Math+ - Hyperbola Optics: UNIZOR.COM - Math+ 4 All - Geometry

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

Geometry+ Hyperbola Optics

Though the word "optics" sounds very much like a topic of Physics, we will consider strictly mathematical aspect of it.
One of the properties of hyperbola is that if its contour is reflective, a ray of light emitted from one of its foci will be reflected along a line that crosses another focus regardless of the direction it was sent.

Consider a hyperbola with foci F₁(−c,0) and F₂(c,0), point P(x,y) on this hyperbola and tangential AR to a hyperbola at this point P(x,y).

Here F₂P is an incident ray, PS is a reflected ray. The optical property of a hyperbola that we want to prove is that reflected ray PS lies on a line that crosses a focus point F₁ regardless of a position of point P on a hyperbola.

As mentioned in the lecture Ellipse Optics of this course, reflection of the curve at some point occurs exactly as if the ray of light reflects of the tangential to this curve at the point of incidence.

The segment F₁P lies on a continuation of a reflected ray. Angle of F₂P with X-axis is α.
Angle of F₁P with X-axis is β.
Angle of a tangent to a hyperbola at point P(x,y) with X-axis is γ.
Angle between a tangent and incident ray F₂P is φ.
Angle between a tangent and reflected ray F₁P is ψ.

To prove that the light emitted from focus F₂(c,0) will reflect at point P(x,y) on a hyperbola and will lie on a line that will hit point F₁(−c,0), it is sufficient to prove that angles φ and ψ are equal.

Simple considerations, based on the theorem that the sum of angles of any triangle equals to π, lead us to the following equalities
φ = α − γ
ψ = γ − β

All the above angles are in the range from 0 to π. In this interval equality of angles follows from equality of their tangents since in this interval tangent is a monotonic function.
So, let's determine tangents of φ and ψ and prove that they are equal.

In both cases we will use the formula for tangent of a difference between angles.

tan(φ) =

tan(α)−tan(γ) 1+tan(α)·tan(γ)

tan(ψ) =

tan(γ)−tan(β) 1+tan(γ)·tan(β)

Tangents of α and β are easy to calculate since we know coordinates of segments F₁P and F₂P.

tan(α) = (y−0)/(x−c) = y/(x−c)
tan(β) = (y−0)/(x+c) = y/(x+c)

The issue with tan(γ) is a bit more complicated since we know that this is a slope of a tangential line to a hyperbola at point P(x,y), and we don't have coordinates of a second point A on this line to calculate a slope using the same method as with angles α or β.

However, we know that the same slope is a derivative of variable y, as a function of variable x, describing our hyperbola.

Using the equation of hyperbola in Cartesian coordinates, we can calculate it as follows.
1. Equation of hyperbola is
x²/a² − y²/b² = 1
where parameter a is a semi-major axis of a hyperbola and equals to a distance from the origin of coordinates to a point of intersection of a hyperbola with X-axis,
parameter c is a semi-focal distance and is equal to a distance from the origin of coordinates to a focus point and
parameter b is called a semi-minor axis and is defined by an equation b²=c²−a².
2. Differentiate both sides by x, taking into consideration that y is a function of x
2x/a² − (2y/b²)·y'(x) = 0
(where y'(x) is a derivative of y by x).
3. Find the slope of our tangential to hyperbola line at point P(x,y)
y' = x·b²/(y·a²) = tan(γ)

Now we have all components to calculate tan(φ) and tan(ψ) in terms of calculated tangents of angles α, β, γ.
If we prove that tan(φ)=tan(ψ), we can say that a light ray emitted from focus F₂ towards any point P(x,y) on a hyperbola will be reflected from point P(x,y) along a line that crosses focus F₁.

So, here are the calculations of tan(φ) and tan(ψ).

We start with calculations of tan(φ) based on values of tan(α) and tan(γ).
We will use the original relation between x and y that reflects the fact that point P(x,y) lies on a hyperbola
x²/a²−y²/b²=1,
from which follows
x²·b²−y²·a²=a²·b²

1. tan(α) − tan(γ) =
= y/(x−c) − x·b²/(y·a²) =
= (y²·a²−x²·b²+c·x·b²)/(x·y·a²−c·y·a²) =
= (−a²·b²+c·x·b²)/(x·y·a²−c·y·a²)

2. 1 + tan(α)·tan(γ) =
= 1 + [y/(x−c)]·[x·b²/(y·a²)] =
= 1 + x·y·b²/(x·y·a²−c·y·a²) =
= (x·y·a²−c·y·a²+x·y·b²)/(x·y·a²−c·y·a²) =
[since a²+b²=c²]
= (x·y·c²−c·y·a²)/(x·y·a²−c·y·a²)

3. tan(φ) = [tan(α)−tan(γ)] / [1+tan(α)·tan(γ)] =
= (−a²·b²+c·x·b²)/(x·y·c²−c·y·a²)

Now let's calculate the value of tan(ψ) based on values of tan(β) and tan(γ).

