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C4 Chapter 2 CO-ORDINATE GEOMETRY IN THE (x,y) PLANE PARAMETRIC EQUATIONS Coordinates x and y are expressed as functions of a parameter. x = f(t) and y = g(t) Example Draw the curve given by the parametric equations: 2 x = t + 3, y = t for – 3 t 3 Draw up a table of values t x y -3 0 9 -2 1 4 -1 2 1 0 3 0 1 4 1 2 5 4 3 6 9 y 10 8 6 4 2 -2 JMcC 0 2 4 6 x 1 C4 Chapter 2 To find the Cartesian equation of the curve, you need to eliminate the parameter between the two equations Example 1 Find the Cartesian equation of the curve given by the parametric equations: 2 x = t + 3, y = t Make t the subject t = x – 3 2 Substitute into the y equation y = (x – 3) Check: this is the parabola y = x2 translated 3 units to the right (see previous example) Example 2 Find the Cartesian equation of the curve given by the parametric equations: 1 2 x = ,y = , t – 3, t 5 t + 3 t – 5 Make t the subject: x = 1 t + 3 t + 3 = 1 x t = 1 x – 3 = 1 – 3x x Substitute into the y equation: t – 5 = 1 – 3x x y = 2 – 5 = 1 – 8x x 1 – 8x x y = 2x 1 – 8x Ex 2A p 10 JMcC 2 C4 Chapter 2 USING PARAMETRIC EQUATIONS TO SOLVE PROBLEMS Example 1 Find the coordinates of the point where the curve with parametric equations: x = √t, y = 4t – 25 meets the x-axis At the x – axis, y = 0 x = t = 25 4 = 4t – 25 = 0 t = 25 4 5 2 5 Points of intersection are , 0 2 and 5 – , 0 2 Example 2 Find the coordinates of the points of intersection of the curve with parametric equations: x = 2t, y = t2 and the straight line y=x+3 Substitute the parameters into the equation of the line: 2 2 t = 2t + 3 t – 2t – 3 = 0 (t – 3)(t + 1) = 0 t = 3, – 1 when t = 3 x = 6 and y = 9 when t = – 1 x = – 2 and y = 1 Points of intersection are (6, 9) and ( – 2, 1) Ex 2B p 13 JMcC 3 C4 Chapter 2 PARAMETRIC EQUATIONS INVOLVING TRIGONOMETRIC PARAMETERS You will need to use some of the Trigonometric Identities from C3 Example 1 Find the Cartesian equation of the curve given by: x = cos t, y = sin 2t y = sin 2t = 2 sin t cos t = 2x sin t we need to find sin t in terms of x or y 2 2 sin t = 1 – cos t sin t = 2 2 1 – cos t = 1 – x 2 y = 2x 1 – x Example 2 Find the Cartesian equation of the curve given by: x = sec t, y = 4 sin t y = 4 sin t and x = sec t = sin t = 2 1 – cos t = 1 cos t 2 cos t = 1 – (1 x ) = 1 x 2 x – 1 2 x 2 = (x – 1) x 2 y = 4 (x – 1) x Ex 2C p 15 JMcC 4 C4 Chapter 2 AREA UNDER A CURVE GIVEN BY PARAMETRIC EQUATIONS Since Area = y dx, Chain Rule y dx = y dx dt dt Example The diagram shows the curve with parametric equations: 2 x = t , y = t(5 – t) The region R is bounded by the curve and the x-axis. Find the area of R. y 6 4 R 2 0 5 10 15 20 25 x Curve crosses x – axis at (25, 0) t(5 – t) = 0 t = 5 Area = = 5 0 25 y dx = 0 5 y 0 dx dt dt = 3 10t (10t – 2t ) dt = 3 2 3 5 t(5 – t) 2t dt 0 4 5 2t – = 104·17 4 o Ex 2D p 18 & Mixed Ex 2E p 19 JMcC 5