I have a well-behaved curve $f:\mathbb{R}\rightarrow \mathbb{R}^2$ which has exactly one linear asymptote passing through points $p$ and $q$ in $\mathbb{R}^2$. I would like to find a projective transform $\pi:\mathbb{R}^2\rightarrow \mathbb{R}^2$ such that the resulting curve $\pi \circ f$ has no asymptotes. How should I solve for a suitable projective transform?
(Because the general case seems complicated, I have been considering the special case where $f$ is simply a hyperbola.)
Note: by a projective transform, I mean a transform with parameters $x_0$, $y_0$, $z_0$ which sends
$$\begin{bmatrix}x\\y\end{bmatrix} \mapsto \begin{bmatrix} (z_0 x)/(x - x_0)\\ (y_0x - x_0 y)/(x - x_0) \end{bmatrix}$$
In general I'd write a projective transformation in homogeneous coordinates as
$$\begin{pmatrix}x\\y\\z\end{pmatrix}\mapsto \begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\cdot \begin{pmatrix}x\\y\\z\end{pmatrix}$$
or, in inhomogeneous coordinates, as
$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}(ax+by+c)/(gx+hy+i)\\(dx+ey+f)/(gx+hy+i)\end{pmatrix}$$
So you are for some reason restricting transformation matrices to
$$\begin{pmatrix} z_0 & 0 & 0 \\ y_0 & -x_0 & 0 \\ 1 & 0 & -x_0 \end{pmatrix}$$
This is quite a severe restriction, and I'm far from sure whether it's intentional (you need a transformation of this type) or accidential (you just got the wrong formula for a projective transformation).
So you want to get rid of a given asymptote. What is an asymptote? At least in the context of conics, an asymptote is a tangent at a point at infinity. So if your conic intersects the line at infinity in two points, you have two asymptotes (i.e. a hyperbola), while if the line at infinity does not intersect the conic (except in complex points) you have no (real) asymptotes (i.e. an ellipse). As the limit between these two, if the line at infinity touches the conic, you get a parabola where the line at infinity itself is the tangent, leading to a well-defined asymptotic direction but no well-defined position for the asymptotes.
A point lies on a line if the dot product between their homogeneous coordinate vectors is zero. Or in other words, if the row vector of the line times the column vector of the point is zero: $l^T\cdot p=0$. If after transformation with a matrix $M$ the transformed point $M\cdot p$ lies on the line $l$, then you have $l^T\cdot M\cdot p=0$. But you can also read that as $p$ itself lying on $l^T\cdot M$. So the preimage of a line under this transformation is $l^T\cdot M$. The line at infinity is $l=(0,0,1)$. In the case of your transformation matrix, its preimage is $(1,0,-x_0)$. This corresponds to the equation $x=x_0$. So $y_0$ and $z_0$ don't matter at all (except for the condition that the whole matrix remains invertible), and you should try to choose $x_0$ in such a way that it does not intersect your conic.
You could take the center of the conic, and use the vertical line though that as the preimage. If there are any vertical lines which don't intersect the conic, the one through the center will be among them. If not, then there is no solution following your specific form of transformation, wile you might of course still find a line $gx+hy-i=0$ which does not intersect the conic, and which can be used as the foundation for a solution with a generic transformation.
The same holds for your arbitrary well-behaved curve: Find a (vertical resp. arbitrary) line that does not intersect the curve, and you have found a key component to the transformation matrix. Choose everything else to avoid a determinant of zero and you are done.