0 n The expression for the tangent MT mav be found by means of the right-angled triangle PMT. 33. Any straight line E nCN which is drawn through the center of the BM ellipse, and terminated N at two opposite points of the curve, is called a P diameter, and if DCd be ! Κ α ΙΑΣΗ a straight line parallel to the tangent at N, the diameters DCd, nCN are said to be conjugale to each other; lines, as MP, parallel to the tangent at G N, are the ordinates of of the diameter CN, and the parts CP are the abscissæ. Lastly, the parameter or latus rectum of any diameter, is a third proportional to that diameter and its conjugate. 34. From the extremities D and N draw the two ordinates NQ. DI to the transverse axis Aa, call CQ=2, DI=u; because of the similar triangles DIC, NQT, we shall have NQʻ: QT?::DI : IC", or a> x* ::: aa ba bz Hence u = and therefore CQ (7): DI (u)::a:b; in the aa a' very same manner we shall find CI : NQ:: 0 :6. Consequently CQ : DI ::C1 : NQ; therefore the triangles DIC CNQ, are equal in surfact. Hence 1°, DI= b* CQ* =66–NQ', or DI'+NQʻ=56 11°, Cr=NQ*=aa-CQ*, or CIP+CQʻ=a* III', a? +.6-CI +CQ*+NQʻ+DI'=CD+CN'; that is, in the ellipse, the sum of the squares of any two conjugate diameters is always equal to the sum of the squares of the troo ares. IV°. If we draw ND, the surface of the triangle NCD will be expressed by + (DI+NQ) (CI+CQ)- CIxID-CQxNQ=> CIxNQ+ b 2)= + ab. ab sin P sin P sin P ") 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* + and ma +n?_2mn =a? +63 2ab ; sin P 2ab therefore m+n=(a* +*+ 5), and mắn=v(a'+b 2ab .). sin P 2ab Consequently m= $ v(a*+b++ sin) + } v(a?+b' — 2ab Qab sin :) 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? a? sin P CQ 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 d al r2 + becomes 62 m2 at2+t? ma? Illy. the similar triangles PNQ, DCI give NT: : 62 ? CD2 :: QT2 : CI? :: NQ?: D1%, or q%;? :: 82 : q?-? :: 92: whence, q? a? g2 t2 and al g2 therefore %_t?+t? therefore as + q? before y? = -(m?—?) 82 n? 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 +69=m* +ne", 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=} (m? 7. n* +2mn sin P)-4V(m+ 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. OF THE HYPERBOLA. 39. The equation yy = sin A cx sin B + xx sin (A+B B –180)} 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 16 KE aa sin (A+B-180°) N perbola M'am'. B 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 67 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?, аа e tyy=—2ax +xx, if we make aP=we shall aa |