Geometrically construct conic through four points and a tangent I'm working on a class project about conic surfaces, and I'm reading the book: "History of the conic sections and quadric surfaces" by Julian Lowell Coolidge, and while talking about Newton, it mentions how to get a conic from four points and a tangent, with the following argument, but I can't manage to understand and apply it.
Here's the book explanation.

Does somebody has a complete explanation on the topic? Can you help me?
Thanks a lot!
 A: You can see below the figure provided by Newton himself (Book I of Principia, Sec. V Prop. XXIII - Case 2). We want to construct the conic passing through points $CPBD$ and tangent to line $HI$. To this end we need to find tangency point $A$ on line $HI$.
Suppose lines $CP$, $BD$ meet at $G$ and intersect the tangent at $I$, $H$ respectively. Construct the line through $H$ parallel to $PC$ and suppose it intersects the conic at $X$ and $Y$.
We can now use what is called "Newton's product theorem" in your book: "the ratio of the products of the distances from a point to two pairs of points of the curve, lying in given directions therefrom, is independent of the point chosen" (note that Newton doesn't mention this theorem, he just writes "by the properties of the conic sections").
If we consider lines $ICP$, $IAH$ from $I$ and their parallels $HYX$, $HAI$ from $H$, we obtain then:
$$
{IA^2\over IC\cdot IP}={HA^2\over HY\cdot HX},
$$
that is:
$$
{HA^2\over IA^2}={HY\cdot HX\over IC\cdot IP}
={HY\cdot HX\over HD\cdot HB}\ {HD\cdot HB\over IC\cdot IP}.
$$
But, by the same theorem:
$$
{HY\cdot HX\over HD\cdot HB}={GP\cdot GC\over GB\cdot GD}
$$
and substituting this into the previous equality we finally get:
$$
{HA^2\over IA^2}
={GP\cdot GC\over GB\cdot GD}\ {HD\cdot HB\over IC\cdot IP}.
$$
From there, the position of $A$ can be computed: note that two solutions are possible, depending on whether point $A$ lies inside or outside segment $HI$.

The recourse to points $X$, $Y$ is a dirty trick, as we have no guarantee that line $HXY$ will intersect the curve. However, Newton knew very well that the relation thus found for $HA/IA$ would retain its validity in any case.
Once tangency point $A$ has been found, the conic can be constructed (see Case 1 of the same proposition in Newton's Principia).
