Download Classifying Conics

8.6 Translate and
Classify Conic Sections
What is the general 2nd degree equation for
any conic?
What information can the discriminant tell
you about a conic?
The equation of any
conic can be written
in the form2
2
Ax  Bxy  Cy  Dx  Ey  F  0
Called a general 2nd degree
equation
Identify the line(s) of symmetry for each conic
section in Examples 1 – 4.
SOLUTION
For the circle in Example
1, any line through the
center (2, – 3) is a line of
symmetry.
For the hyperbola in
Example 2 x = – 1
and y = 3 are lines
of symmetry
Identify the line(s) of symmetry for each conic
section in Examples 1 – 4.
SOLUTION
For the parabola in
Example 3, y = 3 is a
line of symmetry.
For the ellipse in
Example 4, x = 4
and y = 2 are lines
of symmetry.
Identify the line(s) of symmetry for the conic section.
(x – 5)2
7.
+
64
(y)2
16
= 1
ANSWER
For the ellipse the lines of symmetry are x = 5
and y = 0.
Identify the line(s) of symmetry for the conic section.
8. (x + 5)2 = 8(y – 2).
ANSWER
For parabola the lie of symmetry are x = – 5
Identify the line(s) of symmetry for the conic section.
9. (x – 1)2
(y – 2)2
–
= 1
49
121
ANSWER
For horizontal lines of symmetry are x = 1 and y = 2.
Circles
( x  1)  ( y  2)  16
2
2
Can be multiplied out to look like
this….
x  y  2 x  4 y  11  0
2
2
Ellipse
( x  1)
2
 ( y  1)  1
4
2
Can be written like this…..
x  4 y  2x  8 y 1  0
2
2
Parabola
( y  6)  4( x  8)
2
Can be written like this…..
y  12 y  4 x  4  0
2
Hyperbola
( y  4)
( x  4) 
1
9
2
2
Can be written like this…..
9 x  y  72 x  8 y  1  0
2
2
How do you know which conic it is
when it’s been multiplied out?
• Pay close attention to whose squared
and whose not…
• Look at the coefficients in front of
the squared terms and their signs.
Circle
Both x and y are
squared
And their coefficients
are the same number
and sign
x  y  2 x  4 y  11  0
2
2
Ellipse
• Both x and y are
squared
• Their coefficients are
different but their
signs remain the
same.
x  4 y  2x  8 y 1  0
2
2
Parabola
• Either x or y is
squared but not both
y  12 y  4 x  4  0
2
Hyperbola
• Both x and y are
squared
• Their coefficients are
different and so are
their signs.
9 x  y  72 x  8 y  1  0
2
2
1. x 2  4y 2  2x  3  0
2. 2x 2  20x  y  41  0
You Try!
1. Ellipse
2. Parabola
3. 5x 2  3y 2  30  0
3. Hyperbola
4. x 2  y 2  12x  4y  31  0
4. Circle
5.  x 2  y 2  6x  6y  4  0
5. Hyperbola
6. x 2  8x  4y  16  0
7. 3 x  3 y  30 x  59  0
2
2
8. x  2 y  8 x  7  0
2
2
9. 4 x  y  16 x  6 y  3  0
2
2
10. 3 x  y  4 y  3  0
2
2
6. Parabola
7. Circle
8. Ellipse
9. Hyperbola
10.Ellipse
When you want to be sure…
of a conic equation, then find the type of conic
using discriminate information:
Ax2 +Bxy +Cy2 +Dx +Ey +F = 0
B2 − 4AC < 0, B = 0 & A = C
B2 − 4AC < 0 & either B≠0 or A≠C
B2 − 4AC = 0
B2 − 4AC > 0
Circle
Ellipse
Parabola
Hyperbola
Classify the Conic
2x2 + y2 −4x − 4 = 0
Ax2 +Bxy +Cy2 +Dx +Ey +F = 0
A=2
B=0
C=1
B2 − 4AC = 02 − 4(2)(1) = −8
B2 − 4AC < 0, the conic is an ellipse
Write the equation in standard form
by completing the square
2
x  4 y  2x  8 y 1  0
2
1


