umber is a constant. If you used the Pythagorean Theorem as A2 = B2 + C2, then you could get any of the parts of an ellipse that you needed.The next formula that can be derived would be the central equation and the general equation for the ellipse. Lets say that the major axis, the commonly places horizontal one, was on the X-axis, and the center was at the origin. Then you can deride a formula that would give the ellipse a central equation for any point on the perimeter of the ellipse allowing for it then to be drawn. The way this was brought about was by letting the point on the perimeter named P be labeled (X, Y), and the points of focus equal (-C, 0), and (C, 0). Now, by using the distance formula, one could add these two together to get the A, which would be doubled in the equation so that the square root of a square root does not have to be used. Well for example in this case, the distance formula would give you:With PF1 and PF2 being the two distances to the foci, this formula can be rewritten several times. First one would set the sum of the two distances equal to 2a like above. Then you would subtract PF1 from one side to the other and then square both sides. One would result in the following equation: X2 V 2CX + C2 + Y2 = 4A2 V 4A ((X + C)2 + Y2) + X2 + 2CX + C2 + Y2 After being simplified, the following equation would result:A ((X + C)2 + Y2) = A2 + CXSquaring both sides again results in the equation as follows (After simplifying):Eq. #1 (A2 - C2)X2 + A2Y2 = A4 + 2A2CX + C2X2Since after playing around a bit with the Pythagorean Theorem you can come up with equations for each the A2, the B2, and the C2, you can substitute them for their equivalent as follows:As for C: C2 = A2 - B2, or for B: B2 = A2 - C2, or for A: A2 = B2 + C2 In substituting B2 for (A2 V C2) in Eq. #1, you will come up with the following formula with its simplified form located below it.B2X2 + A2Y2 = A2B2, which would then simplify itself to (...
