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”.
CONIC SECTIONS - PARABOLAS
The line is called the directrix and
the point is called the focus.
Focus
Directrix
CONIC SECTIONS - PARABOLAS
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
CONIC SECTIONS - PARABOLAS
FINDING THE FOCUS AND DIRECTRIX
6.5 Parabolas
CONIC SECTIONS - 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
CONIC SECTIONS - PARABOLAS
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
CONIC SECTIONS - PARABOLAS
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
CONIC SECTIONS - PARABOLAS
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
CONIC SECTIONS - PARABOLAS
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
CONIC SECTIONS - PARABOLAS
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
CONIC SECTIONS - PARABOLAS
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
( ?, ?)
CONIC SECTIONS - PARABOLAS
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)
CONIC SECTIONS - PARABOLAS
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)
CONIC SECTIONS - PARABOLAS
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.
CONIC SECTIONS - PARABOLAS
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
CONIC SECTIONS - PARABOLAS
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
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