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Ellipse

\begin{figure}\begin{center}\BoxedEPSF{Ellipse.epsf scaled 750}\end{center}\end{figure}

A curve which is the Locus of all points in the Plane the Sum of whose distances $r_1$ and $r_2$ from two fixed points $F_1$ and $F_2$ (the Foci) separated by a distance of $2c$ is a given Positive constant $2a$ (left figure). This results in the two-center Bipolar Coordinate equation

\begin{displaymath}
r_1+r_2 = 2a,
\end{displaymath} (1)

where $a$ is the Semimajor Axis and the Origin of the coordinate system is at one of the Foci. The ellipse can also be defined as the Locus of points whose distance from the Focus is proportional to the horizontal distance from a vertical line known as the Directrix (right figure).


The ellipse was first studied by Menaechmus, investigated by Euclid, and named by Apollonius. The Focus and Directrix of an ellipse were considered by Pappus. In 1602, Kepler believed that the orbit of Mars was Oval; he later discovered that it was an ellipse with the Sun at one Focus. In fact, Kepler introduced the word ``Focus'' and published his discovery in 1609. In 1705 Halley showed that the comet which is now named after him moved in an elliptical orbit around the Sun (MacTutor Archive).


A ray passing through a Focus will pass through the other focus after a single bounce. Reflections not passing through a Focus will be tangent to a confocal Hyperbola or Ellipse, depending on whether the ray passes between the Foci or not. Let an ellipse lie along the x-Axis and find the equation of the figure (1) where $F_1$ and $F_2$ are at $(-c,0)$ and $(c,0)$. In Cartesian Coordinates,

\begin{displaymath}
\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2a.
\end{displaymath} (2)

Bring the second term to the right side and square both sides,
\begin{displaymath}
(x+c)^2+y^2 = 4a^2-4a\sqrt{(x-c)^2+y^2} + (x-c)^2+y^2.
\end{displaymath} (3)

Now solve for the Square Root term and simplify
$\sqrt{(x-c)^2+y^2}$
$ = - {1\over 4a}(x^2+2xc+c^2+y^2-4a^2 -x^2+2xc-c^2-y^2)$
$ = - {1\over 4a} (4xc-4a^2) = a - {c\over a} x.\quad$ (4)
Square one final time to clear the remaining Square Root,
\begin{displaymath}
x^2-2xc+c^2+y^2 = a^2-2cx+{c^2\over a^2} x^2.
\end{displaymath} (5)

Grouping the $x$ terms then gives
\begin{displaymath}
x^2{a^2-c^2\over a^2} +y^2 = a^2-c^2,
\end{displaymath} (6)

which can be written in the simple form
\begin{displaymath}
{x^2\over a^2} + {y^2\over a^2-c^2} = 1.
\end{displaymath} (7)

Defining a new constant
\begin{displaymath}
b^2\equiv a^2-c^2
\end{displaymath} (8)

puts the equation in the particularly simple form
\begin{displaymath}
{x^2\over a^2} + {y^2\over b^2} = 1.
\end{displaymath} (9)

The parameter $b$ is called the Semiminor Axis by analogy with the parameter $a$, which is called the Semimajor Axis. The fact that $b$ as defined above is actually the Semiminor Axis is easily shown by letting $r_1$ and $r_2$ be equal. Then two Right Triangles are produced, each with Hypotenuse $a$, base $c$, and height $b\equiv \sqrt{a^2-c^2}$. Since the largest distance along the Minor Axis will be achieved at this point, $b$ is indeed the Semiminor Axis.


