10x^2+5 y^2-2xy-28x-6y+41=0 and 3x^2-2y^2+5xy-17x-6y+20

Answer 1

Take a look at wiki pedia for a good article on rotation of axes.

In this case, the discriminant B^2 - $AC < 0, so we have a rotated ellipse.

The angle t is such that tan(2t) = B/(A-C)

Letting c == cos(t) and s =sin(t) the new coordinates in x' and y' can be found by letting

x' = xx + ys
y' = -sx + yc

So, with A,B,C,D,E,F = 10 -2 5 -28 -6 41, we have a new ellipse (if my math is right)

(x-1.2931)² / 2.1926² + (y-1.1635)²/3.1926² = 1

Answer 2

Typos: B^2 - 4AC

x' = xc + ys
y' = -xs + yc