and, by theorem 3, cor. 4, PN PM :: HL: HI... PM × HL= PN × HI. X Therefore, by multiplication, we shall find HL=HO, which is impossible; therefore TM does not cut the ellipse, and it must, in consequence, be a tangent at M. ELLIPSE.- THEOREM 7.. 278. In the straight line, Ta, of the axis major, the semi-axis, CA, is a mean proportional between the abscissa, CP, and the distance, CT, from the centre to the intersection of the tangent. For the triangle CNT (see the preceding diagram) is right-angled at N, since PN meets the circle at N; and, because PN is perpendicular to TC, we shall have, by the similar triangles PCN, NCT, .... CP x CT = CN2 and by the circle Therefore, by multiplication, CP x CT = CA2. CN2 CA2. If CT=u, this conclusion, analytically expressed, is ux-a2; or, if s be the subtangent, then will u=s+x; and, therefore, (s+x) x=a2, that is, sx + x2=a2. 279. COROLLARY 1.-Whence, if x be any other abscissa, and S its subtangent, and if s+x=v, then vx=a2; and, since ux-a2, therefore ux=vz. 280.-COROLLARY 2.-When the abscissa becomes equal to the excentricity, a becomes ; and, consequently, ux-a becomes uɛ=a2, or Sε+2=a2. ELLIPSE. THEOREM 8. 281. The line bisecting the angle at any point in the curve, formed by one of the lines drawn from one of the focii, and the prolongation of the line drawn from the other focus, will be a tangent to the curve. Let MT bisect the angle ƒMK; then, if MT is not a tangent, it will cut the ellipse; let p be the point where it meets the curve, and, making MK=Mƒ, join fK, fp, Fp, Kp, and let MT cut Kf, in z. 1 Then, in the triangles ƒMx, KMs, fM being = KM, M≈ common, and the included angles ƒMx, KMx, equal; the base fx is = Kx, and the angle fx M=K≈M. Again, in the triangles fpz, Kpx, fx being = Kx, pz common, and the angles føp, Køp, equal, the base fp is =Kp. Therefore Fp+pK=Fp+pf = FM+Mƒ=FM+MK=FK; wherefore the sum of the two sides, Fp, pK, of the triangle FpK, is equal to the third FK; which is impossible; therefore as MT cannot cut the ellipse: it must be a tangent. CN f a 282. COROLLARY.-Hence, because the angle KMƒ is bisected, the angle FMƒ will also be bisected by the normal MN. For since the angle TMƒ is =TMK, and TMK equal to the opposite angle FML; therefore TMƒ is FML: and, because the angles TMN and LMN are each a right angle; therefore, taking away the angles TMƒ, LMF, which are equal, from the right angles TMN, LMN, there will remain the angle NMƒ equal to the angle NMF. ELLIPSE. THEOREM 9. 283. If there be any tangent meeting the four perpendiculars to the transverse axis, the one at the extremity of the transverse, next to the point of intersection, will be to that which terminates in the point of contact, as that which passes through the centre is to the remaining one at the other extremity of the transverse. AL: PM :: CŊ : αO Let AL=k, PM-1, CN=m, and a0=n. Then, if we can show that the distances TA, TP, TC, T, are four proportionals, AL, PM, CN, αO, being the homologous sides of similar triangles, will likewise be proportionals. M T P By theorem 7, (278,) sx+x2=a2; therefore, by transposition, sx+x2 — a2=0; to each side of this equation add s2+sx, and we have s2+2sx+x2—a2=s2+sx ; that is, because s2+2sx+x2 is a complete square, and that s2+sx=s(8+x) (s+x)3—a2=s(s+x), that is, (s+ x + a) (s+x− a) =s(s+x). Whence s+x-a: s::s+x: s+x+a. ELLIPSE. THEOREM 10. 284. Every parallelogram, UVWX, circumscribing an ellipse, having its sides parallel to two conjugate diameters, is equal to the rectangle of the For, multiplying the first four equations, a2-x=x; or, by transposition, a2-2x2. Multiply this and the three remaining equations and pn=ab. ELLIPSE. THEOREM 11. 286. The sum of the squares of any pair of conjugate diameters is equal to the sum of the squares of the two axes. Let CM=m, CI=n. See the preceding figure. Therefore, by adding these together, we have a2+b2 m2 +n2. N.B. All these theorems concerning the Ellipse, and their demonstrations, are in the very same words as the corresponding number of those for the Hyperbola, next following; having sometimes only the word sum changed for the word difference. ELLIPSE. THEOREM 12. 287. If a chord be bisected, the tangent at the extremity of the diameter, passing through the point of bisection, will be parallel to that chord. Through M, m, draw LN, In, to meet the circumference of the circle described on the axis DE, in the points, N, n. Produce Mm and DE, to meet each other in S, and join D Sn, SN, and let the chord Mm be bisected in P. Then, because Sl: Im :: SL : LM ... LC therefore eliminating LM, Im, ... Sl: In :: SL : LN. Therefore Sn and SN are in a straight line. Through P draw qr, meeting DE in q, and Nn in r; then Nn is bisected in r. Join Cr, and produce it to meet the circumference in R, and draw the tangent RT, meeting DE produced in T; then RT is parallel to NS. Draw RG, perpendicular to DE, meeting DE in G, and CA in A. And, therefore, the point A is in the curve of the ellipse. Then drawing AT, AT is a tangent by the preceding proposition. 288. The rectangle of the squares of any semi-diameter, and of an ordinate to it, is equal to the rectangle of the square of the semi-conjugate and the difference of the squares of the semi-diameter of the abscissa, and of the abscissa itself. CA2 × PM2 = BC2 × (CA2 – CP2). Draw rK parallel to PM, cutting MN in K, and draw CG parallel to Nn, cutting the circumference in G, and CB parallel to PM, cutting the ellipse in B, and join BG. Let CR=r, Cr=u, rN=rn=v, CA=a, CB=b, CP=x, PM=rK=y. By the equation of the circle..... ..r3y2=b2v2 ....rx2=a2u2. R E Multiply the first and second equations together, and r2y2=b2r2 — b2u2; or, by transposition, b2 u2=r2 (b2-y2). Multiply this and the third equation, and b2u2±r2 b2x2=a2 (b2—y2); or, by transposition, a2 y2=b2 (a2 — x2). ELLIPSE. THEOREM 14. 289. The semi-ordinate, together with its prolongation to meet a tangent at the extremity of a latus rectum, is equal to the radius vector through the same focus with that of the latus rectum. |