If, instead of being centered at (0, 0), the Center of the ellipse is at ($x_0$, $y_0$), equation (9) becomes

\begin{displaymath}
{(x-x_0)^2\over a^2} + {(y-y_0)^2\over b^2} = 1.
\end{displaymath} (10)


As can be seen from the Cartesian Equation for the ellipse, the curve can also be given by a simple parametric form analogous to that of a Circle, but with the $x$ and $y$ coordinates having different scalings,

$\displaystyle x$ $\textstyle =$ $\displaystyle a\cos t$ (11)
$\displaystyle y$ $\textstyle =$ $\displaystyle b\sin t.$ (12)

The unit Tangent Vector of the ellipse so parameterized is
$\displaystyle x_T(t)$ $\textstyle =$ $\displaystyle -{a\sin t\over\sqrt{b^2\cos^2t+a^2\sin^2 t}}$ (13)
$\displaystyle y_T(t)$ $\textstyle =$ $\displaystyle {b\cos t\over\sqrt{b^2\cos^2t+a^2\sin^2 t}}.$ (14)

A sequence of Normal and Tangent Vectors are plotted below for the ellipse.

\begin{figure}\begin{center}\BoxedEPSF{EllipseNormalTangent.epsf}\end{center}\end{figure}


\begin{figure}\begin{center}\BoxedEPSF{RotationCurve.epsf scaled 700}\end{center}\end{figure}

For an ellipse centered at the Origin but inclined at an arbitrary Angle $\theta$ to the x-Axis, the parametric equations are

$\displaystyle \left[\begin{array}{c}x\\  y\end{array}\right]$ $\textstyle =$ $\displaystyle \left[\begin{array}{cc}\cos\theta & \sin\theta\\  -\sin\theta & \...
...ta\end{array}\right] \left[\begin{array}{c}a\cos t\\  b\sin t\end{array}\right]$  
  $\textstyle =$ $\displaystyle \left[\begin{array}{c}a\cos\theta\cos t+b\sin\theta\sin t\\  -a\sin\theta\cos t+b\cos\theta\sin t\end{array}\right].$ (15)


\begin{figure}\begin{center}\BoxedEPSF{EllipseCenter.epsf}\end{center}\end{figure}

In Polar Coordinates, the Angle $\theta'$ measured from the center of the ellipse is called the Eccentric Angle. Writing $r'$ for the distance of a point from the ellipse center, the equation in Polar Coordinates is just given by the usual

$\displaystyle x$ $\textstyle =$ $\displaystyle r'\cos\theta'$ (16)
$\displaystyle y$ $\textstyle =$ $\displaystyle r'\sin\theta'.$ (17)

Here, the coordinates $\theta'$ and $r'$ are written with primes to distinguish them from the more common polar coordinates for an ellipse which are centered on a focus. Plugging the polar equations into the Cartesian equation (9) and solving for $r'^2$ gives
\begin{displaymath}
r'^2 = {b^2a^2\over b^2\cos^2\theta'+a^2\sin^2\theta'}.
\end{displaymath} (18)

Define a new constant $0\leq e<1$ called the Eccentricity (where $e=0$ is the case of a Circle) to replace $b$
\begin{displaymath}
e\equiv \sqrt{1-{b^2\over a^2}}\,,
\end{displaymath} (19)

from which it also follows from (8) that
$\displaystyle a^2e^2$ $\textstyle =$ $\displaystyle a^2-b^2 \equiv c^2$ (20)
$\displaystyle c$ $\textstyle =$ $\displaystyle ae$ (21)
$\displaystyle b^2$ $\textstyle =$ $\displaystyle a^2(1-e^2).$ (22)

Therefore (18) can be written as
\begin{displaymath}
r'^2 = {a^2(1-e^2)\over 1-e^2\cos^2\theta'}
\end{displaymath} (23)


\begin{displaymath}
r' = a\sqrt{1-e^2\over 1-e^2\cos^2\theta'}.
\end{displaymath} (24)

