Question

Show that if b≠0, the cross product term can be eliminated from the quadratic form ax^2 +2bxy+cy^2 by rotating the coordinate ax is through an angle that satisfies the equation: cot 2θ =a-c/2b

Answer 1

Im not getting the 2 in the denominator >.<

Answer 2

Oh, there was a 2bxy, not just bxy. Alright. When rotating the plane by an angle, you use the chanve of variables: \[x=x^\prime\cos(\theta)+y^\prime\sin(\theta)\]\[y=-x^\prime \sin(\theta)+y^\prime \cos(\theta)\] Substituting this into the equation yields something really nasty, but we are only concerned about the terms that contain:\[x^\prime y^\prime\]So im only going to post those:\[x^\prime y^\prime\left(2a\sin(\theta)\cos(\theta)+2b(\cos^2(\theta)-\sin^2(\theta))+2c\sin(\theta)\cos(\theta)\right)=0\]We want this term completely eliminated, so we want the coefficient to be 0. This yields:\[2a\sin(\theta)\cos(\theta)+2b(\cos^2(\theta)-\sin^2(\theta))+2c\sin(\theta)\cos(\theta)=0\]\[\iff a\sin(2\theta)+2b\cos(2\theta)+c\sin(2\theta)=0\]\[\iff (a+c)\sin(2\theta)+2b\cos(2\theta)=0\]Divide sin2theta to get:\[(a+c)+2b\cot(2\theta)=0\iff \cot(2\theta)=-\frac{a+c}{2b}\]