= Eccentricity of a conic
{title2=$e$}
= Eccentricity
{synonym}
The parameter $e\geq0$ in a focus-centred <polar coordinates> equation $r=\ell/(1-e\cos\theta)$ for a <conic section>. It distinguishes an <ellipse> ($e<1$), a <parabola> ($e=1$), and a <hyperbola> ($e>1$).
For a noncircular <conic section>, the <eccentricity of a conic> is the constant ratio
$$
e=\frac{\operatorname{dist}(X,F)}{\operatorname{dist}(X,L)},
$$
where $F$ is a <focus of a conic>, $L$ the corresponding <directrix of a conic>, and $X$ a point on the curve. For an <ellipse> with <semi-major axis> $a$ and distance $c$ from the focal <midpoint> to either <focus of a conic>, $e=c/a=\sqrt{1-b^2/a^2}$. Its <circle> case has $e=0$; a finite directrix does not give that circle through the displayed ratio. The circle is the limit as $c\to0$ and its <directrices of a conic> move arbitrarily far away.
The <eccentricity of a conic> is intrinsic to the cutting <plane>, and is not generally its slope in a chosen three-dimensional drawing. For the family $z=1+mx$ in <conic section>, with $0<m<1$, completing the square gives
$$
(1-m^2)\left(x-\frac{m}{1-m^2}\right)^2+y^2=\frac1{1-m^2}.
$$
A <displacement> $dx$ along this <plane> has <Euclidean norm> $\sqrt{1+m^2}|dx|$. Thus the intrinsic semiaxis ratio satisfies $b^2/a^2=(1-m^2)/(1+m^2)$, giving
$$
e=\frac{\sqrt2m}{\sqrt{1+m^2}}.
$$
For $m=3/5$, this is $3/\sqrt{17}$, rather than $3/5$. This normalization separates the curve's geometry from a distorted projection of it onto the $xy$ plane.
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