circle hyperbola circle · 8/3/2016 · of the axis of symmetry with the parabola. focus directrix...

WARM UP

Find the standard form of the equation by completing the square.

Then identify and graph the conic.

d.

circle

hyperbola

circle

Find the standard form of the equation by completing the square.

Then identify and graph the conic.

d.

6-5 PARABOLAS

Objective: To find equations of

parabolas and to graph them.

CONIC SECTIONS - PARABOLAS

The parabola has the characteristic shape shown

above. A parabola is defined to be the “set of

points the same distance from a point and a line”.


The line is called the directrix and

the point is called the focus.

Focus

Directrix


The line perpendicular to the directrix passing through the focus

is the axis of symmetry. The vertex is the point of intersection

of the axis of symmetry with the parabola.

Focus

Directrix

Axis of

Symmetry

Vertex

The definition of the parabola is the set of points the same

distance from the focus and directrix. Therefore, d1 = d2 for

any point (x, y) on the parabola.

Focus

Directrix

d1

d2


FINDING THE FOCUS AND DIRECTRIX

6.5 Parabolas


We know that a parabola has a basic equation y = ax2. The

vertex is at (0, 0). The distance from the vertex to the focus

and directrix is the same. Let’s call it p.

Focus

Directrix

p

p

y = ax2

Find the point for the focus and the equation of the

directrix if the vertex is at (0, 0).

Focus

( ?, ?)

Directrix ???

p

p( 0, 0)

y = ax2


The focus is p units up from (0, 0), so the

focus is at the point (0, p).

Focus

( 0, p)

Directrix ???

p

p( 0, 0)

y = ax2


The directrix is a horizontal line p units below the

origin. Find the equation of the directrix.

Focus

( 0, p)

Directrix ???

p

p( 0, 0)

y = ax2


The directrix is a horizontal line p units below the origin or

a horizontal line through the point (0, -p). The equation is

y = -p.

Focus

( 0, p)

Directrix

y = -p

p

p( 0, 0)

y = ax2


The definition of the parabola indicates the distance d1 from

any point (x, y) on the curve to the focus and the distance d2

from the point to the directrix must be equal.

Focus

( 0, p)

Directrix

y = -p

( 0, 0)

( x, y)

y = ax2

d1

d2


However, the parabola is y = ax2. We can substitute for

y in the point (x, y). The point on the curve is (x, ax2).

Focus

( 0, p)

Directrix

y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2


What is the coordinates of the point on the directrix

immediately below the point (x, ax2)?

Focus

( 0, p)

Directrix

y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2

( ?, ?)


The x value is the same as the point (x, ax2) and the y

value is on the line y = -p, so the point must be (x, -p).

Focus

( 0, p)

Directrix

y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2

( x, -p)


d1 is the distance from (0, p) to (x, ax2). d2 is the

distance from (x, ax2) to (x, -p) and d1 = d2. Use the

distance formula to solve for p.

Focus

( 0, p)

Directrix

y = -p

( 0, 0)

( x, ax2)

y = ax2

d1

d2

( x, -p)


d1 is the distance from (0, p) to (x, ax2). d2 is the distance

from (x, ax2) to (x, -p) and d1 = d2. Use the distance

formula to solve for p.


d1 = d2

2 2 4 2 2 2 4 2 2

2 2

2 2 2 2 2 2( 0) ( ) ( ) ( )

2 2 2 2 2( ) ( ) ( )

2 2

4

1 4

1

4

x ax p x x ax p

x ax p ax p

x a x ax p p a x ax p p

x ax p

ap

pa

Therefore, the distance p from the vertex to the focus

and the vertex to the directrix is given by the formula


𝒑 =𝟏

𝟒𝒂

𝒂 =𝟏

𝟒𝒑𝒐𝒓 −

𝟏

𝟒𝒑

depending on the

direction of the

parabola.

