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I have three points: A(85, 85, 0), B(-85, -85, 0) and C(0, 0, 30). I must find the equation of the arch that starts from A, finishes in B and goes through C.

Could you help me? I found something called catenary but it's something that overtakes my basic knowledge. Thank you.

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    There are lots of smooth arch-like curves that pass through your three points. Do you just need the equation of some simple curve that looks like an arch, or specifically that of a catenary? If the former, the parametric form of a parabola $(-85t, -85t, 30(1-t^2))$ for $t \in [-1, 1]$ should do.2012-06-01
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    A simple curve that looks like an arch is fine. However I need an implicit function in y that should be digited in a software called Imp View 3D, I dont know how to change this parametric equation.2012-06-02
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    What do you mean by an implicit function in $y$? Can you give an example?2012-06-02
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    For example: (0.75-sqrt(x^2+y^2))^2+z^2 < 0.25^2 - torus with major radius 0.75 and min radius 0.25 or cos(sqrt(x^2+z^2)*5)/22012-06-03

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Your two examples are surfaces, not curves. In fact (roughly speaking) any implicit equation of the form $f(x,y,z)=0$ will yield a surface.

One arch-like surface that will work is a parabolic one related to the previous answer:

$$ z = 30 \left(1-\frac{y^2}{7225}\right)$$

This is an arch-shaped surface that runs parallel to the x-axis. Similarly

$$ z = 30 \left(1-\frac{x^2}{7225}\right)$$ is an arch-shaped surface that runs parallel to the x-axis. Finally, the arch surface

$$z = 30 \left(1-\frac{(x+y)^2}{28900}\right)$$

is somehow "nicer" because it's symmetric in $x$ and $y$.

Taking the intersection of any of these three with the plane $x=y$ will give a curve that meets your specifications.