An elementary treatise on cubic and quartic curves, by A. B. Basset.

28

THEORY OF CURVES.

where Un is a polynomial in y. The axis of x joins the origin
with the point of contact of the line at infinity with the curve.
By proceeding in a similar manner, we can find the Cartesian
equation of a curve with which the line at infinity has a higher
contact than the first.
Imaginary Singularities.
51. It frequently happens that a curve has imaginary singularities. Thus in Chapter V. it will be shown that every
anautotomic cubic has six imaginary points of inflexion, whilst
a quartic may have a pair of imaginary nodes or cusps; but in
order that a curve may be real, it is necessary that the number of
imaginary singularities of any proposed kind shall be even. We
shall now explain a method for determining the conditions for
these singularities.
Let ABC be the triangle of reference, and let us construct a
subsidiary triangle of reference by taking any two points B', C' on
BC. Let (a, 3, ry) and (a, 3', y') be the trilinear coordinates of a
point referred to ABC and AB'C'. Then 3'=0, 7'=0 will be
the equations of AC', AB' referred to ABC, and will be linear
functions of /3 and y.
Let there be two singularities of the same kind at B' and C',
and write down the trilinear equation of a curve referred to
AB'C' having these singularities at B', C'. If the singularities are
imaginary, B' and C' will be imaginary points, and the lines AB',
AC' will also be imaginary; but in order that the curve may be
real, it is necessary that AB', AC' should be a pair of conjugate
imaginary lines, and their equations must accordingly be of the
form /3 + tkly = 0 and / - tk7 = 0, where kc is a real constant. We
must therefore substitute these values of /', y' in the equation of
the curve, and replace the imaginary constants by new real
constants, and the resulting equation will represent a real curve
having a pair of conjugate imaginary singularities on the line BC
or a= 0.
The Cartesian equation of the curve may be obtained by
writing
/3=x, 7=y, a=Ax+By+C;