We can finish this paper with the famous integral for the ellipse arc length, which give the name to all elliptic group. Arc of ellipse: ds=∫dz√(A-x2)/√(1-x2) A=m/(m-1)>1 The transformation of the integrand with application of projective variable change of Moebius x=(az+b)/(cz+d) (1) will give I=√[(z2+2mz+n)/√[(z-q)] denominator of first degree (*). It will be no more elliptic if: A) the green trinomial has a double root, id est n=m2, and (z-m) comes out of radical. B) both trinomials have a common root, id est q has to be root for the green trinomial, then q2+2mq+n=0 Only one condition: and we have 2 freedom degrees. In my blog there is the B case (paper 5) Here we expose the A case ********************** There is not any periodicity double or alternate, we go from x= 0 to 1. It is a normal curve, the only problem is an improper point for x=1 (vertical tangent). Then when we want to calculate one quart of ellipse, the definite integration, we have to make from x=0 to x=1/2 (by instance), and a second integral at derivative to y from x=1/2, to x=1 See figure, the second tintegration will be from x=1/2 to x=1 (y=0) at the derivative respect to Y;ds/dy (it is in page end). We’ll do a double try … ***************** ds=∫dz√(A-x2)/√(1-x2) The projective transformation of Moebius

x=(z+b)/(cz+d) dx=dz/(z+d/c)2 A>1 is a data calling d/x=h gives

x2=[z2+2bz+b2]/[z2+2hz+h2] A-x2=[z2(A-1)+2z(Ah+b)+Ah2-b2]/(z+h)2

1-x2=(2z(h-b)+h2-b2]//z+h)2=(z+q)/(z+h)2

I=∫dz√(A-x2)/√(1-x2)/(z+h)-2 the two (z+h) compensate.

The blue is (z+q) q=(h+b)/2 b=2q-h

we do d=0 h=0 q=b/2 b is 2q the numerator is z2(A-x2)=()z+2bz-b2 the roots are z=-b+/-b√[1+(A-1)]/(A-1)=b(-1+/-√A)/(A-1) one have to be q remind that A-1=√A-1)(√A+1) then z=1/(√(A+1)=q b(+)=2q b=2/(1+√A)

b’(-)=2/(√A-1) is the other root I=∫dz√[(z+2(1-√A)]/z2 we do z=t2-b dz=tdt

I=∫t2dt·/(t2-b’)2=∫dt/(t2-b’)+b’∫dt·/(t2-b’)2 now we shall do t=√b’·chG dt=b’1/2shGdG

I= b’-1/2∫dG/shG+b’∫dG·/sh3G=b’-1/2∫dG/shG+b’-3/2∫dG·/sh3G=wikipedia solves

I=b’-1/2[ln[th(G/2)]+b’-1[chG/sh2G-ln(th(G/2)] being G=argch[√(1+z/b)] b and b’>0

I=b’-1·chG/sh2G+(b’-1/2-b’-1)ln[th(G/2)] b=2/(1+√A) b’=2/(√A-1)

Deriving respect y d’arc=√(dx2+dy2)=dy√(1+x’2) The ellipse is x2+my2=1 m=a2/b2>1, are semi-axes x=√(1-my2) x’=-y/√(1-my2) x’2=y2/(1-my2) 1+x’2=[(1-m)y2+1]/(1-my2) (A-x2)/(1-x2)=[1-(m-1)y2/(1-my2]/(1-x2) A=1/(m-1)>1 Arc=S=∫ dy√(A-y2)/√(1-my2)