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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.
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.
CA? : CB? :: PA x Pa : PM. For *-aixto)
(x-a) (x+a) Therefore a’ : 62 :: (x - a) (x +a) : yo.
308. COROLLARY 1.-Hence every pair of opposite hyperbolas has two focii at an equal distance from the centre; because yo = (a – xo): 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.
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 CB’ : CA' :: CB*+PM’: CP?.
For, (theorem 2,) aʻyo = box — a'b'; and, therefore, by transposition, a’(y+)=b*x*, consequently, 6° : a :: 4° +6°: x. (See figure, theorem 2.)
311. The square of the distance of the focus from the centre is equal to the sum of the squares of the two axes.
CF=AC? + BC
Then, by the equation of the co-ordinates, a’y*=6(** – a”).
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 b? = =—a' =(-a)(8+a).
314. COROLLARY 3.—Hence, if srca, then will y’=a’ – x*+cox – c*a*; for, (theorem 2,) a*y*=6*(x*—a”). But, by this theorem, we have b=&-a; then, multiplying these two equations, we have a’y=(8? – a) (x* — aʼ); or, substituting ca fore, we shall have yo = (c? – 1)(-x? – aʻ) =a? – xo+co.z? – cʻa’.
HYPERBOLA. THEOREM 5.
315. The difference of two lines, drawn from the focii to meet any point in the curve, is equal to the transverse axis.
Let FM=R, FM=r, FC=fC=e=ca.
By Geometry, (prop. 62,).. FM’=PM+FP), Ro=yo +2° +2acx+cod.
v=a-+c2-c^4, wherefore, eliminating y, by adding these equations together, we have the equation
Ro=co x* +2acxta'. then, extracting the roots of each side of this equation, we have R=cx+a =x+a. In like manner will be found r=cx- a=-x-a; therefore R-r=2a.
316. The line bisecting the angle, at any point in the curve formed by the two lines drawn from that point to each focus, is a tangent. The tangent MT at M will bisect the angle FM
f. In MF take MK =Mf, and in MT take any point L; join f L, and let it meet the curve in Q; join also KL, FL, FQ. Then, by hypothesis, the angle KML=fML, KM=fM, and LM is common, the base LK is =Lf; but the difference of any two sides of a triangle is less than the third ;* therefore FL-LK, or FL - Lf, is less than FK, or Ff, or FM - FM, or FQ-fQ. Hence f L is greater than fQ; for, since FL-fL is less than FQ-FQ, if f L were less than fQ, FL +LQ would be less than FQ; which is impossible. Therefore every point, L, in MT, except M, is without the curve of the hyperbola; and MT touches it at M.
* It is shown by every writer of Elementary Geometry, that the sum of every two sides of a triangle are greater than the third. Let a, b, c, be the three sides of a triangle; then, a + bc, a torb, b+cra: therefore, by transposition, arc-b,arb-c, brc-a, br-a-c,c=b-a, cra-b.
Ta f P
317. In the line of the axis major, the rectangle contained by the distance between the centre and the intersection of the tangent, and the distance between the centre and the ordinate, is equal to the square of the semi-axis major.
R= x + a
318. The semi-transverse axis is a mean proportional between the two distances in the line of the transverse axis; the one from the centre to the ordinate, and the other from the centre to the intersection of the tangent.
For, by the preceding proposition, ux ra’, therefore x: a :: a : u. (See figure, theorem 7.)
Or, if the subtangent PT =s, then will CT=CP-PT=x-s. Whence (x—s)x=a’ or x? — 8x =a? ; expressed as in the same proposition of the ellipse.
319. If there be any tangent, and four perpendiculars to the line of axis, contained between the tangent and the line of axis, the rectangle under two, which
passes through the vertices, will be equal to the rectangle of the third, : which passes through the centre, and the fourth which is the ordinate.
: aG XAINCH X PM
or aG : PM :: CH: AI. PT being = s.
See the preceding theorem. Then...... aP=CP-CA= -a
u=CT=CP-PT = x -8
AT=CA +CP-PT=a + x-S
aT=PT- AP= -x+ a. Now, by theorem 8, a' = x (x --s) = x2 — sx. Therefore, by transposition, a*+sx — xo=0; add sx - q? to each side of this equation, and a’ – x2 +25% — go = sx - s?, or a' – (x* – 25x+82) = s(x – s); and, since the difference of two squares is equal to the rectangle of the sum and difference of their roots, (a + x - s) (a-x+8)=8(x – 8). Whence
....AT xaT-PTXCT And, by similar triangles,
TCH, TAI. CT x AI =CH x AT
(TaG, TPM.. aG TP=aT PM. Therefore, eliminating AT, aT,PT, CT, by multiplying these equations and aG AI=CH X PM.
320. Given the transverse axis of an hyperbola and an ordinate, to find the conjugate axis and assymtotes, (which are two straight lines, such as, if
produced indefinitely with the curve, will never meet each other,) and thence to describe the curve itself.
Let Aa, (fig. 1, pl. V,) be the transverse axis, and let PM be an ordinate. Make PD equal to AP. Then on aD describe the semi-circle aND. Produce PM to N. Draw AR perpendicular to CD, and make AR equal to CA. Join NR, and produce NR and DA, if necessary, to meet each other in S; and draw MS, cutting AR in Q. Produce QA to T, and make AT equal to AQ. Then QT will be the conjugate axis, or AQ, AT, will each be the semi-conjugate axis. Through the points C, T, draw JH; and through the points C, Q, draw IK: then JH and IK are the assymtotes, by which the curve may be described.