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,

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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**.

** x ^{2}**/

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 = ½r^{2}t = ½r^{2}Θ (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^{-1}x = sin^{-1}y.

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^{-1}x = sinh^{-1}y.

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