The Parabola

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. Directrix Parabola Vertex Focus Axis of Symmetry

x Directrix x = -p y 2 = 4px Vertex Focus (p, 0) y x Directrix y = -p x 2 = 4pyVertexFocus (p, 0) y Standard Forms of the Parabola The standard form of the equation of a parabola with vertex at the origin is = 4px or x 2 = 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.

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

Find the focus and directrix of the parabola given by x 2 = -8y. The graph the parabola. SolutionThe given equation is in the standard form x 2 = 4py, so 4p = -8. x 2 = -8y 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 x 2 = -8y. The 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 x 2 = -8y, we assign y a value that makes the right side a perfect square. If y = -2, then x 2 = -8(-2) = 16, so x is 4 and –4. The parabola passes through the points (4, -2) and (-4, -2) (4, -2) (-4, -2) Vertex (0, 0) Directrix: y = 2 Focus (0, -2) Text Example cont.

Find the standard form of the equation of a parabola with focus (5, 0) and directrix x = -5. SolutionThe 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 y 2 = 4px. We need to determine the value of p. Recall that the focus, located at (p, 0), is p units from the vertex, (0, 0). Thus, if the focus is (5, 0), then p = 5. We substitute 5 for p into y 2 = 4px to obtain the standard form of the equation of the parabola. The equation is y 2 = 4 5x or y 2 = 20x Focus (5, 0) Directrix: x = -5 Text Example cont.

Find the vertex, focus, and directrix of the parabola given by y 2 + 2y + 12x – 23 = 0. Then graph the parabola. SolutionWe 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 y 2 + 2y + 12x – 23 = 0 This is the given equation. y 2 + 2y = -12x + 23 Isolate the terms involving y. y 2 + 2y + 1 = -12x Complete the square by adding the square of half the coefficient of y. Text Example

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) Solution 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. We see that h = 2 and k = -1. Thus, the vertex of the parabola is (h, k) = (2, -1). Because 4p = -12, p = -3. Based on the standard form of the equation, the axis of symmetry is horizontal. With a negative value for p and a horizontal axis of symmetry, the parabola opens to the left. We locate the focus and the directrix as follows. 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. Text Example cont.

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. Solution (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.

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. Solution Focus (-1, -1) Directrix: x = 5 Vertex (2, -1) (-1, -7) (-1, 5) Text Example cont.

The Latus Rectum and Graphing Parabolas The latus rectum of a parabola is a line segment that passes through its focus, is parallel to its directrix, and has its endpoints on the parabola.

The Parabola