#math/calculus Standard forms place a conic at the origin, but a curve's location should not change its shape. A coordinate translation separates those two ideas: replacing $x$ by $x-p$ and $y$ by $y-q$ moves the center or vertex to $(p,q)$: $ \frac{(x-p)^2}{a^2}+\frac{(y-q)^2}{b^2}=1, \qquad y-q=A(x-p)^2. $ To reveal the shift, `complete the square`. For example, $ x^2-4y^2-2x+16y=19 $ becomes $ (x-1)^2-4(y-2)^2=4, \qquad \frac{(x-1)^2}{4}-(y-2)^2=1. $ It is therefore a hyperbola centered at $(1,2)$ with vertices $(1\pm2,2)$. The general axis-aligned quadratic $ Ax^2+Cy^2+Dx+Ey+F=0 $ can be translated into $AX^2+CY^2+G=0$. A remaining $xy$ term signals that translation is not enough; the axes must be [[14B2 - Rotating Axes and Classifying Conics|rotated]].