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An arithmetic series adds the sequence terms of an arithmetic progression. Its $n$-th partial sum is $S_n={\sum }_{k=1}^n{a}_k$, with the empty sum convention $S_0=0$. For $n\ge 1$,Thus the total is the number of sequence terms multiplied by the arithmetic mean of the two endpoints. The formula also covers decreasing and constant arithmetic progressions.^[ref]

For the eight-entry example, the sum is $8(7+28)/2=140$, and the arithmetic mean of all eight entries is $17.5$. A decreasing example $11,9,7,5,3,1$ gives $6(11+1)/2=36$. Negative values cause no algebraic difficulty: $3,1,-1,-3,-5$ has sum $5(3-5)/2=-5$.

The triangular number $T_n=1+2+\cdots +n$ is the special case $a_1=d=1$, giving $n(n+1)/2$. More generally, an equal increase in each successive allocation makes the total grow as a quadratic function of the number of allocations. The eight-entry example hasIts first four entries add to $46$, its first eight to $140$, and its first twelve to $282$. The sequence terms lie on a straight line when plotted against their sequence indices, while the partial sums follow a quadratic function.

The fixed straight line of a nonidentity three-dimensional rotation. For a rotation matrix $R$, its direction vector is an eigenvector with eigenvalue $1$. A unit vector along it also specifies the positive sense of the angle through the right-hand convention.

Choose an ideal absorbing chemical species with $\epsilon =8000\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}\mathrm{c}{\mathrm{m}}^{-1}$ at a fixed wavelength, and a cuvette with geometric length $l=1\mathrm{cm}$. These are stipulated model parameters. A micromolar molar concentration is $1\mu \mathrm{M}=10^{-6}\mathrm{mol}{\mathrm{L}}^{-1}$, so each additional $25\mu \mathrm{M}$ gives $\Delta A=8000\times 1\times 25\times 10^{-6}=0.2$. Five chosen standards therefore predictThe absorbance graph is a straight line of slope $0.008(\mu \mathrm{M}{)}^{-1}$, while the transmittance graph shows exponential decay. Equal increments of molar concentration multiply transmittance by $10^{-0.2}\approx 0.63096$; they do not subtract a fixed transmitted fraction.

Suppose a measurement returns the stipulated blank-normalized reading $T=0.40$ under those same conditions. The inferred absorbance and molar concentration areThis lies slightly below the $50\mu \mathrm{M}$ standard because $0.40$ is slightly above $10^{-0.4}\approx 0.39811$. A calibration determined experimentally by linear regression may instead have $A=mc+b$. Then use $c=(A-b)/m$, keeping the slope $m$ and intercept $b$ in their measured units. A nonzero intercept calls for checking the reference blank and sample preparation; it does not uniquely identify the cause.

The Beer-Lambert law relates absorbance $A$ to an absorbing species' concentration $c$ and path length $l$ by $A=\epsilon lc$ in a homogeneous medium using sufficiently narrow-band light. The coefficient $\epsilon$ depends on the species, wavelength and chemical solution conditions. This makes Ellman's reagent useful for quantitative thiol assays once its released colored species has been calibrated. ^[ref]

Here $c$ is molar concentration, $l$ is the geometric length through the sample, and $\epsilon$ is the molar absorption coefficient. The ideal beam has parallel light rays, a single wavelength, and a medium with uniform composition and isotropy; finite spectral width must be small relative to features in the optical spectrum.^[ref]

For a spectrophotometer, distinguish the source's output from the reference that reaches the photodetector. Let $P_0$ be the received radiant power through a reference blank and $P$ the received radiant power through the sample, after subtracting the photodetector's dark response. Both refer to the same selected wavelength band and collection geometry. Define the blank-normalized transmittance and absorbance byThe reference blank includes the solvent and other assay ingredients except the absorbing target. Matching cuvettes and conditions makes their common losses divide out. Thus this $T$ is relative to the reference blank; it is not the absolute transmittance of the entire cuvette and its contents. Both radiant powers can be expressed in watts, while their ratio, $T$, and its logarithm, $A$, are dimensionless. An instrument displaying $40\%$ transmission means $T=0.40$, not $T=40$.

