Example 1. In the parabola HOI, suppose the abscissa mo= 2, and its ordinate H m = 6, what is the length of the curve H OI?

2

√ Hm2 + 4 m

2

62 + ( x 22) =

✓1⁄2 (108 + 16) = 3√93 = 6·429105, half the length of the curve nearly*.

(H M2 + 3 MO2) ×

4

Hm

=

62 + (3 × 22) ×

12.8336, the length of the

257; then (6·429105 × 9) — 25332·084167,

hence 32.084167

curve H o I required.

2. What is the length of the parabolic curve, whose abscissa is 3, and ordinate 8?

PROBLEM XXIX.

Ans. 17.384.

In a Parabola, any three of the four following Terms being given, viz. any two Ordinates and their two Abscissas, to find a fourth.

RULE t. 1. Any abscissa is to the square of its ordinate, as any other abseissa is to the square of its ordinate.

2. The square root of any abscissa is to its ordinate, as the square root of any other abscissa is to its ordinate.

Example 1. If the abscissa mo = 9, and the ordinate m 6, what is the ordinate a D, when the abscissa Do is 16?

mo: Hm2:: DO: A D2, viz. 9: 36 :: 16:64 the square root of which is 8, the ordinate A d. Or, mo: Hm:: DO: A D, viz. 3 : 6 :: 4 8 the ordinate A D.

2. Required the abscissa corresponding to the ordinate 8, that corresponding to the ordinate 6 being 9. Ans. 16.

* Hence 6.429105 × 2 = 12·858210 the length of the curve nearly. See the 5th property, p. · 78.

**

3. If

3. If the two abscissas be 4 and 9, and their corresponding ordinates 8 and 12; any three of these being given, it is required to find the fourth.

PROBLEM XXX.

Given two Ordinates perpendicular to the Axis, and their Distance to find the corresponding Abscissas.

RULE. The difference of the squares of the ordinates is to their distance, as the square of either of them is to its corresponding abscissa *.

Example 1. In the parabola A O B, given the ordinate c & 5, the ordinate H m = 3, and their distance Gm = 4, to find the two abscissas G O, and mo.

2

C G2- Hm2: Gm :: C G 4: 52: 6.25: =GO.

C G2

: GO,

viz. 52 — 32 :

H m2 : & m ¦¦ μm2 : m 0, viz. 5a — 32 ; 32: 2.25 =mo.

Hence the two abscissas are 6·25 and 2·25. Ans. 2. If the greater ordinate of a parabola be 10, the less 6, and their distance 6; what are the two abscissas ? Ans. 9.375 and 3·375.

3. OF THE HYPERBOLA.

Some general Properties of the Hyperbola.

1. The difference between any two straight lines, drawn from the foci to any point in the curve, is equal to the transverse axis, or axis major.

Viz. PFPA B the axis major.

2. The two foci are equidistant from the centre of the hyperbola.

Viz. fo Fo, where o is the centre.

3. The square of the distance of the focus from the centre, is equal to the sum of the squares of the semi-transverse and seini-conjugate.

*This Rule follows from the demonstration of the rule to Problem XXVII

V12. 0 F2 A 02 + o c2, this is obvious for oF

AC by

the construction of the hyperbola, see the definitions, &c. page 8.

5. The rectangle of the distance of either focus from the two vertices, is equal to the square of the semi-conjugate.

VIZ. FAX FB An2 = 0 D2; or, fв x ƒ A = 0 D2. 6. The transverse axis is to the conjugate, as the conjugate is to the latus rectum.

Viz. AB: CD (or mn) :: CD: LR; hence LR CD2

A B.

7. The square of the transverse axis is to the square of the conjugate, as the rectangle of the distance of the ordinate from. each vertex is to the square of the ordinate.

Viz. A B C D2 EA XE B E P2. Hence if transverse axis, the conjugate,

y the ordinate EP, then will

the

the abscissa E A, and the other abscissa

B, and 2: c2:: x × (t + x) : y2 any three of these quantities being given, the fourth may be found..

