. Plane and solid analytic geometry; an elementary textbook. these may be obtained by multiplying the coeffi-cient of every term which contains x by the exponent of x,decreasing that exponent by unity, and leaving out all termswhich do not contain x. The second may be formed in asimilar way, using y. In this case they are 10 x0 + 2y0 - 12 = 0, and 2 x0 + 10y0-12 = 0. From these the coordinates of the centre are found to be (1,1).From equation (24), F =— 12. The equation, referred tothe point (1,1) as origin, is then 5x2 + 2xy + 5y2-12 = 0. Next revolve the axes through an angle 0, such that ta
. Plane and solid analytic geometry; an elementary textbook. these may be obtained by multiplying the coeffi-cient of every term which contains x by the exponent of x,decreasing that exponent by unity, and leaving out all termswhich do not contain x. The second may be formed in asimilar way, using y. In this case they are 10 x0 + 2y0 - 12 = 0, and 2 x0 + 10y0-12 = 0. From these the coordinates of the centre are found to be (1,1).From equation (24), F =— 12. The equation, referred tothe point (1,1) as origin, is then 5x2 + 2xy + 5y2-12 = 0. Next revolve the axes through an angle 0, such that tan 2* = -^=^. We have decided to use the acute value of 6, which is here -• 4 To determine A and C, we use the equations (18) and (22) or A + C = A + C = 10, and <LAC = 4;AC-B2 = 96. Solving, we have A = 6 or 4, and C = 4 or 6. But sincewe chose the acute value of 6, we must choose A and C so thatA — C has the same sign as B. This is positive. Hence thefinal form of the equation is +4?/2 = l2. Ch. XIII, § 93] EQUATION OF THE SECOND DEGREE 175. But this is the equation of the curve referred to axeswith origin at the point(1,1), and making an angle of - with the original axes. 4 Constructing these axesand plotting the equation 6ar9 + 4?/2 = 12with respect to them, wehave the locus of the origi-nal equation, referred to theoriginal axes. 2. Determine the char- FlG- 90- acter of the loci of the following equations, reduce them totheir simplest forms, and plot: (a) 2x2 + 2y2-Ax-Ay + 1 = 0, (6) x2 + tf + 2x + 2 = 0, (c) 4 xy -235 + 2 = 0, (rf) y2 — + 6 x2 — 14 x + 5 y -f 4 = 0, B2 - 4 AC = 0. 93. Removal of the term in xy. — We have seen that, ifW — 4 A C = 0, it is not possible to transform to a neworigin such that the terms in x and y shall disappear. Inthis case we shall first revolve the axes through an angle 0. Proceeding as in Art. 90, we obtain the equation Ax2 + Bxy 4- Cif + Dx + Ey + F=0,where (13) A = A cos2 0 + B sin 0 cos 0 + Csin2 0, (14) B = (
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