Physics+ Kepler's 1st Law: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Kepler's First Law

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 experimentally based on many years of observations.

Based on our knowledge of Physics, we will prove analytically a more general theorem about objects moving in the central gravitational field:
The trajectory of an object moving in the central gravitational field is a flat curve of the second order (ellipse, parabola or hyperbola) with the source of the gravitational field in the focal point of this curve.

The previous lecture Planet Orbits of this section of this course was about why the trajectory of an object in a central gravitational field is a flat curve with the center of a gravitational field lying within this plane.
So, we assume this topic is covered.

Also covered were the properties of ellipse, hyperbola and parabola, including their canonical equations in Cartesian and polar coordinates (see UNIZOR.COM > Math+ 4 All > Geometry).

Assume, a point-mass M is the source of a gravitational field and is located at point O in space.

Assume further that a point-mass m is moving in the gravitational field of point-mass M with no other forces involved, and at time t it's located at point P(t). Then its position relative to the source of gravity is vector OP(t)=r(t).

The velocity vector of an object is v(t)=r'(t) - a derivative of the position vector by time (here we will use a single apostrophe to signify the first derivative of a function by time and a double apostrophe - to signify its second derivative by time).

Let's choose some moment of time as the beginning of motion and consider it as an initial time t=0.
The initial position of our object moving in a central gravitational field will be r(0) and initial velocity will be v(0).

As we have proven in the previous lecture, an entire trajectory of the motion of point-mass m will be within the plane of motion going through the center of gravity O and vectors r(0) and v(0).
Consequently, at any moment of time t vectors of position r(t) and velocity v(t) will lie in the same plane.

Let's introduce a system of coordinates OXY on this plane with an origin O at the center of gravity and axis OX directed to a position of our object r(0) at time t=0.
Let's fix the directions of axes OX and OY within the plane of motion for now, but keep in mind that we might rotate axes in the future to simplify the final equations of a trajectory.

In such a system of XY-coordinates with point O as an origin the vector OP(t)=r(t) can be represented as a pair
r(t) = {x(t),y(t)}.

In these XY-coordinates the velocity vector v(t) can be represented as
v(t) = r'(t) = {x'(t),y'(t)}.

Our task is to analytically describe the trajectory of an object in a gravitation field as a curve inside a plane OXY.

The first foundation for theoretical derivation of the Kepler's First Law is the Second Law of Newton.

Provided the force F (a vector) acts on an object of mass m, this Law states that F = m·a
where a is a vector of object's acceleration, that is the first derivative from the object's velocity vector v or, equivalently, the second derivative of object's position vector r
a(t) = v'(t) = r"(t)

The second component is the Newton's Universal Law of Gravitation.

It assumes that an object of mass M is fixed in space at the origin of coordinates and is the only source of gravitation.

It further assumes that an object of mass m is moving in the gravitational field produced by an object of mass M and at time t is at position defined by a position vector r(t) that stretches along the line between the source of gravitation (the origin of coordinates) and the object's location.

Then the Law of Gravitation states that
(a) the vector of gravitational force F(t) produced by object of mass M acting on an object of mass m at any time t is collinear with the position vector r(t) and directed towards the object of mass M and
(b) the magnitude F(t) of the gravitational force F(t) is
F(t) = G·M·m/r²(t)
where
G is the universal gravitational constant whose value is 6.67430·10⁻¹¹ N·m²/kg² and
r(t) is the magnitude of the position vector r(t) - the distance of an object from the origin of coordinates.

Combining conditions (a) and (b) together, we can express the force of gravity in vector form as
F(t) = −G·M·m·r(t)/r³(t)

The Newton's Second Law, which connects a force and an acceleration, and the Universal Law of Gravitation, which defines the value of a gravitational force, allow to establish the differential equation that defines the motion of our object in a gravitational field
F(t) = m·a(t) = m·r"(t) =
= −G·M·m·r(t)/r³(t)
or, canceling mass m,
r"(t) = −G·M·r(t)/r³(t)

The fact that mass m can be canceled is quite remarkable. It says that different objects would follow the same trajectory in the gravitational field, if at some moment their positions and velocities are the same.
This is not supposed to be a surprise. Recall Galileo's experiments with different objects dropped from the Tower of Pisa. They would spend the same time to hit the ground regardless of their mass.
If, in addition to dropping them down, you give them the same horizontal speed, they would spend the same time going horizontally and fall on the same distance from a vertical, thus going along the same trajectory, regardless of mass.