Why take a logarithm? Consider two consecutive identical absorbing layers. Each transmits a fraction $t$ of the radiant power entering it, so together they transmit $t^2$. Their absorbances add because $-{\log }_{10}(t^2)=2(-{\log }_{10}t)$. Doubling the sample length doubles absorbance while squaring transmittance. This turns multiplicative loss into an additive scale and makes a straight line useful for inferring molar concentration.

A short model derivation makes the assumption explicit. Suppose a thin layer of thickness $dx$ removes the fraction $kcdx$ of its incoming radiant power, where $k$ is a fixed constant of proportionality and $c$ is a uniform molar concentration. The corresponding ordinary differential equation, initial condition, and solution areTaking the base-ten logarithm gives $A=kcl/\ln 10$, so $\epsilon =k/\ln 10$. The exponential decay follows from equal layers removing equal fractions of what remains. Proportionality to molar concentration additionally assumes that the absorbing chemical species retains the same response as its molar concentration changes; the geometric argument alone does not establish that.

An actual teaching experiment prepares a Kool-Aid stock by weighing powder, transferring it quantitatively into a volumetric flask, and filling to the mark with water. A Genesys 30 spectrophotometer, blanked with water in a cuvette, scans for the peak absorbance wavelength. A Genesys 10 then measures absorbance at that wavelength, with a water reference blank. Each serial dilution transfers $50\mathrm{mL}$ of the preceding chemical solution to a $100\mathrm{mL}$ volumetric flask and fills to the mark; the protocol records the blank and sample readouts and repeats the halving sequence. These are prescribed experimental steps, not a published numerical data set.^[ref]

The experiment tests whether absorbance tracks the prepared concentration. Its concentration is powder mass per final volume, so the calibration does not by itself supply a single dye's molar absorption coefficient. The manual identifies loss of discrimination for very concentrated, very dilute, or very similar samples.^[ref]

To interpret such a calibration, hold cuvette orientation fixed with its clear faces in the light path, keep fingerprints and bubbles out of that path, and keep the geometric length unchanged. Mix each dilution completely. Re-read the reference blank during the sequence and re-read one standard to test drift. A separate preparation of a standard checks errors propagated through serial dilution; repeating only its instrumental reading does not test the transferred volumes. These controls follow from keeping $P_0$, $l$, and $c$ comparable between readings.

Three mechanisms explain why a straight line can fail. First, scattering of light can divert radiant power away from the photodetector, so cloudiness can raise apparent absorbance without the intended absorbing solute. A reference blank removes only losses shared with the sample. Second, stray light reaching the photodetector can make a strongly absorbing sample appear too transmissive. In an original model with extra detected radiant power $s{P}_0$ in both readings,As the ideal transmittance approaches zero, the reported absorbance approaches ${\log }_{10}((1+s)/s)$ for $s>0$. This finite ceiling is a model prediction, not measured instrument performance.

Third, chemical equilibrium can redistribute the absorbing chemical species when dilution, pH, or temperature changes. Even if each chemical species separately follows the Beer-Lambert law, their combined absorbance is $A=l{\sum }_i{\epsilon }_i{c}_i$, where $c_i$ is each species' molar concentration; changing those fractions need not preserve one effective molar absorption coefficient. Thus an unknown should be compared with standards under matching chemical solution conditions, within the tested concentration range. A good linear regression alone does not establish chemical selectivity.

Calculus connects local change with accumulated change. A derivative asks how fast a function changes near one input; an integral adds its contributions across an interval. The fundamental theorem of calculus explains why these operations fit together: accumulating a continuous rate produces a quantity whose derivative is that rate.

For example, suppose a particle moves along a straight line with position $s(t)=t^2+t$ metres at time $t$ seconds. Its velocity is $s^{\prime }(t)=2t+1$ metres per second. Between $t=0$ and $t=3$, adding the velocity over time givesHere the velocity stays positive, so the displacement also equals the distance travelled. If the particle reverses direction, the integral of its signed velocity still gives displacement, while the integral of its speed gives total distance.