If $e\ll 1$, then
\begin{displaymath}
r' = a\{1-{\textstyle{1\over 2}}e^2\sin^2\theta'-{\textstyle{1\over 16}}e^4[5+3\cos(2\theta')]\sin^2\theta'+\ldots\},
\end{displaymath} (25)

so
\begin{displaymath}
{\Delta r'\over a} \equiv {a-r'\over a} \approx {\textstyle{1\over 2}}e^2\sin^2\theta'.
\end{displaymath} (26)


\begin{figure}\begin{center}\BoxedEPSF{EllipseFocus.epsf}\end{center}\end{figure}

If $r$ and $\theta$ are measured from a Focus instead of from the center, as they commonly are in orbital mechanics, then the equations of the ellipse are

$\displaystyle x$ $\textstyle =$ $\displaystyle c+r\cos\theta$ (27)
$\displaystyle y$ $\textstyle =$ $\displaystyle r\sin\theta,$ (28)

and (9) becomes
\begin{displaymath}
{(c+r\cos\theta)^2\over a^2} + {r^2\sin^2\theta\over b^2} = 1.
\end{displaymath} (29)

Clearing the Denominators gives
\begin{displaymath}
b^2(c^2+2cr\cos\theta +r^2\cos^2\theta)+a^2r^2\sin^2\theta = a^2b^2
\end{displaymath} (30)


\begin{displaymath}
b^2c^2+2rcb^2\cos\theta+b^2r^2\cos^2 \theta+a^2r^2-a^2r^2\cos^2\theta = a^2b^2.
\end{displaymath} (31)

Plugging in (21) and (22) to re-express $b$ and $c$ in terms of $a$ and $e$,
$a^2(1-e^2)a^2e^2+2aea^2(1-e^2)r\cos\theta +a^2(1-e^2)r^2\cos^2\theta $
$ +a^2r^2-a^2r^2\cos^2\theta = a^2[a^2(1-e^2)].\quad$ (32)
Simplifying,
\begin{displaymath}
-r^2+[er\cos\theta-a(1-e^2)]^2 = 0
\end{displaymath} (33)


\begin{displaymath}
r = \pm [er\cos\theta-a(1-e^2)].
\end{displaymath} (34)

The sign can be determined by requiring that $r$ must be Positive. When $e=0$, (34) becomes $r = \pm (-a)$, but since $a$ is always Positive, we must take the Negative sign, so (34) becomes
\begin{displaymath}
r = a(1-e^2)-er\cos\theta
\end{displaymath} (35)


\begin{displaymath}
r(1+e\cos\theta) = a(1-e^2)
\end{displaymath} (36)


\begin{displaymath}
r = {a(1-e^2)\over 1+e\cos\theta}.
\end{displaymath} (37)

The distance from a Focus to a point with horizontal coordinate $x$ is found from
\begin{displaymath}
\cos\theta = {c+x\over r}.
\end{displaymath} (38)

Plugging this into (37) yields
\begin{displaymath}
r+e(c+x) = a(1-e^2)
\end{displaymath} (39)


\begin{displaymath}
r = a(1-e^2)-e(c+x).
\end{displaymath} (40)

Summarizing relationships among the parameters characterizing an ellipse,
$\displaystyle b$ $\textstyle =$ $\displaystyle a\sqrt{1-e^2} = \sqrt{a^2-c^2}$ (41)
$\displaystyle c$ $\textstyle =$ $\displaystyle \sqrt{a^2-b^2} = ae$ (42)
$\displaystyle e$ $\textstyle =$ $\displaystyle \sqrt{1 - {b^2\over a^2}}={c\over a}.$ (43)

The Eccentricity can therefore be interpreted as the position of the Focus as a fraction of the Semimajor Axis.


