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How do you use rotation of axes to identify and sketch the curve of #16x^2-8sqrt2xy+2y^2+(8sqrt2-3)x-(6sqrt2+4)y=7#?

1 Answer
Nov 1, 2016

#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#