Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/279

 SINGULAR POINTS Examine the following curves for asymptotes : . p cos = a cos 2 0. Ans. There is an asymptote perpendicular to the initial line at a distance a to the left of the origin. . p = a tan 6. Ans. There are two asymptotes perpendicular to the initial line and at a dis- tance a from the origin, on either side of it. . The lituus p6% = a. Ans. The initial line. . p = a sec 20. Ans. There are four asymptotes at the same distance - from the origin, and inclined 45 to the initial line. . (p - a) sin = b. Ans. There is an asymptote parallel to the initial line at the distance b above it. . p = a (sec 2 + tan 2 0}. Ans. Two asymptotes parallel to, at distance a on each side of origin. . Show that the initial line is an asymptote to two branches of the curve p 2 sin B = a 2 cos 2 6. . Parabola p = costf 154. Singular points. Given a curve whose equation is Ans. There is no asymptote. Any point on the curve for which = and ^ = dx dy is called a singular point of the curve. All other points are called ordinary points of the curve. Since by (57 a), p. 199, we have . dy dx it is evident that at a singular point the direction of the curve (or tangent) is indeterminate, for the slope takes the form In the noxt section it will be shown how tangents at such points may be found. . Determination of the tangent to an algebraic curve at a given point by inspection. If we transform the given equation to a new set of parallel coordinate axes having as origin the point in question on the curve, we know that the new equation will have no constant term. Hence it may be written in the form f(x, y) = ax + ly + (ex* + dxy + ey 2 )