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How to get a parabola shape according to a moved point and nearest points

Being not very good at math, I have a problem with my project.
The objective is boundary correction on 3D files.
In my application, the user moves a 3D point on X-axis in order to correct or modify the boundary of the object.
I want to move the nearest boundary points in the same direction but decreasingly. I mean no point should move more than the main point. The nearest points move most and, the farthest points should move less.
On the image, the red dots represent the initial status of points. And the user pulls the P0 in the x-direction. And the other points follow it. The last status of the points is represented by violet dots.
Here is what I tried.
//On point moved event
//Get nearest boundary Points (Uses Geometry3D to get boundary points).
(var clothDMesh, _) = Utilities3D.BuildDMesh(baseMesh);
CreateClothModel(clothDMesh);
var boundryVertices = nodes.Where(ro => ro.Value.isBorder).Select(ro => ro.Value.vertex).ToList();
var refPoint = CustomPoint.FromPoint3D(movedPoint);
//Gets the delta X.
var deltaX = p.X - initialX;
//Gets nearest country points, so 15 points above and 15 points below to move only a given number of points (I know maybe this should be calculated according to delta).
var nearestPoints = refPoint.GetNearesPoints(boundryVertices, 30);
foreach (var item in nearestPoints)
{
//This is only one of what I tried as a function. None of them worked correctly.
item.X += deltaX - (deltaX * 1/ Math.Pow(item.Distance, 2));
}
Any help will be appreciated.
Thanks in advance.

Here's the math part:
I call "a" your "deltaX".
We also need a second parameter: "b", the maximum height of the red dots. I assume it is symetrical and "-b" would be the minimum height of the red dots.
So, if you look for the value X, horizontal move, in fonction of the coordinate Y of the dot:
X = a - a * Y * Y / (b * b);
You can verify that for Y = 0, you obtain X = a and for Y = b (or -b) you get X = 0.
You have your parabola (X is function of Y^2).

2D Rotated rectangle contains point calculation

My issue is that I've been trying to check if a rectangle that is rotated by a certain amount of degrees contain a point, but I wasn't able to calculate that after many attempts with the help of some code samples and examples that I've found online.
What I got is the rectangle (X, Y, Width, Height, Rotation) and the point (X, Y) and I've been trying to create a simple function that lets me instantly calculate that, which would be something something like this:
public bool Contains(Rect Rectangle, float RectangleRotation, Point PointToCheck);
But as I mentioned, I wasn't able to do so, those mathematical calculations that include some formulas I found online are way too much for me to understand.
Could someone help me with calculating this? If you could provide the calculation in C# code form (not formulas) then that would be great! Thanks.
PS: Using the 2D Physics Engine that is available in Unity3D is not a option, my rectangle is not associated with a gameobject that I could attach a 2D collision component to, I need to do this mathematically without the involvement of gameobjects or components.
Edit: I forgot to mention, the rectangle is being rotated by the middle of the rectangle (center/origin).

Rather than checking if the point is in a rotated rectangle, just apply the opposite of the rotation to the point and check if the point is in a normal rectangle. In other words, change your perspective by rotating everything by -RectangleRotation, so that the rectangle does not appear rotated at all.
public bool Contains(Rect rect, float rectAngle, Point point)
{
// rotate around rectangle center by -rectAngle
var s = Math.Sin(-rectAngle);
var c = Math.Cos(-rectAngle);
// set origin to rect center
var newPoint = point - rect.center;
// rotate
newPoint = new Point(newPoint.x * c - newPoint.y * s, newPoint.x * s + newPoint.y * c);
// put origin back
newPoint = newPoint + rect.center;
// check if our transformed point is in the rectangle, which is no longer
// rotated relative to the point
return newPoint.x >= rect.xMin && newPoint.x <= rect.xMax && newPoint.y >= rect.yMin && newPoint.y <= rect.yMax;
}

Determining the grids cut by a plane,formed from 3 vertices

I found out an equation of a plane,from three vertices.
Now,if I have a bounding box(i.e. a large cube),How can I determine the grid positions(small cubes),where the plane cuts the large cube.
I am currently following this approach:
For each small cube center, say(Xp, Yp, Zp), calculate perpendicular distance to the plane i.e., (aXp + bYp + c*Zp + d)/ (SquareRoot Of (a^2 + b^2 + c^2)). This should be less than or equal to (length of smallCube * SquareRoot(3))/2.
If this criteria,gets satisfied,then I assume my plane to cut the large cube at this small cube position.
a,b,c,d are coefficients of the plane,of the form ax+by+cz+d = 0.
I would be really glad,if someone can let me know,if I am doing something wrong (or) also,any other simple approach.

