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

Printable View

- Jan 9th 2007, 02:27 AM^_^Engineer_Adam^_^Strange Parabola
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 - Jan 9th 2007, 04:36 AMearboth
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 - Jan 9th 2007, 04:39 AMticbol
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. - Jan 9th 2007, 08:48 AMSoroban
Hello, ^_^Engineer_Adam^_^!

Let's start from the beginning . . .

Quote:

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: