Ellipse

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Fig. 1. Ellipse (black closed curve). The sum of the lengths of the two red line segments (ending in P₁) is equal to the sum of the lengths of the two blue line segments (ending in P₂).

In mathematics, an ellipse is a planar locus of points characterized by having a constant sum of distances to two given fixed points in the plane. In figure 1, the two fixed points are F₁ and F₂, these are the foci of the ellipse. Consider an arbitrary point P₁ on the ellipse that has distance F₁P₁ to F₁ and distance F₂P₁ to F₂, and let d be the sum of distances of P₁ to the foci,

$d:=\mathrm {F} _{1}\mathrm {P} _{1}+\mathrm {F} _{2}\mathrm {P} _{1},\,$

then for all points of the ellipse the sum of distances is also d. Thus, for another arbitrary point P₂ on the ellipse with distance F₁P₂ to F₁ and distance F₂P₂ to F₂, by definition, the sum of distances of P₂ to the foci is equal to d,

$\mathrm {F} _{1}\mathrm {P} _{2}+\mathrm {F} _{2}\mathrm {P} _{2}=d.\,$

The horizontal line segment between S₁ and S₂ in figure 1, going through the foci, is known as the major axis of the ellipse. Traditionally, the length of the major axis is indicated by 2a. The vertical dashed line segment, drawn halfway between the foci and perpendicular to the major axis, is referred to as the minor axis of the ellipse; its length is usually indicated by 2b. The major and minor axes are distinguished by a ≥ b. When a = b the ellipse is a circle—a special case of an ellipse. The foci of a circle coincide with the center of the circle. Clearly both ellipse axes are symmetry axes, reflection about either of them transforms the ellipse into itself. Basically, this is a consequence of the fact that reflection preserves (sums of) distances. The intersection of the axes is the center of the ellipse.

The two foci and the points S₁ and S₂ are connected by reflection about the minor axis. Hence the distance S₂F₂ =: p is, by symmetry, equal to the distance S₁F₁. The distance of S₂ to F₁ is equal to 2a − p. By the definition of the ellipse their sum is equal to d, hence

$d=(2a-p)+p=2a.\;$

The sum d of distances from any point on the ellipse to the foci is equal to the length of the major axis.

Conic section

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Fig. 2. Upper shaded (green) section: ellipse; lower shaded (red) section: circle.

In the work of the Greek mathematician Apollonius (c. 262–190 BC) the ellipse arose as the intersection of a plane with a cone. Apollonius gave the ellipse its name, though the term ἔλλειψις (elleipsis, meaning "falling short") was used earlier by Euclid (c. 300 BC) in the construction of parallelograms with areas that "fell short". Apollonius applied the word to the conic section that at present we call ellipse. See Ref. for the—in modern eyes—complicated reasoning by which Apollonius tied the shape of certain conic sections to Euclid's concept of deficient areas.

In figure 2 a cone with a circular base is shown. It has a vertical symmetry axis, an axis of revolution. A cone can be generated by revolving around the axis a line that intersects the axis of rotation under an angle α (between 0 and 90 degree). A horizontal plane (plane perpendicular to the axis of the cone) — that does not contain the vertex — intersects the cone in a circle (a special ellipse). If the angle between the plane is greater than α (but smaller than a right angle) then it intersects the cone in an ellipse. (Otherwise, the intersection is either a parabola or a hyperbola.)

Eccentricity

The eccentricity e of an ellipse (usually denoted by e or ε) is the ratio of the distance OF₂ (cf. figure 3) to the length a (half the major axis), that is, e := OF₂ / a. Let ${\vec {a}}$ be a vector of length a along the x-axis, then

$e{\vec {a}}:={\overrightarrow {\mathrm {OF} }}_{2}.$

The following two vectors have common endpoint at P, see figure 3,

${\vec {r}}:={\overrightarrow {\mathrm {OP} }}={\begin{pmatrix}x\\y\end{pmatrix}}\quad {\hbox{and}}\quad {\vec {g}}:={\overrightarrow {\mathrm {F} _{2}\mathrm {P} }}={\overrightarrow {\mathrm {F} _{2}\mathrm {O} }}+{\overrightarrow {\mathrm {O} \mathrm {P} }}=-e{\vec {a}}+{\vec {r}}.$