In Pedal Coordinates with the Pedal Point at the Focus, the equation of the ellipse is

\begin{displaymath}
{b^2\over p^2}={2a\over r}-1.
\end{displaymath} (44)


To find the Radius of Curvature, return to the parametric coordinates centered at the center of the ellipse and compute the first and second derivatives,

$\displaystyle x'$ $\textstyle =$ $\displaystyle -a\sin t$ (45)
$\displaystyle y'$ $\textstyle =$ $\displaystyle b\cos t$ (46)


$\displaystyle x''$ $\textstyle =$ $\displaystyle -a\cos t$ (47)
$\displaystyle y''$ $\textstyle =$ $\displaystyle -b\sin t.$ (48)

Therefore,
$\displaystyle R$ $\textstyle =$ $\displaystyle {(x'^2+y'^2)^{3/2}\over x'y''-x''y'}$  
  $\textstyle =$ $\displaystyle {(a^2\sin^2t+b^2\cos^2 t)^{3/2}\over -a\sin t(-b\sin t)-(a\cos t)(b\cos t)}$  
  $\textstyle =$ $\displaystyle {(a^2\sin^2t+b^2\cos^2t)^{3/2}\over ab(\sin^2t+\cos^2t)}$  
  $\textstyle =$ $\displaystyle {(a^2\sin^2t+b^2\cos^2t)^{3/2}\over ab}.$ (49)

Similarly, the unit Tangent Vector is given by
\begin{displaymath}
\hat{\bf T} = \left[{\matrix{-a\sin t\cr b\cos t\cr}}\right] {1\over \sqrt{a^2\sin ^2t+b^2\cos ^2t}}.
\end{displaymath} (50)


The Arc Length of the ellipse can be computed using

$\displaystyle s$ $\textstyle =$ $\displaystyle \int\sqrt{x'^2+y'^2}\,dt = \int\sqrt{a^2\cos^2 t+b^2\sin^2t}\,dt$  
  $\textstyle =$ $\displaystyle a\int\sqrt{(1-\sin^2 t)+{b^2\over a^2}\sin^2 t}\,dt$  
  $\textstyle =$ $\displaystyle a\int\sqrt{1-\left({1-{b^2\over a^2}}\right)\sin^2t}\,dt$  
  $\textstyle =$ $\displaystyle a\int\sqrt{1-e^2\sin^2 t}\,dt = aE(t,e),$ (51)

where $E$ is an incomplete Elliptic Integral of the Second Kind. Again, note that $t$ is a parameter which does not have a direct interpretation in terms of an Angle. However, the relationship between the polar angle from the ellipse center $\theta$ and the parameter $t$ follows from
\begin{displaymath}
\theta=\tan^{-1}\left({y\over x}\right)=\tan^{-1}\left({{b\over a}\tan t}\right).
\end{displaymath} (52)

\begin{figure}\begin{center}\BoxedEPSF{EllipseFunction.epsf}\end{center}\end{figure}

This function is illustrated above with $\theta$ shown as the solid curve and $t$ as the dashed, with $b/a=0.6$. Care must be taken to make sure that the correct branch of the Inverse Tangent function is used. As can be seen, $\theta$ weaves back and forth around $t$, with crossings occurring at multiples of $\pi/2$.

\begin{figure}\begin{center}\BoxedEPSF{EllipseInfo.epsf scaled 750}\end{center}\end{figure}

The Curvature and Tangential Angle of the ellipse are given by

$\displaystyle \kappa$ $\textstyle =$ $\displaystyle {ab\over (b^2\cos^2t+a^2\sin^2 t)^{3/2}}$ (53)
$\displaystyle \phi$ $\textstyle =$ $\displaystyle -\tan^{-1}\left({{b\over a}\cos t}\right).$ (54)


The entire Perimeter $p$ of the ellipse is given by setting $t=2\pi$ (corresponding to $\theta=2\pi$), which is equivalent to four times the length of one of the ellipse's Quadrants,

\begin{displaymath}
p=aE(2\pi,e)=4aE({\textstyle{1\over 2}}\pi,e)= 4aE(e),
\end{displaymath} (55)