2
2
x  2 x  4( y  2 y)  1
 2 
2
2
x  2 x  ___  4( y  2 y  ___)  1  ___  ___
2
2
( x  2 x  1)  4( y  2 y  1)  1  1  (4)(1)
2
2
( x 1)  4( y 1)  4
2
2
( x  1) 2 4( y  1) 2 4


4
4
4
2
2
( x  1) ( y  1)

1
4
1
Steps to Complete the Square
1. Group x’s and y’s. (Boys with the boys and
girls with the girls) Send constant numbers to
the other side of the equal sign.
2. The coefficient of the x2 and y2 must be 1. If
not, factor out.
3. Take the number before the x, divide by 2 and
square. Do the same with the number before y.
4. Add these numbers to both sides of the
equation. *(Multiply it by the common factor in
#2)
5. Factor
Graph the Conic
2x2 + y2 −4x − 4 = 0
Complete the Square
2x2 −4x + y2 = 4
2(x2 −2x +___)+ y2 = 4 + ___
(−2/2)2= 1
2(x2 −2x +1)+ y2 = 4 + 2(1)
2(x−1)2 + y2 = 6
2( x  1)
y

1
6
6
2
2
( x  1)
y

1
3
6
V(1±√6), CV(1±√3)
2
2
10. Classify the conic given by x2 + y2 – 2x + 4y + 1 = 0.
Then graph the equation.
SOLUTION
Note that A = 1, B = 0, and C = 1, so the value of the
discriminant is: B2 – 4AC = 02 – 4(1)(1) = – 4
Because B2 – 4AC < 0 and A = C, the conic is an circle.
To graph the circle, first complete the square in both
x and y simultaneity .
x2 + y2 – 2x + 4y + 1 = 0
x2 – 2x +1+ y2 + 4y + 4 = 4
(x – 1)2 +( y + 2)2 = 4
ANSWER
From the equation, you can see that (h, k) = (– 1, 2),
r = 2 Use these facts to draw the circle.
Physical Science
In a lab experiment, you record
images of a steel ball rolling past
a magnet. The equation
16x2 – 9y2 – 96x + 36y – 36 = 0
models the ball’s path.
• What is the shape of the path ?
• Write an equation for the path in standard form.
• Graph the equation of the path.
SOLUTION
Identify the shape. The equation is a general
STEP 1
second-degree equation with A = 16, B = 0, and
C = – 9. Find the value of the discriminant.
B2 – 4AC = 02 – 4(16)(– 9) = 576
Because B2 – 4AC > 0, the shape of the path is a hyperbola.
STEP 2
Write an equation. To write an equation of the
hyperbola, complete the square in both x and y
simultaneously.
16x2 – 9y2 – 96x + 36y – 36 = 0
(16x2 – 96x) – (9y2 – 36y) = 36
16(x2 – 6x + ? ) – 9(y2 – 4y + ? ) = 36 + 16( ? ) – 9( ? )
16(x2 – 6x + 9) – 9(y2 – 4y + 4) = 36 + 16(9) – 9(4)
16(x – 3)2 – 9(y – 2)2 = 144
(x – 3)2
(y –2)2
–
= 1
16
9
STEP 3 Graph the equation. From the equation, the
transverse axis is horizontal, (h, k) = (3, 2),
a=
9 = 3 and b = 16. = 4
The vertices are at (3 + a, 2), or (6, 2) and (0, 2).
See page 530
What is the general 2nd degree equation for any
conic?
Ax  Bxy  Cy  Dx  Ey  F  0
2
2
What information can the discriminant tell you
about a conic?
B2- 4AC < 0, B = 0, A = C
Circle
B2- 4AC < 0, B ≠ 0, A ≠ C
Ellipse
B2- 4AC = 0,
Parabola
B2- 4AC > 0
Hyperbola
8.6 Assignment
Page 531, 23-43 odd