1. tan(γ) − tan(β) =
= x·b²/(y·a²) − y/(x+c) =
= (x²·b²+c·x·b²−y²·a²)/(x·y·a²+c·y·a²) =
= (a²·b²+c·x·b²)/(x·y·a²+c·y·a²)

2. 1 + tan(γ)·tan(β) =
= 1 + [x·b²/(y·a²)]·[y/(x+c)] =
= 1 + x·y·b²/(x·y·a²+c·y·a²) =
= (x·y·a²+c·y·a²+x·y·b²)/(x·y·a²+c·y·a²) =
[since a²+b²=c²]
= (x·y·c²+c·y·a²)/(x·y·a²+c·y·a²)

3. tan(ψ) = [tan(γ)−tan(β)] / [1+tan(γ)·tan(β)] =
= (a²·b²+c·x·b²)/(x·y·c²+c·y·a²)

What remains is to show that two calculated values, tan(φ) and tan(ψ) are the same.
Both are fractions with numerator and denominator. Equality between two fractions P/Q=R/S is equivalent to equality P·S=Q·R.
Let's check equality of our fractions using this method.

Numerator of tan(φ) multiplied by denominator of tan(ψ) is
(−a²·b²+c·x·b²)·(x·y·c²+c·y·a²) =
= −x·y·a²·b²·c² + x²·y·b²·c³ −
− y·a⁴·b²·c + x·y·a²b²·c² =
= x²·y·b²·c³ − y·a⁴·b²·c

Denominator of tan(φ) multiplied by numerator of tan(ψ) is
(x·y·c²−c·y·a²)·(a²·b²+c·x·b²) =
= x·y·a²·b²·c² − y·a⁴·b²·c +
+ x²·y·b²·c³ − x·y·a²·b²·c² =
= − y·a⁴·b²·c + x²·y·b²·c³

Final expressions for two cases above are identical, which proves that tan(φ)=tan(ψ).

This, as we mentioned above, is a sufficient condition for a ray of light emitted from one focus point of a hyperbola to any direction after reflecting from the hyperbola to end up at the other focus point.

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

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

Geometry+ Ellipse Optics

Though the word "optics" sounds very much like a topic of Physics, we will consider strictly mathematical aspect of it.
One of the properties of ellipse is that if its contour is reflective, a ray of light emitted from one of its foci will come to another foci regardless of the direction it was sent.

Let's start with a mathematical meaning of reflection of the contour of ellipse, which is not a straight line.
We do know what a reflection of the straight line is. Simply speaking, angles of incidence and reflection are equal to each other. Reflection of a curve occurs exactly as if, instead of a curve at the point of incidence, there was a tangential line to a curve, and reflection was of that tangential straight line.

Consider an ellipse with foci F₁(−c,0) and F₂(c,0), point P(x,y) on this ellipse and tangential to an ellipse at this point P(x,y).
Angle of F₁P with X-axis is α.
Angle of F₂P with X-axis is β.
Angle of a tangent to an ellipse at point P(x,y) with X-axis is γ.
Angle between a tangent and F₁P is φ.
Angle between a tangent and F₂P is ψ.

To prove that the light emitted from focus F₁(−c,0) will reflect at point P(x,y) on an ellipse and will hit point F₂(c,0), it is sufficient to prove that angles φ and ψ are equal.

Simple considerations, based on the theorem that the sum of angles of any triangle equals to &pi, lead us to the following equalities
φ = α − γ
ψ = γ − β + π

All the above angles are in the range from 0 to π. In this interval equality of angles follows from equality of their tangents since in this interval tangent is a monotonic function.
So, let's determine tangents of φ and ψ and prove that they are equal.

In both cases we will use the formula for tangent of a difference between angles and, in case of angle ψ, will take into consideration that tangent is periodic with a period π.

tan(φ) =

tan(α)−tan(γ) 1+tan(α)·tan(γ)

tan(ψ) =

tan(γ)−tan(β) 1+tan(γ)·tan(β)

As we know, the tangent of an angle from a positive direction of the X-axis and a straight line connecting to points on a plane A(a_x,a_y) and B(b_x,b_y) (usually called a slope of a line) equals to (b_y−a_y)/(b_x−a_x).

