Conics

1
Parabolas
Circles
Conics
Circles
Ellipses
Hyperbolas
2
Circles
General form (x - h)² (y - k)² r²
h
k
r
Center (h, k) radius r
3
Given Center and radius
k
h
Ex. 1 C(5, 2) r 7
5
2
7
(x - )² (y - )² ²
5
2
7
(x - 5)² (y - 2)² 49
4
(x - h)² (y - k)² r²
h
k
-3
4
(x - )² (y - )² ²
-3
4
(x 3)² (y - 4)² 20
5
(x - h)² (y - k)² r²
Given Center Another Point
h
k
-7
Ex. 3 C(4, -7) (5, 3)
4
5
3
(x - )² (y - )² ²
-7
4
( - 4)² ( 7)² r²
5
3
To find r2, you can plug in the point or
use the distance formula
(1)² (10)² r²
101 r²
(x - 4)² (y 7)² 101
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(x - h)² (y - k)² r²
Ex. 4 C origin (-5, 2)
(x - )² (y - )² ²
0
0
To find r2, you can plug in the point or
use the distance formula
x² y² 29
7
Given Endpoints of diameter
Ex. 5 (2, 8) (-4, 6) are endpoints of the
diameter.
(-1, 7)
C
Lets use C (-1, 7) and (2, 8)
Thenchoose either endpoint and finish like
before.
8
k
h
C (-1, 7) and (2, 8)
-1
7
2
8
(x - )² (y - )² ²
-1
7
( 1)² ( - 7)² r²
2
8
(3)² (1)² r²
10 r²
(x 1)² (y - 7)² 10
9
More Circle Fun Next Time!!
10
Circles Day 2
Putting into standard form and graphing.
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Remember how to
Complete the Square
?!?
If a quadratic equation isnt in
Standard form for a Circle (x - h)² (y - k)²
r²
you will need to
Complete the Square
to get it in the correct form.
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Heres how to do it
x2 y2 16x 22y 20 0
x2 16x ( )
y2 22y ( )
20 ( ) ( )

• Rewrite the problem

• Group your xs and leave a space.

• Group your ys and leave a space.

• Move the constant and leave 2 spaces.

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x2 16x ( ) y2 22y ( ) 20 ( )
( )
8 2
64
(x 8)2 (y 11)2 205
Complete the square

• Half the linear term and square it.

• Add to both sides.

• Do this for both x and y.

• Factor and simplify.

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Now you try it
x2 y2 - 12x 8y 32 0
x2 - 12x ( ) y2 8y ( ) -32 ( ) (
)
(x - 6)2 (y 4)2 20
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Graphing Circles
Ex. 1 (x)² (y)² 36 Center (0, 0)
radius 6
Center (0, 0)
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Ex. 2 (x - 3)² (y - 4)² 25 Center (3, 4)
radius 5
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That's all for circles!!
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Parabolas
Circles
Conics
Ellipses
Ellipses
Hyperbolas
20
Ellipses
(0, 0)
name of ellipse
center
vertical
4
b
a
5
b was under the x2, so you move b units from the
center in a x direction.
major axis
10
focus (0, 3)
minor axis
8
a was under the y2, so you move a units from the
center in a y direction.
(0, 5), (0, -5), (4, 0), (-4, 0)
center (0, 0)
vertices
focus (0, -3)
(0, 3), (0, -3)
foci
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x2 y2 1 9 20
(0, 0)
name of ellipse
center
vertical
3
b
a
2v5
major axis
4v5
minor axis
6
center (0, 0)
(0, 2v5) (3, 0)
vertices
foci
(0, v11)
22
__ __ 1
25
9
5
3
Where is the center of this ellipse?
How many units from the center to the curve in an
x direction?
How many units from the center to the curve in an
y direction?
23
__ __ 1
36
16
4
6
Where is the center of this ellipse?
How many units from the center to the curve in an
x direction?
How many units from the center to the curve in an
y direction?
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x2 y2 1 10 1
x2 10y2 10
SF center vertices foci
SF
(0, 0)
center
(v10, 0) (0, 1)
vertices
foci
(3, 0)
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x2 y2 1 3 24
24x2 3y2 72
SF center vertices foci
SF
(0, 0)
center
(0, 2v6) (v3, 0)
vertices
foci
(0, v21)
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bye-bye
ellipses
ellipses
ellipses
ellipses
ellipses
ellipses
ellipses
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Parabolas
Circles
Conics
Hyperbolas
Ellipses
Hyperbolas
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x2 - y2 1 9 16
x2 - y2 1 9 16
Hyperbolas
(0, 0)
center a b vertices foci
center
4
a
b
3
vertices
(3, 0) (-3, 0)
foci
(5, 0) (-5, 0)
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Divide each term by 100 to get into form.
(3, -1)
center a b vertices foci
center
2
a
b
5
vertices
(3, -6) (3, 4)
(3, -1v29)
foci
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Getting it into Standard Form
16x2 - 9y2 54y 63 0
16x2 (-9y2 54y ( )) -63 ( )
Factor the 9 out of the y terms.
16x2 -9(y2 - 6y ( )) -63 -9( )
Remember Put the 9 on the right too.
16x2 -9(y - 3)2 -144
Divide each term by -144.
(y -3)2 x2 1 16 9
Why did the x and y terms trade places?
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(0, 3)
center a b vertices foci
center
3
a
b
4
vertices
(0, 7) (0, -1)
foci
(0, 8) (0, -2)
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9x2 - 4y2 54x 8y 41 0
(9x254x( ))(-4y28y( )) -41 ( ) ( )
9(x26x( )) -4(y2-2y( )) -41 9( )
-4( )
9(x 3)2 4(y - 1)2 36
(x 3)2 (y 1)2 1 4 9
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(-3, 1)
center a b vertices foci
center
3
a
b
2
vertices
(-5, 1) (-1, 1)
foci
(-3v13, 1)
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