In mathematics, an ellipse is a plane curve surrounding two focal points, such that for all points on the curve, the sum of the two distances to the focal points is a constant. It generalizes a circle, which is the special type of ellipse in which the two focal points are the same. The elongation of an ellipse is measured by its eccentricity$e$, a number ranging from $e=0$ (the limiting case of a circle) to $e=1$ (the limiting case of infinite elongation, no longer an ellipse but a parabola).
Analytically, the equation of a standard ellipse centered at the origin with width $2a$ and height $2b$ is:
${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.$
Assuming $a\geq b$, the foci are $(\pm c,0)$ for ${\textstyle c={\sqrt {a^{2}-b^{2}}}}$. The standard parametric equation is:
$(x,y)=(a\cos(t),b\sin(t))\quad {\text{for}}\quad 0\leq t\leq 2\pi .$
Ellipses are the closed type of conic section: a plane curve tracing the intersection of a cone with a plane (see figure). Ellipses have many similarities with the other two forms of conic sections, parabolas and hyperbolas, both of which are open and unbounded. An angled cross section of a right circular cylinder is also an ellipse.
An ellipse may also be defined in terms of one focal point and a line outside the ellipse called the directrix: for all points on the ellipse, the ratio between the distance to the focus and the distance to the directrix is a constant. This constant ratio is the above-mentioned eccentricity:
$e={\frac {c}{a}}={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}.$
Ellipses are common in physics, astronomy and engineering. For example, the orbit of each planet in the Solar System is approximately an ellipse with the Sun at one focus point (more precisely, the focus is the barycenter of the Sun–planet pair). The same is true for moons orbiting planets and all other systems of two astronomical bodies. The shapes of planets and stars are often well described by ellipsoids. A circle viewed from a side angle looks like an ellipse: that is, the ellipse is the image of a circle under parallel or perspective projection. The ellipse is also the simplest Lissajous figure formed when the horizontal and vertical motions are sinusoids with the same frequency: a similar effect leads to elliptical polarization of light in optics.
The name, ἔλλειψις (élleipsis, "omission"), was given by Apollonius of Perga in his Conics.
Definition as locus of points
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An ellipse can be defined geometrically as a set or locus of points in the Euclidean plane:
Given two fixed points $F_{1},F_{2}$ called the foci and a distance $2a$ which is greater than the distance between the foci, the ellipse is the set of points $P$ such that the sum of the distances $|PF_{1}|,\ |PF_{2}|$ is equal to $2a$: $E=\left\{P\in \mathbb {R} ^{2}\,\mid \,\left|PF_{2}\right|+\left|PF_{1}\right|=2a\right\}.$
The midpoint $C$ of the line segment joining the foci is called the center of the ellipse. The line through the foci is called the major axis, and the line perpendicular to it through the center is the minor axis. The major axis intersects the ellipse at two vertices$V_{1},V_{2}$, which have distance $a$ to the center. The distance $c$ of the foci to the center is called the focal distance or linear eccentricity. The quotient $e={\tfrac {c}{a}}$ is the eccentricity.
The case $F_{1}=F_{2}$ yields a circle and is included as a special type of ellipse.
The equation $\left|PF_{2}\right|+\left|PF_{1}\right|=2a$ can be viewed in a different way (see figure):
If $c_{2}$ is the circle with center $F_{2}$ and radius $2a$, then the distance of a point $P$ to the circle $c_{2}$ equals the distance to the focus $F_{1}$: $\left|PF_{1}\right|=\left|Pc_{2}\right|.$
$c_{2}$ is called the circular directrix (related to focus $F_{2}$) of the ellipse.^{[1]}^{[2]} This property should not be confused with the definition of an ellipse using a directrix line below.
Using Dandelin spheres, one can prove that any section of a cone with a plane is an ellipse, assuming the plane does not contain the apex and has slope less than that of the lines on the cone.
In Cartesian coordinates
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Standard equation
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The standard form of an ellipse in Cartesian coordinates assumes that the origin is the center of the ellipse, the x-axis is the major axis, and:
the foci are the points $F_{1}=(c,\,0),\ F_{2}=(-c,\,0)$,
the vertices are $V_{1}=(a,\,0),\ V_{2}=(-a,\,0)$.
For an arbitrary point $(x,y)$ the distance to the focus $(c,0)$ is ${\textstyle {\sqrt {(x-c)^{2}+y^{2}}}}$ and to the other focus ${\textstyle {\sqrt {(x+c)^{2}+y^{2}}}}$. Hence the point $(x,\,y)$ is on the ellipse whenever:
${\sqrt {(x-c)^{2}+y^{2}}}+{\sqrt {(x+c)^{2}+y^{2}}}=2a\ .$
Removing the radicals by suitable squarings and using $b^{2}=a^{2}-c^{2}$ (see diagram) produces the standard equation of the ellipse:^{[3]}${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1,$
or, solved for y:
$y=\pm {\frac {b}{a}}{\sqrt {a^{2}-x^{2}}}=\pm {\sqrt {\left(a^{2}-x^{2}\right)\left(1-e^{2}\right)}}.$
The width and height parameters $a,\;b$ are called the semi-major and semi-minor axes. The top and bottom points $V_{3}=(0,\,b),\;V_{4}=(0,\,-b)$ are the co-vertices. The distances from a point $(x,\,y)$ on the ellipse to the left and right foci are $a+ex$ and $a-ex$.
It follows from the equation that the ellipse is symmetric with respect to the coordinate axes and hence with respect to the origin.
In principle, the canonical ellipse equation ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ may have $a<b$ (and hence the ellipse would be taller than it is wide). This form can be converted to the standard form by transposing the variable names $x$ and $y$ and the parameter names $a$ and $b.$
Linear eccentricity
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This is the distance from the center to a focus: $c={\sqrt {a^{2}-b^{2}}}$.
Eccentricity
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The eccentricity can be expressed as:
$e={\frac {c}{a}}={\sqrt {1-\left({\frac {b}{a}}\right)^{2}}},$
assuming $a>b.$ An ellipse with equal axes ($a=b$) has zero eccentricity, and is a circle.
Semi-latus rectum
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The length of the chord through one focus, perpendicular to the major axis, is called the latus rectum. One half of it is the semi-latus rectum$\ell$. A calculation shows:^{[4]}$\ell ={\frac {b^{2}}{a}}=a\left(1-e^{2}\right).$
An arbitrary line $g$ intersects an ellipse at 0, 1, or 2 points, respectively called an exterior line, tangent and secant. Through any point of an ellipse there is a unique tangent. The tangent at a point $(x_{1},\,y_{1})$ of the ellipse ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ has the coordinate equation:
${\frac {x_{1}}{a^{2}}}x+{\frac {y_{1}}{b^{2}}}y=1.$
A vector parametric equation of the tangent is:
${\vec {x}}={\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+s\left({\begin{array}{r}-y_{1}a^{2}\\x_{1}b^{2}\end{array}}\right),\quad s\in \mathbb {R} .$
Proof:
Let $(x_{1},\,y_{1})$ be a point on an ellipse and ${\textstyle {\vec {x}}={\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+s{\begin{pmatrix}u\\v\end{pmatrix}}}$ be the equation of any line $g$ containing $(x_{1},\,y_{1})$. Inserting the line's equation into the ellipse equation and respecting ${\textstyle {\frac {x_{1}^{2}}{a^{2}}}+{\frac {y_{1}^{2}}{b^{2}}}=1}$ yields:
${\frac {\left(x_{1}+su\right)^{2}}{a^{2}}}+{\frac {\left(y_{1}+sv\right)^{2}}{b^{2}}}=1\ \quad \Longrightarrow \quad 2s\left({\frac {x_{1}u}{a^{2}}}+{\frac {y_{1}v}{b^{2}}}\right)+s^{2}\left({\frac {u^{2}}{a^{2}}}+{\frac {v^{2}}{b^{2}}}\right)=0\ .$
There are then cases:
${\frac {x_{1}}{a^{2}}}u+{\frac {y_{1}}{b^{2}}}v=0.$ Then line $g$ and the ellipse have only point $(x_{1},\,y_{1})$ in common, and $g$ is a tangent. The tangent direction has perpendicular vector${\begin{pmatrix}{\frac {x_{1}}{a^{2}}}&{\frac {y_{1}}{b^{2}}}\end{pmatrix}}$, so the tangent line has equation ${\textstyle {\frac {x_{1}}{a^{2}}}x+{\tfrac {y_{1}}{b^{2}}}y=k}$ for some $k$. Because $(x_{1},\,y_{1})$ is on the tangent and the ellipse, one obtains $k=1$.
${\frac {x_{1}}{a^{2}}}u+{\frac {y_{1}}{b^{2}}}v\neq 0.$ Then line $g$ has a second point in common with the ellipse, and is a secant.
Using (1) one finds that ${\begin{pmatrix}-y_{1}a^{2}&x_{1}b^{2}\end{pmatrix}}$ is a tangent vector at point $(x_{1},\,y_{1})$, which proves the vector equation.
If $(x_{1},y_{1})$ and $(u,v)$ are two points of the ellipse such that ${\textstyle {\frac {x_{1}u}{a^{2}}}+{\tfrac {y_{1}v}{b^{2}}}=0}$, then the points lie on two conjugate diameters (see below). (If $a=b$, the ellipse is a circle and "conjugate" means "orthogonal".)
Shifted ellipse
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If the standard ellipse is shifted to have center $\left(x_{\circ },\,y_{\circ }\right)$, its equation is
${\frac {\left(x-x_{\circ }\right)^{2}}{a^{2}}}+{\frac {\left(y-y_{\circ }\right)^{2}}{b^{2}}}=1\ .$
The axes are still parallel to the x- and y-axes.
General ellipse
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In analytic geometry, the ellipse is defined as a quadric: the set of points $(x,\,y)$ of the Cartesian plane that, in non-degenerate cases, satisfy the implicit equation^{[5]}^{[6]}$Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$
provided $B^{2}-4AC<0.$
To distinguish the degenerate cases from the non-degenerate case, let ∆ be the determinant$\Delta ={\begin{vmatrix}A&{\frac {1}{2}}B&{\frac {1}{2}}D\\{\frac {1}{2}}B&C&{\frac {1}{2}}E\\{\frac {1}{2}}D&{\frac {1}{2}}E&F\end{vmatrix}}=ACF+{\tfrac {1}{4}}BDE-{\tfrac {1}{4}}(AE^{2}+CD^{2}+FB^{2}).$
Then the ellipse is a non-degenerate real ellipse if and only if C∆ < 0. If C∆ > 0, we have an imaginary ellipse, and if ∆ = 0, we have a point ellipse.^{[7]}^{: 63 }
The general equation's coefficients can be obtained from known semi-major axis $a$, semi-minor axis $b$, center coordinates $\left(x_{\circ },\,y_{\circ }\right)$, and rotation angle $\theta$ (the angle from the positive horizontal axis to the ellipse's major axis) using the formulae:
${\begin{aligned}A&=a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta &B&=2\left(b^{2}-a^{2}\right)\sin \theta \cos \theta \\[1ex]C&=a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta &D&=-2Ax_{\circ }-By_{\circ }\\[1ex]E&=-Bx_{\circ }-2Cy_{\circ }&F&=Ax_{\circ }^{2}+Bx_{\circ }y_{\circ }+Cy_{\circ }^{2}-a^{2}b^{2}.\end{aligned}}$
These expressions can be derived from the canonical equation
${\frac {X^{2}}{a^{2}}}+{\frac {Y^{2}}{b^{2}}}=1$
by a Euclidean transformation of the coordinates $(X,\,Y)$:
${\begin{aligned}X&=\left(x-x_{\circ }\right)\cos \theta +\left(y-y_{\circ }\right)\sin \theta ,\\Y&=-\left(x-x_{\circ }\right)\sin \theta +\left(y-y_{\circ }\right)\cos \theta .\end{aligned}}$
Conversely, the canonical form parameters can be obtained from the general-form coefficients by the equations:^{[3]}
where atan2 is the 2-argument arctangent function.
Parametric representation
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Standard parametric representation
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Using trigonometric functions, a parametric representation of the standard ellipse ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ is:
$(x,\,y)=(a\cos t,\,b\sin t),\ 0\leq t<2\pi \,.$
With the substitution ${\textstyle u=\tan \left({\frac {t}{2}}\right)}$ and trigonometric formulae one obtains
$\cos t={\frac {1-u^{2}}{1+u^{2}}}\ ,\quad \sin t={\frac {2u}{1+u^{2}}}$
and the rational parametric equation of an ellipse
${\begin{cases}x(u)=a\,{\dfrac {1-u^{2}}{1+u^{2}}}\\[10mu]y(u)=b\,{\dfrac {2u}{1+u^{2}}}\\[10mu]-\infty <u<\infty \end{cases}}$
which covers any point of the ellipse ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ except the left vertex $(-a,\,0)$.
For $u\in [0,\,1],$ this formula represents the right upper quarter of the ellipse moving counter-clockwise with increasing $u.$ The left vertex is the limit ${\textstyle \lim _{u\to \pm \infty }(x(u),\,y(u))=(-a,\,0)\;.}$
Alternately, if the parameter $[u:v]$ is considered to be a point on the real projective line${\textstyle \mathbf {P} (\mathbf {R} )}$, then the corresponding rational parametrization is
$[u:v]\mapsto \left(a{\frac {v^{2}-u^{2}}{v^{2}+u^{2}}},b{\frac {2uv}{v^{2}+u^{2}}}\right).$
A parametric representation, which uses the slope $m$ of the tangent at a point of the ellipse
can be obtained from the derivative of the standard representation ${\vec {x}}(t)=(a\cos t,\,b\sin t)^{\mathsf {T}}$:
${\vec {x}}'(t)=(-a\sin t,\,b\cos t)^{\mathsf {T}}\quad \rightarrow \quad m=-{\frac {b}{a}}\cot t\quad \rightarrow \quad \cot t=-{\frac {ma}{b}}.$
With help of trigonometric formulae one obtains:
$\cos t={\frac {\cot t}{\pm {\sqrt {1+\cot ^{2}t}}}}={\frac {-ma}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\ ,\quad \quad \sin t={\frac {1}{\pm {\sqrt {1+\cot ^{2}t}}}}={\frac {b}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}.$
Replacing $\cos t$ and $\sin t$ of the standard representation yields:
${\vec {c}}_{\pm }(m)=\left(-{\frac {ma^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}},\;{\frac {b^{2}}{\pm {\sqrt {m^{2}a^{2}+b^{2}}}}}\right),\,m\in \mathbb {R} .$
Here $m$ is the slope of the tangent at the corresponding ellipse point, ${\vec {c}}_{+}$ is the upper and ${\vec {c}}_{-}$ the lower half of the ellipse. The vertices$(\pm a,\,0)$, having vertical tangents, are not covered by the representation.
