Recognizing conic section from the general equation.

Question

What is the proof from the general equation $Ax^2 +Cy^2+Dx+Ey+F =0$ of a conic section that $AC>0$ is an ellipse $AC =0$ is a parabola $AC<0$ is a hyperbola?

Answer 1

First of all, this assertion is constrained by conditions (you can see one of them in (3)). Unless these conditions are fullfilled, you have neither a parabola, nor an ellipse or hyperbola...

For example $x^2+y^2+2x+2y+3=0$ which can be written $(x+1)^2+(y+1)^2+1=0$ has evidently no real solution, i.e., is the void set...

Let us consider $$Ax^2+Cy^2+Dx+Ey+F=0\tag{1}$$

• if $AC=0$, let us assume that it is $C=0$. Then (1) becomes $Ax^2+Dx+Ey+F=0$ giving a form $y=ux^2+vx+w$ if $E\neq0$ which is the equation of a parabola. If $E=0$, the locus is reduced to 2 lines, or one line or is the void set... (again !)

• if $AC \neq 0$ we can assume $A>0$ WLOG because one can multiply equation (1) by any non-zero quantity. Then :

a) If $C>0$ (thus if $AC>0$) : one can write (1) under the "canonical" form :

$$A\left(\underbrace{x+\dfrac{D}{2 A}}_{X}\right)^2-\dfrac{D^2}{4 A^2}+C\left(\underbrace{y+\dfrac{E}{2 C}}_{Y}\right)^2-\dfrac{E^2}{4C^2}+F=0 \tag{2}$$

$$AX^2+CY^2=\underbrace{\dfrac{D^2}{4 A^2}+\dfrac{E^2}{4C^2}-F}_H$$

which is a non-void locus if and only if $$ \text{condition 1} : \ \ H:=\dfrac{D^2}{4 A^2}+\dfrac{E^2}{4C^2}-F \geq 0 \tag{3}$$

finally giving :

$$AX^2+CY^2=H \ \tag{4}$$

Taking into account (2), we recognize in (4), if $H>0$, the equation of the ellipse

$$\begin{cases}\text{with semiaxes } \frac{1}{A\sqrt{H}} \ \text{and} \ \frac{1}{C\sqrt{H}}\\ \text{centered in } (-\dfrac{D}{2 A},-\dfrac{E}{2 C})\end{cases} .$$

b) If $C<0$ : Let $C'=-C$ and ...up to you... You should find in this case a form $AX^2-C'Y^2-H'=0$ characteristic of an hyperbola, under another condition expressing that $H' \geq 0$.

Answer 2

First, consider the special case of a circle, defined by $$\text{distance from center}= \text{radius} \tag{1}$$ If the center is $(h,k)$, and radius is $r$, we can square $(1)$ and write $$(x-h)^2 + (y-k)^2 - r^2 = 0 \tag{2}$$ Comparing to the "general form" $Ax^2+Bxy+C y^2+D x+Ey+F=0$, we have $A=C$, so that $AC>0$, as expected for an "ellipse"; this situation is unchanged if $(2)$ is multiplied-through by any non-zero constant.

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For a non-circle conic —that is, one with non-zero eccentricity— we can use the focus-directrix property $$\text{distance from focus} = \text{eccentricity}\cdot\text{distance from directrix} \tag{3}$$ If the eccentricity is $e\neq 0$, the focus is $(h,k)$, and the directrix (in, say, "normal form") is $x\cos\theta +y\sin\theta=d$, then we can square $(3)$ and write $$(x-h)^2+(y-k)^2=e^2 \left( x \cos\theta + y\sin\theta - d \right)^2 \tag{4}$$ so that $$x^2\left(1-e^2\cos^2\theta\right) - 2 x y e^2\cos\theta\sin\theta + y^2\left(1-e^2\sin^2\theta\right) + \cdots = 0 \tag{5}$$ Again comparing to the general form, $A=1-e^2\cos^2\theta$, $B=-2e^2\cos\theta\sin\theta$, $C=1-e^2\sin^2\theta)$, whereupon $$\begin{align} 4AC-B^2 &= 4(1-e^2\cos^2\theta)(1-e^2\sin^2\theta)-4e^4\cos^2\theta\sin^2\theta \\ &= 4\left( 1-e^2\left(\cos^2\theta+\sin^2\theta\right)+e^4\sin^2\theta\cos^2\theta-e^4\cos^2\theta\sin^2\theta \right) \\ &= 4\left( 1 - e^2\right) \tag{6} \end{align}$$ We see that the sign of $4AC-B^2$ is directly associated with how eccentricity compares to $1$ (a value that distinguishes the type of conic). This association is unchanged if $(5)$ is multiplied-through by any non-zero constant.

$$\begin{align} 4AC-B^2 > 0 &\quad\iff\quad e < 1 \quad\iff\quad \text{ellipse} \\ 4AC-B^2 = 0 &\quad\iff\quad e = 1 \quad\iff\quad \text{parabola} \\ 4AC-B^2 < 0 &\quad\iff\quad e > 1 \quad\iff\quad \text{hyperbola} \end{align} \tag{7}$$

In those special circumstances when the conic equation has no $xy$-term (i.e., the directrix of the conic is parallel to a coordinate axis), $B=0$, so that the signs are determined by the product $AC$ alone. $\square$