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PROBLEM III.

To find the area of an ellipse.

RULE

Multiply the tranfverse by the conjugate, and this product again by :7854 for the area.

EXAMPLE. I.

Required the area of an ellipse, whose two axes are 30 and

40 feet.

30 40

I 200 -7854

942.4800

Ex. 2. Required the area of an ellipse, whose transverse and conjugate are 20 and 10 feet.

Anf. 157.08. Ex. 3. Required the area of an ellipse, whose diameters are 48 and 30 yards.

Anf. 1357.1712. Ex. 4. Required the area of an ellipse, whose two axes are 14 and 12 feet.

Anf. 131.9472.

PROBLEM IV.

The transverse, conjugate, and ordinate being given, to find the abfciffa.

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RULE.

As the conjugate
Is to the transverse,
So is the square root of the difference of the squares of the

ordinate and semi-conjugate
To the distance between the ordinate and centre.

Note. This dist ce is to be added to or subtracted from the semi-transverse, according as the abscissa is greater or less than the semi-transverse.

EXAMPLE I.

The transverse AB is 60, the conjugate CD 20, and the ordinate Fx 8. It is required to find the abscissa.

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20)360

18 distance between the ordinate and centre.
30 semi-transverse.
48 the absciffa x B.
12 the abscissa A x.

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Ex. 2. The transverse go, the conjugate 30, and the ordinatë 12, required the abscissas.

Anf:.72 and 18. Ex. 3. Tlie transverse 105, the conjugate 35, and the ordinale 14, required the abfciflas.

Ans. 84 and 21.

Ex. 4. The transverse diameter is 3, the conjugate i, and the ordinate 4, required the abfcifras. Anf. . and 21..

PROBLEM V.

The conjugate, ordinate, and abscisa being given, to find the trans

verse.

RULE.

Find the square root of the difference of the squares of the semi-conjugate and ordinate, and, according as the greater or less obscilla is proposed, add this root to or subtract it from the semi-conjugate. Then use the following proportion

As the square of the ordinate
Is to the product of the conjugate and abfciffa,
So is the sum or difference, found as above,
To the transverse.

EXAMPLE. I.

The conjugate CD is 20, the ordinate Fx is 8, and the abscissa Ax 12, required the transverse AB.

io=100
8 = 64

366 root of the differences of the square of the fe36 10 semi-conjugate.

[mi conj. and ord. 16 fum.

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As 64*. 12 20 ::16
64 : 240 ::16

16

64)3840(60 the transverse.

384

1

Anf. 90.

Ex. 2. The conjugate 30, the ordinate 12, and the abfciffa 18, required the transverse,

Ex. 3. The conjugate 35, the ordinate 14, and the abscifra 84, required the transverse.

Ex. 4., The conjugate 1, the absciffa 24, and the ordinate 7, required the transverse.

Anf. 105

Anf. 3.

PROBLEM VI.

The transverse, ordinate, and abscisa being given, to find the core

jugate.

RULE.

As the square root of the products of the two abscissas
Is to the ordinate,
So is the transverse
To the conjugate.

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The transverse AB is 60, the ordinate Fx 8, and the absciffi 12, required the conjugate.

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1

Anf. 30.

Ex. 2. The transverse 90, the ordinate 12, and the abscilla 18, required the conjugate diameter.

Ex. 3. The transverse 105, the ordinate 14, and the abfciffa 84, it is required to find the conjugate.

Anf. 35Ev, 4. The transverse 3, the ordinate, and the abfciffa 276, required the conjugate.

Anf. 1.

PROBLEM VII.

The abscissa, transverse, and conjugate being giren, to find the or

dinate.

RULE.

As the transverse
Is to the conjugate,
So is the square root of the product of the two abfciflas
To the ordinate.

EXAMPLE I.

The transverse AB 60, the abscissas 12, 48, and the conjugate 20, required the ordinate.

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Ex. 2. The transverse 90, the abfciffa 18, and the conjugate 30, required the ordinate.

Anf. 12.

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