The equation of the tangent at point ${\vec {c}}_{\pm }(m)$ has the form $y=mx+n$. The still unknown $n$ can be determined by inserting the coordinates of the corresponding ellipse point ${\vec {c}}_{\pm }(m)$:
$y=mx\pm {\sqrt {m^{2}a^{2}+b^{2}}}\,.$
This description of the tangents of an ellipse is an essential tool for the determination of the orthoptic of an ellipse. The orthoptic article contains another proof, without differential calculus and trigonometric formulae.
Any ellipse is an affine image of the unit circle with equation $x^{2}+y^{2}=1$.
Parametric representation
An affine transformation of the Euclidean plane has the form ${\vec {x}}\mapsto {\vec {f}}\!_{0}+A{\vec {x}}$, where $A$ is a regular matrix (with non-zero determinant) and ${\vec {f}}\!_{0}$ is an arbitrary vector. If ${\vec {f}}\!_{1},{\vec {f}}\!_{2}$ are the column vectors of the matrix $A$, the unit circle $(\cos(t),\sin(t))$, $0\leq t\leq 2\pi$, is mapped onto the ellipse:
${\vec {x}}={\vec {p}}(t)={\vec {f}}\!_{0}+{\vec {f}}\!_{1}\cos t+{\vec {f}}\!_{2}\sin t\,.$
Here ${\vec {f}}\!_{0}$ is the center and ${\vec {f}}\!_{1},\;{\vec {f}}\!_{2}$ are the directions of two conjugate diameters, in general not perpendicular.
Vertices
The four vertices of the ellipse are ${\vec {p}}(t_{0}),\;{\vec {p}}\left(t_{0}\pm {\tfrac {\pi }{2}}\right),\;{\vec {p}}\left(t_{0}+\pi \right)$, for a parameter $t=t_{0}$ defined by:
$\cot(2t_{0})={\frac {{\vec {f}}\!_{1}^{\,2}-{\vec {f}}\!_{2}^{\,2}}{2{\vec {f}}\!_{1}\cdot {\vec {f}}\!_{2}}}.$
(If ${\vec {f}}\!_{1}\cdot {\vec {f}}\!_{2}=0$, then $t_{0}=0$.) This is derived as follows. The tangent vector at point ${\vec {p}}(t)$ is:
${\vec {p}}\,'(t)=-{\vec {f}}\!_{1}\sin t+{\vec {f}}\!_{2}\cos t\ .$
At a vertex parameter $t=t_{0}$, the tangent is perpendicular to the major/minor axes, so:
$0={\vec {p}}'(t)\cdot \left({\vec {p}}(t)-{\vec {f}}\!_{0}\right)=\left(-{\vec {f}}\!_{1}\sin t+{\vec {f}}\!_{2}\cos t\right)\cdot \left({\vec {f}}\!_{1}\cos t+{\vec {f}}\!_{2}\sin t\right).$
Expanding and applying the identities $\;\cos ^{2}t-\sin ^{2}t=\cos 2t,\ \ 2\sin t\cos t=\sin 2t\;$ gives the equation for $t=t_{0}\;.$
Area
From Apollonios theorem (see below) one obtains:
The area of an ellipse $\;{\vec {x}}={\vec {f}}_{0}+{\vec {f}}_{1}\cos t+{\vec {f}}_{2}\sin t\;$ is
$A=\pi \left|\det({\vec {f}}_{1},{\vec {f}}_{2})\right|.$
Semiaxes
With the abbreviations
$\;M={\vec {f}}_{1}^{2}+{\vec {f}}_{2}^{2},\ N=\left|\det({\vec {f}}_{1},{\vec {f}}_{2})\right|$ the statements of Apollonios's theorem can be written as:
$a^{2}+b^{2}=M,\quad ab=N\ .$
Solving this nonlinear system for $a,b$ yields the semiaxes:
${\begin{aligned}a&={\frac {1}{2}}({\sqrt {M+2N}}+{\sqrt {M-2N}})\\[1ex]b&={\frac {1}{2}}({\sqrt {M+2N}}-{\sqrt {M-2N}})\,.\end{aligned}}$
Implicit representation
Solving the parametric representation for $\;\cos t,\sin t\;$ by Cramer's rule and using $\;\cos ^{2}t+\sin ^{2}t-1=0\;$, one obtains the implicit representation
$\det {\left({\vec {x}}\!-\!{\vec {f}}\!_{0},{\vec {f}}\!_{2}\right)^{2}}+\det {\left({\vec {f}}\!_{1},{\vec {x}}\!-\!{\vec {f}}\!_{0}\right)^{2}}-\det {\left({\vec {f}}\!_{1},{\vec {f}}\!_{2}\right)^{2}}=0.$
$x^{2}+2cxy+d^{2}y^{2}-e^{2}=0\ ,$ with $\;d^{2}-c^{2}>0\;,$
of an ellipse centered at the origin is given, then the two vectors
${\vec {f}}_{1}={e \choose 0},\quad {\vec {f}}_{2}={\frac {e}{\sqrt {d^{2}-c^{2}}}}{-c \choose 1}$
point to two conjugate points and the tools developed above are applicable.
Example: For the ellipse with equation $\;x^{2}+2xy+3y^{2}-1=0\;$ the vectors are
${\vec {f}}_{1}={1 \choose 0},\quad {\vec {f}}_{2}={\frac {1}{\sqrt {2}}}{-1 \choose 1}.$
Rotated Standard ellipse
For ${\vec {f}}_{0}={0 \choose 0},\;{\vec {f}}_{1}=a{\cos \theta \choose \sin \theta },\;{\vec {f}}_{2}=b{-\sin \theta \choose \;\cos \theta }$ one obtains a parametric representation of the standard ellipse rotated by angle $\theta$:
${\begin{aligned}x&=x_{\theta }(t)=a\cos \theta \cos t-b\sin \theta \sin t\,,\\y&=y_{\theta }(t)=a\sin \theta \cos t+b\cos \theta \sin t\,.\end{aligned}}$
Ellipse in space
The definition of an ellipse in this section gives a parametric representation of an arbitrary ellipse, even in space, if one allows ${\vec {f}}\!_{0},{\vec {f}}\!_{1},{\vec {f}}\!_{2}$ to be vectors in space.
