Removal of the xy term By transforming to parallel axes through a properly chosen point (h,k) I was asked to prove that the following equation can be reduced to one containing only terms of the 2nd degree: $$12x^2-10xy+2y^2+11x-5y+2=0$$.
What I did was shifted the origin to (h,k) and then isolated the $x^2$,$y^2$,$x$,$y$ and $xy$ terms. I was able to remove the $x$ and $y$ terms by taking the coefficient equal to 0 but was unable to do it for $xy$ term. Please help
 A: As Arthur points out in his comment to your question, the term in $xy$ is a second-degree term, so there’s no need to eliminate it. Indeed, that can’t be done with a translation (per dxiv’s comment), anyway.  
The given equation represents a conic section. For a conic section that has a center, translating the origin to the conic’s center eliminates the linear terms from its equation. There are several ways to find this center. One is what you did, namely translate the origin to the unknown point $(h,k)$, isolate the linear terms and solve the resulting system of linear equations. There are other methods, though.
It turns out that the center is the point at which both partial derivatives vanish, so we can find the center of this conic by solving the system $$\begin{align}24x-10y+11&=0\\-10x+4y-5&=0.\end{align}$$ This system might look familiar to you, as it’s the same one that you get with the first method.  
If we write this equation in matrix form as $(x,y,1)A_Q(x,y,1)^T=0$, where $$A_Q=\pmatrix{12&-5&\frac{11}2\\-5&2&-\frac52\\\frac{11}2&-\frac52&2}$$ we find that $\det A_Q=0$ and the determinant of its principal minor is negative, so the conic section is degenerate—a pair of intersecting lines. This means that it can be factored into a product of linear terms, one for each line. This gives a third way to find the center, which will be at the intersection of the lines. Factoring the equation produces $$(3x-y+2)(4x-2y+1)=0,$$ which yields a different system of linear equations for the center than the one above. All of these methods produce the same answer, which you can verify for yourself.  
If you do wish to eliminate the (quadratic) cross term, you’ll have to find the conic’s principal axes and rotate the coordinate system to align with them.
A: This is a second degree general equation.
To remove $xy$ term, express it as a quadratic in variable $x$ taking $y$ as a constant.To remove the required term, convert the equation into a new coordinate system taking 
$$x=x'-\frac{g+hy}a$$ to get an equation in $x'$ and $y$.Hence the term is eliminated.
