1. To describe the curve whose equation is x4 a2* + a*y* = 0, we have, by differentiating (2.2 - a’x) dx + a’ydy = 0 .... (a) And (6.x2 - a)dx + a’dy' + a’ydy = 0....(6) First let us determine the limits or maxima and minima of y; dy aRx - 2x SO dr a'y 2.2 = 0 2 2 0 .. e'y = 0, or y so VOL II. a ar And substituting in the given equation there will result dy 2x3 = 9, and it being found, by equation (b), aʻy = 0 that x = 0, y = 0, are the only values deduced from the numerator and denominator, which satisfy the given equation, we have a double point at A. Also, since when x and y each = 0, dy = - 1= I tan. the tangents at the d.x a point A are inclined to Bb at an angle of 45 degrees. Again, to find the points of inflexion, we have dy a' - 6x2 - a dx2 1 O or d.x? 2 4 a’y .. we have a point of inflexion at the origin A. Since there are no infinite branches, the curve cannot have an asymptote. This curve which has been called the lemniscata, is the locus of the extremity of that part of the ordinate of a circle whose radius is (a), which is constantly taken = a sin. 8. cos. 8, 8 being the angle at the centre subtended by that ordinate. For from the given equation x. And AM'B and Am'b are equal to either of these. 4a. :: the whole area A M B M' Ambm'A = 3 N. B. Unless the branches of an oval intersect, it is not perfectly quadrable. See Waring, Vince, &c. (2) Let E F (Fig. 3) = y, AE = AD, the diameter of the generating O AHD, = a. Then AH being joined, by the property of the cycloid, we have y= 2 AH = 2 JAE. AD, by similar A, = 2 ar. :: y = 4ax, the equation to the curve, which is therefore a parabola, whose vertex is A, and focus D. Now the area of a cycloid is known to be triple that of its gene 2 rating O, and the area of a parabola , of its circumscribing rectangle; hence AFGD : ABCD :: ŞAD X DG : 3A HD But if x = the circumference of a circle whose diameter is 1, = its area. 3 1 х Hence AFGD : ABCD :: A AD" : 84 AD :: 32 :9. RN + rn (3) Let CB or CD or CR (Fig. 4) = r, Cb=a, PM =4, CP = 4. Then since M bisects Rr in M, we have RN y = PM = + 2 2 :: RN = ту ✓ x2 + y2 Hence y = ry + and by involution, &c. ma - a'- 4y + 4ay which is the equation to the 2y curve referred to rectangular co-ordinates. To find the equation referred to polar co-ordinates, we have (putting the MCP = 0 and CM = ) RN + rn 2 r'. sin. O a PM = + 2 2 (6) which is more simple than a + 2 sin. A the former. To find the asymptotes, let a Then 2y – a = 0, or y = 2 if Cb be bisected in v, and vt be drawn parallel to AD, it will be an asymptote to the branch «M. Let y = a Then x = V – a? = Fb, or the curve passes through F, the intersection of br with the circle. Similar properties belong to each of the branches generated by the other quadrants. (4) To construct a’y — xoy – a = 0 a3 or y = a? - x2 Let the origin of abscissæ be at A, (Fig. 5), « being then = 0, and . y = a = AC. Again, let x = $a = AB or Ab. Then y = co = BN or BN, which are :. asymptotes to the branches CR, Cr respectively. When = xis > a, the va of y are negative, and the corresponding branches are Ss, Tt; which, since when + x is infinite, |