Given the equation:
#x^2+4xy-2y^2-6 = 0#
we can classify the conic by calculating the invariants:
#I_3 = det((1,4,0),(4,-2,0),(0,0,-6)) = 108#
#I_2 = det((1,4),(4,-2)) = -18#
As #I_3 != 0# the conic is not degenerate, and as #I_2 < 0# the conic is an hyperbole.
Determine now the eigenvalues of the quadratic matrix associated to the equation:
#det((1-lambda, 4),(4, -2-lambda)) = 0#
#-(1-lambda)(2+lambda) -16 = 0#
#-2-lambda+2lambda+lambda^2-16 = 0#
#lambda^2+lambda -18 = 0#
#lambda_1 = (-1-sqrt73)/2#
#lambda_2 = (-1+sqrt73)/2#
The equation will therefore be in the form:
#-(1+sqrt73)/2xi^2-(1-sqrt73)/2eta^2 + t =0#
To determine #t# we compute the cubic invariant for this form:
#I_3 = det( ( -(1+sqrt73)/2,0,0),(0, -(1-sqrt73)/2,0), (0,0,t))#
#I_3 =((1+sqrt73)/2)((1-sqrt73)/2)t#
#I_3 =((1-73)/4)t = -18t#
So:
#-18t = 108 => t =-6#
Then the equation of the conic is:
#-(1+sqrt73)/2xi^2-(1-sqrt73)/2eta^2 -6 =0#
or:
#(sqrt(73)+1)xi^2 -(sqrt(73)-1)eta^2 +12 = 0#
and in canonic form:
#(sqrt(73)+1)/(73-1)xi^2 -(sqrt(73)-1)/(73-1)eta^2 +12/72 = 0#
#xi^2/(sqrt73-1) -eta^2/(sqrt73+1) = -1/6#
graph{x^2+4xy-2y^2-6 = 0 [-10, 10, -5, 5]}
graph{x^2/(sqrt73-1) -y^2/(sqrt73+1) = -1/6 [-10, 10, -5, 5]}
