Just if the line doesn't intersect the conic section in the usual sense, these points have coordinates which are complex numbers, not real.

For instance the circle (x-1)² + (y-2)² = 3 "intersects" line y = 0 in points with coordinates (1-i, 0) and (1+i, 0)

We deal here with the geometric construction of the real and imaginary parts of these "intersection" points.

We set a coordinate system in which the given line is axis Ox, we have then to construct the values a and b
of intersection points with complex, not real
abscissa OA = a + ib and OB = a - ib

In the case of a circle, we get a quite simple construction :

Because of symmetry, the perpendicular diameter to Ox intersects Ox in M, midpoint of AB,
even if the points are imaginary,

we get then OM = (OA + OB)/2 = a

For constructing b, we may consider the power of M with respect to the circle.

Let MT, MT' the tangents from M to the circle.

The power of M is MA.MB = (ib)(-ib) = b² = MT² = MT'²

b is then the length of the tangents from M (M is outside as the intersections are not real).

We then care of the general case of any conic section, in any position versus Ox.

The trick of the power doesn't work anymore...

Consider then the polar of the point at infinity on Ox.

It is really the "conjugate diameter" of direction Ox, locus of the midpoints m of secants CD parallel
to Ox, for (∞, m, C, D) = -1 ⇔ m midpoint of CD

This polar intersects Ox at point M and (∞, M, A, B) = -1

That is M is the midpoint of AB, OM = ((a+ib) + (a-ib))/2 = a,
as in the circle case, just the conjugate diameter is not always perpendicular,
it is only when the axes of the conic section are parallel resp. perpendicular to Ox.

Remains to find b.

Let's consider the polar of any point on Ox, for instance O.

However other than M, otherwise we don't get a new relation,
but the same : M midpoint of AB.

Let U the intersection of polar of O with Ox. (O,U,A,B) = -1, and changing origin into
midpoint M : MA² = MB² = (±ib)² = -b² = MO.MU < 0

The construction of b is then a classical one :

Draw the circle with diameter OU. The half perpendicular chord through M = b.

Recall the classical construction of the polar of P (here P = O) :

Draw any two secants PCD and PEF.

CE and DF intersect in I, CF and DE intersect in J

IJ is the polar of P.

The intersection points of lines through these 5 points are then easily constructed, with straightedge only, using the Pascal theorem.

The secants PC'D, PE'F and the parallels to Ox CD and EF are then choosen as going through some of the points defining the conic sections (points 1 and 3).

Of course the 5 points as well as O are draggable.

The applet doesn't handle degenerate cases which occur depending on specific positions of O or the 5 points. For instance if O lies on line DF, the two secants OC'D and OE'F are the same line.

In practical, we should then just choose as points D and F some other points than points 1 and 3.

Then the previous properties are still valid :

M is still the midpoint of AB, and MA² = MB² = MO.MU

But here MA and MB are real, hence MO.MU > 0

MA = MB is then the length of the tangent from M to the circle with diameter OU, M being then outside the circle.

The above applet automatically handles this case, depending on wether M is outside (real intersections) or inside (imaginary intersections) the segment OU.

Other constructions are possible, using an homology transforming the conic section into a circle, or using the Desargues involution on the line from the pencil of conic sections defined by 4 of the given points. See Conic sections for details.