It is [well documented](https://en.wikipedia.org/wiki/String_art) that certain string-art patterns generate quadratic Bezier curves: let $x, y\_1, y\_2$ be three points in $\mathbb{E}^2$, consider the family of line segments joining $x + (1-t) (y\_1 - x)$ to $x + t (y\_2 - x)$ for $t\in [0,1]$, its envelope forms a quadratic Bezier curve (and hence a parabola). Furthermore, at the parameter $t$, the intersection of the line segment with the parabolic curve is at
$$ (1-t)[x + (1-t)(y\_1 - x)] + t[x + t(y\_2 - x)] $$
the linear interpolant with parameter $t$ on the line segment.


It is also not hard to verify that given the same three points, considering the family of line segments joining $x + e^t(y\_1 - x)$ to $x + e^{-t}(y\_2 - x)$, its envelope forms a hyperbola. The intersection of the line segments with the hyperbola always occur at the midpoint.


In both cases the construction changes appropriately under affine transformations of $\mathbb{E}^2$, and in the case $x, y\_1, y\_2$ being colinear, recovers the degenerate conics of the appropriate types.


Given that the parabolas can be thought of as a limit of hyperbolas.



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> *Question 1*: Is there a way to take a "limit" of the construction above for the hyperbolas to get the one for the parabola?
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> *Questions 2*: Can this be taken further to include elliptical arcs as part of the family of constructions?
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Conics, string art, and Bezier-like curves
You could do the usual string-art construction in 3D (giving a 3D parabola), and then do a central projection $(x,y,z) \mapsto (x/z,y/z,1)$ down onto the plane $z=1$. This will give you any conic section curve you like. 


Suppose the three 3D points are $\mathbf{P}\_i = (x\_i, y\_i,z\_i)$ for $i=1,2,3$. We can assume WLOG that $z\_1 = z\_3 = 1$. If $z\_2= 1$, you get a a parabola, if $z\_2 < 1$, you get an ellipse, and if $z\_2 > 1$, you get a hyperbola. 


This is related to rational quadratic Bezier curves. 


