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