Now choose P as the intersection P₁ of the positive y-axis with the ellipse; then its position vector is:

${\vec {r}}_{1}={\begin{pmatrix}0\\b\end{pmatrix}}.$

By symmetry, the distance of this point P₁ to either focus is equal, thus the length of the corresponding vector ${\vec {g}}_{1}$ (with endpoint on the y-axis) is equal to the length a of the semi-major axis. For the following two inner products (indicated by a centered dot) we find,

${\vec {r}}_{1}\cdot (e{\vec {a}})=0\quad {\hbox{and}}\quad {\vec {r}}_{1}\cdot {\vec {r}}_{!}=b^{2}.$

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Fig. 3. An ellipse situated such that the major and minor axis are along Cartesian axes. The center of the ellipse coincides with the origin O.

Hence, (in fact the Pythagoras theorem applied to POF₂),

$|{\vec {g}}_{1}\;|^{2}=a^{2}=({\vec {r}}_{1}-e{\vec {a}}\;)\cdot ({\vec {r}}_{1}-e{\vec {a}}\;)=b^{2}+e^{2}a^{2},$

so that the eccentricity is given by

$e={\sqrt {\frac {a^{2}-b^{2}}{a^{2}}}}\quad {\hbox{with}}\quad 0\leq e\leq 1.$

Algebraic form

Consider an ellipse that is located with respect to a Cartesian frame as in figure 3 (major axis on x-axis, minor axis on y-axis). For a point P=(x,y) of the ellipse it holds that

${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.$

Note that this equation is reminiscent of the equation for a unit circle. An ellipse may be seen as a unit circle in which the x and the y coordinates are scaled independently, by 1/a and 1/b, respectively.

Proof

Introduce the vectors

${\begin{aligned}{\overrightarrow {\mathrm {F} _{1}\mathrm {P} }}={\overrightarrow {\mathrm {F} _{1}\mathrm {O} }}+{\overrightarrow {\mathrm {O} \mathrm {P} }}=&\;e{\vec {a}}+{\vec {r}}\\{\overrightarrow {\mathrm {F} _{2}\mathrm {P} }}={\overrightarrow {\mathrm {F} _{2}\mathrm {O} }}+{\overrightarrow {\mathrm {O} \mathrm {P} }}=&-e{\vec {a}}+{\vec {r}}={\vec {g}}\\\end{aligned}}$

By definition of ellipse, the sum of the lengths is 2a

$|{\vec {r}}+e{\vec {a}}|+|{\vec {r}}-e{\vec {a}}|=2a\qquad \qquad \qquad \qquad (1)$

Multiply Eq. (1) by

$|{\vec {r}}+e{\vec {a}}|-|{\vec {r}}-e{\vec {a}}|$

and work out the left-hand side:

$|{\vec {r}}+e{\vec {a}}|^{2}-|{\vec {r}}-e{\vec {a}}|^{2}=4e{\vec {r}}\cdot {\vec {a}}$

Hence

$4e{\vec {r}}\cdot {\vec {a}}=2a(|{\vec {r}}+e{\vec {a}}|-|{\vec {r}}-e{\vec {a}}|)$

Using

${\frac {{\vec {r}}\cdot {\vec {a}}}{a}}=x$

one obtains

$|{\vec {r}}+e{\vec {a}}|-|{\vec {r}}-e{\vec {a}}|=2ex\qquad \qquad \qquad \qquad (2)$

Add and subtract Eqs (1) and (2) and we find expressions for the distance of P to the foci,

