I wish to share verification/completion of the original construction of the oblique hyperbola by Descartes. At first I was looking at the construction of right branch only and temporarily concluded that the sketch was in error as it was not included in original. But then realized that the incomplete sketch defined one branch only. And now, I have an approximate construction to include either branch of the inclined asymptote (not shown) hyperbola. Hopefully it gives a fuller exposition to what Descrates may have had in his mind on this when he introduced co-ordinate geometry.
HyperbolaEqunSketch_by_Descartes
In the above link the equation of hyperbola has been derived using Similar triangle relations in the above original by Descartes and now by Steven Gregory and also MvG using homogeneous coordinates. A Mathematca plot with rough lines added by hand is:
{a, c, b} = {15, 2, 1};
slope = c/b;
cp = ContourPlot[{y (1/a + 1/c) - y^2/(a c) - x y/(a b) == 1, y == c,
y == -c}, {x, -10, 30}, {y, a, -10}, GridLines -> Automatic,
ContourStyle -> {Red, Thick}, AspectRatio -> .5]
I request anyone here to please draw and upload a better sketch using Geogebra, GSP, Z.u.L with Java etc. by using bigger $LNK$ generating triangle of the oblique hyperbola... as the small moving rigid triangle generates this hyperbola as locus.
(The region for my discomfiture prompting the post could be stated only after a night's sleep! ). There are three regions in the plot generaton. In the region X hypotenuse line and rotating line through G cut on the moving rigid but tiny triangle. In the region A it is above ( in Descrates plot) and is below in region B. Region A alone he gave is insufficient to visualize at first instance the generation of the entire plot, imho.
In Summary,
If a right triangle slides with its short side $b$ on a line with minimum distance $a$ from a fixed point $G$ connecting right angle vertex to it by a ray $R$,then the intersection of $R$ and hypotenuse $h$ completly defines an oblique hyperbola of axes angle $ \cot^{-1}(b/c). $
If you're just looking for a better diagram, here's one based on a GeoGebra sketch I made while considering your earlier question.
Note the triangle $\triangle O_\star A_\star B_\star$ such that $\overline{O_\star C} \parallel \overline{AB}$. As the point ostensibly constructed where $\overleftrightarrow{O_\star C}$ meets $\overleftrightarrow{A_\star B_\star}$ is "at infinity", we may conclude that $\overleftrightarrow{A_\star B_\star}$ is actually the hyperbola's oblique asymptote.
Observation. If $\overline{OC}$ is extended to meet the oblique asymptote at, say, $D$, then $\square CDB_\star O_\star$ is a parallelogram, so that $\overline{OB} \cong \overline{CD}$. Thus, we have a general property that if a secant line $\ell$ to a hyperbola is perpendicular to an asymptote, then the segments of $\ell$ "between" the hyperbola and the asymptotes are congruent. As a special case: If a tangent line at $P$ is perpendicular to an asymptote, then $P$ is the midpoint of the line segment bounded by the asymptotes. Neat.
Edit. As @Narasimham comments, the bisected-tangent-segment property is true for any tangent. The secant-segment property is also true for secants that are not-necessarily-perpendicular to an asymptote. So, things are even neater than I first thought. :) Still, I like that the Descartes construction makes these properties nearly-obvious for lines that are perpendicular to the asymptotes.