The line through the midpoints M and N of AA' and BC is a diameter of the conic section.

Repeating this construction, we get a second diameter PQ, hence the center O of the conic section.

(If these directions are parallel, it is a parabola and we just get the direction of axis)

The direction BC is conjugate of MN, and we can construct a pair of conjugate diameters MN, M'N' by drawing the parallel to BC from O.

The problem is then to get a pair of

The conjugate relation between diameters defines an involution of the pencil of lines O* into itself.

This involution is determined by two pairs of conjugate diameters : the ones just constructed (MN,M'N') and (PQ,P'Q') .

Choose a circle centered in any point o and going through O. Let's project the pencil O* on this circle, this results into an involution on the circle, of which we construct the center ω : intersection of mm' and pp'

The diameter uu' going through ω defines two lines Ou and Ou' corresponding in the involution on O* and perpendicular : hence they are the searched axes.

We can then construct the vertices of the conic section, as intersections with the just constructed axes.
As this is the intersection with "any" line
(not going through the given points), we'll prefer to construct directly the focii. So we won't require this complicated construction.

Construct a tangent in one given point by the usual method (Pascal), and draw the perpendicular (normal).

They intersect the focal axis in I and J, then (F,F',I,J) is an harmonic ratio
(as tangent and normal are the angle bisectors of the focus rays BF, BF'),
hence OF² = OF'² = OI.OJ.

This gives a direct construction of OF as the length of the tangent from O to circle with diameter IJ.

The focal axis is therefore the one for which the points I and J are on the same side from O.

The perpendicular projection of a focus on the tangent lies on the principal circle,
which then gives the vertices on the focal axis.

The chord of this circle perpendicular in F to axis is equal to the minor axis (ellipse).

For an hyperbola, the perpendicular bisector of OJ and the normal intersect in K.
The circle centered in K and going through O intersect the tangent in the intersection
points of this tangent with the asymptotes, hence the asymptotes
(not constructed in the applet).

Construct the tangent in P4 = O (with Pascal, as usual).

Choose any circle (o) tangent in O to the conic section, image of the conic section by an homology.

The images p1, p2, p3, p5 are drawn by intersecting the circle with lines OP1 etc.

Construct the axis of this homology : U = P2P3∩p2p3, V = P1P5∩p1p5, axis is UV.

Construct the image of the line at infinity : a parallel to P2P3 from O intersects p2p3 in s, image of point at infinity in direction of P2P3.

The image d

The pole of line at infinity is the center W of the conic section, hence the pole of d

The pole of d

This might give at once the center of the conic section, but let it for the moment.

Let x the intersection of d

The circle centered in x and going through O is orthogonal to circle (o).

Its diameter yy' on d

OY = Oy and OY' = Oy' are then two conjugate directions, and being perpendicular, they are the directions of the conic section axes.

We really construct the axes themselves as images of lines wy and wy'.

wy intersects the circle in a and b, images of vertices A and B of the conic section.

ap2 intersect axis UV in Ia. IaP2 and Oa intersect in A image of a.

Similarily for B, C, D. The center is then the common midpoint of AB and CD.

If the conic section is an hyperbola, one of the two (A,B) (C,D) is imaginary

Once the axes and vertices are constructed, constructing the focii is a child's play,
with c² = a² - b², or as above.

In the case of hyperbola (only two vertices) the asymptotes are directly obtained
through the intersections with line at infinity.

As we have already the image of the line at infinity in the homology,
the intersections of image d_{∞} with the circle
give at once the directions of asympotes.

The asymptotes themselves are parallels to these directions from the center.

(not constructed in the applet, as confused with the "fake asymptotes" joining two
"far" points of the locus drawn as conic section)

Note that the focii of a conic section are projectively defined as intersection points of tangent from the cyclic points (homogenous coordinates (1,±i,0). This is inpracticable as these points and tangents are ... imaginary !

The conic sections of this pencil which are tangent to this line (hence parabolas), touch this line in the fixed points of the involution.

We can in the same way construct a conic section through 4 given points and tangent to a given line (at finite distance).

Let's first study this case, conic section from 4 points and a tangent :

The degenerate conic sections (AB, CD) and (AC, BD) intersect (d) in M,M' and N,N',
which determine the Desargues involution.

These points are projected into m,m',n,n' on any circle.

The involution axis is IJ, with I = mn'∩m'n and J = mn∩m'n'.

The intersections u and u' with the circle (fixed points) are projected into U and U' on (d),
which are the contact points of conic sections with the line.

Having now a 5th point, the sequel is as previously (conic through 5 points).

Of course the tangent choosen for homology constructions is choosen as the given line d.

In the case of parabola, the practical construction is just slightly different, as points on the given line are replaced by directions of parallel lines (points at infinity).

We just get the direction of axis. The focus is then constructed :

all rays parallel to axis reflect on the parabola toward the focus,
hence obtained as intersection of two reflected rays on tangent lines.

The axis itself is then constructed as going through the focus.

The perpendicular projection of focus on a tangent gives the tangent at apex, hence the apex.

Another construction of parabola here, due to Newton.

Let T the contact point with AE.

Segments AT and TE can be considered as two sides of the tangent (circumscribed) hexagon ABCDET, both tangent in vertex T to the conic section.

Brianchon says then : AD, BE and CT concurrent.

This gives an immediate construction of T.

Repeating this construction with the other tangents, we get the other contact points.

The construction can also give a current tangent MN, M on DE and N on AE :

Choosing any point M on DE, we construct point N with AD, BM, CN concurrent.