The above equation is a vector differential equation that can be transformed into a system of two differential equations for Cartesian coordinates {x(t),y(t)} of vector r(t) with
r"(t) = {x"(t),y"(t)} and
r(t) = √x²(t)+y²(t)

In coordinate form this system of two differential equations with two unknown functions x(t) and y(t) looks like
x"(t)=−G·M·x(t)/r³
y"(t)=−G·M·y(t)/r³

Directly solving this system of differential equations is not an easy task to do.
But let's keep in mind that its solution will give us more than we asked for. We wanted to prove that objects in a gravitational field move along ellipses, hyperbolas or parabolas. We just needed a shape of trajectories, not a time-dependent position at every moment in time.

Having this smaller task in mind, we do not have to derive explicit functions x(t) and y(t) that are solutions to the above system of equations, we just have to concentrate on geometrical characteristics of trajectories.

Consider a position vector r from a fixed center of gravity (point O in our system of coordinates) to a position of an object at any moment of time.
Apparently, having a dependency of its length r=|r| on its angle θ from OX-axis is sufficient to determine geometric characteristics of its trajectory.
Basically, we can derive the geometric characteristics of a trajectory by analyzing its equation in polar coordinates {r,θ}. Deriving a function r=r(θ) is a much easier task than solving the above system of differential equations.
Let's concentrate on this smaller task.

Consider an angle ∠θ(t) from the positive direction of X-axis to position vector r(t).
Using this, coordinates of position are
x(t) = r(t)·cos(θ(t))
y(t) = r(t)·cos(θ(t))

At this point let's concentrate not on time-dependency of coordinates, but on their dependency on the angle θ. That will allow to derive the shape of a trajectory without knowing exactly the coordinates of an object moving along this trajectory at any moment of time.
Now the coordinates of an object's position, as functions of θ, are
x(θ(t)) = r(θ(t))·cos(θ(t))
y(θ(t)) = r(θ(t))·cos(θ(t))
Now we are not interested in function θ(t) and the above equations can be viewed as x(θ) = r(θ)·cos(θ)
y(θ) = r(θ)·cos(θ)

Using this representation, the original system of differential equations is
x"(t) = −G·M·cos(θ)/r²(θ)
y"(t) = −G·M·sin(θ)/r²(θ)

Recall from the lecture Planet Orbits of this part of the course that vector of Angular Momentum L=m·r·v of an object moving in a central gravitational field is a constant.
This vector L is perpendicular to both r and v and, therefore, is directed along the third dimension OZ-axis that is perpendicular to a plane of object motion.

The length L=|L| (a constant) of an Angular Momentum vector equals to L=m·r²·θ'.

Proof

In three dimensional XYZ space the position vector r and velocity vector v=r' have Z-coordinate equal to zero:
r = {r·cos(θ), r·sin(θ), 0}.
v = {r'·cos(θ)−r·sin(θ)·θ', r'·sin(θ)+r·cos(θ)·θ', 0}.

According to the rules of vector product in three-dimensional Cartesian coordinates, their vector product is
L/m = r⨯v = {0, 0, r·cos(θ)·[r'·sin(θ)+r·cos(θ)·θ'] −
− r·sin(θ)·[r'·cos(θ)−r·sin(θ)·θ']} =
= {0, 0, r²·[cos²(θ)+sin²(θ)]·θ'} =
= {0, 0, r²·θ'}
Therefore,
L = m·r²·θ'

Now we can express r² in terms of θ' and constants
r² = L/(m·θ')
Therefore, the system of our differential equations can be expressed only in terms of θ(t) as follows
x"(t) = −θ'·cos(θ)·(G·M·m/L)
y"(t) = −θ'·sin(θ)·(G·M·m/L)
In these equations θ(t) is an unknown function of time, while x"(t) and y"(t) are components of an acceleration vector and are the first derivatives of the components of a velocity vector {x'(t),y'(t)}.

