f(x,y,z,p,q)=pq−x=0fx=−1,fy=0,fz=0,fp=q,fq=p
fpdx=fqdy=pfp+qfqdz=−(fx+pfz)dp=−(fy+qfz)dq
qdx=pdy=2pqdz=1dp=0dq
dx/q=dp
dq=0⟹q=c
p=x/c+c1
z=x2/(2c)+c1x+c3
dy=pdp
p2=2y+c2
(x/c+c1)2=2y+c2
F(z−x2/(2c)−c1x,(x/c+c1)2−2y)=0
we have y=1/2 and z=x, so
F(x(1−c1)−x2/(2c),(x/c+c1)2−1)=0
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