Seems you want to get a list of small cubes (grid voxels) intersected by given plane.
The simplest approach:
Find intersection of the plane with any cube edge. For example, intersection with vertical edge of AAB (X0,Z0 are constant) might be calculated by solving this equation for unknown Y:
aX0 + bY + c*Z0 + d = 0
and checking that Y is in cube range. Get small cube coordinates (0, ky=Floor(Y/VoxelSize), 0) and then check neighbor voxels in order (account for plane coefficients to check only real candidates).
candidates:
0,ky,0
1,ky,0
0,ky-1,0
0,ky+1,0
0,ky,1
There are more advanced methods to generate voxel sequence for ray case (both 2d and 3d) like Amanatides/Woo algorithm. Perhaps something similar exists also for plane voxelization
Here is AABB-plane intersection test code from this page (contains some explanations)
// Test if AABB b intersects plane p
int TestAABBPlane(AABB b, Plane p) {
// Convert AABB to center-extents representation
Point c = (b.max + b.min) * 0.5f; // Compute AABB center
Point e = b.max - c; // Compute positive extents
// Compute the projection interval radius of b onto L(t) = b.c + t * p.n
float r = e[0]*Abs(p.n[0]) + e[1]*Abs(p.n[1]) + e[2]*Abs(p.n[2]);
// Compute distance of box center from plane
float s = Dot(p.n, c) - p.d;
// Intersection occurs when distance s falls within [-r,+r] interval
return Abs(s) <= r;
}
Note that e and r remain the same for all cubes, so calculate them once and use later.

I have a list of points and would like to determine if they form a circle

I'm using C#, I have a list of Vector points and would like to try and approximate how close they look like a circle.
Any ideas how to implement this or if someone else has?

Based on the "gestures" tag I guess you not only want to know how close are these points to the smallest-circle (search for "Smallest-circle problem"), but you have to be concerned also about their order and spread:
I would start with the distance from the smallest-circle. If they are too far, you're done, it's not a circle.
If they are close enough to your configured threshold, compute the angle between the vector defined by the circle center, first point and each other point (picture bellow)
Check that each angle is greater than the previous.
Check that difference between any two angles next to each other is not over some configured threshold.
Check that the last point is close enough to the first one.
You will probably think of some other checks eventually, so make it simple to extend.

Another possibility:
Find the centroid
(http://en.wikipedia.org/wiki/Centroid#Of_a_finite_set_of_points) of
the points
Determine the distance (radius) of each point from the location of
the centroid
Calculate the distribution and determine if the points
are within acceptable tolerances (e.g. standard deviation < ±0.05 × mean radius or something like that)
Without knowing more about the source of the points, it's hard to suggest the best solution.
These might be of use: http://link.springer.com/article/10.1007%2FBF02276879#page-1 and http://www.dtcenter.org/met/users/docs/write_ups/circle_fit.pdf. Those methods will give you the best fitting circle through the points, but you are still going to need to determine whether your data points are close enough for your purposes.
UPDATE: based on the 'gesture' tag, somebody has already implemented it: http://depts.washington.edu/aimgroup/proj/dollar/

1) Pick any three points from that list, find the center of their appropriate circle
We can do this, using triangle circumcircle construction method, you find the medians of all three sides (two are sufficient) and their intersection is the center of the circle. Something like this:
public PointF findCenter(PointF a, PointF b, PointF c)
{
float k1 = (a.Y - b.Y) / (a.X - b.X) //Two-point slope equation
float k2 = (a.Y - c.Y) / (a.X - c.X) //Same for the (A,C) pair
PointF midAB = new PointF((a.X + b.X) / 2, (a.Y + b.Y) / 2) //Midpoint formula
PointF midAC = new PointF((a.X + c.X) / 2, (a.Y + c.Y) / 2) //Same for the (A,C) pair
k1 = -1*k1; //If two lines are perpendicular, then the product of their slopes is -1.
k2 = -1*k2; //Same for the other slope
float n1 = midAB.Y - k1*midAB.X; //Determining the n element
float n2 = midAC.Y - k2*midAC.Y; //Same for (A,C) pair
//Solve y1=y2 for y1=k1*x1 + n1 and y2=k2*x2 + n2
float x = (n2-n1) / (k1-k2);
float y = k1*x + n1;
return new PointF(x,y);
}
2) Check if the other points are equivalently distanced from this center, if yes, you have a circle, if no, you don't.
P.S. I haven't tested the code, so be prepared to debug. Ask if you need anything else

Take any three points from your point set.
If the points are co-linear then, your point set doesn't all lie on a circle.
Find the circumcircle of the triangle. The diameter is given by: d = (a*b*c)/2*area. The center of the circle is the point of intersection of the perpendicular bisectors of the three sides.
Now for every remaining point in the point set, if the distance from the center is not equal to the radius then the points are not on a circle. You can speed up the calculations by comparing the square of the radius against the square of the distance between the given point and the center.