where $E(e)$ is a complete Elliptic Integral of the Second Kind with Modulus $k$. The Perimeter can be computed numerically by the rapidly converging Gauss-Kummer Series
$\displaystyle p$ $\textstyle =$ $\displaystyle \pi(a+b)\sum_{n=0}^\infty {{\textstyle{1\over 2}}\choose n}^2 h^{2n}$  
  $\textstyle =$ $\displaystyle \pi(a+b)(1+{\textstyle{1\over 4}}h^2+{\textstyle{1\over 64}}h^4+{\textstyle{1\over 256}}h^6+\ldots),$ (56)

where
\begin{displaymath}
h\equiv {a-b\over a+b}
\end{displaymath} (57)

and ${n\choose k}$ is a Binomial Coefficient. Approximations to the Perimeter include
$\displaystyle p$ $\textstyle \approx$ $\displaystyle \pi \sqrt{2(a^2+b^2)}$ (58)
  $\textstyle \approx$ $\displaystyle \pi[3(a+b)-\sqrt{(a+3b)(3a+b)}]$ (59)
  $\textstyle \approx$ $\displaystyle \pi(a+b)\left({1+{3t\over 10+\sqrt{4-3t}}}\right),$ (60)

where the last two are due to Ramanujan (1913-14),
\begin{displaymath}
t\equiv \left({a-b\over a+b}\right)^2,
\end{displaymath} (61)

and (60) is accurate to within $\sim 3\cdot 2^{-17}t^5$.


The maximum and minimum distances from the Focus are called the Apoapsis and Periapsis, and are given by

$\displaystyle r_+$ $\textstyle =$ $\displaystyle r_{\rm apoapsis}= a(1+e)$ (62)
$\displaystyle r_-$ $\textstyle =$ $\displaystyle r_{\rm periapsis}= a(1-e).$ (63)


The Area of an ellipse may be found by direct Integration

$\displaystyle A$ $\textstyle =$ $\displaystyle \int_{-a}^a \int^{b\sqrt{a^2-x^2}/a}_{-b\sqrt{a^2-x^2}/a} dy\,dx =\int_{-a}^a {2b\over a}\sqrt{a^2-x^2}\,dx$  
  $\textstyle =$ $\displaystyle {2b\over a}\left\{{{1\over 2} \left[{x\sqrt{a^2-x^2}+a^2\sin^{-1}\left({x\over \vert a\vert}\right)}\right]}\right\}^a_{x=-a}$  
  $\textstyle =$ $\displaystyle ab[\sin^{-1}1-\sin^{-1}(-1)] = ab\left[{{\pi\over 2}-\left({-{\pi\over 2}}\right)}\right]=\pi ab.$  
      (64)

The Area can also be computed more simply by making the change of coordinates $x'\equiv (b/a) x$ and $y'\equiv y$ from the elliptical region $R$ to the new region $R'$. Then the equation becomes
\begin{displaymath}
{1\over a^2}\left({{a\over b}x'}\right)^2+{y'^2\over b^2}=1,
\end{displaymath} (65)

or $x'^2+y'^2=b^2$, so $R'$ is a Circle of Radius $b$. Since
\begin{displaymath}
{\partial x\over\partial x'}=\left({\partial x'\over\partial x}\right)^{-1} = \left({b\over a}\right)^{-1} = {a\over b},
\end{displaymath} (66)

the Jacobian is
\begin{displaymath}
\left\vert\matrix{\partial (x,y)\over \partial (x',y')\cr}\r...
...rt\matrix{{a\over b} & 0\cr 0 & 1\cr}\right\vert = {a\over b}.
\end{displaymath} (67)

The Area is therefore
$\displaystyle \int\!\!\!\int _R dx\,dy$ $\textstyle =$ $\displaystyle \int\!\!\!\int _{R'} \left\vert{\partial (x,y)\over \partial (x',y')}\right\vert\,dx'\,dy'$  
  $\textstyle =$ $\displaystyle {a\over b}\int\!\!\!\int _{R'} dx'dy' = {a\over b}(\pi b^2) = \pi ab,$ (68)

as before. The Area of an arbitrary ellipse given by the Quadratic Equation
\begin{displaymath}
ax^2+bxy+cy^2=1
\end{displaymath} (69)

is
\begin{displaymath}
A={2\pi\over\sqrt{4ac-b^2}}.
\end{displaymath} (70)