8. If a straight line be drawn parallel to the conjugate axis, so as to meet the asymptotes of the curve; the rectangle conzained by the segments of this line, between the curve and the asymptotes, is equal to the square of the semi-conjugate axis.

Viz. If o M and oN be the asymptotes to the curve TBS,. mвn the conjugate axis, and м w N a straight line parallel to it; then M S X SN ==B n2.

9. If a straight line be drawn through the curve, meeting th asymptotes, the parts of this line between the curve and the asymptotes, are equal.

Viz. MTS N, and s NT.

Note. All the foregoing properties are demonstrated in Emerson's Conic Sections.

PROBLEM XXXI.

Given the transverse Axis of an Hyperbola, the Conjugate, and the Abscissa, to find the Area.

RULE *. 1. To the product of the transverse and abscissa, add of the square of the abscissa, and multiply the square-root of the sum by 21.

2. Add 4 times the square-root of the product of the transverse and abscissa, to the preceding product, and divide the sum by 75.

3. Divide 4 times the product of the conjugate and abscissa by the transverse; this quotient, multiplied by the former quotient, will give the area of the hyperbola nearly.

verse axis A B =

Example 1. In the hyperbola H A K, if the trans100, the conjugate C D = 50, what is the area?

the abscissa I A =

21

60, and

First, (A B X I ^) + § 1 Aa × 21 (100 × 50) + (§ × 502) × 21=82·375447 × = 1729 884387.

Second, 4AB XIA = 4100 x 50 = 40 50 282.842712, then (282 842712 + 1729° 884387) 75 = 26·83636132 quotient.

Third, (CD XIA X 4) - A B = (60 × 50 X 4) 100 = = 120 quotient, then 26 83636132 × 1203220-3633584, area of the hyperbola HAK,

nearly.

2. Required the area of the hyperbola PA G, the abscissa E A being 10, the transverse axis A B = 30, and the conjugate axis c D or mn = 18. Ans. 15168.. 3. What is the area of the hyperbola T B S, the transverse axis AB being 50, the conjugate axis mвn 30, and the abscissa B W 25? Ans. 805.0909.

This Rule is Sir Isaac Nervton's; it is given in Robertson's Mensuration, 2d edition, p. 102; and demonstrated by Dr. Hutton, in the quarto edition of his Mensuration, p. 376.

PRO

PROBLEM XXXII.

To find the Length of any Arc of an Hyperbola, beginning at the Vertex.

RULE*.

1. Divide the square of the conjugate axis by the transverse, and the quotient will give the latus rectum or parameter +.

2. To 19 times the transverse axis, add 21 times the parameter, and multiply the sum by the ab

scissa.

Again. To 9 times the transverse axis, add 21 times the parameter, and multiply the sum by the abscissa.

To each of these products, add 15 times the square of the conjugate axis.

3. Then the less sum is to the greater, as the ordinate is to half the length of the curve.

Example 1. In the hyperbola PAG, the transverse diameter A B = = 80, the conjugate C D = 60, the ordinate PE= 10, and the abscissa E ▲ = 2.1637, what is the length of the arc ALP?

First. C D2 ÷ A B = 602 ÷ 80

latus rectum or parameter.

45 LR the

Second. (19 A B+ 21 L R) × EA = (19 × 80) + (21 × 45) × 2·1637 =5333 5205.

(9 A B + 21 L R) X E A = (9 × 80) + (21 × 45) x 2.1637: 3602.5605.

Then 5333 5205 + (15 × 602) =59333.5205. And 3602-5605 + (15 × 602) = 57602·5605. Third. 57602·5605 : 59333·5205 :: 10: 10.3005 the length of the arc A LP, the double of which, viz. 20-601, is the length of the whole curve PL AR G.

2. The transverse diameter A B of the hyperbola PAG is 60, the conjugate C D 36, the ordinate FE

This Rule is deduced from Dr. Hutton's Mensuration, quarto edition, p. 364.

+ By the 6th property, p. 84.