ON THE ASYMPTOTES OF THE HYPERBOLA.

DEFINITION.—An Asymptote is a diameter which approaches nearer to meet the curve, the farther it is produced, but which, being produced ever so far, does never actually meet it.

PROP. XXI.

If tangents be drawn at the vertices of the axes, the diagonals of the rectangle so formed are asymptotes to the four curves.

Let MP meet CE in Q;

Then, MQ CM :: AE : AC

E.

Q

P

Now, as CM increases, the ratio of CM2 to CM3 — CA2 continually approaches to a ratio of equality; but CM3 — CA2 can never become actually equal to CM3, however much CM may be increased.

M

Hence, MP is always less than MQ, but approaches continually nearer to an equality with it.

In the same manner it may be proved, that CQ is an asymptote to the conjugate hyperbola BP'.

Cor. 1. The two asymptotes make equal angles with the axis major and with the axis minor.

Cor. 2. The line AB joining the vertices of the conjugate axes is bisected by one asymptote and is parallel to the other.

Cor. 3. All lines perpendicular to either axis and terminated by the asymplotes are bisected by the axis.

PROP. XXII.

All the parallelograms are equal, which are formed between the asymptotes and curve, by lines drawn parallel to the asymptotes.

That is, the lines GE, EK, AP, AQ, being parallel to the asymptotes CH, Ch, then the parallelogram CGEK = parallelogram CPAQ.

For, let A be the vertex of the curve, or extremity of the semi-transverse axis AC, perpendicular to which draw AL or Al, which will be equal to the semi-conjugate, by definition xix. Also, draw HEDeh parallel to Ll.

P

L

K

CA: AL:: CD2 - CA2 : DE,
CA AL:: CD: DH2;

Then, and by parallels, therefore, by subtract. CA: consequently, the square AL

-

Q

D

AL2 :: CA2 : DH2. DE or rect. HE. Eh; the rectangle HE. Eh.

But, by similar trian. PA: AL:: GE: EH,

and, by the same,

QA

:

Al :: EK: Eh;

therefore, by comp. PA. AQ: AL2 :: GE.EK : HE.Eh;

and, because AL HE. Eh, therefore PA. AQ GE. EK.

But the parallelograms CGEK, CPAQ, being equiangular, are as the rectangles GE. EK and PA. AQ.

And therefore the parallelogram GK the parallelogram PQ.

That is, all the inscribed parallelograms are equal to one another. Q. ED. Corol. 1. Because the rectangle GEK or CGE is constant, therefore GE is reciprocally as CG, or CG: CP:: PA: GE. And hence the asymptote continually approaches towards the curve, but never meets it; for GE decreases continually as CG increases; and it is always of some magnitude, except when CG is supposed to be infinitely great, for then GE is infinitely small or nothing. So that the asymptote CG may be considered as a tangent to the curve at a point infinitely distant from C.

Corol. 2. If the abscissas CD, CE, CG, &c., taken on the one asymptote, be in geometrical progression increasing; then shall the ordinates DH, EI, GK, &c., parallel to the other asymptote, be a decreasing geometrical progression, having the same ratio. For, all the rectangles CDH, CEI, CGK, &c., being equal, the ordinates DH, EI, GK, &c., are reciprocally as the abscissas CD, CE, CG, &c., which are geometricals. And the reciprocals of geometricals are also geometricals, and in the same ratio, but decreasing, or in converse order.

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PROF. XXIII.

The three following spaces, between the asymptotes and the curve, are equal; namely, the sector or trilinear space contained by an arc of the curve and two radii, or lines drawn from its extremities to the centre; and each of the two quadrilaterals, contained by the said arc, and two lines drawn from its extremities parallel to one asymptote, and the intercepted part of the other asymptote.

That is,

The sector CAE = PAEG = QAEK, all standing on the same arc AE.

For, as has been already shown, CPAQ= CGEK;
Subtract the common space CGIQ,

So shall the paral. PI = the paral. IK;

To each add the trilineal IAE,

Then is the quadril. PAEG = QAEK.

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P

Again, from the quadrilateral CAEK, take the equal triangle CAQ, CEK, and there remains the sector CAE QAEK.

Therefore, CAE = QAEK = PAEG.

Q. E., D.

APPLICATION OF ALGEBRA

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GEOMETRY.

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WHEN it is proposed to resolve a geometrical problem algebraically, or by algebra, it is proper, in the first place, to draw a figure that shall represent the several parts or conditions of the problem, and to suppose that figure to be the true one. Then, having considered attentively the nature of the problem, the figure is next to be prepared for a solution, if necessary, by producing or drawing such lines in it as appear most conducive to that end. This done, the usual symbols or letters, for known and unknown quantities, are employed to denote the several parts of the figure, both the known and unknown parts, or as many of them as necessary, as also such unknown line or lines as may be easiest found, whether required or not. Then proceed to the operation, by observing the relations that the several parts of the figure have to each other; from which, and the proper theorems in the foregoing elements of geometry, make out as many equations independent of each other, as there are unknown quantities employed in them the resolution of which equations, in the same manner as in arithmetical problems, will determine the unknown quantities, and resolve the problem proposed.

