40 Proceedings of Royal Society of Rdinhurgh. [sess. 
gular system, the axis being aa? + /3y + A = 0, the tangent at vertex 
will be 
(P -a cos o))x -(a- j3 cos o))y + />(, = 0 , 
since it is perpendicular to the axis, and the equation to the para- 
bola consequently 
(ax + py + xy = ^k{(/ 3 - a cos 0 ))x - (a - /? COS (x))y + fx}, 
in which the parameter is 2k sin^w/y, which may be seen by writing 
equation to curve in the form already indicated. 
We find X, g, k by writing equation to curve in form 
(ax + fyy + 2{aA - (;8- acos a))/c}a?-f 2{/5A-f (a - yScos 
+ — 2k^ “ 0 , 
and comparing with the given equation, when we have 
^ aX — (^ - a cos (o)k /3X + (a — j3 COS a>)/c A^ — 2Kfx 
9 ^ 7 
and therefore 
1 V 
U.(J + 13/- (a/ +/3g) cos m af-Pg’ 
giving A and k. Thus the parameter is 
2{af-j3g)sio?oij-f- 
and since 2/c/x = A^ - c, the tangent at vertex is 
(/? - a cos (j))x - (a - /3 cos to)y + (A^ — c)y^j2(af- pg) = 0 . 
(3.) The axis may also be found by simply expressing the con- 
dition that the diameter ax-\- Py = 0 shall be at right angles to the 
tangent at any point, when we get 
a\ax' -1- Pf) + ag + P\ax' -h pi/) -f Pf 
— ^^aP(ax' -1- Pf) + af -p aP(ax' -P Pf) + ^^}cos o) = 0 
that is 
(ax -f- Py)y‘^ + ag + Pf- (af+ Pg) cos to = 0 , 
which is the equation to the axis. 
2. To find the co-ordinates of the veidex. 
As in the case of rectangular axes, we have 
and 
Kaf-pg)x = X(Xp-2f) + cp 
2(a/- Pg)y = X(Xa- 2g) -p Ca . 
