1. ## Parabola help please

I'm not sure if this falls under calculus as its been a while since I took it.

I'm writing a program and I need help with finding projectile parabola equations.

Rather than explain the whole situation I drew a quick, but poor, picture in paint.net.

Basically I need know, that given x1,y1,x2,y2, and y0, can I find x0.

I already solved another situation where x0,y0 is known but another point is missing. I just cannot seem to do it when the center point x-value is not known.

It seems like it should be simple, but maybe I'm just tired. I could use some help please.

2. A parabola has a general equation

$(x - x_0)^2 = 4a(y-y_0)$

If $x_0$ is unknown, simply substitute $(x_1,y_1)$ and $(x_2,y_2)$ into the equation separately.

This gives:
$(x_1 - x_0)^2 = 4a(y_1-y_0)$
$(x_2 - x_0)^2 = 4a(y_2-y_0)$

Dividing one by the other:

$\left(\frac{x_1 - x_0}{x_2 - x_0}\right)^2 = \left(\frac{y_1 - y_0}{y_2 - y_0}\right)$

Since we know the values $y_1, y_2,$ and $y_0$, $\frac{y_1 - y_0}{y_2 - y_0}$ is a constant. Lets call it $C^2$ for convenience's sake.
$\left(\frac{x_1 - x_0}{x_2 - x_0}\right)^2 = C^2$

$x_1 - x_0 = (x_2 - x_0) C = x_2C - x_0 C$

$x_0 C - x_0 = x_2C - x_1$

$x_0 (C - 1) = x_2C - x_1$

$x_0 = \frac{x_2C - x_1}{C-1}$

We know that $x_1, x_2$ and $C$ are constants, so you'll be able to solve for $x_0$

Overall, the equation should be:

$x_0 = \frac{x_2 \sqrt{\frac{y_1 - y_0}{y_2 - y_0}} - x_1}{\sqrt {\frac{y_1 - y_0}{y_2 - y_0}} - 1 }$

You may want to check my algebra. If you have any questions please ask.

3. I appreciate the response but it does not seem to work. First I don't know where you got that general form because I was always taught the general form of a parabola is y=ax2+bx+c, and the standard form y=a(x-h)2+k.

I tested a simple case anyway.
Imagine...
(x1,y1) = (0,0)
(x2,y2) = (6,1)
(x0,y0) = (x,4)
obviously it is a downward pointing parabola and the missing x value should lie somewhere between 0 and 6, but most likely close to the midpoint since the y values of the side points only differ by 1.

I plug in the (y1-y0)/(y2-y0) = (0-4)/(1-4) = 4/3
so plugging in the rest...
6 sqrt(4/3) / (sqrt(4/3) - 1) = 6.9282/.1547 = 44.78 which is way off.

4. I think we may have a slight misunderstanding. I assumed $(x_0,y_0)$ is the vertex of the parabola. Is this so? If not my previous post is irrelevant. You cannot find an equation for a parabola, and consequently $x_0$ with just two points.

Originally Posted by gyan1010
I appreciate the response but it does not seem to work. First I don't know where you got that general form because I was always taught the general form of a parabola is y=ax2+bx+c, and the standard form y=a(x-h)2+k.
$(x - x_0)^2 = 4a(y-y_0)$
$y - y_0 = \frac{1}{4a} (x - x_0)^2$
$y = \frac{1}{4a} (x - x_0)^2 + y_0$ Which is in your from y =a(x-h)2+k.

I plug in the (y1-y0)/(y2-y0) = (0-4)/(1-4) = 4/3
so plugging in the rest...
6 sqrt(4/3) / (sqrt(4/3) - 1) = 6.9282/.1547 = 44.78 which is way off.
Note that $C = \pm \sqrt {\frac{y_1 - y_0}{y_2 - y_0}}$. Given two points, you have no way of knowing which one is pointing up or down. Substitute $C = -\sqrt {\frac{y_1 - y_0}{y_2 - y_0}}$ and you should be getting a value of $x_0$ for a downward pointing parabola. Without the negative it's an upward pointing parabola.

5. My apologies, must have been tired because I didnt consider the negative square root. Since all of my parabolas are downward it was always wrong until I used the negative. Thanks so much for your help.