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).