This is useful because local rules are often easier to specify than whole trajectories. A differential equation can prescribe a particle's velocity or a population's growth rate; integration then reconstructs its evolution, subject to an initial condition. In the other direction, a derivative extracts local information from a known function, and optimization uses that information to locate or approach promising inputs. The same two operations therefore connect measurement, prediction and design.

The visual connection is developed in derivative, integral and fundamental theorem of calculus. The definitions depend on limits, which turn progressively finer finite measurements into exact mathematical statements. A graph with a corner need not have a derivative there, and the conditions permitting integration or interchange of limits matter; calculus is more than a collection of symbolic rules.

Historically, Isaac Newton described changing quantities and their rates in the language of fluents and fluxions; his manuscript History of the Method of Fluxions records that approach.^[ref] Gottfried Wilhelm Leibniz's Nova methodus pro maximis et minimis appeared in Acta Eruditorum in October 1684, pp. 467–473; the original article is preserved in the HKUST Library collection.^[ref] The modern limit definitions used here should not be read back into those seventeenth-century formulations.

A circular arc is a connected portion of a circle. For radius $R>0$ and subtended central angle $0<\theta <2\pi$ measured in radians, its arc length is $R\theta$. Two distinct endpoints determine two circular arcs: when they are not opposite, the minor arc has $\theta <\pi$ and the major arc has $\theta >\pi$. Opposite endpoints determine two semicircles. The straight line segment joining the endpoints is a chord, rather than the curved circular arc itself.

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 second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the number ${\Delta }_2=B^2-4AC$ is the discriminant of a binary quadratic form associated with its quadratic part. For a nonempty, nondegenerate real conic section, a negative value gives an ellipse, zero a parabola, and a positive value a hyperbola. It is invariant under a rotation of orthonormal coordinates.^[ref]

A short explanation uses the symmetric matrixAn orthogonal matrix $R$ changes this matrix to $R^{\mathsf{T}}HR$, whose determinant equals $\det H$. The spectral theorem for real symmetric matrices then removes the mixed term. Equal-sign, opposite-sign and one-zero eigenvalues give the three types after the remaining linear terms and offsets are considered.

The qualification matters. The equations $x^2+y^2=0$, $y^2=0$ and $xy=0$ describe a point, a repeated straight line, and two crossing straight lines, respectively. Their quadratic discriminants have the signs expected of ellipse, parabola and hyperbola types, but all three are degenerate conics. Also, $x^2+y^2=-1$ has an empty set of real solutions despite its negative quadratic discriminant.

One of the straight lines through the origin parallel to a standard basis vector in a Cartesian coordinate system. The positive direction corresponds to positive values of that coordinate.

For a fixed unit vector $g\in {\mathbb{R}}^3$, the linear map $r\mapsto r\times g$ has kernel $\mathbb{R}g$ and image the perpendicular plane $g^{\perp }$. Thus $r\times g=f$ is solvable exactly when $f\cdot g=0$, and then its full solution set isFor the proof, a cross product with $g$ must be perpendicular to $g$. Conversely the vector triple product identity gives $g\times (r\times g)=r-(r\cdot g)g$, which determines the perpendicular orthogonal projection of $r$. The remaining parallel component is arbitrary. Finally $(g\times f)\times g=f$ when $f\cdot g=0$, verifying sufficiency.

Geometrically, the solution set is a straight line with direction $g$. Its nearest point to the origin is $g\times f$ and its distance from the origin is $|f|$. For a general nonzero fixed vector $a$, the corresponding formula is $r=(a\times f)/|a{|}^2+ta$, again subject to $f\cdot a=0$.

A current-voltage characteristic describes the relation between terminal electric potential difference and electric current for specified physical conditions. A resistor obeying Ohm's law gives a straight line; a diode has an asymmetric relation.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

For a twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

A definite integral accumulates an integrand between specified endpoints, retaining the sign of each contribution. For a real-valued function on a finite real interval, a finite definite integral is a scalar; an indefinite integral describes a family of antiderivatives. The lower and upper limits of integration $a,b$ specify where accumulation starts and ends. The integration variable is a placeholder: ${\int }_a^{b}f(x)dx={\int }_a^{b}f(u)du$.

This is useful whenever a varying rate must be converted into a total change. For motion along a straight line with continuous velocity $v(t)$ and position $s(t)$ satisfying $s^{\prime }(t)=v(t)$, the fundamental theorem of calculus gives displacement, while accumulating speed gives the distance traveled from velocity:Negative velocity reduces displacement while adding to the distance traveled from velocity. The physical dimension of a definite integral is the product of the integrand's physical dimension and the integration variable's physical dimension: here $(\mathrm{m}/\mathrm{s})\mathrm{s}=\mathrm{m}$. An area on a rate graph need not be a spatial area.^[ref]

As an explicitly synthetic motion model, let $v(t)=(2\mathrm{m}{\mathrm{s}}^{-2})t-2\mathrm{m}{\mathrm{s}}^{-1}$ for $0\le t\le 3\mathrm{s}$, with $s(0)=0$. Using numerical values in seconds and metres, an antiderivative gives $s(t)=t^2-2t$. The velocity changes sign at $t=1$: the model first moves $1$ m in the negative direction, then $4$ m in the positive direction. Its displacement is $-1+4=3$ m, and its distance traveled from velocity is $1+4=5$ m. The initial position fixes the constant of integration, but does not change either accumulated quantity.

Orientation matters independently of whether the integrand is positive or negative. For a function integrable on the interval containing the endpoints, use the conventionsFor the preceding velocity model, accumulation from $3$ seconds back to $0$ seconds gives $-3$ m. Reversing the bounds of the speed integral gives $-5$ m, which is an oriented integral, not a negative distance traveled from velocity; physical elapsed-time totals use the earlier endpoint first. The splitting rule holds even when $b$ is outside the interval between $a$ and $c$, provided integrability holds on the interval containing all three endpoints.^[ref]

Existence needs hypotheses. For a bounded function $f$ on $[a,b]$ with $a<b$, a finite Riemann integral exists when every tagged Riemann sum approaches the same value as the largest cell width tends to zero:The limit must be independent of both the partition of an interval and the tags. Merely making the number of cells large can leave a wide cell unresolved. On a finite closed interval, every continuous function, and every bounded function that is continuous except at finitely many points, is Riemann integrable.^[ref] Boundedness alone is insufficient: the Dirichlet function on $[0,1]$ is $1$ on the rational numbers and $0$ on the irrational numbers. Tags can all be chosen rational or all irrational, giving Riemann sums of $1$ or $0$ at arbitrarily small cell widths.^[ref]

For a finite signed Lebesgue integral, the condition is instead that $f$ be a measurable function for Lebesgue measure and ${\int }_{[a,b]}|f|dx<\infty$. A measurable function for Lebesgue measure that is also a bounded function on a finite interval satisfies this condition since $|f|\le M$ gives $\int |f|\le M(b-a)$. The Dirichlet function therefore has Lebesgue integral zero: the rational numbers form a null set, so it equals zero Lebesgue almost everywhere. The Riemann integral and Lebesgue integral agree for the continuous functions used in the diagrams.^[ref] An unbounded integrand or infinite endpoint requires further analysis, for example the limits defining an improper integral; the notation alone does not establish convergence.

For $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the homogeneous quadratic coefficient matrix isThe conic is degenerate when $Q$ is a singular matrix. Examples over the real numbers include the point $x^2+y^2=0$, the repeated line $y^2=0$, and the two straight lines in $xy=0$. The real solution set alone need not reveal the algebraic distinction: $x^2+y^2=-1$ has no real points, but its matrix is nonsingular. This determinant test is different from the conic type discriminant, which only uses the quadratic part in $x,y$.

