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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.

For a two-dimensional positive-definite covariance matrix $\Sigma$, the centered covariance ellipse is the image of the unit circle under its positive definite square root of a matrix. Equivalently, it satisfies $x^T{\Sigma }^{-1}x=1$. Its semi-major and semi-minor axes have lengths given by the square roots of the larger and smaller eigenvalues, along the corresponding eigenvectors. The ellipse encodes a second-moment geometry; a probability-coverage interpretation needs additional information about the probability distribution and the chosen scale.

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.

For a real positive-definite matrix $A$ in finite dimension, write $A=Q\Lambda Q^T$ by the spectral theorem for real symmetric matrices. In the coordinates in an orthonormal basis $w=Q^{T}x$, the positive level set $q(x)=k$ becomesThis is an ellipsoid, with semi-major and semi-minor axes in two dimensions forming an ellipse; each semi-axis has length $r_j=\sqrt{k/{\lambda }_j}$ and direction given by the corresponding eigenvector. In the opening example, the shorter semi-axis is $\sqrt{k}/2$ along $e_{+}$, while the longer is $\sqrt{k/2}$ along $e_{-}$. A stiffer direction, meaning a larger eigenvalue, reaches the same energy at a smaller displacement. The ratio of longest to shortest semi-axis is $\sqrt{{\lambda }_{\max }/{\lambda }_{\min }}$, the square root of the spectral condition number of a positive-definite matrix.^[ref]

Strict positivity matters because ${\lambda }_{\min }>0$ givesHence a fixed positive level set cannot extend arbitrarily far. The Cholesky decomposition $A=L{L}^T$ gives another view: $q(x)=\Vert L^{T}x{\Vert }^2$, so the ellipsoid is the inverse image of a sphere under an invertible linear map. A singular positive semidefinite matrix has a nonzero kernel; adding a kernel vector preserves its quadratic form, so any nonempty positive level set is unbounded. These statements concern finite dimension; the positive definite symmetric operator entry explains why strict positivity alone need not give a uniform lower bound on an infinite-dimensional Hilbert space.

For an elliptically polarized wave, the tip of its transverse vector traces an ellipse at a fixed point during each wave period. For polarization of an electromagnetic wave, this is the electric field vector. With the real components in polarization of waves, assume $A_x,A_y>0$ and $\sin \delta \ne 0$. Eliminating time givesA short derivation uses $\varphi =\omega t$ and the cosine addition formula to obtain $E_y/A_y=(E_x/A_x)\cos \delta -\sin \varphi \sin \delta$; squaring and using the sine and cosine identity ${\sin }^2\varphi =1-{\cos }^2\varphi$ gives the displayed equation. Outside phase quadrature, the mixed term gives a tilted ellipse; $A_x,A_y$ are component wave amplitudes and need not be its semi-major and semi-minor axes.

In phase quadrature, $\delta =\pm \pi /2$, the mixed term vanishes. For $(A_x,A_y)=(2,1)V/m$, the ellipse is $E_x^2/4+E_y^2=1$, its semi-major and semi-minor axes have lengths $2$ and $1V/m$, and its electric field magnitude varies between $1$ and $2V/m$. At $\varphi =\pi /4$ with $\delta =\pi /2$, the electric field is $(\sqrt{2},-1/\sqrt{2})V/m$ and has magnitude $\sqrt{5/2}V/m$. Equal wave amplitudes instead give circular polarization, a special case of elliptical polarization with constant magnitude. As $\delta$ approaches $0$ or $\pi$, the ellipse collapses to a line segment and gives linear polarization; this limit is degenerate rather than a nondegenerate ellipse.

Changing $\delta$ from $\pi /2$ to $-\pi /2$ reverses traversal of the same ellipse. In the displayed $x,y$ plot, $\delta =\pi /2$ moves from $(A_x,0)$ toward negative $y$ as time increases. A label such as right- or left-handed circular polarization additionally requires a stated viewing direction and naming convention: viewing along propagation and viewing against propagation reverse the apparent rotation. ^[ref].

The ellipse records both the relative wave amplitudes and the phase difference of two transverse oscillations. This separates a change of polarization of waves from a change of overall wave amplitude: multiplying both components by the same positive factor enlarges the ellipse while preserving its shape and orientation.

A rotating ideal linear polarizer can probe this geometry. For the $2:1$ ellipse in phase quadrature, orthogonal projection onto a pass axis at angle $\theta$ gives $E_{\mathrm{out}}(t)=2\cos \theta \cos \omega t-\sin \theta \sin \omega t$, in $V/m$. Averaging its square over a wave period gives $(4{\cos }^2\theta +{\sin }^2\theta )/2$, while the incoming squared electric field magnitude averages to $5/2$. Thus the transmitted fraction of light intensity is $(4{\cos }^2\theta +{\sin }^2\theta )/5$: it is $4/5$ at $0^{\circ }$, $1/2$ at $45^{\circ }$ and $1/5$ at $90^{\circ }$. There is no single incident line of linear polarization to insert into Malus's law; projecting the electric field and averaging its square is the appropriate generalization. This measurement cannot distinguish $\delta =\pi /2$ from $\delta =-\pi /2$, because reversing the traversal leaves all these squared averages unchanged.

Microtomy prepares thin physical sections with a microtome, exposing interior structure for microscopy. The knife separates a layer from a supported specimen while a controlled feed determines the nominal depth increment. Physical section thickness, section orientation and cutting-induced deformation all enter the interpretation of an image. Ultramicrotomy is the specialized thin-section technique used in applications such as transmission electron microscopy.

The preparation method also serves materials research. Yale's instrument description lists a Leica Ultra UC7 with a trimming mill and glass knife maker, and identifies polymers, material fibres and soft metals among suitable specimen materials.^[ref] Manufacturer guidance discusses polymer blends and composite materials, along with the possibility of section compression and cutting artifacts.^[ref]

A useful teaching experiment would compare cuts through a straight, known circular material fibre at different section orientations. A normal vector to the cut aligned with the fibre axis gives a circular profile. An oblique cut gives an ellipse even when the fibre has suffered no deformation. Image the recovered sections with an optical microscope, measure profile dimensions with an established image stack axis calibration, and record the angle between the section's normal vector and the fibre axis. This tests the geometry of the section rather than assuming that every measured elongation is a physical distortion.

For an original ideal calculation, model the fibre boundary as an infinite circular cylinder of radius $R$, with axis along the unit vector $\mathbf{u}=(\sin \theta ,0,\cos \theta )$, where $0\le \theta <\pi /2$. Parallel section planes have equation $z=z_0$. The squared Euclidean distance to the axis is $(x\cos \theta -z\sin \theta {)}^2+y^2$, obtained by subtracting the squared axial orthogonal projection from $x^2+y^2+z^2$. At the section plane, the boundary therefore satisfiesThis oblique section of a circular cylinder is an ellipse with semi-major and semi-minor axes and areaThe area of an ellipse gives the last expression. For chosen $R=20\mu \mathrm{m}$ and $\theta =45^{\circ }$, the semi-major and semi-minor axes are approximately $28.28\mu \mathrm{m}$ and $20.00\mu \mathrm{m}$. The profile area is $\sqrt{2}$ times the perpendicular-profile area, an increase of about 41.4 percent entirely explained by section orientation.

The center of an ellipse in this model has $x$ coordinate $x_c(z_0)=z_0\tan \theta$. In an ordered image stack, two sampled section planes separated by $h$ consequently have lateral centre displacement $\Delta x=h\tan \theta$. At the chosen $h=20\mu \mathrm{m}$ and $\theta =45^{\circ }$, this displacement is $20\mu \mathrm{m}$. Here $h$ is the spacing between sampled planes, which can differ from the physical section thickness if some sections are not imaged.

The calculation assumes a circular, straight fibre, known section geometry and negligible physical section thickness relative to resolved features. One elliptical profile alone cannot distinguish an oblique circular fibre from a differently shaped fibre or an unknown deformation. Comparing known reference features and following their positions through an image stack provides additional evidence. Image registration can align images, but an inferred shift may contain real specimen geometry as well as instrument motion; removing every shift would erase the tilt information in this example.

A mathematical plane cutting an infinite circular cylinder at a nonzero angle to its axis produces an ellipse, unless it is parallel to that axis. More precisely, let the cylinder radius be $R>0$, and let $\theta$ be the acute angle between the cylinder axis and the plane's normal vector, with $0\le \theta <\pi /2$. The semi-major and semi-minor axes and area areThe angle is measured from the normal, not from the cutting plane itself. At $\theta =0$, the profile is a circle.

To prove the result, choose Cartesian coordinates so the cylinder is $x^2+y^2=R^2$, with axis along $z$. Translate the cut so it crosses the axis at $(0,0,h)$. After rotating about the axis if necessary, an orthonormal basis in the cutting plane isA point in the plane is $(0,0,h)+u{\mathbf{e}}_1+v{\mathbf{e}}_2$. Substitution into the cylinder equation gives $u^2{\cos }^2\theta +v^2=R^2$, orBecause the basis is orthonormal, $u$ and $v$ measure actual lengths within the plane. The area of an ellipse then gives the stated area.

For an original example, $R=2$ and $\theta =60^{\circ }$ give $a=4$, $b=2$ and $A=8\pi$, twice the perpendicular-profile area. As $\theta$ approaches $\pi /2$, the ellipse grows without bound. At the excluded parallel orientation, a plane through the axis instead meets the lateral cylinder in two generator lines; the limiting formula does not describe a finite ellipse there.

Microtomy applies this geometry to thin physical sections of a circular material fibre. Its worked example and figure show how an elongated profile can arise from section orientation without deformation. Applying the ideal lemma requires accounting for finite thickness, a finite specimen and departures from a straight circular cylinder.

The quadratic form isThe symmetric matrix has eigenvalues $2,8$, with respective orthogonal eigenvectors $(1,-1{)}^T,(1,1{)}^T$. Normalize them to obtain the orthonormal basisThe given change of basis meansSubstitution, or orthogonal diagonalization of a real symmetric matrix, givesThus the semi-major and semi-minor axes have lengths $2\sqrt{2}$ and $\sqrt{2}$. The eccentricity and enclosed area areThe orthogonal change of basis preserves area, so this is also the area in the original Cartesian coordinates.

Read the squared semiaxis lengths from the denominators in the standard equation obtained in (b). With the positive choice of $C$ there,Since $C^2=A^2-B^2<A^2$, the horizontal semiaxis $a$ is the larger. Thus these are the semi-major and semi-minor axes, respectively.

The semi-major and semi-minor axes are the larger and smaller half-axis lengths of an ellipse. In principal-axis coordinates the ellipse has equation $x^2/a^2+y^2/b^2=1$ with $a\ge b>0$. The same terminology is used for a hyperbola $x^2/a^2-y^2/b^2=1$, where $a$ is half its transverse axis and $b$ half its conjugate axis. Orbital semiaxes apply these geometric quantities to conic sections.

The semi-major axis is one half of an ellipse's longest diameter. For a Kepler ellipse with apsidal distances $r_1,r_2$, it is $a=(r_1+r_2)/2$.

This is one of the semi-major and semi-minor axes; the two-axis geometry is the broader concept.

For an ellipse of eccentricity $e$, the semi-minor axis is $b=a\sqrt{1-e^2}$.

This is one of the semi-major and semi-minor axes; the two-axis geometry is the broader concept.