Notice that expressions on the right side of both differential equations are full derivative by time of simple functions of θ, which is easy to integrate.
Therefore, integrating by time left and right sides of equations, we obtain
x'(t) = −sin(θ)·(G·M·m/L) + c₁
y'(t) = cos(θ)·(G·M·m/L) + c₂
or, substituting for simplicity a constant K for G·M·m/L,
x'(t) = −sin(θ)·K + c₁
y'(t) = cos(θ)·K + c₂
where constants c₁ and c₂ are defined by initial conditions at time t=0, when θ(0)=0:
x'(0) = −sin(0)·K + c₁
y'(0) = cos(0)·K + c₂
c₁ = x'(0)
c₂ = y'(0)−K
The initial components of velocity at x'(0) and y'(0) are given and K=G·M·m/L is a known constant.

The remaining task is to find the length of a position vector r(θ), as a function of angle θ from the above expressions for components of the velocity.

The components of a position vector, as functions of θ, are
x = r·cos(θ)
y = r·sin(θ)
differentiating these by time, we obtain
x'(t) = r'·cos(θ) − r·sin(θ)·θ'
y'(t) = r'·sin(θ) + r·cos(θ)·θ'

Comparing these with the expressions for components of a velocity vector above, give
r'·cos(θ) − r·sin(θ)·θ' =
= −sin(θ)·K + c₁
r'·sin(θ) + r·cos(θ)·θ' =
= cos(θ)·K + c₂

The next step is to get rid of r' and θ'.
The way to do it is to multiply the first equation by sin(θ), the second - by cos(θ) and subtract the first from the second.
r·cos²(θ)·θ' + r·sin²(θ)·θ' =
= cos²(θ)·K + c₂·cos(θ) +
+ sin²(θ)·K − c₁·sin(θ)
Now we can use trigonometric identity sin²(θ)+cos²(θ)=1 getting
r·θ' = K + c₂·cos(θ) − c₁·sin(θ)

Also, recall that L=m·r²·θ' and, therefore, θ'=L/(m·r²).
Using this result in the following
L/(m·r)=K+c₂·cos(θ)−c₁·sin(θ)

Resolving for r results in the expression of r as a function of θ - exactly what we are looking for
r = 1/[A+B·cos(θ)−C·sin(θ)]
where all coefficients are known constants:
A = K·m/L
B = c₂·m/L = [y'(0)−K]·m/L
C = c₁·m/L = x'(0)·m/L
K = G·M·m/L

The above looks like a complete solution r(θ), but we can improve it a little.
The expression B·cos(θ)−C·sin(θ) can be transformed into D·cos(θ+φ) as follows
B·cos(θ)−C·sin(θ) =
= √B²+C²·[cos(θ)·B/√B²+C² −
− sin(θ)·C/√B²+C²]
We can always find an angle φ such that
B/√B²+C² = cos(φ)
C/√B²+C² = sin(φ)
which results in
B·cos(θ)−C·sin(θ) =
= √B²+C²·[cos(θ)·cos(φ)−
+sin(θ)·sin(φ)] =
= √B²+C²·cos(θ+φ)

Therefore,
r = 1/[A+√B²+C²·cos(θ+φ)]
Let's do a small correction in the direction of the X-axis.
Originally, we placed the beginning of coordinates point O at the point where the source of gravitation is located. Then we chose OX axis directed to the initial position of our object at time t=0.
Now, knowing angle φ we can change the direction of the X-axis by turning it by an angle φ.
As a result, in this new coordinate system the equation of the trajectory looks even simplier:
r = 1/[A+√B²+C²·cos(θ)]
which, depending on the values of all constants involved, is an equation in polar coordinates of either an ellipse or a hyperbola, or a parabola, as explained in the previous lecture dedicated to 2-nd order curves.

End of proof.

Physics+ 2nd Order Curves: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
2nd-order Curves

We talk here about second-order curves because they describe the trajectories that objects move along in a central gravitational field.
These curves were discussed in the UNIZOR.COM course Math+ 4 All in the part Geometry, where you can find lectures on ellipse, hyperbola and parabola, their defining characteristics and equations in Cartesian and polar coordinates.

