<s>[INST] Finding an equation of a parabola I was asked to show that a$\pm\frac{cx}{b}=\sqrt{x^2+y^2}$ is a parabola if $b=c$
$$\text{If }b=c\rightarrow a\pm x=\sqrt{x^2+y^2}\\
 \left(a\pm x\right)^2=x^2+y^2\\
 \left(a\pm x\right)^2-x^2=y^2\\
 \pm\sqrt{a^2+2ax+x^2-x^2}=y \text{ or } \pm\sqrt{a^2-2ax+x^2-x^2}=y\\
\pm\sqrt{a^2\pm2ax}=y$$
Can I go from here to a standard form of a parabola? How?
How can I show that if $b>c$ it's an ellipse and if $b<c$ it's a hyperbola?
*Edit: I am also attaching my work for $b>c$. $$\text{If } b>c\rightarrow a\pm\frac{cx}{b}=\sqrt{x^2+y^2}\\
 \left(a\pm\frac{cx}{b}\right)^2=x^2+y^2\\
 a^2\pm2a\frac{cx}{b}+\frac{\left(cx\right)^2}{b^2}=x^2+y^2\\
 \pm2\frac{cx}{ab}+\frac{\left(cx\right)^2}{\left(ab\right)^2}=\frac{x^2}{a^2}+\frac{y^2}{a^2}\\
\frac{cx}{ab}\left(\pm2+\frac{cx}{ab}\right)=\frac{x^2}{a^2}+\frac{y^2}{a^2}\\
\pm2+\frac{cx}{ab}=\frac{b}{cx}\left(\frac{x^2}{a}+\frac{y^2}{a}\right)\\
\pm2=\frac{b}{cx}\left(\frac{x^2}{a}+\frac{y^2}{a}\right)-\frac{cx}{ab}\\
\pm2=\frac{b^2x^2+b^2y^2}{abcx}-\frac{\left(cx\right)^2}{abcx}\\
\pm2=\frac{b^2x^2+b^2y^2-c^2x^2}{abcx}\\
\pm1=\frac{\left(b^2-c^2\right)x^2+b^2y^2}{abcx}$$
 [/INST] All conic sections follow the following general form
$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $, with $A, B, C, D, E, F$ all real and $A,B,C$ non-zero.
(see: https://en.wikipedia.org/wiki/Conic_section#cite_ref-Protter_1970_326_15-0):
The discriminant $B^2 - 4AC$ determines ellipse, circle, parabola, hyperbola, etc.
(1) if $B^2 - 4AC = 0$, the equation is a parabola.
(2) if $B^2 - 4AC < 0$, the equation is an ellipse, and so on.
From $\left(a\pm \frac{cx}{b}\right)^2=x^2 + y^2$,
$\big(\frac{c^2 - b^2}{b^2}\big)x^2 - y^2 \pm2\frac{ac}{b}x + a^2 = 0$,
Then, $A = \big(\frac{c^2 - b^2}{b^2}\big)$, $B = 0$, $C = -1$, $D =\pm2\frac{ac}{b}$, $E = 0$, $F = a^2$.
If $b = c$, then $B^2 - 4AC = 0$, so it is a parabola.
For the ellipse, the discriminant is negative if $b > c$, and similarly for hyperbola.
Alternatively,
$ y^2 = a^2 \pm 2ax$ from your last step,
then, use a substitution $x = x' - \frac{a}{2}$:
$ y^2 = a^2 \pm 2a\big(x' - \frac{a}{2}\big) = \pm 2ax'$, which is an equation of parabola.
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