350] 
ON THE CLASSIFICATION OF CUBIC CURVES. 
381 
81. The equation of the harmonic conic, making the foregoing transformation in 
/X — V v — 
the equation 1 
X y 
— + ——- = 0, becomes 
z 
that is 
which is 
or, what is the same thing, 
X — v + /id X — v — /xi /xi _ Q 
x + yi x — yi z ’ 
oft -f- xft 
— 2 (A — v) yi + 2/xi x — /xi ——= 0, 
y (X 2 + y 2 ) - 25 {fix -(x-v)y}= 0, 
or developing, 
/x (x 2 + y 2 — 2zx) + 2 (X — v) yz = 0, 
which, putting for z its value = 1 — x, is 
¡x (Sx 2 + y 2 — 2«) + 2 (A — v) y (1 — x) = 0, 
3/xx 2 —2 — v) xy + fxy 2 — 2fix + 2 (A — v) y — 0, 
either of which is the equation of the harmonic conic corresponding to the satellite 
line (X — v) x + fxy + v = 0, or, since the direction is alone material, to the satellite line 
— v) x + fxy = 0. The second form shows that the conic is 
an ellipse for S/x 2 > (X — v) 2 , 
a parabola „ S/x 2 = (X — v) 2 , 
a hyperbola „ S/x 2 < (X — v) 2 . 
82. The first form shows that the conic passes through the four points which 
are the intersection of the ellipse Sx 2 + y 2 — 2x = 0, (or, as the equation may also be 
written, 9 (x — ^) 2 + Sy 2 = 1), with the pair of lines (x— l)y = 0: this is right, for the 
points in question are the three points (x=0, y = 0), (x = 1, y = i), (x=l, y = — i), 
which are the vertices of the triangle formed by the asymptotes (x 2 + y 2 )(x — 1) = 0; 
and the point x = f, y = 0, which is the harmonic point. 
83. Putting for shortness \ — v = k, so that the equation of the satellite line is 
kx + /xy + v = 0, and that of the corresponding harmonic conic is 
/x (Sx 2 + y 2 — 2x) + 2/cy (1 — x) = 0, 
the coordinates of the centre are found from the formulae 
l (oc + yi) : | (x - yi) : z = (k + /ii) 2 : ( K -/xi) 2 : -4>/x 2 , 
whence, since x+z = 1, we have for the coordinates of the centre 
/x 2 — A 
x = 
— 2 /x/c 
and it is easy to verify that these belong to a point on the twofold centre conic 
Sx 2 — — y 2 + 1 = 0.