Calling #p = {x,y}#, #f_1={-6,0}#, #f_2={6,0}#, #v_1={-5,0}# and #v_2={5,0}# the hyperbole is deffined as the geometric place where
#norm(p-f_1)-norm(p-f_2) = norm(v_1-f_2)-norm(v_1-f_1) = delta#
or
#norm(p-f_1) = delta+norm(p-f_2)#
Squaring both sides
#<< p-f_1,p-f_1>> = << p-f_2,p-f_2>> +2delta norm(p-f_2)+delta^2#
or
#<< p-f_1,p-f_1>> - << p-f_2,p-f_2>> -delta^2 = 2delta norm(p-f_2)#
simplifying
#<< f_1,f_1>> - << f_2,f_2>>+2<< p,f_2-f_1>>-delta^2 = 2delta norm(p-f_2) #
Calling #c = << f_1,f_1>> - << f_2,f_2>>-delta^2# we have
#2<< p,f_2-f_1>>+c=2delta norm(p-f_2) #
Squaring both sides
#(2<< p,f_2-f_1>>+c)^2=4delta^2<< p-f_2,p-f_2>>#
Making the pertinent substitutions we get at
#11/25x^2-y^2-11=0#
