Transcript PPT Review 10.5-10.7-0

C
Review 10.5-10.7
Conic Sections
H
General Form of a Conic
Equation
We usually see conic equations written in
General, or Implicit Form:
Ax  Bxy  Cy  Dx  Ey  F  0
2
2
where A, B, C, D, E and F are integers and A, B and C are
NOT ALL equal to zero.
Please Note:
A conic equation written in General Form doesn’t
have to have all SIX terms! Several of the
coefficients A, B, C, D, E and F can equal zero, as
long as A, B and C don’t ALL equal zero.
If A, B and C all equal zero, what kind of
equation do you have?
... T H I N K...
Dx  Ey  F  0
Linear!
So, it’s a conic equation if...
• the highest degree (power) of x and/or y is 2
(at least ONE has to be squared)
• the other terms are either linear, constant, or the
product of x and y
• there are no variable terms with rational exponents
(i.e. no radical expressions) or terms with negative
exponents (i.e. no rational expressions)
The values of the coefficients in the conic
equation determine the TYPE of conic.
Ax  Bxy Cy Dx  Ey  F  0
2
2
What values form an Ellipse?
What values form a Hyperbola?
What values form a Parabola?
Ellipses...
Ax  Cy  Dx  Ey  F  0
2
2
where A & C have the SAME SIGN
NOTE: There is no Bxy term, and D, E & F may equal zero!
For example:
2x  y  8x  0
2
2
 x  2x  3y  6y  0
2
2
2x  2y  8x  6  0
2
2
The General Form of the equations can be
converted to Standard Form by
completing the square and dividing so that
the constant = 1.
Ellipses...
2x  y  8x  0
2
2
2x  y  8x  0
2
2
This is an ellipse since x & y are
both squared, and both quadratic
terms have the same sign!
Center (-2, 0)
2(x  4x  4)  y  8
2
2
2(x  2)  y  8
2
2
2(x  2)2 y 2 8


8
8 8
(x  2)
y

1
4
8
2
2
Hor. Axis = 2
Vert. Axis = √8
In this example, x2 and y2 are both
negative (still the same sign), you can see
in the final step that when we divide by
negative 4 all of the terms are positive.
Ellipses...
x  3y  2x  6y  0
2
2
 x  2x  3y  6y  0
2
2
Vert. axis = 2/√3
(x  2x  1)  3(y  2y  1)  1  3
2
2
(x  1)2  3(y  1)2  4
(x  1)2 3(y  1)2

1
4
4
(x  1) (y  1)

1
4
4
3
2
center (-1, 1)
2
Hor. axis = 2
Ellipses…a special case!
When A & C are the same value as well as the same
sign, the ellipse is the same length in all directions …
it is a ...
Circle!
Radius = √5
2x  2y  8x  6  0
2
2
2(x 2  4x  2)  2y 2  6  4
2(x  2)  2y  10
2
2
(x  2)2 y 2

1
5
5
Center (2, 0)
Hyperbola...
Ax  Cy  Dx  Ey  F  0
2
2
where A & C have DIFFERENT signs.
NOTE: There is no Bxy term, and D, E & F may equal zero!
For example:
9x  4y  36x  8y  4  0
2
2
x  y  6y  5  0
2
2
x  10x  4y  8y  5  0
2
2
Hyperbola...
The General, or Implicit, Form of the
equations can be converted to Graphing
Form by completing the square and
dividing so that the constant = 1.
9x 2  4 y 2  36x  8y  4  0
9x 2  4y2  36x  8y  4  0
This is a hyperbola since x & y are both
squared, and the quadratic terms have
different signs!
9(x 2  4x)  4(y2  2y)  4
9(x 2  4x  4)  4(y 2  2y  1)  4  36  4
9(x  2)2  4(y  1)2  36
9(x  2)
4(y  1)
36


36
36
36
(x  2)2 (y  1)2

1
4
9
2
2
x-axis=2
y-axis=3
Center (2,-1)
Hyperbola...
In this example, the signs change, but
since the quadratic terms still have
different signs, it is still a hyperbola!
x  y  6y  5  0
2
2
x 2  y 2  6y  5  0
x 2  (y 2  6y)  5
x  (y  6y  9)  5  9
2
2
x 2  (y  3)2  4
x 2 (y  3)2 4


