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CONIC SECTIONS.

Ex. 3. The transverse 105, the abscissa 84, and the conjugate 35, required the ordinate. Ans. 14.

Ex. 4. The transverse 36, the abscissa 28, and the conju

gate 12, required the ordinate.

Ans. 4.8.

PROBLEM VIII.

To find the area of an elliptic fegment, whose base is parallel to either of the axis.

RULE.

Divide the height of the fegment by that axe of the ellipfe of which it is a part, and find, in the table of circular segments, an area, whose versed sine shall be equal to this quotient. 'Then multiply the area so found, and the two axes continually, and the last product will give the area of the segment required.

EXAMPLE I.

Required the area of the elliptic segment ECF, whose height

is GC 20, and the axes CD and AB 70 and 50.

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Ex. 2. Required the area of an elliptic segment, cut off parallel to the conjugate, at the distance of 18 from the centre, the axis being 60 and 20. Ans. 134.1876.

Ex. 3. Required the area of an elliptical segment, cut off parallel to the transverse, whose height is 6, the diameters being 30 and 20. Anf. 118 9008.

Ex. 4. Required the area of an elliptical segment, cut off parallel to the transverse, whose height is 10, the diameters being 70 and 50. Af. 391.3829.

PROBLEM IX.

To defcribe à parabola, the abscissa and ordinate to the axle being gi

ven.

RULE.

Bifect the given ordinate BA in G, join VG, and draw GD at right angles to VG, meeting the axis in D, and make VO, OF, each equal to BD, and F will be the focus of the parabola.' Take any number of points, x, x, &c. in the axis, and through these points draw double ordinates of an indefinite length.

Then with the radii VF, Vx, &c. and centre F, defcribe the arches c, c, &c. and through all the points of interfection the curve may be drawn.

Note. The line cFc is called the parameter.
For other methods of construction, See Gunnery.

PROBLEM X.

Any three of the four following particulars being given, viz. any two ordinates and their two abfciffas, to find the fourth.

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As any abscissa

RULE.

Is to the square of its ordinate,

So is any other abfciffa

To the square of its ordinate.

EXAMPLE Ι.

Let the abscissa VC be 6, and its ordinate AC 5, required

the ordinate DF, whose abscissa VF is 12.

6:25::12

12

6)300

50=DF*

and 50 = 7.071 Anf.

Ex. 2. The ordinates are 6 and 8, and the less abscissa 9, re

Anf. 16.

Ex. 3. The ordinate is 18, and its abscissa 27, the other ab

quired the greater.

seissa is 48, required its corresponding ordinate.

Ans. 24.

PROBLEM XI.

To find the length of an arch of a parabolic curve, cut off by a double ordinate.

RULE.

To the square of the ordinate add + of the square of the abscissa, multiply this sum by 4, and the square root of the product will be the length of the curve required.

EXAMPLE EXAMPLE I.

Let the abscissa VF be 4, and its ordinate DF 12, required the length of the arch DAVBE.

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Ex. 2. Required the length of the curve, when the abscissa

is 8, and the ordinate 16.

Anf. 36.951.

Ex. 3. Required the length of the curve, when the abscisla

is 15, and ordinate 12.

Ff2

Anf. 21.071.

PROBLEM

PROBLEM XII.

To find the area of a parabola, the base and height being given,

RULE.

Multiply the base by the height, and the product will be the area required.

Note. Every parabola is equal to of the circumfcribing parallelogram.

1

EXAMPLE I.

Required the area of a parabola, whose base is 16, and height 20.

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Ex. 2. Required the area of a parabola, whose base is 30, Anf. 400.

and height 20.

Ex. 3. Required the area of a parabola, whose base is 9, and Anf. 84.

height 14.

Ex. 4. Required the area of a parabola, whose base is 12, Anf. 96.

and height 12.

Ex. 5. Required the area of a parabola, whose base and altiAnf. 220.

tude are 15 and 22.

Ex. 6. Required the area, when the base and altitude are

3 and 4.

Anf. 8.

PROBLEM

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