38 DR. A. JOHNSON ON SYMMETRICAL 
As the same formulae are applicable, we may here consider the question :— 
To find the equation of the quadric 
avtbytepy+t2lyz+2mer4+2nzy=—1 
referred to its axes. 
Taking the same system of rectangular co-ordinate planes as before (3), (a, 6), c;, how- 
ever, not being given) and applying the same formulae of transformation we get 
(ay by Cy) 
Ts 

b 9 co 9 3 b 3 2 
Ge) eo 4 7, pt OB à 



(hip: d fr a fe yz GES d fo TNT D 
+ (ai + no) VE + (524 E+ ah) 
d ay db, "dC, Gacy) Wain 


d fo pdt Afyn Ty 
RME ar Le ae) : 
d ay 1 To 
Hence we get, in addition to conditions (6), the two following 

aie ae (20) 

d fr d fs @ fo — 
Gi Tia ba eae 
From (16) and (17) we see that these will be fulfilled by putting 4 = 0 in equations 
(7) or (8). Hence equations (8) become 
(a-k) a; nb; + me, = 0 
bok) D, Hie, = 0 (21) 
Mm dz + 1 bs + (c-k) c; = 0 
From these we get the determinant 
a-k n m 
n b-k l = 10 (22) 
m L c-k 
giving the well-known cubic to determine the values of &. 
Proceeding in the same manner as before, we can show that & has the values 
fo (@ bi G1) f (dz Ds C3) fa (as Ds C3) 
Tn 9 T2 ? Fr 9 

and hence we see that the three roots of (22) are the coefficients of 2’, y, 2°, respectively in 
the transformed equation. 
Geometrical interpretation of h = 0. 
The geometrical interpretation of 

is à a te ee 
Nae Ge yt TE 
(or h = 0) is that the normal to the quadric at the end of the line whose direction cosines 
are proportional to 4, bs, ¢;, 18 perpendicular to the line whose direction cosines are propor- 
tional to a, b,, ¢. The other analytical conditions prove, as has been shown above, that the 
normal must also be in the same plane with these two lines. Hence. as these two are at 
