ellipse ellipse
Family of ellipses with eccentricities {0.25, 0.36, 0.46, 0.57, 0.67, 0.78, 0.88, 0.99} in order of light to dark shade. The left family share vertexes, the right are confocal.

Mathematica Notebook for Ellipse


See the History section of Conic Sections page.


Ellipse is a family of curves of one parameter. Together with hyperbola and parabola, they make up the conic sections. Ellipse is also a special case of hypotrochoid.

Ellipse is commonly defined as the locus of points P such that the sum of the distances from P to two fixed points F1, F2 (called foci) are constant. That is, distance[P,F1] + distance[P,F2] == 2 a, where a is a positive constant.

ellipse ellipseGen1
Ellipse as sum of 2 line lengths. Ellipse Family

The eccentricity of a ellipse, denoted e, is defined as e := c/a, where c is half the distance between foci. Eccentricity is a number that describe the degree of roundness of the ellipse. For any ellipse, 0 ≤ e ≤ 1. The smaller the eccentricy, the rounder the ellipse. If e == 0, it is a circle and F1, F2 are coincident. If e == 1, then it's a line segment, with foci at the two end points.


Let a ellipse's sum of distances be “2*a”, and center to a focus be c, and semiminor axis be b, and essentricity be e. These values are related by the formula: b^2+c^2==a^2, e*a==c, and center to a vertex is “a”. This is a very useful formula when trying to solve for a unknown.


The formula for ellipse can be derived in many ways. By the definition of sum of distances distance[P,F1] + distance[P,F2] == 2*a, setting focuses to be {±c,0} we easily find the Cartesian equation to be:

Sqrt[(x-c)^2+(y-0)^2] + Sqrt[(x- (-c))^2+(y-0)^2] == 2 *a

After getting rid of square roots, we have:

16*a^4 - 16*a^2*c^2 + (-16*a^2 + 16*c^2)*x^2 - 16*a^2*y^2 == 0

which has the form of 2nd degree polynomial. Now minus both sides by a constant we have the form:


for some A and B, where the vertexes of the ellipse happens to be {±A,0} and {±B,0}. To derive the parametric formula, noticed that the x component or y component oscillates. So, if we relpace y by B*Sin[t] and solve for x, we easily found a parametric formula corresponding to x^2/A^2+y^2/B^2==1 to be:

{B* Cos[t], A* Sin[t]}


Similarly, we can derive other forms. The following gives ellipses with eccentricity as parameter, with Vertexes at {±1,0}, and focuses at {±e, 0}.


Point Construction

This method is obvious from the paramteric formula for ellipses {a Cos[t], b Sin[t]}, where a and b are the radiuses of the circles. Figure below showing a ellipse with a:=5, b:=3, and e==8/10.

ellipse ellipseGenPointwise
Point-wise Construction Point-wise Construction
ellipse point-wise
Ellipse's Property of inscribed and circumscribed circles and center

From the point of view of a given ellipse, this is a property that relates a point on the ellipse, to its circumscribed circle, inscribed circle, and center. That is: given a ellipse and a point P on the curve. Let there be lines passing P and perpendicular to the ellipse's major and minor axes. These 2 pendicular lines will intersect the circimscribed and inscribed circles at 2 points, if we consider only intersections that lie in the same quadrant as P. Now, a line, passing these two points will intersect the ellipse's center.

This theorem leads to the Trammel of Archimedes below.

Trammel of Archimedes

Ellipse is the glissette of a line of constant length with endpoints on other two mutually orthogonal lines. That is, the trace of fixed point on a line of constant length with its endpoints freely glides on two mutually orthogonal lines. This mechanism is called Trammel of Archimedes. The trace of the line itself will generate a astroid.

ellipse ellipse ellipseTrammel
Trammel in Motion
ellispe as Trammel
ellipse trammel proof

