Memoirs presented to the Cambridge philosophical society on the occasion of the jubilee of Sir George Gabriel Stokes, bart., Hon. LL. D., Hon. SC. D., Lucasian professor.

DR TAYLOR, THE GEOMETRY OF KEPLER AND NEWTON. 215
Then AM: AF=BQ: BI, and FI, MQ meet on the diameter AB.
We can now solve the problem as follows.
F M A L
D
H
Complete the parallelogram IKLM by drawing the tangent KL parallel to IM.
Then IL, KM cross at the centre of the conic. Conversely, from five given tangents
we can determine the conic.
Case 1. Let four of the tangents be the sides of a parallelogram, as in the figure,
Its diagonals by their intersection give the centre, and FI, MQ also intersect on the
chord of contact AB. The diameter AB being known, the conjugate radius is a mean
proportional to AM, BI.
Case 2. Let the tangents at A, B only be parallel. These with FQ, MI determine
a point (FI, MQ) on the chord of contact AB; and with IM, KL they determine a
point (IL, KM) on AB.
Case 3. When none of the tangents are parallel, the same construction determines
AB; for one pair of them, or two pairs, can be transformed into parallels by Lemma
xxII. All the points of contact can be found in this way, and the conic can then
be traced by various methods.
LEMMA XXV. Corol. 1. If IEM, IQK be fixed tangents to a conic and MK
rectangle KQ.ME, or (IK - IQ) (IM-IE) is constant. This leads to a tangential
equation of the form,
a. IE. IQ + b. IE + c. IQ + d = O.