# How do I find the equation of a curve formed by a series of lines tangent to a curve?

When I was young I used to draw a sequence of straight lines on graph paper which made a curve after I finished. On a coordinate plane, the lines would be equivalent to starting at $y=9$ on the $y$ axis and ending at $x=1$ on the $x$ axis. With each line, I would decrease the $y$ by one unit and increase the $x$ by one unit.

Here is a desmos graph that illustrates. https://www.desmos.com/calculator/u4ea8swmfg

Here is a similar example where the angle between the lines is 60 degrees. https://drive.google.com/open?id=0B5QHq_oPha0ybGdrbFNhUHRPOGc

I think each line is basically a tangent line along the curve produced by taking a first derivative.

Are these types of curves parabolas or possibly hyperbolae? And how could I find the equation of such a curve?

• What you're looking for is the envelope of a family of curves. See in particular Example 2 of the Wikipedia article. Jan 27, 2017 at 21:15
• @ Christopher Gaston $\sqrt x + \sqrt y = \sqrt a$ is the equation of a parabola with each tangential intercept = $a.$ Jan 27, 2017 at 22:00
• I have suppressed the tag "hyperbolic geometry" which deals with one of the non-euclidean geometry, which is not the case here. Jan 27, 2017 at 23:04

The curve you have described is in fact a parabola; specifically, it is the parabola generated as the quadratic Bézier curve with the control points as the ends of your lines.

The equation of this parabola is, for three points $a, b, c$:

$$f(t)=(1-t)^2a+t(1-t)b+t^2c$$

This is a parametric formula; you enter a $t$ value and it gives the $x$ and $y$ values of a point on the curve, as opposed to entering an $x$ value and getting the $y$ value out.

In the first case, with the right angle, the parabola is simply

$$f(t)=\left(10t^2,10(1-t)^2\right)$$

shown in red on this figure.

The $60^\circ$ angle looks like this.

• [+1] Very good answer. Jan 27, 2017 at 22:04

With points $A_n(0,n)$ on $y-$axis and $B_n(10-n,0)$ on $x-$axis the equations of all lines pass through are $$y=\dfrac{-n}{10-n}x+n$$ with $C$ as the variable of curves family, we write $$y=\dfrac{-C}{10-C}x+C$$ for envelope of curves family, $\dfrac{\partial f}{\partial C}=0$ where $$f=y+\dfrac{C}{10-C}x-C$$ after deletion $C$ between $f$ and $\dfrac{\partial f}{\partial C}=0$ we get $$10x=\Big(10-\frac{y\sqrt{10}}{\sqrt{10}-\sqrt{x}}\Big)^2$$