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Parabola Example Problems.Pdf Conic Sections: Parabolas Example 1 Consider the equation y2 = 4x + 12. a. Find the coordinates of the focus and the vertex and the equations of the directrix and the axis of symmetry. b. Graph the equation of the parabola. a. First, write the equation in the form (y – k)2 = 4p(x – h). y2 = 4x + 12 y2 = 4(x + 3) Factor. (y – 0)2 = 4(1)(x + 3) 4p = 4, so p = 1. In this form, we can see that h = -3, k = 0, and p = 1. We can use this to find the desired information. Vertex: (-3, 0) (h, k) Directrix: x = -4 x = h - p Focus: (-2, 0) (h + p, k) Axis of Symmetry: y = 0 y = k The axis of symmetry is the x-axis. Since p is positive, the parabola opens to the right. b. Graph the directrix, the vertex, and the focus. To determine the shape of the parabola, graph several other ordered pairs that satisfy the equation and connect them with a smooth curve. Example 2 Consider the equation y2 = -12x. a. Find the coordinates of the focus and the vertex and the equations of the directrix and the axis of symmetry. b. Graph the equation of the parabola. a. First, write the equation in the form (y – k)2 = 4p(x – h). y2 = -12x (y – 0)2 = 4(-3)(x – 0) In this form, we can see that h = 0, k = 0, and p = -3. We can use this to find the desired information. Vertex: (0, 0) Directrix: x = 3 Focus: (-3, 0) Axis of Symmetry: y = 0 The axis of symmetry is the x-axis. Since p is negative, the parabola opens to the left. b. Graph the directrix, the vertex, and the focus. To determine the shape of the parabola, graph several other ordered pairs that satisfy the equation and connect them with a smooth curve. Example 3 Consider the equation 4x2 + 40x + y + 106 = 0. a. Write the equation in standard form. b. Find the coordinates of the vertex and focus and the equations for the directrix and the axis of symmetry. c. Graph the equation of the parabola. a. Since x is squared, the directrix of this parabola is parallel to the x-axis. 4x2 + 40x + y + 106 = 0 4x2 + 40x = -y – 106 Isolate the x terms and the y terms. 4(x2 + 10x + ?) = -y – 106 + ? The GCF of the x terms is 4. 4(x2 + 10x + 25) = -y – 106 + 100 Complete the square. 4(x + 5)2 = -(y + 6) Simplify and factor. 1 (x + 5)2 = - (y + 6) Divide each side by 4. 4 1 The standard form of the equation is (x + 5)2 = - (y + 6). 4 1 1 b. Since 4p = - , p = - . 4 16 vertex: (-5, -6) (h, k) 95 directrix: y = - y = k - p 16 97 focus: -5, - (h, k + p) 16 axis of symmetry: x = -5 x = h c. Now sketch the graph of the parabola using the information found in part b. Example 4 SPORTS A ball is thrown upward into the air with an initial velocity of 80 feet per second. The path of the ball can be modeled by the equation y = 80x – 16x2 which gives the height above the ground after x seconds. What is the maximum height the ball reaches? The graph of the equation is a parabola which opens downward. Therefore, the vertex of the parabola would give us maximum height of the ball. To find the vertex, write the equation in standard form. 2 y = 80x – 16x 2 y = -16(x – 5x) 2 25 y – 100 = -16 xx5 4 1 2 - (y – 100) = x 5 16 2 1 The standard form of the equation is = - (y – 100). 16 5 The vertex of the equation is , 100. 2 Therefore, the ball reaches a maximum height of 100 feet. This occurs 2.5 seconds after the ball is thrown. .
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