The Parabola 10.1

1. The Parabola10.1

2. Parabola Directrix Focus Axis of Symmetry Vertex Definition of a Parabola • A Parabola is the set of all points in a plane that are equidistant from a fixed line (the directrix) and a fixed point (the focus) that is not on the line.

3. The standard form of the equation of a parabola with vertex at the origin is y2= 4px or x2 = 4py. The graph illustrates that for the equation on the left, the focus is on the x-axis, which is the axis of symmetry. For the equation of the right, the focus is on the y-axis, which is the axis of symmetry. y y 2 = 4px Focus (0, p) x 2 = 4py Directrix x = -p Focus (p, 0) Vertex x Vertex Directrix y = -p Standard Forms of the Parabola y x

4. Example • Find the focus and directrix of the parabola given by: Solution: 4p = 16p = 4Focus (0,4) and directrix y=-4

5. This is 4p. We can find both the focus and the directrix by finding p. 4p = -8 p = -2 The focus, on the y-axis, is at (0, p) and the directrix is given by y = - p. Text Example Find the focus and directrix of the parabola given by x2 = -8y. Then graph the parabola. Solution The given equation is in the standard form x2 = 4py, so 4p = -8. x2 = -8y

6. 5 Directrix: y = 2 4 Vertex (0, 0) 3 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 -1 (-4, -2) (4, -2) -2 -3 -4 -5 Focus (0, -2) Text Example cont. Find the focus and directrix of the parabola given by x2 = -8y. Then graph the parabola. Solution Because p < 0, the parabola opens downward. Using this value for p, we obtain Focus: (0, p) = (0, -2) Directrix: y = - p; y = 2. To graph x2 = -8y, we assign y a value that makes the right side a perfect square. If y = -2, then x2 = -8(-2) = 16, so x is 4 and –4. The parabola passes through the points (4, -2) and (-4, -2).

7. 7 6 Directrix: x = -5 5 4 3 Focus (5, 0) 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 -1 -2 -3 -4 -5 -6 -7 Text Example cont. Find the standard form of the equation of a parabola with focus (5, 0) and directrix x = -5. Solution The focus is (5, 0). Thus, the focus is on the x-axis. We use the standard form of the equation in which x is not squared, namely y2 = 4px. y2 = 4 • 5x or y2 = 20x.

8. y2 + 2y + 12x – 23 = 0 This is the given equation. y2 + 2y = -12x + 23 Isolate the terms involving y. y2 + 2y + 1 = -12x + 23 + 1 Complete the square by adding the square of half the coefficient of y. Text Example Find the vertex, focus, and directrix of the parabola given by y2 + 2y + 12x – 23 = 0. Then graph the parabola. Solution We convert the given equation to standard form by completing the square on the variable y. We isolate the terms involving y on the left side. (y + 1)2 = -12x + 24

9. Text Example cont. Solution To express this equation in the standard form (y – k)2 = 4p(x – h), we factor –12 on the right. The standard form of the parabola’s equation is (y + 1)2 = -12(x – 2) We use this form to identify the vertex, (h, k), and the value for p needed to locate the focus and the directrix. (y – (-1))2 = -12(x – 2) The equation is in standard form. Focus: (h + p, k) = (2 + (-3), -1) = (-1, -1) Directrix: x = h – p x = 2 – (-3) = 5 Thus, the focus is (-1, -1) and the directrix is x = 5.

10. (y + 1)2 = -12(-1 – 2) Substitute –1 for x. (y + 1)2 = 36 Simplify. y + 1 = 6 or y + 1 = -6 Write as two separate equations. y = 5 or y = -7 Solve for y in each equation. Text Example cont. Solution To graph (y + 1)2 = -12(x – 2), we assign x a value that makes the right side of the equation a perfect square. If x = -1, the right side is 36. We will let x = -1 and solve for y to obtain points on the parabola.

11. Directrix: x = 5 7 6 5 (-1, 5) 4 3 2 1 -5 -4 -3 -2 1 3 4 6 7 -2 -3 Focus (-1, -1) -4 Vertex (2, -1) -5 -6 (-1, -7) -7 Text Example cont. Solution Because we obtained these values of y for x = -1, the parabola passes through the points (-1, 5) and (-1, -7). Passing a smooth curve through the vertex and these two points, we sketch the parabola below.