2-8y-y%5E2%2B16y ... x%5E2-9x%2B57=8y

so y=x%5E2%2F8-9x%2F8%2B57%2F8 ........THIS ANSWER HAS BEEN CORRECTED (with my apologies)