Proof: Assume the ellipse to be in a position where its major axis is aligned with the x axis and minor axis aligned with the y axes. Let A be the center of ellipse. Let P be a point on the ellipse. Let there be a circumscribed circle r, and a inscribed circle q. Now, let there be a line thru P, parallel to minor axis. Let the intersection of this line and the circumscribed circle be Q (pick the intersection that lies in the same quadrant as P). Let there be a line thru P, parallel to the major axis. Let the intersection of the line and the inscribed circle be R. By the point-wise construction theorem, line ARQ are collinear. Let F be the midpoint of RQ. Let F be the center of a rectangle, and lower left corner at A. Label the upper left corner D, upper right C, lower right E. Now, consider triangle[Q,P,R]. Note that dist[Q,R] is constant, angle[Q,P,R]==90°, F bisects QR. Therefore, then distance[F,P] is constant. Now, consider the rectangle DCEA. The diagnals of the rectangle is constant, since AF is constant. (AF is constant because F is the bisector of RQ, and R and Q lies on the circum/in-scribed circles, and ARQ is collinear.) The distance[D,E] is constant because AC twice of AE, and AE is constant because F is mid of the radiuses of the circumscribed circle and inscribed circle. P lies on DE by symmetry arguments with center F. (this proof is badly worded and verbose. Exercise: write a better proof)

Tangent Construction

ellips tangent construction
Ellipse Tangent Construction

Givens: A circle k with center F1, a point F2 inside the circle, and a point P on the circle. Now, Let t be a line that bisects Line[P,F2]. Let Q be the intersection between b and Line[P,F1]. As P moves around the circle, the traces of Q is a ellipse with focus F1 and F2 and dist[F1,P] being its distance sum, and the line t is its tangent at Q.

Proof: We want to prove that dist[F1,Q]+dist[F2,Q]==dist[F1,P]. Since Q lies on the line t, and t bisects line[P,F2], thus dist[Q,P]==dist[Q,F2]. Since Q also lies on segment[P,F1], so dist[F1,Q] + dist[Q,P] == dist[F1,P]. Combine the above equation together shows dist[F1,Q]+dist[F2,Q]==dist[F1,P]. To show that t is the tangent at Q, note that line perpendicular to t and passing Q bisects the angle[F1,Q,F2]. (detail omitted here)

Note, that the midpoint of P and F2 traces out a circle. This leads to the following theorem that the pedal of a ellipse with respect to a focus is a circle.


The pedal of a ellipse with respect to a focus is a circle, conversely, the negative pedal of a circle with respect to a point inside the circle is a ellipse. This fact can be used to draw ellipses by envelope of lines.

  1. Start with a circle and a point F1 inside the circle.
  2. Draw a line j from a point P on the circle to F1.
  3. Draw a line k perpendicular to j and passing P.
  4. Repeat step 2 and 3 for other points P on the cricle.
  5. The envelope of lines m is a ellipse with a focus at F1 and two vertexes touching the circle.
Ellipse as Circle's Negative Pedal

Ellipse as Hypotrochoid

Ellipse can be generated as a hypotrochoid. Let the parameters of the hypotrochoid be {A,B,H}, where A is the radius of the fixed circle, B is the radius of the rolling circle, and H is the distance from the tracing point to the center of the rolling circle. If A/2 == B and H ≠ B, then it's a ellipse with semimajor axis a==B+H and semiminor axis b==Abs[B-H]. Eccentricity is then e:=c/a == Sqrt[a^2-b^2]/a == Sqrt[(B+H)^2-(B-H)^2]/(B+H)

Given A, B:=1/2 A, and e, we may want to find H such that the hypotrochoid {A,B,H} generate a ellipse with eccentricity e. This is easily solved with the above information. The solution is:

h == ( -B (-2+e^2) ± 2 B Sqrt[(1-e)*(1+e)] ) / e^2

If we let A:=1, B:=1/2, e:=8/10, we get h==1/8 or 2. Two ellipses of eccentricity 8/10 as hypotrochoids with parameters {1,1/2,1/8} and {1,1/2,2} are shown below.

ellipseGenHypo1 ellipseGenHypo1
Ellipse as hypotrochoid
ellipseGenHypo2 ellipseGenHypo2
Ellipse as hypotrochoid

Optical Property

Lightrays from one focus will reflect to the ther focus. If the radiant point is not at a focus, a caustic curve forms.

ellipse ellipse
Ellipse's Caustic


Ellipse's inversion with respect to a focus is a dimpled limacon of Pascal.

Ellipse Inversion

Cylinder Slice

The intersection of a right circular cylinder and a plane is a ellipse. One can see this by tilting a cup or a cone shaped paper cup filled with liquid. Let r be the radius of the base circle of the cylinder, α be the angle formed by the cutting plane and the plane of the base circle of the right circular cylinder. The intersection will be a ellipse with semi-major axis r/Cos[α] and semi-minor axis r.


graphics code

Related Web Sites

See: Websites on Plane Curves, Printed References On Plane Curves.

Robert Yates: Curves and Their Properties.


See: Websites on Conic Sections.