Find the equation of the parabola
Axis Vertical; pasess Through (-1,0),(5,0) & (1,8)
could u please help me solve using this equation substituting the x and y values
Because i always end up with this answer hence not a parabola
Hello,
(1. the x-value of the vertex must be the mean of -1 and 5, that means 2, because the first two points have the same y-value, they are situated symmetrical to the axis)
plug in the x- and y-values into the given equation:
This is a system of simultaneous equations. I don't know which method you use. But the solution is: d = -4, E = 1, F = -5
Thus the equation of the parabola is:
EB
x^2 +Dx +Ey +F is not a parabola.
x^2 +Dx +Ey +F = 0 is. -----------(i)
At (-1,0),
(-1)^2 +D(-1) +E(0) +F = 0
1 -D +F = 0 ------------------(1)
At (5,0),
5^2 +D(5) +E(0) +F
25 +5D +F = 0 -------------(2)
(2) minus (1),
24 +6D = 0
So, D = -24/6 = -4 ----***
Substitute that into (1),
1 -(-4) +F = 0
So, F = -5 ------------***
At (1,8),
1^2 +D(1) +E(8) +F = 0
1 +D +8E +F = 0 ----------------(3)
Plug D and F in,
1 +(-4) +8E +(-5) = 0
8E -8 = 0
So, E = 8/8 = 1 -----***
Therefore, the parabola is x^2 -4x +y -5 = 0 --------answer.
Hello, ^_^Engineer_Adam^_^!
Let's start from the beginning . . .
Find the equation of the parabola with a vertical axis
and passes through
The general equation of a "vertical" parabola is: .
Subtract (1) from (2): .
Subtract (3) from (2): .
Subtract (5) from (4): .
Substitute into (4): .
Substitute into (3): .
Therefore, the equation of the parabola is: .
Note: . is not a parabola.
. . . . . It is an equation . . . with two solutions: