36 
Proceedings of Eoyal Society of Edinburgh. [sess. 
(2.) We may also proceed as follows : — 
Since the axis is -f + A. = 0, the tangent at vertex will be 
/Sx - ay + fjL = 0j and the equation to the parabola consequently 
{ax-[-l3y + Xy = 2 k{I3x -ay + fx), 
in which the parameter is obviously 2k j sjd^ + 
We -find X, /x, k by comparing this equation with the given one. 
Writing it in the form 
(ax + j3yY + 2(aX - I3k)x + 2(/3X + aK)y + — 2k/x = 0 
we have 
^ aX — f^K fSX + aK A,^-2k/x 
1 = = -Z = J 
g / 
and therefore 
^g^Pf cLf-/3g 
giving X and k. Thus the parameter is 
2{af-(Sg)/{a^ + l3^)h 
and since 2 k/x = A,^ - c, the tangent at vertex is 
2(«/- fSgX/Sx - ay) + (a^ + - c) = 0 . 
(3.) The axis may also be found immediately, by simply express- 
ing the condition that the diameter ax-\- Py = 0 shall be at right 
angles to the tangent at any point 
{ax 4- Ptj){ax' + (By) -f g{x ^x')+f{y-hy') + c = 0, 
when we get 
a\ax -h Pf) ^ag-\- P\ax' -P Pf) -P = 0 , 
that is, 
{a? + P^^){ax -P Py) + ag + Pf= 0 , 
which is the equation to the axis. 
2. To find the co-ordinates of the vertex. 
These may be found in several ways. By regarding the vertex as 
intersection of the axis and tangent at vertex, or rather more simply, 
as the intersection of the axis and curve, we have to solve the * 
simultaneous equations 
ax + Py +X = 0 I 
2gx -P 2fy = -p c = 0 J ’ 