Here is a quick recap of this material.

Ellipse
Ellipse is a locus of points on a plane that satisfy the following condition.
The sum of distances from each such point P to two given points F₁ and F₂ (called foci) is equal to a given positive real number called its length of major axis (2a on a graph below), that is supposed to be greater than the distance between its two foci (2c on a graph below).

The value 2b, where b²=a²−c², is called the length of minor axis.
The equation of an ellipse in Cartesian coordinates (x,y) with X-axis coinciding with the line between the foci and origin of coordinates is a midpoint between foci is
x²/a² + y²/b² = 1
The equation in polar coordinates (r,θ) with an origin at one of the ellipse' foci and a base axis coinciding with the line between the foci is
r= a·(1−e²)/[1−e·cos(θ)]
where the ratio e=c/a is called eccentricity of an ellipse and it's always less than 1.

Hyperbola
Hyperbola is a locus of points on a plane that satisfy the following condition.
The difference of distances from each such point P from two given points F₁ and F₂ (called foci) is equal by absolute value to a given positive real number called its length of major axis (2a on a graph below), that is supposed to be less than the distance between its two foci (2c on a graph below).

The value 2b, where b²=c²−a², is called the length of minor axis.
The equation of a hyperbola in Cartesian coordinates (x,y) with X-axis coinciding with the line between the foci and origin of coordinates is a midpoint between foci is
x²/a² − y²/b² = 1
The equation in polar coordinates (r,θ) with an origin at one of the hyperbola' foci and a base axis coinciding with the line between the foci is
r= a·(e²−1)/[1−e·cos(θ)]
where the ratio e=c/a is called eccentricity of a hyperbola and it's always greater than 1.

Parabola
Parabola is a locus of points on a plane that satisfy the following condition.
Each such point P is equidistant from a given point F (called focus) and a given straight line d (called directrix) with the distance between a focus and a directrix being a given positive number (2c on a graph below).
The equation of an ellipse in Cartesian coordinates (x,y) with X-axis coinciding with the perpendicular line from a focus to a directrix and origin of coordinates is a midpoint between focus and directrix
y² = 4c·x
The equation in polar coordinates (r,θ) with an origin at parabola' focus and a base axis coinciding with the perpendicular line from a focus to a directrix is
r= 2c/[1−cos(θ)]

Generally speaking, a second-order curve is defined in Cartesian coordinates as a locus of (x,y) points on a plane that satisfy a general equation of a second order for two variables
Ax²+Bxy+Cy²+Dx+Ey+F=0

We can show that any second-order curve belongs to one of the three types presented above, it's either ellipse or hyperbola, or parabola.

If coefficient B at xy is zero, the expression above can be easily converted into one of the three canonical equations by shifting and stretching coordinates (which shifts and stretches the graph without changing its type, an ellipse will remain an ellipse etc.) and/or exchanging places of x and y (which changes the orientation of a graph from horizontal to vertical).
If coefficient B is not zero, we can turn the system of coordinates in such a way that the equation of the same curve in the new system will not contain a non-zero coefficient at xy.
Details of the transformation of a general equation into one of three canonical forms are at the end of these notes.

The main purpose of the above recap of the properties of second-order curves is to analyze the motion of a point-mass object in the central gravitational field of another point-mass object fixed at the origin of an inertial reference frame.

More precisely, we will prove in the next lecture that a trajectory of an object in a central gravitational field is some second-order curve (ellipse, hyperbola or parabola) depending on its position and velocity relative to the center of gravitational field at initial moment of time t=0.

To prove it, we will, primarily, rely on representation of second-order curves in polar coordinates.
The reason for this is that all polar coordinate representations of three types of curves (ellipse, hyperbola or parabola) are very similar and differ only in certain parameters.
If we derive the equation of a trajectory in polar coordinates as
r(θ) = A/[B+C·cos(θ)],
where A, B and C depend only on position and velocity of our object relative to a center of the gravitational field at the beginning of motion, we can say that a trajectory would be either ellipse or hyperbola, or parabola.

