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common equation of conic sections ax^2 +bxy+cy^2 +dx+ey+f=0 if b≠0 we rotate tan 2α =(b/(a−c)) [if a=c ⇒ α=45°] { ((x=x′cos α −y′sin α)),((y=x′sin α +y′cos α)) :} we now have [using x, y again instead of x′, y′] Ax^2 +Cy^2 +Dx+Ey+F=0 now complete the squares A(x+(D/(2A)))^2 +C(y+(E/(2C)))^2 +(F−(D^2 /(4A^2 ))−(E^2 /(4C^2 )))=0 { ((x=x′−(D/(2A)))),((y=y′−(E/(2C)))) :} we now have [using x, y again instead of x′, y′] one of these { ((Ax^2 +By^2 +C=0)),((Ax+By^2 +C=0)),((Ax^2 +By+C=0)) :} a, c, A, C, A, B, C ≠0 in all other cases use your brain to interprete which kind of curve (if any) we have

$$\mathrm{common}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{conic}\:\mathrm{sections} \\ $$$${ax}^{\mathrm{2}} +{bxy}+{cy}^{\mathrm{2}} +{dx}+{ey}+{f}=\mathrm{0} \\ $$$$\mathrm{if}\:{b}\neq\mathrm{0}\:\mathrm{we}\:\mathrm{rotate} \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha\:=\frac{{b}}{{a}−{c}}\:\left[\mathrm{if}\:{a}={c}\:\Rightarrow\:\alpha=\mathrm{45}°\right] \\ $$$$\begin{cases}{{x}={x}'\mathrm{cos}\:\alpha\:−{y}'\mathrm{sin}\:\alpha}\\{{y}={x}'\mathrm{sin}\:\alpha\:+{y}'\mathrm{cos}\:\alpha}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{now}\:\mathrm{have}\:\left[\mathrm{using}\:{x},\:{y}\:\mathrm{again}\:\mathrm{instead}\:\mathrm{of}\:{x}',\:{y}'\right] \\ $$$${Ax}^{\mathrm{2}} +{Cy}^{\mathrm{2}} +{Dx}+{Ey}+{F}=\mathrm{0} \\ $$$$\mathrm{now}\:\mathrm{complete}\:\mathrm{the}\:\mathrm{squares} \\ $$$${A}\left({x}+\frac{{D}}{\mathrm{2}{A}}\right)^{\mathrm{2}} +{C}\left({y}+\frac{{E}}{\mathrm{2}{C}}\right)^{\mathrm{2}} +\left({F}−\frac{{D}^{\mathrm{2}} }{\mathrm{4}{A}^{\mathrm{2}} }−\frac{{E}^{\mathrm{2}} }{\mathrm{4}{C}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\begin{cases}{{x}={x}'−\frac{{D}}{\mathrm{2}{A}}}\\{{y}={y}'−\frac{{E}}{\mathrm{2}{C}}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{now}\:\mathrm{have}\:\left[\mathrm{using}\:{x},\:{y}\:\mathrm{again}\:\mathrm{instead}\:\mathrm{of}\:{x}',\:{y}'\right] \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{these} \\ $$$$\begin{cases}{\mathcal{A}{x}^{\mathrm{2}} +\mathcal{B}{y}^{\mathrm{2}} +\mathcal{C}=\mathrm{0}}\\{\mathcal{A}{x}+\mathcal{B}{y}^{\mathrm{2}} +\mathcal{C}=\mathrm{0}}\\{\mathcal{A}{x}^{\mathrm{2}} +\mathcal{B}{y}+\mathcal{C}=\mathrm{0}}\end{cases} \\ $$$$ \\ $$$${a},\:{c},\:{A},\:{C},\:\mathcal{A},\:\mathcal{B},\:\mathcal{C}\:\neq\mathrm{0} \\ $$$$\mathrm{in}\:\mathrm{all}\:\mathrm{other}\:\mathrm{cases}\:\mathrm{use}\:\mathrm{your}\:\mathrm{brain}\:\mathrm{to}\:\mathrm{interprete} \\ $$$$\mathrm{which}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{curve}\:\left(\mathrm{if}\:\mathrm{any}\right)\:\mathrm{we}\:\mathrm{have} \\ $$

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