${\begin{aligned}|{\overrightarrow {\mathrm {F} _{1}\mathrm {P} }}|&=|{\vec {r}}+e{\vec {a}}|&=a+ex&\\|{\overrightarrow {\mathrm {F} _{2}\mathrm {P} }}|&=|{\vec {r}}-e{\vec {a}}|&=a-ex&\qquad \qquad \qquad \quad (3)\\\end{aligned}}$

Square both equations

${\begin{aligned}r^{2}+e^{2}a^{2}+2e{\vec {r}}\cdot {\vec {a}}&=a^{2}+e^{2}x^{2}+2eax\\r^{2}+e^{2}a^{2}-2e{\vec {r}}\cdot {\vec {a}}&=a^{2}+e^{2}x^{2}-2eax\\\end{aligned}}$

Adding, using the earlier derived value for e², and reworking gives

$r^{2}+e^{2}a^{2}=a^{2}+e^{2}x^{2}\Longrightarrow r^{2}+a^{2}-b^{2}=a^{2}+{\frac {x^{2}}{a^{2}}}(a^{2}-b^{2})\Longrightarrow r^{2}=x^{2}+y^{2}=b^{2}+x^{2}-x^{2}{\frac {b^{2}}{a^{2}}}\Longrightarrow y^{2}=b^{2}-x^{2}{\frac {b^{2}}{a^{2}}}$

Division by b² gives finally

${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.$

Second degree equation

Under certain conditions the following general equation of second degree in x and y represents an ellipse:

$f(x,y):=Ax^{2}+2Bxy+Cy^{2}+2Dx+2Ey+F=0,\,$

(all variables are real).

To state the conditions we need the solutions t₁ and t₂ of the following system of linear equations:

${\begin{matrix}At_{1}+Bt_{2}&=-D\\Bt_{1}+Ct_{2}&=-E.\end{matrix}}$

These equations have a unique solution if and only if AC − B² ≠ 0. The first condition for f(x,y) = 0 to represent an ellipse requires more, namely that AC − B² > 0.

Define f_t := f(t₁, t₂), then the second condition is: f_t ≠ 0.

The third condition is:

If f_t > 0 then A+C < 0

If f_t < 0 then A+C > 0

• The equation f(x,y) = 0 represents an ellipse if and only if these three conditions are satisfied.

Proof

In order to find the conditions that a quadratic equation represents an ellipse, we switch to matrix-vector notation and write f(x,y) as

$f(\mathbf {r} )=\mathbf {r} ^{\mathrm {T} }\mathbf {Q} \mathbf {r} +\mathbf {a} ^{\mathrm {T} }\mathbf {r} +\mathbf {r} ^{\mathrm {T} }\mathbf {a} +F,$

with

$\mathbf {r} :={\begin{pmatrix}x\\y\\\end{pmatrix}},\quad \mathbf {Q} :={\begin{pmatrix}A&B\\B&C\\\end{pmatrix}},\quad \mathbf {a} :={\begin{pmatrix}D\\E\\\end{pmatrix}}.$

The superscript T stands for transposition (row vector becomes column vector and vice versa). The expression can be rewritten by introducing the inverse of the matrix Q. This gives a condition: in order that the matrix be invertible its determinant det(Q) := AC−B² ≠ 0.

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Fig. 4. r′ = r − t

Then

$f(\mathbf {r} )=\left(\mathbf {r} +\mathbf {Q} ^{-1}\mathbf {a} \right)^{\mathrm {T} }\mathbf {Q} \left(\mathbf {r} +\mathbf {Q} ^{-1}\mathbf {a} \right)-\mathbf {a} ^{\mathrm {T} }\mathbf {Q} ^{-1}\mathbf {a} +F.$

Note that this uses

$\mathbf {Q} ^{\mathrm {T} }=\mathbf {Q} \quad \Longrightarrow \quad \left(\mathbf {Q} ^{-1}\right)^{\mathrm {T} }=\mathbf {Q} ^{-1},$

i.e., that both the matrix Q and its inverse are symmetric.

