difference of the distances from the foci to the line always gives a constant. The diagram below shows the hyperbola and the method for which a constant is achieved.The central equation for the hyperbola is derived from using several distance formulas and many substitutions of these formulas to derive a final and simple equation. This equation is based upon the center of the hyperbola being located at the origin. We will let 2A equal the length of the major axis, and let 2C equal the distance from one focus to the other focus. After doing so, you can rotate the C, after taking the distance from the center of the hyperbola to a focus, and make a right triangle when creating B. Also, you would take just half of the 2A and make it just a single A. (This is shown in the diagram on the next page, just imagine a circle with a perimeter on the foci, then the point where C and B meet would also be on that perimeter.) Now, by assuming that the hyperbola is centered at the origin with the foci on the X-axis, you can then once again apply distance formulas with multiple simplifications and substitutions derive the final equation of the hyperbola. Since the distance from point F2, or (-C, 0), to the point (X, Y) on the graph would be equal to ((X-C)2 + (Y-0)2) which in turn would be equal to ((X-C)2 + Y2), and the distance from point F1, or (C, 0), is equal to the equation,((X+C)2 + (Y-0)2), which is reduced to ((X+C)2 + Y2), one then could subtract them and set it equal to 2A, since we know the constant is the same length as the major axis. This would result in the early stage of a formula, and look like this:((X-C)2 + Y2) - ((X+C)2 + Y2) = 2AAfter moving the distance formula for the larger distance over, squaring both sides, and reducing two times, one will come up with the following:(C2 V A2)X2 V A2Y2 = A2(C2 V A2)With the previously assessed knowledge of B2 being equivalent to C2 V A2, we can substitute the B2 in both cases that its e...
