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GeoGebra Scheme for Said 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).

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  • $\begingroup$ can you express the distance from any point $(x,y)$ to any point $(x',e^{x'}-1)$? $\endgroup$ – aradarbel10 Sep 18 '20 at 23:10
  • $\begingroup$ 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), $\endgroup$ – sammy gerbil Sep 21 '20 at 17:16
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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 $z(t)$ :

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

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