Therefore,
tan(α) = (y−0)/(x+c) = y/(x+c)
tan(β) = (y−0)/(x−c) = y/(x−c)

The issue with tan(γ) is a bit more complicated since we know that this is a slope of a tangential line to an ellipse at point P(x,y), and we don't have a second point on this line to calculate a slope using the same method as with angles α or β.

However, we know that the same slope is a derivative of variable y, as a function of variable x, describing our ellipse.

Using the equation of ellipse in Cartesian coordinates, we can calculate it as follows.
1. Equation of ellipse is
x²/a² + y²/b² = 1
2. Differentiate both sides by x, taking into consideration that y is a function of x
2x/a² + (2y/b²)·y'(x) = 0
(where y'(x) is a derivative of y by x).
3. Find the slope of our tangential to ellipse line at point P(x,y)
y' = −x·b²/(y·a²) = tan(γ)

Now we have all components to calculate tan(φ) and tan(ψ) in terms of calculated tangents of angles α, β, γ.
If we prove that tan(φ)=tan(ψ), we can say that a light ray emitted from focus F₁ towards any point P(x,y) on an ellipse will be reflected from point P(x,y) towards focus F₂.

So, here are the calculations of tan(φ) and tan(ψ).

We start with calculations of tan(φ) based on values of tan(α) and tan(γ).
We will use the original relation between x and y that reflects the fact that point P(x,y) lies on an ellipse
x²/a²+y²/b²=1,
from which follows
x²·b²+y²·a²=a²·b²

1. tan(α) − tan(γ) =
= y/(x+c) + x·b²/(y·a²) =
= (y²·a²+x²·b²+c·x·b²)/(x·y·a²+c·y·a²) =
= (a²·b²+c·x·b²)/(x·y·a²+c·y·a²)

2. 1 + tan(α)·tan(γ) =
= 1 − x·y·b²/(x·y·a²+c·y·a²) =
= (x·y·a²+c·y·a²−x·y·b²)/(x·y·a²+c·y·a²)

3. tan(φ) = [tan(α)−tan(γ)] / [1+tan(α)·tan(γ)] =
= (a²·b²+c·x·b²)/(x·y·a²+c·y·a²−x·y·b²)

Now let's calculate the value of tan(ψ) based on values of tan(β) and tan(γ).

1. tan(γ) − tan(β) =
= −x·b²/(y·a²) − y/(x−c) =
= (−x²·b²+c·x·b²−y²·a²)/(x·y·a²−c·y·a²) =
= (−a²·b²+c·x·b²)/(x·y·a²−c·y·a²)

2. 1 + tan(γ)·tan(β) =
= 1 + [−x·b²/(y·a²)]·[y/(x−c)] =
= (x·y·a²−c·y·a²−x·y·b²)/(x·y·a²−c·y·a²)

3. tan(ψ) = [tan(α)−tan(γ)] / [1+tan(α)·tan(γ)] =
= (−a²·b²+c·x·b²)/(x·y·a²−c·y·a²−x·y·b²)

What remains is to show that two calculated values, tan(φ) and tan(ψ) are the same.
Both are fractions with numerator and denominator. Equality between two fractions P/Q=R/S is equivalent to equality P·S=Q·R.
Let's check equality of our fractions using this method.

Numerator of tan(φ) multiplied by denominator of tan(ψ) is
(a²·b²+c·x·b²)·(x·y·a²−c·y·a²−x·y·b²) =
= x·y·a⁴·b² + x²·y·a²·b²·c −
− y·a⁴·b²·c − x·y·a²b²·c² −
− x·y·a²·b⁴ − x²·y·b⁴·c
Notice that
x·y·a⁴·b² − x·y·a²·b⁴ =
= x·y·a²·b²·(a²−b²) =
= x·y·a²·b²·c²
which cancels the analogous term with a minus sign in the above expression.
This leaves our expressions equal to
(a²·b²+c·x·b²)·(x·y·a²−c·y·a²−x·y·b²) =
= x²·y·a²·b²·c − y·a⁴·b²·c − x²·y·b⁴·c
Next simplification is
x²·y·a²·b²·c − x²·y·b⁴·c =
= x²·y·b²·c·(a²−b²) = x²·y·b²·c³
That leaves our expression to
x²·y·b²·c³ − y·a⁴·b²·c

