Writing equations of conics in vertex form

Post on 15-Feb-2016

18 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

DESCRIPTION

Writing equations of conics in vertex form. MM3G2. Write the equation for the circle in vertex form :. Example 1 Step 1: Move the constant to the other side of the equation & put your common variables together. Example 1. - PowerPoint PPT Presentation

Transcript

WRITING EQUATIONS OF CONICS IN VERTEX FORMMM3G2

Write the equation for the circle in vertex form:

Example 1

Step 1: Move the constant to the other side of the equation & put your common variables together

Example 1

Step 2: Identify the coefficients of the squared terms and divide everything by that coefficient. Both coefficients are 1 so divide

everything by 1

Example 1

Step 3: Group the x terms together and the y terms together using parenthesis.

Example 1

Step 4: Complete the square for the x terms

Then for the y terms

22=1 12=1 −42 =−2 (−2)2=4

Example 1

Step 5: Write the factored form for the groups.

What is the center of this circle?

What is the radius?

Write the equation for the circle in vertex form:

Example 2

Step 1: Move the constant to the other side of the equation & put your common variables together

Example 2

Step 2: Identify the coefficients of the squared terms and divide everything by that coefficient. Both coefficients are 2 so divide

everything by 2

Example 2

Step 3: Group the x terms together and the y terms together using parenthesis.

Example 2

Step 4: Complete the square for the x terms

Then for the y terms

62=3 32=9 42=2 22=4

Example 2

Step 5: Write the factored form for the groups.

What is the center of this circle?

What is the radius?

Write the equation for the circle in vertex form:

Example 3

Step 1: Move the constant to the other side of the equation & put your common variables together

Example 3

Step 2: Identify the coefficients of the squared terms and divide everything by that coefficient. Both coefficients are 4 so divide

everything by 4

Example 3

Step 3: Group the x terms together and the y terms together using parenthesis.

Example 3

Step 4: Complete the square for the x terms

Then for the y terms

62=3 32=9 82=4 42=16

Example 3

Step 5: Write the factored form for the groups.

What is the center of this circle?

What is the radius?

Write the equation for the circle in vertex form:

Example 4

Step 1: Move the constant to the other side of the equation & put your common variables together

Example 4

Step 2: Identify the coefficients of the squared terms and divide everything by that coefficient. Both coefficients are 5 so divide

everything by 5

Example 4

Step 3: Group the x terms together and the y terms together using parenthesis.

Example 4

Step 4: Complete the square for the x terms

Then for the y terms

−162 =−8¿ 4

2=2 22=4

Example 4

Step 5: Write the factored form for the groups.

What is the center of this circle?

What is the radius?

Recall: The equation for a circle does not have

denominators The equation for an ellipse and a

hyperbola do have denominators The equation for a circle is not equal to

one The equation for an ellipse and a

hyperbola are equal to one We have a different set of steps for

converting ellipses and hyperbolas to the vertex form:

Write the equation for the ellipse in vertex form:

Example 5

Step 1: Move the constant to the other side of the equation and move common variables together

Example 5

Step 2: Group the x terms together and the y terms together

Step 3: Factor the GCF (coefficient)from the x group

and then from the y group

Example 5

Step 4: Complete the square on the x group (don’t forget to multiply by the GCF before you add to the right side.)

Then do the same for the y terms

22=1 12=1

62=3 32=9

4(𝑥¿¿2+2𝑥+1)+9 ( 𝑦2+6 𝑦+9 )=36¿

9 ( 𝑦2+6 𝑦+9 ) +81

Example 5

Step 5: Write the factored form for the groups.

**Now we have to make the equation equal 1 and that will give us our denominators

Example 5 Step 6: Divide by the constant.

Example 5 Step 7: simplify each fraction.

Now the equation looks like what we are used to!!

9 41

(𝑥+1)2

9+

(𝑦+3 )2

4=1

What is the center of this ellipse?

What is the length of the major axis?

What is the length of the minor axis?

Example 6: Ellipse

Step 2:

Step 1:

Step 3:

Example 6

− 82=−4 −42=16 − 62=−3 −3 2=9

4(𝑥¿¿2−8𝑥+16)+25 (𝑦 2−6 𝑦+9 )=100¿

25 ( 𝑦2−6 𝑦+9 ) +225

4 (𝑥−4 )2+25 (𝑦−3 )2=100

Step 4:

Step 5:

Example 6

25

41

Step 6:

(𝑥−4)2

25+

(𝑦−3 )2

4=1

What is the center of this ellipse?

What is the length of the major axis?

What is the length of the minor axis?

Example 7: Ellipse

Step 2:

Step 1:

Step 3:

Example 7

42=2 22=4−

102 =−5 −52=25

9 (𝑥¿¿ 2+4 𝑥+4)+4 ( 𝑦2−10 𝑦+25 )=324 ¿

4 ( 𝑦2−10 𝑦+25 ) +100

9 (𝑥+2 )2+4 (𝑦−5 )2=324

Step 4:

Step 5:

Example 7

36

811

Step 6:

(𝑥+2)2

36+

(𝑦−5 )2

81=1

What is the center of this ellipse?

What is the length of the major axis?

What is the length of the minor axis?

Example 8: Hyperbola

Step 2:

Step 1:

Step 3:

Example 8

22=1 12=1

62=3 32=9

(𝑥¿¿2+2 𝑥+1)−9 ( 𝑦2+6 𝑦+9 )=18 ¿

−9 (𝑦2+6 𝑦+9 ) −81

(𝑥+1 )2−9 (𝑦+3 )2=18

Step 4:

Step 6:

Example 8

21

Step 6:

(𝑥+1)2

18− (𝑦+3 )2

2=1

What is the center of this hyperbola?

What is the length of the transverse axis?

What is the length of the conjugate axis?

Example 9: Hyperbola

Step 2:

Step 1:

Step 3:

Example 9

42=2 22=4 − 82=−4−42=16

4(𝑦¿¿ 2+4 𝑦+4)−9 (𝑥2−8 𝑥+16 )=36¿

−9 (𝑥2−8 𝑥+16 ) −144

4 (𝑦+2 )2−9 (𝑥−4 )2=36

Step 4:

Step 5:

Example 9

9 41

Step 6:

(𝑦+2)2

9− (𝑥−4 )2

4=1

What is the center of this hyperbola?

What is the length of the transverse axis?

What is the length of the conjugate axis?

You Try! Write the equation of each conic section

in vertex form:

Identify the center of each conic section as well as the length of the major/minor or

transverse/conjugate axis.

top related