Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση
[graphic]
[ocr errors]
[ocr errors]
[graphic]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]
[ocr errors]

ab sin P

sin P

sin P

sin :)

") a perfectly agreeing with the equation to the two axes.*

From the above equation it follows that any diameter NCn, bisects the ordinates MPm, and consequently the whole ellipse. And also that every diameter Nu is bisected at the centre C; for at the points N and n we have z?=mo; and therefore 2=

=+ m. 37. Problem I. Given the two semi-axes a and b, to find two diameters which shall make a given angle. P=DCn. (See preceding figure.) We have mi +n=a +6", mn = therefore m’ +n + 2mn=

2ab +6* +

2ab and ma +n?_2mn =a? +63

;

2ab therefore m+n=(a* +*+ 5), and mắn=v(a'+b

2ab

.).

sin P 2ab

2ab Consequently m= $ v(a*+b++ sin) + } v(a?+b' —

2ab and n = v(a+64 +

Qab

P We have now only to determine the direction of one of the diameters, or the angle ACN which call C. The triangle CNT gives sin (P-C):m:: sin P : CT= aa

m sin P

; hence CQ =

sin (P-C) ao sin (P-C)

; therefore in the right angled triangle CNQ we have m sin P 1:m:: cos C

a* sin (P-C), which gives mcos C sin P=a* sin (P

m sin P --C) = a* sin P cos Cm-asin C cos P, or.

am?

sin P cos C=sin

a?
therefore
tang

a?

sin P

sin P

CQ

[ocr errors]

C cos P,

C=a?—ma

tang P.

m2

m?

* Without sapposing z' or y'=0, we may equally arrive at the final equation. For I° since a? 72

al r2
2-12, the coefficient of 2" which is
62

+

becomes

62 m2 at2+t? Illy. the similar triangles PNQ, DCI give NT: :

62 ? CD2 :: QT2 : CI? :: NQ?: D1%, or q%;? :: 82 : q?-? :: 92: whence,

q? a? g2 t2

al g2
and
therefore

%_t?+t?

therefore as +

q? before y? = -(m?—?)

82

a ?

[ocr errors]
[ocr errors]

38. Problem II. The two diarneters m and n being given, and also the angle P which they make with one another; to find the axes and their direction. From the equations mn sin P=ab, and a +b=m* +n?, by a similar process to that in the foregoing problems, we deduce a=iv(m+ n° +2mn sin P) +*n (m*+n--2mn sin P) and b=1 /(m + n° +2mn sin P)-4v (mo+n' -2mn sin P.)

The angle C which determines the direction of the two axes, may be found as in the preceding problem.

The hgure of wrich we have now treated, is one of the most useful of all the conic sections. The forms of the orbit in the planatary system are elliptical, and hence a knowledge of its properties is essential in astronomy.

The theory of the ellipse is necessary to establish the projection of the sphere, and the representation of a circle whether othographical or schenographical.

As walls and other objects are frequently built upon elliptical plans, and arches constructed of this figure, to understand the different methods of describing it, under various circumstances, is as necessary to the architect, as it is to the workman employed under him.

It would be impossible to enumerate all the uses to which this beautiful figure can be applied, not only in the research of principles, but in fanciful decoration, which receives much of its elegance from the introduction of the ellipse.

[graphic]

OF THE HYPERBOLA.

sin A 39. The equation yy =

cx sin B + xx sin (A+B {

B –1809} shews that that the hyperbola meets its axis AP in two points, one of which is at A, where :=o: the other at a where : c sin B

; therefore supposing that Aa is equal to sin (A+B -180°)

c sin B

the point

The points

C

13

KE

aa

sin (A+B-180°)
a will be in the opposite hy-

N perbola M'am'.

B
D

1 A, a, are called the vertices of the hyperbola, the line Aa (2a) is its axis, its middle point fja

T P C is the centre; lastly, a right line Bb = 2 CB = 26, such

a that bb_sin A sin(A+B-180°)

cos 2} B drawn perpendicular to the axis, and passing through the centre C, is called the second, less, or conjugate axis.

50. These values of a and 6 being substituted in the equation of the hyperbola, gives yy=

66

(2ax +xx). This equation shews that the curve has two branches, AM and Am, equal and infinite in a positive direction. But if x be negative there will be no curve as long as x <- 2a; if x 7 2a the ordinates become real, and the curve will have two other branches indefinitely extended.

It is easy to prove that these two branches are equal to those of the positive hyperbola MAm. For since, calling AP (-2), P'm' =

a P'M' (y), we have

аа

e tyy=—2ax +xx, if we make aP=we shall

have s= 2ą +', and consequently --2ax' + x'x'; an equa

62 tion perfectly similar to that of the hyperbola MAm.

60 41. Since yy =

(2ax + r.r), we have PM?: AP x Pa :: CB? : CA?,

aa

« ΠροηγούμενηΣυνέχεια »