Polar forms
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Polar form relative to center
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In polar coordinates, with the origin at the center of the ellipse and with the angular coordinate $\theta$ measured from the major axis, the ellipse's equation is^{[7]}^{: 75 }$r(\theta )={\frac {ab}{\sqrt {(b\cos \theta )^{2}+(a\sin \theta )^{2}}}}={\frac {b}{\sqrt {1-(e\cos \theta )^{2}}}}$
where $e$ is the eccentricity, not Euler's number.
Polar form relative to focus
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If instead we use polar coordinates with the origin at one focus, with the angular coordinate $\theta =0$ still measured from the major axis, the ellipse's equation is
$r(\theta )={\frac {a(1-e^{2})}{1\pm e\cos \theta }}$
where the sign in the denominator is negative if the reference direction $\theta =0$ points towards the center (as illustrated on the right), and positive if that direction points away from the center.
The angle $\theta$ is called the true anomaly of the point. The numerator $\ell =a(1-e^{2})$ is the semi-latus rectum.
Eccentricity and the directrix property
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Each of the two lines parallel to the minor axis, and at a distance of ${\textstyle d={\frac {a^{2}}{c}}={\frac {a}{e}}}$ from it, is called a directrix of the ellipse (see diagram).
For an arbitrary point $P$ of the ellipse, the quotient of the distance to one focus and to the corresponding directrix (see diagram) is equal to the eccentricity: ${\frac {\left|PF_{1}\right|}{\left|Pl_{1}\right|}}={\frac {\left|PF_{2}\right|}{\left|Pl_{2}\right|}}=e={\frac {c}{a}}\ .$
The proof for the pair $F_{1},l_{1}$ follows from the fact that ${\textstyle \left|PF_{1}\right|^{2}=(x-c)^{2}+y^{2},\ \left|Pl_{1}\right|^{2}=\left(x-{\tfrac {a^{2}}{c}}\right)^{2}}$ and $y^{2}=b^{2}-{\tfrac {b^{2}}{a^{2}}}x^{2}$ satisfy the equation
$\left|PF_{1}\right|^{2}-{\frac {c^{2}}{a^{2}}}\left|Pl_{1}\right|^{2}=0\,.$
The second case is proven analogously.
The converse is also true and can be used to define an ellipse (in a manner similar to the definition of a parabola):
For any point $F$ (focus), any line $l$ (directrix) not through $F$, and any real number $e$ with $0<e<1,$ the ellipse is the locus of points for which the quotient of the distances to the point and to the line is $e,$ that is: $E=\left\{P\ \left|\ {\frac {|PF|}{|Pl|}}=e\right.\right\}.$
The extension to $e=0$, which is the eccentricity of a circle, is not allowed in this context in the Euclidean plane. However, one may consider the directrix of a circle to be the line at infinity in the projective plane.
(The choice $e=1$ yields a parabola, and if $e>1$, a hyperbola.)
Proof
Let $F=(f,\,0),\ e>0$, and assume $(0,\,0)$ is a point on the curve. The directrix $l$ has equation $x=-{\tfrac {f}{e}}$. With $P=(x,\,y)$, the relation $|PF|^{2}=e^{2}|Pl|^{2}$ produces the equations
$(x-f)^{2}+y^{2}=e^{2}\left(x+{\frac {f}{e}}\right)^{2}=(ex+f)^{2}$ and $x^{2}\left(e^{2}-1\right)+2xf(1+e)-y^{2}=0.$
The substitution $p=f(1+e)$ yields
$x^{2}\left(e^{2}-1\right)+2px-y^{2}=0.$
This is the equation of an ellipse ($e<1$), or a parabola ($e=1$), or a hyperbola ($e>1$). All of these non-degenerate conics have, in common, the origin as a vertex (see diagram).
If $e<1$, introduce new parameters $a,\,b$ so that $1-e^{2}={\tfrac {b^{2}}{a^{2}}},{\text{ and }}\ p={\tfrac {b^{2}}{a}}$, and then the equation above becomes
${\frac {(x-a)^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1\,,$
which is the equation of an ellipse with center $(a,\,0)$, the x-axis as major axis, and
the major/minor semi axis $a,\,b$.
Construction of a directrix
Because of $c\cdot {\tfrac {a^{2}}{c}}=a^{2}$ point $L_{1}$ of directrix $l_{1}$ (see diagram) and focus $F_{1}$ are inverse with respect to the circle inversion at circle $x^{2}+y^{2}=a^{2}$ (in diagram green). Hence $L_{1}$ can be constructed as shown in the diagram. Directrix $l_{1}$ is the perpendicular to the main axis at point $L_{1}$.
General ellipse
If the focus is $F=\left(f_{1},\,f_{2}\right)$ and the directrix $ux+vy+w=0$, one obtains the equation
$\left(x-f_{1}\right)^{2}+\left(y-f_{2}\right)^{2}=e^{2}{\frac {\left(ux+vy+w\right)^{2}}{u^{2}+v^{2}}}\ .$
(The right side of the equation uses the Hesse normal form of a line to calculate the distance $|Pl|$.)
Focus-to-focus reflection property
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An ellipse possesses the following property:
The normal at a point $P$ bisects the angle between the lines ${\overline {PF_{1}}},\,{\overline {PF_{2}}}$.
Proof
Because the tangent line is perpendicular to the normal, an equivalent statement is that the tangent is the external angle bisector of the lines to the foci (see diagram).
Let $L$ be the point on the line ${\overline {PF_{2}}}$ with distance $2a$ to the focus $F_{2}$, where $a$ is the semi-major axis of the ellipse. Let line $w$ be the external angle bisector of the lines ${\overline {PF_{1}}}$ and ${\overline {PF_{2}}}.$ Take any other point $Q$ on $w.$ By the triangle inequality and the angle bisector theorem, $2a=\left|LF_{2}\right|<{}$$\left|QF_{2}\right|+\left|QL\right|={}$$\left|QF_{2}\right|+\left|QF_{1}\right|,$ therefore $Q$ must be outside the ellipse. As this is true for every choice of $Q,$$w$ only intersects the ellipse at the single point $P$ so must be the tangent line.
Application
The rays from one focus are reflected by the ellipse to the second focus. This property has optical and acoustic applications similar to the reflective property of a parabola (see whispering gallery).
Additionally, because of the focus-to-focus reflection property of ellipses, if the rays are allowed to continue propagating, reflected rays will eventually align closely with the major axis.
Conjugate diameters
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Definition of conjugate diameters
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A circle has the following property:
The midpoints of parallel chords lie on a diameter.
An affine transformation preserves parallelism and midpoints of line segments, so this property is true for any ellipse. (Note that the parallel chords and the diameter are no longer orthogonal.)
Definition
Two diameters $d_{1},\,d_{2}$ of an ellipse are conjugate if the midpoints of chords parallel to $d_{1}$ lie on $d_{2}\ .$
From the diagram one finds:
Two diameters ${\overline {P_{1}Q_{1}}},\,{\overline {P_{2}Q_{2}}}$ of an ellipse are conjugate whenever the tangents at $P_{1}$ and $Q_{1}$ are parallel to ${\overline {P_{2}Q_{2}}}$.
Conjugate diameters in an ellipse generalize orthogonal diameters in a circle.
