Conic Sections
Conic Sections
Conic sections - circles, parabolas, ellipses, and hyperbolas - are best handled by converting general-form equations into standard form by completing the square.
Standard forms to know. Circle: $(x-h)^2+(y-k)^2=r^2$ with centre $(h,k)$. Parabola: $(y-k)^2=4p(x-h)$ opens right for $p>0$, left for $p<0$; the version $(x-h)^2=4p(y-k)$ opens up or down. The vertex is $(h,k)$ and the focal length is $|p|$. Ellipse: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$ with centre $(h,k)$; the larger denominator marks the major axis, and $c^2=a^2-b^2$ gives the foci. Hyperbola: the same left side set equal to 1 but with a minus sign; the transverse axis follows the positive term, and $c^2=a^2+b^2$. Asymptotes of a centred hyperbola are $y=\pm\tfrac{b}{a}x$ when the $x$-term is positive.
Traps. Students confuse which denominator is $a^2$ when the ellipse is taller than it is wide - always compare denominators. In hyperbolas the relation is $c^2=a^2+b^2$, not a difference as with ellipses. Completing the square requires factoring out the leading coefficient from both $x$ terms before adding the balancing constant, and that constant must be multiplied by the factored coefficient when added to the other side. Also check whether the conic is degenerate (a point or no graph) after completing the square.
Rule of thumb. Group $x$ terms and $y$ terms, factor out coefficients, complete each square, and divide to reach a right-hand side of 1.
Worked example. $9x^2+4y^2-18x+16y-11=0$. Group: $9(x^2-2x)+4(y^2+4y)=11$. Complete: $9(x-1)^2-9+4(y+2)^2-16=11$, so $9(x-1)^2+4(y+2)^2=36$, giving $\dfrac{(x-1)^2}{4}+\dfrac{(y+2)^2}{9}=1$: an ellipse centred at $(1,-2)$ with vertical major axis.