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)
