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
