# Conic section and Bézier curves

I'd like to prove or disprove the following:

given a compact arc of a conic section $$\mathcal{T} \subset \mathbb{R}^2$$ such that for any $$P \neq Q \in \mathcal{T}$$ the tangent vectors $$\tau_P, \tau_Q$$ aren't parallel, then i can parametrize it using a quadratic Rational Bézier curve.

First of all I can find the three points: the endpoints of the arc and the intersection of the tangent lines at the endpoints (here i use the hypotesis making this intersection nonempty, as a counterexample we can pick a semicircumference). Then I can, without loss of generality, normalize the weights $$w_0=w_2=1$$. Now, how can I find $$w_1$$? I searched on NURBS,Farin, online and even on Stack Exchange. I didn't find a general case (only the one with circular arcs) and moreover it seems that you can only approximate an arc of conic section (why?).

Edit1: I found these slides https://public.vrac.iastate.edu/~oliver/courses/me625/week11.pdf but I don't understand nor the direct approach (because who grants me that of the three equations at least two are linearly independent) nor the geometric one (that one I just don't get it at all).

• I think it’s $w_1$ that you’re trying to find, not $w_2$. Commented May 10, 2022 at 22:06
• Yep typo, thanks Commented May 10, 2022 at 22:11

Choose a point $$P$$ on the conic and a line parameterized by $$L(t)= A + tB$$. For any given value of $$t$$, construct a line through $$P$$ and the line point $$L(t)$$. Since the conic is a curve of degree 2, this line must intersect the conic in two points. By construction, one of these points is $$P$$; call the other one $$C(t)$$. If you work through the algebra, you’ll find that $$C(t)$$ is a rational quadratic function of $$t$$.

Alternatively: use the geometry of the given conic to figure out its “shape factor”, $$s$$. You’ll have to find the intersection of the conic with the line joining its tangent intersection point $$P_1$$ and the mid-point $$M$$ of the chord $$P_0P_2$$. In other words, you have to find $$s$$ such that this intersection occurs at $$(1-s)M + sP_1$$. Then compute the weight of the middle control point from this shape factor: $$w_1 = s/(1-s)$$, as explained in the slides you found.

• I'd like to use the first method since i didn't quite get how to find the shape factor. $A$ and $B$ are the end points of the arc?If so, how comes the line through$L(t)$ and $P$ intesects the conic twice for any given $t$? Commented May 10, 2022 at 18:14
• You can use any line you like as long as it doesn’t pass through the point P. To make sure I wasn’t talking nonsense, I worked through an example with P equal to the origin and A and B chosen to give me the line $x=1$, so A=(1,0,0) and B=(0,1,0). Commented May 10, 2022 at 21:58
• The second method is easier. The shape factor $s$ is simple enough. The conic has to intersect the line segment $MP_1$ somewhere, right. Call the intersection point $Q$. So we can find $s$ such that $Q = (1-s)M + sP_1$. In fact $s=|MQ|/|MP_1|$. Commented May 10, 2022 at 22:04
• Ok I think that for what I need the first method might not do the trick. I'm trying to parametrize an arc of the conic, and in some examples the point $C(t)$ is on the conic but not on the arc. I should find suitable $A$ and $B$ ecc... Meanwhile i understood the relationship between $w_1$ and $s$ but i don't know how to find $s$. If I compute it as you suggested, why would it be the shape parameter of the conic? As an alternative I could also use the method on page 432 of the slides but that's way too unformal: is that a projective map?points in general position? ecc... Commented May 10, 2022 at 22:19
• ‘Why would it be the shape parameter” — that ratio is the definition of the shape parameter. Commented May 10, 2022 at 23:16

We can do this from first principals. From the given conic arc, we can compute the Bézier control points $$P_0, P_1, P_2$$. And we know we can assume that the weights for the first and last control points are both $$1$$. So, then, the equation of the Bézier curve is: $$C(t) = \frac {(1-t)^2P_0 + 2t(1-t)wP_1 + t^2P_2} {(1-t)^2+ 2t(1-t)w+ t^2}$$ We just need to find the middle weight, $$w$$.

Let $$M = \tfrac12(P_0 + P_2)$$ be the mid-point of the chord, and let $$Q$$ be the point where the conic intersects the line segment $$MP_1$$. After computing $$Q$$, we define $$k = |MQ|/|MP_1|$$. Then $$Q = (1-k)M + kP_1 = \tfrac12(1-k)P_0 + kP_1 + \tfrac12(1-k)P_2.$$ Our strategy is to choose $$w$$ so that $$C(\tfrac12) = Q$$. Then the Bézier curve will pass through the points $$P_1, Q, P_2$$ and it will have the same end tangents as the original conic. A conic is fully determined by three points and two tangents, so if we accept that our rational quadratic curve is a conic, then it must be the same conic as the original one.

But $$C(\tfrac12) = Q$$ gives us $$\frac {\tfrac14P_0 + \tfrac12wP_1 + \tfrac14P_2} {\tfrac14 + \tfrac12w+ \tfrac14} = \tfrac12(1-k)P_0 + kP_1 + \tfrac12(1-k)P_2.$$ Equating coefficients of $$P_1$$. We get $$\frac{w}{1+w}= k$$ And rearranging gives $$w = \frac{k}{1-k}$$

• Thank you very much! Commented May 11, 2022 at 0:53