Example 1. When the transverse axis A B, of the ellipsis A D B C, is 52, the conjugate C D 26, and the ordinate Gm 12, what are the two abscissas a m and Bm?

√262 — (122 × 4) =

100 10, then 26: 52 :: 5 : 10=om the distance of the ordinate from the centre o of the ellipsis, then 26 + 10 = AM, and 26 — 10 — 16 = BM.`

36

=

2. When the transverse axis is 80, the conjugate 60, and the ordinate 24; what are the two abscissas? Ans. 64 and 16.

CASE III.

When the conjugate axis, ordinate, and abscissas, are given, to find the transverse.

RULE. Find the square root of the difference of the squares of the semi-conjugate axis and the ordinate, which add to the semi-conjugate, if the less abscissa be given, but subtract it if the greater ab* scissa be given.

Then, the square of the ordinate is to the rectangle of the conjugate axis and the abscissa; as the sum, or difference found above, is to the transverse required.

Example 1. In the ellipsis A D & C there are given the conjugate axis C D = 26, the ordinate &m= 12, and the less abscissa B m = 16, required the transverse axis A B.

✓132 · 122 = √25 = 5, then 13 + 5 =

sum.

18,

Then 12 26 × 16 :: 18: 52 the transverse axis A B.

2. If the conjugate axis be 60, the ordinate 24, and the greater abscissa 64; what is the length of the transverse axis ? Ans. 80.

CASE IV.

When the transverse axis, ordinate, and abscissus, are given to find the conjugate axis.

RULE. The square root of the product of the

two abscissas is to the ordinate, as the transverse axis is to the conjugate.

Example 1. In the ellipsis A D B C, there are given the transverse axis A B = 120, the ordinate G m = 16, and the abscissa mв 24, to find the conjugate axis c D.

120-24 =

96 the abscissa a m.

Then 96 x 24: 16 :: 120: 40 c n the conjugate axis.

2. If the transverse axis be 80, the greater abscissa 64, and the ordinate 24; what is the length of the conjugate axis?

2. OF THE PARABOLA.

Ans. 60.

Some general Properties of the Parabola.

The distance from the focus to any point in the curve, is equal to the perpendicular distance of that point from the directrix.

Thus, if F be the focus, TP the directrix, and n any point in the curve, then F HHT.

4. The square of any ordinate to the axis, is equal to the rectangle of the latus rectum, and the abscissa to that ordinate.

Viz. C G2 LR X OG.

Hence, if any abscissa, y its ordinate, and the parameter or latus rectum, then p2px, any two of which being given, the third is easily found.

5. The abscissas are to each other, as the squares of their ordinates.

6. The latus rectum is to the sum of any two ordinates, as the difference of these ordinates is to the difference between their abscissas.

VIZ. LR сG + Hm :: C G - H MOG-Om

7. All parabolas are similar to each other, and any two para bolas, having equal parameters and abscissas, are equal to each other.

8. The area of a parabola is equal to of the area of its circumscribing parallelogram.

9. The distance of the focus from the vertex, is equal to the square of any ordinate divided by four times its abscissa.

VIZ. OF C G2 ÷ 40 G.

10. The latus rectum is equal to the square of any ordinate divided by its abscissa.

VIZ. LRC G2 ÷ 0 G.

Note. The preceding, and various other properties of the parabola, aie demonstrated in Emerson's Conic Sections.

PROBLEM XXVI

To find the Area of a Parabola, its Base* and Height being given.

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

Example 1. If the base A B, or greatest double ordinate of a parabola ▲ O в, be 53.75, and the abscissa Do, or height of the parabola be 39.25, what is the area?

(AB X DO) X = (53.75 x 39.25) × 3 × 3 = 1406-4583, the area required.

2. What is the area of a parabola, whose height or abscissa is 10; and the base, or double ordinate, 16? Ans. 1063.

* Any double ordinate A B or C E of a parabola may be called the base, and the axis o D, or any part of it oG, the height; hence the base is any double ordinate, and the height is an abscissa to that double ordinate.

3. Required the area of a parabola, whose base, or double ordinate, is 38, and the height or abscissa, 12? Ans. 301.

PROBLEM XXVII.

To find the Area of the Zone of a Parabola, or of the Space included between two parallel double Ordi

nates.

RULE. If the two double ordinates, their dis tance, and the altitude of the whole parabola be given; find the area of the whole parabola, and also of the upper segment, the difference will be the area of the zone.

But if the two double ordinates, and their distance only, be given. To the sum of the squares of the two double ordinates, add their product, divide the result by the sum of the two double ordinates, the quotient multiplied by of the altitude of the zone will give the area.

Example 1. In the parabolic zone A BIH, there are given the double ordinate a B == 10, the double ordinate HI= 6, their distance Dm4, and the

*The first part of the Rule is self-evident. By the 5th pro perty, p. 78, D.0: mo :: A D2: Hm; now, if D A B the greater double ordinate, dн the less, and Dm a their distance, then Do: mo: D2: 4 d2, or Do: mo :: D2 : d2, and by division, (EUCLID 17 of V.) Domo: mo :: da: d2, viz. Dm mo :: D2 - · d2 : d2, or a : mo:: · d2 : d2, hence mo — a d2 (b2 — d2), consequently a d2

D2.

D2

a dr

Now D X

a t

area

area of the whole parabola A OB, and d x

of the segment Ho 1, their difference

D2 + D d + ď2

Xa, is the area of the zone A HIB, which

altitude

altitude D O of the whole parabola = 6·25, required

Difference = 32 area of the zone A H I B.

When the altitude Do, of the whole of the parabola,

is not given.

(102 +62) + (10 × 6)

10+6

= 323 area as above.

12, and 12 X 4 X 3

2. Suppose the double ordinate e E of the zone CEIH to be 25.298, H I = 16, and their distance Gm = 9, what is the area of the zone? Ans. 188.98.

3. Suppose the double ordinate A B = 24, CE= 20, and their distance DG 55, what is the area of the zone ABEC Ans. 121

PROBLEM XXVIII.

To find the Length of any Arc of a Parabola, cut off by a double Ordinate.

of

RULE. 1. To the square of the ordinate add the square of the abscissa, and the square root of the sum will be half the length of the curve nearly.

2. To the square of the ordinate add of the square of the abscissa, and divide four times this sum by the ordinate; subtract the quotient from nine times the half length of the curve found above, and of the remainder will be the whole length of the curve extremely neur *.

This rule is derived from a series which ceases to converge when the abscissa exceeds half the ordinate, or when the double ordinate exceeds the latus rectum; in such cases it ought not to be used. The first part of the rule, which has generally been given, with improper examples as exercises, is derived from the same series.

Example