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 (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 = 112 = 1

6/2 = 332 = 9

4(𝑥¿¿2+2𝑥+1)+9 ( 𝑦2+6 𝑦+9 )=36¿

9 ( 𝑦2+6 𝑦+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

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Example 2: Ellipse

Example 2

-8/2 = -4-42 = 16

-6/2 = -3-32 = 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

Example 2

25

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Example 3: Ellipse

Example 3

4/2 = 222 = 4

-10/2 = -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

Example 3

36

811

Example 4: Hyperbola

Example 4

2/2 = 112 = 1

6/2 = 332 = 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

Example 4

21

Example 5: Hyperbola

Example 5

4/2 = 222 = 4

-8/2 = -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

Example 5

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