Question 1021847
You can find the rotation angle of the general ellipse equation,
{{{Ax^2+2Bxy+Cy^2+D=0}}}
using,
{{{tan(2*theta)=(2B)/(A-C)}}}
So in this case,
{{{tan(2*theta)=(1)/(2-2)=1/0}}}
{{{2*theta=0 +- 90}}}
{{{theta=0 +- 45}}}
Let's graph it to see which it is.
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*[illustration fq6.JPG].
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{{{theta=-45}}}