Define

$\mathbf {t} :=-\mathbf {Q} ^{-1}\mathbf {a} \quad \Longrightarrow \quad \mathbf {Q} \mathbf {t} =-\mathbf {a} ,$

and

$\mathbf {r} ':=\mathbf {r} -\mathbf {t} .$

The minus sign in the definition of t is introduced to get the translation of the origin as depicted in figure 4. Substitute r′ in the expression for f:

$f(\mathbf {r} )=\left(\mathbf {r} -\mathbf {t} \right)^{\mathrm {T} }\mathbf {Q} \left(\mathbf {r} -\mathbf {t} \right)-\mathbf {a} ^{\mathrm {T} }\mathbf {Q} ^{-1}\mathbf {a} +F=\left(\mathbf {r} '\right)^{\mathrm {T} }\mathbf {Q} \mathbf {r} '+f_{t}$

with

$f_{t}:=f(\mathbf {t} )=-\mathbf {a} ^{\mathrm {T} }\mathbf {Q} ^{-1}\mathbf {a} +F.$

By translation of the origin over t   the linear terms in f(r) have been eliminated, only two quadratic terms (in x′ and y′), one bilinear term, and one constant term (f_t) appear in the equation for f. The price paid for it is the requirement det(Q) ≠ 0.

The next step is rotation of the x′ and y′ axis (with origin in O'), this will eliminate the bilinear term and decouple x′ and y′, the components of r′. Let us recall that a real symmetric matrix may be diagonalized by an orthogonal matrix. For the 2×2 case:

$\mathbf {R} ^{\mathrm {T} }\mathbf {Q} \mathbf {R} ={\begin{pmatrix}\alpha _{1}&0\\0&\alpha _{2}\end{pmatrix}}\quad {\hbox{with}}\quad \mathbf {R} ^{\mathrm {T} }\mathbf {R} =\mathbf {R} \mathbf {R} ^{\mathrm {T} }=\mathbf {I} ,$

where the last matrix on the right is the identity matrix I. Now

$f=\left(\mathbf {r} '\right)^{\mathrm {T} }\;\mathbf {R} \mathbf {R} ^{\mathrm {T} }\;\mathbf {Q} \;\mathbf {R} \mathbf {R} ^{\mathrm {T} }\;\mathbf {r} '+f_{t}=\left(\mathbf {r} ''\right)^{\mathrm {T} }{\boldsymbol {\alpha }}\mathbf {r} ''+f_{t},$

with

${\boldsymbol {\alpha }}:={\begin{pmatrix}\alpha _{1}&0\\0&\alpha _{2}\end{pmatrix}},\quad {\hbox{and}}\quad \mathbf {r} '':=\mathbf {R} ^{\mathrm {T} }\mathbf {r} '.$

An ellipse is obtained if the parameters α₁, α₂, and f_t are non-zero and if the signs of α₁ and α₂ are equal and opposite to the sign of f_t in the following expression:

$\left(\mathbf {r} ''\right)^{\mathrm {T} }{\boldsymbol {\alpha }}\mathbf {r} ''+f_{t}=\alpha _{1}(x'')^{2}+\alpha _{2}(y'')^{2}+f_{t}=0.$

It is known that the determinant of a matrix is invariant under a matrix similarity transformation, hence if (and only if) the signs of α₁ and α₂ are equal it follows that

$\det(\mathbf {Q} )=AC-B^{2}=\alpha _{1}\alpha _{2}>0.$

So, the condition det(Q) ≠ 0 is narrowed to det(Q) > 0.

The trace A+C of the matrix is also invariant under a similarity transformation. There are two possibilities in order that the equation f(x,y) = 0 represents an ellipse

${\begin{aligned}\alpha _{1},\alpha _{2}<0&\Longrightarrow A+C=\alpha _{1}+\alpha _{2}<0\quad {\hbox{then}}\quad f_{t}>0\\\alpha _{1},\alpha _{2}>0&\Longrightarrow A+C=\alpha _{1}+\alpha _{2}>0\quad {\hbox{then}}\quad f_{t}<0.\\\end{aligned}}$

Clearly, in order to determine a priori whether the quadratic equation represents an ellipse, a diagonalization of Q is not necessary. It is sufficient to determine the sign of f_t = f(t) by solving the equation given for the vector t.

