16 
Now we have 
== 
The expressions for # and y are of the form: 
aw = (ae + ale—*) z, 
y = (Ber + Pe) z. 
In order to have F = a,,#? + 2a,,ey + a,,y? + a,,2? = 0, we must put: 
BO ne eo Ee A,,), 
B = 6a,, ’ p= og, 
with the condition 
As 
—-4a,,A,, 
! 
OOR 
In the case of the real ellipse we have A,, >0 an de <00: We 
di, 
then can put: 
N= a 
2 Oan 
So we find 
1 
od oman ien Us (er EE iVA,,. (err Se) em 
2 a, fe Re 
he ’ 
== ——— .(— a,, cos t — WA, sin t) z, > (88) 
a, ae 
—a,,a 
83 en De 11” 38 
a, Ce 2— COST. 2. 
A = A 
11 33 
We can use the same expression if we have to deal with a 
a 
hyperbola not intersecting the z-axis. For then A,,<0 and —°>0, 
ay, 
so o=o' real. We prefer to write —\/—A,, . sh (it) for VA,,.sint 
= — ip -A,,.sint and ch (it) cos for t. Then real points of the hyperbola 
correspond to purely imaginary values of r. 
If the hyperbola does intersect the x-axis we have A,, <0 ane 
<0, so = 0' imaginary. 
a 
11 
I 
We then put = —0 => geet van get in this manner 
7 das 
1 \ 
nn ane kn VA es a )e= 
2 ran 
= —=—— .{—a,, sh(it) + WA, ch(iz)} 2, (88)! 
Gist 33 
vlg Taas a, (ei: Dn ie) sake Vee 11 A33 . sh(it) . 5 
11 ae a3 
