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.