4
4
4
(y  3)2 x 2

1
4
4
x-axis=2
Center (0,3)
y-axis=2
Parabola...
A Parabola can be
oriented 2 different ways:
A parabola is vertical if the equation has an x squared term AND a
linear y term; it may or may not have a linear x term & constant:
Ax  Dx  Ey  F  0
2
A parabola is horizontal if the equation has a y squared term AND
a linear x term; it may or may not have a linear y term & constant:
Cy  Dx  Ey  F  0
2
Parabola …Vertical
The following equations all represent vertical parabolas in
general form; they all have a squared x term and a linear y
term:
x  4x  y  7  0
2
4x  8x  y  0
2
x  y70
2
x y0
2
Parabola …Vertical
To write the equations in Standard Form, complete the square for
the x-terms. There are 2 popular conventions for writing parabolas
in Graphing Form, both are given below:
0  x 2  4x  y  7
0  (x 2  4x  4)  y  7  4
0  (x  2)2  y  3
y  (x  2)2  3
or
y  3  (x  2)2
Vertex (2,3)
Parabola …Vertical
In this example, the signs must be changed at the end so that the y-term is
positive, notice that the negative coefficient of the x squared term makes
the parabola open downward.
0  4x 2  8x  y
0  4(x  2x  1)  y  4
2
0  4(x  1)  y  4
2
 y  4(x  1)  4
2
y  4(x  1)2  4
or
y  4  4(x  1)2
Vertex (-1,4)
Parabola …Horizontal
The following equations all represent horizontal parabolas in
general form, they all have a squared y term and a linear x
term:
y  8y  2x  18  0
2
x  y 3 0
2
3y  6y  x  2  0
2
y x0
2
Parabola …Horizontal
To write the equations in Standard Form, complete the square for
the y-terms. There are 2 popular conventions for writing parabolas
in Standard Form, both are given below:
0  y 2  8y  2x  18
0  (y 2  8y  16)  2x  18  16
0  (y  4)  2x  2
2
2x  (y  4) 2  2
1
x  (y  4)2  1
2
or
0  (y  4) 2  2(x  1)
2(x  1)  (y  4) 2
Vertex (1,-4)
Parabola …Horizontal
In this example, the signs must be changed at the end so that the x-term is
positive; notice that the negative coefficient of the y squared term makes
the parabola open to the left.
0  x  y2  3
0  y2  x  3
 x  y2  3
x   y2  3
or
(x  3)   y 2
Vertex (3,0)
What About the term Bxy?
Ax  Bxy  Cy  Dx  Ey  F  0
2
2
None of the conic equations we have looked at so far included
the term Bxy. This term leads to a hyperbolic graph:
4xy 8  0
or, solved for y:
8
2
y

4x x
What
About
term
Bxy?
If there
is athe
Bxy
term:
Ax  Bxy  Cy  Dx  Ey  F  0
2
2
You need to find the discriminant and use that to determine the
conic section.
B  4 AC  0
The graph is a circle (A = C) or an ellipse
(A ≠ C)
B  4 AC  0
The graph is a parabola
B  4 AC  0
The graph is a hyperbola
2
2
2
Summary ...
General Form of a Conic Equation:
Ax 2  Bxy  Cy 2  Dx  Ey  F  0
where A, B, C, D, E and
F are integers and A, B
and C are NOT ALL
equal to zero.
Identifying a Conic Equation:
Conic
Parabola
Circle
Ellipse
Hyperbola
Equation Stats
A = 0 or C = 0, but not both.
If A = 0, then the
If C = 0, then the
parabola is horizontal. parabola is vertical.
A=C
A & C have the same sign.
A & C have different signs.
Practice ...
Identify each of the following equations as a(n):
(a) ellipse
(b) circle
(d) parabola
(e) not a conic
(c) hyperbola
See if you can rewrite each equation into its Graphing Form!
1) x 2  4y 2  2x  24y  33  0
2) 4x 2  4y 2  9  0
3) x 2  4x  y  0
4) x 2  y2  2x  8  0
5) 9x 2  25y 2  54x  50y  119  0
6) x 2  x  0
7) y 2  8y  9x  52  0
8) x 2  2x  y 2  4y  7  0
Answers ...
(a) ellipse (b) circle (c) hyperbola (d) parabola (e) not a conic
1) x  4y  2x  24y  33  0
2
2
2) 4x  4y  9  0
2
2
(x  1)2 (y  3)2
- -- > (a)

