# controling Parabola intersection

• Jul 2nd 2012, 04:56 AM
controling Parabola intersection
Hello all, madlysavage here and I (obviously) need help. I'm working on a making a video game(yes at 14), and i need to have two things spread out in 2 opposite parabolas on the x and z axis. the problem is i have a start point and an end point, both parabolas need to intersect where they hit (its an ion cannon :D) and at the cannon itself, and my problem is, i need to figure out how to control the intersection points of two parabolas
• Jul 2nd 2012, 05:15 AM
Prove It
Re: controling Parabola intersection
Quote:

Hello all, madlysavage here and I (obviously) need help. I'm working on a making a video game(yes at 14), and i need to have two things spread out in a parabola on the x and z axis. the problem is i have a start point and an end point, both parabolas need to intersect where they hit (its an ion cannon :D) and at the cannon itself, and my problem is, i need to figure out how to control the intersection points of two parabolas

What do you mean by control the intersection points?
• Jul 2nd 2012, 05:20 AM
Re: controling Parabola intersection
i need to be able to set the intersection points, i need to be able to say "I need you to intersect here...and here"
• Jul 2nd 2012, 05:23 AM
emakarov
Re: controling Parabola intersection
Quote:

Originally Posted by Prove It
What do you mean by control the intersection points?

Yes, and I would also like to ask the OP to explain the following terms:

"spread out in a parabola on the x and z axis"

"a start point and an end point" (of what?)

"both parabolas" (before you were talking about one parabola)

"where they hit."
• Jul 2nd 2012, 05:30 AM
Re: controling Parabola intersection
I need to be able to set the intersections of two parabolas to two separate locations, lets say on a grid, i need the intersections to be at (19,5) and (10, 24)... its as simple as that, sorry about any confusion, im trying to explain it the to the fullest of my capabilities, im terrible with words hahaha
• Jul 2nd 2012, 05:39 AM
emakarov
Re: controling Parabola intersection
Quote:

lets say on a grid, i need the intersections to be at (19,5) and (10, 24)

So, you need to find two parabolas that intersect at these two points? There are infinitely many parabolas that pass through these points. For example, parabolas ax˛ - a for all real numbers a pass through (-1, 0) and (1, 0). Note also that all these parabolas have a vertical axis of symmetry. In general, to fix a parabola with a vertical axis, one needs three points.

Also, you were talking about x- and z-axes. Does it mean you also have a y-axis and therefore a three-dimensional space?
• Jul 2nd 2012, 05:51 AM
Re: controling Parabola intersection
yes it is three dimensional, but i can figure out how to turn it on its side, so just ignore that and say its on the x and y axis. and yes, i know that there are many parabolas that go through that space. ok i will try to explain the best i can. lets say i need two parabolas that pass through lets say (0, 10) and (0,-10). i need the parabolas to both be the same shape, but on is upside down(negative) but it has to be in the "eye" shape (yes i will explain that too). so i would use the two parabolas y=.025x^2-2.5 and -.025x^2+2.5 (i got this by trial and error).the space between the two parabolas forms a kind of pointy oval(eye shape), go to this site and type in those two equations for a visual.
• Jul 2nd 2012, 06:14 AM
emakarov
Re: controling Parabola intersection
And what if the two points have different y-coordinates? For example, consider (-1, 0) and (2, 3). All parabolas passing through them have equations y(x) = ax˛ + (1 - a)x + (1 -2a) for different a. But they don't really give the "eye shape" (e.g., try a = 1 and a = -1).
• Jul 2nd 2012, 06:17 AM
HallsofIvy
Re: controling Parabola intersection
I think you are saying that you have $z= ax^2$ and $x= bz^2$ for some numbers a and b (y= 0). They will intersect when $z^2= a^2x^4= a^2(bz^2)^4= a^2b^4z^8$. Obviously z= 0 satisfies that but is not the point we want. If z is not 0, we can divide both sides by $z^2$ to get $a^2b^4z^6= 1$ and then $z= \frac{1}{\sqrt[3]{ab^2}}$. We can do essentially the same thing, focusing on x rather than z, to get $x= \frac{1}{\sqrt[3]{a^2b}}$. Set those equal to the x and z coordinates of the point you want as intersection and you have two equations to solve for a and b.
• Jul 2nd 2012, 06:25 AM
emakarov
Re: controling Parabola intersection
Quote:

Originally Posted by HallsofIvy
I think you are saying that you have $z= ax^2$ and $x= bz^2$ for some numbers a and b (y= 0).

Hmm, I think the OP means (post #7) that one parabola, roughly speaking, should be mapped into the other one under the rotation by 180 degrees, not 90 degrees.