PARABOLA EQUATION6.5

Parabolas

PARABOLA EQUATIONCENTER AT THE ORIGIN (0,0)

𝒑 =𝟏

𝟒𝒂

𝟏. 𝒚 =𝟏

𝟒𝒑𝒙𝟐 𝟐. 𝒚 = −

𝟏

𝟒𝒑𝒙𝟐 𝟑. 𝒙 =

𝟏

𝟒𝒑𝒚𝟐 𝟒. 𝒙 = −

𝟏

𝟒𝒑𝒚𝟐

𝒂 =𝟏

𝟒𝒑𝒐𝒓 −

𝟏

𝟒𝒑

p

p

p

p

(p, 0)

p

p)

p

CONIC SECTIONS - PARABOLASUsing transformations, we can shift the parabola

y=ax2 horizontally and vertically.

2( )y a x h k

The vertex is shifted from (0, 0) to (h, k). Recall that

when “a” is positive, the graph opens up. When “a”

is negative, the graph reflects about the x-axis and

opens down.

𝒚 = −𝟏

𝟒𝒑𝒙 − 𝒉 𝟐 + 𝒌

𝒚 =𝟏

𝟒𝒑𝒙 − 𝒉 𝟐 + 𝒌

𝒙 =𝟏

𝟒𝒑𝒚 − 𝒌 𝟐 + 𝒉

𝒙 = −𝟏

𝟒𝒑𝒚 − 𝒌 𝟐 + 𝒉

PARABOLA EQUATIONCENTER AT (H,K)

EXAMPLES GRAPH A PARABOLA.FIND THE VERTEX, FOCUS AND

DIRECTRIX.

6.5

Parabolas

EXAMPLE 1 FIND THE FOCUS AND DIRECTRIX OF EACH PARABOLA

a. 𝑦 = 2𝑥2 b. x =1

20𝑦2

𝑝 =1

4(2)

𝒑 =𝟏

𝟒𝒂

p =1

8

Focus: 0,1

8 Directrix: 𝑦 = −1

8

𝑝 =1

4(120)

p = 5

Focus: 5,0 Directrix: x = −5

EXAMPLE 2 FIND AN EQUATION OF THE PARABOLA WITH VERTEX (0,0) AND DIRECTRIX X = 2.

Sketch the information. 𝑥 = 2

𝒙 = −𝟏

𝟒𝒑𝒚𝟐

𝒙 = −𝟏

𝟒(𝟐)𝒚𝟐

𝒙 = −𝟏

𝟖𝒚𝟐

21

2 38

xy

EXAMPLE 3 GRAPH THE PARABOLA. FIND THE VERTEX, FOCUS, AND DIRECTRIX.

The vertex is (-2, -3). The parabola opens up.

1

4p

a

The focus and directrix are “p” units from the vertex

where

21

2 38

xy

1 1

2114 28

p

The focus and directrix are 2 units from the vertex.

Find the focus and directrix.

Focus: (-2, -1) Directrix: y = -5

2 Units

Find the focus and directrix.

Plot the known points. What can be

determined from

these points?

EXAMPLE 4 WRITE THE EQUATION OF A PARABOLA WITH VERTEX AT (3, 2) AND FOCUS AT (-1, 2).

The parabola opens the

the left and has a model

of x = −𝟏

𝟒𝒑(y – k)2 + h.

EXAMPLE 4 WRITE THE EQUATION OF A PARABOLA WITH VERTEX AT (3, 2) AND FOCUS AT (-1, 2).

x = −𝟏

𝟒𝒑(y – k)2 + h.

x = −𝟏

𝟒𝒑(y – 2)2 + 3

The distance from the vertex

to the focus is 4, so p = 4.

x = −𝟏

𝟒(𝟒)(y – 2)2 + 3

x = −𝟏

𝟏𝟔(y – 2)2 + 3

HOMEWORKPage 240

#1-25 odds

circle hyperbola circle · 8/3/2016 · of the axis of symmetry with the parabola. focus directrix...

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