Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/277

 ASYMPTOTES FOURTH STEP. Substitute these values of m and k in y = mx -f~ k. This gives the required asymptotes. ILLUSTRATIVE EXAMPLE 2. Examine y 3 = 2 ax 2 x 3 for asymptotes. Solution. Since none of the terms involve both A y x and y, it is evident that there are no asymptotes parallel to the coordinate axes. To find the oblique asymptotes, eliminate y between the given equation and y = mx + k. This gives (mx + k) 3 = 2 ax 2 x 3 ; and arranging the terms in powers of x, (1 + m 3 ) x 3 + (3 m"k - 2 a) x 2 + 3 k 2 mx + k 3 = 0. Placing the first two coefficients equal to zero, + m 3 = and 3 m?k 2a = Q. Solving, we get m 1, k = Substituting in y = mx + fc, we have y = x +  3 the equation of asymptote AB. EXAMPLES Examine the first eight curves for asymptotes by the method of 150, and the remaining ones by the method of 151 : . y = e?. Ans. y = 0. 2. y e-**. Ans. y = Q. . y = logx. Ans. x = 0. x y = e, x= 1. n being any odd integer, x = x = 0, y 0. y = x + 2. . y = tanx. i . ?/ = e*-l. . y 3 = . Show that the parabola has no asymptotes. . y 3 = a 3 - x 3 . . The cissoid y 2 = . y*a = y z x + x 3 . . y 2 (x 2 a) = x 3 a 3 . . x 2 ?/ 2 = a 2 (x 2 + y 2 )- . y(i . The folium x 3 + y 3 - 3 axy = 0. . The witch x 2 y = 4 a 2 (2 a - y). . xy 2 + x 2 y = a 3 . 20. y + x = 0. x = 2r. x = a. . y=x. x = 2a, y -(x + a). x =, y = a. x = 6, x-26, y + 3a y = c, X = b. y + x + a = 0. x = 0, y = 0, x + y = 0. L, x--2y = 0, x + y = 1, x y 1.