I'm curious as to how the graph of x^2 + y^2 = xy will look in the 2nd and 4th quadrent of the cartesian plane (ONLY THOSE quadrents).
Any help with this problem is much appreciated.
First this is how we would graph it
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$\displaystyle x^2-xy+y^2=0$
$\displaystyle \cot(2\theta)=\frac{1-1}{-1} \implies 2\theta=\frac{\pi}{2}\iff \theta=\frac{\pi}{4}$
$\displaystyle A'=A\cos^2(\theta)+C\sin^2(\theta)+\frac{1}{2}B\si n(2\theta)$
$\displaystyle C'=C\cos^2(\theta)+A\sin^2(\theta)-\frac{1}{2}B\sin(2\theta)$
$\displaystyle A'=\cos^2\left(\frac{\pi}{4} \right)+\sin^2\left(\frac{\pi}{4} \right)+\frac{1}{2}\sin\left(\frac{\pi}{2}\right)= \frac{3}{2}$
$\displaystyle C'=\cos^2\left(\frac{\pi}{4} \right)+\sin^2\left(\frac{\pi}{4} \right)-\frac{1}{2}\sin\left(\frac{\pi}{2}\right)=\frac{1} {2}$
So we get
$\displaystyle \frac{3}{2}x^2+\frac{1}{2}y^2=0$
So we get the degenerate hyperbola consisting only of the one point (0,0)
$\displaystyle
x^2+y^2=xy
$
You see LHS is always positive,whereas RHS would be negative in 2nd$\displaystyle (x<0,y>0)$ and 4th$\displaystyle (x>0,y<0)$ quadrants.
Therefore,the plot of its graph can only exist only in the 1st and the 3rd quadrant,needless to say it will also pass through the point $\displaystyle (0,0)$.