1- -e 1+e √ {1+(1+e)2v3} × { 1+ (1 − e)22} Let (1+e) v = u,, (1 − e) v = b ̧ u ̧, or b1 = 1 + b, also let √(1+u2)(1+b2u3) = U, √ (1 + u,2) (1 + b2u ̧3) ≈ U1 du Then d U i+b1. 2 Hence it is evident that by repeating the foregoing operatious, we shall obtain results symmetrical with the former scil. which, by aid of the formula (4), which evidently expands into Consequently the difficulty of rectifying an ellipse is reduced to that of integrating the expressions which, however, can only be effected by approximation. Now since in every case of eccentricity, and more especially when b is small, or e nearly 1, the values of become rapidly less and less, we shall soon arrive at b, which may be consideredo without materially affecting the result, .. equation (6) becomes, by substitution and integration and to obtain the whole quadrant E, of the ellipse, b nearly, (m being the index at which b+ may first be taken Hence, by substituting in equation (7) and making the neces sary reductions, and putting P B1 B§ .... b B, Q= bb, the approximate value of the elliptic quadrant. In the problem we have But since 4 = .000006313 = .0000063129 differs so little from b2, we make For a very elaborate and profound discussion of this and kindred subjects, the reader is referred to a paper in the Transactions of the Royal Society for 1804, by Mr. Woodhouse. He will also find it ably treated by Mr. Ivory, in the 4th volume, and by Mr. Wallace in the 5th volume of the Edinburgh Transactions. Euler, moreover, in his Animadversiones in Rectificationem Ellipseos Opuscul. Vol. II., may be consulted with advantage. 173. Let the rectangular co-ordinates (x, y) of the hyperbola (H) be referred to its centre, then, 1 and b being its semi-axes .. denoting by H, the arc whose abscissa = x, where e the eccentricity = √1+ 6o. Now since e is > 1, to transform d II, into algebraical and |