1. ## parable

Finding the equation of the parable and build it

Axis parallel to y = 0 and goes through A(-2,4) B(-3,2) C(-11,-2)

$y^2-8y+4x+24=0$

2. I hope you mean "parabola"

There are two forms of parabolas. What are they?

Once you know which of the two you need, how can you determine the specifics of the equation?

3. The equation is:

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

but with points and I think the equation

4. All right. We have an equation now. That's a step in the right direction.
A parabola is defined by three points. Since we have three points here, we can define it.
In our equation $(y-k)^2=2p(x-h)$, we can use $(x,y)=(-2,4)$ to get an equation in three unknowns: $h,k,p$. Doing the same with the other two points will give us three equations in three unknowns, and from there, we can solve for $h,k,p$.

5. Thank you. Do you have any means of online communication, for example, msn

6. No, I really don't. My computer is so old, I can barely run Itunes and Firefox at the same time.
However, I will be checking this site frequently, so a private message to me would probably be easiest.
I'm glad to have been of assistance.

7. I not find the answer and you?

8. Originally Posted by Apprentice123
Finding the equation of the parable and build it

Axis parallel to y = 0 and goes through A(-2,4) B(-3,2) C(-11,-2)

$y^2-8y+4x+24=0$
Originally Posted by Apprentice123
The equation is:

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

but with points and I think the equation
So
$(4 - k)^2 = 2p(-2 - h)$

$(2 - k)^2 = 2p(-3 - h)$

$(-2 - k)^2 = 2p(-11 - h)$

So we get
$k^2 - 8k + 16 = -4p - 2ph$ (1)

$k^2 - 4k + 4 = -6p - 2ph$ (2)

$k^2 + 4k + 4 = -22p - 2ph$ (3)

Subtract equation (2) from equation (1) and then subtract equation (3) from equation (1)
$-4k + 12 = 2p$

$-12k + 12 = 18p$

You can find values of k and p from these and then use one of the original equations to find h.

Another (perhaps simpler in this case) way to approach this is to use the "standard" form for this kind of parabola:
$ay^2 + by + c = x$
and plug the points into that. This gives you a linear system in a, b, and c.

-Dan

9. Originally Posted by topsquark

Another (perhaps simpler in this case) way to approach this is to use the "standard" form for this kind of parabola:
$ay^2 + by + c = x$
and plug the points into that. This gives you a linear system in a, b, and c.

-Dan
$(-2,4), (-3,2), (-11,-2)$

$ay^2+by+c=x$

Substituting each ordered pair into the standard equation, we get:

$1.\ \ a(4)^2+b(4)+c=-2$
$16a+4b+c=-2$

$2. \ \ a(2)^2+b(2)+c=-3$
$4a+2b+c=-3$

$3. \ \ a(-2)^2+b(-2)+c=-11$
$4a^2-2b+c=-11$

Solve the system using matrix equation (or any method you like):

$\left[\begin{array}{cccc}16 & 4 & 1 & \\ 4 & 2 & 1 & \\ 4 & -2 & 1 \\ \end{array}\right] \cdot \left[\begin{array}{c}a \\ b \\ c \end{array}\right] = \left[\begin{array}{c}-2 \\ -3 \\ -11 \end{array}\right]$

$\left[\begin{array}{cccc}16 & 4 & 1 & \\ 4 & 2 & 1 & \\ 4 & -2 & 1 \\ \end{array}\right]^{-1} \cdot \left[\begin{array}{c}-2 \\ -3 \\ -11 \end{array}\right] = \left[\begin{array}{c}a \\ b \\ c \end{array}\right]$

$\left[\begin{array}{c}a \\ b \\ c \end{array}\right]= \left[\begin{array}{c}-\frac{1}{4} \\ 2 \\ -6 \end{array}\right]$

Substituting back into your original equaton:

$ay^2+by+c=x$

$-\frac{1}{4}y^2+2y-6=x$

$\boxed{y^2-8y+4x+24=0}$

10. Thank you very much