1. The Parabola 10.1

2. 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

3. Standard Forms of the Parabola
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. x
Directrix x = -p
y 2 = 4px
Vertex
Focus (p, 0) y y
Focus (0, p)
x 2 = 4py
Vertex x
Directrix y = -p

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

5. 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
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.

6. 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. -5 -4 -3 -2 -1 1 2 3 4 5
(4, -2)
(-4, -2)
Vertex (0, 0)
Directrix: y = 2
Focus (0, -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. 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. -1 -5 -4 -3 -2 1 2 3 4 5 6 7 -6 -7
Focus (5, 0)
Directrix: x = -5

8. 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.
y2 + 2y + 12x – 23 = 0
This is the given equation.
y2 + 2y = -12x + 23
Isolate the terms involving y.
y2 + 2y + 1 = -12x
Complete the square by adding the square of half the coefficient of y.
(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. 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 = or y = -7
Solve for y in each equation.

11. 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. -5 -4 -3 -2 1 3 4 6 7 5 2 -6 -7
Focus (-1, -1)
Directrix: x = 5
Vertex (2, -1)
(-1, -7)
(-1, 5)