SOLUTION: Find the equation of the parabola with vertex on Oy, axis parallel to Ox, and passing through (2, 2) and (8, -1)

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Question 1190797: Find the equation of the parabola with vertex on Oy, axis parallel to Ox, and passing
through (2, 2) and (8, -1)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

axis parallel to Ox=>your parabola opens right/left
Vertex form:
%28y-k%29%5E2=4p%28x-h%29
vertex on Oy=> h=0
%28y-k%29%5E2=4p%28x-0%29
%28y-k%29%5E2=4px............passing through (2, 2) , use it to calculate k and p
%282-k%29%5E2=4p%2A2
k%5E2+-+4+k+%2B+4=8p
p=+%28k%5E2+-+4+k+%2B+4%29%2F8..........1)

passing through (8, -1)

%28-1-k%29%5E2=4p%2A8
k%5E2+%2B+2+k+%2B+1=32p
p=%28k%5E2+%2B+2+k+%2B+1%29%2F32..........2)

from 1) and 2) we have

%28k%5E2+-+4+k+%2B+4%29%2F8=%28k%5E2+%2B+2+k+%2B+1%29%2F32
32%28k%5E2+-+4+k+%2B+4%29=8%28k%5E2+%2B+2+k+%2B+1%29
32+k%5E2+-+128+k+%2B+128-8k%5E2+-+16+k+-8=0
24+k%5E2+-+144+k+%2B+120=0
24+%28k%5E2+-+6k+%2B+5%29=0
24+%28%28k+-+1%29+%28k+-+5%29%29=0

=> k=1+or k=5

then, if k=1
p=+%281%5E2+-+4%2A1+%2B+4%29%2F8..........1)
p=+1%2F8
or, if k=5
p=+%285%5E2+-+4%2A5+%2B+4%29%2F8
p=9%2F8

and your equations are:
%28y-1%29%5E2=4%281%2F8%29x
%28y-1%29%5E2=%281%2F2%29x.........1
or
%28y-5%29%5E2=4%289%2F8%29x
%28y-5%29%5E2=%289%2F2%29x.........2