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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.

Anf 134.1876. Ex. 3. Required the area of an elliptical segment, cut off parallel to the transverse, whose height is 6, the diameters being

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.

Aif. 391.3829

30 and 20.

PROBLEM IX.

To describe a parabola, the abscissa and ordinate to the axle being giá

ven.

RULE.

Bilect the given ordinate BA in G, jön 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 intersection 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

trw ordinates and their iwo abc://as, to find the fourth.

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RULE.

As any abfciffa
Is to the square of its ordinate,
So is any other abscissa
To the square of its ordinate.

EXAMPLE I.

Let the abscissa VC be 6, and its ordinate AC 5, required the ordinate DF, whose absciffa VF is 12.

6:25 :: 12

12

6)300

and 50 = 7.091 Anf.

50=DF

Ex. 2. The ordinates are 6 and 8, and the less absciffa 9, required the greater.

Anf. 16. Ex. 3. The ordinate is 18, and its abfciffa 27, the other ab seissa is 48, required its corresponding ordinate.

Anf. 24.

PROBLEM XI.

To find the length of an arch of a parabolic curve, cut off by e

double ordinate.

RULE.

To the square of the ordinate add 4 of the square of the abseilla, 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 abfciffa VF be 4, and its ordinate DF 12, required the length of the arch DAVBE.

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144 sq. of the ordinate.

16 sq. of the absciffa. 21.33

4 165.33

3)64
4

21.33
661.33(25.7162 the length of the arch.
4

45)261

225 507)3633

3549

5141)8333

5141

51426)319233

308556 514322)1067623

1028644

38978 &c.

Ex. 2. Required the length of the curve, when the abscista is 8, and the ordinate 16.

Anf. 36.951. Ex. 3. Required the length of the curve, when the abscislà is 152 and ordinate 12.

Anf. 21.071.

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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 circumscribing parallelogram.

EXAMPLE I.

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

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Ex. 2. Required the area of a parabola, whose base is 30 and height 20.

Anf. 400. Ex. 3. Required the area of a parabola, whose base is

9,

and height 14

Anf. 84. Ex. 4. Required the area of a parabola, whose base is 12, and height 12.

A::. 96. Ex. 5. Required the area of a parabola, whose base and altitude are 15 and 22.

Anf. 220. Ex. 6. Required the area, when the base and altitude are 3 and 4.

Anf. 8.

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PROBLEM XIII.

To find the area of the fruftum of a parabola.

RULE.

Divide the difference of the cubes of the two ends of the frustum by the difference of their squares, multiply this quotient by the altitude, and the product will be the area required.

EXAMPLE I.

In the parabolic frustum DABE, the two parallel ends DE, AB, are 12 and 2c, and the altitude FC 6, required the area.

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Ex. 2. The greater end of a frustum is 20, the less 10, and their distance 12, required the area.

Anf. 186.

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