Parametrization of curve in $\Bbb R^3$

Question

I’m trying to find the parametrization of a curve in $\Bbb R^3$ satisfying $p_1=(0,1,1)$ and $p_2=(1,0,0)$ that parallel projects to the x-y plane onto $\log(x)\log(y)=1.$ I think that maybe I can do a rotation on the x-y plane curve and then scale its length to satisfy the endpoint conditions? Also, I must note that $p_1,p_2$ are undefined for the function $\log(x)\log(y)=1.$ So I don’t require that the function has to be defined exactly at the two points $p_1,p_2.$ Any help is appreciated! Thanks.

Answer 1

HINT:

Let

$$ \log x\log y=\text{ a constant =} \log p\log q \to \log_p x \log_ q y = 1$$

from which again let

$$ \log_p x =u,\; \log_ q y = 1/u $$

$z$ can be arbitrarily chosen. So a parameterization can be

$$(x,y,z)= (p^u,q^{1/u}, v \frac13 \cos u ) $$

> p = 0.8; q = 1.25; ParametricPlot3D[{p^u,    q ^(1/u), (v Cos[u]/3)},
> {u, -4, 2}, {v, -2, 4},   PlotPoints -> {40, 60}, PlotStyle -> Yellow]

We have yet to choose constants to make the surface contain the given points.