This is definitely pushing the limits for my trig knowledge.

Is there a formula for calculating an intersection point between a quadratic bezier curve and a line?

Example:

in the image below, I have P1, P2, C (which is the control point) and X1, X2 (which for my particular calculation is just a straight line on the X axis.)

What I would like to be able to know is the X,Y position of T as well as the angle of the tangent at T. at the intersection point between the red curve and the black line.

After doing a little research and finding this question, I know I can use:

t = 0.5; // given example value
x = (1 - t) * (1 - t) * p[0].x + 2 * (1 - t) * t * p[1].x + t * t * p[2].x;
y = (1 - t) * (1 - t) * p[0].y + 2 * (1 - t) * t * p[1].y + t * t * p[2].y;

to calculate my X,Y position at any given point along the curve. So using that I could just loop through a bunch of points along the curve, checking to see if any are on my intersecting X axis. And from there try to calculate my tangent angle. But that really doesn't seem like the best way to do it. Any math guru's out there know what the best way is?

I'm thinking that perhaps it's a bit more complicated than I want it to be.

This is definitely pushing the limits for my trig knowledge.

Is there a formula for calculating an intersection point between a quadratic bezier curve and a line?

Example:

in the image below, I have P1, P2, C (which is the control point) and X1, X2 (which for my particular calculation is just a straight line on the X axis.)

What I would like to be able to know is the X,Y position of T as well as the angle of the tangent at T. at the intersection point between the red curve and the black line.

After doing a little research and finding this question, I know I can use:

t = 0.5; // given example value
x = (1 - t) * (1 - t) * p[0].x + 2 * (1 - t) * t * p[1].x + t * t * p[2].x;
y = (1 - t) * (1 - t) * p[0].y + 2 * (1 - t) * t * p[1].y + t * t * p[2].y;

to calculate my X,Y position at any given point along the curve. So using that I could just loop through a bunch of points along the curve, checking to see if any are on my intersecting X axis. And from there try to calculate my tangent angle. But that really doesn't seem like the best way to do it. Any math guru's out there know what the best way is?

I'm thinking that perhaps it's a bit more complicated than I want it to be.

• javascript
• html
• canvas
• trigonometry

• 3 Maybe Pomax can offer you some help. – Edward J Beckett Commented Dec 27, 2014 at 5:31
• 2 I'm pretty sure I covered it in the Primer, in the section on curve intersection. Simply rotate your curve and line so that the line "lines up" with the x-axis, and now you've reduced "finding the intersection" to "root finding" (with aligning explained here) – Mike 'Pomax' Kamermans Commented Mar 11, 2016 at 7:01

3 Answers 3

Quadratic curve formula:

y=ax^2+bx+c // where a,b,c are known

Line formula:

// note: this `B` is not the same as the `b` in the quadratic formula ;-)

y=m*x+B  // where m,B are known.

The curve & line intersect where both equations are true for the same [x,y]:

Here's annotated code and a Demo:

// canvas vars
var canvas=document.getElementById("canvas");
var ctx=canvas.getContext("2d");
var cw=canvas.width;
var ch=canvas.height;

// linear interpolation utility
var lerp=function(a,b,x){ return(a+x*(b-a)); };

// qCurve & line defs
var p1={x:125,y:200};
var p2={x:250,y:225};
var p3={x:275,y:100};
var a1={x:30,y:125};
var a2={x:300,y:175};

// calc the intersections
var points=calcQLintersects(p1,p2,p3,a1,a2);

// plot the curve, line & solution(s)
var textPoints='Intersections: ';
ctx.beginPath();
ctx.moveTo(p1.x,p1.y);
ctx.quadraticCurveTo(p2.x,p2.y,p3.x,p3.y);
ctx.moveTo(a1.x,a1.y);
ctx.lineTo(a2.x,a2.y);
ctx.stroke();
ctx.beginPath();
for(var i=0;i<points.length;i++){
  var p=points[i];
  ctx.moveTo(p.x,p.y);
  ctx.arc(p.x,p.y,4,0,Math.PI*2);
  ctx.closePath();
  textPoints+=' ['+parseInt(p.x)+','+parseInt(p.y)+']';
}
ctx.font='14px verdana';
ctx.fillText(textPoints,10,20);
ctx.fillStyle='red';
ctx.fill();