Moreover, analyzing the values of position and velocity of our object relative to a center of the gravitational field at the beginning of motion, we will be able to predict the type of trajectory it will take.


Transformation of General Second-Order Equation into One of Three Canonical Forms

The general form of a second-order curve on an XY-plane as an equation in Cartesian coordinates is
A·x²+B·x·y+C·y²+D·x+E·y+F=0
It's called 'second-order' because X- and Y-coordinates participate in this equation as polynomial with combined exponent 2.

Our task is to analyze the shape of a curve defined by such an equation depending on coefficients A, B. C etc.

Let's start with a simpler case of B=0.
Then our equation looks like
A·x²+C·y²+D·x+E·y+F=0

In this form we will analyze the shape of a curve in two different cases:
1. Either A or C is zero.
2. Both A and C are not equal to zero.
Both of them cannot be equal to zero because then the equation would not be of a second order.

1. CASE Either A or C is zero

We will consider two separate subcases.

Subcase A≠0 but C=0

In this case we can easily transform the original equation into a form
A·(x+½D/A)²+E·y=K
where
K = D²/(4A) − F

Let's shift our coordinate system parallel to itself moving the origin to point (½D/A,0).
This transformation will not change the shape of a curve, but will allow to express it in a simple form in new coordinates
A·x² + E·y = K
or
E·y = −A·x² + K
which is a parabola for any E≠0.
For example, for E=1, A=−1, K=−4 this parabola looks like this:
Physical trajectory described by this equation would be when an object moves towards but slightly off the source of gravity. The gravitational force will turn it around the origin, but it will not be sufficient to keep it on an orbit around a center, and an object will move away from the center of gravitation to infinity.

Subcase A=0 but C≠0

In this case we can easily transform the original equation into a form
C·(y+½E/C)²+D·x=K
where
K = E²/(4C) − F
Let's shift our coordinate system parallel to itself moving the origin to point (0,½E/C).
This transformation will not change the shape of a curve, but will allow to express it in a simple form in new coordinates
C·y²+D·x=K
or
D·x= −C·y² + K
which is a parabola for any D≠0.
For example, for D=1, C=−1, K=−4 this parabola looks like this:


2. CASE A≠0 and C≠0

In this case we can easily transform the original equation into a form
A·(x+½D/A)²+C·(y+½E/C)²=K
where
K = D²/(4A) + E²/(4C) − F

Let's shift our coordinate system parallel to itself moving the origin to point (½D/A,½E/C).
This transformation will not change the shape of a curve, but will allow to express it in a simple form in new coordinates
A·x²+C·y²=K

Subcase A and C are
both positive or both negative

If K is of the same sign as A and C, it's an ellipse.
For example, if A=4, C=9 and K=36, this ellipse looks like this
which, from the physical standpoint, would be an orbit of an object moving around a central point-mass.

If K=0, the only point (0,0) fits the equation. You can consider it as a degenerate ellipse with zero dimension in any direction.
Physical meaning of this is that this is a trajectory of an object in central gravitational field with its initial position at its center and initial velocity zero.
So, it's stuck at the center where the source of gravity is located and will not escape without any initial velocity, which is meaningless for our analysis.

Finally, if K is of the opposite sign to A and C, there are no points that satisfy the equation.
In a physical sense it means that there is no object in the gravitational field.

Subcase A and C are
of opposite signs

The curve will be a hyperbola

(if signs of A and K are the same)
or

(if signs of C and K are the same)
From the physical standpoint the trajectory like that will be of an object moving fast enough to overcome the force of gravitation and, after its trajectory has been curved by this force, still manage to fly away from the source of gravitation force to infinity.

If K=0 we will have two straight lines defined by equation
A·x² = C·y²
which can be simplified to
y=±√(A/C)x
with the graph looking like this The physical meaning of this trajectory corresponds to an object moving straight towards the center of gravity or directly from it.

All the above cases assumed that in the original equation of a second-order curve
A·x²+B·x·y+C·y²+D·x+E·y+F=0
the coefficient B=0.

What if it's not?

