C OF ELLIPSES. Fig. 16. 000 b To find the Circumference of an Ellipse, fig. 15. Fig. 15. α b Fig. 17. d RULE.-Square the two axes, a b and c d, and multiply the square root of half that sum by 3.1416; the product will be the circumference nearly. To find the Area of an Ellipse, fig. 15. RULE.-Multiply the transverse diameter, a b, by the conjugate c d, and the product by 7854. To find the Area of an Elliptic Segment, fig. 16. RULE.-Divide the height of the segment, a p, by the axis a b, of which it is a part, and find, in the table of circular segments at the end of Mensuration of Solids, a circular segment having the same versed sine as this quotient. Then, multiply the segment thus found and the two axes of the ellipse continually together, and the product will give the area required. When the transverse, a b, the conjugate, c d, and the abscissæ, a p and p b, are given, to find the ordinate, p q, fig. 17. RULE.-Multiply the abscissæ into each other, and extract the square root of the product; this will give the mean between them. Then, as the transverse diameter is to the conjugate diameter, so is the mean to the ordinate required. When the transverse, a b, the conjugate, c d, and the ordinate, p q, are given, to find the abscissa, fig. 17. RULE. From the square of half the conjugate, take the square of the ordinate, and extract the square root of the remainder. Then, as the conjugate diameter is to the transverse, so is that square root to half the difference of the two abscissæ. Add this half difference to half the transverse, for the greater abscissa; and subtract it for the less. When the transverse, a b, the ordinate, p q, and the two abscissæ, a p and p b, are given, to find the conjugate, c d. RULE.-As the square root of the product of the two abscissæ is to the ordinate, so is the transverse diameter to the conjugate. Note.-In the same manner the transverse diame ter may be found from the conjugate, using the two abscissæ of the conjugate, and their ordinate perpendicular to the conjugate. When the conjugate, c d, ordinate, p q, and abscissæ, a p and b p, are given, to find the transverse diameter. RULE.-From the square of half the conjugate subtract the square of the ordinate, and extract the root of the remainder. Add this root to the half conjugate if the less abscissa be given; but subtract it when the greater abscissa is given. Then, as the square of the ordinate is to the rectangle of the abscissa and conjugate, so is the reserved sum, or difference, to the transverse diameter. OF PARABOLAS. Fig. 18. Fig. 19. Fig. 20. v Ä Ä Ä α C C b d d f e e e To find the Area of a Parabola. RULE.-Multiply the base by the height, and two-thirds of the product will be the area. To find the Area of a Frustum of a Parabola, fig. 19. RULE.-Multiply the difference of the cubes of the two ends of the frustum, a c d f, by twice its altitude, b e, and divide the product by thrice the difference of their squares. To find the Abscissa or Ordinate of a Parabola, fig. 18. RULE.-The abscissæ, v b and b e, are to each other as the squares of their ordinates, a b and d e, that is, as any abscissa is to the square of its ordinate, so is any other abscissa to the square of its ordinate. Or, as the square root of any abscissa is to its ordinate, so is the square root of another abscissa to its ordinate. To find the Length of a Parabolic Curve, cut off by a Double Ordinate, fig. 20. RULE.-To the square root of the ordinate, a b, add of the square of the abscissa, v b; the square root of that sum, multiplied by 2, will give the length of the curve nearly. e Fig. 22. v b C e d To find the Area of a Hyperbola, fig. 21. RULE.-To five-sevenths of the abscissa, v b, add the transverse diameter, ve; multiply the sum by the abscissa, and extract the square root of the product. Then multiply the transverse diameter by the abscissa, and extract the square root of that product. Then, to 21 times the first root add 4 times the second root; multiply the sum by double the product of the conjugate and abscissa, and divide by 75 times the transverse; this will give the area nearly. To find the Length of a Hyperbolic Curve, fig. 22. RULE.—To 21 times the square of the conjugate, a b, add 9 times the square of the transverse; also, to 21 times the square of the conjugate add 19 times the square of the transverse, and multiply each of these sums by the abscissa, v b. To each of the two products add 15 times the product of the transverse and square of the conjugate. |