///////////////////////////////////////////////////

function calcQLintersects(p1, p2, p3, a1, a2) {
  var intersections=[];

  // inverse line normal
  var normal={
    x: a1.y-a2.y,
    y: a2.x-a1.x,
  }

  // Q-coefficients
  var c2={
    x: p1.x + p2.x*-2 + p3.x,
    y: p1.y + p2.y*-2 + p3.y
  }

  var c1={
    x: p1.x*-2 + p2.x*2,
    y: p1.y*-2 + p2.y*2,
  }

  var c0={
    x: p1.x,
    y: p1.y
  }

  // Transform to line 
  var coefficient=a1.x*a2.y-a2.x*a1.y;
  var a=normal.x*c2.x + normal.y*c2.y;
  var b=(normal.x*c1.x + normal.y*c1.y)/a;
  var c=(normal.x*c0.x + normal.y*c0.y + coefficient)/a;

  // solve the roots
  var roots=[];
  d=b*b-4*c;
  if(d>0){
    var e=Math.sqrt(d);
    roots.push((-b+Math.sqrt(d))/2);
    roots.push((-b-Math.sqrt(d))/2);
  }else if(d==0){
    roots.push(-b/2);
  }

  // calc the solution points
  for(var i=0;i<roots.length;i++){
    var minX=Math.min(a1.x,a2.x);
    var minY=Math.min(a1.y,a2.y);
    var maxX=Math.max(a1.x,a2.x);
    var maxY=Math.max(a1.y,a2.y);
    var t = roots[i];
    if (t>=0 && t<=1) {
      // possible point -- pending bounds check
      var point={
        x:lerp(lerp(p1.x,p2.x,t),lerp(p2.x,p3.x,t),t),
        y:lerp(lerp(p1.y,p2.y,t),lerp(p2.y,p3.y,t),t)
      }
      var x=point.x;
      var y=point.y;
      // bounds checks
      if(a1.x==a2.x && y>=minY && y<=maxY){  
        // vertical line
        intersections.push(point);
      }else if(a1.y==a2.y && x>=minX && x<=maxX){
        // horizontal line
        intersections.push(point);
      }else if(x>=minX && y>=minY && x<=maxX && y<=maxY){
        // line passed bounds check
        intersections.push(point);
      }
    }
  }
  return intersections;
}

body{ background-color: ivory; padding:10px; }
#canvas{border:1px solid red;}

<h4>Calculate intersections of QBez-Curve and Line</h4>
<canvas id="canvas" width=350 height=350></canvas>

If you only need an intersection with a straight line in the x-direction you already know the y-coordinate of the intersection. To get the x-coordinate do something like this:

• The equation for your line is simply y = b
• Setting it equal to your y-equation of the beziér function y(t) gets you:
  b = (1 - t) * (1 - t) * p[0].y + 2 * (1 - t) * t * p[1].y + t * t * p[2].y
• Solving* for t gets you:
  t = (p[0].y - p[1].y - sqrt(b*a + p[1].y*p[1].y - p[0].y*p[2].y)) / a
  with a = p[0].y - 2*p[1].y + p[2].y
• Insert the resulting t into your x-equation of the beziér function x(t) to get the x-coordinate and you're done.

You may have to pay attention to some special cases, like when no solution exists, because the argument of the square root may then become negative or the denominator (a) might become zero, or something like that.

Leave a comment if you need more help or the intersection with arbitrary lines.

(*) I used wolfram alpha to solve the equation because I'm lazy: Wolfram alpha solution.

calculate line's tangθ with x-coordinate

then intersection of the curve's (x, y) should be the same tangθ

so solution is

a = line's x distance from (line.x,0) to (0,0)

(curve.x + a) / curve.y = tangθ (θ can get from the line intersection with x-coordidate)