996 CHAPTER 10 Analytic Geometry EXAMPLE 4 Writing Equations of Parabolas (Vertex at the Origin) Write an equation for each parabola with vertex at the origin. (a) focus A2 3 , 0B (b) vertical axis of symmetry, through the point 1-2, 122 SOLUTION (a) The focus A2 3 , 0B and the vertex 10, 02 are both on the x-axis, so the parabola is horizontal. It opens to the right because p = 2 3 is positive. See Figure 10. The equation will have the form y2 = 4px. y2 = 4 a 2 3b x, or y2 = 8 3 x (b) The parabola will have an equation of the form x2 = 4py because the axis of symmetry is vertical and the vertex is 10, 02. Because the point 1-2, 122 is on the graph, it must satisfy the following equation. x2 = 4py Parabola with vertical axis of symmetry 1-222 = 4p1122 Let x = -2 and y = 12. 4 = 48p Apply the exponent, and multiply. p = 1 12 Solve for p. Then, x2 = 4 a 1 12b y Let p = 1 12 in the form x 2 = 4py. x2 = 1 3 y, or y = 3x2. S Now Try Exercises 35 and 39. x y 2 2 –2 V(0, 0) y2 = x 8 3 F( , 0) 2 3 Figure 10 The equations x2 = 4py and y2 = 4px can be extended to parabolas having vertex 1h, k2 by replacing x and y with x - h and y - k, respectively. Equation Forms for Translated Parabolas A parabola with vertex 1h, k2 has an equation of the following form. 1 x −h22 =4p1y −k2 Vertical axis of symmetry or 1 y −k22 =4p1x −h2 Horizontal axis of symmetry The focus is distance 0 p 0 from the vertex. x y = 3 0 y F(–1, 3) V(1, 3) (y – 3)2 = –8(x – 1) Figure 11 EXAMPLE 5 Writing an Equation of a Parabola Write an equation for the parabola with vertex 11, 32 and focus 1-1, 32, and graph it. Give the domain and range. SOLUTION Because the focus is to the left of the vertex, the axis of symmetry is horizontal and the parabola opens to the left. See Figure 11. The directed distance between the vertex and the focus is -1 - 1, or -2, so p = -2 (because the parabola opens to the left).
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