# Parabola: Vertex =(?,0) and know 2 arbitrary points. How solve x?

Summary:: I have a upwards opening parabola where I know the Y vertex point = 0. I also know 2 arbitrary points separated by a step size on the parabola. How can I solve for x at the vertex point?

X=horizontal plane
Y=vertical plane

I have a upwards opening parabola where I know the Y vertex point = 0. I also know 2 arbitrary points on the parabola separated by a small step size. How can I solve for x at the vertex point?

[Moderator's note: Moved from a technical forum and thus no template.]

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
phinds
Gold Member
2019 Award
Start with the generic equation for a parabola and see what you can fill in.

y = A (x + xo) ^2 + yo
yo=0
√(y/A) = x+xo
Can you explain how this helps find vertex x position?
I don't know x or the x offset.
I was hoping if I knew a slope of a point on the parabola and y vertex=0 there was a formula for that.

berkeman
Mentor
y = A (x + xo) ^2 + yo
yo=0
How can I solve for x at the vertex point?
Can you explain how this helps find vertex x position?
What happens when you fill in your known points into the equation of the curve? How many equations with how many unknowns do you end up with?

Sorry I meant A and xo are unknown. Plugging in my known points does not seam to get me any closer.
For example: This is all the data I have to go off.
2 pointes (x,y) = (-6 , 54) and (-5 , 36)
vy = 0

working the problem, I still seam to have the same unknowns. am i missing something?

Mark44
Mentor
y = A (x + xo) ^2 + yo
yo=0
Sorry I meant A and xo are unknown. Plugging in my known points does not seam to get me any closer.
For example: This is all the data I have to go off.
2 pointes (x,y) = (-6 , 54) and (-5 , 36)
vy = 0
Substitute the known points (-6, 54) and (-5, 36) for x and y in the equation at the top. You will get two equations in the unknowns A and x0, so you should be able to solve for A and x0.

Ok, so here are the 2 solutions.
1st (-6, 54)
A = 54 / (-6-X0)2

X0 =

2nd (-5, 36)
A = 36 / (-5-X0)2

X0 =

Ok, I think I get what you are saying.
54 / (-6-X0)2 = 36 / (-5-X0)2
X0 =-3+√6 and X0 =-3-√6
Why are there 2 values for X0 when Y0 = 0 ?

Last edited:
mfb
Mentor
One of them is an artifact from taking the square root, it's not an actual solution. If you plug that back into the original equation (for one of the two points) you'll see that only one fits.

Thank you guys for your help. I got a function that works now.
I wrote a function where v1 = (x,y) coordinate on the parabola, and v2 = another (x,y) coordinate a step apart. It returns the 2 results for the x vertex. It appears the x2 value is the correct one where x1 is wrong.

x1 = (v1.y * v2.x - v1.x * v2.y
- v2.x * Math.Sqrt(v1.y * v2.y)
+ v1.x * Math.Sqrt(v1.y * v2.y)) / (v1.y - v2.y);

x2 = (v1.y * v2.x - v1.x * v2.y
+ v2.x * Math.Sqrt(v1.y * v2.y)
- v1.x * Math.Sqrt(v1.y * v2.y)) / (v1.y - v2.y);

Now, my next question is why is this not used in gradient descent?
Seems like it gives an instant answer if vertex Y=0.

Last edited:
Mark44
Mentor
v2 = another (x,y) coordinate a step apart.
What do you mean "a step apart"?
why is this not used in gradient descent?
What do you mean by "gradient descent"?

phinds
Gold Member
2019 Award

mfb
Mentor
Now, my next question is why is this not used in gradient descent?
Seems like it gives an instant answer if vertex Y=0.
For parabolas that have a vertex at y=0, but not for anything else. If you know the function is a parabola with that property you don't use gradient descent.

Ok, thanks for the help. For my solver my goal is minimum error which would be zero. So, this will probably work better than gradient descent.
Thanks for the help
Keep up the good work.