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A directrix of a conic is the fixed line in its focus-directrix description: distance to the focus of a conic is a constant multiple of distance to this line. For a parabola the multiple is one. The directrix of a conic belongs to the geometry of the conic section, rather than being a tangent or an asymptote.

The parameter $e\ge 0$ 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 ratiowhere $F$ 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 $c\to 0$ 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 givesA displacement $dx$ along this plane has Euclidean norm $\sqrt{1+m^2}|dx|$. Thus the intrinsic semiaxis ratio satisfies $b^2/a^2=(1-m^2)/(1+m^2)$, givingFor $m=3/5$, 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.

An ellipse is the locus of points whose Euclidean distances from two fixed foci of a conic have constant sum $2a$. If the distance between the foci of a conic is $d<2a$, the semi-major and semi-minor axes are $a$ and $\sqrt{4a^2-d^2}/2$. Writing $c=d/2$ gives the conventional relation $b^2=a^2-c^2$.^[ref]

For an original calculation, choose $a=5$, $b=3$, and hence $c=4$. The ellipsehas foci of a conic $F_{-}=(-4,0)$ and $F_{+}=(4,0)$. The point $P=(3,12/5)$ lies on it since $9/25+16/25=1$. Its focal distances areIts eccentricity of a conic is $e=4/5$. The matching directrix of a conic for $F_{+}$ is $x=a/e=25/4$, and the perpendicular distance from $P$ to that straight line is $13/4$. Therefore the ratio of focal distance to directrix distance is $(13/5)/(13/4)=4/5$, agreeing with $e$.

The constant sum can be turned into a construction. Fix a pin at each focus of a conic and pull a closed string loop taut with a pencil. In this example the segment between the pins has length $8$, so a loop of length $18$ leaves $10$ for the two pencil-to-pin segments. Moving the pencil while keeping that loop taut traces the ellipse. This is a geometric prediction from the length constraint, not an assertion about a measured drawing.

The parametrization $x=5\cos t$, $y=3\sin t$ maps the unit circle onto this ellipse. It makes plotting easy, but is an affine map rather than an isometry: horizontal and vertical lengths have different scale factors. An ellipse is not a circle under the original Euclidean metric merely because this map exists. Its two semiaxes and eccentricity of a conic retain the metric information lost by that equivalence.

A useful proof outline begins with $|X-F_{-}|+|X-F_{+}|=2a$. Isolate one square root, square, and then repeat; the resulting quadratic equation is $x^2/a^2+y^2/(a^2-c^2)=1$. Since $a>c$, the quantities isolated before squaring have the required signs, so the reverse implication introduces no extra branch. The condition $a>c$ is essential: at equality the focal-distance locus becomes the line segment between the foci of a conic.

In the focus-directrix description of a noncircular conic section, the focus is the fixed point $F$ for which every point $X$ of the curve satisfies $|X-F|=e\mathrm{dist}(X,L)$, where $L$ is a directrix of a conic and $e$ is the eccentricity of a conic. For $y^2=4ax$ with $a>0$, the focus of a conic is $(a,0)$, the directrix of a conic is $x=-a$, and $e=1$.

A parabola is a plane curve whose points are equally distant from a focus and a directrix. In suitable coordinates it has equation $y=a{x}^2$, with $a\ne 0$. Its quadratic geometry can also describe a contour deformation through a saddle point.

For example, take a focus of a conic at $(1,0)$ and a directrix of a conic $x=-1$. Equality of squared Euclidean distances gives $(x-1{)}^2+y^2=(x+1{)}^2$, hence $y^2=4x$. The point $(1,2)$ has distance $2$ to both the focus and the directrix. Its eccentricity of a conic is one. Unlike an ellipse, a parabola is unbounded and has only one matched focus-directrix description.

Write $z=x+iy$. Dividing by two givesThe left side is nonnegative, so squaring is reversible only when $x+a\ge 0$. Squaring yields $y^2=4ax$.

If $a>0$, this equation implies $x\ge 0$, so the sign condition holds automatically. The locus is the parabolawith vertex at the origin, focus of a conic $(a,0)$ and directrix of a conic $x=-a$. Indeed the original equation equates distance from the focus of a conic with distance from the directrix of a conic on its right side.

If $a<0$, the squared equation forces $x\le 0$, making $x+a<0$. This violates the sign condition. Thus the locus is empty for $a<0$; squaring without checking that condition would introduce a spurious parabola.

For a fixed point $F$ not on a fixed line $D$, the ellipse with focus of a conic $F$, directrix of a conic $D$ and eccentricity $0<e<1$ is the locus of points $P$ satisfyingSet $c=\sqrt{a^2-b^2}$ and $e=c/a$. One matched focus of a conic and directrix of a conic pair isIndeed the squared distance condition isExpanding and cancelling the terms linear in $x$ giveswhich is exactly $x^2/a^2+y^2/b^2=1$. Both sides of the unsquared distance condition are nonnegative, so squaring introduces no extra points. Reflecting across the $y$-axis gives the other matched pair: $F=(-c,0)$ and $D:x=-a^2/c$, with the same eccentricity.

A hyperbola with focus of a conic $F$, directrix of a conic $D$ not containing $F$, and eccentricity $e>1$ is the locusPut $c=\sqrt{a^2+b^2}$ and $e=c/a>1$. For $F=(c,0)$ and $D=\{x=a^2/c\}$, the defining equality becomesBoth sides are nonnegative, so squaring is reversible. Expanding and cancelling the linear terms givesThus the two matched focus of a conic and directrix of a conic choices, related by reflection, areThe signs in each pair must agree.

On the hyperbola, $y=\pm (b/a)\sqrt{x^2-a^2}$. As $|x|\to \infty$,Hence the perpendicular distance to one of the two straight lines tends to zero on each end of a branch. The asymptotes are therefore

A conic section with a focus of a conic at the origin has the polar coordinates equation $r=\ell /(1+e\cos \theta )$, where $\ell >0$ is the semilatus rectum and $e\ge 0$ the eccentricity. The directrix of a conic definition $r=e(d-r\cos \theta )$ gives the formula with $\ell =ed$ when $e>0$; $e=0$ is the circle limit. The cases $0<e<1$, $e=1$, and $e>1$ give an ellipse, parabola, and hyperbola. The physical radius requires a positive denominator. For an attractive inverse-square central force, the Laplace-Runge-Lenz vector gives $\ell =m{h}^2/k$ and $e=A/(mk)$.