In the parametric equation for a general ellipse given above,
${\vec {x}}={\vec {p}}(t)={\vec {f}}\!_{0}+{\vec {f}}\!_{1}\cos t+{\vec {f}}\!_{2}\sin t,$
any pair of points ${\vec {p}}(t),\ {\vec {p}}(t+\pi )$ belong to a diameter, and the pair ${\vec {p}}\left(t+{\tfrac {\pi }{2}}\right),\ {\vec {p}}\left(t-{\tfrac {\pi }{2}}\right)$ belong to its conjugate diameter.
For the common parametric representation $(a\cos t,b\sin t)$ of the ellipse with equation ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ one gets: The points
$(x_{1},y_{1})=(\pm a\cos t,\pm b\sin t)\quad$ (signs: (+,+) or (−,−) )
In case of a circle the last equation collapses to $x_{1}x_{2}+y_{1}y_{2}=0\ .$
Theorem of Apollonios on conjugate diameters
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For an ellipse with semi-axes $a,\,b$ the following is true:^{[9]}^{[10]}
Let $c_{1}$ and $c_{2}$ be halves of two conjugate diameters (see diagram) then
$c_{1}^{2}+c_{2}^{2}=a^{2}+b^{2}$.
The triangle$O,P_{1},P_{2}$ with sides $c_{1},\,c_{2}$ (see diagram) has the constant area ${\textstyle A_{\Delta }={\frac {1}{2}}ab}$, which can be expressed by $A_{\Delta }={\tfrac {1}{2}}c_{2}d_{1}={\tfrac {1}{2}}c_{1}c_{2}\sin \alpha$, too. $d_{1}$ is the altitude of point $P_{1}$ and $\alpha$ the angle between the half diameters. Hence the area of the ellipse (see section metric properties) can be written as $A_{el}=\pi ab=\pi c_{2}d_{1}=\pi c_{1}c_{2}\sin \alpha$.
The parallelogram of tangents adjacent to the given conjugate diameters has the ${\text{Area}}_{12}=4ab\ .$
Proof
Let the ellipse be in the canonical form with parametric equation
${\vec {p}}(t)=(a\cos t,\,b\sin t).$
The two points ${\textstyle {\vec {c}}_{1}={\vec {p}}(t),\ {\vec {c}}_{2}={\vec {p}}\left(t+{\frac {\pi }{2}}\right)}$ are on conjugate diameters (see previous section). From trigonometric formulae one obtains ${\vec {c}}_{2}=(-a\sin t,\,b\cos t)^{\mathsf {T}}$ and
$\left|{\vec {c}}_{1}\right|^{2}+\left|{\vec {c}}_{2}\right|^{2}=\cdots =a^{2}+b^{2}\,.$
The area of the triangle generated by ${\vec {c}}_{1},\,{\vec {c}}_{2}$ is
$A_{\Delta }={\tfrac {1}{2}}\det \left({\vec {c}}_{1},\,{\vec {c}}_{2}\right)=\cdots ={\tfrac {1}{2}}ab$
and from the diagram it can be seen that the area of the parallelogram is 8 times that of $A_{\Delta }$. Hence
${\text{Area}}_{12}=4ab\,.$
Orthogonal tangents
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For the ellipse ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ the intersection points of orthogonal tangents lie on the circle $x^{2}+y^{2}=a^{2}+b^{2}$.
This circle is called orthoptic or director circle of the ellipse (not to be confused with the circular directrix defined above).
Drawing ellipses
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Ellipses appear in descriptive geometry as images (parallel or central projection) of circles. There exist various tools to draw an ellipse. Computers provide the fastest and most accurate method for drawing an ellipse. However, technical tools (ellipsographs) to draw an ellipse without a computer exist. The principle was known to the 5th century mathematician Proclus, and the tool now known as an elliptical trammel was invented by Leonardo da Vinci.^{[11]}
For any method described below, knowledge of the axes and the semi-axes is necessary (or equivalently: the foci and the semi-major axis). If this presumption is not fulfilled one has to know at least two conjugate diameters. With help of Rytz's construction the axes and semi-axes can be retrieved.
de La Hire's point construction
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The following construction of single points of an ellipse is due to de La Hire.^{[12]} It is based on the standard parametric representation$(a\cos t,\,b\sin t)$ of an ellipse:
Draw the two circles centered at the center of the ellipse with radii $a,b$ and the axes of the ellipse.
Draw a line through the center, which intersects the two circles at point $A$ and $B$, respectively.
Draw a line through $A$ that is parallel to the minor axis and a line through $B$ that is parallel to the major axis. These lines meet at an ellipse point (see diagram).
Repeat steps (2) and (3) with different lines through the center.
de La Hire's method
Animation of the method
Pins-and-string method
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The characterization of an ellipse as the locus of points so that sum of the distances to the foci is constant leads to a method of drawing one using two drawing pins, a length of string, and a pencil. In this method, pins are pushed into the paper at two points, which become the ellipse's foci. A string is tied at each end to the two pins; its length after tying is $2a$. The tip of the pencil then traces an ellipse if it is moved while keeping the string taut. Using two pegs and a rope, gardeners use this procedure to outline an elliptical flower bed—thus it is called the gardener's ellipse. The Byzantine architect Anthemius of Tralles (c. 600) described how this method could be used to construct an elliptical reflector,^{[13]} and it was elaborated in a now-lost 9th-century treatise by Al-Ḥasan ibn Mūsā.^{[14]}
This representation can be modeled technically by two simple methods. In both cases center, the axes and semi axes $a,\,b$ have to be known.
Method 1
The first method starts with
a strip of paper of length $a+b$.
The point, where the semi axes meet is marked by $P$. If the strip slides with both ends on the axes of the desired ellipse, then point $P$ traces the ellipse. For the proof one shows that point $P$ has the parametric representation $(a\cos t,\,b\sin t)$, where parameter $t$ is the angle of the slope of the paper strip.
A technical realization of the motion of the paper strip can be achieved by a Tusi couple (see animation). The device is able to draw any ellipse with a fixed sum $a+b$, which is the radius of the large circle. This restriction may be a disadvantage in real life. More flexible is the second paper strip method.
Ellipse construction: paper strip method 1
Ellipses with Tusi couple. Two examples: red and cyan.
A variation of the paper strip method 1 uses the observation that the midpoint $N$ of the paper strip is moving on the circle with center $M$ (of the ellipse) and radius ${\tfrac {a+b}{2}}$. Hence, the paperstrip can be cut at point $N$ into halves, connected again by a joint at $N$ and the sliding end $K$ fixed at the center $M$ (see diagram). After this operation the movement of the unchanged half of the paperstrip is unchanged.^{[15]} This variation requires only one sliding shoe.
Variation of the paper strip method 1
Animation of the variation of the paper strip method 1
Method 2
The second method starts with
a strip of paper of length $a$.
One marks the point, which divides the strip into two substrips of length $b$ and $a-b$. The strip is positioned onto the axes as described in the diagram. Then the free end of the strip traces an ellipse, while the strip is moved. For the proof, one recognizes that the tracing point can be described parametrically by $(a\cos t,\,b\sin t)$, where parameter $t$ is the angle of slope of the paper strip.
