conic sections: the parabolaanitakoen.weebly.com/.../lp38_conic_sections_parabolas.pdf7. sketch the...

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Conic Sections: The Parabola

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+Parabola

n Review: The geometric definition relies on a cone and a plane intersecting it.

n Algebraic definition: All points that are equidistant from a given line (the directrix)and a fixed point not on the directrix (the focus).

Parabola

Focus

Directrix

x

y

Any point on the parabola is equidistant to the focus and the directrix. Example: Point A: d1=d2 Point B: d1=d2

A

d1

d2

d1 d2 B

Items referenced on the graph of a parabola:

Vertex

Focus

Directrix

Axis of Symmetry

x

y

+Facts: Parabola Equations

q There are many ways the equation of a parabola can be written. We will get the quadratic part (variable that is squared) on the left of the equal sign and the linear part (variable is to the first power) on the right of the equal sign.

q One variable is squared and one is not. (How does this differ from linear equations?)

q Equation: (x - h)2 = c(y – k) OR (y - k)2 = c(x – h)

+ (x - h)2 = c(y – k) OR (y - k)2 = c(x – h) where the vertex is at (h,k) and |c| is the width at the focus.

To graph:

1. Put in standard form (above) – squared term on left

2. Decide which way the parabola opens.

Look at the right side. If y: + c → opens up

If y: - c → opens down

If x: + c → opens right

If x: - c → opens left

+(x - h)2 = c(y – k) OR (y - k)2 = c(x – h) where the vertex is at (h,k) and |c| is the width at the focus. To graph:

3. Plot the vertex (h,k) Note what happens to the signs.

4. Plot the focus: move │¼ c │ from the vertex in the direction that the parabola opens. Mark with an f.

5. Draw the directrix: │¼ c │ from the vertex in the opposite direction of the focus (Remember that the directrix is a line.)

+(x - h)2 = c(y – k) OR (y - k)2 = c(x – h) where the vertex is at (h,k) and |c| is the width at the focus.

To graph: 6. Plot the endpoints of the latus rectum/focal

chord (width at the focus). The width is the │c│ at the focus.

7. Sketch the parabola by going through the vertex and the endpoints of the latus rectum. (Be sure to extend the curve and put arrows.)

8. Identify the axis of symmetry. (The line that goes through the vertex dividing the parabola in half.)

+ Ex. 1: Graph (x - 5)2 = 12(y – 6) To graph:

1. Put in standard form– squared term on left

Done

2. Decide which way the parabola opens.

Look at the right side. If y: + c → opens up

If y: - c → opens down

If x: + c → opens right

If x: - c → opens left

Up because y is on the right and 12 is positive

+ Ex. 1: Graph (x - 5)2 = 12(y – 6)

To graph:

3. Plot the vertex (h,k) Note what happens to the signs.

(5,6)

4. Plot the focus: move │¼ c │ from the vertex in the direction that the parabola opens. Mark with an f.

(5,9): found by moving up 3 from the vertex

5. Draw the directrix: │¼ c │ from the vertex in the opposite direction of the focus (Remember that the directrix is a line.)

y = 3: found by moving down 3 from the vertex

+ Ex. 1: Graph (x - 5)2 = 12(y – 6) To graph: 6. Plot the endpoints of the latus rectum/focal

chord (width at the focus). The width is the │c│ at the focus. L.R. = 12 with endpoints at (-1,9) & (11,9)

7. Sketch the parabola by going through the vertex and the endpoints of the latus rectum. (Be sure to extend the curve and put arrows.)

8. Identify the axis of symmetry. (The line that goes through the vertex dividing the parabola in half.) x = 5

+ Ex. 2: Graph (y + 3)2 = -4(x – 2) To graph: 1. Put in standard form

Done 2. Decide which way the parabola opens.

Left because x is on the right and 4 is negative 3. Plot the vertex (h,k)

(2,-3) 4. Plot the focus:

(1,-3): found by moving left 1 from the vertex 5. Draw the directrix:

x = 3: found by moving right 1 from the vertex

+ Ex. 2: Graph (y + 3)2 = -4(x – 2) To graph:

6. Plot the endpoints of the latus rectum

L.R. = 4 with endpoints at (1,-1) & (1,-5)

7. Sketch the parabola

8. Identify the axis of symmetry.

y = -3

+ Ex. 3: Graph x = y2 - 4y + 1 n  What’s the first step?

Put in standard form. y2 - 4y + 1 = x y2 - 4y + 4 = x - 1 + 4 Complete the square. (y – 2)2 = x + 3 (y – 2)2 = 1(x + 3)

n  Now you try graphing the parabola and labeling all the parts.

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−8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4

−8

−7

−6

−5

−4

−3

−2

−1

1

2

3

x

y

Ex. 4: Given the Graph, write the equation.

Vertex: (2,-3)

Focus: (1,-3)

Directrix: x = 3

L.R.: 4

Axis: y = -3

(y + 3)2 = -4(x – 2)

+ Ex. 5: Given the Graph, write the equation.

Vertex: (5,6) Focus: (5,9) Directrix: y = 3 L.R.: 12 Axis: x = 5 Graph (x - 5)2 = 12(y – 6)

f

+ Challenge Question

Given the following information, write the equation of the parabola.

Vertex is (0,0) and Focus is at (0,2)

+Review 1)  How can you tell the graph of an equation

will be a parabola?

2)  What’s the standard form of a parabola?

3)  What are the steps for graphing a parabola?

4)  What are common errors people make when graphing parabolas?