I was just testing to see if I could derive the equation of an ellipse (and consequently a hyperbola) with the least amount of information to remember. The small amount of information I chose to use is that of the how the eccentricity relates to the length of the line from the directrix to a point on the ellipse/hyperbola and the length of the line from the point on the ellipse/hyperbola to the focus.The reason I wanted to try this was because most of derivations I've looked at used things like letting the focus be the point $(a e,0)$ or other such things that make the algebra simpler but means you have to remember more! Hopefully it's correct but there are one or two things that I'm not too sure are correct in a mathematical sense. But the equation derived matches up with the usual one quoted and after using an online graphing calculator gives the equation of an ellipse and a hyperbola (though not a circle nor a parabola).
If we have an ellipse with general point: $P(x,y)$, directrix $x=l$ and focus $(f,0)$. The line PD is the line from $P(x,y)$ to the directrix and similarly for the focus PF. So knowing that:
$\bar{\text{PF}}=e\; \bar{\text{PD}}$ therefore converting this into an equation using the above parameters:
$\sqrt{(x-f)^2+y^2}=e(l-x)$ which after squaring both sides and expanding gives: $x^2-2xf+f^2+y^2=e^2x^2-2e^2lx+e^2l^2$ this is where in order for it to work I need the $x $ terms to cancel thus invoking the condition that $-2f=-2e^2l\rightarrow f=e^2l$ which seems reasonable and indeed the equation works given this condition but does this a cause a loss in generality (I would assume so but I can't spot quite what it is).
Next, a bit of manipulation leads to: $x^2(1-e^2)+y^2+f^2-e^2l^2=0$ now using the condition that $f=e^2l$ simplifies to: $x^2(1-e^2)+y^2-e^2l^2(1-e^2)=0$ and thus dividing everything by $e^2l^2(1-e^2)$ gives the equation in the more tradition looking format:
$\frac{x^2}{e^2l^2}+\frac{y^2}{e^2l^2(1-e^2)}=1$ which implies we take $a=le$ and $b=\sqrt{a^2(1-e^2)}$ giving: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
I'm just curious if there's anything wrong with the way I've done it (particularly about just stating that $f=e^2l$, I just spotted that if this were the case we could simply it but I'm not entirely sure how it's effected the derivation) and if there's any preference to how one derives it since there seems to be differing ways of doing it (like Spivak uses the definition that the sum of the distances between the foci is constant though even then you need to remember that the constant is $\pmb{2} a$ else the algebra is harder, and it's more time consuming than the above one anyway). I apologise if it's wrong or blatantly obvious I've only really been briefly looking at the topic.
Thanks in advance.
$f=e^2l$ if and only if the ellipse is centered at the origin. But a general ellipse with focus at $(f, 0)$ and line at $x=l$ need not be centered at the origin.