Given the cartesian conic expression:
$$ A x^2 + B x y + C y^2 + D x + E y + F = 0$$
I (Mathematica) derived the corresponding polar equation (1):
The expression was transformed by multiplying the numerator & denominator by the "root conjugate" of the numerator resulting in (2):
One can "verify" by trial that in general both expressions render the same conic.
But when substituting $F=0$ in (1) we get a polar equation which clearly is not equal to 0 and renders a non-degenerate conic.
Whereas substituting $F=0$ in (2) we get $0$:
Obviously (1) and (2) are not equal expressions!
Is the above multiply by the "root conjugate" not a legal transform?
Any enlightenment is welcome.
BTW why am I not able to embed the linked images?
Regards
Robert
EDIT1:
When $F=0$ the conic passes through the origin. Setting $F=0$ and we have
$$ r= \dfrac{-(D \cos \phi + E \sin \phi)}{A \cos^2 \phi +B \cos \phi\sin \phi + C \sin^2 \phi }. $$
with $r=0$ is a root for the first expression.
The second expression for $F=0$ has indeterminate form $\frac00$, should be reduced by L'Hospital's Rule to the above form.
EDIT1:
This is algebra of the radical sign, not of conics in particular. The quadratic equation
$$ y^2-2xy + G = 0$$
has two roots for $y$ whose sum is $2x$ and product $G$
First root
$$ x + \sqrt{x^2-G}= \frac{G}{x - \sqrt{x^2-G}} \tag1 $$
Second Root
$$ x - \sqrt{x^2-G}= \frac{G}{x + \sqrt{x^2-G}} \tag2 $$
When $G$ vanishes $ y\,(y-2 x)=0 \rightarrow y=0 \, y= 2x \ $ are the roots seen on the LHS of (1), (2)
and and their reciprocals are seen on the RHS of (1), (2) as
$$ \frac{1}{2x} ,\, \infty $$
but only when you consider the negative sign in symbol $\pm$ .
Note that it is not plus or minus but plus and minus!!