SOLUTION: Find the equation of the parabola with axis parallel to 0x and passing through (5,4),(11,-2),(21,-4).

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation of the parabola with axis parallel to 0x and passing through (5,4),(11,-2),(21,-4).      Log On


   

Question 942047: Find the equation of the parabola with axis parallel to 0x and passing through (5,4),(11,-2),(21,-4).
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
So the general equation would be,
x=ay%5E2%2Bby%2Bc
5=16a%2B4b%2Bc
11=4a-2b%2Bc
21=16a-4b%2Bc
Use Cramer's rule to solve for a,b,c.
A=%28matrix%283%2C3%2C16%2C4%2C1%2C4%2C-2%2C1%2C16%2C-4%2C1%29%29
abs%28A%29=96
.
.
A%5Ba%5D=%28matrix%283%2C3%2C5%2C4%2C1%2C11%2C-2%2C1%2C21%2C-4%2C1%29%29
abs%28A%5Ba%5D%29=48
.
.
A%5Bb%5D=%28matrix%283%2C3%2C16%2C5%2C1%2C4%2C11%2C1%2C16%2C21%2C1%29%29
abs%28A%5Bb%5D%29=-192
.
.
A%5Bc%5D=%28matrix%283%2C3%2C16%2C4%2C5%2C4%2C-2%2C11%2C16%2C-4%2C21%29%29
abs%28A%5Bc%5D%29=480
So then,
a=abs%28A%5Ba%5D%29%2Fabs%28A%29=%2848%29%2F96=1%2F2
b=abs%28A%5Bb%5D%29%2Fabs%28A%29=%28-192%29%2F96=-2
c=abs%28A%5Bc%5D%29%2Fabs%28A%29=%28480%29%2F96=5
So,
x=%281%2F2%29y%5E2-2y%2B5