Considering
#{(-a/(a^2-p)=2a),(sqrt(a^2+(p-a)^2)=1):}#
after dividing by #a# the first equation and squaring both terms into the second, we get
#{(-1/(a^2-p)=2),(a^2+(p-a)^2=1):}#
or
#{(-1=2(a^2-p)),(2a^2-2ap+p^2=1):}#
or
#{(-1=2a^2-2p),(2a^2-2ap+p^2=1):}#
now adding term to term both equation results in
#2a^2-2ap+p^2-1=2a^2-2p+1#
after simplifications
#p^2-2(a-1)p-2=0#
now solving for #p#
#p = (2(a-1)pm sqrt(4(a-1)^2+8))/2=a-1pm sqrt((a-1)^2+2)#
Now taking the first equation
#2a^2=2p-1#
or
#a^2=p-1/2# and substituting
#a^2=2(a-1pm sqrt((a-1)^2+2))-1/2#
or
#a^2-2a+1=-1pm2sqrt((a-1)^2+2)-1/2#
or
#(a-1)^2=pm2sqrt((a-1)^2+2)-3/2#
calling now #a-1=b# we have
#b^2=pm2sqrt(b^2+2)-3/2#
or
#b^2+2/3=pm2sqrt(b^2+2)#
squaring both sides
#(b^2+2/3)^2=4(b^2+2)#
or
#(b^2)^2+4/3b^2+4/9=4b^2+8#
or
#(b^2)^2-8/3b^2-68/9=0#
now solving for #b^2#
#b^2=2/3 (2 pm sqrt[21])#
and
#b=pmsqrt(2/3 (2 pm sqrt[21]))#
or
#a=1pmsqrt(2/3 (2 pm sqrt[21]))#
The next step is the determination of #p# but this is left as an exercise to the reader.
(Don't forget that #p = (2a^2+1)/2#)
