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Neil Klagge, Chief Operator

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                          Klagge's Theorem (or unified "formula")


Although I discovered the formula, A = ½abt, on my own, I have always wondered if others had discovered it before me. This formula can be used to find the AREA of the SECTOR of an ELLIPSE, an HYPERBOLA, or a CIRCLE, thus the reference below to it as a UNIFIED formula.  I have searched through several texts but only found references to A = ½t for the hyperbolic sector. In the summer of 2008, I did an internet search and found a page that helps you calculate many things about a hyperbola including the area of a sector. This is the first time I have seen “my formula" but I have yet to find it applied to the ellipse as I have done in the following theorem.  If anyone can find this formula, A = ½abt, in a publication prior to about 1991, I would like to know about it. I can be reached at "w0yse" at "msn" dot "com". Thanks for letting me share this with you.

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The following theorem unifies the area of a sector of a hyperbola, an ellipse and a circle and is stated as follows:

1)    For a hyperbola with a semi-transverse axis of “a”, a semi-conjugate axis “b”, and the parameter t = cosh-1(x/a) OR = sinh-1(y/b), the area of the hyperbolic sector whose vertices are the point P(x,y) on the hyperbola, and O(0,0) and Q(a,0) is equal to ½abt.


      x2/a2 – y2/b2 = 1                 x2/a2 + y2/b2 = 1

 

2) For an ellipse with a semi-major axis of “a”, a semi-minor axis of “b”, and the parameter t = cos-1(x/a) = sin-1(y/b), the area of the elliptic sector whose vertices are the point P(x,y) on the ellipse, and O(0,0) and Q(a,0) is equal to ½abt.

3) For a circle of radius r, and the parameter t = cos-1(x/r) = sin-1(y/r), the area of the circular sector whose vertices are the point P(x,y) on the circle, and O(0,0) and Q(r,0) is equal to ½abt = ½rrt = ½r2t = ½r2Θ (where a = b = r and Θ = t = the central angle of the sector in circular radians).

 

When the parameter t = 1 in each of the above cases, “t” can be referred to as a hyperbolic radian, or an elliptic radian or a circular radian, respectively. It must be noted that only the circular radian is DIRECTLY proportional to the actual size of the angle when measured with a common protractor.

 

A History of the Above Discovery

 Several years ago (during the 1991-1992 school year) while teaching a pre-calculus class in Walden Colorado, I came across an interesting similarity between the area of a sector of the unit circle and the area of a sector of the unit hyperbola as follows:

 A.     If we take a circle of radius 1 (where x² + y² = 1) with center, O, at the origin, and a point P(x, y) on the circle and Q(1,0), the area of the sector POQ = ½t where t = Θ (theta) = cos-1x = sin-1y.

B.     Similarly, If we take a unit hyperbola (where a = b = 1 and x² - y² = 1) with center, O, at the origin, and a point P(x,y) on the hyperbola and Q(1,0), the area of the sector POQ = ½t where t = cosh-1x = sinh-1y.

 I then wondered whether there was a way to find the area of the hyperbolic sector if “a” and “b” were not equal to 1 and also if a ≠ b. I figured that the formula, if one existed, would have to involve “a”, the semi-transverse axis, and “b”, the semi-conjugate axis, and “t”.  After graphing several hyperbolas on graph paper (and counting little squares to determine the approximate area) and then using calculus integration and subtraction of areas, I verified that A = ½abt as stated in part 1 of the above theorem. Then t = cosh-1(x/a) = sinh-1(y/b).

 

My curiosity led me to wonder if A = ½abt would work for an ellipse.
I discovered by similar investigation that it did indeed work with t = cos-1(x/a) = sin-1(y/b) as with the circle.

 

The case for circles of radius, r, the formula A = ½abt reduces to A = ½r²t where a = b = r.

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If you can find this formula, A = ½abt, in a publication prior to about 1991, please contact me. I can be reached at "w0yse" at "msn" dot "com". (Please note that the second character of "w0yse" is the numeral "zero".) Again, thanks for letting me share this with you.  (Back to Theorem)

 
Neil Klagge