Q.Find the point to which the origin is to be shifted so as to remove the linear terms in the equation $2x^2+3y^2-4x+6y-1=0$.

The general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ can represent a circle, parabola, ellipse, hyperbola or a degenerate case, but its type is hidden by the linear terms $2gx,\,2fy$ and the cross term $2hxy$. Translation and rotation are used, in that order, to strip these away.

Step 1 — remove the linear terms by translation. Substitute $x=x'+x_0,\,y=y'+y_0$ for an as-yet-unknown point $(x_0,y_0)$, expand, and collect the coefficients of the new linear terms $x'$ and $y'$. Setting each of these coefficients to zero — equivalently, setting $\partial/\partial x=0$ and $\partial/\partial y=0$ on the original expression — gives the simultaneous linear system

Solving these two equations for $x_0,y_0$ gives the exact point to which the origin must be shifted; after this translation the equation has no linear terms, only $aX^2+2hXY+bY^2+c'=0$.

Step 2 — remove the $xy$ term by rotation. Rotate the once-translated axes through an angle $\theta$ using $X=X'\cos\theta-Y'\sin\theta,\ Y=X'\sin\theta+Y'\cos\theta$, substitute into $aX^2+2hXY+bY^2+c'=0$, and collect the coefficient of $X'Y'$. Setting that coefficient to zero gives …