Denominator of tan(φ) multiplied by numerator of tan(ψ) is
(x·y·a²+c·y·a²−x·y·b²)·(−a²·b²+c·x·b²) =
= −x·y·a⁴·b² − y·a⁴·b²·c + x·y·a²·b⁴ +
+ x²·y·a²·b²·c + x·y·a²·b²·c² − x²·y·b⁴·c
Notice that
−x·y·a⁴·b² + x·y·a²·b⁴ =
= −x·y·a²·b²·(a²−b²) =
= −x·y·a²·b²·c²
which cancels the analogous term with a plus sign in the above expression.
This leaves our expressions equal to
(x·y·a²+c·y·a²−x·y·b²)·(−a²·b²+c·x·b²) =
= − y·a⁴·b²·c + x²·y·a²·b²·c − x²·y·b⁴·c
Next simplification is
x²·y·a²·b²·c − x²·y·b⁴·c =
= x²·y·b²·c·(a²−b²) = x²·y·b²·c³
That leaves our expression to
x²·y·b²·c³ − y·a⁴·b²·c
Final expressions for two cases above are identical, which proves that tan(φ)=tan(ψ).

This, as we mentioned above, is a sufficient condition for a ray of light emitted from one focus point of an ellipse to any direction after reflecting from the ellipse to end up at the other focus point.

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

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

Geometry+ Hyperbola

Hyperbola is another class of curves on a plane with the following defining properties.

For each curve of this class (that is, for each hyperbola) there are two specific points called foci (plural of focus) or focuses and a specific positive real number called its major axis that is supposed to be smaller than the distance between its two foci, such that this hyperbola consists of all points on a plane (or is a locus of all points on a plane) with the difference of their distances to its two foci equaled to this hyperbola's major axis, a constant for all points on a hyperbola.

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

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 smaller than the length 2c of a focal line F₁F₂ between foci.
Consider a hyperbola on a Cartesian coordinate plane with foci to be at points F₁(−c,0) and F₂(0,c).

The following drawing represents a hyperbola defined by these two parameters, a half of focal distance c and a half of major axis a.

Point O is a midpoint of segment F₁F₂.
Points A and B are located on the focal line F₁F₂ on a distance a from point O.

Point A lies on this hyperbola because the distance from A to F₁ is c−a and the distance from A to F₂ is a+c. The difference of these distances is
(a+c)−(c−a)=2a,
which is a condition of any point on a hyperbola.

Similarly, point B lies on a hyperbola because BF₁=a+c, BF₂=c−a and
BF₁−BF₂=(a+c)−(c−a)=2a,
which is a condition of any point on a hyperbola.

Segment AB constitutes the major axis of this hyperbola.

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

The definition of hyperbola requires that difference between distances from point P(x,y) to 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 hyperbola 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 parameter b is called a minor axis and is defined by an equation b²=c²−a².

The transformation into this simple form is straightforward but, unfortunately, lengthy and boring.
√ (x+c)²+y² − √ (x−c)²+y² = 2a
√ (x+c)²+y² = 2a + √ (x−c)²+y²
Square both sides of an equation
x²+2xc+c²+y² =
= 4a²+4a√ (x−c)²+y²+x²−2xc+c²+y²
Cancel equal terms on both sides of an equation
2xc = 4a²+4a√ (x−c)²+y²−2xc
Separate a square root
into the left side of an equation
−4a√ (x−c)²+y² = 4a²−4xc
Divide both sides by 4 and square the equation
a²x²−2a²xc+a²c²+a²y² =
= a⁴−2a²xc+x²c²
Cancel equal terms on both sides of an equation and combine expressions for variables x and y in the right side of an equation
a²(c²−a²) = (c²−a²)x²−a²y²,
Using relation b²=c²−a²,
it can be simplified
b²x²−a²y² = a²b²
Divide both sides by a²b²
to get a canonical equation for hyperbola
x²/a² − y²/b² = 1

Our next task is to represent a hyperbola 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 one of the foci of a hyperbola. Then the distance from any point P(r,θ) of a hyperbola to this focus is r.
The difference between this value and a distance from P(r,θ) to another focus should be equal to 2a by absolute value.

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 a hyperbola because a distance to one of the foci in this case would be just equal to r.

Another reason for choosing a focus as an origin of polar coordinates is that, when we will analyze the trajectory of an object in the central gravitational field, we will derive the equation of this trajectory in polar coordinates originated at the source of gravity - a single focus point, the only fixed point we would know.

For definitiveness, let's concentrate on the right branch of a hyperbola and place the origin of polar coordinates at focus F₂.