This method is the base for several ellipsographs (see section below).
Similar to the variation of the paper strip method 1 a variation of the paper strip method 2 can be established (see diagram) by cutting the part between the axes into halves.
Most ellipsograph drafting instruments are based on the second paperstrip method.
Approximation by osculating circles
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From Metric properties below, one obtains:
The radius of curvature at the vertices $V_{1},\,V_{2}$ is: ${\tfrac {b^{2}}{a}}$
The radius of curvature at the co-vertices $V_{3},\,V_{4}$ is: ${\tfrac {a^{2}}{b}}\ .$
The diagram shows an easy way to find the centers of curvature $C_{1}=\left(a-{\tfrac {b^{2}}{a}},0\right),\,C_{3}=\left(0,b-{\tfrac {a^{2}}{b}}\right)$ at vertex $V_{1}$ and co-vertex $V_{3}$, respectively:
mark the auxiliary point $H=(a,\,b)$ and draw the line segment $V_{1}V_{3}\ ,$
draw the line through $H$, which is perpendicular to the line $V_{1}V_{3}\ ,$
the intersection points of this line with the axes are the centers of the osculating circles.
(proof: simple calculation.)
The centers for the remaining vertices are found by symmetry.
Given two pencils$B(U),\,B(V)$ of lines at two points $U,\,V$ (all lines containing $U$ and $V$, respectively) and a projective but not perspective mapping $\pi$ of $B(U)$ onto $B(V)$, then the intersection points of corresponding lines form a non-degenerate projective conic section.
For the generation of points of the ellipse ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$ one uses the pencils at the vertices $V_{1},\,V_{2}$. Let $P=(0,\,b)$ be an upper co-vertex of the ellipse and $A=(-a,\,2b),\,B=(a,\,2b)$.
$P$ is the center of the rectangle $V_{1},\,V_{2},\,B,\,A$. The side ${\overline {AB}}$ of the rectangle is divided into n equal spaced line segments and this division is projected parallel with the diagonal $AV_{2}$ as direction onto the line segment ${\overline {V_{1}B}}$ and assign the division as shown in the diagram. The parallel projection together with the reverse of the orientation is part of the projective mapping between the pencils at $V_{1}$ and $V_{2}$ needed. The intersection points of any two related lines $V_{1}B_{i}$ and $V_{2}A_{i}$ are points of the uniquely defined ellipse. With help of the points $C_{1},\,\dotsc$ the points of the second quarter of the ellipse can be determined. Analogously one obtains the points of the lower half of the ellipse.
Steiner generation can also be defined for hyperbolas and parabolas. It is sometimes called a parallelogram method because one can use other points rather than the vertices, which starts with a parallelogram instead of a rectangle.
As hypotrochoid
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The ellipse is a special case of the hypotrochoid when $R=2r$, as shown in the adjacent image. The special case of a moving circle with radius $r$ inside a circle with radius $R=2r$ is called a Tusi couple.
Inscribed angles and three-point form
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Circles
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A circle with equation $\left(x-x_{\circ }\right)^{2}+\left(y-y_{\circ }\right)^{2}=r^{2}$ is uniquely determined by three points $\left(x_{1},y_{1}\right),\;\left(x_{2},\,y_{2}\right),\;\left(x_{3},\,y_{3}\right)$ not on a line. A simple way to determine the parameters $x_{\circ },y_{\circ },r$ uses the inscribed angle theorem for circles:
For four points $P_{i}=\left(x_{i},\,y_{i}\right),\ i=1,\,2,\,3,\,4,\,$ (see diagram) the following statement is true:
The four points are on a circle if and only if the angles at $P_{3}$ and $P_{4}$ are equal.
Usually one measures inscribed angles by a degree or radian θ, but here the following measurement is more convenient:
In order to measure the angle between two lines with equations $y=m_{1}x+d_{1},\ y=m_{2}x+d_{2},\ m_{1}\neq m_{2},$ one uses the quotient: ${\frac {1+m_{1}m_{2}}{m_{2}-m_{1}}}=\cot \theta \ .$
Inscribed angle theorem for circles
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For four points $P_{i}=\left(x_{i},\,y_{i}\right),\ i=1,\,2,\,3,\,4,\,$ no three of them on a line, we have the following (see diagram):
The four points are on a circle, if and only if the angles at $P_{3}$ and $P_{4}$ are equal. In terms of the angle measurement above, this means: ${\frac {(x_{4}-x_{1})(x_{4}-x_{2})+(y_{4}-y_{1})(y_{4}-y_{2})}{(y_{4}-y_{1})(x_{4}-x_{2})-(y_{4}-y_{2})(x_{4}-x_{1})}}={\frac {(x_{3}-x_{1})(x_{3}-x_{2})+(y_{3}-y_{1})(y_{3}-y_{2})}{(y_{3}-y_{1})(x_{3}-x_{2})-(y_{3}-y_{2})(x_{3}-x_{1})}}.$
At first the measure is available only for chords not parallel to the y-axis, but the final formula works for any chord.
Three-point form of circle equation
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As a consequence, one obtains an equation for the circle determined by three non-collinear points $P_{i}=\left(x_{i},\,y_{i}\right)$: ${\frac {({\color {red}x}-x_{1})({\color {red}x}-x_{2})+({\color {red}y}-y_{1})({\color {red}y}-y_{2})}{({\color {red}y}-y_{1})({\color {red}x}-x_{2})-({\color {red}y}-y_{2})({\color {red}x}-x_{1})}}={\frac {(x_{3}-x_{1})(x_{3}-x_{2})+(y_{3}-y_{1})(y_{3}-y_{2})}{(y_{3}-y_{1})(x_{3}-x_{2})-(y_{3}-y_{2})(x_{3}-x_{1})}}.$
For example, for $P_{1}=(2,\,0),\;P_{2}=(0,\,1),\;P_{3}=(0,\,0)$ the three-point equation is:
${\frac {(x-2)x+y(y-1)}{yx-(y-1)(x-2)}}=0$, which can be rearranged to $(x-1)^{2}+\left(y-{\tfrac {1}{2}}\right)^{2}={\tfrac {5}{4}}\ .$
Using vectors, dot products and determinants this formula can be arranged more clearly, letting ${\vec {x}}=(x,\,y)$:
${\frac {\left({\color {red}{\vec {x}}}-{\vec {x}}_{1}\right)\cdot \left({\color {red}{\vec {x}}}-{\vec {x}}_{2}\right)}{\det \left({\color {red}{\vec {x}}}-{\vec {x}}_{1},{\color {red}{\vec {x}}}-{\vec {x}}_{2}\right)}}={\frac {\left({\vec {x}}_{3}-{\vec {x}}_{1}\right)\cdot \left({\vec {x}}_{3}-{\vec {x}}_{2}\right)}{\det \left({\vec {x}}_{3}-{\vec {x}}_{1},{\vec {x}}_{3}-{\vec {x}}_{2}\right)}}.$
The center of the circle $\left(x_{\circ },\,y_{\circ }\right)$ satisfies:
${\begin{bmatrix}1&{\dfrac {y_{1}-y_{2}}{x_{1}-x_{2}}}\\[2ex]{\dfrac {x_{1}-x_{3}}{y_{1}-y_{3}}}&1\end{bmatrix}}{\begin{bmatrix}x_{\circ }\\[1ex]y_{\circ }\end{bmatrix}}={\begin{bmatrix}{\dfrac {x_{1}^{2}-x_{2}^{2}+y_{1}^{2}-y_{2}^{2}}{2(x_{1}-x_{2})}}\\[2ex]{\dfrac {y_{1}^{2}-y_{3}^{2}+x_{1}^{2}-x_{3}^{2}}{2(y_{1}-y_{3})}}\end{bmatrix}}.$
The radius is the distance between any of the three points and the center.
