Ex. 12. If the lengths of two tangents be a, b, and the angle between them w; the parameter. find Draw the diameter bisecting the chord of contact; then the parameter of that dia meter is p' y2 = and the principal parameter is p = y2 sin20 w2y? = = 4.x3 (where w is the length of the perpendicular on the chord from the intersection of the tangents). But = ab sin W1 and 2wy 4ab sin w 16x2 = a2 + b2 + 2ab cosw; hence p = (see p. 180). (a2 + b2 + 2ab cos w) Ex. 13. Show, from the equation of the circle circumscribing three tangents to a parabola, that it passes through the focus. The equation of the circle circumscribing a triangle being (Art. 105) By sin A + ya sin B + aß sin C = 0, the absolute term in this equation is found (by writing at full length for a, a cosa + y sin a -p, &c.) to be p'p" sin (3 − y) + p′′p sin (y - a) +pp'sin (a - ẞ). But if the line a be a tangent to a parabola, and the origin the focus, we have (Art. 224) p cos a + sin (ya) cosß + sin (a – B) cosy}, which vanishes identically. Ex. 14. Find the locus of the intersection of tangents to a parabola, being given either (1) the product of sines, (2) the product of tangents, (3) the sum or (4) difference of cotangents of the angles they make with the axis. Ans. (1) a circle, (2) a right line, (3) a right line, (4) a parabola. 233. We add a few miscellaneous examples. Ex. 1. If an equilateral hyperbola circumscribe a triangle, it will also pass through the intersection of its perpendiculars (Brianchon & Poncelet: Gergonne, Annales, xi. 205; Walton, p. 283). The equation of a conic meeting the axes in given points is (Ex. 1, p. 137) bb'x2+Bxy+aa'y2 bb' (a + a') x - aa' (b + b) y + aa'bb' = 0. And if the axes be rectangular, this will represent an equilateral hyperbola (Art. 178) if aa =- bb. If, therefore, the axes be any side of the given triangle, and the perpendicular on it from the opposite vertex, the portions a, a, b, are given, therefore b' is also given; or the curve meets the perpendicular in the fixed point y (Ex. 7, p. 35) the intersection of the perpendiculars of the triangle. Ex. 2. Given a triangle, such that any vertex is the pole of the opposite side with respect to an equilateral hyperbola; the circle circumscribing the triangle passes through the centre of the curve. [This is a particular case of a theorem to be proved in the next Chapter (Brianchon & Poncelet, Gergonne, xi. 210; Walton, p. 304).] Take two sides of the triangle for axes; now the pole of the axis of x, with regard to a conic given by the general equation, lies on the diameter bisecting chords parallel to that axis (2Ar + By + D = 0), and also on the polar of the origin (Dx + Ey + 2F = 0). If, then, we have DE = 2BF, both these lines will meet the axis of y in the same point, and the pole of the axis of x will be the point y case the pole of the axis of y will be the point on the axis of x, x = B' or The equation of the circle through the origin and through these two points is B(x2+ 2xy cosw + y2) + Ex + Dy = 0, x (2Cy+ Bx + E) + y (2Ax + By + D) − 2 (A + C − B cos w) xy = = 0, an equation which will evidently be satisfied by the co-ordinates of the centre, provided we have A + C = B cosw, that is to say, provided the curve be an equilateral hyperbola (Arts. 70, 178). If DE be not 2BF, we have still proved that the circle passes == through the centre, which is described through the origin and through the points E 이 that is to say, through the points where each axis is met by the diameter bisecting chords parallel to the other. Hence, a circle described through the centre of an equilateral hyperbola, and through any two points, will also pass through the intersection of lines drawn through each of these points parallel to the polar of the other. Ex. 3. If on any tangent to a conic there be taken points A, B, such that AB may be constant; find the locus of the intersection of tangents from A and B (see the section on the Anharmonic Properties of Conics). The points where a pair of tangents to a conic, given by the general equation, meet the axis of x are found (Art. 150) from the equation. {(4AC — B2) y'2 + (4AE − 2BD) y' + 4AF − D2 } x2 + 2 { (BD – 2AE) x'y' + (2CD – BE) y2+(D2-4AF) x' + (DE - 2BF) y'} + {(4AF - D2) x2+(4BF - 2DE) x'y' +(4CF - E2) y'2} = 0. Forming the difference of the roots of this equation, and putting it equal to a constant, we obtain the equation of the locus which will be in general of the fourth degree; but if D2 = 4AF, the axis of a will touch the given conic, and the equation of the locus will become divisible by y2, and will reduce to the second degree. We could, by the help of the same equation, find the locus of the intersection of tangents; if the sum, product, &c., of the intercepts on the axis be given. 