# Tangent line of curve
**URL:** https://discourse.processing.org/t/tangent-line-of-curve/19600
**Category:** Coding Questions
**Created:** [April 9, 2020, 9:59pm UTC](https://discourse.processing.org/t/tangent-line-of-curve/19600 "2020-04-09T21:59:10Z")
**Posts on this page:** 3
**Page:** 1
### Author:  [@Martaf2001](https://discourse.processing.org/u/Martaf2001)
#### Post date: [April 9, 2020, 9:59pm UTC](https://discourse.processing.org/t/tangent-line-of-curve/19600/1 "2020-04-09T21:59:10Z")
Hi everyone!!!
I’m new at coding and I have a problem, I would like to draw a program where there is a bicorn curve and every time the mouse is on the curve, it can be generated a tangent line, so on infinite points, but for now I just have the static drawing but I don’t know how to generate a tangent line on the curve as the mouse move…
This is all I have for now
```auto
void setup(){
size(500,500);
}
void draw(){
background(0);
translate(width/2, height/2);
noFill();
stroke(255);
beginShape();
for(float a=0; a < TWO_PI; a+=0.01) {
float r=200;
float x=(r)*sin (a);
float y=(-r)*(pow(cos(a),2)/(2-cos(a)));
vertex(x,y);
}
endShape();
}
```
THANK YOU SO MUCH !!! 😄
---
### Author:  [@glv](https://discourse.processing.org/u/glv)
#### Post date: [April 10, 2020, 9:13am UTC](https://discourse.processing.org/t/tangent-line-of-curve/19600/2 "2020-04-10T09:13:45Z")
Hello,
Welcome to the forum.
Please format your code:
[https://discourse.processing.org/faq#format-your-code](https://discourse.processing.org/faq#format-your-code)
---
### Author:  [@jeremydouglass](https://discourse.processing.org/u/jeremydouglass)
#### Post date: [April 11, 2020, 9:57pm UTC](https://discourse.processing.org/t/tangent-line-of-curve/19600/3 "2020-04-11T21:57:16Z")
One method:
Currently you draw your bicorn with:
```auto
float x=(r)*sin (a);
```
Say you have your mouseX as input. Your mouseX = 50. What is a solution for `a` for a point on your bicorn that also has x = 50?
If you can solve for a, then you know r (constant) and x (the mouse) – so now you can solve for y on your bicorn. Now you know a point on your bicorn that is x-aligned with the mouse.
To test, add this to the end of your draw:
```auto
float mx = mouseX-width/2;
float my = mouseY-height/2;
ellipse(mx, my, 10, 10);
mouseCurvePoint(mx);
```
…and create a mouseCurvePoint function to draw a circle.
```auto
void mouseCurvePoint(float x) {
// solve 'a' here given x, r
// solve 'y' here given x, r, a
println(x, a);
ellipse(x, y, 10, 10); // mark your solution point
}
```
Now that you have a point, you need to solve for the slope at that point in order to know the slope of your tangent line. You can then create a line segment of that slope that passes through your point using the slope-intercept form of a line – and you have your tangent!