The Area of an Ellipse with semiaxes $a$ and $b$ with respect to a Pedal Point $P$ is
\begin{displaymath}
A={\textstyle{1\over 2}}\pi(a^2+b^2+\vert OP\vert^2).
\end{displaymath} (71)


The ellipse Inscribed in a given Triangle and tangent at its Midpoints is called the Midpoint Ellipse. The Locus of the centers of the ellipses Inscribed in a Triangle is the interior of the Medial Triangle. Newton gave the solution to inscribing an ellipse in a convex Quadrilateral (Dörrie 1965, p. 217). The centers of the ellipses Inscribed in a Quadrilateral all lie on the straight line segment joining the Midpoints of the Diagonals (Chakerian 1979, pp. 136-139).


The Area of an ellipse with Barycentric Coordinates $(\alpha,\beta,\gamma)$ Inscribed in a Triangle of unit Area is

\begin{displaymath}
\Delta=\pi\sqrt{(1-2\alpha)(1-2\beta)(1-2\gamma)}.
\end{displaymath} (72)

(Chakerian 1979, pp. 142-145).


The Locus of the apex of a variable Cone containing an ellipse fixed in 3-space is a Hyperbola through the Foci of the ellipse. In addition, the Locus of the apex of a Cone containing that Hyperbola is the original ellipse. Furthermore, the Eccentricities of the ellipse and Hyperbola are reciprocals. The Locus of centers of a Pappus Chain of Circles is an ellipse. Surprisingly, the locus of the end of a garage door mounted on rollers along a vertical track but extending beyond the track is a quadrant of an ellipse (the envelopes of positions is an Astroid).

See also Circle, Conic Section, Eccentric Anomaly, Eccentricity, Elliptic Cone, Elliptic Curve, Elliptic Cylinder, Hyperbola, Midpoint Ellipse, Parabola, Paraboloid, Quadratic Curve, Reflection Property, Salmon's Theorem, Steiner's Ellipse


References

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, pp. 126 and 198-199, 1987.

Casey, J. ``The Ellipse.'' Ch. 6 in A Treatise on the Analytical Geometry of the Point, Line, Circle, and Conic Sections, Containing an Account of Its Most Recent Extensions, with Numerous Examples, 2nd ed., rev. enl. Dublin: Hodges, Figgis, & Co., pp. 201-249, 1893.

Chakerian, G. D. ``A Distorted View of Geometry.'' Ch. 7 in Mathematical Plums (Ed. R. Honsberger). Washington, DC: Math. Assoc. Amer., 1979.

Courant, R. and Robbins, H. What is Mathematics?: An Elementary Approach to Ideas and Methods, 2nd ed. Oxford, England: Oxford University Press, p. 75, 1996.

Dörrie, H. 100 Great Problems of Elementary Mathematics: Their History and Solutions. New York: Dover, 1965.

Lawrence, J. D. A Catalog of Special Plane Curves. New York: Dover, pp. 72-78, 1972.

Lee, X. ``Ellipse.'' http://www.best.com/~xah/SpecialPlaneCurves_dir/Ellipse_dir/ellipse.html.

Lockwood, E. H. ``The Ellipse.'' Ch. 2 in A Book of Curves. Cambridge, England: Cambridge University Press, pp. 13-24, 1967.

MacTutor History of Mathematics Archive. ``Ellipse.'' http://www-groups.dcs.st-and.ac.uk/~history/Curves/Ellipse.html.

Ramanujan, S. ``Modular Equations and Approximations to $\pi$.'' Quart. J. Pure. Appl. Math. 45, 350-372, 1913-1914.



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© 1996-9 Eric W. Weisstein
1999-05-25