WRITING EQUATIONS OF CONICS IN VERTEX FORMMM3G2
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 1
Step 1: move the constant to the other side of the equation and move common variables together
Example 1
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 1
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 = 112 = 1
6/2 = 332 = 9
4(π₯ΒΏΒΏ2+2π₯+1)+9 ( π¦2+6 π¦+9 )=36ΒΏ
9 ( π¦2+6 π¦+9 ) +81
Example 1
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 1
Step 6: Divide by the constant.
Example 1
Step 7: simplify each fraction.
Now the equation looks like what we are used to
9 41
Example 2: Ellipse
Example 2
-8/2 = -4-42 = 16
-6/2 = -3-32 = 9
4(π₯ΒΏΒΏ2β8π₯+16)+25 (π¦ 2β6 π¦+9 )=100ΒΏ
25 ( π¦2β6 π¦+9 ) +225
4 (π₯β4 )2+25 (π¦β3 )2=100
Example 2
25
41
Example 3: Ellipse
Example 3
4/2 = 222 = 4
-10/2 = -5-52 = 25
9 (π₯ΒΏΒΏ 2+4 π₯+4)+4 ( π¦2β10 π¦+25 )=324 ΒΏ
4 ( π¦2β10 π¦+25 ) +100
9 (π₯+2 )2+4 (π¦β5 )2=324
Example 3
36
811
Example 4: Hyperbola
Example 4
2/2 = 112 = 1
6/2 = 332 = 9
(π₯ΒΏΒΏ2+2 π₯+1)β9 ( π¦2+6 π¦+9 )=18 ΒΏ
β9 (π¦2+6 π¦+9 ) β81
(π₯+1 )2β9 (π¦+3 )2=18
Example 4
21
Example 5: Hyperbola
Example 5
4/2 = 222 = 4
-8/2 = -4-42 =16
4(π¦ΒΏΒΏ 2+4 π¦+4)β9 (π₯2β8 π₯+16 )=36ΒΏ
β9 (π₯2β8 π₯+16 ) β144
4 (π¦+2 )2β9 (π₯β4 )2=36
Example 5
9 41
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