We will show that with a proper turning of our system of Cartesian coordinates (which would not change the shape of a curve, only its representation as a second-order equation in Cartesian coordinates) we can change the equation into a form with B=0.
That would reduce a general task of analysis of the shape of any second-order curve to a case analyzed above, when B=0.

Assume, we can turn our Cartesian system of coordinates by angle β.
Recall that the new system will have coordinates u and v that are related to old coordinates x and y in the original system as
u = x·cos(β) + y·sin(β)
v = −x·sin(β) + y·cos(β)
or in vector form

u v

=

x y

⨯

cos(β) sin(β) −sin(β) cos(β)

Old coordinates x and y can be expressed in terms of new ones u and v by turning the new system of coordinates by angle −β:
x = u·cos(β) − v·sin(β)
y = u·sin(β) + v·cos(β)
or in vector form

x y

=

u v

⨯

cos(β) −sin(β) sin(β) cos(β)

Now we can substitute new coordinates u and v into original second-order equation for a curve getting another second order equation for the same curve in terms of new coordinates.

Can we find such an angle β that all mixed terms that contain a product u·v will cancel each other?
If yes, that would prove that in some coordinate system the equation of a second-order curve contains no terms with product of different coordinates, like x·y.

In terms of u and v the new equation will contain terms with u², u·v, v², u, v and a constant.
Let's collect only terms with a product of different coordinates u·v that we want to nullify.

These are:
A·x² → −2·A·u·v·cos(β)·sin(β)
B·x·y → B·u·v·(cos²(β)−sin²(β))
C·y² → 2·C·u·v·sin(β)·cos(β)
We would like the sum of them to be zero.
After obvious usage of the formulas for sin and cos of double angle, it means that we have to find such an angle β that
A·sin(2β)=B·cos(2β)+C·sin(2β)

The above trigonometric equation can be easily solved by dividing by cos(2β) getting
tan(2β) = B/(A−C)
β = ½arctan[B/(A−C)]
It works only for A≠C. In case A=C the equation for β looks like
B·cos(2β) = 0
from which follows
β=π/4

Therefore, by rotating the system of Cartesian coordinates we can find the one where the second-order equation of our curve will have no terms with a product of different coordinates x·y.

After that we can use the technique presented above to shift the origin of our system of coordinates to simplify this equation even further to a form
A·x² + E·y = K (parabola open along Y-axis) or
C·y² + D·x = K (parabola open along X-axis) or
A·x² + C·y² = K (ellipse if all multipliers are positive or hyperbola if A and C are of opposite signs)

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

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

Geometry+ Parabola 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 parabola is that if its contour is reflective, a ray of light emitted from its focus will be reflected parallel to its axis of symmetry regardless of the direction it was sent.

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 a parabola with focus F(c,0), directrix d, point P(x,y) on this parabola and tangential BC to a parabola at point P(x,y).
The ray of light is emitted from a focus F(c,0), goes to point P(x,y) and reflects along PS.
Because of the laws of reflection, angles ∠FPB and ∠SPC are equal, they are marked as α.
If we prove that these angles α are also equal to angle ∠ψ the tangential at point P line BC makes with the axis of symmetry of parabola BX, it will prove the parallelism of reflected ray PS with line BC, regardless of position of point P on a parabola.

Our plan to prove it is to prove that tan(α)=tan(ψ), from which the equality α=ψ follows because function tan() is monotonic for these angles.

Since sum of angles of a triangle ΔBPF equals to π,
α + ψ + (π−φ) = π
Therefore,
α = φ − ψ
We can calculate tan(α) using a formula for tangent of difference between angles, we can express tan(α) in terms of tan(φ) and tan(ψ).
tan(α) =
= [tan(φ)−tan(ψ)]/[1+tan(φ)·tan(ψ)]

Angle ∠φ=∠XFP.
Knowing coordinates of points P(x,y) and F(c,0), it's easy to calculate
tan(φ) = y/(x−c)

As we know, a tangent of an angle between a tangential line to function f(x) at some point and X-axis is a function's derivative f'(x) at that point.
Therefore, tan(ψ)=y'(x)
Since an equation of a parabola is y²=4c·x, differentiating this equation we get
2y·y' = 4c
Hence, y' = 2c/y = tan(ψ).