Need some math - Projecting Slope

I have a rectangle.
Its height (RH) is 400.
Its width (RW) is 500.
I have circle.
Its height (CH) is 10.
Its width (CW) is 10.
Its starting location (CX1, CY1) is 20, 20.
The circle has moved.
Its new location (CX2, CY2) is 30, 35.
Assuming my circle continues to move in a straight line.
What is the circle's location when its edge reaches the boundary?
Hopefully you can provide a reusable formula.
Perhaps some C# method with a signature like this?
point GetDest(size itemSize, point itemPos1, point itemPos2, size boundarySize)
I need to calculate what that location WILL be once it arrives - knowing that it is not there yet.
Thank you.
PS: I need this because my application is watching the accelerometer on my Windows Phone. I am calculating the target necessary to animate the motion of the circle inside the rectangle as the user is tilting their device.

It is 1 radius away from the boundar(y/ies).

The answer is X=270 Y=395
first define the slope V as dy/dx =(y2-y1)/(x2-x1). In your example: (35-20)/(30-20)=1.5
the line equation is
y = V * (x-x1) + y1. You are interested in the horizontal locations x at:
y= CH/2 OR y= H-CH/2
so (not code, just math)
if (y2-y1)<0:
x=(CH/2 -y1)/V +x1 10 for your example. OR
if (y2-y1)>0:
x=(H-CH/2 -y1)/V +x1 270 for your example
else (that is: y2==y1)
the upper or lower lines were not hit.
if CH/2 <= x <= W-CH/2 the circle did hit the that upper or lower side: since V>0, we use x=270 and that is within CH/2 and W-CH/2.
So the answer to your question is y=H-CH/2 = 395 , X=270
For the side lines it's similar:
(if (x2-x1)<0)
y=(CH/2 -x1)*V +y1
(if (x2-x1)>0)
y=(W-CH/2 -x1)*V +y1
else (that is: x2==x1)
the side lines were not hit.
if CH/2 <= y <= H-CH/2 the circle did hit that side at that y.
be careful with the trivial cases of completely horizontal or vertical movement so that you don't divide by zero. when calculating V or 1/V. Also deal with the case where the circle did not move at all.
Since you now asked, here's metacode which you should easily be able to convert to a real method. It deals with the special cases too. The input is all the variables you listed in your example. I here use just one symbol for the circle size, since it's a circle not an ellipse.
method returning a pair of doubles getzy(x1,y1,W,H,CH){
if (y2!=y1){ // test for hitting upper or lower edges
Vinverse=(x2-x1)/(y2-y1)
if ((y2-y1)<0){
xout=(CH/2 -y1)*Vinverse +x1
if (CH/2 <= y <= H-CH/2) {
yout=CH/2
return xout,yout
}
}
if ((y2-y1)>0){
xout=(H-CH/2 -y1)*Vinverse +x1
if (CH/2 <= y <= H-CH/2) {
yout=H-CH/2
return xout,yout
}
}
}
// reaching here means upper or lower lines were not hit.
if (x2!=x1){ // test for hitting upper or lower edges
V=(y2-y1)/(x2-x1)
if ((x2-x1)<0){
yout=(CH/2 -x1)*V +y1
if (CH/2 <= x <= W-CH/2) {
xout=CH/2
return xout,yout
}
}
if ((x2-x1)>0){
yout=(H-CH/2 -x1)*V +y1
if (CH/2 <= x <= W-CH/2) {
xout=H-CH/2
return xout,yout
}
}
}
// if you reach here that means the circle does not move...
deal with using exceptions or some other way.
}

It's easy; no calculus required.
Your circle has radius R = CW/2 = CH/2, since the diameter of the circle D = CW = CH.
In order to have the circle touch the vertical edge of the rectangle at a tangent point, you have to move the circle to the right by a distance (W - (CX1 + CW/2))
Likewise, the circle will touch the bottom edge of the rectangle at a tangent point when you move it down by a distance (H - (CY1 + CH/2)).
If you do this in two separate translations (e.g., first to the right by the amount given, then down by the amount given or visa versa), you'll see that the circle will touch both the right hand vertical and the bottom horizontal edges at tangent points.

When the moving circle arrives at a wall (boundary) then it will be tangent at one of four points on the circle, call them N, S, E, and W. You know their initial coordinates.
The points travel in a line with a slope known to you: m=(y2-y1)/(x2-x1); where in your example (x1, y1) - (20,20) and (x2, y2)= (30, 35).
Your problem is to find the trajectory of the first point N, S, E, or W which reaches any wall. The trajectory will be a line with slope m.
You can do this by adding (or subtracting) the direction vector for the line to the point N, S, E, or W, scaled by some t.
For example, N is (20, 15). The direction vector is (x2-x1,y2-y1) = (10, 15). Then (20, 15) + t * (10, 15) will hit the boundary lines at different t's. You can solve for these; for example 20 + t*10 = 0, and 20 + t*10 = 400, etc.
The t that is smallest in magnitude, over all four trajectories, gives you your tangent point.

Not sure its calculus..wouldn't it just be the following:
if y >= 390 then it reached the top edge of the rectangle
if x >= 490 then it reached the right edge of the rectangle
if y <= 0 then it reached the bottom edge of the rectangle
if x <= 0 then it reached the left edge of the rectangle