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## Pixar in a Box

### Course: Pixar in a Box>Unit 11

Lesson 2: Calculating parabolas

# Bonus: Completing the proof

One way we can prove that we are calculating the actual touching point.

## Want to join the conversation?

• I understand all "mechanical" part of this, but I do not get it how this proves that point P lies on the parabola. • I'm confused.
How and why do you prove that the point is always on the curve?
I thought the curve is defined by the points
If the curve is defined before the points are defined, how did you define the curve? • Yes, you're right - the point is always one the curve because the curve is defined by the points. This proof is demonstrating that, given that the point is one the curve, then the equation for the curve is P = (1 - t)Q + tR, that is that it is t along the line QR.
• Isn't this proof backwards? It proves that if
``P = (1 - s)Q + sR``

then
``P = (1 - t)Q' + tR'``

and therefore P lies on the intersection of `QR` and `Q'R'`. It does not appear to prove that if P is the intersection of `QR` and `Q'R'` then `P = (1 - s)Q + sR`. • The proof is fine, though the video is not so clear.
First of all, let's take an example:
The point `P` lies on a line called `XY` and thus, we are able to represent the point P by the formula:
`P = (1 - t)X + tY`
The fact that you can represent a point on a line with this kind of formula is reversible - it also shows that, given this formula, the point `P` must be somewhere along the line `XY`.
So, back to the video, you pick a point on the `QR` line which corresponds to:
`P = (1 - s)Q + sR`
According to the proof, it is then showed that:
`P = (1 - t)Q' + tR'`
This actually means that the point `P` is not only on the line `QR` (as shown in the first formula), but it is also on the line `Q'R'`.
When `t` is not equal to `s`, the lines `QR` and `Q'R'` have only one intersection point and it is `P`.
• I think I got the point of everything he explained so far, but I virtually understood how he substituted the formula for Q & R( Q = ( 1 - t) A + tB and R = (1 - t)B + tC ) into the formula for P ( P = (1 - s)Q + sR).
I substituted it myself but got something a little bit different. Please can someone explain how he substituted his? • It's possible you got the same answer but multiplied out some of the expressions.
`P = (1 - s)Q + sR`
Substitute for Q and R:
`P = (1 - s)((1 - t)A + tB) + s((1 - t)B + tC)`
Distribute the (1 - s) and s:
`P = (1 - s)(1 - t)A + (1 - s)tB + s(1 - t)B + stC`
Factor the B out:
`P = (1 - s)(1 - t)A + (s(1 - t) + (1 - s)t)B + stC`
• Is there somewhere you can direct me to see the steps for simplifying the equation P=(1-s)Q +sR? I am getting lost in the translation provided in the video.
Thanks,
Scott. • P = (1 - s)Q + s R -------> (1 - t)Q + t R
( Here they substituting or rearranging he equations for Q{= (1 - t)A + t B} and R{= (1 - t)b + t c} )
So, the equation becomes
P = (1-s)Q + s R.
P = (1-s) ((1-t)A + t B) + s ((1-t)B + t C).
P = (1-s)(1-t)A + (1-s)(t B) + s(1-t)B + st C.
P = (1-s)(1-t)A + (s(1-t) + t(1-s))B + st C.
This is the derivation of the equation given in the video. Hope you understand it.
• I would love to know what software he is using to make that tutorial.
(1 vote) • To me I think that it seems to be a lot easier for you guys to make a movie with more people, but have you ever had a period or chunk of time where their were not enough people to construct parts of the movie so you had to push back the date for putting a movie out into theaters? • Where do the equations for parabolas come in? • At Dr. DeRose states that p = (1-s)Q + sR approaches (1- t)Q + tR. Is that equivalent to saying that the magenta line will super impose on the pink line? I was able to follow the substitutions of Q and R. I also rearranged the expression using partial factoring and understood the substitutions of Q prime and R prime arriving at Dr. Derose's final expression for p. I think I understand that since p can be written equivalent to a weighted average involving Q and R as well as a weighted average involving Q prime and R prime it is on both lines at the same time. That sounds like an intersection to me. I suppose I am a little confused because it seems that p is "s" along QR but it is "t" along Q prime R prime.?! • Why didn't you show the steps between
``P = (1 - s)(1 - t) * A + (s * (1 - t) + t * ( 1 - s)) * B + s * t * C``

and
``P = (1 - t) * Q' + t * R'``

? 