Areas of Conics

Most of us know area formulas for basic geometrical objects, such as circles, squares, rectangles, and triangles.  (The area of a circle is A = πr², the area of a square is A = s², the area of a rectangle is A = lw, and the area of a triangle is A = ½bh.)  Others of us may even remember learning area formulas of other geometrical objects in high school geometry, such as parallelograms, trapezoids, and equilateral triangles.  (The area of a parallelogram is A = bh, the area of a trapezoid is ½(b₁ + b₂)h, and the area of an equilateral triangle is ¼√3s².)

Other than the circle, the area of conics are neglected in most geometry textbooks, even though the Cartesian equation for each conic are included in most algebra textbooks.  The area of a parabola is A = ²/₃bh (where b and h are the base and height of the rectangle that contains it).

The area of an ellipse is A = πab (where a and b are the major and minor axes of the ellipse). 

The area of a hyperbola is A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|) (where a and b are the axes of the hyperbola and h is the height).
 

The area formulas for the parabola and ellipse are relatively simple, the equation for the hyperbola is not.

To prove the area of a parabola, we can use integration.  The Cartesian equation for a parabola with its vertex at the origin is y = ax², which means another point on the parabola would be (b, ab²). 

The area of a rectangle with opposite vertices on the origin and (b, ab²) is A = b∙ab² = ab³.  The area of the parabola inside this rectangle is A = ∫₀^b ab² – ax² dx = ab²x – ¹/₃ax³] ₀^b = ab²b – ¹/₃ab³ = ab³ – ¹/₃ab³ = ²/₃ab³.  (The left half of the parabola would be the same by symmetry.)  Therefore, the area of the parabola is ²/₃ of the area of the rectangle that contains it, or A = ²/₃bh.

To prove the area of an ellipse, we can compare its Cartesian equation to the Cartesian equation of a circle.  The Cartesian equation for an ellipse centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 + ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(a² – x²), and the Cartesian equation for a circle centered at the origin where a is a radius is x² + y² = a², which rearranged is y = √(a² – x²), so the difference in heights is a factor of ^b/_a. 

    

So if the area of a circle is A = πr², or in this case A = πa², the area of an ellipse must be A = ^b/_aπa² = πab.  (Technically speaking, the area of an ellipse is 4∫₀^{a b}/_a√(a² – x²)dx = ^b/_a 4∫₀^a √(a² – x²)dx = ^b/_aπa² = πab.)

To prove the area of the hyperbola, we can use integration once again.  The Cartesian equation for a horizontal hyperbola centered at the origin where a and b are the major and minor axes is ^{x^2}/_a^2 – ^{y^2}/_b^2 = 1, which rearranged is y = ^b/_a√(x² – a²). 

The area is then A = 2∫_a^{h+a b}/_a√(x² – a²)dx.  Using a right triangle with legs a and √(x² – a²) and hypotenuse x, tan q = ^{√(x^2 – a^2)}/_a or √(x² – a²) = a tan q, and sec q = ^x/_a or x = a sec q, which means dx = a sec q tan q dq. 

Using substitution, A = 2∫_sec-1(a)^{sec-1(h+a) b}/_a a tan q a sec q tan q dq = 2ab∫₀^{sec-1((h+a)/a)} tan² q sec q dq = 2ab∫₀^{sec-1((h+a)/a)} (sec² q – 1)sec q dq = 2ab∫₀^{sec-1((h+a)/a)} sec³ q – sec q dq.  Integrating results in 2ab[½ sec q tan q + ½ ln|sec q + tan q| – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}  = ab[sec q tan q – ln|sec q + tan q|] ₀^{sec-1((h+a)/a)}.  Since sec(sec^-1(x)) = x and tan(sec^-1(x)) = √(x² – 1), A = ab(^h+a/_a√(( ^h+a/_a)² – 1) – ln|^h+a/_a + √(( ^h+a/_a)² – 1)|).

Although the area of a hyperbola is not an easy formula, both area formulas for a parabola and an ellipse are relatively simple.  The area of an ellipse can also be used to show the special case of a circle where a = b = r (A = πab = π∙r∙r = πr²), and the area formulas of conics nicely correspond with the Cartesian equations of conics learned in algebra.  It makes one wonder why the area formulas for the parabola and ellipse are typically not included in a high school geometry curriculum or textbooks.

Trapezoidal Prime Numbers

In Karl Sagan’s Contact, aliens communicate with human beings by sending out a sequence of sounds in the pattern of prime numbers.  This idea is based upon the principle that two isolated intelligent races ought to use mathematics, the common language of the universe, to communicate.  However, our mathematics is in base ten because humans have ten fingers, but an alien race with more or less fingers may be in a different base.  To overcome this difference, primes can be used, because they are the same in any numerical base.

