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### Chapter 10: Parametric Equations and Polar Coordinates

#### Learning module LM 10.5: Conic Sections:

Slicing a cone
Ellipses
Hyperbolas
Parabolas and directrices
Completing the square

# Completing the square

Completing the Square

The most general conic section has an equation of the form $$Ax^2 + By^2 + Cxy + Dx + E y + F = 0.$$ So far we have taken $C=D=E=0$, except for parabolas. Here we'll see how to adjust for nonzero values of $D$ and $E$. We'll still assume that $C=0$, which means that our curves will point in the direction of the coordinate axes. $C \ne 0$ describes rotated conic sections.

If $A \ne 0$, then we can always absorb the $Dx$ term by completing the square: \begin{eqnarray*} Ax^2 + Dx & = & A \left (x^2 + \frac{D}{A}x \right) \cr \cr & = & A\left (x + \frac{D}{2A}\right )^2 - \frac{D^2}{4A}\end{eqnarray*} Likewise, if $B \ne 0$ then we can absorb the $Ey$ term: $$\displaystyle{By^2 + Ey = B\left (y+\frac{E}{2B}\right )^2-\frac{E^2}{4B}}.$$ If $B=0$ then we cannot absorb the $Ey$ term, and the equation of a parabola winds up looking like $\displaystyle{y-k = \frac{-A}{E} (x-h)^2}$.

 Example : Identify the center, foci and eccentricity of the ellipse $$9x^2 + 25y^2 -18 x + 50y - 191 = 0.$$ Solution: Rewrite $9x^2 -18x$ as $9(x-1)^2 -9$, and rewrite $25 y^2 + 50 y$ as $25 (y+1)^2 -25$. This makes the equation $$9(x-1)^2 + 25(y+1)^2 -225=0,$$ or equivalently $$\frac{(x-1)^2}{25}+\frac{(y+1)^2}{9}=1.$$ This is an ellipse centered at $(1,-1)$ with $a=5$ and $b=3$, hence $c=\sqrt{5^2-3^2}=4$. The foci are 4 units the the right and left of the center, at $(5,-1)$ and $(-3,-1)$. This ellipse has eccentricity $c/a=4/5=0.8$.