9.1 Conic Sections

Conic sections – curves that result from the intersection of a right circular cone and a plane.

Circle Ellipse Parabola Hyperbola

9.2 Quadratic Function: Any function of the form

y = ax2 + bx + c a, b, c are real numbers & a 0

A quadratic Equation: y = x2 + 4x + 3 a = _____ b = _____ c = ______1 4 3

x y-2 -1-1 0-3 0 0 3-4 3

Where is the vertex?Where is the axis of symmetry?

Formula forVertex:

X = -b 2a

Plug x in toFind y

The graph is a parabola.

9.2 Parabola

Directrix

Parabola

Vertex

Focus

Axis of Symmetry

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.

khxay 2)(

hkyax 2)(

(opens up or down)

(opens right or left)

x

Directrix

Parabola

Vertex

Focus

Vertex = (h, k)

Focus & Directrix = +/- 1 from vertex 4a

Plotting Points of the parabola can be done by: • finding x/y intercepts and/or • choosing values symmetric to the vertex

Vertex Form of a Parabola

Parabola Graphing

khxay 2)(

hkyax 2)(

(opens up or down)

(opens right or left)

Step 1: Place equation in one of two vertex forms:

Step 2: Find the vertex at : (h, k) & determine which direction the parabola opensStep 3: Use an x/y chart to plot points symmetric to the vertex & draw the parabolaStep 4: Find the focus/directrix displacement with 1/(4a)

Try the following Examples:

#1) y2 = 4x

#2) x2 = -16y

#3) (y + 4)2 = 12(x + 2)

#4) x2 – 2x – 4y + 9 = 0

1.1 Circles

h is 2.

(x – 2)2 + (y – (-3))2 = 32

k is –3. r is 3.

Graph the circle: (x – 2)2 + (y + 3)2 = 9

Step 1: Find the center & radius

Step 2: Graph the center point

Step 3: Use the radius to find 4 points directly north, south, east, west of center.

Standard circle equation (x – h)2 + (y – k)2 = r2

center = (h, k) radius = r

Center = (2, -3)Radius = 3

General Form of a Circle Equation(Changing from General to Standard Form)

Standard circle equation : (x – h)2 + (y – k)2 = r2 center = (h, k) radius = r General Form of a circle : x2 + y2 + Dx + Ey + F = 0

Change from General Form to Standard FormGeneral Form: x2 + y2 –4x +6y +4 = 0

Move constant : x2 + y2 –4x +6y = -4

X’s & Y’s together : (x2 – 4x ) + (y2 + 6y ) = -4

Complete the Square: (x2 – 4x +4 ) + (y2 + 6y +9 )= -4 +4 +9

Factor (x – 2) (x –2) + (y + 3) (y + 3) = 9 Standard Form (x – 2)2 + (y + 3)2 = 9

Setup to Complete Square½ (-4) = -2 ½ (6) = 3

(-2)2 = 4 (3)2 = 9

9.3 Ellipse

F1 F2

P P

Ellipse - the set of all points in a plane the sum of whose distances from

two fixed points, is constant. These two fixed points are called the foci.

The midpoint of the segment connecting the foci is the center of the ellipse.

A circle is a special kind of ellipse. Since we know about circles, theequation of a circle can be used to help us understand the equationand graph of an ellipse.

Comparing Ellipses to Circlescircle: (x – h)2 + (y – k)2 = r2 Ellipse: (x – h)2 + (y – k)2 = 1

a2 b2

center = (h, k) center = (h, k)

radius = r radiushorizontal = a radiusvertical = b

We do not usually call ‘a’ and ‘b’ radiisince it is not the same distancefrom the center all around the ellipse,but you may think of them as such for the purpose of locatingvertices located at : (h + a, k) (h – a, k) (h, k + b) (h, k – b)

Example1: Graph the ellipse: 25x2 + 16y2 = 400

(0, 5)

(0, -5)

(0, 3)

(0, -3)

(0, 0) (4, 0)(-4, 0)

Major and Minor axis is determinedby knowing the longest and shortesthorizontal or vertical distance across the ellipse.

Finding Ellipse Foci: (foci-radius)2 = (major-radius)2 – (minor-radius)2

Example2: Calculate/Find the foci above.

