Parabola Through Four Points 

Any four points in the plane determine two conjugate parabolas, each of which passes through all four points. Of course, the axis of these parabolas aren't necessarily parallel to the y axis. Given the points P_{j} with coordinates (x_{j},y_{j}) for j = 1,2,3,4, we are free to translate the coordinate system so that x_{3} = y_{3} = 0. In terms of this coordinate system we can fit a second degree polynomial to the three points P_{1},P_{3},P_{4}. Since P_{3} is at (0,0) this polynomial has the form 

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Similarly we can fit a second degree polynomial to the three points P_{2},P_{3},P_{4}. This polynomial will be of the form 

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The coefficients c_{2} and d_{2} are given by 

_{} 

The numerators of these coefficients are invariant under rotations but the denominators are not. Thus, if we rotate the coordinate system relative to the configuration of points, we can vary the values of c_{2} and d_{2}. There exists a particular orientation in which c_{2} = d_{2}, and for this orientation the difference between the two parabolas is Dy’ = (c_{1} – d_{1})x’, which is a straight line that passes through zero x_{3}’ and x_{4}’, and hence is identically zero. Therefore, the two parabolas are the same, so this is the orientation for which a single polynomial of degree 2 passes through all four points. 

The original coordinates x,y are related to the rotated coordinates x’,y’ by a transformation of the form 

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for some angle q. Substituting for x and y into the expressions for c_{2} and d_{2}, and setting c_{2} = d_{2}, we arrive at a quadratic equation of the form 

_{} 

If we choose our (arbitrary) initial orientation and scale factor such that x_{4} = 1 and y_{4} = 0 then the coordinates of this quadratic are simply 

_{} 

The two roots of the quadratic give the two orientations in which the four points lie on a single second degree polynomial. (Incidentally, this shows that if a parabola with vertical axis intersects a horizontal line at points P,Q, and a second parabola intersects the first at the points P,Q,R,S, then the slope of the second parabola's axis is half the slope of the line through R,S.) 

In general, given the coordinates of any four distinct points, we can express the configuration in purely relative terms (since the position and orientation don’t matter), e.g., as the lengths of the three segments 

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and the angles 

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If we let s_{j} and c_{j} denote the sine and cosine of q_{j}, then the angle theta that the ray P_{3}  P_{4} must make with the positive x axis in order for the four given points to lie on a single parabola must satisfy the quadratic 

_{} 

where 
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As an example, consider the four points (4,4), (3,2), (0,0), and (3,1). This relative configuration of points can be represented by 

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To make a single parabola pass through all four points we must place P_{3} at the origin and orient the set of points such that the segment P_{3}  P_{4} makes an angle of either 0.574133 or 0.750499 radians with the positive x axis. In our original coordinate system we had P_{4} at (3,1), so the angle between P_{3}  P_{4} and the x axis was atan(1/3), which equals 0.32175055, so this means we have to rotate those four points from that configuration by either 0.252382 radians or 1.072250 radians. (Note that the signs of the angles are negated, because q was defined as the angle from the transformed coordinates to the original coordinates, whereas now we are rotating from the original to the transformed.) In the first case the coordinates of the four points become 

_{} 

and the parabola that passes through those points is 

_{} 

In the second case the coordinates of the four points become 

_{} 

and the parabola that passes through those points is 

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We can transform both of these parabolas back to the original coordinate system, and show how both general parabolas pass through all four points. This is illustrated in the figure below. 


This method can evidently be applied to give the parabola passing through any four distinct points, and yet one may wonder how it's possible for a single parabola to pass through each of the four points in a configuration such as shown below: 


In general, if any of the points falls inside the triangle formed by the other three, then we can still find the parabola that contains all four points, but in such cases the parabola's coefficients are complex. As an example, consider the four points with the following coordinates 

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Here the point P_{3} falls inside the triangle formed by P_{1},P_{2},P_{4}. Using the formula given above, this leads to the equation 

_{} 

where q is the angle through which the xy coordinate axes must be rotated (counter clockwise) to give a coordinate system in which all four points lie on a single polynomial of degree 2. Solving this for theta gives the two complex solutions 

_{} 

Both of these roots correspond to 

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In the rotated coordinate system x'y', the coordinates of the four points are 

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The second degree polynomial that passes through all four points is 

_{} 

where 
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In the original xy coordinate system we have the following equation for this conic 

_{} 

where 
_{} 

with 
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Since b^{2} = 4ac, this conic is formally a parabola. For our example, the real and imaginary parts of the conic coefficients are as shown below. 

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It's interesting that the real and imaginary parts of these parabola coefficients, taken separately, give two real conics (hyperbolas), each of which passes through all four points as shown below. 


For more on this topic, see Revisiting the FourPoint Parabola. 
