Orthopoles and congruent ellipses

Here, we consider the standard orthopole. Then we have the following known result
concerning the locus C(Z) = {X : X is the orthopole of a line through Z}.
In this page, H,O,L have their usual meanings, and D is the isotomic conjugate of H.

Theorem 1
For any point Z, C(Z)
(1) is an ellipse,
(2) contains the vertices of the pedal triangle of Z,
(3) has centre the mid-point of ZH,
(4) has axes parallel to the asymptotes of the rectangular
hyperbola through the isogonal conjugate of Z (unless Z = O).

Proof notes
The first part is easiest by computation. Any line L through Z can be described by its
direction, that is by an infinite point 1:t:-1-t. We can then compute the orthopole of
L. Eliminating t, we get an equation for the locus. This shows that we have a conic.
It is an ellipse since orthopoles are defined as the intersection of lines perpendicular
to the sidelines. These are not parallel, so the orthopole is always finite.
Part (2) is best seen by considering the orthopole of the line through Z parallel to an
altitude. For the altitude AH, the points B, C project to the foot of the intersection with
BC. This is then the orthopole of AH, and is the vertex of the pedal triangle on BC.
Part (3) is a routine calculation using the matrix of the conic.
Part (4) is also done by calculation. The directions of the axes are the unique pair of
orthogonal points at infinity which are conjugate with respect to the conic. The algebra
leads to the same results as finding the infinite points on the given hyperbola.
There is an alternative proof below when we look at the major and minor axes.

There are two further results we require later. These are less well-kown.

Theorem 2
(1) Perpendicular lines through Z give antipodal points on C(Z).
(2) If Z,W,O are collinear, then C(Z), C(W) have parallel axes.

Proof notes
Part (1) is another routine calculation.
Part (2) is just the observation that the points Z and W give the same hyperbola,
which is the isogonal conjugate of the line through Z, W and O.

Corollary 2.1
The axes of C(Z) are parallel to the Simpson lines of the intersections of the line ZO
with the Circumcircle.

Proof notes
The theorem shows that all C(W) with W on ZO have parallel axes.
Taking W on the circumcircle gives a degenerate "conic" - the portion of the Simpson
Line within the Steiner Deltoid. By continuity, this must be in the direction of an axis.

We investigate part (2) further below. The axes are parallel, but the major axes may
be perpendicular. We shall see exactly when this occurs.

The case Z = O is special. From Theorem 1, C(O) is the conic with centre N, the
mid-point of OH. It passes through the vertices of thepedal triangle of O, the medial
triangle. This is the nine-point circle.

We need a well-known general result about orthopoles.

Theorem 3
If L and M are parallel lines, then their orthopoles P, Q are such that the vector PQ
is equal to the vector perpendicular to L with length the distance between the lines.

Proof notes
This follows easily from the definition of orthopole. The figure for the construction
of the orthopole of M is just the translation of that for L by the stated vector.

To determine the major and minor axes, we adopt the following stategy. Choose
X = 1:t:-1-t, the general point at infinity. Then Y, the orthopole of ZX, is on C(Z).
We can compute D(Z,t), the square of the distance from Y to the centre of the conic.
The extreme values of D(Z,t) correspond to the end-points of the axes.

Theorem 4
With the above notation, for any point Z, and any constant k ≠ -1, put W = kZ + O.
(1) The derivative D'(Z,t) has the form f(t)/g(t), with f(t) of degree 4.
(2) (k+1)D'(W,t) = kD'(Z,t),
(3) (k+1)D(W,t) - kD(Z,t) = (R2-k(k+1)ZW2)/4, where R denotes the radius of the
Circumcircle of ΔABC.

This is purely computational. Since we have an ellipse, part (1) indicates that f(t) has
four real zeros, and that these must correspond to the required end-points.
Part (2) can be used to show (again) that all points on the same line through O have
loci with parallel axes.