Polar representation relative to focus

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Fig. 5. Polar representation

The length g of the vector (cf. figure 5) with endpoint on the ellipse

${\overrightarrow {\mathrm {F} _{2}\mathrm {P} }}=:{\vec {g}}=(ea+g\cos \theta ,\;g\sin \theta )$

is given by the polar equation of an ellipse

$g={\frac {\ell }{1+e\cos \theta }}\quad {\hbox{with}}\quad \ell :={\frac {b^{2}}{a}},$

where 2ℓ is known as the latus rectum (lit. erect side) of the ellipse; it is equal to 2g for θ = 90^° (twice the length of the vector ${\vec {g}}$ when it makes a right angle with the major axis).

Proof

Earlier [Eq. (3)] it was derived for the distance from the right focus F₂ to P that

$|{\overrightarrow {\mathrm {F} _{2}\mathrm {P} }}|=|{\vec {g}}|=g=a-ex.$

Expressing x from

$x=ea+g\cos \theta =ea+(a-ex)\cos \theta \,$

gives

$x={\frac {ea+a\cos \theta }{1+e\cos \theta }},$

so that

$g=a-ex=a-{\frac {e^{2}a+ea\cos \theta }{1+e\cos \theta }}={\frac {a(1-e^{2})}{1+e\cos \theta }}.$

Substitute

$a(1-e^{2})=a(1-{\frac {a^{2}-b^{2}}{a^{2}}})={\frac {b^{2}}{a}}=:\ell$

and the polar equation for the ellipse follows.

Trammel construction

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Fig. 6. A trammel in theory

Before drafting was done almost exclusively by the aid of computers, draftsmen used a simple device for drawing ellipses, a trammel. Basically, a trammel is a rigid bar of length a (semi-major axis). In the top drawing of figure 6 the bar is shown as a blue-red line segment bounded by a black and a blue bead. On this bar a segment of length b (semi-minor axis) is marked; this is the red segment on the bar. Two beads fixed to the rigid bar move back and forth along the x-axis and y-axis, respectively. The blue bead fixed at one end of the bar moves along the y-axis, the red bead, which marks the beginning of the red segment of length b, moves along the x-axis. The endpoint of the bar (the black bead in figure 6) moves along an ellipse with semi-major axis a and semi-minor axis b and typically has a pen fixed to it.

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Fig. 7. A trammel in practice

The fact that the trammel construction works is proved very easily, cf. the bottom drawing in figure 6,

$x=a\;\cos \theta \quad {\hbox{and}}\quad y=b\;\sin \theta .$

Hence

${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=\cos ^{2}\theta +\sin ^{2}\theta =1,$

which indeed is the equation for an ellipse.

A device called a trammel point is used to guide a woodworking router in making elliptical cuts.

Gardener's construction

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Fig. 8. Gardener's construction

It is possible to construct an ellipse of given major and minor axis by the aid of a compass, a ruler, three thumbtacks, and a piece of string, see figure 8.

First draw the major axis AB, and then obtain with the compass its perpendicular bisector intersecting AB in the midpoint E. Along the bisector one measures off the length of the minor axis CD. Given that the distances CF and CG are the semi-major axis (AB/2), one can determine the foci by drawing an arc with the compass using C as center and AB/2 as radius. One now pins the thumbtacks in the foci and the point C and fixes a piece of string around the triangle F₁F₂C (i.e, its length equals the perimeter of the triangle). Removing the thumbtack at C, and keeping the string taut, one draws the ellipse by moving the pencil from C to A, D, B, and back to C.

Clearly this procedure can be used in the garden to create an elliptic lawn or flowerbed, which is why the procedure is sometimes referred to as the gardener's construction.

Notes

Figures 7 and 8 are from George Watson Kittredge, The New Metal Worker Pattern Book, David Williams Company, New York, (1901) Online