# Parabola?

• May 11th 2011, 04:14 PM
homeylova223
Parabola?
Write the equation of the parabola that meets each set of conditions.

The parabola has a horizontal axis and passes through the origin and points at (-1,2) and (3,-2).

Now I am not sure how to do this would I use the midpoint formula to find the vertex but If I do I end up with (1,0)(Crying)
• May 12th 2011, 02:48 AM
jgv115
The general formula for a parabola is $y=ax^2 + bx + c$

Since it goes through the origin, you can sub in (0,0) to the equation

So it becomes: $0=c$ So we know that so far the equation is $y= ax^2 + bx$.

Now you have 2 variables "a" and "b" and you have 2 points. Sub those 2 points in to find a and b.
• May 12th 2011, 03:07 AM
earboth
Quote:

Originally Posted by homeylova223
Write the equation of the parabola that meets each set of conditions.

The parabola has a horizontal axis and passes through the origin and points at (-1,2) and (3,-2).

Now I am not sure how to do this would I use the midpoint formula to find the vertex but If I do I end up with (1,0)(Crying)

1. Let V(h, k) denote the vertex of the parabola and p the distance between the vertex and the focus then the genaral equation of the parabola is:

$(y-k)^2=4p(x-h)$

2. You know 3 points of the parabola: (0, 0), (-1, 2), (3, -2) and you have to determine 3 unknowns: h, k, p

3. Plug in the coordinates of the points into the general equation and solve for (h,k,p).
• May 12th 2011, 06:05 AM
Soroban
Hello, homeylova223!

Quote:

Write the equation of the parabola.

The parabola has a horizontal axis and passes through the origin
and the points at (-1,2) and (3,-2).

Now I am not sure how to do this.
Would I use the midpoint formula to find the vertex? . What?

A horizontal parabola has the form: . $x \;=\;ay^2 + by + c$

We know three points on the parabola.
Substitute their values and solve the system of equations.

$(0,0)\!:\;\;a(0^2) + b(0) + c \:=\:0 \quad\Rightarrow\quad \boxed{c \:=\:0}$

$(\text{-}1,2)\!:\;\;a(2^2) + b(2) + 0 \:=\:\text{-}1 \quad\Rightarrow\quad 4a + 2b \:=\:\text{-}1$ .[1]

$(3,\text{-}2)\!:\;\;a(\text{-}2)^2 + b(\text{-}2) + 0 \:=\:3 \quad\Rightarrow\quad 4a - 2b \:=\:3$ .[2]

Add [1] and [2]: . $8a \,=\,2 \quad\Rightarrow\quad \boxed{a \:=\:\tfrac{1}{4}}$

Substitute into [1]: . $4(\tfrac{1}{4}) + 2b \:=\:-1 \quad\Rightarrow\quad \boxed{b \:=\:\text{-}1}$

The equation is: . $x \;=\;\tfrac{1}{4}y^2 - y$