Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I convert a segment of parabola to a cubic Bezier curve?

The parabola segment is given as a polynomial with two x values for the edges.

My target is to convert a quadratic piecewise polynomial to a Bezier path (a set of concatenated Bezier curves).

share|cite|improve this question
The title does not reflect the question. – lhf Mar 20 '13 at 19:36
Fixed that. Thanks. – Ben-Uri Mar 20 '13 at 20:17
See also – lhf Mar 20 '13 at 22:13
Just minor correction - I had submitted an edit but it was rejected for not being "substantive," and I don't have the rep to comment. To calculate the first control point the formula should be: $C=(\frac{x_1+x_2}{2},f(x_1)+f'(x_1)\cdot \frac{x_2-x_1}{2})$ Note the minor difference in computing the point's Y component, without which the formula results in an incorrect control point for segments of the parabola where $x1 \neq 0$. The conversion to a cubic Bezier works fine. – Roland Nov 19 '13 at 19:03
up vote 3 down vote accepted

You can do this in two steps, first convert the parabola segment to a quadratic Bezier curve (with a single control point), then convert it to a cubic Bezier curve (with two control points).

Let $f(x)=Ax^2+Bx+C$ be the parabola and let $x_1$ and $x_2$ be the edges of the segment on which the parabola is defined.

Then $P_1=(x_1,f(x_1))$ and $P_2=(x_2,f(x_2))$ are the Bezier curve start and end points and $C=(\frac{x_1+x_2}{2},f(x_1)+f'(x_1)\cdot \frac{x_2-x_1}{2})$ is the control point for the quadratic Bezier curve.

Now you can convert this quadratic Bezier curve to a cubic Bezier curve by define the two control points as: $C_1=\frac{2}{3}C+\frac{1}{3}P_1$ and $C_2=\frac{2}{3}C+\frac{1}{3}P_2$.

share|cite|improve this answer
How would you go about proving your formula for C if you don't mind me asking? – user2662833 Jan 6 '15 at 13:43

Let $f(x)=ax^2+bx+c$ be the parabola and let $[x_1, x_2]$ be the interval of the segment that you want to convert to cubic Bezier curve. The first step is to convert that segment into a quadratic Bezier curve as follows:

1) The start point is $P_1=(x_1,f(x_1))$ and the end point $P_2=(x_2,f(x_2))$.
2) The control point $C$ should be the intersection point of the tangent lines at $P_1$ and $P_2$. These two tangent lines can be found as

$y=(2ax_1+b)(x-x_1)+y_1$, and

Solving for the intersection point of above two line equations result in $C=(C_x,C_y)$ as


Once you have the quadratic Bezier curve, the cubic Bezier curve control points can be found as $C_1=\frac13P_1+\frac23C$ and $C_2=\frac23C+\frac13P_2$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.