$r={\sqrt {\left(x_{1}-x_{\circ }\right)^{2}+\left(y_{1}-y_{\circ }\right)^{2}}}={\sqrt {\left(x_{2}-x_{\circ }\right)^{2}+\left(y_{2}-y_{\circ }\right)^{2}}}={\sqrt {\left(x_{3}-x_{\circ }\right)^{2}+\left(y_{3}-y_{\circ }\right)^{2}}}.$
Ellipses
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This section considers the family of ellipses defined by equations ${\tfrac {\left(x-x_{\circ }\right)^{2}}{a^{2}}}+{\tfrac {\left(y-y_{\circ }\right)^{2}}{b^{2}}}=1$ with a fixed eccentricity $e$. It is convenient to use the parameter:
${\color {blue}q}={\frac {a^{2}}{b^{2}}}={\frac {1}{1-e^{2}}},$
and to write the ellipse equation as:
$\left(x-x_{\circ }\right)^{2}+{\color {blue}q}\,\left(y-y_{\circ }\right)^{2}=a^{2},$
where q is fixed and $x_{\circ },\,y_{\circ },\,a$ vary over the real numbers. (Such ellipses have their axes parallel to the coordinate axes: if $q<1$, the major axis is parallel to the x-axis; if $q>1$, it is parallel to the y-axis.)
Like a circle, such an ellipse is determined by three points not on a line.
For this family of ellipses, one introduces the following q-analog angle measure, which is not a function of the usual angle measure θ:^{[16]}^{[17]}
In order to measure an angle between two lines with equations $y=m_{1}x+d_{1},\ y=m_{2}x+d_{2},\ m_{1}\neq m_{2}$ one uses the quotient: ${\frac {1+{\color {blue}q}\;m_{1}m_{2}}{m_{2}-m_{1}}}\ .$
Inscribed angle theorem for ellipses
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Given four points $P_{i}=\left(x_{i},\,y_{i}\right),\ i=1,\,2,\,3,\,4$, no three of them on a line (see diagram).
The four points are on an ellipse with equation $(x-x_{\circ })^{2}+{\color {blue}q}\,(y-y_{\circ })^{2}=a^{2}$ if and only if the angles at $P_{3}$ and $P_{4}$ are equal in the sense of the measurement above—that is, if ${\frac {(x_{4}-x_{1})(x_{4}-x_{2})+{\color {blue}q}\;(y_{4}-y_{1})(y_{4}-y_{2})}{(y_{4}-y_{1})(x_{4}-x_{2})-(y_{4}-y_{2})(x_{4}-x_{1})}}={\frac {(x_{3}-x_{1})(x_{3}-x_{2})+{\color {blue}q}\;(y_{3}-y_{1})(y_{3}-y_{2})}{(y_{3}-y_{1})(x_{3}-x_{2})-(y_{3}-y_{2})(x_{3}-x_{1})}}\ .$
At first the measure is available only for chords which are not parallel to the y-axis. But the final formula works for any chord. The proof follows from a straightforward calculation. For the direction of proof given that the points are on an ellipse, one can assume that the center of the ellipse is the origin.
Three-point form of ellipse equation
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A consequence, one obtains an equation for the ellipse determined by three non-collinear points $P_{i}=\left(x_{i},\,y_{i}\right)$: ${\frac {({\color {red}x}-x_{1})({\color {red}x}-x_{2})+{\color {blue}q}\;({\color {red}y}-y_{1})({\color {red}y}-y_{2})}{({\color {red}y}-y_{1})({\color {red}x}-x_{2})-({\color {red}y}-y_{2})({\color {red}x}-x_{1})}}={\frac {(x_{3}-x_{1})(x_{3}-x_{2})+{\color {blue}q}\;(y_{3}-y_{1})(y_{3}-y_{2})}{(y_{3}-y_{1})(x_{3}-x_{2})-(y_{3}-y_{2})(x_{3}-x_{1})}}\ .$
For example, for $P_{1}=(2,\,0),\;P_{2}=(0,\,1),\;P_{3}=(0,\,0)$ and $q=4$ one obtains the three-point form
${\frac {(x-2)x+4y(y-1)}{yx-(y-1)(x-2)}}=0$ and after conversion ${\frac {(x-1)^{2}}{2}}+{\frac {\left(y-{\frac {1}{2}}\right)^{2}}{\frac {1}{2}}}=1.$
Analogously to the circle case, the equation can be written more clearly using vectors:
${\frac {\left({\color {red}{\vec {x}}}-{\vec {x}}_{1}\right)*\left({\color {red}{\vec {x}}}-{\vec {x}}_{2}\right)}{\det \left({\color {red}{\vec {x}}}-{\vec {x}}_{1},{\color {red}{\vec {x}}}-{\vec {x}}_{2}\right)}}={\frac {\left({\vec {x}}_{3}-{\vec {x}}_{1}\right)*\left({\vec {x}}_{3}-{\vec {x}}_{2}\right)}{\det \left({\vec {x}}_{3}-{\vec {x}}_{1},{\vec {x}}_{3}-{\vec {x}}_{2}\right)}},$
where $*$ is the modified dot product${\vec {u}}*{\vec {v}}=u_{x}v_{x}+{\color {blue}q}\,u_{y}v_{y}.$
Pole-polar relation
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Any ellipse can be described in a suitable coordinate system by an equation ${\tfrac {x^{2}}{a^{2}}}+{\tfrac {y^{2}}{b^{2}}}=1$. The equation of the tangent at a point $P_{1}=\left(x_{1},\,y_{1}\right)$ of the ellipse is ${\tfrac {x_{1}x}{a^{2}}}+{\tfrac {y_{1}y}{b^{2}}}=1.$ If one allows point $P_{1}=\left(x_{1},\,y_{1}\right)$ to be an arbitrary point different from the origin, then
point $P_{1}=\left(x_{1},\,y_{1}\right)\neq (0,\,0)$ is mapped onto the line ${\tfrac {x_{1}x}{a^{2}}}+{\tfrac {y_{1}y}{b^{2}}}=1$, not through the center of the ellipse.
This relation between points and lines is a bijection.
line $y=mx+d,\ d\neq 0$ onto the point $\left(-{\tfrac {ma^{2}}{d}},\,{\tfrac {b^{2}}{d}}\right)$ and
line $x=c,\ c\neq 0$ onto the point $\left({\tfrac {a^{2}}{c}},\,0\right).$
Such a relation between points and lines generated by a conic is called pole-polar relation or polarity. The pole is the point; the polar the line.