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

This allows to get an equation of a hyperbola in polar coordinates with a focus F₂ as the origin of coordinates:
√r²+4c²+4r·c·cos(θ) − r = 2a

Now we have to transform it into a form of distance r of point P from an origin of polar coordinates (focus F₂) 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) = r + 2a
r²+4c²+4r·c·cos(θ) =
= r²+4a·r+4a²
4a·r−4r·c·cos(θ) = 4c²−4a²
r·[a−c·cos(θ)] = c²−a²
Therefore,
r= (c²−a²)/[a−c·cos(θ)] =
= a²(c²/a²−1)/[a−c·cos(θ)]

The ratio e=c/a is called eccentricity of a hyperbola and its greater than 1. Using it, the formula can be transformed to

r= a·(e²−1)/[1−e·cos(θ)]

The above is an equation of a hyperbola 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.

The distance r from focus F₂ is not always positive for all values of polar angle θ, which means that the ray from F₂ will not intersect a hyperbola in every direction.

For example, for θ=0 cos(θ)=1, and the denominator in the equation above will be negative, which means there is no point of a hyperbola in that direction.
On the other hand, for θ=π cos(θ)=−1, and
r= a·(e²−1)/[1+e] = a·(e−1) =
= c − a

In general, r is positive and defines a particular point on a hyperbola when 1−e·cos(θ) is positive, which is true for angle θ to be outside of interval from −arccos(1/e) to arccos(1/e), which is the same as from −arccos(a/c) to arccos(a/c).

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.

Physics+ Newton Laws, Planet Orbits: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Planet Orbits

Kepler's First Law states that all planets move around the Sun on elliptical orbits with the Sun in one of the two focus points of their orbits.
Kepler had come up with this law based on many years of observation, that is, experimentally.

On our quest to analytically prove the First Kepler's Law we begin with the proof that the trajectory of a point-mass moving in a central gravitational field produced by another point-mass fixed at some location in space is a flat curve, that is a curve all points of which belong to the same plane.

Consider a gravitational field produced by a point-mass M fixed at some point O in space.
Assume, a point-mass m is moving in this field and at some moment of time t is at point P.

Gravitational field is a central force field, and in the Central Force Field lecture of this chapter of this course we have proven the Angular Momentum Conservation Law.
Here is a short recap.

The point O, the fixed location of the source of gravitational field, is chosen as the origin of coordinates.
Let r(t) be a position vector OP of an object moving in this field at time t.
Let v(t) be the derivative of this position vector by time, that is a vector of velocity of our point-mass m:
v(t) = (dr/dt)(t) = r'(t).
Let a(t) be the derivative of a velocity vector by time, that is a vector of acceleration of our point-mass m, that is the second derivative of a position vector:
a(t) = v'(t) = r"(t).

We assume that our object does not move directly towards or from the center of gravity O, that is vectors r and v are not collinear.
If vectors of position and velocity are collinear, an object would move along a straight line towards or away from the source of gravity, which is a trivial case that we will not consider here.

The Angular Momentum Conservation Law states that vector L(t)=m·r(t)⨯v(t) is a constant of motion, that is L(t) does not depend on time, which is equivalent to its derivative by time is a zero-vector:
dL(t)/dt = L'(t) = 0

The proof is simple and is based on the fact that a vector product of collinear vectors is a zero-vector.
L'(t) = d[m·r(t)⨯v(t)]/dt =
= m·[r(t)⨯v'(t) + r'(t)⨯v(t)] =
= m·[r(t)⨯a(t) + v(t)⨯v(t)]
The first vector product is a zero-vector because, according to the Second Law of Newton acceleration of an object is collinear with the force, which is central, that is collinear with position vector r(t). So, r(t) and a(t) are collinear and, therefore, their vector product is a zero-vector.
The second vector product in the above expression is a zero-vector because the vector product by itself is always a zero-vector, since any vector is collinear to itself.

Since the derivative L'(t) is zero-vector, the Angular Momentum vector L(t) is independent of time and is fixed in space, which allows us to omit (t) from its value.

Recall that the result of a vector product of two vectors is a vector perpendicular to each of them.
Therefore,
L⊥r(t) and L⊥v(t).

Let's construct a plane in space going through vector r(t) (that is, points O and P) and vector v(t) (that is, through endpoint of this vector, thus having three points that define a plane). This plane will be perpendicular to a constant vector L.

Let's define a Cartesian system of coordinates in space with XY-plane to be the plane that we just constructed based on vectors r and v.
The Z-axis in our system of coordinates can be chosen coinciding with vector L that is perpendicular to XY plane.

Both vectors r(t) and v(t) lie in the constructed XY plane, and their Z-coordinate is zero.
Hence, the incremented position of our object at infinitesimally incremented time t+dt, that is a vector
r(t+dt) = r(t) + v(t)·dt
will also lie in the same plane, it Z-coordinate is zero, which means that the movement from point P will continue within the same XY plane.

