Presentation is loading. Please wait.

Presentation is loading. Please wait.

Writing equations of conics in vertex form

Similar presentations


Presentation on theme: "Writing equations of conics in vertex form"β€” Presentation transcript:

1 Writing equations of conics in vertex form
MM3G2

2 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:

3 STEPS: Writing equations of ellipses and hyperbolas in vertex form
Step 1: move the constant to the other side of the equation and move common variables together 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 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. Step 5: Write the factored form for the groups.

4 STEPS: Step 6: Divide by the constant. Step 7: simplify each fraction.

5 Write the equation for the ellipse in vertex form:
Example 1 4π‘₯ 2 + 9𝑦 2 +8π‘₯+54𝑦+52=3 Step 1: move the constant to the other side of the equation and move common variables together 4π‘₯ 2 +8π‘₯+ 9𝑦 2 +54𝑦=3βˆ’52 4π‘₯ 2 +8π‘₯+ 9𝑦 2 +54𝑦=βˆ’49

6 Example 1 4(π‘₯ 2 +2π‘₯ ) + 9(𝑦 2 +6𝑦 )=βˆ’49 4π‘₯ 2 +8π‘₯+ 9𝑦 2 +54𝑦=βˆ’49
Step 2: Group the x terms together and the y terms together (4π‘₯ 2 +8π‘₯ )+ (9𝑦 2 +54𝑦 )=βˆ’49 Step 3: Factor the GCF (coefficient)from the x group and then from the y group 4(π‘₯ 2 +2π‘₯ ) + 9(𝑦 2 +6𝑦 )=βˆ’49

7 Example 1 4 (π‘₯ 2 +2π‘₯ +1)+ =βˆ’49+4 9 𝑦 2 +6𝑦 +9 +81
4(π‘₯ 2 +2π‘₯ )+ 9(𝑦 2 +6𝑦 )=βˆ’49 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 2/2 = 1 6/2 = 3 12 = 1 32 = 9 4 (π‘₯ 2 +2π‘₯ +1) =βˆ’49+4 9 𝑦 2 +6𝑦 +9 +81 4 (π‘₯ 2 +2π‘₯ +1)+9 𝑦 2 +6𝑦 +9 =36

8 Example 1 4 (π‘₯ 2 +2π‘₯ +1)+9 𝑦 2 +6𝑦 +9 =36 Step 5: Write the factored form for the groups. 4(π‘₯+1) 𝑦+3 2 =36 Now we have to make the equation equal 1 and that will give us our denominators

9 Example 1 Step 6: Divide by the constant. 4(π‘₯+1) 2 +9 𝑦+3 2 =36 36
4(π‘₯+1) 𝑦+3 2 =36 36 4(π‘₯+1) 𝑦 = 36 36

10 Example 1 Step 7: simplify each fraction.
4(π‘₯+1) 𝑦 = 36 36 (π‘₯+1) 𝑦 =1 Now the equation looks like what we are used to 1 9 4

11 Example 2: Ellipse ( 4π‘₯ 2 βˆ’32π‘₯ )+(25 𝑦 2 βˆ’150𝑦 )=βˆ’189
4π‘₯ 𝑦 2 βˆ’32π‘₯βˆ’150𝑦+189=0 4π‘₯ 2 βˆ’32π‘₯+25 𝑦 2 βˆ’150𝑦=βˆ’189 ( 4π‘₯ 2 βˆ’32π‘₯ )+(25 𝑦 2 βˆ’150𝑦 )=βˆ’189 4( π‘₯ 2 βˆ’8π‘₯ )+25( 𝑦 2 βˆ’6𝑦 )=βˆ’189

12 Example 2 4( π‘₯ 2 βˆ’8π‘₯ )+25( 𝑦 2 βˆ’6𝑦 )=βˆ’189 4 (π‘₯ 2 βˆ’8π‘₯ +16)+ =βˆ’189+64
4( π‘₯ 2 βˆ’8π‘₯ )+25( 𝑦 2 βˆ’6𝑦 )=βˆ’189 -8/2 = -4 -6/2 = -3 -42 = 16 -32 = 9 4 (π‘₯ 2 βˆ’8π‘₯ +16) =βˆ’189+64 25 𝑦 2 βˆ’6𝑦 +9 +225 4 (π‘₯ 2 βˆ’8π‘₯ +16)+25 𝑦 2 βˆ’6𝑦 +9 =100 4 π‘₯βˆ’ π‘¦βˆ’3 2 =100

13 Example 2 4 π‘₯βˆ’ π‘¦βˆ’3 2 = 4(π‘₯βˆ’4) π‘¦βˆ’ = (π‘₯βˆ’4) π‘¦βˆ’ =1 1 25 4

14 Example 3: Ellipse ( 9π‘₯ 2 +36π‘₯ )+(4 𝑦 2 βˆ’40𝑦 )=188
9π‘₯ 2 + 4𝑦 2 +36π‘₯βˆ’40π‘¦βˆ’100=88 9π‘₯ 2 +36π‘₯+4 𝑦 2 βˆ’40𝑦=188 ( 9π‘₯ 2 +36π‘₯ )+(4 𝑦 2 βˆ’40𝑦 )=188 9( π‘₯ 2 +4π‘₯ )+4( 𝑦 2 βˆ’10𝑦 )=188

