490 
ON A TRIANGLE IN-AND-CIRCUMSCRIBED TO A QUARTIC CURVE. 
[367 
and the coordinates of a, c, e, B, D, F are as follows : 
Coordinates of a are 0, 
^2 0> 
c, e, „ ± \/2, 
» 
D, 
B, F 
0, — V^0, 
0 2 -1 2 \ 0 
- V2 (0 2 + 1) ’ V2 (0 2 + 1) ' 
It is easy to verify that the points a, c, e, D are points of the curve, and it is 
obvious that the tangent at D is the horizontal line ce. It only remains to be shown 
that B and F are points of the curve, and that the tangents at these points are the 
lines ac and ea respectively. It is sufficient to consider one of the two points, say 
the point F ; and taking its coordinates to be 
0 2 — 1 2 \/(f) 
v/lT^Tl)’ v ~ V2 (0 2 + 1) ’ 
we have to show that (£, 77) is a point of the curve, and that the equation of the 
tangent at this point is X V0 + Y= V£0, where (X, Y) are current coordinates. 
First, to show that (£, 77) is a point of the curve, the equation to be verified may 
be written 
/t. _ + („2 _ P + 4< ^> 2 ~ 1V _ (0 4 +60 2 + iy 
VÇ ' {V 40 (0 2 + 1) y ~~ 160 2 (0 2 + l) 2 
and we have 
¿2 _ -J _ _ 0 4 + 60 2 + I _ 0 4 + 40 2 — 1 _ (0 2 — 1)(0 4 + 60 2 + 1) 
2 (0 2 -f l) 2 V 4(6 (6- + 1 ) “ 4(6 ( c6 2 4- 1 ) 2 ’ 
so that the equation becomes 
(0 4 + 60 2 +l) 2 (0 2 -1) 2 (0 4 + 60 2 + l) 2 _ (0 4 + 60 2 + l) 2 
4 (0 2 + l) 4 160 2 (0 2 + l) 4 160 2 (0 2 + l) 2 5 
that is 
40 2 + (0 2 — l) 2 = (0 2 + l) 2 , 
which is right. 
Next, the equation of the tangent at the point (£, 77) is 
(I 2 ~ (fZ - a 2 ) + (77 s - 6 2 ) (77 Y - b 2 ) - c 4 = 0; 
that is 
(F-I)(fX-l) + 
0 4 + 4 0 2 — 1 
40 (0 2 + I) 
0 4 + 40 2 — 1\ 
40 (0 2 + 1) J 
(0 4 + 60 2 + l) 2 
160 2 (0 2 TT) 2 ;