This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1911 edition. Excerpt: ...be moved (the new axes being parallel to the old) in order that the new equation of the locus 2z2-5a-3?/2-2a; + 13y-12 = 0 shall have no terms of first degree? Solution. Let the new origin be (h, k); then x = x'-f h, y = y' + fc, and the new equation is 2 (V + h)2-5(x' + h)(y' + k)-3(y'+ k)2-2(x' + 7i) + 130/ ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1911 edition. Excerpt: ...be moved (the new axes being parallel to the old) in order that the new equation of the locus 2z2-5a-3?/2-2a; + 13y-12 = 0 shall have no terms of first degree? Solution. Let the new origin be (h, k); then x = x'-f h, y = y' + fc, and the new equation is 2 (V + h)2-5(x' + h)(y' + k)-3(y'+ k)2-2(x' + 7i) + 130/ + k)-12 = 0, i.e., 2 x'2-5 x'y'-3y'2 + (ih-5k-2)x'-(5 h +6 k-13)2/ + 2 A2-5 hk-3 k2-2 h + 13 k-12 = 0; but it is required that the coefficients of x' and y' shall be 0; i.e., h and k are to be determined so that 4A-5&-2 = 0, and 5A + 6fc-13=0; hence h--y-and k = %. Therefore the new origin must be at the point (y, ), and the new equation is 2x'2-5x'y'-3y'2-8 = 0. 6. The new axes being parallel to the old, determine the new origin so that the new equation of the locus x2 + 2xy-y2 + 8x + iy-S = 0 shall have no terms of first degree. 7. Transform the equations x + y--4 = 0 and 3x--2y + 5 = 0 to parallel axes having the point of intersection of these lines as origin. 8. Transform the equation---= 1 to new rectangular 6 4 axes through the point (3, 2), and making the angle tan1 ( ) with the old axes. 9. Through what angle must the axes be turned that the new equation of the line 3 x + y--10 = 0 shall have no?/-terin? Show this geometrically, from a figure. 10. Through what angle must the axes be turned in order that the new equation of the line 3 x + y = 10 shall have no aterm? Show analytically. Cf. also examples 8 and 9. Solution. Let 6 be the required angle; then the equations of transformation are x = x' cos 6--y' sin 6 and y = x' sin 6 + y' cos 6; and the new equation is (6 cos 6 + 4 sin 6)x'--(6 sin 6-4 cos %' = 24; but it is required that the coefficient of x be 0, 6 cos 6 + 4 sin 6 = 0, i.e., tan 6 =--; whence 6 = tan-1...
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