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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?

First, consider the special case of a circle, defined by
$$textdistance from center= textradius tag1$$
If the center is $(h,k)$, and radius is $r$, we can square $(1)$ and write
$$(x-h)^2 + (y-k)^2 - r^2 = 0 tag2$$
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.

For a non-circle conic —that is, one with non-zero eccentricity— we can use the focus-directrix property
$$textdistance from focus = texteccentricitycdottextdistance from directrix tag3$$
If the eccentricity is $eneq 0$, the focus is $(h,k)$, and the directrix (in, say, "normal form") is $xcostheta +ysintheta=d$, then we can square $(3)$ and write
$$(x-h)^2+(y-k)^2=e^2 left( x costheta + ysintheta - d right)^2 tag4$$
so that
$$x^2left(1-e^2cos^2thetaright) - 2 x y e^2costhetasintheta + y^2left(1-e^2sin^2thetaright) + cdots = 0 tag5$$
Again comparing to the general form, $A=1-e^2cos^2theta$, $B=-2e^2costhetasintheta$, $C=1-e^2sin^2theta)$, whereupon
$$beginalign
4AC-B^2 &= 4(1-e^2cos^2theta)(1-e^2sin^2theta)-4e^4cos^2thetasin^2theta \
&= 4left( 1-e^2left(cos^2theta+sin^2thetaright)+e^4sin^2thetacos^2theta-e^4cos^2thetasin^2theta right) \
&= 4left( 1 - e^2right) tag6
endalign$$
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.

$$beginalign
4AC-B^2 > 0 &quadiffquad e < 1 quadiffquad textellipse \
4AC-B^2 = 0 &quadiffquad e = 1 quadiffquad textparabola \
4AC-B^2 < 0 &quadiffquad e > 1 quadiffquad texthyperbola
endalign tag7$$

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. $square$

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 $$Ax^2+Cy^2+Dx+Ey+F=0tag1$$

• 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 $Eneq0$ 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 :

$$Aleft(underbracex+dfracD2 A_Xright)^2-dfracD^24 A^2+Cleft(underbracey+dfracE2 C_Yright)^2-dfracE^24C^2+F=0 tag2$$

$$AX^2+CY^2=underbracedfracD^24 A^2+dfracE^24C^2-F_H$$

which is a non-void locus if and only if
$$ textcondition 1 : H:=dfracD^24 A^2+dfracE^24C^2-F geq 0 tag3$$

finally giving :

$$AX^2+CY^2=H tag4$$

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

$$begincasestextwith semiaxes frac1AsqrtH textand frac1CsqrtH\ textcentered in (-dfracD2 A,-dfracE2 C)endcases .$$

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 $H' geq 0$.