Transcript INTRO TO CONIC SECTIONS

INTRO TO
CONIC
SECTIONS
IT ALL DEPENDS ON
HOW YOU SLICE IT!
Start with a cone:
IF WE SLICE THE CONE, PARALLEL
TO THE BASE, WHAT DO WE GET?
A Circle!
IF WE SLICE THE CONE AT AN
ANGLE, WHAT DO WE GET NOW?
An Ellipse!
IF WE JUST TAKE A SLICE FROM THE
LATERAL FACE OF THE CONE, WHAT
DO WE GET?
A Parabola!
FINALLY, LET’S TAKE A SLICE FROM
THE LATERAL FACE, PERPENDICULAR
TO THE BASE
A Hyperbola!
THESE SHAPES ARE CALLED CONIC
SECTIONS. WE CAN USE ALGEBRA TO
DESCRIBE THE EQUATIONS AND GRAPHS
OF THESE SHAPES.
10.2 - PARABOLAS
REVIEW: NO CALCS!
Find the distance between the given points.
1. (2, 3) and (4, 1)
2. (4, 6) and (3, –2)
HMMM. . . Remember something called the
Distance Formula?
3. (–1, 5) and (2, –3)
SOLUTIONS
LET’S REVIEW PARTS OF A
PARABOLA
The Vertex Form of a Parabola looks like
this:
‘a’ describes
y = a(x-h)² + k how wide or
narrow the
parabola will be.
This is the y-intercept,
It is where the parabola
crosses the y-axis
These are the roots
Roots are also
called:
-zeros
-solutions
- x-intercepts
This is the vertex, V
(h, k)
This is the called
the axis of symmetry, a.o.s.
Here a.o.s. is the line x = 2
NEW
VOCABULARY
A parabola is a set of
all points that are the
same distance form a
fixed line and a fixed
point not on the line
The fixed point is
called the focus of the
parabola. C is the distance between
the focus and the vertex
The fixed line is called
the directrix.
The focus 
• A point in the arc of the parabola
such that all points on the
parabola are equal distance
away from the focus and the
directrix
The line segment through a focus of a
parabola, perpendicular to the major axis ,
which has both endpoints on the curve is
called the Latus Rectum.
EXAMPLE
Write an equation for a parabola with a vertex at the origin and a focus
at (0, –7).
Step 1: Determine the orientation of the parabola.
Does it point up or down?
Make a sketch.
Since the focus is located below
the vertex, the parabola must
open downward. Use y = ax2.
Step 2: Find a.
WRITE AN EQUATION FOR A PARABOLA WITH A VERTEX AT THE
ORIGIN AND A FOCUS AT (0, –7).
To find a, we
use the
1
formula: | a | = 4c
1
= 4(7)
Note: c is the distance from the
vertex to the focus.
Since the focus is a distance of
7 units from the vertex, c = 7.
1
= 28
Since the parabola opens downward, a is negative.
So a = – 1 .
28
An equation for the parabola is y = – 1 x2.
28
WRITING EQUATIONS GIVEN …
WRITE THE EQ GIVEN VERTEX(-2,4) AND FOCUS(-2,2)
Draw a graph with given info
Use given info to get measurements
• C = distance from Vertex to Focus, so
c = 2 (4-2 of the vertical coordinates)
• Also, parabola will open down because the
focus is below the vertex
Use standard form
• Y = a(x – h)² + k
• Need values for h,k, and a
• (h , k) = (-2 , 4)
• To find a, use formula a = 1/4c
• Therefore a = 1/8
Plug into formula
• Y = -1/8 (x + 2)² + 4
V(-2,4)
C=2
F(-2,2)
WRITING EQUATIONS GIVEN …
WRITE THE EQ GIVEN F(3,2) AND DIRECTRIX IS X = -5
Draw a graph with given info
Use given info to get measurements
• Vertex is in middle of directrix and
focus. The distance from the
directrix to the focus is 8 units.
