The Khan Academy, Pixar and Parabolas

     The Khan Academy is a treasure which I recommend to everyone. Recently they added a short course called Pixar in a Box (link at bottom of entry) in which the first section describes how parabolic arcs are used to make the incredible fields of grass or fur on bears or hairs on heads, etc. It is a fun read and in about a half hour you will get the idea of how these things are done. However, I was left a bit dissatisfied about the explanation of why the construction used really defines a parabolic arc and so I worked out a proof that satisfied myself and perhaps might be of interest to others.
   
     The idea is that if you choose any three distinct points A, B, and C, and divide segments BA and BC into n equal subdivisions each, and connect the dividing points in reverse order, the envelope of the drawn segments will form a polygon and as n increases to infinity, the polygon will approach a parabolic arc. The Khan Academy lesson shows this in pictures which I recommend viewing.

  
     What I will do here is prove the general statement of this property.

    Statement to be Proved: Given any three distinct points A, B, C, define the sets of points 
Q(t) = (1-t)A + tB, R(t) = (1-t)B + tC, for t in [0,1]. The set of points 
P(t) = (1-t)Q(t) + tR(t) for t in [0,1] define a parabolic arc.

Proof: We know that we have a parabolic arc if we can show that there is a point F and a line L such that for every point P in the set, the distance from P to F equals the distance from P to the line L. 

Define the two-dimensional rectilinear coordinate system {A,B,C} as the system with origin B, a “vertical” v-axis being the line through B and the midpoint M of segment AC where M is on the positive half of the axis, and a “horizontal” u-axis being the perpendicular to line BM through B such that the unit positive u-vector is the cross product of the unit positive v-vector and the unit vector k pointed upward from the uv-plane.

In {A,B,C}, let the points have coordinates A(a₁, a₂), C(c₁, c₂), B(0, 0), 

M( (a₁₊ c₁)/2, (a₂₊ c₂)/2 ) where by definition (a₁₊ c₁)/2 = 0, so c₁ = -a₁ .

Then P(t) = (1-t)Q(t) + tR(t) = (1-t)[(1-t)A + tB] + t[(1-t)B + tC]

= t² (A-2B+C) +2t(B-A) + A, which we can write componentwise as the pairs:

P(t) = [t² (a₁ - 2b₁ + c₁) +2t(b₁ - a₁) + a₁) , t² (a₂ - 2b₂ + c₂) +2t(b₂ - a₂) + a₂) ] 

= [ a₁(1-2t) , t² (a₂ + c₂) + a₂(1-2t) ] for t in [0,1].

Let u = a₁(1-2t) so t = (a₁ - u)/2a₁ and P(u) = [ u , au² + bu + c] for u in [a₁, -a₁] where 

a = (a₂ + c₂)/4a₁² , b = (a₂ - c₂)/2a₁ , c = a a₁² . This shows P(u) forms a parabolic arc (i.e. a quadratic in the {A,B,C} coordinate system).

In particular, we know that for a parabola v = au² + bu + c ,

the vertex V is ( -b/2a , c - b²/4a), the focus F is V+(0, p), and the directrix L is the line v = v₂ - p, where p = 1/(4a). Thus we have proved the statement.

Example: Find an equation for the parabolic arc through A(3,9) and C(2,4) as defined by P(t) above with B(5/2,6).

Let the original rectilinear coordinate sytem which defines A, B, and C for the problem be called the standard xy-coordinate system. In this system the midpoint of segment AC is M(5/2, 13/2). We define the {A,B,C} rectilinear coordinate system as that having origin B(0,0), “vertical” axis as the line through B and M, and the “horizontal” axis as the perpendicular to line BM through B. In this case the “vertical” axis is vertical with respect to the standard coordinates because B and M have the same first coordinate in that system. This tells us that there is no rotation involved in the change of coordinates. Therefore the change of coordinate systems involves only a translation sending point (x,y) in the standard system to the point (x-5/2, y-6) in the new system. 

Thus A(3,9) becomes A(1/2, 3), B(5/2,6) becomes B(0,0), and C(2,4) becomes C(-1/2,-2). Using these translated coordinates, we can write, using the formulas derived in the proof above, P(t)  = [(1/2 - t), t²  - 6t + 3], for t in [0,1] so letting u = (1/2 - t), get P(u) = [ u , u² + 5u + 1/4], for u in [1/2, -1/2]. 

Thus p=1/4, V(-5/2,-6), F(-5/2, -6 +1/4), L is v= -6 - 1/4. These values can be translated back to the standard coordinate system by adding (5/2,6) to get the values for the desired focus and directrix given the original points as F(0, 1/4) and y= -1/4 as well as the vertex (0,0) which of course correspond to the parabola y = x².

More work would be involved in an example where the line through B and M was rotated with respect to the standard y-axis but the theory is the same and of course the proof does not depend on the particulars of the orientation of the {A,B,C} system.

https://www.khanacademy.org/partner-content/pixar/environment-modeling-2/mathematics-of-parabolas2-ver2/a/parabolas-lesson-brief