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For $0\le k\le 1$, define $E(k)={\int }_0^{\pi /2}\sqrt{1-k^2{\sin }^2\theta }d\theta$. An ellipse with semi-major and semi-minor axes $a\ge b>0$ and eccentricity of a conic $e=\sqrt{1-b^2/a^2}$ has perimeter $4aE(e)$. Differentiating its parametrization gives the speed $a\sqrt{1-e^2{\sin }^2\theta }$, and fourfold symmetry reduces the full arc length to a quadrant. The binomial series gives $E(k)=(\pi /2)(1-k^2/4-3k^4/64+O(k^6))$ near zero.

A conic section is obtained by intersecting a plane with a double circular cone. Cuts away from the vertex produce an ellipse, a parabola or a hyperbola; a perpendicular cut is the circle case of an ellipse. A cut through the vertex can instead produce a degenerate conic. One simple surface therefore connects several familiar plane curves.^[ref]

Consider the original familySubstituting the plane equation into the double circular cone equation givesThus $m=0$ gives a circle, $m=3/5$ an ellipse, $m=1$ a parabola, and $m=7/5$ a hyperbola. The last cut meets both sheets of the double circular cone, producing its two branches. These parameters specify a mathematical family, rather than measurements of a physical object.

The Cartesian coordinate system descriptions of these curves are second-degree equationsA rotation and a translation expose their principal directions and offsets. The conic type discriminant identifies the ordinary nondegenerate real type, while a separate degenerate conic check prevents confusing two crossing straight lines with a hyperbola.

This is useful because a difficult-looking equation can be understood through a small set of geometric features: foci of a conic, directrices of a conic, eccentricity of a conic and the semi-major and semi-minor axes. The same ellipse can be traced by a constant-length string or parametrized by $x=a\cos t$, $y=b\sin t$. These are different descriptions of one curve, with different advantages for construction and calculation.

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 hyperbola is a conic section with two unbounded branches. In suitable coordinates its equation is $x^2/a^2-y^2/b^2=1$ with $a,b>0$. The unit hyperbola is parametrized by $x=\cosh t$, $y=\sinh t$ on its right branch, relating it to the hyperbolic functions.

For example, $x^2/9-y^2/16=1$ has foci of a conic at $(\pm 5,0)$, because $5^2=3^2+4^2$. Its focal-distance difference has absolute value $6$, while its eccentricity of a conic is $5/3>1$. At the point $(3,0)$, the focal distances are $8$ and $2$, verifying that difference. The asymptotes are $y=\pm 4x/3$; a distant point approaches those directions while remaining on one of the two branches. The focal-distance difference, rather than a constant sum, distinguishes this hyperbola from an ellipse.

A Newtonian binary chirp is a leading-order model in which loss of orbital energy to gravitational radiation makes an almost circular binary system shrink and its gravitational-wave frequency rise. Here is a derivation with explicit SI units. Let the constant component masses $m_1,m_2$ be in kilograms, $M=m_1+m_2$, $\mu =m_1{m}_2/M$, separation $a$ in metres, and orbital angular frequency $\Omega$ in radians per second. Newtonian gravity, circular motion and the quadrupole formula giveThe factor $\pi$ rather than $2\pi$ appears because the dominant mass quadrupole moment repeats twice per orbit. Since $dE/da>0$, radiative energy loss implies $\dot{a}<0$, henceThe quantity $G\mathcal{M}/c^3$ is a time in seconds. Thus $K$ has units ${\mathrm{s}}^{5/3}$ and $K{f}^{11/3}$ has units ${\mathrm{s}}^{-2}$, as required for frequency change per second. This model assumes weak gravity, slow relative motion, negligible eccentricity of a conic, no spin corrections and adiabatic evolution over an orbit. It does not describe merger or ringdown.

Separating variables gives a finite formal coalescence time $t_c$ and accumulated wave phase $\Phi$:Indeed $d\Phi /dt=2\pi f$; the formal divergence at $t_c$ signals extrapolation outside the approximation. Over a finite frequency interval, the number of gravitational wave cycles is $N=3(f_0^{-5/3}-f_1^{-5/3})/(5K)$. This accumulated phase is why a chirp mass can be constrained even when distance and inclination remain entangled.

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.

Differentiating the parametrization gives $dx=a\cos \theta d\theta$ and $dy=-b\sin \theta d\theta$. Thus the arc length element isAll four quadrants give the same contribution. Consequently the ellipse perimeter isThe final factor is the complete elliptic integral of the second kind, evaluated at the eccentricity of a conic $e$.

The distance from a focus to a conic section along a line perpendicular to its symmetry axis. In focus-centred polar coordinates, an ellipse may be written $r(1-e\cos \theta )=\ell$, where $e$ is its eccentricity of a conic.

The distance from a focus of a conic section to the curve measured perpendicular to its principal axis. In the focus-based polar coordinates equation $r=p/(1+e\cos \varphi )$ it is the length $p$, where $e$ is the eccentricity. For an ellipse, $p=a(1-e^2)=b^2/a$. For a Newtonian Kepler orbit, $p=j^2/(GM)$ with $j$ the specific angular momentum.