# The Parabola Definition of a Parabola A Parabola

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. Parabola Directrix Focus Axis of Symmetry Vertex

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

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

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

Text Example cont. Find the focus and directrix of the parabola given by x 2 = -8 y. The graph the parabola. Solution Because p < 0, the parabola opens downward. Using this value for p, we obtain 5 Directrix: y = 2 Focus: (0, p) = (0, -2) Vertex (0, 0) 4 3 Directrix: y = - p; y = 2. To graph x 2 = -8 y, 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). 2 1 -5 -4 -3 -2 -1 (-4, -2) -1 -2 -3 -4 -5 1 2 3 4 5 (4, -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. Solution The focus is (5, 0). Thus, the focus is on the x-axis. We use the standard form of the equation in Directrix: x = -5 which x is not squared, namely y 2 = 4 px. Focus (5, 0) 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 = 4 px to obtain the standard form of the equation of the parabola. The equation is y 2 = 4 • 5 x or y 2 = 20 x. 7 6 5 4 3 2 1 -5 -4 -3 -2 -1 -1 -2 -3 -4 -5 -6 -7 1 2 3 4 5 6 7

Text Example Find the vertex, focus, and directrix of the parabola given by y 2 + 2 y + 12 x – 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 2 + 2 y + 12 x – 23 = 0 y 2 + 2 y = -12 x + 23 This is the given equation. y 2 + 2 y + 1 = -12 x + 23 + 1 Complete the square by adding the square of half the coefficient of y. (y + 1)2 = -12 x + 24 Isolate the terms involving y.

Text Example cont. Solution To express this equation in the standard form (y – k)2 = 4 p(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. We see that h = 2 and k = -1. Thus, the vertex of the parabola is (h, k) = (2, -1). Because 4 p = -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. 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. (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 Solve for y in each equation. or y = -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. Directrix: x = 5 7 6 (-1, 5) 5 4 3 2 1 -5 -4 -3 -2 1 3 4 6 7 -2 Focus (-1, -1) (-1, -7) -3 -4 -5 -6 -7 Vertex (2, -1)

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

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