Με

Multiply the first and second of these equations by e, and the results are

uε - e2=eq, and wɛ=uɛ + ex. In each of these two equations, for us substitute its equal a from the second, and we have a -eq and we=a2 + ex. Multiply these two equations, and the remaining two of the given equations, and the final result is ava2 + ex, therefore v=a+a.

But, by theorem 5, we have R=a+x, therefore R=v.

OF THE HYPERBOLA.

DEFINITIONS RELATIVE TO THE HYPERBOLA.

290. That portion of the primary line between the vertices of the two opposite curves, is called the transverse axis.

291. A straight line, drawn perpendicularly to the transverse axis, between the transverse axis and the curve, is called an ordinate.

292. The middle of the transverse axis is called the centre of the figure.

293. In the annexed diagram, Aa is the transverse axis,

C, the middle of Aa, the centre; PM the ordinate, and
CP the abscissa.

HYPERBOLA.-THEOREM 1.

294. The squares of the ordinates of the axis are to each other as the rectangles of the two segments of the axis, from each ordinate to each of the two vertices of the opposite curves.

Let VRQ, passing through the common line of axis of two opposite cones, be a plane perpendicular to the cutting plane of the opposite sections; and let AMII'M' be one of the sections, HA the common line of section of the two planes, and let HA cut the two conic surfaces in A, X; then AX will be the primary axis; and let OMNM' be a section of the cone parallel to its base, RIQIʻ.

Because the plane of the base, RIQI', is perpendicular to the plane VRQ, the plane OMNM' will also be perpendicular to the plane VRQ; and, since each of the two planes AMII'M', RIQI', and AMII'M', OMNM', are perpendicular to the plane VRQ, their common sections II', MM', are perpendicular to the plane VRQ; therefore II', MM', are perpendicular to the lines RQ, ON, AH,

M

in the plane VRQ; and, because the plane VRQ passes through the axis of the cone, it will divide all the circles parallel to the base into two equal parts; therefore RQ, ON, will be diameters of the two circles; and, since the chords II', MM', are at right angles to the diameters RQ, ON, the chords II', MM', will be bisected in H and P. Therefore HI-HI', and PM=PM'. Let CA=CX=a, CP=CX=x, PM=y, CH=x, HI=7, PN=t, PO=u, HQ=v, and HR = w.

Therefore, eliminating t, u, v, w, by multiplying the given equations, there will result 2 (x− a) (x+a) = y2(x−a)(x+a), or, by actual multiplication,

y2

-

y2(x2 — a2) = y2(x2 —a); therefore y2: y2 :: (≈ − a) (≈ + a) : (x − a) (x + a).

2

y2

295. COROLLARY 1.---Hence is a constant quantity.

2

296. COROLLARY 2.-Hence if ≈ be made constant, y will be constant also. Therefore, when ≈2—a2 becomes a2, let y=b, and, consequently, y2

b2
a2=

297. COROLLARY 3.-Hence every chord, perpendicular to the transverse axis, is bisected by the same axis.

298. COROLLARY 4.-Hence the tangent at the extremity of the transverse axis is bisected by the transverse axis.

299. COROLLARY 5.-Hence ay2=b2 (x2 - a2)=b2x2 — a2b2.

300. The constant value b of 7, when 2-a2 becomes a2, is called the semiconjugate axis, or twice b the conjugate axis.

301. A third proportional to the transverse and conjugate axes, is called the parameter, or latus rectum.

Thus a and b being the semi-transverse and semi-conjugate axes, 2a:2b::2b:p, the parameter; therefore, ap = 262, or if ƒ=p, we shall have af = b2, therefore

302. That point in the axis cut by an ordinate which is equal to half the parameter is called the focus.

303. If there be two opposite hyperbolas, and two others having the same centre, and their common line of axis at right angles to that of the two former, the transverse equal to the conjugate of the two former, and the conjugate equal to the transverse of the two former; these four hyperbolas are called conjugate hyperbolas.

304. Any straight line, drawn through the centre, and terminated by opposite curves, is called a diameter.

A diameter which is parallel to a tangent, at the extremity of another diameter, is called a conjugate diameter to that other diameter.

305. A straight line, parallel to a tangent, meeting the diameter in one extremity, and the curve in the other, is called an ordinate to that diameter.

306. The distance between the centre and an ordinate is called the abscissa.

In the diagram here annexed, the straight line Aa, drawn through the centre, C, is a diameter; and, if AT is a tangent at A, then Bb, parallel to AT, passing through the centre C, is the conjugate diameter; PM parallel to AT, is the ordinate to Aa, and CP is the abscissa.

HYPERBOLA.—THEOREM 2.

307. The square of the transverse axis is to the square of the conjugate axis as the rectangle of the two distances from the ordinate to the vertex of each curve.

CA2: CB2 :: PA x Pa : PM.

308. COROLLARY 1.-Hence every pair of opposite hyperbolas has two focii at an equal distance from the centre; because y2=(a-x2): and, since the origin of the abscissa commences at the centre, therefore the ordinate y must be the same at the same distance on each side of the centre.

309. COROLLARY 2.-Hence the tangent at the vertex of either curve is parallel to the ordinates, and, consequently, perpendicular to the transverse axis.

HYPERBOLA. THEOREM 3.

310. The square of the conjugate axis is to the square of the transverse axis, as the sum of the squares of the semi-conjugate axis; and that of the ordinate is to the square of the abscissa as CB2 : CA2 :: CB2+PM2 : CP2.

For, (theorem 2,) a2y' = b2x2-a2b2; and, therefore, by transposition, a2(y2+b2) = b2x2, consequently, b3: a2 :: y3 + b2 : x3. (See figure, theorem 2.)

HYPERBOLA.—THEOREM 4.

311. The square of the distance of the focus from the centre is equal to the sum of the squares of the two axes.

CF2=AC2+ BC2

Let the ordinate FG, which passes through the focus, be denoted by f, and CF, the distance of the focus from

the centre, by ε.

Then, by the equation of the co-ordinates, .. a'y2=b2(x2 — a2).

Now, in the first of these equations, when the abscissa x becomes equal to ε, the ordinate y will become f; and, consequently, a2ƒ2 = b2(e-a2); whence, b'—b2 (e2—a2) or b2=-a2, and, by transposition, a2+b2.

312. COROLLARY 1.-The two semi-axes, and the distance of a focus from the centre, are the sides of a right-angled triangle, ACB, and the hypothenuse, AB, is equal to the distance of the focus from the centre.

313. COROLLARY 2.-The conjugate axis, CB, is a mean proportional between AF and Fa, or fa, fA, the distances between either focus and the two vertices; for b2 = ε2 — a2 = (e — a) (e+a).

314. COROLLARY 3.-Hence, if e=ca, then will y2=a2 − x2+c2x2 - c2a3; for, (theorem 2,) a2y=b2(x-a2). But, by this theorem, we have b'e-a2; then, multiplying these two equations, we have a2y2 = (è2 — a2) (x2 — a2); or, substituting ca for e, we shall have y2= (c2 — 1) (x2 — a2) =a3 — x2 + c2x2 — c2 a2.