Repeating this logic to consecutive incremental positions, we see that an entire trajectory will lie in our XY plane.
In other words, Z-coordinate of vector r(t) is zero at any time t.

This concludes the proof that a trajectory of an object in the central gravitational field is a curve lying in one plane.

Physics+ Newton Laws, Gravity Two Objects: UNIZOR.COM - Physics+ - Laws ...

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Gravitation of Two Objects

We recommend to refresh your knowledge of the gravitational field in three-dimensional space and the concept of field potential (scalar defined at each point of a field) using the material presented in this chapter in lectures
Field & Potential,
Work Lemmas,
Potential Theorem and
Problem 1.

The Problem 1 in this chapter of this course presented a proof that gravitation field of a point-mass M located at some point Q in three-dimensional space is conservative.

The work performed by this field when it moves some test object of mass m from point A to point B depends only on the location of these endpoints of its trajectory.

The potential of the field at point P is an amount of work needed by a field force to move a test object of a unit mass from point P to infinity.

Also, the gradient of a potential equals to the field intensity - the force (vector) acting on a test object of unit mass at any point.

Earlier in this course the gravitational force F(P) at some point P of the field was presented as
F(P) = F(r) = −G·M·m·r/r³
where r=QP is the relative position vector from the source of gravity Q to point P,
r is the magnitude (scalar) of a relative position vector r,
M is a mass of an object that is the source of gravitation,
m is a mass of a test object and
G is the Universal Gravitational Constant.

Of course, r²=x²+y²+z²
where (x,y,z) are Cartesian coordinates of vector QP (point Q is the source of gravity).

In coordinate form, components of the gravitational force vector F(P)=F(r) are:
F_x(P)=−G·M·m·x/(x²+y²+z²)^3/2
F_y(P)=−G·M·m·y/(x²+y²+z²)^3/2
F_z(P)=−G·M·m·z/(x²+y²+z²)^3/2

The potential of a gravitational field was determined as a scalar function defined at any point P of the field as
U(P) = U(r) = G·M/r =
= G·M/(x²+y²+z²)^1/2 =
= G·M·(x²+y²+z²)^−1/2

Its gradient, a vector of partial derivatives by coordinates, is
∇U(r)={∂U/∂x, ∂U/∂y, ∂U/∂z}
which should be equal to a vector of gravitational field intensity (the force per unit of test object's mass).

Indeed,
∂U(P)/∂x =
= −½G·M·(x²+y²+z²)^−3/2·2x =
= −G·M·x/(x²+y²+z²)^3/2 =
= (1/m)·F_x(P)

Similarly, partial derivatives by y and z produce the corresponding components of the gravitational field intensity - the force acting on a unit of mass of a test object.
∂U(P)/∂y =
= −G·M·y/(x²+y²+z²)^3/2 =
= (1/m)·F_y(P)
∂U(P)/∂z =
= −G·M·z/(x²+y²+z²)^3/2 =
= (1/m)·F_z(P)

We came to conclusion that
∇U(P)={∂U/∂x,∂U/∂y,∂U/∂z}=
= (1/m)·(F_x(P),F_y(P),F_z(P)) =
= E(P)
where E(P) is a vector of field intensity (the force per unit of test object's mass) at point P.

Let's determine a potential of the gravitational field produced by two point-masses M_a and M_b fixed at positions A and B correspondingly within some inertial frame of reference.

At any point P in space we can consider two positional vectors:
r_a=AP originated at A (first source of gravitation) and ending at point P and
r_b=BP originated at B (second source of gravitation) and ending at the same point P.

Assume, coordinates of all points involved are
A(x_a,y_a,z_a)
B(x_b,y_b,z_b)
P(x_p,y_p,z_p)
Then the coordinates of vectors r_a and r_b are
r_a = (x_p−x_a, y_p−y_a, z_p−z_a)
r_b = (x_p−x_b, y_p−y_b, z_p−z_b)

According to a principle of superposition, the combined gravitational force of our two sources acting on a test object of mass m positioned at point P should be a vector sum of two separate forces, that is
F(P) = −(G·M_a·m)·(r_a/r_a³) −
− (G·M_b·m)·(r_b/r_b³)

The field intensity (the force per unit of mass of a test object) at point P, therefore, is
E(P) = −(G·M_a)·(r_a/r_a³) −
− (G·M_b)·(r_b/r_b³)
Notice, the above formula involves the addition of vectors.