Now we have all the components to calculate tan(α):
tan(α) =
= [y/(x−c)−2c/y]/[1+(y/(x−c))·(2c/y)]
Let's simplify this expression.
Its numerator equals to
(y²−2c·x+2c²)/(x·y−c·y)
But for each point of a parabola y²=4c·x.
Use it in the formula above, getting the same numerator as
(4c·x−2c·x+2c²)/(x·y−c·y) =
= 2c·(x+c)/(x·y−c·y)
The denominator in the formula above can be simplified as
1 + (y/(x−c))·(2c/y) =
= 1 + 2c·y/((x−c)·y) =
= (x·y−c·y+2c·y)/((x−c)·y) =
= y·(x+c)/(x·y−c·y)

Dividing the numerator
2c·(x+c)/(x·y−c·y)
by denominator
y·(x+c)/(x·y−c·y)
we get the value of tan(α) as
tan(α) = 2c/y

But this is the same value as tan(ψ), which proves that angles ∠α and ∠ψ are equal, which, in turn, proves that lines PS (reflected ray of light) and FX (axis of symmetry of a parabola) are parallel regardless of the position of point P on a parabola.

Therefore, all the rays from the parabola's focus directed in any direction will be reflected in one direction - parallel to the axis of symmetry of this parabola.

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

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

Geometry+ Parabola

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

For each curve of this class (that is, for each parabola) there is one specific point called its focus and a specific straight line called its directrix not going through a focus, such that this parabola consists of all points on a plane (or is a locus of all points on a plane) equidistant from its focus and directrix.

As we see, the position of a focus point F and a directrix d (should not go through focus F) uniquely identifies a parabola.

Obviously, a midpoint of a perpendicular from focus F onto directrix d belongs to a parabola defined by them.
From this point we can draw a parabola point by point maintaining equality of the distance from each point to both a focus and a directrix.

Choosing an X-axis perpendicular to a directrix with an origin of coordinates 0 at midpoint between a focus and a directrix, we will have the following picture of a parabola on a coordinate plane.
We can derive an equation that defines this parabola using the main characteristic property of every point on this parabola to be equidistant from focus and a directrix.

The distance from any point P(x,y) on a parabola to a focus F that has coordinates (c,0) can be calculated using the known formula of a distance between two points.
The distance from point P to a directrix is calculated along a perpendicular from point P to a directrix that is parallel to X-axis.

If the distance between a focus and a directrix is 2c, as on a drawing above, the equality of the distances to a focus and a directrix is
√(x−c)²+(y−0)² = x−(−c)

We can simplify this as follows
(x−c)²+(y−0)² = (x+c)²
y² = (x+c)²−(x−c)²
y² = 4c·x

The only parameter that defines the shape of a parabola is c (half of a distance between a focus and a directrix), which is called the focal distance (or focal length) of a parabola.

The perpendicular from a focus to a directrix is an axis of symmetry of a parabola.
The midpoint of this perpendicular is called a vertex of a parabola.

Another item of interest is parabola's focal width. This is the length of a segment drawn through a focus parallel to a directrix with its endpoints being its intersections with a parabola.
Using a formula of a parabola y²=4c·x we determine that if x=c then y²=4c² and y=2c, which is a half of a segment described above.
Therefore, parabola's focal width is 4c.

Let's derive a formula r=r(θ) for a parabola in polar coordinates with an origin at its focus and base axis perpendicular to a directrix.
The distance from point (r,θ) to a focus is, obviously, r.
The distance from this point to a directrix is
AB = AF + FB = 2c+r·cos(θ).
From this the equation of a parabola in polar coordinates is

r = 2c/[1−cos(θ)]

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

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

Geometry+ Hyperbola Asymptotes

Let's start with a definition of an asymptote.
From less rigorous standpoint an asymptote to a curve is a straight line to which a curve goes infinitesimally close as a point of observation on our straight line moves to infinity.

More precisely, we assume that there is a plane with Cartesian coordinates on it.
There is some relationship between coordinates f(x,y)=0 that defines a curve on a coordinate plane that graphically represents it.
This relationship can be expressed as a function of one coordinate of another, like y=1/x or not in a functional way, like x·y=1.