Converting to Ellipse Standard FormExample: Graph the ellipse: 9x2 + 4y2 – 18x + 16y – 11 = 0

Move constant : 9x2 + 4y2 –18x +16y = 11

X’s & Y’s together : (9x2 – 18x ) + (4y2 + 16y ) = 11

9(x2 – 2x ) + 4(y2 + 4y ) = 11

Complete the Square: 9(x2 – 2x + 1 ) + 4(y2 + 4y + 4) = 11 + 9 + 16

Factor 9(x – 1) (x –1) + 4(y + 2) (y + 2) = 36

9 (x – 1)2 + 4(x + 2)2 = 36

Standard Form 9(x – 1)2 + 4(y + 2)2 = 1

36 36

(x – 1)2 + (y + 2)2 = 1

4 9

Can yougraph this?

(h, k – a)

(h, k + a)

(h, k – c)

(h, k + c)

Parallel to the y-axis, vertical

(h, k)

b2 > a2 and a2 = b2 – c2

(h – a, k)

(h + a, k)

(h – c, k)

(h + c, k)

Parallel to the x-axis, horizontal

(h, k)

a2 > b2 and b2 = a2 – c2

VerticesFociMajor AxisCenterEquation

(h, k)Major axis

x

y

(h, k)

Major axis

x

Focus (h – c, k) Focus (h + c, k)

Vertex (h – a, k) Vertex (h + a, k)

Focus (h + c, k) Vertex (h + a, k)

Focus (h – c, k) Vertex (h + a, k)

Standard Forms of Equations of Ellipses Centered at (h,k)

9.4 Hyperbola

x

y

Transverse axis VertexVertex

FocusFocusCenter

A hyperbola is the set of points in a plane the difference whose distances

from two fixed points (called foci) is constant

x

y

Can you find the traverse axis, center,vertices and foci inthe hyperbola above?

Traverse Axis - line segment joining the vertices.

When you graph a hyperbola you must first locate the center and direction of the traverse access -- parallel or horizontal

Graphing HyperbolaHyperbola: (x – h)2 - (y – k)2 = 1 or (y – k)2 - (x – h)2 = 1

a2 b2 a2 b2

transverse axis: Horizontal transverse axis: Vertical

center = (h, k) – BE Careful! (Foci-displacement)2 = a2 + b2

Verticies at center traverse-coordinate +a Asymptotes: y – k = m (x – h)

center traverse-coordinate –a m = +/- y-displacement

x-displacement

Example1 :

Center: (3, 1) Traverse Axis: Horizontal

Foci-displacementF2 = 4 + 25 = 29F = 29 = 5.4

C

Graphing Hyperbola (Cont.)Hyperbola: (x – h)2 - (y – k)2 = 1 or (y – k)2 - (x – h)2 = 1

a2 b2 a2 b2

transverse axis: Horizontal transverse axis: Vertical

center = (h, k) – BE Careful! (Foci-displacement)2 = a2 + b2

Verticies at center traverse-coordinate +a Asymptotes: y – k = m (x – h)

center traverse-coordinate –a m = +/- y-displacement

x-displacement

Example2 :

Center: (-1, 2) Traverse Axis: Vertical

Foci-displacementF2 = 9 + 16 = 25F = 25 = 5

C(y – 2)2 – (x + 1)2 = 1 9 16

Graphing Hyperbola (Completing the Square)Hyperbola: (x – h)2 - (y – k)2 = 1 or (y – k)2 - (x – h)2 = 1

a2 b2 a2 b2

transverse axis: Horizontal transverse axis: Vertical

center = (h, k) – BE Careful! (Foci-displacement)2 = a2 + b2

Verticies at center traverse-coordinate +a Asymptotes: y – k = m (x – h)

center traverse-coordinate –a m = +/- y-displacement

x-displacement

Example3 : 4x2 – y2 + 32x + 6y +39 = 0

(h, k – a)

(h, k + a)

(h, k – c)

(h, k + c)

Parallel to the y-axis, vertical

(h, k)

b2 = c2 – a2

(h – a, k)

(h + a, k)

(h – c, k)

(h + c, k)

Parallel to the x-axis, horizontal

(h, k)

b2 = c2 – a2

VerticesFociTransverse AxisCenterEquation

(h, k)

x

y

(h, k)

x

Focus (h – c, k) Focus (h + c, k)

Vertex (h – a, k) Vertex (h + a, k)

Focus (h + c, k)

Vertex (h + a, k)

Focus (h – c, k)

Vertex (h + a, k)

Standard Forms of Equations of Hyperbolas Centered at (h,k)