From the fact that field potential is an amount of work needed to move a test object of a unit mass from its initial location P to infinity and that the work is, generally speaking, an additive quantity intuitively follows that potential of two fields should be equal to a sum of potentials.

More rigorously, work is an integral of a scalar product of force by differential of distance
W = ∫[P,∞](F·dl)

If F=F_a+F_b, the work of a combined force is the sum of work of its components
W = ∫[P,∞]((F_a+F_b)·dl) =
= ∫_[P,∞](F_a·dl) + ∫_[P,∞](F_b·dl) =
= W_a + W_b
which proves additive characters of work and, consequently, of field potential in case a field is generated by more than one source.

The sum of potentials involves the addition of scalars, which is easier than addition of vectors of forces, so dealing with potentials is a preferable way.

Now consider a different scenario.

Assume, our two objects of mass M_a and M_b are not fixed at particular points A and B in space, but can move freely, and their corresponding position vectors are r_a and r_b.
Assume further that there are no external forces that can influence the movement of our system of two objects, and the only force that somehow affects their movement is the force of gravitation of one to another.
We will prove that in this case a center of mass of these two objects moves uniformly in space along a straight line with constant velocity.

The only force acting on the a-object with mass M_a is the gravitation force sourced at b-object with mass M_b.
This force is directed along a straight line connecting these two objects and its magnitude is
F_ab(t) = G·M_a·M_b/d²(t).
where d=|r_a − r_b| is a time-dependent distance between these objects.

In a more appropriate vector form it looks like
F_ab(t) =
= G·M_a·M_b·[r_a(t)−r_b(t)]/d³(t).

Similarly, the only force acting on the b-object with mass M_b is the gravitation force sourced at a-object with mass M_a.
This force is directed along a straight line connecting these two objects and its magnitude is
F_ba(t) = G·M_b·M_a/d²(t).

In a more appropriate vector form it looks like
F_ba(t) =
= G·M_a·M_b·[r_b(t)−r_a(t)]/d³(t).

So, the magnitudes of these forces, F_ab(t) and F_ba(t), are the same, but directions are opposite to each other, F_ab(t) is directed from a-object to b-object, while F_ba(t) is directed from b-object to a-object:
F_ab(t) = −F_ba(t)

Using the Newton's Second Law, knowing the forces and masses, we obtain equations for accelerations of these two objects, second derivatives of position vectors r_a" and r_b".
r_a" · M_a = F_ab
r_b" · M_b = F_ba
In the equations above and following it is assumed that all variables except masses are time-dependent functions. We just skip (t) for brevity.

If masses M_a and M_b are positioned at r_a and r_b, the center of mass is at position
r = (M_a·r_a + M_b·r_b)/(M_a+M_b)
The acceleration of this point is it's second derivative of position
r" = (M_a·r_a" + M_b·r_b)"/(M_a+M_b) =
= (F_ab + F_ba)/(M_a+M_b) = 0
because, as we stated above, F_ab(t)=−F_ba(t)

Since acceleration of the center of mass is zero, it velocity is a constant vector, which means that the center of mass in the absence of external forces moves along straight line with the same velocity.

Physics+ Newton Laws, Problem 5: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 5

Problem A

A point mass M is uniformly rotating within a plane with XY Cartesian coordinates on radius R and angular speed of rotation ω on a weightless unstretchable thread around a point that is the center of coordinates on this plane.
Determine the force F with which a thread holds this point mass on its fixed orbit (centripetal force) and express this force in Cartesian coordinates [F_x,F_y].
What is the direction and magnitude of this force?

Solution A

Position of a rotating point mass, as a function of time t, is a vector r(t) with the following coordinates
X(t) = R·cos(ω·t)
Y(t) = R·sin(ω·t)

Coordinates of the velocity vector V=[V_x,V_y] are, by definition, derivatives of coordinates of position
V_x(t) = −R·ω·sin(ω·t)
V_y(t) = R·ω·cos(ω·t)

Vector of velocity is perpendicular to a position vector. To confirm it, let's form a scalar product of these two vectors and check if it's equal to zero.
X(t)·V_x(t) + Y(t)·V_y(t) =
= −R²·ω·cos(ω·t)·sin(ω·t) +
+ R²·ω·sin(ω·)·cos(ω·t) = 0

Coordinates of the acceleration vector a=[a_x,a_y] are, by definition, derivatives of coordinates of the velocity vector
a_x(t) = −R·ω²·cos(ω·t)
a_y(t) = −R·ω²·sin(ω·t)

Since the centripetal force and acceleration are related, according to the Newton's Second Law, as F=M·a,
F_x(t) = M·a_x = −M·R·ω²·cos(ω·t)
F_y(t) = M·a_x = −M·R·ω²·sin(ω·t)

As we see, the vector of force is collinear and opposite in direction to the vector of position, that is, it's directed towards the center of rotation.

