82 DR. A. JOHNSON ON SYMMETRICAL 
To find the equation (referred to the axes), of the section of the central quadric 
ax’ + by + c + 2 y 2+2mezx+2nxy—= 1, (1) 
Fale aie) a 
made by the plane 
act+bhyt+aqz=0 (2) 
Transform the equation of the quadric to a new rectangular system of co-ordinate planes, 
consisting of (2) and two others perpendicular to it and to one another, viz: 
&L+ by +a2—=0 [or X¥=0] 
a & + by + 20 [or Y=0] (3) 
a3 X + sy + Gz—=0 [or Z=—0] 
Let these be the planes of Y Z, ZX, X Y, respectively. The conditions of mutual 
perpendicularity will give three equations connecting the co-efficients. 
Then the formule of transformation are : 
a Guan a et EY, 
Ta F5 rs 
b, b, bs 
= — = A 
= = XY + 2 Y+ À (4) 
Cy Co C3 
2 Æ À: = 1625 2% 
= se = =" = 
wheren=Var + b2 + ¢2; m= Va? +b) + ce; r=Vaf+0?+ cf. For, the direction cosines 
: Die 
of the new axis of X (which is the perpendicular to &x+by+a2—0) are =, Pa = 
1 1 
similarly for the axes of Y and Z. 
To find the equation of the section made by &axz+hy+az—0, or Y=—0, we must 
substitute the values given by (4) in (1), and then make X— 0, or, which is the same 
thing, substitute in (1) 

— ae Yo us Z 
T2 Ts 
by b 
= — Vy = 
y To ei T3 
= ws Y+ = Z 
2 T3 
The result is 
a fo d fo d fy 
2 ae be ats C2 
Hs ye + fe Cas bs 6) = 3) yo 4 (CE PET LE Ue = MZ 1 (5) 
2 3 278 
dfs : : : 
where the SE &c. are the differential co-efficients of fz (4s bs ¢s) with regard to a; &c. 
