853 8.8 Parametric Equations, Graphs, and Applications Now substitute this result into the first equation to eliminate the parameter t. x = t2 x = a y - 3 2 b 2 Substitute for t. x = 1y - 322 4 A a bB 2 = a2 b2 1y - 322 = 4x Multiply by 4 and rewrite. or y2 - 4x - 6y + 9 = 0 Square the binomial; Subtract 4x. This is the equation of a horizontal parabola opening to the right, which agrees with the graph given in Figure 74. Because t is in 3-3, 34, x is in 30, 94 and y is in 3-3, 94. The rectangular equation must be given with restricted domain as x = 1 41y - 322, for x in 30, 94. S Now Try Exercise 9(b). Parametric representations of a curve are not unique. In fact, there are infinitely many parametric representations of a given curve. If the curve can be described by a rectangular equation y = ƒ1x2, with domain X, then one simple parametric representation is x = t, y = ƒ1t2, for t in X. Add corresponding sides of the two equations. EXAMPLE 3 Graphing a Plane Curve Defined Parametrically Graph the plane curve defined by x = 2 sin t, y = 3 cos t, for t in 30, 2p4. SOLUTION To convert to a rectangular equation, it is not productive here to solve either equation for t. Instead, we use the fact that sin2 t + cos2 t = 1. x = 2 sin t y = 3 cos t Given equations x2 = 4 sin2 t y2 = 9 cos2 t Square each side. x2 4 = sin2 t y2 9 = cos2 t Solve for sin2 t and cos2 t. x2 4 + y2 9 = sin2 t + cos2 t x2 4 + y2 9 = 1 sin2 t + cos2 t = 1 This is an equation of an ellipse. See Figure 76. S Now Try Exercise 31. y x 3 –3 2 –2 0 x = 2 sin t y = 3 cos t for t in [0, 2P] x2 4 y2 9 + = 1 −3.1 −5 3.1 5 Parametric graphing mode Figure 76
RkJQdWJsaXNoZXIy NjM5ODQ=