The case k = -1 corresponds to W at infinity.
The case k = 0 gives W = O. Then (2) shows that we have a circular locus.
Values of k between -1 and 0 correspond to Z,W on opposite sides of O. Then part (2)
shows that the major axis of one locus is perpendicular to the minor axis of the other.
Part (3) relates the length of the axes since, after part(2), the same t give the required
end-points for both conics. We observe that the length ZW is twice the distance between
the centres after Theorem 1(3).

The case k = -1/2 gives W as the reflection of Z in O. This allows us to determine the
lengths of the axis of C(Z). As a corollary, we find that C(Z) and C(W) are congruent.

Theorem 5
Let d denote the distance between Z and O. Let W be the reflection of Z in O.
(1) The conic C(Z) has major axis of length R+d, minor axis of length |R-d||.
It has eccentricity 2(Rd)½/(R+d).
(2) The conics C(Z) and C(W) are congruent, with perpendicular axes.

Proof notes
This is quite a lengthy proof.
We begin with case k = -1/2 of Theorem 4, so that W = kZ+O, the reflection of Z in O.
Theorem 4(2) shows that the major axis of C(Z) corresponds to the minor axis of C(W)
and vice versa. Let C(Z) have axes of length 2a, 2b, and C(W) axes 2a', 2b'. Let P1,P2
be the end-points of the major axis of C(Z), Q1,Q2 those of the minor axis of C(W).
As usual, we write R for the circumradius. We also write d for the length of ZO.
Theorem 4(3) then shows that we have
(*) a2 +b'2 = (R2 + d2)/2 = a'2 +b2.
It also shows that P1 and P2 arise as orthopoles of perpendicular lines L, M through Z,
and that L', M', the lines through W parallel to L,M give Q1, Q2. The lines L,M,L',M' form
a rectangle with diagonal ZW. Say L, L' are l apart, and M,M' are m apart.
Now Theorem 3 shows that P1Q1 is parallel to M and has length l, while P2Q2 is parallel
to L, and has length m. Note that P1Q1 and P2Q2 are therefore perpendicular.
Theorem 2(2) and Theorem 4 prove P1P2 and Q1Q2 parallel.
Although we are about to prove that a = a' (and b = b'), for the moment we will assume
that a ≥ a', so that a > b'. Otherwise, we could interchange Z and W.
Now we have two possible pictures ; either
(a) P1Q1 and P2Q2 meet between P1P2 and Q1Q2, (right-hand figure),or
(b) P1Q1 and P2Q2 meet beyond Q1Q2,(left-hand figure).

In case (a), we imagine P1Q1 moved 2b' up.
We get a right-angled triangle with sides l,m 2(a+b').
In case (b) we imagine P1Q1 moved down by 2b'.
We get a right-angled triangle with sides l,m,2(a-b').
The remaining analysis is similar in both cases. Note that we already know that l, m are
sides of aright-angled triangle with hypotenuse |ZW| = 2d.
In case (a), we then have (*) and now
(**) a+b' = d.
Then (a-b')2 = 2(a2+b'2) - d2 = R2.
Now a > b', so a-b' - R. With (**), we get 2a = R+d, 2b' = d-R (so we must have d ≥ R).
A similar argument for (b) gives 2a = R+d, 2b' = R-d (so here we must have R ≥ d).
The argument for a' and b is similar, though we cannot yet say a' > b, so we might get
2a' = |R-d|, 2b = R+d. But then C(Z) has equal axes, so it is a circle. Theorem 4 shows
that this occurs only for Z = O.
Note also that R = d only when Z is on the circumcircle, when C(Z) degenerates as a line.
Otherwise, we have either R > d or R < d. The two conics have the same lengths of axes,
viz major axis of length R+d, minor axis of length |R-d|.
Finally, the eccentricity is an easy calculation from the lengths of the axes.

Corollary 5.1
Conics C(Z), C(W) are congruent if and only if Z, W are on the same circle centre O.

This is a trivial consequence of the formulae in Theorem 5 : the values of d must agree.