the beginning of chapter 8: conic sections (8.1a) parabolas!!!
TRANSCRIPT
The Beginning of Chapter 8:The Beginning of Chapter 8:Conic Sections (8.1a)Conic Sections (8.1a)
Parabolas!!!Parabolas!!!
Imagine two non-perpendicular lines intersecting at point V.
V
Rotating one of the lines (thegenerator) around the other (theaxis) yields a pair of rightcircular cones…
A conic section is formed by theintersection of a plane and thesecones.
Basics: Parabola, Ellipse, Hyperbola
Degenerates: Point, Line, IntersectingLines
Check out the diagram on p.631…Check out the diagram on p.631…
Conic SectionsAll conic sections can be defined algebraically as thegraphs of second-degree (quadratic) equationsin two variables…in the form:
2 2 0Ax Bxy Cy Dx Ey F where A, B, and C are not all zero.
Today, we focus on parabolas!!!Today, we focus on parabolas!!!
Definition: ParabolaA parabola is the set of all points in a plane equidistant from aparticular line (the directrix) and a particular point (the focus)in the plane.
Axis
Directrix
Dist. todirectrix
Point on the parabolaDist. tofocus
Focus
Vertex
Deriving the equation of a parabola
Focus F(0, p)
p
p
P(x, y)
Directrix: y = –p
D(x, –p)
Let’s equate theseLet’s equate thesetwo distances:two distances:
22 2 20x y p x x y p
Deriving the equation of a parabola
22 2 20x y p x x y p
22 2 20x y p x x y p
2 22 0x y p y p 2 2 2 2 22 2x y py p y py p
2 4x pyStandard form of the equation of an up- or down-openingparabola. If p > 0, it opens up, if p < 0, it opens down.
Deriving the equation of a parabola
Focus F(0, p)
p
p
P(x, y)
Directrix: y = –p
D(x, –p)
2 4x py
The value p is the focal length of the parabola.
A segment with endpoints on a parabola is a chord.
The value |4p| is the focal width.
Deriving the equation of a parabola
Focus F(0, p)
p
p
P(x, y)
Directrix: y = –p
D(x, –p)
2 4x py
Parabolas that open right or left are inverse relations of theupward or downward opening parabolas…standard form:
2 4y px
Parabolas with Vertex (0, 0)• Standard Equation 2 4x py 2 4y px• Opens Upward or
downwardTo the right orto the left
• Focus 0, p ,0p• Directrix y p x p• Axis y-axis x-axis
• Focal Length p p
• Focal Width 4p 4p
Guided PracticeGuided PracticeFind the focus, the directrix, and the focal width of the givenparabola. Then, graph the parabola by hand.
21
2y x
Focus: (0, –1/2),Focus: (0, –1/2),Directrix: y = 1/2,Directrix: y = 1/2,Focal Width: 2Focal Width: 2
Guided PracticeGuided PracticeFind an equation in standard form for the parabola whosedirectrix is the line x = 2 and whose focus is the point (–2, 0).
y = –8xy = –8x
Would a graph help???
22
Guided PracticeGuided PracticeFind an equation in standard form for the parabola whosevertex is (0, 0), opens downward, and has a focal width of 4.
x = – 4yx = – 4y
Would a graph help???
22
Whiteboard Practice …
Find an equation in standard form for the parabola with vertex(0, 0), opening upward, with focal width = 3.
, 0,0h k 4 3p 3
4p (since parabola
opens upward)
Standard Form: 24x h p y k
2 3x y
20 3 0x y
Translations Translations of Parabolasof Parabolas
We have only considered parabolas with the vertex on theorigin…………………..................what happens when it’s not???
V (h, k)
F (h, k + p) (h, k) F (h + p, k)
V
Such translations do not change the focal length, the focalwidth, or the direction the parabola opens!!!
Parabolas with Vertex (Parabolas with Vertex (hh, , kk))• Standard Equation 2
4x h p y k • Opens Upward or downward
• Focus ,h k p• Directrix y k p • Axis x h• Focal Length p
• Focal Width 4p
Parabolas with Vertex (Parabolas with Vertex (hh, , kk))• Standard Equation 2
4y k p x h • Opens To the right or to the left
• Focus ,h p k• Directrix x h p • Axis y k
• Focal Length p
• Focal Width 4p
Practice ProblemsPractice ProblemsFind the standard form of the equation for the parabola withvertex (3, 4) and focus (5, 4).
Which general equation do we use? 24y k p x h
What are the values of h and k? 24 4 3y p x
Now, how do we find p? 24 8 3y x
Practice ProblemsPractice ProblemsUse a function grapher to graph the given parabola.
24 8 3y x
4 8 3y x
4 8 3y x
First, we must solve for y!!!
Now, plug these two equationsinto your calculator!!!
Practice ProblemsPractice ProblemsProve that the graph of the given equation is a parabola, thenfind its vertex, focus, and directrix.
2 6 2 13 0y x y
2 2 1 6 13 1y y x
2 2 6 13y y x We need to complete the square…
The CTS step!!!The CTS step!!!
21 6 2y x
21 6 12y x We have h = 2, k = –1,and p = 6/4 = 1.5
Vertex: (2, –1), Focus: (3.5, –1), Directrix: x = 0.5Vertex: (2, –1), Focus: (3.5, –1), Directrix: x = 0.5
Practice ProblemsPractice ProblemsFind an equation in standard form for the parabola that satisfiesthe given conditions.
Vertex (–3, 3), opens downward, focal width = 20
, 3,3h k
4 20p 5p (since parabolaopens downward)
Standard Form: 23 20 3x y
Practice ProblemsPractice ProblemsFind an equation in standard form for the parabola that satisfiesthe given conditions.
Vertex (2, 3), opens to the right, focal width = 5
, 2,3h k
4 5p 5
4p (since parabola
opens to the right)
Standard Form: 23 5 2y x