• That means V = (-1 , 2)
• C = distance from V to F, so c = 4
• Also, parabola will open right
Use standard form
• x = a(y – k)² + h
• Need values for h,k, and a
• (h , k) = (-1 , 2)
• To find a, use formula a = 1/4c
• Therefore a = 1/16
Plug into formula
• x = 1/16 (y – 2)² – 1
X = -5
C=4
F(3,2)
Distance = 8
Vertex is in middle
of directrix and F
So V = (-1 , 2)
LET’S TRY ONE
A parabolic mirror has a focus that is located 4 in. from the vertex of
the mirror. Write an equation of the parabola that models the cross
section of the mirror.
The distance from the vertex to the focus is 4 in., so
c = 4. Find the value of a.
a =
1
4c
1
= 4(4)
=
1
16
Since the parabola opens upward, a is positive.
The equation of the parabola is y = 1 x2.
16
LET’S TRY ONE
Write an equation for a graph that is the set of all points in the plane
that are equidistant from point F(0, 1) and the line y = –1.
LET’S TRY ONE
Write an equation for a graph that is the set of all points in the plane
that are equidistant from point F(0, 1) and the line y = –1.
An equation for a graph that is the set of all points in the plane that are
equidistant from the point F (0, 1) and the line y = –1 is y = 1 x2.
4
EXAMPLE: GIVEN THE EQUATION OF A
PARABOLA, GRAPH AND LABEL ALL PARTS
Start with the vertex
• V = (h,k) = (3,-1)
Find the focus.
• C = 1/4a
• Since a = 1/8, then C = 2
• Focus is at (3, -1 + 2)  (3,1)
Label the directrix and A.O.S.
• Directrix is at y = -3 (the y
coordinate of the vertex, minus the
value of c)
• A.O.S.  x = 3
Find the latus rectum
• The length of the l.r. is | 1/a |
• Since a = 1/8, the l.r. =8
Plot and label everything
1
y  (x  3) 2  1
8
X=3
F(3,1)
Y = -3
V(3,-1)
GRAPH AND LABEL ALL
PARTS
Start with the vertex
• V = (h,k) = (2,-1)
Find the focus.
• C = 1/4a
• Since a = ¼ , then C = 1
• Focus is at (2+1, -1)  (3,-1)
Label the directrix and A.O.S.
• Directrix is at x = 1
• A.O.S.  y = -1
Find the latus rectum
• The length of the l.r. is | 1/a |
• Since a = ¼ , the l.r. =4
Plot and label everything
x
1
( y  1)2  2
4
X=1
V(2,-1)
Y = -1
F(3,-1)
EXAMPLE
Identify the focus and directrix of the graph of the
equation x = – 1 y2.
8
The parabola is of the form x = ay2, so the vertex is at the origin
and the parabola has a horizontal axis of symmetry. Since a < 0,
the parabola opens to the left.
|a| = 1
4c
1
| – 1 | =
8
4c
4c = 8
c = 2
The focus is at (–2, 0). The equation of the directrix is x = 2.
EXAMPLE
Identify the vertex, the focus, and the directrix of the graph of the
equation x2 + 4x + 8y – 4 = 0. Then graph the parabola.
x2 + 4x + 8y – 4 = 0
8y = –x2 – 4x + 4
Solve for y, since y is the only term.
8y = –(x2 + 4x + 4) + 4 + 4
Complete the square in x.
y = – 1 (x + 2)2 + 1
8
vertex form
The parabola is of the form y = a(x – h)2 + k, so the vertex is at (–2, 1)
and the parabola has a vertical axis of symmetry. Since a < 0 (negative),
the parabola opens downward.
EXAMPLE (CONT)
1
| a | = 4c
1
| –1 | =
8
4c
4c = 8
Substitute – 1 for a.
8
Solve for c.
c = 2
The vertex is at (–2, 1) and the focus is
at (–2, –1). The equation of the directrix
is y = 3.
Locate one or more points on the
parabola. Select a value for x such as –6.
The point on the parabola with an x-value of –6 is (–6, –1). Use the
symmetric nature of a parabola to find the corresponding point (2, –1).