Equation of a rotated parabola knowing only 3pts the equation has to satisfy

[arpace]

I am trying to determine the tangent of the vertex and the axis of symmetry of a rotated parabola, not knowing the equation of the parabola, and only knowing 3pts the equation has to satisfy.

Can anyone help out with this? I have three points, the second is always the vertex, and pt1 and point 2 are on opposite sides of the parabola; yet, they are unequal in distance from the vertex.

for example:

pt1 is something like 1,2
pt2 , the vertex, is 2,8
pt3, is something like 9,7

I already know the math to get tangent, I just need the process to find the equations of the parabola that is rotated knowing only three points.

This relates to quadratic bezier curves cause it will enable me to find the control point with some additional steps; it is just getting over this hurdle that bites.

 

[tiny-tim]

Welcome to PF!

Hi arpace ! Welcome to PF!

Hint: first, simplify it by changing coordinates so that the vertex is at (0,0).

Then you know the equation for a parabola with vertex (0,0) must be ax + by = … ?

 

[HallsofIvy]

In general, you can't. Given any 3 points, there exist a unique parabola with vertical axis of symmetry passing through those 3 points. There exist an infinite number of parabolas with other axes passing through those same three points.

 

[kts123]

^ I think the TC may have implicitly implied that the focus and directrix are positioned at
(0,p) and (0,-p) respectively, and then rotated from there.

 

[arpace]

I know that I would have to rotate it back to 0 degrees; yet, in order to do that don't you need to know either the tangent or the axis of symmetry to find the degree it needs to be rotated by?

The second point is always the vertex; yet, the points 1 and 3 are not an equal distance from the vertex; thus, this is where my hard time starts.

If it is not possible, then is there another way to determine the control point for a quadratic bezier curve, knowing the same information?

 

[kts123]

Try brute algebra force? That's my only suggestion.

Observe that for a parabola symetric between the Y axis (I forget what that's called) for every point, (x,y,) there is a corrisponding point, (-x,y), such as that (x,y) and (-x,y) are equal in distance to the y-axis (the point x=0.) When rotated, each point must still contain this symetry. You could probably use this fact in tandem with some serious trig to figure it out (I'd see what I could do, but I'm at work, I can't be jotting down equations willy-nilly. =p)

 

[mr articulate]

How do i calculate the equation of a parabola with a horizontal directrix given three random points along the line. I tried to translate the graph around the origin but that still didnt work. The points are (1/2,6), (3/2,16), (-3/2,-2).

 

[HallsofIvy]

How do i calculate the equation of a parabola with a horizontal directrix given three random points along the line. I tried to translate the graph around the origin but that still didnt work. The points are (1/2,6), (3/2,16), (-3/2,-2).

A parabola with horizontal directrix has vertical axis of symmetry and so is of the form y= ax²+ bx+ c. Put the x and y values of the three points into that equation and solve for a, b, and c.