At the same time, the vector of force, being collinear to the vector of position, is perpendicular to the vector of velocity.

The magnitude of the centripetal force is
|F(t)| = M·R·ω²


Problem B

In the context of the Problem A above, consider that the plane of rotation is made of some frictionless material and positioned perpendicularly to the force of gravity. There is a hole in the center of rotation and a thread that holds the rotating point mass M goes down through that hole and is held by hand in a fixed position.
The radius of rotation is the same as above R and the angular speed is ω.
What mass m₀ should be attached to the bottom of a thread to replace the holding hand to maintain original radius and angular speed of rotation?
How the rotation will change (in terms of radius and angular speed) if we add another mass m₁ to the mass m₀ needed to maintain the original rotation?

Solution B

The results of the Problem A above regarding the direction and magnitude of the centripetal force F are:
1. The centripetal force is directed towards the center of the rotation, opposite to a position vector of the rotating object;
2. The magnitude of this force is |F(t)| = M·R·ω²

The condition of the Problem B is that the source of the centripetal force holding an object on a fixed orbit of rotation is the holding hand. If we replace it with the gravitation force of an object of mass m₀, the magnitude of the gravitational force must be equal to the one calculated above:
m₀·g = M·R·ω²
Therefore,
m₀ = M·R·ω²/g

Adding another mass m₁ to the above will change the picture. Both radius and angular frequency of rotation will change to correspond to a changed magnitude in the centripetal force.
At the same time, all forces acting on a rotating object are central, and that is a sufficient condition for the Law of Conservation of Angular momentum.
The above is sufficient to calculate a new radius and angular speed of a rotating object.

New magnitude of a centripetal force is (m₀+m₁)·g.
Therefore,
(m₀+m₁)·g = M·R_new·ω²_new
This is one equation with two variables to find:
R_new and ω_new.

Taking into account the Law of Conservation of the Angular Momentum will produce another equation, and that should be sufficient to find both unknown variables.

The magnitude of the original angular momentum of an object rotating on radius R with the angular speed ω is
L = |L| = |r⨯ p|
where
r is a position vector of a rotating object and
p is a linear momentum of a rotating object.

Linear speed v of an object rotating on a radius R with an angular speed ω is R·ω.
Since r is a radial vector and p=M·v is tangential to an orbit, these two vectors are perpendicular to each other, and the magnitude of their vector product is a product of their magnitudes.
Therefore,
L = M·R·v = M·R²·ω

After we changed the centripetal force the angular momentum must be the same. Therefore,
L = M·R²·ω =
= L_new = M·R²_new·ω_new
or
R²·ω = R²_new·ω_new

This produces the second equation we need to determine new radius and angular velocity. Hence, the system of two equations with two unknows is
(m₀+m₁)·g = M·R_new·ω²_new
R²·ω = R²_new·ω_new

Resolving the second equation for ω_new and substituting it into the first produces
ω_new = R²·ω/R²_new
(m₀+m₁)·g =
= M·R_new·(R²·ω/R²_new)² =
= M·R⁴·ω²/R³_new

From the above follows the value for a new radius of rotation
R³_new = M·R⁴·ω²/[(m₀+m₁)·g]
R_new =
= {M·R⁴·ω²/[(m₀+m₁)·g]}^⅓
The new angular speed can be derived from the above and the condition ω_new=R²·ω/R²_new
ω_new = R²·ω·
·{(m₀+m₁)·g/[M·R⁴·ω²]}^⅓

This rather complex formula needs to be verified somehow.
Let's consider a simple case of a radius shortened by half and check what additional mass m₁ is needed.
In a formula for R³_new above we will set R_new=½R.
(½R)³ = M·R⁴·ω²/[(m₀+m₁)·g]
from which follows
1 = 8M·R·ω²/[(m₀+m₁)·g]
(m₀+m₁)·g = 8M·R·ω² =
= 8m₀·g
Therefore,
m₁ = 7m₀.
In other words, to decrease the radius by half we have to add 7 times the mass that used to hold a rotating object on its original distance from a center of rotation, that is we have to increase the mass by factor 8.

As we know (see Problem A above), the magnitude of the centripetal force is
|F(t)| = M·R·ω²
Also known is that, when the radius of rotation shortens by half, the angular velocity increases by factor 4.
From these two statements follows that a centripetal force in this case increases by a factor of 8, which corresponds to our calculations above.