Question 1161851
There must be an easier, much more elegant way, but this works:
 
{{{z[2]=a(cos(theta)+i*sin(theta))}}} describes the circle of radius {{{a}}}
 
{{{z[1]=z[2]+1/z[2]}}}
{{{z[1]=a(cos(theta)+i*sin(theta))+1/(a(cos(theta)+i*sin(theta))) }}}
{{{z[1]=a(cos(theta)+i*sin(theta))+(1/a)(cos(-theta)+i*sin(-theta))) }}}
{{{z[1]=a(cos(theta)+i*sin(theta))+(1/a)(cos(theta)+i*(-sin(theta))) }}}
{{{z[1]=a(cos(theta)+i*sin(theta))+(1/a)(cos(theta)-i*sin(theta))) }}}
{{{z[1]=a*cos(theta)+i*a*sin(theta)+(1/a)*cos(theta)-i*(1/a)*sin(theta)}}}
{{{z[1]=((a^2+1)/a)*cos(theta)+i*((a^2-1)/a)*sin(theta)}}}
The point representing {{{z[1]=x+i*y}}} has {{{system(x=((a^2+1)/a)*cos(theta),y=((a^2-1)/a)*sin(theta))}}}
or {{{system(x=((1+a^2)/a)*cos(theta),y=(-(1-a^2)/a)*sin(theta))}}}
Then,
{{{x/(1+a^2)}}}{{{"="}}}{{{x*(1/(1+a^2))=((1+a^2)/a)*cos(theta)*(1/(1+a^2))=(1/a)*cos(theta)}}}
and
{{{(x/(1+a^2))^2=x^2/(1+a^2)^2=(1/a)^2*cos^2(theta)=(1/a^2)*cos^2(theta)}}} meaning that {{{highlight(x^2/(1+a^2)^2=(1/a^2)*cos^2(theta))}}} 
Similarly,
{{{y/(1-a^2)}}}{{{"="}}}{{{y*(1/(1-a^2))=(-(1-a^2)/a)*sin(theta)*(1/(1-a^2))=(-1/a)*sin(theta)}}}
and
{{{(y/(1-a^2))^2=y^2/(1-a^2)^2=((-1/a)*sin(theta))^2=(-1/a)^2*sin^2(theta)=(1/a^2)*sin^2(theta)}}} meaning that {{{highlight(y^2/(1-a^2)^2=(1/a^2)*sin^2(theta))}}}
Adding up the equations highlighted above,
{{{x^2/(1+a^2)^2+y^2/(1-a^2)^2=(1/a^2)*cos^2(theta)+(1/a^2)*sin^2(theta)=(1/a^2)(cos^2(theta)+sin^2(theta))=(1/a^2)*1}}}
meaning that
{{{highlight(x^2/(1+a^2)^2+y^2/(1-a^2)^2=1/a^2)}}}