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 an equal sum of distances to two given fixed points in the plane. In figure 1 two fixed points F₁ and F₂ are shown, these are the foci of the ellipse. An arbitrary point P₁ on the ellipse has distance |F₁P₁| to F₁ and distance |F₂P₁| to F₂. Let d be the sum of distances of P₁ to the foci,

$d\equiv |\mathrm {F} _{1}\mathrm {P} _{1}|+|\mathrm {F} _{2}\mathrm {P} _{1}|,$

then all points of the ellipse have the sum of distances d. Thus, another arbitrary point P₂ on the ellipse has 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 S₁–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 the foci 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. Clearly both axes are symmetry axes, reflection in either of them transforms the ellipse into itself. Basically, this a consequence of the fact that a reflection conserves (sums of) distances. The intersection of the axes is the center of the ellipse.

Clearly, the two foci and the points S₁ and S₂ are connected by reflection in 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. The distance of S₂ to F₂ is equal to p. By the definition of the ellipse the 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.

Eccentricity

The eccentricity e of an ellipse (usually denoted by e or ε) is the ratio of the distance |OF₂| to half the length a of the major axis, that is, e ≡ |OF₂|/a. Introducing the vector along the x-axis,

${\vec {a}}\equiv {\begin{pmatrix}a\\0\end{pmatrix}}$

one finds that e satisfies the vector relation,

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

Introduce the following two vectors (cf. figure 2),

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

Make now the special choice that the point P is on the positive y-axis,

${\vec {r}}={\begin{pmatrix}0\\b\end{pmatrix}},$

then the distance of P to either focus is equal to a and equal to the length of the vector ${\vec {g}}$. For two inner products (indicated by a dot) we find,

${\vec {r}}\cdot {\vec {a}}=0\quad {\hbox{and}}\quad {\vec {r}}\cdot {\vec {r}}=b^{2}.$

Hence, (in fact by the Pythagoras theorem),

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

so that the eccentricity is given by

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

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Fig. 2. An ellipse situated such that the major and minor axis are along Cartesian axes. The center of the ellipse coincides with the origin O.

Algebraic form

For a point P=(x,y) on an ellipse that is located with respect to a Cartesian frame as in figure 2, 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

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

By definition, 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

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

Use

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

and 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

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

Division by b² gives finally

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

Polar representation

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Fig. 3. Polar representation

The length g of the vector (cf. figure 3)

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

is given by the polar equation

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

where ℓ is known as the semi-lactus rectum of the ellipse.

Proof

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

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

Elimination of 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

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

and the polar equation for the ellipse follows.