right lines, according as ab-h is negative or positive. Hence, as we have already seen, p. 72, the condition that the general equation of the second degree should represent two right lines, is abc+2fgh — af2 — bg2 — ch2 = 0. For it must plainly be fulfilled, in order that when we transfer the origin to the point of intersection of the right lines, the absolute term may vanish. Ex. 1. Transform 3x2 + 4xy + y2 — 5x – 6y − 3 = 0 to the centre (7, — 4). Ex. 2. Transform x2 + 2xy - y2 + 8x + 4y - 8 = 0 to the centre (- 3, − 1). Ans. x2+2xy - y2 = 22. 154. We have seen (Art. 136) that when satisfies the condition Ө a cos2+2h cos sin ✪ + b sin20 = 0, the radius vector meets the curve at infinity, and also meets the curve in one other point, whose distance from the origin is C p=- g cose+fsin @" But if the origin be the centre, we have g=0, f=0, and this distance will also become infinite. Hence two lines can be drawn through the centre, which will meet the curve in two coincident points at infinity, and which therefore may be considered as tangents to the curve whose points of contact are at infinity. These lines are called the asymptotes of the curve; they are imaginary in the case of the ellipse, but real in that of the hyperbola. We shall show hereafter, that though the asymptotes do not meet the curve at any finite distance, yet the further they are produced the more nearly they approach the curve. Since the points of contact of the two real or imaginary tangents drawn through the centre are at an infinite distance, the line joining these points of contact is altogether at an infinite distance. Hence, from our definition of poles and polars (Art. 89), the centre may be considered as the pole of a line situated altogether at an infinite distance. This inference may be confirmed from the equation of the polar of the origin, gx+fy+c=0, which, if the centre be the origin, reduces to c=0, an equation which (Art. 67) represents a line at infinity. 155. We have seen that by taking the centre for origin, the coefficients g and ƒ in the general equation can be made to vanish; but the equation can be further simplified by taking a pair of conjugate diameters for axes, since then (Art. 143) h will vanish, and the equation be reduced to the form ax2 + by2+ c = 0. It is evident, now, that any line parallel to either axis is bisected by the other; for if we give to x any value, we obtain equal and opposite values for y. Now the angle between conjugate diameters is not in general right; but we shall show that there is always one pair of conjugate diameters which cut each other at right angles. These diameters are called the axes of the curves and the points where they meet it are called its vertices. We have seen (Art. 143) that the angles made with the axis by two conjugate diameters are connected by the relation b tane tane+h (tan 0 + tan 0′) + a = = 0. But if the diameters are at right angles, tan0' = — (Art. 25). Hence h tan30+ (a - b) tan ✪ – h = 0. 1 tan 0 We have thus a quadratic equation to determine . Multiplying by p3, and writing x, y, for p cose, p sine, we get hx3-(a - b) xy-hy" = 0. This is the equation of two real lines at right angles to each other (Art. 74); we perceive, therefore, that central curves have two, and only two, conjugate diameters at right angles to each other. On referring to Art. 75 it will be found that the equation which we have just obtained for the axes of the curve is the same a that of the lines bisecting the internal and external angles between the real or imaginary lines represented by the equation ax2+2hxy+by2 = 0. The axes of the curve, therefore, are the diameters which bisect the angles between the asymptotes; and (note, p. 71) they will be real whether the asymptotes be real or imaginary; that is to say, whether the curve be an ellipse or a hyperbola. 156. We might have obtained the results of the last Article by the method of transformation of coordinates, since we can thus prove directly that it is always possible to transform the equation to a pair of rectangular axes, such that the coefficient of xy in the transformed equation may vanish. Let the original; axes be rectangular; then, if we turn them round through any angle e, we bave (Art. 9) to substitute for x, x cos y sine, and for y, x sine+y cose; the equation will therefore become a (x cose-y sin 0)+2h (x cos- y sin 0) (x sin 0 + y cose) + b (x sino + y cos 0)2 + c = 0 or, arranging the terms, we shall have the new aa cos2+2h cos sin 0 + b sin20; the new hb sin cos 0 + h (cos2 0 — sin" 0) — a sin cose; the new ba sin20-2h cose sine + b cos2 0. Now, if we put the new h=0, we get the very same equation as in Art. 155, to determine tane. This