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Writing equations of conics in vertex form
MM3G2
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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:
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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 (donβt 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.
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STEPS: Step 6: Divide by the constant. Step 7: simplify each fraction.
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Write the equation for the ellipse in vertex form:
Example 1 4π₯ 2 + 9π¦ 2 +8π₯+54π¦+52=3 Step 1: move the constant to the other side of the equation and move common variables together 4π₯ 2 +8π₯+ 9π¦ 2 +54π¦=3β52 4π₯ 2 +8π₯+ 9π¦ 2 +54π¦=β49
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Example 1 4(π₯ 2 +2π₯ ) + 9(π¦ 2 +6π¦ )=β49 4π₯ 2 +8π₯+ 9π¦ 2 +54π¦=β49
Step 2: Group the x terms together and the y terms together (4π₯ 2 +8π₯ )+ (9π¦ 2 +54π¦ )=β49 Step 3: Factor the GCF (coefficient)from the x group and then from the y group 4(π₯ 2 +2π₯ ) + 9(π¦ 2 +6π¦ )=β49
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Example 1 4 (π₯ 2 +2π₯ +1)+ =β49+4 9 π¦ 2 +6π¦ +9 +81
4(π₯ 2 +2π₯ )+ 9(π¦ 2 +6π¦ )=β49 Step 4: Complete the square on the x group (donβt 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) =β49+4 9 π¦ 2 +6π¦ +9 +81 4 (π₯ 2 +2π₯ +1)+9 π¦ 2 +6π¦ +9 =36
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Example 1 4 (π₯ 2 +2π₯ +1)+9 π¦ 2 +6π¦ +9 =36 Step 5: Write the factored form for the groups. 4(π₯+1) π¦+3 2 =36 Now we have to make the equation equal 1 and that will give us our denominators
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Example 1 Step 6: Divide by the constant. 4(π₯+1) 2 +9 π¦+3 2 =36 36
4(π₯+1) π¦+3 2 =36 36 4(π₯+1) π¦ = 36 36
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Example 1 Step 7: simplify each fraction.
4(π₯+1) π¦ = 36 36 (π₯+1) π¦ =1 Now the equation looks like what we are used to 1 9 4
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Example 2: Ellipse ( 4π₯ 2 β32π₯ )+(25 π¦ 2 β150π¦ )=β189
4π₯ π¦ 2 β32π₯β150π¦+189=0 4π₯ 2 β32π₯+25 π¦ 2 β150π¦=β189 ( 4π₯ 2 β32π₯ )+(25 π¦ 2 β150π¦ )=β189 4( π₯ 2 β8π₯ )+25( π¦ 2 β6π¦ )=β189
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Example 2 4( π₯ 2 β8π₯ )+25( π¦ 2 β6π¦ )=β189 4 (π₯ 2 β8π₯ +16)+ =β189+64
4( π₯ 2 β8π₯ )+25( π¦ 2 β6π¦ )=β189 -8/2 = -4 -6/2 = -3 -42 = 16 -32 = 9 4 (π₯ 2 β8π₯ +16) =β189+64 25 π¦ 2 β6π¦ +9 +225 4 (π₯ 2 β8π₯ +16)+25 π¦ 2 β6π¦ +9 =100 4 π₯β π¦β3 2 =100
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Example 2 4 π₯β π¦β3 2 = 4(π₯β4) π¦β = (π₯β4) π¦β =1 1 25 4
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Example 3: Ellipse ( 9π₯ 2 +36π₯ )+(4 π¦ 2 β40π¦ )=188
9π₯ 2 + 4π¦ 2 +36π₯β40π¦β100=88 9π₯ 2 +36π₯+4 π¦ 2 β40π¦=188 ( 9π₯ 2 +36π₯ )+(4 π¦ 2 β40π¦ )=188 9( π₯ 2 +4π₯ )+4( π¦ 2 β10π¦ )=188
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Example 3 9( π₯ 2 +4π₯ )+4( π¦ 2 β10π¦ )=188 9 (π₯ 2 +4π₯ +4)+ =188+36
9( π₯ 2 +4π₯ )+4( π¦ 2 β10π¦ )=188 4/2 = 2 -10/2 = -5 22 = 4 -52 = 25 9 (π₯ 2 +4π₯ +4) =188+36 4 π¦ 2 β10π¦ +25 +100 9 (π₯ 2 +4π₯ +4)+4 π¦ 2 β10π¦ +25 =324 9 π₯ π¦β5 2 =324
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Example 3 9 π₯ π¦β5 2 = 9(π₯+2) π¦β = (π₯+2) π¦β =1 1 36 81
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Example 4: Hyperbola π₯ 2 +2π₯ + β9 π¦ 2 β54π¦ =98
π₯ 2 β9 π¦ 2 +2π₯β54π¦β98=0 π₯ 2 +2π₯β9 π¦ 2 β54π¦=98 π₯ 2 +2π₯ β9 π¦ 2 β54π¦ =98 π₯ 2 +2π₯ β9( π¦ 2 +6π¦ )=98
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Example 4 π₯ 2 +2π₯ β9( π¦ 2 +6π¦ )=98 (π₯ 2 +2π₯ +1) =98+1 β9 π¦ 2 +6π¦ +9
π₯ 2 +2π₯ β9( π¦ 2 +6π¦ )=98 2/2 = 1 6/2 = 3 12 = 1 32 = 9 (π₯ 2 +2π₯ +1) =98+1 β9 π¦ 2 +6π¦ +9 β81 (π₯ 2 +2π₯ +1)β9 π¦ 2 +6π¦ +9 =18 π₯+1 2 β9 π¦+3 2 =18
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Example 4 π₯+1 2 β9 π¦+3 2 =18 18 (π₯+1) 2 18 β 9 π¦+3 2 18 = 18 18
π₯+1 2 β9 π¦+3 2 =18 18 (π₯+1) β 9 π¦ = 18 18 (π₯+1) β π¦ =1 1 2
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Example 5: Hyperbola 4π¦ 2 +16π¦ + β9 π₯ 2 +72π₯ =164
4π¦ 2 β9 π₯ 2 +16π¦+72π₯β164=0 4π¦ 2 +16π¦β9 π₯ 2 +72π₯=164 4π¦ 2 +16π¦ β9 π₯ 2 +72π₯ =164 4 π¦ 2 +4π¦ β9( π₯ 2 β8π₯ )=164
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Example 5 4 π¦ 2 +4π¦ β9( π₯ 2 β8π₯ )=164 4(π¦ 2 +4π¦ +4) =164+16
4 π¦ 2 +4π¦ β9( π₯ 2 β8π₯ )=164 4/2 = 2 -8/2 = -4 22 = 4 -42 =16 4(π¦ 2 +4π¦ +4) =164+16 β9 π₯ 2 β8π₯ +16 β144 4(π¦ 2 +4π¦ +4)β9 π₯ 2 β8π₯ +16 =36 4 π¦+2 2 β9 π₯β4 2 =36
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Example 5 4 π¦+2 2 β9 π₯β4 2 =36 36 4(π¦+2) 2 36 β 9 π₯β4 2 36 = 36 36
4 π¦+2 2 β9 π₯β4 2 =36 36 4(π¦+2) β 9 π₯β = 36 36 (π¦+2) 2 9 β π₯β =1 1 9 4
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