EXAMPLE. If the axes of an ellipfe be 35 and 25, what is the area? :7854 × 35 × 25 = 687.225 the area required. RULE II. Multiply continually together any two conjugate diameters, the natural fine of their included angle, and the number 7854. That is, dens the area; putting d and c = any two conjugate diameters, s fine of their included angle, and n = 7854. E X Corol. 3. If the conjugate diameters be equal to each other, it will follow that, As radius is to the fine of the angle made by the equal conjugate diameters; fo is the circle defcribed on one of thofe diameters, to the ellipfe. Corol. 4. The ellipfe is a mean circles defcribed on the two axes. other a right angle, whofe fine is proportional between the two For they make with each to the radius. Corol. 5. As radius is to the fine of the angle made by any two conjugate diameters; fo is the circle whofe diameter is a mean proportional between the conjugate diameters, to the ellipfe. This follows from corollary 2. Corol. 6. The ellipfe is equal to a circle whofe diameter is a mean proportional between the two axes.— -From corollary 4. Corol. 7. As an ellipfe is to the rectangle of its two axes, or to the rectangle of any two conjugate diameters drawn into the fine of their included angle, the radius being I :: fo is any circle to the fquare of its diameter.-Any two like fegments or zones of the ellipfe and circle arc alfo in the same proportion. Corol. 8. Ellipfes, and their like fegments, are to one another as the rectangles of their axes, or as the rectangles of any conjugate diameters forming the fame angle in each. Corol. 9. Similar ellipfes are to one another as the fquares of their like diameters. Corol. EXAMPLE. If two conjugate diameters of an ellipfe be 28 and 32, and their included angle 77° 34 required its area. The fine of 77° 34′ is 9765625; therefore •9765625 × 32 × 28 ×·7854 the area. 687-225 PROBLEM VI. To find the Area of the Segment of an Ellipfe cut off by a Double Ordinate to either Axe, that is, by a Line Perpendicular to that Axe. RULE I. Find the correfponding fegment of the circle defcribed upon the fame axe to which the cutting line, or base of the fegment, is perpendicular. Corol. 10. From corollary 7 comes alfo the following conftruction. a dDh H E F Let ADE be an oblique fegment of the ellipfe AFBGA, cut off by a double ordinate to the diameter AB, FG being the conjugate. Through the center c draw ap perpendicular to FG and meeting Aa and BP, in a and P, both parallel to FG; then about the axes ap, FG, defcribe the ellipfe aFpGa, mecting the ordinate produced in e and d Then will the right elliptic fegment dae be equal to the oblique fegment DAE, as well as the whole ellipfe аFPGA equal to the ellipfe AFBGA; moreover their correfponding ordinates de, DE, parallel to the common diameter FG, are every where equal, as are the like parts or zones contained between any two of fuch ordinates. And the fame may be faid of all ellipfes contained between the parallels Aa, BP, infinitely produced in which property they refemble parallelograms of the fame bafe and between the fame parallels. B P Then as this axe is to the other axe :: fo is the circular fegment: to the elliptic fegment. This follows from corollary 7 to the laft problem. RULE II. Find the tabular circular fegment whofe verfed fine or height is equal to the quotient of the height of the elliptic fegment divided by its axe. Then multiply continually together this fegment and the two axes of the ellipfe, for the area of the fegment required. This rule follows from the former. EXAMPLE. What is the area of an elliptic fegment cut off by a line parallel to, and at the diftance of, 7 from the less axe, the axes being 35 and 25 ? 2 Here 171-710= the height of the fegment. And to 352 ÷ 7 = 2857+ = the tabular verfed fine; whole fegment is 1851669. Then 1851669 × 35 × 25 = 162·0210375 = the area of the lefs fegment. If the greater fegment had been required, Then 78539816185166960023126. And 60023126 X 25 X 35 = 525 2023525 = the area of the greater fegment. EXAMPLE II. What is the area of the elliptic fegment cut off by a double ordinate perpendicular to the conjugate axe at the diftance of 7 from the center, the axes being 35 and 25? Here 124-75 the altitude of the fegment. And 525 = 1 ÷ 5 = •2 = the tabular verfed fine; whole correfponding fegment is 1118238. Hence 1118238 X 25 X 35 = 97.845825 the area of the lefs fegment. Again, 78539816 111823867357436. And 67357436 X 25 X 35 = 589 377565 the greater fegment. PROBLEM VII. To find the Area of an Elliptic Segment cut off by a Double Ordinate to any Diameter; that is, by a Line Oblique to the Axes. Divide the abfcifs AF (fig. to prob. 5) of the double ordinate, by its diameter AB, and find the tabular circular fegment whofe verfed fine is the quotient. Then multiply continually together the tabular area and the two axes. Or the tabular area, the diameter AB to which the bafe of the fegment is a double ordinate, its conjugate diameter CH, and the fine of their included angle, for the area of the elliptic fegment required.* EXAMPLE. The axes of an ellipfe being 35 and 25, it is required to find the area of a fegment whofe bafe is a double ordinate to a diameter whofe length is 33, it being divided by the double ordinate into the two abfciffes 7 and 26. Here FA AB = 733 2121 the tabular verfed fine; to which correfponds the area 12162869. U 4 Hence * DEMONSTRATION. For, by cor. 7 prob. 5, as AB2: to circular fegment LAK :: fo is AB X DE X s. Lc = rectangle of the two axes: to elliptic fegment GAH; but circular fegment LAK = AB2 X tabular circular fegments whofe verfed fine is FAAB; there fore AB AB2 × 5 :: Is rectangle of the axes to elliptic fegment. 2. E, D. Hence 12162869 X 25 × 35 = 106.425104 = the fegment GAH. Moreover, 78539816-1216286966376947. And 66376947 X 25 X 35 = 580.79828625 = the greater fegment GBH. PROBLEM VIII. To find the Trilineal Area ABQ, included by either Axe, a Line drawn from any Point in it, and their Intercepted Arc. Draw the ordinate DB meeting the circle defcribed upon the faid axe in c, and draw Ac, oc.* For, by the investigation and corollaries to prob. 5, :: circular fegment DCQ: elliptic fegment DBQ :: (because of the common bafe AD) triangle ACD: triangle ABD :: (by compofition) the trilineal ACQ: trilineal ABQ. Q. E.D. Corol. 1. Since, by rule 2 prob. 6 feet, 1 part 2, putting oq=r, and DC = y, the circ, arc oc is = y X (1+ + 314 we have the fect. ocoryx (1+ 3.282 5.2.4747 3.575 7.2.4.675 &c), 314 3.2r2 + +-3.576 5.2.4r4' 7.2.4.6r6` &c); which being increafed or diminished by OA X DC = the tri angle Aco, we fhall have + 3·575 3.2r 5.2.4r3 7.2.4.6r5 &c), for the general value of the trilineal ACQ; and confequently y2 334 &c) for the elliptic trilineal ABQ; r being the radius of the circle, or femi-axe o of the ellipfe, and the other femi-axe. And fincerc :: y: BD = 2, we fhall have y = = and confequently the value of the elliptic trilineal ABQ, expreffed in terms of its own ordinate z or BD, and femi-axes r, c, will be |