equation gives us a simple expression for the angle made with the given axes by either axis of the curve, namely, 157. When it is required to transform a given equation to the form ax+by+c=0, and to calculate numerically the value of the new coefficients, our work will be much facilitated by the following theorem: If we transform an equation of the second degree from one set of rectangular axes to another, the quantities a+b and ab-h' will remain unaltered. The first part is proved immediately by adding the values of the new a and b (Art. 156), when we have a+b=a+b. To prove the second part, write the values in the last article 2a = a+b+2h sin 20+ (a - b) cos 20, Hence But 26′ = a+b-2h sin 20 — (a - b) cos 20. 4a'b' = (a + b) - {2h sin 20+ (a - b) cos 20}. 4h"= {2h cos 20 (a - b) sin 20}"; therefore 4 (a'b' — h2) = (a + b)2 — 4h" — (a - b)2 = 4 (ab — h3). When, therefore, we want to form the equation transformed to the axes, we have the new h=0, a'+b'=a+b, a'b' = ab - h3. Having, therefore, the sum and the product of a' and b′, we can form the quadratic which determines these quantities. Ex. 1. Find the axes of the ellipse 14x2 - 4xy + 11y2 = 60, and transform the equation to them. The axes are (Art. 155) 4x2 + 6xy — 4y2 = 0, or (2x − y) (x + 2y) = 0. We have a' + b′ = 25 ; a'b' = 150; a′ = 10; b′ = 15; and the transformed equation is 2x2+3y2 = 12. Ex. 2. Transform the hyperbola 11x2 + 84xy – 24y2 = 156 to the axes. a' + b' - 13, a'b' =- - 2028; a' 39; b' = - 52. Transformed equation is 3x2 - 4y2 = 12. Ex. 3. Transform ax2 + 2hxy + by2 = c to the axes. Ans. (a+b-R) x2 + (a + b + R) y2 = 2c, where R2 = 4h2 + (a - b)2. *158. Having proved that the quantities a+b and ab-h3 remain unaltered when we transform from one rectangular system to another, let us now inquire what these quantities become if we transform to an oblique system. We may retain the old axis of x, and if we take an axis of y inclined to it at an angle w, then (Art. 9) we are to substitute x+y cos w for x, and y sin w for y. We shall then have If, then, we transform the equation from one pair of axes to any We may, by the help of this theorem, transform to the axes an equation given in oblique coordinates, for we can still express the sum and product of the new a and b in terms of the old coefficients. Ex. 1. If cos∞ =, transform to the axes 10x2 + 6xy + 5y2 = 10. a+b= 285, ab = 1025, a = 5, b= 205. Ans. 16x2 +41y2 = 32. Ans. x2-15y2 = 3. Ex. 2. Transform to the axes x2 − 3xy + y2 + 1 = 0, where w = 60°. Ex. 3. Transform ax2 + 2hxy + by2 = c to the axes. Ans. (a+b2h cos w - R) x2 + (a + b — 2h cos w + R) y2 = 2c sin2 w, where R2 = {2k (a + b) cos w}2 + (a - b)2 sin2 w. *159. We add the demonstration of the theorems of the last two articles given by Professor Boole (Cambridge Math. Jour., III. 1, 106, and New Series, VI. 87). Let us suppose that we are transforming an equation from axes inclined at an angle w, to any other axes inclined at an angle ; and that, on making the substitutions of Art. 9, the quantity ax+2hxy+by" becomes a X2+2h'XY+UY. Now we know that the effect of the same substitution will be to make the quantity x2 + 2xy cosw+y" become X+2XY cos + Y2, since either is the expression for the square of the distance of any point from the origin. It follows, then, that ax2 +2hxy +by3 +λ (x2 + 2xy cosw + y2) = a'X2+2h'XY+ b′ Y2 + λ (X2+2XY cos + Y2). And if we determine λ so that the first side of the equation may be a perfect square, the second must be a perfect square also. But the condition that the first side may be a perfect square is (a + λ) (b + λ) = (h + λ cos w)2, or λ must be one of the roots of the equation λ2 sin3w+(a + b − 2h cos w) λ + ab — h2 = 0. We get a quadratic of like form to determine the value of x, which will make the second side of the equation a perfect square; but since both sides become perfect squares for the same values of A, these two quadratics must be identical. Equating, then, the coefficients of the corresponding terms, we have, as before, a+b-2h cos w a+b'-2h' cos ab-h2 a'b' - h'2 sinΩ Ex. 1. The sum of the squares of the reciprocals of two semi-diameters at right angles to each other is constant. Let their lengths be a and ß; then making alternately x = 0, y = 0, in the equation of the curve, we have aa2c, b82c, and the theorem just stated is only the geometrical interpretation of the fact that a + b is constant. Ex. 2. The area of the triangle formed by joining the extremities of two conjugate semi-diameters is constant. Ex. 3. The sum of the squares of two conjugate semi-diameters is constant. a+b-2h cos w Since 1 sin2w is constant, (2 + 2) = ap's sin3w sin2w since a's' sin w is constant, so must a22 + ß22, is constant; and |