Karl Sagan’s Contact

But are prime numbers necessarily universal?  Recall that a prime number is a number that is only divisible by one and itself.  Thinking about it geometrically, a prime number of objects can be arranged into a rectangle in exactly one way.  For example, 5 objects can be arranged in a rectangle in exactly one way (5 x 1 or 1 x 5) so 5 is a prime number, but 6 objects can be arranged in a rectangle in more than one way (6 x 1 and 3 x 2) so 6 is not prime. 

5 objects can only make a 5 x 1

rectangle, so 5 is prime

 
6 objects can make a 6 x 1 rectangle and

a 3 x 2 rectangle, so 6 is not prime

Could an alien race develop a different mathematical system not based on rectangles but rather on a different shape?

When we were little, my grandparents used to collect their pennies in a jar and give them to us as a treat.  We would gleefully dump the penny jar and make groups of 50 so that we could roll them up and take them to the bank.  To make a group of 50, we would usually arrange the pennies so that there were 5 rows of 10 pennies each. 

50 pennies in the shape of a rectangle

But instead of making a rectangular grid of coins, it saved more space to offset each row a little bit, so that the bottoms of the coins of one row interlocked the tops of the coins of the next row, and this formed a parallelogram instead. 

50 pennies in the shape of a parallelogram

It also felt natural to form trapezoids with the coins, where each successive row was one coin longer than the next row, and a trapezoid with a top row of 8 and a bottom row of 12 also added to 50.

50 pennies in the shape of a trapezoid

Let’s try to build a numbering system based on trapezoids instead of rectangles.  We can define x * y as a trapezoid with a top row of x units and a bottom row of y units, where each row is one longer than the next row (if the top row is shorter than the bottom row), or one shorter than the next row (if the top row is longer than the bottom row).  Therefore, 2 * 4 = 2 + 3 + 4 = 9, and 6 * 3 = 6 + 5 + 4 + 3 = 18. 

2 * 4 = 2 + 3 + 4 = 9

6 * 3 = 6 + 5 + 4 + 3 = 18

With this in mind, we can construct the following x * y table:

	1	2	3	4	5	6	7	8	9	10
1	1	3	6	10	15	21	28	36	45	55
2	3	2	5	9	14	20	27	35	44	54
3	6	5	3	7	12	18	25	33	42	52
4	10	9	7	4	9	15	22	30	39	49
5	15	14	12	9	5	11	18	26	35	45
6	21	20	18	15	11	6	13	21	30	40
7	28	27	25	22	18	13	7	15	24	34
8	36	35	33	30	26	21	15	8	17	27
9	45	44	42	39	35	30	24	17	9	19
10	55	54	52	49	45	40	34	27	19	10

From the table, we can identify some properties that the trapezoidal system has and doesn’t have.  It has an identity property of itself, because x * x = x.  It also has the commutative property because x * y = y * x.  However, it does not have the associative or distributive properties like the rectangular system.  We can show that it is not associative by giving a counter-example of (2 * 3) * 4 ≠ 2 * (3 * 4), because (2 * 3) * 4 = 5 * 4 = 9 but 2 * (3 * 4) = 2 * 7 = 27.  We can also show that it is not distributive by giving a counter-example of 2 * (3 + 4) ≠ 2 * 3 + 2 * 4, because 2 * (3 + 4) = 2 * 7 = 27 but 2 * 3 + 2 * 4 = 5 + 9 = 14.  Admittedly, the trapezoidal system is probably not one to build a civilization upon, because it’s hard to imagine a system without these properties to advance in science.

The most amazing thing about the trapezoidal system, however, are its prime numbers.  In the rectangular system, a prime number is a number that can be arranged into a rectangle in exactly one way, so in the trapezoidal system, a prime number is a number that can be arranged into a trapezoid in exactly one way.  Using the chart, we see that 2 is a trapezoidal prime, because 2 objects can be arranged in a trapezoid in exactly one way (2 * 2).  The number 3 is not a trapezoidal prime, because 3 objects can be arranged in more than one way (3 * 3 and 1 * 2). 

2 objects can only make a 2 * 2 trapezoid,

so 2 is a trapezoidal prime

3 objects can make a 3 * 3 trapezoid and a 1 * 2 trapezoid,

so 3 is not a trapezoidal prime

As we go through this process, we see that the first few trapezoidal primes are 2, 4, 8, 16, 32, etc. In other words, all trapezoidal primes are powers of 2.  So rectangular prime numbers follow no known pattern, but trapezoidal prime numbers do follow a pattern!

The fact that all trapezoidal primes are powers of 2 is easier to observe than to prove.  First of all, we need to be able to convert the trapezoidal system in terms of the rectangular system.  If x ≤ y we can place an x * y trapezoid and a y * x trapezoid beside each other (a trapezoid and an upside-down identical trapezoid), and this makes a parallelogram with a height of y – x + 1 and a base of x + y. 

x * y = ½(y – x + 1)(x + y) if x ≤ y

Since one x * y trapezoid makes up half of the parallelogram with a height of y – x + 1 and a base of x + y, x * y = ½(y – x + 1)(x + y) if x ≤ y.  A similar argument can be made that x * y = ½(x – y + 1)(x + y) if x ≥ y. 

