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Fig. 6-5 EXAMPLES
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(a) The ellipse graphed in Fig. 6-4(b) is symmetric with respect to the origin because x2 + y2 = 1 4 ( x)2 + ( y)2 = 1 4
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(b) The hyperbola xy = 1 (see Fig. 6-6) is symmetric with respect to the origin, for if xy = 1, then ( x)( y) = 1. (c) If y = ax, then y = a( x). Hence, any straight line through the origin is symmetric with respect to the origin.
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6.1 Determine whether the line y = x (see Fig. 6-7) is symmetric with respect to: (c) the origin. (a) the x-axis; (b) the y-axis;
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(a) The line is not symmetric with respect to the x-axis, since ( 1, 1) is on the line, but ( 1, 1), the re ection of ( 1, 1) in the x-axis, is not on the line. (b) The line is not symmetric with respect to the y-axis, since ( 1, 1) is on the line, but (1, 1), the re ection of ( 1, 1) in the y-axis, is not on the line. (c) The line is symmetric with respect to the origin, by example (c) above.
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SYMMETRY
Fig. 6-7
Fig. 6-8
6.2 Determine whether the parabola x = y2 (see Fig. 6-8) is symmetric with respect to: (c) the origin.
(a) The parabola is symmetric with respect to the x-axis, since x = y2 implies x = ( y)2 .
(a) the x-axis; (b) the y-axis;
(b) It is not symmetric with respect to the y-axis, since (1, 1) is on the parabola, but ( 1, 1) is not. (c) It is not symmetric with respect to the origin, since (1, 1) is on the parabola, but ( 1, 1) is not.
6.3 Show that if the graph of an equation f (x, y) = 0 is symmetric with respect to both the x-axis and the y-axis, then it is symmetric with respect to the origin. (However, the converse is false, as is shown by Problem 6.1.)
Assume that f (x, y) = 0; we must prove that f ( x, y) = 0. Since f (x, y) = 0 and the graph is symmetric with respect to the x-axis, f (x, y) = 0. Then, since f (x, y) = 0 and the graph is symmetric with respect to the y-axis, f ( x, y) = 0.
6.4 Let points P and Q be symmetric with respect to the line L : y = x. If P has coordinates (a, b), show that Q has coordinates (b, a).
Let Q have coordinates (u, v), and let B be the intersection point of L and the line PQ (see Fig. 6-9). B bisects PQ; hence, its coordinates are given by (2.2) as a+u b+v , 2 2 from which, since B lies on L , a+u b+v = 2 2 b+v =a+u v u=a b Furthermore, the perpendicular lines PQ and L have respective slopes b v a u (1)
SYMMETRY
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so that, by Theorem 4.2, b v (1) = 1 a u b v =u a b+a=v+u v+u=a+b (2)
To solve (1) and (2) simultaneously for u and v, rst add the two equations, yielding 2v = 2a, or v = a. Then subtract (1) from (2), yielding 2u = 2b, or u = b. Thus, Q has coordinates (b, a).
Fig. 6-9 6.5 If the graph of 3x 2 + xy = 5 is re ected in the y-axis (that is, each point on the graph is replaced by the point symmetric to it with respect to the y-axis), nd an equation of the new graph.
A point (x, y) lies on the new graph if and only if ( x, y) is on the original graph; that is, if and only if 3( x)2 + ( x)y = 5. This is equivalent to 3x 2 xy = 5. Note that the new equation is obtained from the old equation by replacing x by x.
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