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## Pixar in a Box

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

Lesson 2: Calculating parabolas- Start here!
- Weighted average of two points
- Weighted averages
- 2. Where is the touching point?
- Exploring the parabola construction
- 3. Compute the touching point
- Calculating the touching point
- 4. How can we prove this?
- Touching point
- Bonus: Completing the proof

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# 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.(54 votes)
- You are right. What was shown in the proof is that the calculated point is always on the curve. Not that the curve is a segment of a parabola.(22 votes)

- 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?(20 votes)- 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.(12 votes)

- 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`

.(5 votes)- 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`

.(11 votes)

- 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?(2 votes)- 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`

(5 votes)

- 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.(2 votes)- 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.(4 votes)

- 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?(2 votes)
- Where do the equations for parabolas come in?(2 votes)
- At2:10Dr. 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.?!(2 votes)
- Yes the lines will intersect when s equals t. You are also correct that P is s along QR and t along Q'R'.(0 votes)

- 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'`

?(2 votes)- Can you teach me how to learn the computer animations like spookiz(1 vote)

## Video transcript

(clunking) - Thanks for hanging in there. I know this discussion is getting a little bit technical, but we finally have all the tools we need to complete the derivation of a formula for the touching point on a parabola. But, before we continue, let's back up just little bit, and remind ourselves why we're doing this again. Well, we need that touching point formula so that shots like this in Brave can be computed really efficiently. Right, because that touching point will allow us to write computer programs to draw each blade of grass, without having to draw all the individual string on lines. So to turn this into formulas, let's again label things. So this light magenta line is controlled by the parameter t, So I'm going to label, as before, this point Q and this point R. The dark magenta line is controlled by the parameter S, so I'm gonna call this point Q prime, and this point R prime. Now, let's start writing down a few things we know. Well, we know that Q is a fraction t along the line segment AB, which means I can write Q as one minus t times A plus t times B. Similarly, R is a fraction t along the line segment BC, so it can be written as one minus T times B plus t times C. Similarly, Q prime is s along the way from A to B, so I can write Q prime as one minus s times A, plus s times B, and I can write R prime, finally, as one minus s times B plus s times C. Okay, now this intersection point here that we're looking for, P, is somewhere on the line segment QR. But where on the line segment is it? Well, I'm gonna prove in a second that it's at fraction s. That is, I claim that P can be written as one minus s, times Q, plus s times R. Now, if that's true, something nice happens, because as s approaches t, this expression here approaches one minus t, times Q, plus t times R. And that's the thing I ultimately want to prove. So the only thing left to show is that the intersection point can be written this way. Why should that be the case? What I'm going to do, is I'm gonna substitute this expression for Q in here, this expression for R in here, and if I do that and rearrange, I'll leave that rearrangement to you, but the result is that P can be written as one minus s, times one minus t, times A, plus s times one minus t, plus t times one minus s, times B, plus s, t, times C. And now, if I rewrite this, using these expressions for Q prime, R prime, I see I can write P as one minus t, times Q prime, plus t times R prime. Well, this expression says that P is somewhere on the line segment Q prime R prime, and this expression says that P is somewhere on the line segment QR. And the only point that can be on both line segments is the intersection point. So our proof is complete. (bow stretching) (arrow thumps) Bullseye!