The derivative measures the instantaneous rate at which a function changes. For a real function $f$ defined near $a$, it exists when the difference quotient has a finite two-sided limit:The quotient is the slope of the straight line joining two nearby points of the graph. Its limit is the slope of the tangent line. Thus a derivative describes a local rate without choosing a particular finite measurement interval.

For $f(x)=x^2$, expanding the numerator givesAt $a=1$, the increments $h=1$, $1/2$ and $1/10$ give secant line slopes $3$, $2.5$ and $2.1$. The limit is $2$, and the tangent line is $y=1+2(x-1)$. The picture makes the limiting process visible.

The same calculation shows why the derivative is useful for approximation:This is a local linear approximation. Near $a=1$, replacing $(1+h{)}^2$ by $1+2h$ makes the exact error $h^2$, smaller than the leading change when $h$ is small. A curved graph can therefore be studied locally using a straight line.

If the function records position against time, its derivative is velocity. If it records an objective function, its derivative tests local changes that matter in optimization; in several variables, the gradient collects these directional rates and guides gradient descent. The interpretation comes from the function and its units, rather than from a new rule for each application.

Existence matters. For the absolute value function $f(x)=|x|$, the one-sided derivatives at zero are $-1$ and $1$, so there is no single tangent line described by a finite derivative there. Nor does a zero derivative alone identify a maximum or minimum: $f(x)=x^3$ has $f^{\prime }(0)=0$ but keeps increasing through zero.

The directional derivative tests a function along a chosen straight line, including lines that follow neither coordinate axis. For a fixed vector $v$ and a point $a$ in an open subset of ${\mathbb{R}}^n$, defineif the limit exists. With a unit vector $v$, the parameter $t$ measures signed distance in the input Euclidean space, so the directional derivative is change per unit distance. For a general $v$, it is change per unit parameter; the norm of $v$ also sets the rate of travel.^[ref]

For a differentiable real-valued function, the multivariable chain rule givesThus one gradient predicts the local response in every direction. This formula requires differentiability, rather than merely separate partial derivatives.^[ref]

Return to $f(x,y)=x^2+xy+2y^2$ at $a=(1,1)$. Its gradient is $(3,5)$. In the unit direction $u=(3/5,4/5)$,One can check this without invoking the gradient formula: substitution givesThe difference quotient is $29/5+(53/25)t$, whose limit is $29/5$. This calculation separates the exact change along a finite path from its first-order rate at the starting point.

Changing the travel speed changes the parameter rate. Taking $v=(3,4)=5u$ gives $D_{v}f(a)=29$, because the input travels five distance units per parameter unit. For a differentiable function, $D_{cv}f(a)=c{D}_{v}f(a)$ for any real $c$: negative $c$ reverses the orientation as well as scaling the rate. Normalizing $v$ before comparing different slopes avoids confusing direction with speed.

The Cauchy-Schwarz inequality bounds the rate in any unit direction by $\Vert \nabla f(a)\Vert$. Here the largest rate is $\sqrt{34}\approx 5.831$, attained in the direction $(3,5)/\sqrt{34}$; the chosen $(3/5,4/5)$ gives the slightly smaller rate $5.8$. The perpendicular unit direction $(-5,3)/\sqrt{34}$ gives zero. That means no first-order height change, rather than no change at all: along this direction the example has $f(a+tu)=4+(28/34)t^2$.

This is useful for following a height or temperature scalar field, choosing a local search direction in mathematical optimization, and distinguishing a coordinate sensitivity from a response along an actual path. For a curved path, derivative along a curve uses its instantaneous tangent vector. Even existence of all directional derivatives does not guarantee one common linear approximation; directional derivatives do not imply differentiability gives a concrete counterexample.

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.

A rotation through an angle of $\pi$. In three dimensions, a half-turn about a straight line with unit vector $\mathbf{n}$ fixes its parallel component and negates its orthogonal complement, giving the rotation matrix $2\mathbf{n}{\mathbf{n}}^T-I$ by orthogonal projection.