By calculation one can confirm the following properties of the pole-polar relation of the ellipse:
For a point (pole) on the ellipse, the polar is the tangent at this point (see diagram: $P_{1},\,p_{1}$).
For a pole $P$outside the ellipse, the intersection points of its polar with the ellipse are the tangency points of the two tangents passing $P$ (see diagram: $P_{2},\,p_{2}$).
For a point within the ellipse, the polar has no point with the ellipse in common (see diagram: $F_{1},\,l_{1}$).
The intersection point of two polars is the pole of the line through their poles.
The foci $(c,\,0)$ and $(-c,\,0)$, respectively, and the directrices $x={\tfrac {a^{2}}{c}}$ and $x=-{\tfrac {a^{2}}{c}}$, respectively, belong to pairs of pole and polar. Because they are even polar pairs with respect to the circle $x^{2}+y^{2}=a^{2}$, the directrices can be constructed by compass and straightedge (see Inversive geometry).
Pole-polar relations exist for hyperbolas and parabolas as well.
Metric properties
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All metric properties given below refer to an ellipse with equation
${\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1$
(1)
except for the section on the area enclosed by a tilted ellipse, where the generalized form of Eq.(1) will be given.
Area
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The area$A_{\text{ellipse}}$ enclosed by an ellipse is:
$A_{\text{ellipse}}=\pi ab$
(2)
where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively. The area formula $\pi ab$ is intuitive: start with a circle of radius $b$ (so its area is $\pi b^{2}$) and stretch it by a factor $a/b$ to make an ellipse. This scales the area by the same factor: $\pi b^{2}(a/b)=\pi ab.$^{[18]} However, using the same approach for the circumference would be fallacious – compare the integrals${\textstyle \int f(x)\,dx}$ and ${\textstyle \int {\sqrt {1+f'^{2}(x)}}\,dx}$. It is also easy to rigorously prove the area formula using integration as follows. Equation (1) can be rewritten as ${\textstyle y(x)=b{\sqrt {1-x^{2}/a^{2}}}.}$ For $x\in [-a,a],$ this curve is the top half of the ellipse. So twice the integral of $y(x)$ over the interval $[-a,a]$ will be the area of the ellipse:
${\begin{aligned}A_{\text{ellipse}}&=\int _{-a}^{a}2b{\sqrt {1-{\frac {x^{2}}{a^{2}}}}}\,dx\\&={\frac {b}{a}}\int _{-a}^{a}2{\sqrt {a^{2}-x^{2}}}\,dx.\end{aligned}}$
The second integral is the area of a circle of radius $a,$ that is, $\pi a^{2}.$ So
$A_{\text{ellipse}}={\frac {b}{a}}\pi a^{2}=\pi ab.$
An ellipse defined implicitly by $Ax^{2}+Bxy+Cy^{2}=1$ has area $2\pi /{\sqrt {4AC-B^{2}}}.$
The area can also be expressed in terms of eccentricity and the length of the semi-major axis as $a^{2}\pi {\sqrt {1-e^{2}}}$ (obtained by solving for flattening, then computing the semi-minor axis).
So far we have dealt with erect ellipses, whose major and minor axes are parallel to the $x$ and $y$ axes. However, some applications require tilted ellipses. In charged-particle beam optics, for instance, the enclosed area of an erect or tilted ellipse is an important property of the beam, its emittance. In this case a simple formula still applies, namely
where $y_{\text{int}}$, $x_{\text{int}}$ are intercepts and $x_{\text{max}}$, $y_{\text{max}}$ are maximum values. It follows directly from Apollonios's theorem.
Circumference
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The circumference $C$ of an ellipse is:
$C\,=\,4a\int _{0}^{\pi /2}{\sqrt {1-e^{2}\sin ^{2}\theta }}\ d\theta \,=\,4a\,E(e)$
where again $a$ is the length of the semi-major axis, ${\textstyle e={\sqrt {1-b^{2}/a^{2}}}}$ is the eccentricity, and the function $E$ is the complete elliptic integral of the second kind,
$E(e)\,=\,\int _{0}^{\pi /2}{\sqrt {1-e^{2}\sin ^{2}\theta }}\ d\theta$
which is in general not an elementary function.
The circumference of the ellipse may be evaluated in terms of $E(e)$ using Gauss's arithmetic-geometric mean;^{[19]} this is a quadratically converging iterative method (see here for details).
The exact infinite series is:
${\begin{aligned}C&=2\pi a\left[{1-\left({\frac {1}{2}}\right)^{2}e^{2}-\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{2}{\frac {e^{4}}{3}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right)^{2}{\frac {e^{6}}{5}}-\cdots }\right]\\&=2\pi a\left[1-\sum _{n=1}^{\infty }\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}{\frac {e^{2n}}{2n-1}}\right]\\&=-2\pi a\sum _{n=0}^{\infty }\left({\frac {(2n-1)!!}{(2n)!!}}\right)^{2}{\frac {e^{2n}}{2n-1}},\end{aligned}}$
where $n!!$ is the double factorial (extended to negative odd integers in the usual way, giving $(-1)!!=1$ and $(-3)!!=-1$).
This series converges, but by expanding in terms of $h=(a-b)^{2}/(a+b)^{2},$James Ivory,^{[20]}Bessel^{[21]} and Kummer^{[22]} derived an expression that converges much more rapidly. It is most concisely written in terms of the binomial coefficient with $n=1/2$:
${\begin{aligned}{\frac {C}{\pi (a+b)}}&=\sum _{n=0}^{\infty }{{\frac {1}{2}} \choose n}^{2}h^{n}\\&=\sum _{n=0}^{\infty }\left({\frac {(2n-3)!!}{(2n)!!}}\right)^{2}h^{n}\\&=\sum _{n=0}^{\infty }\left({\frac {(2n-3)!!}{2^{n}n!}}\right)^{2}h^{n}\\&=1+{\frac {h}{4}}+{\frac {h^{2}}{64}}+{\frac {h^{3}}{256}}+{\frac {25\,h^{4}}{16384}}+{\frac {49\,h^{5}}{65536}}+{\frac {441\,h^{6}}{2^{20}}}+{\frac {1089\,h^{7}}{2^{22}}}+\cdots .\end{aligned}}$
The coefficients are slightly smaller (by a factor of $2n-1$), but also $h$ is numerically much smaller than $e$ except at $h=e=0$ and $h=e=1$. For eccentricities less than 0.5 ($h<0.005$), the error is at the limits of double-precision floating-point after the $h^{4}$ term.^{[23]}
Srinivasa Ramanujan gave two close approximations for the circumference in §16 of "Modular Equations and Approximations to $\pi$";^{[24]} they are
$C\approx \pi {\biggl [}3(a+b)-{\sqrt {(3a+b)(a+3b)}}{\biggr ]}=\pi {\biggl [}3(a+b)-{\sqrt {3(a+b)^{2}+4ab}}{\biggr ]}$
and
$C\approx \pi \left(a+b\right)\left(1+{\frac {3h}{10+{\sqrt {4-3h}}}}\right),$