9.1 Parabola
Parabola: the locus of points equidistant from a fixed point call the focus and a fixed line called the directrix
(See Calculus in Motion: Conics, Definitions and Graphs, parabola 1)
2
Vertex: (h, k)
1.5
Focus:
(h, k + p)
1
Directrix: y = k – p
0.5
Line of Symmetry: x = h
Equation:
4 py = x 2
FOCUS (h, k + p)
-2
-1
VERTEX1 (h, k)
2
-0.5
Equation (vertex not at origin):
4 p ( y − k ) = ( x − h) 2
-1
DIRECTRIX Y = k - p
-1.5
“p” is the distance between the
vertex and the focus
-2
Vertex: (h, k)
2
Focus: (h + p, k)
1.5
Directrix: y = h – p
1
Line of Symmetry: y = k
0.5
Equation:
VERTEX (h, k)
4 px = y 2
-2
-1
1
2
FOCUS (h
+ p, k) 3
Equation (vertex not at origin):
-0.5
4 p ( x − h) = ( y − k ) 2
“p” is the distance between the
vertex and the focus
DIRECTRIX
X=h-p
-1
-1.5
-2
Homework Examples:
Given the equation, determine the vertex, focus, directrix and sketch the graph.
1.
y = −4x 2
2.
Homework: Day 1, p. 637 (1-6 vocabulary, 1-6, 7-25 odd)
y 2 − 4 y − 4x = 0
Homework Examples:
Based on the given information, determine the standard form of the equation of the parabola and sketch the
graph.
⎛5 ⎞
Focus: ⎜ ,0 ⎟
⎝2 ⎠
Vertex: (0,0)
Directrix: y = 3
Vertex: (0,0)
Vertex: (3, -3)
9⎞
⎛
Focus: ⎜ 3,− ⎟
4⎠
⎝
http://www.ies.co.jp/math/java/conics/focus/focus.html
Vertex: (-2, 1)
Directrix: x = 1
See this applet for a demonstration of how beams
reflect off of a parabola and converge at one point,
the focus. This phenomenon occurs for any
parabola.
Homework: Day 2, p. 637 (25-47 odd, 53,59)