Math+ Ellipse: UNIZOR.COM - Math+ 4 All - Geometry

Notes to a video lecture on http://www.unizor.com

Geometry+ Ellipse

Certain curves on a plane have common properties that allow to combine them into classes.

For example, a circle is a class of curves with the following property.
For each curve of this class (that is, for each circle) there is one specific point called its center and a specific positive real number called its radius, such that this circle consists of all points on a plane located on the distance equaled to this circle's radius from its center.

Ellipse is another class of curves on a plane with the following defining properties.
For each curve of this class (that is, for each ellipse) there are two specific points called foci (plural of focus) or focuses and a specific positive real number called its length of major axis that is supposed to be greater than the distance between its two foci, such that this ellipse consists of all points on a plane (or is a locus points on a plane) with the sum of their distances to its two focuses equaled to this ellipse's length of major axis.

As we see, the position of two focus points F₁, F₂ and a length of major axis 2a (should be greater than the length of segment F₁F₂) uniquely identify an ellipse.

Using this definition of ellipse, we can draw it on the board using two nails and a thread by fixing a thread on both ends at the nails, tightening a thread using a pencil and moving a pencil around keeping a thread tightened.
The result of this procedure will be an ellipse.

Below is a drawing of such an ellipse and the prove that, if a sum of distances from any point on an ellipse to its two foci is 2a, the length of its major axis - a segment P₄P₃ between intersections of a focal line F₁F₂ with an ellipse - is also 2a.

Assume, we fix the position of foci F₁ and F₂ with the distance between them being 2c. The parameter c is called a focal distance.
Assume further that we have chosen the length of a major axis 2a greater than the length 2c of a focal line F₁F₂ between focus points.
The following drawing represents an ellipse defined by these two parameters, half a major axis a and half of focal distance c.
Let point O be a midpoint of segment F₁F₂.
Point C is an intersection of an ellipse with a midpoint perpendicular to F₁F₂, so OC⊥F₁F₂.
Points A and B are intersection of the focal line F₁F₂ with an ellipse.

Consider an ellipse on a Cartesian coordinate plane with foci to be at points F₁(−c,0) and F₂(0,c).

Point A(−a,0) lies on the X-axis and on this ellipse because the distance from A to F₁ is a−c and the distance from A to F₂ is a+c. The sum of these distances is 2a, which is a condition of any point on an ellipse.

Similarly, point B(0,a) lies on the X-axis and on this ellipse because the distance from B to F₁ is a+c and the distance from B to F₂ is a−c. The sum of these distances is 2a, which is a condition of any point on an ellipse.

Segment AB constitutes the major axis of this ellipse.


Consider points C(0,b) and D(0,−b), where b is a positive real number that satisfies the Pythagorean equation b²=a²−c² and, therefore, a²=b²+b².
Point C lies on the Y-axis and on this ellipse because the distance from C to F₁ is √(0+c)²+(b−0)²=a and the distance from C to F₂ is √(0−c)²+(b−0)²=a. The sum of these distances is 2a, which is a condition of any point on an ellipse.

Similarly, point D lies on the Y-axis and on this ellipse because the distance from D to F₁ is √(0+c)²+(−b−0)²=a and the distance from D to F₂ is √(0−c)²+(−b−0)²=a. The sum of these distances is 2a, which is a condition of any point on an ellipse.

Our goal is to find an equation for x and y that describes the condition on point P(x,y) to lie on an ellipse with focal distance 2c and major axis length of 2a.

The definition of ellipse requires that sum of distances from point P(x,y) to both foci is 2a.
Therefore, an equation on x and y that is a necessary and sufficient condition for point P(x,y) to lie on our ellipse is
√(x+c)²+y² + √(x−c)²+y² = 2a

Fortunately, this cumbersome equation can be converted into a much simpler form

x²/a² + y²/b² = 1

where b² substitutes a²−c².

The transformation into this simple form is straightforward but, unfortunately, lengthy and boring. We have decided to omit it from this lecture and recommend you to do it yourself by repeatedly separating one of the square roots into one side of an equation and squaring both sides.

Alternatively, you can look at any Web page that presents this transformation in details.
For example, detail steps of such a transformation can be found at
https://courses.lumenlearning.com/odessa-collegealgebra/chapter/deriving-the-equation-of-an-ellipse-centered-at-the-origin/

Our next task is to represent an ellipse in polar coordinates r (a distance from a center of polar coordinates to a point under consideration) and θ (angle from a polar axis to a direction from a center to a point under consideration).

It makes sense to set a polar base axis coinciding with the line between foci with an origin of polar coordinates positioned at the focus F₁ of an ellipse and direction of a base axis to be towards another focus. Then the distance from any point P(r,θ) of an ellipse to this focus is r.
This value, summed with a distance from P(r,θ) to another focus F₂, should be equal to 2a.

While it seems more natural to take a midpoint between the foci as an origin of polar coordinates, the equation is simpler to derive if this origin is at the focus of an ellipse.

Consider a triangle ΔPF₁F₂ formed by two foci and a point P(r,θ) on the ellipse.
We can use the Law of Cosines to determine PF₂ by two other sides PF₁=r and F₁F₂=2c.
(PF₂)²=r²+4c²−4r·c·cos(θ)

This allows to get an equation of an ellipse in polar coordinates with a focus as an origin of coordinates as
r + √r²+4c²−4r·c·cos(θ) = 2a

Now we have to transform it into a form of distance r of a point P from an origin of polar coordinates (focus F₁) as a function of a polar angle θ.

Straight forward way is to separate a square root in one side of an equation and square both sides.
√r²+4c²−4r·c·cos(θ) = 2a − r
r²+4c²−4r·c·cos(θ) =
= 4a²−4a·r+r²
4a·r−4r·c·cos(θ) = 4a²−4c²

Using a²=b²+c², the equation for an ellipse would be
r·[a−c·cos(θ)] = b²
Therefore,
r= b²/[a−c·cos(θ)]

The ratio e=c/a is called eccentricity of an ellipse. Using it, the formula can be transformed to
r= (b²/a)/[1−(c/a)·cos(θ)]
or, using again formula a²=b²+c²,

r= a·(1-e²)/[1−e·cos(θ)]

The above is an equation of an ellipse in polar coordinates, when one of the foci is taken as an origin of coordinates and polar base axis is directed from it towards another focus.
This equation is expressed in terms of semi-major axis a and eccentricity e=c/a, where c is half of a distance between foci.

Let's fix the major axis of an ellipse and change the distance between foci.
The smaller the distance between the foci (that is, the smaller c with fixed a) - the more our ellipse resembles a circle. When e=0 the equation of an ellipse looks like an eqiation of a circle in polar coordinates:
r = a
(that is, constant distance from the origin)

If we increase the eccentricity, that is if e increases to its maximum value of 1, its foci get closer and closer to endpoints of a major axis and the ellipse's size in a direction perpendicular to its major axis (that is, 2b) decreases to zero.