SOLUTION: find the equation of a parabola through (-1,-3) and (2,3/2), whose axis is parallel to the y-axis and latus rectum equal to 6. Thanks for the help. :)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the equation of a parabola through (-1,-3) and (2,3/2), whose axis is parallel to the y-axis and latus rectum equal to 6. Thanks for the help. :)      Log On


   

Question 891029: find the equation of a parabola through (-1,-3) and (2,3/2), whose axis is parallel to the y-axis and latus rectum equal to 6.
Thanks for the help. :)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Using the value of the latus rectum,
y-k=%281%2F6%29%28x-h%29%5E2
6%28y-k%29=%28x-h%29%5E2
Now use both points,
6%28-3-k%29=%28-1-h%29%5E2
-18-6k=-%28h%5E2%2B2h%2B1%29
h%5E2%2B2h%2B1%2B6k=-18
1.h%5E2%2B2h%2B6k=-19
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6%283%2F2-k%29=%282-h%29%5E2
9-6k=4-4h%2Bh%5E2
2.h%5E2-4h%2B6k=5
Subtracting eq. 2 from eq. 1,
h%5E2%2B2h%2B6k-h%5E2%2B4h-6k=-19-5
6h=-24
h=-4
Then,
%28-4%29%5E2%2B2%28-4%29%2B6k=-19
16-8%2B6k=-19
6k=-27
k=-9%2F2
So then the equation is,
y%2B9%2F2=%281%2F6%29%28x%2B4%29%5E2
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