132. Conjugate Hyperbolas.-The equation (x-2y-1)(x+3y-2)=0 represents two intersecting straight lines, and (x−2y−1)(x+3y−2)+c=0 represents, for any constant value of c, some hyperbola meeting these lines only at infinity; i. e., having them as asymptotes. If c is positive, the hyperbola will lie in one pair of vertical angles formed by the asymptotes, and if c is negative, it will lie in the other pair. The two hyperbolas determined by and (x-2y-1)(x+3y-2)+a=0 (x-2y-1)(x+3y-2)-a=0 are said to be conjugate to each other. For instance, the hyperbola we have been discussing was x2+xy-6y2-3x+y-4=0, its asymptotes -6. For so that the value of c for the original hyperbola was -6. its conjugate, it is +6; the conjugate is therefore x2+xy-6y2-3x+y+8=0. Examples. 1. Trace 2y2+xy — 6x2 - 10y—27x-24-0 and its conjugate. 2. Trace y2+xy−2x2 −y+7x=0 and its conjugate. 3. Find the equation of the locus of points as far from (-2, 1) as from y=2x, and trace the curve. 133. An equation ax2+bxy+cy2+dx+ey+f=0, if b2-4ac<0, ordinarily represents an ellipse; if b=0 and a=c, it represents a circle, and it may happen that y is real for only one real value of x, or for none; in the first case the equation represents a single point, or null ellipse, in the second, an imaginary ellipse. If b2—4ac=0, the equation ordinarily represents a parabola, but it may happen to represent two parallel or two coincident straight lines. If b2-4ac>0, the equation ordinarily represents a hyperbola, but it may happen to represent two intersecting straight lines. Miscellaneous Examples. Trace the following: 1. y2-2xy+x2-2y-x+4=0. 2. 4x2-12xy+13y2+24x−52y+16=0. 3. x2-xy-3 and y2+xy=10, showing intersections. 4. x2-y2=0 and 3x2-4xy+5y=9, showing intersections. 5. 5x2+4xy-y2+4y-8x-5=0. 6. 4y2-4xy+x2+8y-12x-12=0. 7. 5x2+4y2+4xy+4y-8x-5=0. 8. 4x2+49y2-28xy+12x-42y+9=0. 9. 4y2-4xy+17x2-8y+132x+196=0. 10. 4x2+49y2—28xy+16x-56y+15=0. 11. 2x2+xy-6y2 — 6x+16y—8=0. 12. 4x2-4xy+2y2 −8x+2y-4=0. 16. 5x2+8xy+4y2-21x-20y+19=0. 19. xy+2x-4y=0. 20. xy+x2=3y. 21. xy+x2-2y2-3x=0. 22. 4y2+4xy+x2-x-4=0. 23. 4y2—4xy+5x2+8y−28x+40=0. 24. 2y2+7xy-4x2-6x+3y-5=0. 25. 3x2-5xy-2y2+6x-5y-3=0. 26. 4y2-4xy+65x2+8y−68x−236=0. 27. 2x2-2xy + y2+2x-4y+1=0. 28. y2+xy-2x2-4y-2x+8=0. 29. y2-2xy+x2+20y-21x+75=0. CHAPTER VIII. UNDETERMINED COEFFICIENTS 134. We have several times had occasion to determine a function when we knew its general form and one or more of its values. For instance, in finding the equation of a line through a given point parallel or perpendicular to a given straight line, we knew the general form of the equation to be y=mx+b, and from the given conditions we determined the values of the constants m and b. We used the same method in finding the tangent to a circle from an outside point, and in determining the asymptotes of a hyperbola. (Arts. 117 and 128.) When a function is assumed with coefficients which are to be determined from given conditions, the coefficients are called undetermined coefficients. In the example cited, m and b are undetermined coefficients in the function mx+b. 135. The method of undetermined coefficients is useful even when we do not know precisely the general form of the desired function, for we can apply it to any form that seems probable and then test the results for accuracy. For instance, suppose we think that x+2x3 − 35x2+12x-1 may have factors of the form (x2+ax−1)(x2+bx+1). We assume the identity. - x2+2x3-35x2+12x−1 = (x2+ax−1) (x2+bx+1), which expresses the fact that its two members are different expressions for the same function and are therefore equal for any value of x. On this assumption, we have: If x=1, if x=-1, -21=a(2+b), -49-a (2-b), whence a=7, b=-5 and the factors are (x2+7x−1)(x2−5x+1). Multiplying these together, we find that the assumption was correct. If we treat x+2x3 — 42x2+12x-1 in precisely the same way, we find a=%, b=−6, but the product of the factors (x2+7x−1) (x2 −6x+1) is x+x3-42x2+13x−1; so that, in this case, the assumption is not correct. Examples. 1. Find the equation of the line joining (1, 3) and (-2, 1) by assuming it to be y=mx+b, and determining m and b from the fact that each point lies on the line. 2. Find the tangent to x2+y2=9 through (9,7); first, assuming its equation to be y=mx±3V1+m2; second, assuming its equation to be y=mx+b and determining m and b from the facts that (9,7) is on y=mx+b and that the intersections of the circle and the line are coincident, so that x2+(mx+b)2=9 has equal roots, and thus a zero discriminant. 3. Factor x-3x3+7x-15, assuming the factors to be of the form (x2+ax+3)(x2+bx-5). 4. Factor x+5x3- 4x2-27x+21= (x2+ax−3) (x2+bx−7). 136. Partial Fractions.-It is often necessary to transform a rational fraction into an equivalent sum of simpler rational fractions, called partial fractions. The following rules give assumptions that will lead to correct results: The denominator of a given proper fraction having been factored as completely as possible: (1) Corresponding to every linear factor x-a, assume a partial fraction of the form A |