15 Example 3 9( π‘₯ 2 +4π‘₯ )+4( 𝑦 2 βˆ’10𝑦 )=188 9 (π‘₯ 2 +4π‘₯ +4)+ =188+36
9( π‘₯ 2 +4π‘₯ )+4( 𝑦 2 βˆ’10𝑦 )=188 4/2 = 2 -10/2 = -5 22 = 4 -52 = 25 9 (π‘₯ 2 +4π‘₯ +4) =188+36 4 𝑦 2 βˆ’10𝑦 +25 +100 9 (π‘₯ 2 +4π‘₯ +4)+4 𝑦 2 βˆ’10𝑦 +25 =324 9 π‘₯ π‘¦βˆ’5 2 =324

16 Example 3 9 π‘₯ π‘¦βˆ’5 2 = 9(π‘₯+2) π‘¦βˆ’ = (π‘₯+2) π‘¦βˆ’ =1 1 36 81

17 Example 4: Hyperbola π‘₯ 2 +2π‘₯ + βˆ’9 𝑦 2 βˆ’54𝑦 =98
π‘₯ 2 βˆ’9 𝑦 2 +2π‘₯βˆ’54π‘¦βˆ’98=0 π‘₯ 2 +2π‘₯βˆ’9 𝑦 2 βˆ’54𝑦=98 π‘₯ 2 +2π‘₯ βˆ’9 𝑦 2 βˆ’54𝑦 =98 π‘₯ 2 +2π‘₯ βˆ’9( 𝑦 2 +6𝑦 )=98

18 Example 4 π‘₯ 2 +2π‘₯ βˆ’9( 𝑦 2 +6𝑦 )=98 (π‘₯ 2 +2π‘₯ +1) =98+1 βˆ’9 𝑦 2 +6𝑦 +9
π‘₯ 2 +2π‘₯ βˆ’9( 𝑦 2 +6𝑦 )=98 2/2 = 1 6/2 = 3 12 = 1 32 = 9 (π‘₯ 2 +2π‘₯ +1) =98+1 βˆ’9 𝑦 2 +6𝑦 +9 βˆ’81 (π‘₯ 2 +2π‘₯ +1)βˆ’9 𝑦 2 +6𝑦 +9 =18 π‘₯+1 2 βˆ’9 𝑦+3 2 =18

19 Example 4 π‘₯+1 2 βˆ’9 𝑦+3 2 =18 18 (π‘₯+1) 2 18 βˆ’ 9 𝑦+3 2 18 = 18 18
π‘₯+1 2 βˆ’9 𝑦+3 2 =18 18 (π‘₯+1) βˆ’ 9 𝑦 = 18 18 (π‘₯+1) βˆ’ 𝑦 =1 1 2

20 Example 5: Hyperbola 4𝑦 2 +16𝑦 + βˆ’9 π‘₯ 2 +72π‘₯ =164
4𝑦 2 βˆ’9 π‘₯ 2 +16𝑦+72π‘₯βˆ’164=0 4𝑦 2 +16π‘¦βˆ’9 π‘₯ 2 +72π‘₯=164 4𝑦 2 +16𝑦 βˆ’9 π‘₯ 2 +72π‘₯ =164 4 𝑦 2 +4𝑦 βˆ’9( π‘₯ 2 βˆ’8π‘₯ )=164

21 Example 5 4 𝑦 2 +4𝑦 βˆ’9( π‘₯ 2 βˆ’8π‘₯ )=164 4(𝑦 2 +4𝑦 +4) =164+16
4 𝑦 2 +4𝑦 βˆ’9( π‘₯ 2 βˆ’8π‘₯ )=164 4/2 = 2 -8/2 = -4 22 = 4 -42 =16 4(𝑦 2 +4𝑦 +4) =164+16 βˆ’9 π‘₯ 2 βˆ’8π‘₯ +16 βˆ’144 4(𝑦 2 +4𝑦 +4)βˆ’9 π‘₯ 2 βˆ’8π‘₯ +16 =36 4 𝑦+2 2 βˆ’9 π‘₯βˆ’4 2 =36

22 Example 5 4 𝑦+2 2 βˆ’9 π‘₯βˆ’4 2 =36 36 4(𝑦+2) 2 36 βˆ’ 9 π‘₯βˆ’4 2 36 = 36 36
4 𝑦+2 2 βˆ’9 π‘₯βˆ’4 2 =36 36 4(𝑦+2) βˆ’ 9 π‘₯βˆ’ = 36 36 (𝑦+2) 2 9 βˆ’ π‘₯βˆ’ =1 1 9 4


Download ppt "Writing equations of conics in vertex form"

Similar presentations


Ads by Google