234. It is always advantageous to express the position of a point on a curve, if possible, by a single independent variable, rather than by the two co-ordinates x'y'. We shall, therefore, find it useful, in discussing properties of the ellipse, to make a substitution similar to that employed (Art. 100) in the case of the circle; and shall write x' = a cos p, y = b sin 4, * The use of this angle occurred to me some years ago, as a particular case of the methods given in Chapter XIV. It has, however, been already recommended by Mr. O'Brien in the Cambridge Mathematical Journal, vol. iv. p. 99, and has since been introduced by him, under the name here adopted, into his treatise on Plane Co-ordinate Geometry, p. 111. a substitution, evidently, consistent with the equation of the ellipse The geometric meaning of the angle is easily explained. ф a If we describe a circle on the axis major as diameter, and produce the ordinate at P to meet the circle at Q, then the angle b QCL=p, for CL = CQ cos QCL, or x = a cos; and PL = - QL x' (Art. 166); or, since QL = a sin p, we have y' = b sin p. 235. Some important consequences may be drawn from this construction. If we draw through P a parallel PN to the radius CQ, then B D' a line a be inflected to the minor axis, its intercept to the axis major = b. If the ordinate PQ were produced to meet the circle again in the point Q', it could be proved, in like manner, that a parallel through P to the radius CQ' is cut into parts of a constant length. Hence, conversely, if a line MN, of a constant length, move about in the legs of a right angle, and a point P be taken so that MP may be constant, the locus of P is an ellipse, whose axes are equal to MP and NP. (See Art. 230, Ex. 1.) On this principle has been constructed an instrument for describing an ellipse by continued motion, called the Elliptic Com- . passes. CA, CD', are two fixed rulers, MN a third ruler of a constant length, capable of sliding up and down between them, then a pencil fixed at any point of MN will describe an ellipse. If the pencil be fixed at the middle point of MN it will describe a circle. (O'Brien's Co-ordinate Geometry, p. 112.) 236. The consideration of the angle & affords a simple method of constructing geometrically the diameter conjugate to a given one, for Let the ordinate at the Q Hence we obtain the following construction for drawing the diameter conjugate to any given one. given point P, when produced, meet the semicircle on the axis major at Q, join CQ, and erect CQ'perpendicular to it; then the perpendicular let fall on the axis from Q'will pass through P', a point on the conjugate diameter. P' M' C M Hence, too, can easily be found the co-ordinates of P' given in Art. 176, for, since From these values it appears that the areas of the triangles PCM, PCM', are equal. Ex. 1. To express the lengths of two conjugate semidiameters in terms of the angle 4. Ans. a2 = a2 cos2 + b2 sin20; b'2 = a2 sin2 + b2 cos2. Ex. 2. To express the equation of any chord of the ellipse in terms of p and p' (see p. 93), Ans. a y cos(+)+sin(+4)= cos (p − 4'). Ex. 3. To express similarly the equation of the tangent. Ex. 4. To express the length of the chord joining two points a, ß, D2 = a2 (cosa cos 3)2 + b2(sina - sin 3)2 D = 2 sin (a - ẞ) { a2 sin2 } (a + ẞ) + b2 cos2 } (a + ẞ) } §. But (Ex. 1) the quantity between the parentheses is the semidiameter conjugate to that to the point (a + ẞ); and (Ex. 2, 3) the tangent at the point (a + B) is parallel to the chord joining the points a, ß; hence, if b' denote the length of the semidiameter parallel to the given chords, D 2b' sin (à - ẞ). - Ex. 5. To find the area of the triangle formed by three given points a, ẞ, y. 2Σ = ab { sin (a − ẞ) + sin (ß − y) + sin(y − a)} =ab (2 sin (a-ẞ) cos (a-3)-2 sin (a-ẞ) cos(a+B-2y)} = 4ab sin † ( a − ẞ) sin † (ẞ − y) sin § ( y − a) Σ = 2ab sin (a - ẞ) sin (ẞ − y) sin (y − a). Ex. 6. To find the radius of the circle circumscribing the triangle formed by three given points a, ß, y. def b'b"b"" = ΑΣ = ab If d, e, f be the sides of the triangle formed by the three points, R where b', b", b" are the semidiameters parallel to the sides of the triangle. If c', c", c" c' c" c"" be the parallel focal chords, then (see p. 193) R2= 4p to Mr. MacCullagh, Dublin Exam. Papers, 1836, p. 22; see also Crelle, vol. XL. p. 31.) Ex. 7. To find the equation of the circle circumscribing this triangle. (These expressions are due From this equation the co-ordinates of the centre of this circle are at once obtained. Ex. 8. To find the locus of the intersection of the focal radius vector FP with the radius of the circle CQ. Let the central co-ordinates of P be x'y', of Oxy, then we have, from the similar triangles, FON, FPM, Now, since is the angle made with the axis by the radius vector to the point O, we at once obtain the polar equation of the locus by writing p cos for x, p sin for y, and we find Hence (Art. 199) the locus is an ellipse, of which C is one focus, and it can easily be proved that F is the other. Ex. 9. The normal at P is produced to meet CQ; find the locus of their intersection. The equation of the normal is (Art. 184) but we may, as in the last example, write p cos and p sin for x and y, and the equation becomes or (a - b) p = c2, p = a + b. The locus is, therefore, a circle concentric with the ellipse. |