# What is the equation to produce this Bezier curve?

What is the equation for the first curve in the image?

I have this formula:

But when I substitute values, I get an image that looks like x^2. It isn't the same shape as the image.

• $P_0,P_1,P_2$ is probably the points. I would guess $P_1$ is the cross point. You treat them as vectors [x,y] when you do the calculations. Jun 22, 2018 at 5:10
• So if you want to apply Bezier formula in order to get an expression you can use it to calculate $x$ and $y$ as functions of $t$ and then try to simplify. Jun 22, 2018 at 5:16

The three points $P_0,P_1,P_2$ are the control points of the quadratic Bezier segment. On the images these points are connected with straight lines. $P_0$ and $P_2$ are the endpoints of the curve, $P_1$ (marked with $\times$) usually is not on the curve. The formula \begin{align} B(t)&=(1-t)^2P_0+2(1-t)tP_1+t^2P_2 ,\quad t\in[0,1] \end{align}
is parametric, that is there are two expressions in terms of parameter $t$, that define $x(t)$ and $y(t)$:

Example for $a=3$: \begin{align} B_x(t)&=(1-t)^2P_{0x}+2(1-t)tP_{1x}+t^2P_{2x} ,\\ B_y(t)&=(1-t)^2P_{0y}+2(1-t)tP_{1y}+t^2P_{2y} . \end{align}

So for the first image we can assume that the control points looks like

and for the second one

for some constant $a$.

Note, that coordinates of the points are completely independent of the parametric range $[0,1]$.

The two images demonstrate how the curve changes, when just one endpoint is moved.

Example for $a=3$:

The tangent lines on the images illustrate the process of construction of the points on a quadratic Bezier curve by means of the points on linear Bezier segments:

1. Connecting the control points $P_0-P_1-P_2$, we create two linear Bezier segments: $P_0P_1$ with control points $P_0$ and $P_1$, and $P_1P_2$ with control points $P_1$ and $P_0$,

2. For any $t\in[0,1]$, find points $u_t,v_t$ on the line segments $P_0P_1$ and $P_1P_2$

\begin{align} u_t&=P_0(1-t)+P_1(t) ,\\ v_t&=P_1(1-t)+P_2(t) . \end{align}

Next, find the point $w_t$ on the linear Bezier segment $u_tv_t$ with control points $u_t$ and $v_t$,

\begin{align} w_t&=u_t(1-t)+v_t t , \end{align}

and the point $w_t$ is the point on the quadratic Bezier curve.

In the image, the curve seems to be the envelope of the lines whose $x$ and $y$ intercepts add to $1$. These lines are given by

$$y=y_0\left(1-\frac x{1-y_0}\right)\;,$$

where $y_0$ is the $y$ intercept.

The envelope maximizes $y$ with respect to $y_0$ for given $x$. Setting the derivative of $y$ with respect to $y_0$ to zero yields

$$1-\frac x{1-y_0}-\frac{y_0x}{(1-y_0)^2}=0$$

and thus $y_0=1-\sqrt x$. Substituting this into the equation for the lines yields $y=\left(1-\sqrt x\right)^2$, or in more manifestly symmetric form,

$$\sqrt x+\sqrt y=1\;.$$

This can also be written parametrically as $(x,y)=(\sin^4 t,\cos^4 t)$.

• Jun 22, 2018 at 7:02

Using vector notation $[x,y]^T$ : First we need to write down the points: $$P_0 = [0,1]^T, P_1 = [0,0]^T, P_2 = [1,0]^T$$ Now to the formula: $$B(t) = (1-t)^2[0,1]^T + (1-t)t[0,0]^T + t^2[1,0]^T = ...\\ [t^2,(1-t)^2]^T$$

So it is true you get quadratic functions, but the quadratic functions are in both $x$ and $y$ coordinates.

$$x = t^2, y = 1-2t+t^2$$ and then substitute $x=\sqrt{t}$ based on our interval $x\in[0,1]$, which belongs to positive real numbers,

$$y = 1-2\sqrt{x}+x$$

You can check with this Wolfram alpha plot.