SECTION 10.2 The Parabola 687 Solution The vertex ( ) −2, 3 and focus ( ) 0, 3 both lie on the horizontal line = y 3 (the axis of symmetry). The distance a from the vertex ( ) −2, 3 to the focus ( ) 0, 3 is = a 2. Because the focus lies to the right of the vertex, the parabola opens to the right. Consequently, the form of the equation is ( ) ( ) − = − y k a x h 4 2 where ( ) ( ) = − h k , 2, 3 and = a 2. Therefore, the equation is ( ) ( ) ( ) [ ] ( ) − = ⋅ − − − = + y x y x 3 4 2 2 3 8 2 2 2 Since the vertex is midway between the focus and the directrix, the line = − x 4 is the directrix of the parabola. To find the points that define the latus rectum, substitute = x 0 in the equation ( ) ( ) − = + y x 3 8 2 . 2 Then − = ± y 3 4, so = − y 1 or = y 7. The points ( ) − 0, 1 and ( ) 0, 7 determine the latus rectum. See Figure 13. Figure 13 ( ) ( ) − = + y x 3 8 2 2 x y V 5 (22, 3) (0, 21) (0, 7) 8 24 6 26 Axis of symmetry y 5 3 D: x 5 24 F 5 (0, 3) Now Work PROBLEM 33 Using a Graphing Utility to Graph a Parabola, Vertex Not at Origin Using a graphing utility, graph the equation ( ) ( ) − = + y x 3 8 2 . 2 Solution EXAMPLE 8 For some graphing utilities, we must first solve the equation for y. ( ) ( ) ( ) ( ) − = + − = ± + = ± + y x y x y x 3 8 2 3 8 2 3 8 2 2 Figure 14(a) shows the graphs of the equations ( ) = + + Y x 3 8 2 1 and ( ) = − + Y x 3 8 2 2 using a TI-84 Plus CE. Figure 14(b) shows the graph using Desmos. Use the Square Root Method. Add 3 to both sides. Figure 14 13 27 29 23 Y2 5 3 2 8(x 1 2) Y1 5 3 1 8(x 1 2) (a) (b) Polynomial equations involving two variables define parabolas whenever they are quadratic in one variable and linear in the other. Analyzing the Equation of a Parabola Analyze the equation + − = x x y 4 4 0. 2 Solution EXAMPLE 9 To analyze the equation + − = x x y 4 4 0, 2 complete the square involving the variable x. ( ) ( ) + − = + = + + = + + = + x x y x x y x x y x y 4 4 0 4 4 4 4 4 4 2 4 1 2 2 2 2 The equation is of the form ( ) ( ) − = − x h a y k 4 , 2 with = − = − h k 2, 1, and = a 1. The graph is a parabola with vertex at ( ) ( ) = − − h k , 2, 1 that opens up (is concave up). The focus is at ( ) −2, 0 , and the directrix is the line = − y 2. See Figure 15. Isolate the terms involving x on the left side. Complete the square on the left side. Factor. Now Work PROBLEM 51 Figure 15 + − = x x y 4 4 0 2 x 24 4 23 Axis of symmetry x 5 22 V 5 (22, 21) y 4 D: y 5 22 F 5 (22, 0) (24, 0) (0, 0)
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