EXAMPLES.

1. What is the area of a parabola GVK, whose height VH is 12, and the base or double ordinate GK 16?

1. The abscissa is 12, and the double ordinate or base 38; what is the area?

Ans. 304.

3. What is the area of a parabola whose abscissa is 10, and ordinate 8? Ans. 106.

PROBLEM X.

To find the area of a frustrum of a parabola.

RULE.*

Divide the difference of the cubes of the two ends of the frustrum by the difference of their squares, and this quotient multiplied by 3 of the altitude, will give the area required.

*Demon. Let D=GK, d=EI, and a=FH.

Then by the nature of the curve D2-d2: a :: D2 : AD2

Vн, and D2-d2: a :: d2 :

D2-d2

ad2

=VF.

Note. Any parabolic frustrum is equal to a parabola of the same altitude, whose base is equal to one end of the frustrum, increased

EXAMPLES.

1. In the parabolic frustrum GEIK, the two parallel ends EI and GK are 6 and 10, and the altitude, or part of the abscissa FH, is 3; what is the area?

103 63

Here 10363 (GK3 — EI3) ÷ 102 — 62 (GK2 — EI2) = 1000-216 784 98 49

100-36
2 x 3

12.25.

And 12.25 X

3

2. The greater end of the frustrum is 24, the lesser end is 20, and their distance 5; what is the area?

Ans. 121.3333. 3. Required the area of the parabolic frustrum, the greater end of which is 10, the less 6, and the height 4. Ans. 32.

PROBLEM XI.

To construct an hyperbola, the transverse and conjugate diameters being given.

B F N n

by a third proportional to the sum of the two ends, and the other

end.

* Construction. 1. Make AB the transverse diameter, and CD perpendicular to it, the conjugate.

2. Bisect AB in O, and from O with the radius OC, or OD, describe the circle DfCF, cutting AB produced in F and f, which points will be the two foci.

3. In AB produced take any number of points, n, n, &c. and from F and f, as centres, with the distances Bn, An, as radii, describe arcs cutting each other in s, s, &c.

4. Through the several points s, s, &c. draw the curve sBs, and it will be the hyperbola required.

5. If straight lines be drawn from the point O, through the extremities CD of the conjugate axis, they will be the asymptotes of the hyperbola, whose property it is to approach continually to the curve, and yet never to meet it.

PROBLEM XII.

In an hyperbola, any three of the four following terms being given, viz. the transverse and conjugate diameters, an ordinate, and its abscissa, to find the fourth.

CASE I.

The transverse and conjugate diameters, and the two abscissas being known, to find the ordinate.

RULE.†

As the transverse diameter,

Is to the conjugate;

So is the square root of the product of the two abscissas, To the ordinate required.

* The sum of two lines drawn from the foci of an ellipse to any point in the curve, is equal to its transverse diameter.

In like manner the difference of two lines drawn from the foci of any hyperbola to any point in the curve, is equal to its transverse diameter, as is shown by the writers on conics.

But the arcs intersecting each other in 8, 8, &c. are described from the focif and F, and with the distances An, and Bn, whose difference is AB, and therefore the points s, s, &c. are in the curve of an hyperbola.

Let t=transverse diameter, c=conjugate, x=abscissa, and

EXAMPLES.

1. In the hyperbola GAH, the transverse diameter is 120, the conjugate 72, and the abscissa AF is 40; required the ordinate FH.

2. The transverse diameter is 24, the conjugate 21, and the less abscissa 8; what is its ordinate?

Ans. 14. 3. The transverse diameter of an hyperbola is 50, the conjugate 30, and the less abscissa 12; required the ordiAns. 16.3658.

nate.

CASE II.

The transverse and conjugate diameters and an ordinate being given, to find the two abscissas.

RULE.*

As the conjugate diameter is to the transverse,

So is the square root of the sum of the squares of the ordinate and semi-conjugate,

To the distance between the ordinate and the centre, or half the sum of the abscissas.

y-ordinate. Then the general property of the curve is 2: c2:: X (t+x): y2; and from this analogy all the eases of this problem are deduced.

Note-In the hyperbola, the less abscissa added to the axis, gives the greater.

K

Then the sum of this distance and the semi-transverse will give the greater abscissa, and their difference the less abscissa required.*

EXAMPLES.

The transverse diameter is 120, the conjugate 72, and the ordinate 48; what are the two abscissas?

2. The transverse and conjugate diameters are 24 and 21; required the two abscissas to the ordinate 14.

Ans. 32 and 8.

3. The transverse being 60, and the conjugate 36; required the two abscissas to the ordinate 24.

Ans. 80 and 20.

* This rule in species is√c2+y2±‡t=x,= greater

c

or less abscissa, according as the upper or under sign is used.