# Equation for a Locus I'm wondering how to go about finding the equation that defines this locus. In short, the locus is all the points the same distance from the closest point on $$e^x - 1$$ as $$(0,1)$$. The setup in GeoGebra also lets me change the function it depends on easily which produces very interesting results for almost every function (for a straight line, it produces a parabola).

• can you express the distance from any point $(x,y)$ to any point $(x',e^{x'}-1)$? – aradarbel10 Sep 18 '20 at 23:10
• One of the definitions of the parabola is the locus of a point which is equidistant from a fixed point (the focus) and a straight line (the directrix), – sammy gerbil Sep 21 '20 at 17:16

## 1 Answer

The distance from any point $$P(x, y)$$ on the required locus to the fixed point $$Q(0, 1)$$ is given by $$PQ^2=x^2+(y-1)^2$$ The distance from the same point $$P(x, y)$$ to a point $$T$$ on the given curve $$(t, e^t-1)$$ is given by $$s=PT^2=(x-t)^2+z^2$$ where $$z=y-e^t+1$$ hence $$y-1=z+e^t-2$$.

The point $$T$$ is closest to $$P$$ when $$s$$ is a minimum wrt $$t$$ : $$\frac{ds}{dt}=2(x-t)(-1)+2z(-e^t)=0$$ $$x=t-ze^t$$

The required locus is defined by $$PQ^2=PT^2$$ : $$x^2+(y-1)^2=(x-t)^2+z^2$$

Substitute to eliminate $$x, y$$ leaving an equation in $$z, t$$ : $$(t-ze^t)^2+(z+e^t-2)^2=z^2e^{2t}+z^2$$ $$t^2-2tze^t+e^{2t}+4-2(ze^t-2z-2e^t)=0$$ $$z=\frac{t^2+(e^t-2)^2}{2[(t-1)e^t+2]}$$

Co-ordinates of points on the locus can be obtained in terms of parameters $$t$$ and $$z(t)$$ :

$$x=t-ze^t, y=z+e^t-1$$