1
4
1
x2 y2
- -- > (c) 9  9  1
4
4
3) x 2  4x  y  0
- -- > (d) ( y  4)  (x  2)2
4) x 2  y2  2x  8  0
- -- > (b) (x  1) 2  y2  9
(x  3) 2 (y  1)2
5) 9x  25y  54x  50y  119  0 - - > (a)

1
25
9
6) x 2  x  0
- -- > (e) not a conic
2
2
7) y 2  8y  9x  52  0
--- > (d) 9(x  4)  (y  4)2
8) x  2x  y  4y  7  0
(x  1)2 (y  2)2
- -- > (c)

1
4
4
2
2
Write the general from of the equation
fo the translation of
-6x2 + 24x + 4y – 8 = 0 for T(-1, -2)
6(x 1) 24(x 1)  4(y  2)  8  0
2
6(x 2x 1)  24(x 1)  4(y  2)  8  0
2
6x 12x  6  24 x  24  4 y  8  8  0
2
6x 12x  4 y 18  0
2
Identify the graph of each equation and
then find θ
2x  4 xy  y  3  0
2
Use this
Formula:
2
B
tan 2 
A C
  37.98
4
tan 2 
2 1

tan 1tan2   tan 14

Ellipse
2  75.96

Identify the graph of each equation and
then find θ
x  3xy  3y  3
2
2
Hyperbola
Use this
Formula:
B
tan 2 
A C
3
tan 2 
1 3

3 
tan tan 2   tan  
 2 
1

1
2  56.3

  28.15
Identify the graph of each equation and
then find θ
3x  4 3xy  y 15
2
Use this
Formula:


2
B
tan 2 
A C
4 3
tan 2 
3  (1)


1
1 4 3
tan tan 2   tan 

 4 
2  60

Hyperbola
  30
Solve this system of equations:
2x  y  8
x2  y2  9
Substitution:
Straight
Line and a
circle
Not Factorable

Step 1
Solve for a variable
y  2x  8
Step 2
Plug into other equation
x 2  (2x  8) 2  9
x  4 x  32x  64  9
2
2
5x  32x  55  0
2
NO SOLUTION!!!
Solve this system of equations:
x y 4
y 1
2
2
Substitution:
Hyperbola
And a
straight
line
Step 1
Solve for a variable

x  (1)  4
2
2
x 5
2
Step 3
Plug into step 1 to find the
other variable
y 1
y 1
Step 2
Plug into other equation
x 5

Solution(s):
( 5,1)
Solve this system of equations:
y  2
CRICLE
And a
2
2
4 x  9y  36 ELLIPSE
x2  y2  4
Step 3
Plug into first equation to find
the other variable
ELIMINATION:

Step 1
Make a new system
x 2  (2)2  4
4 x 2  4 y 2  16
x 2  (2)2  4
4 x 2  9y 2  36

Step 2
Combine to eliminate
x 4 4
x2  4  4
x 2 
0
x2  0
5y 2  20
Solution(s):
y 4
2
2



(0,  2)
Solve this system of equations:
Hyperbola
And a
2
2
9x  y 16  0 Ellipse
x  1
5x 2  y 2  30
Elimination:
Step 1
Re-write the system:

Step 3
Plug into step 1 to find the
other variable
5(1)  y  30
2
2
2
5x  y  30
5  y 2  30
9x  y  16
y  25
2
2
2
2
Step 2

Combine to eliminate
14 x  14
2
x 1
2
y  5

Solution(s):
 (1, 5)

(1,  5)
Solve this system of equations:
2y  x  3  0
x 2  16  y 2
Substitution:
circle
And a
straight
line
5y 12y  7  0
2
12  (12) 2  4(5)(7)
y
2(5)

Step 1
Solve for a variable
y

x  2y  3

Step 2
Plug into other equation
2y  3
2
 16 y
2
4 y 2 12y  9  16  y 2

12  284
 0.48
10
y
12  284
 2.89
10
Step 3
Plug into step 1 to find the

other variable
x  2(0.48)  3
x  3.96
x  2(2.89)  3
x  2.78
Solution(s):
(3.96, 0.48) and (2.78,  2.89)