As no general rule can be given for drawing the lines, and selecting the fittest quantities to substitute for, so as always to bring out the most simple conclusions, because different problems require different modes of solution; the best way to gain experience, is to try the solution of the same problem in different ways, and then apply that which succeeds best, to other cases of the same kind when they afterwards occur. The following particular directions, however, may be of

some use.

1st, In preparing the figure, by drawing lines, let them be either parallel or perpendicular to other lines in the figure, or so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular bé opposite to that angle, and to fall from one end of a given line, if possible.

2d, In selecting the quantities proper to substitute for, those are to be chosen, whether required or not, which lie nearest the known or given parts of the figure, and by means of which the next adjacent parts may be expressed by addition and subtraction only, without using surds.

3d, When two lines or quantities are alike related to other parts of the figure or problem, the way is, not to make use of either of them separately, but to substitute for their sum, or difference or rectangle, or the sum of their alternate quotients, or for some line or lines in the figure, to which they have both the same relation.

4th, When the area, or the perimeter, of a figure, is given, or such parts of it as have only a remote relation to the parts required; it is sometimes of use to assume another figure similar to the proposed one, having one side equal to

unity, or some other known quantity. For, hence the other parts of the figure may be found, by the known proportions of the like sides, or parts, and so an equation be obtained. For examples, take the following problems.

PROBLEM I.

In a right-angled triangle, having given the base (3), and the sum of the hypothenuse and perpendicular (9); to find both these two sides.

Let ABC represent the proposed triangle right-angled at B. Put the base AB = 3 = b, and the sum AC + BC of the hypothenuse and perpendicular = 9 = s; also, let x denote the hypothenuse AC, and y the perpendicular BC.

N. B. In this solution, and the following ones, the notation is made by using as many unknown letters, x and y, as there are unknown sides of the triangle, a separate letter for each; in preference to using only one unknown letter for one side, and expressing the other unknown side in terms of that letter and the given sum or difference of the sides; though this latter way would render the solution shorter and sooner; because the former way gives occasion for more and better practice in reducing equations; which is the very end and reason for which these problems are given at all.

PROBLEM 11.

In a right-angled triangle, having given the hypothenuse (5), and the sum of the base and perpendicular (7); to find both these two sides.

Let ABC represent the proposed triangle right-angled at B. Put the given hypothenuse AC = 5 = a, and the sum AB + BC of the base and perpendicular = 7 = s; also let x denote the base AB, and y the perpendicular BC.

By substituting this value for x, gives s2-2sy + 2y = a2,

PROBLEM III.

In a rectangle, having given the diagonal (10), and the perimeter, or sum of all the four sides (28); to find each of the sides severally.

Let ABCD be the proposed rectangle; and put the diagonal AC 10=d, and half the perimeter AB + BC or AD + DC = 14 = a; also put one side AB=x, and the other BC= y.

Hence, by right-angled triangles,

And by the question

Then by transposing y in the 2d, gives
This value substituted in the 1st, gives
Transposing a2, gives

And dividing by 2, gives

By completing the square, it is

And extracting the root, gives

And transposing a, gives

x2 + y2

= d2,

x + y

= a,

= α

- Y,

a2

2y 2
y 2

y'

-

2ay + 2y

D

= d2,

y La

or 6, the values of x and y.

PROBLEM IV.

Having given the base and perpendicular of any triangle; to find the side of a square inscribed in the same.

Let ABC represent the given triangle, and EFGH its inscribed square. Put the base AB=b, the perpendicular CD = a, and the side of the square GF or GH = DI= x; then will CI = CD — DI = a — x.

Then, because the like lines in the similar triangles ABC, GFC, are proportional (by theorem 84, Geom.), AB: CD:: GF: CI, that is, b: a :: x: a — X. Hence, by multiplying extremes and means, ab bxax, and transposing bæ, gives ab ax + bx; then dividing by a + b, gives x =

ab = GF or GH, the a+b side of the inscribed square; which therefore is of the same magnitude, whatever the species or the angles of the triangles may be.

PROBLEM V.

In an equilateral triangle, having given the lengths of the three perpendiculars, drawn from a certain point within, on the three sides; to determine the sides. Let ABC represent the equilateral triangle, and DE,

DF, and DG, the given perpendiculars from the point D. Draw the lines DA, DB, DC, to the three angular points; and let fall the perpendicular CH on the base AB. Put the three given perpendiculars, DE = a, DF b, DG = c, and put x = AH or BH, half the side of the equilateral triangle. Then is AC or BC= 2x, and by right-angled triangles the perpendicular CH = ✓ AC2 — AH2 = √/ 4x3 — 22 = √ 3x2 = x √/ 3.

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