Secondly, we can prove that all numbers can be arranged as a trapezoid in at least one way.  When x = y, x * y = ½(x – x + 1)(x + x) = ½(1)(2x) = x.  This can be seen in the diagonal row of the table. 

	1	2	3	4	5	6	7	8	9	10
1	1	3	6	10	15	21	28	36	45	55
2	3	2	5	9	14	20	27	35	44	54
3	6	5	3	7	12	18	25	33	42	52
4	10	9	7	4	9	15	22	30	39	49
5	15	14	12	9	5	11	18	26	35	45
6	21	20	18	15	11	6	13	21	30	40
7	28	27	25	22	18	13	7	15	24	34
8	36	35	33	30	26	21	15	8	17	27
9	45	44	42	39	35	30	24	17	9	19
10	55	54	52	49	45	40	34	27	19	10

Thirdly, we can prove that all multiples of odd numbers can be arranged as a trapezoid in at least one other way.  The following table shows a string of multiples of the odd number 7.

	1	2	3	4	5	6	7	8	9	10
1	1	3	6	10	15	21	28	36	45	55
2	3	2	5	9	14	20	27	35	44	54
3	6	5	3	7	12	18	25	33	42	52
4	10	9	7	4	9	15	22	30	39	49
5	15	14	12	9	5	11	18	26	35	45
6	21	20	18	15	11	6	13	21	30	40
7	28	27	25	22	18	13	7	15	24	34
8	36	35	33	30	26	21	15	8	17	27
9	45	44	42	39	35	30	24	17	9	19
10	55	54	52	49	45	40	34	27	19	10

It can be shown that any multiple of an odd number follows a similar path, the first leg starting from the middle of the table and going diagonally up and to the right, and the second leg starting from the top row of the table and going diagonally down and to the right.  For each multiple of the odd number 2n + 1, the first leg for 1 ≤ k ≤ n  is

(n – k + 1) * (n + k)

= ½((n + k) – (n – k + 1) + 1)((n – k + 1) + (n + k))

= ½(2k)(2n + 1)

= k(2n + 1)

The second leg for k > n is

(k – n) * (k + n)

= ½((k + n) – (k – n) + 1)((k – n) + (k + n))

= ½(2n + 1)(2k)

= k(2n + 1).  

Therefore, all integers k(2n + 1) for k ≥ 1 can be arranged as a trapezoid in one other way other than x = y, which means any number with an odd factor cannot be a trapezoidal prime.  This leaves only numbers with only factors of 2, or powers of 2, to be possible trapezoidal primes.

Last, we need to prove that a number that is a power of 2 must necessarily be a trapezoidal prime.  We can do this by showing that if x ≠ y and if x and y are both positive integers, then x * y must have an odd factor.  There are four scenarios to consider: x and y are both even, x and y are both odd, x is even and y is odd, and x is odd and y is even. (Since x * y are commutative, arrange the numbers so that x < y.) First, if x and y are both even, and x ≠ y, then y – x + 1 is odd and greater than 1, so x * y = ½(y – x + 1)(x + y) has an odd factor.  Second, if x and y are both odd, and x ≠ y, then again y – x + 1 is odd and greater than 1, so again x * y = ½(y – x + 1)(x + y) has an odd factor.  Third, if x is even and y is odd, then x + y is odd and greater than 1, so x * y = ½(y – x + 1)(x + y) still has an odd factor.  And fourth, if x is odd and y is even, then again x + y is odd and greater than 1, so again so again x * y = ½(y – x + 1)(x + y) has an odd factor. 

color-coded scenarios for odd and even numbers of x * y if x < y:

1) x * y = ½(y – x + 1)(x + y) = ½(y – x + 1)(x + y)

2) x * y = ½(y – x + 1)(x + y) = ½(y – x + 1)(x + y)

3) x * y = ½(y – x + 1)(x + y) = ½(y – x + 1)(x + y)

4) x * y = ½(y – x + 1)(x + y) = ½(y – x + 1)(x + y)

when x < y, x * y necessarily has an odd factor

So either way, if x ≠ y then x * y must have an odd factor, which means the only way to make a trapezoid with a number that is a power of 2 is when x = y, which makes all numbers that are a power of 2 necessarily prime.

It is truly amazing that the trapezoidal system has a pattern for its primes, especially since there is no known pattern for our own rectangular system of primes.  Since trapezoidal primes are powers of 2, and powers of 2 are defined in the rectangular system, the pattern for trapezoidal primes is defined by an outside system.  In the same way, perhaps the key to finding the pattern for rectangular primes is to use some other outside system, but so far nobody has been able to do this.