#f(x,y)=16x^2-8sqrt2xy+2y^2+(8sqrt2-3)x-(6sqrt2+4)y=7#
has the structure
#(x,y)((m_1,m_(12)),(m_(12),m_2))((x),(y))+(b_1,b_2)((x),(y))+c=0#
with
#M=((16,-4sqrt(2)),(-4sqrt(2),2))#
#b=(8sqrt(2)-3,-(6sqrt(2)+4))#
#c=-7#
The #M#'s characteristic polynomial qualifies the conic type.
#p(M)=s^2-18s=s(s-18)# so it is a slanted parabola because #p(M)# has a null and a non null roots.
Calling now #p=(x,y)# and #P=(X,Y)# introducing a rotation such that
#P=R cdot p# we have
#p=R^T cdot P# and
#p cdot M cdot p + b cdot p + c->P cdot R cdot M cdot R^T cdot P+b cdot R^T cdot P+c # or
#P cdot M_R cdot P + b_R cdot P + c=0# is the conic in the rotated new set of coordinates. The rotation matrix #R# is choosed such that
#M_R=R cdot M cdot R^T# is a diagonal matrix. #.^T# indicates transposition which in this case is equivalent to inversion.
As we know #R=((costheta,-sintheta),(sintheta,costheta))# so
#M_R=((m_1 Cos^2theta - 2 m_(12) Costheta Sintheta+ m_2 Sin^2theta,
m_(12) Cos(2theta) + (m_1 - m_2) Costheta Sintheta),(m_(12) Cos2theta + (m_1 - m_2) Costheta Sintheta,
m_2 Cos^2theta + m_1 Sin^2theta + m_(12) Sin2theta))#
Choosing now #theta# such that
#m_(12) Cos2theta+ (m_1 - m_2) CosthetaSintheta=0# we have
#theta=arctan((2 m_(12))/(m_1 + sqrt[4 m_(12)^2 + (m_1 - m_2)^2] - m_2))#
we will have the rotated matrix #M_R#
