2. The major axis of an hyperbola is 15, the minor axis 9, and the less absciss 5; what is the length of the ordinate? Ans. 6. The less absciss being 5, the greater will be 15+5, or 20; then by property 2d:-225: 81: 100 to the square of the ordinate; consequently, the square root of 36, viz. 6, is the required ordinate. 3. The minor and major axes are 48 and 42, and the less absciss 16; what is the ordinate? Ans. 28. 4. The major axis is 25, the minor 15, and the less absciss 83; what is the ordinate? Ans. 10. 5. The major axis is 30,. the parameter 6, and the distances of an ordinate from the two vertexes 12 and 42; what is the length of the ordinate? Ans. 4.1, nearly. 6. If the segments of a line passing through the focus of an hyperbola, and meeting the curve on opposite sides, are 20 and 8, what is the length of the parameter? Ans. 22. 7. In an equilateral hyperbola, the semi-major axis is 10; required the distance of the focus of the hyperbola from its Ans. 4.1421. vertex. The solution of this question is derived from the fifth property of the hyperbola. Thus :-Extract the square root of twice the square of the semi-major axis, and subtract the semi-major axis from this root, and the remainder will be the required distance. 8. In an equilateral hyperbola, the distances of an ordinate from the vertexes are 25 and 49; required the ordinate. Ans. 35. 9. Two hyperbolas have the same major axes, and their abscisses are the same; the minor axis of one of them is 20, and the minor axis and area of the other are 15 and 260; required the area of the other. Ans. 3463. 10. The major and minor axes are 30 and 18, and the ordinate 12; what are the abscisses ? Ans. 10 and 40. 324 900 225 to the square of the distance of the ordinate from the centre, viz. 625, the square root of which is 25, which being added to and subtracted from the semi-major axis, gives the abscisses. 11. The major and minor axes are 24 and 21, and the ordinate 14; what are the abscisses? Ans. 32 and 8. 12. The major axis is 15, and an ordinate 6, and the two abscisses 20 and 5; what is the minor axis? Ans. 9. 13. The major axis is 36, the ordinate 21, and the abscisses are 12 and 48; find the minor axis. Ans. 31.5. 14. The minor axis is 18, the ordinate 12, and the less absciss 10; what is the major axis? Ans. 30. 15. The minor axis is 45, the less absciss 30, and the ordinate 30; required the major axis. Ans. 90. 16. The major and minor axes of an hyperbola are 30 and 18, the ordinate 12, and the less absciss 10; what is the length of the arc? Ans. 15.66. 17. The major and minor axes of an hyperbola are 105 and 63, and a double ordinate 84, and the less absciss 35; what is the length of the whole arc ? Ans. 109.637. 18. In an hyperbola, the major and minor axes are 15 and 9, and the less absciss 5; what is the area? Ans. 37.919. D 19. The major and minor axes are 20 and 12, and the less absciss 6; find the Ans. 67.408. area. To construct an hyperbola :-Draw the rectangle DBCE, making BC equal to the base of the hyperbola, and BD equal to its absciss; then, from the centre of the B HYPERBOLA. A base, and perpendicular to it, draw AF, equal to the height of the cone from which the hyperbola is supposed to be cut; then divide the base, BC, into an even number of equal parts, and from each of these divisions draw lines meeting each other at A. Then divide the sides of the rectangle, DB and EC, each into half as many equal parts as there are divisions in the base, and from each of these divisions draw lines meeting at the centre of the line DE, which point is the vertex of the hyperbola. From said vertex, draw the curve through the points where the lines meeting at A cross those meeting at the vertex, and the curve will be an hyperbola. T81. CYCLOID. C A D B The cycloid is a figure generated by the motion of a point in the circumference of a circle as it rolls along on a plane, or on a straight line. The line AB, which is called the base of the cycloid, is equal to the circumference of the generating circle; and CD, which is called the axis of the cycloid, is equal to the diameter of the generating circle; and the whole length of the cycloidal arc, or curve, ACB, is equal to four times the diameter of the generating circle; and the whole area is equal to three times the area of the generating circle. Therefore, to find the area :— Multiply the square of the diameter by 2.3562. If a heavy body descends (by the force of gravity) the arc of an inverted cycloid, the velocity which it acquires is exactly proportional to the length of the arc; so that, from whatever point in the arc it may begin to fall, it will arrive at the lowest point in precisely the same time.-A body will descend the cycloidal arc, by the force of gravity, so as to accomplish the passage from B to C, in the inverted cycloid, in less time than by moving in a straight line, or in any other path whatever.— A pendulum made to vibrate in the arc of a cycloid, will move with uniform velocity through every point of the arc. EXAMPLES. 1. What is the area of a cycloid, whose generating circle is $7.696 feet in circumference? and what is the length of the curve? Answers, 339.28 acres; and 48 feet. 2. What is the area of a cycloid generated by a circle 3 feet in diameter ? and of one whose generating circle is 4 feet in diameter ? Ans. 21.19; and 37.7. To solve this problem by the sliding rule :-Place 2.3562 on C over 1 on D, and over the diameter of the generating circle found on D will be found the area on C. Thus, over 3 you will find 21.19, and over 4, 37.7 square feet. 3. What is the area of a cycloid, whose generating circle is 13 rods in diameter ? of one whose generating circle is 20 rods in diameter? and of one whose generating circle is 24 rods in diameter ? Ans. 2.5; 5.88; and 8.5 acres. Place 3 on the line C over 14.28 on D, and over the diameter of the generating circle found on D, will be found the area in acres, on the line C. Thus, over 13 you will find 2.5; over 20, 5.88; and over 24, 8.5 acres. T32. IRREGULAR BOUNDARY. To find the area of a field, or piece of land, bounded by an irregular line, as a winding creek: Draw a straight line as near to the irregular boundary as can n m B C D 8 p t be conveniently done; and then from all the points in the irregular boundary nearest and most remote from the right line, DC, measure the length of the perpendiculars, or ordinates, AD, rs, np, mt, &c., falling on the right line, DC; and divide the sum of these perpendiculars by their number, and the quotient will be the mean or average width nearly, which being multiplied by the length of the base, DC, will give the area. If, however, any of the curves be much greater than the others, or should the whole length of the irregular boundary be a curve analogous to the hyperbolic, or parabolic, or cycloidal curve, draw a straight line, as in the above figure, and from the extremities of the curve draw ordinates to the straight line, and between these draw several equidistant ordinates; then the sum of all the ordinates, divided by their number, will give the average breadth, nearly, which, multiplied by the length of the base line, will give the area. When the curve terminates at both extremities on the base line, divide the sum of all the equidistant ordinates by their number less one, and the quotient will be the average breadth of the curvilinear area. EXAMPLES. 1. If the base line DC be 25 rods, and the sum of 11 ordinates be 178 rods, what is the area? Ans. 404 rods. 2. What is the area of a space bounded by a parabolic curve on one side, and the opposite side by a right line, the ordinates at the extremities of the curve being 14 and 18, and two other ordinates, equidistant from these and each other, being 15 and 16, and the base 36? Ans. 567. 3. A curve terminates at its extremities on a right line, and 5 equidistant ordinates are 12, 20, 26, 30, and 24, and their common distance is 14; required the area. Ans, 1568. |