Conics, string art, and Bezier-like curves It is well documented 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.

Question 1: Is there a way to take a "limit" of the construction above for the hyperbolas to get the one for the parabola?
Questions 2: Can this be taken further to include elliptical arcs as part of the family of constructions?

 A: 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. 
A: Here I show how the description given by bubba connects the parabolic case to the hyperbolic case. 
Consider the three triples in $\mathbf{R}^3$:
$$ (x_0, y_0, z_0), (x_0^+, y_0^+, 1), (x_0^-, y_0^-, 1) $$
via the string-art construction we can define the two line segments
$$ (x_+(s), y_+(s), z_+(s)) = (x_0, y_0, z_0) + (1-s)(x_0^+, y_0^+, 1) $$
and
$$ (x_-(s), y_-(s), z_-(s)) = (x_0, y_0, z_0) + s (x_0^-, y_0^-, 1) $$
The envelope of the line-segments by connecting $(x_+(s), y_+(s), z_+(s))$ to $(x_-(s), y_-(s), z_-(s))$ are parabolic. It is an easy exercise in projective geometry that their corresponding central projections
$$ \left( \frac{x_\pm(s)}{z_\pm(s)}, \frac{y_\pm(s)}{z_\pm(s)} \right) $$
give rise to envelopes which are other conics, with, in particular, when $z_0 = 1$ a parabola, when $z_0 \in (0, 1)$ an ellipse, and when $z_0 > 1$ a hyperbola. 
Up to this point is what was written in the accepted answer above. 
It turns out that we can rewrite the formulae for the projected curves in the following form
$$ \frac{x_+(s)}{z_+(s)} = \hat{x}_0 + \frac{1-s}{z_+(s)}(1+z_0) \left(\hat{x}_+ - \hat{x}_0 \right) $$
and
$$ \frac{x_-(s)}{z_-(s)} = \hat{x}_0 + \frac{s}{z_-(s)}(1+z_0) \left(\hat{x}_- - \hat{x}_0\right) $$
where
$$ \hat{x}_{\pm} = \frac{x_{\pm}+ x_0}{1+z_0}, \qquad \hat{x}_0 = \frac{x_0}{z_0}. $$
Remark now that 
$$ \frac{1-s}{z_+(s)}(1+z_0) = \frac{1+z_0}{1 + \frac{s}{1-s} z_0} $$
and
$$ \frac{s}{z_-(s)}(1+z_0) = \frac{1+z_0}{1 + \frac{1-s}{s} z_0} $$
and we can make sense of the construction at the limit $z_0 \to +\infty$ provided we simultaneously rescale to ensure $\hat{x}_0$ and $\hat{x}_\pm$ converges to finite limits. 
At the limit the curves are given by 
$$ \hat{x}_0 + \frac{1-s}{s} (\hat{x}_+ - \hat{x}_0), \qquad \hat{x}_0 + \frac{s}{1-s} (\hat{x}_- - \hat{x}_0) $$
which, up to a simple reparametrization, is exactly the hyperbola construction stated in the question. 
A: Regarding question 1: 
(1) For sake of simplicity let's take $x=0$, $y_1 = (0,1)$, $y_2 = (1,0)$. 
As a "limit" of the construction is wanted, but the construction does not contain any free parameter (for fixed $x, y_1, y_2$; for the case of moving the points, see my comment above), we need to add a parameter, which "interpolates" between the parabola and the hyperbola construction. 
(2) In the following the interpolation is motivated:   


*

*For the line segments connecting $(0,y)$ to $(x,0)$ yielding a parabola, the sum $x+y$ is constant (in the question $(1-t)+t =1$).  

*For the line segments connecting $(0,y)$ to $(x,0)$ yielding a hyperbola, the product $xy$ is constant (in the question $e^t e^{-t} =1$).

*Thus to interpolate it seems to be natural to take the linear combination of the two invariants, i.e. $\alpha (x+y) + (1- \alpha)xy =1$. ($\alpha=0$ giving constant product of the axis intercept and a hyperbola as envelope; $\alpha=1$ giving constant sum of the axis intercept and a parabola as envelope). 
As the two axis intercepts one can chose (by solving the new invariant for $x$) e.g. $t$ and $(1-\alpha t)/(\alpha +(1-\alpha)t)$.  
(3) It remains to show that the intermediate values of $\alpha$ ($ 0 < \alpha < 1$) give also conic sections as envelopes. 
This is indeed the case as can be shown by explicit calculation of the envelope. The equation of the envelope for general $\alpha$ is:
 $\alpha^2 x^2 + \alpha^2 y^2 - (2\alpha^2-4\alpha+4)xy -2\alpha x-2\alpha y+1 =0$, which is indeed a conic. 
(4) Generalisation to general $x, y_1, y_2$ is straightforward: 
Consider the family of line segments joining $x + (1-\alpha t)/(\alpha +(1-\alpha)t) (y_1 - x)$ to $x + t (y_2 - x)$. 
(The limit $\alpha =1$ yields exactly the construction given in the question for the parabola, the limit $\alpha =0$ yields after a simple variable transformation the construction given in the question for the hyperbola.