In most cases we will consider the functional representation of dependency between coordinates, like y=f(x).

Vertical (parallel to Y-axis) asymptote to this function at point x=c where c, is a constant, is a line parallel to Y-axis and crossing X-axis at point x=c such that any one particular or any combination of the conditions below is true
lim_x→c f(x) = +∞ or = −∞
lim_x→c+0 f(x) = +∞ or = −∞
lim_x→c−0 f(x) = +∞ or = −∞
where x→c means that an argument x tends to value c in any possible way,
x→c+0 means that an argument x tends to value c while always being greater than c (tending from the right),
x→c−0 means that an argument x tends to value c while always being less than c (tending from the left).

The value of a function at exact point x=c might or might not be defined and is irrelevant.

Here is an example of a function with a vertical asymptote that satisfies the condition
lim_x→c f(x) = +∞
where f(x)=1/(x−2)² and c=2. Here is an example of a function with a vertical asymptote that satisfies the condition
lim_x→c+0 f(x) = −∞
where f(x)=ln(x−2) and c=2. Here is an example of a function with a vertical asymptote that satisfies the conditions
lim_x→c+0 f(x) = +∞
lim_x→c−0 f(x) = −∞
where f(x)=1/(x−2) and c=2.
In any of the above cases the distance between a point on a vertical asymptote and a point on a curve in the direction perpendicular to an asymptote (that is, along a line parallel to X-axis) tends to zero as an argument tends to point c.

Horizontal (parallel to X-axis) asymptote has a different definition.
First of all, it's about a curve that represents a function getting infinitesimally close to a horizontal line as an argument tends to +∞ or −∞.
Obviously, we assume that this function is defined on an infinite segment.
Line y=c (c is a constant) is an asymptote for a function y=f(x) if
lim_x→+∞f(x) = c or
lim_x→−∞f(x) = c
Here is an example of a function with a horizontal asymptote that satisfies the condition
lim_x→−∞ f(x) = c
where f(x)=2+e^x and c=2. The distance between a point on a horizontal asymptote and a point on a curve in the direction perpendicular to an asymptote (that is, along a line parallel to Y-axis) tends to zero as an argument x tends to infinity.

Oblique (or slanted) asymptote for some function y=f(x) is a straight line L described by an equation y=a·x+b (a≠0) which is not parallel to any of the axes of a coordinate system, such that a distance from a point P[p,q]∈L (that is, q=a·p+b) to a point S[x,y] on a curve representing a graph of our function (that is, y=f(x)) in a direction perpendicular to line L (that is, PS⊥L) will be an infinitesimal variable, as a point P on a line L goes to infinity (that is if p→+∞ or p→−∞).

Here is an example of an oblique asymptote that satisfies a condition of infinitesimal closeness of a graph of a function to an asymptote as a value of an argument goes to infinity (positive or negative).

If a function is represented by a graph, and we expect that some straight line is an asymptote, the distance between any point on an asymptote and a graph of a function in a direction perpendicular to an asymptote must tend to zero as our point moves to infinity along an asymptote.
Let's derive the relationship between a function y=f(x) and its asymptote y=a·x+b.
Since PR=QR·cos(φ),
it will tend or not tend to zero exactly as QR=f(x)−a·x−b.
Therefore, if
lim_x→∞[f(x)−a·x−b]=0
then y=a·x+b is an asymptote.

Let's use this result to find asymptotes of a hyperbola defined by an equation
x²/a² − y²/b² = 1
Let's concentrate on the top right branch of this hyperbola, that we can express as function
y = √x²·b²/a² − b²

Without subtracting a constant b² under a square root, the equation above would look like y=(b/a)·x - straight line.
This prompts us that a straight line y=(b/a)·x might be an asymptote.
Let's check it out.

lim_x→∞[√x²·b²/a²−b²−(b/a)·x]=
multiply and divide by √x²·b²/a²−b²+(b/a)·x
= 0
because we will have a constant −b² in the numerator and a function in the denominator that monotonically tends to infinity.

Analogously, a line y=−(b/a)·x is an asymptote to the top left side of a graph.

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.