the axis forming a discontinuous branch, of the same shape as the continuous branch above the axis. Next consider what happens when x is negative. Let the coordinates of any point P on the branch in the first quadrant be (x, y), then ON=x. Take On - along the negative portion of the axis of x, then, if p be the corresponding point on the curve, we have pn=(-x)-x, PN=2, and therefore pn. PN=(-1)*, which may be = 1, −1, or imaginary, according to the particular value of x. Hence, when the ordinate pn is real, its magnitude is inverse to that of the corresponding ordinate PN. Hence on this curve we have two infinite series of conjugate points, as shown in the figure. For an account of M. Vincent's memoir and criticisms upon it see Dr. Salmon's "Higher Plane Curves," p. 275, or a paper by Mr. D. F. Gregory, “Camb. Math. Journal,” vol. i., pp. 231, 264. 262. Maclaurin's Theorem with regard to Cubics. We conclude the present chapter with an important theorem with regard to cubic curves, which is due to Maclaurin. If a radius vector OPQ be drawn through a point of inflexion (0) of a cubic, cutting the curve again in P and Q, to show that the locus of the extremities of the harmonic means, between OP and OQ, is a straight line. If the origin be taken at the point of inflexion and the tangent at the point of inflexion as the axis of y, the equation of the cubic must assume the form where u is the most general expression of the second and lower degrees, viz., ax2+2hxy+by2+2gx+2fy+c, for it is clear that the axis of y cuts this curve in three points ultimately coincident with the origin. The equation (1) when put into polars takes the form Lr2+Mr+N=0, where L= sin30+(a cos20+2h sin cos 0+b sin20) cos 0, N=c cos 0. r. 2 If r1, r, be the roots of this quadratic, and harmonic mean between them, we have which shows that the Cartesian Equation of the locus of the extremity of the harmonic mean is the straight line gx+fy+c=0. 263. It is obvious from Art. 187 that the equation of the polar conic of the cubic (1) with regard to the origin is x(2gx+2fy)+2cx = 0, x(gx+fy+c)= 0. or Hence the polar conic of a point of inflexion on a cubic breaks up into two straight lines, one of which is the tangent at the point of inflexion, and the other the locus of the extremities of the harmonic means of the radii vectores through the point of inflexion. It appears from this that only three tangents can be drawn from a point of inflexion on a cubic to the curve, viz., one to each of the points in which the line gx+fy+c=0 meets the curve, and consequently also that their three points of contact lie in a straight line. 264. If a Cubic have three real points of Inflexion they are Collinear. It follows immediately from Maclaurin's Theorem above proved that if A and B be two points of inflexion on a cubic, the line AB produced will cut the curve in a third R 2 point C, which is also a point of inflexion on the cubic. For if B, B1, B, be the three ultimately coincident points on the cubic, which lie in a straight line (B being a point of inflexion), let AB, AB,, AB, cut the curve in C, C, C and let AH, AH,, AH, be the harmonic means between AB, AC; AB„, AC,; AB,, AC, respectively, then H, ÍÍ ̧ lie in a straight line by Maclaurin's Theorem, and B, B, B, lie in a straight line; therefore by a theorem in conic sections C, C, C, also lie in a straight line, and they are ultimately coincident points. C is therefore a point of inflexion. 2 1' 2 2 EXAMPLES. 1. Write down the equations of the tangents at the origin for each of the following curves :-- 2. Show that the curve ye* is at every point convex to the foot of the ordinate of that point. 3. Show that for the cubical parabola a2y = (x − b)3 there is a point of inflexion whose abscissa is b. 4. Show that for the semicubical parabola ay2 = x3 the origin is a cusp of the first species. 5. Show that the origin is a cusp of the first species on the a(y-x)2=x3. curve 6. Show that on the curve (ay-x2)2= ba3 there is a cusp of the first species at the origin, and a point of 7. Show that there are points of inflexion at the origin on each of the curves there is a node or a conjugate point according as a and b have like or unlike signs. 10. Show that there is a point of inflexion on the curve has a point of inflexion at the point in which it cuts the axis of x, and show that the tangent at the point of inflexion makes with the axis of x an angle tan-13. 12. Show that the curve b(ay-x2)2=25 has a cusp of the second species at the origin. 13. Show that, if n be greater than 2, the curve has a cusp at the origin of the first or second species according as n is less or greater than 4. 14. Find the two points of inflexion of the curve and draw figures showing the characters of the inflexions. 15. Show that every point in which the curve of sines cuts the axis of x is a point of inflexion on the curve. 16. Show that the points of inflexion on the cubic Show that these three points of inflexion lie on the straight line x = 4y. 17. Find by polars the points of inflexion on the curve 2x(x2 + y2) = a(2x2 + y2). 18. Show that the curve au= On has a point of inflexion 19. Show that if the origin be a point of inflexion on the U1 + U2 + Uz + = 0 curve u2 will contain u1 for a factor. 20. Show that there is a point of inflexion at the origin on the cubic y =axy + by2 + cx3. 21. Show that there is a point of undulation at the origin on the curve y = ax2 + bx2y2 + cya. 22. Show that the origin is a triple point on the curve and that there is a cusp of the first species there. |