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equation 1: 4x² + 9y² - 32x + 36y + 64 = 0

Group like terms.

(4x² - 32x) + (9y² + 36y) + 64 = 0

Factor out numbers if possible. You want the squared term to have a coefficient of 1.

4(x² - 8x) + 9(y² + 4y) + 64 = 0

Add blanks as placeholders.

4(x² - 8x + ______) + 9(y² + 4y + ______) + 64 = 4(______) + 9(______)

Notice the coefficients on the RHS to account for what you factored out.

Take the coefficient of the x term: -8

Divide it by 2: -8 / 2 = -4

Square it: (-4)² = 16

Add to two of the blanks. You want to put it with the 4 coefficient.

4(x² - 8x + 16) + 9(y² + 4y + ______) + 64 = 4(16) + 9(______)

Take the coefficient of the y term: 4

Divide it by 2: 4 / 2 = 2

Square it: (2)² = 4

Add to two of the blanks. You want to put it with the 9 coefficient.

4(x² - 8x + 16) + 9(y² + 4y + 4) + 64 = 4(16) + 9(4)

x² - 8x + 16 is the factored form of a perfect square binomial.

x² - 8x + 16 = (x - 4)²

Apply this to the equation.

4(x - 4)² + 9(y² + 4y + 4) + 64 = 4(16) + 9(4)

y² + 4y + 4 is the factored form of a perfect square binomial.

y² + 4y + 4 = (y + 2)²

Apply this to the equation.

4(x - 4)² + 9(y + 2)² + 64 = 4(16) + 9(4)

Simplify the rest.

4(x - 4)² + 9(y + 2)² + 64 = 4(16) + 9(4)

4(x - 4)² + 9(y + 2)² + 64 = 64 + 36

4(x - 4)² + 9(y + 2)² + 64 = 100

4(x - 4)² + 9(y + 2)² = 100 - 64

4(x - 4)² + 9(y + 2)² = 36

Divide both sides by 36.

[4(x - 4)² / 36] + [9(y + 2)² / 36] = 36 / 36

[(x - 4)² / 9] + [(y + 2)² / 4] = 1 <== equation for an ellipse

Remember that the general ellipse equation is:

[(x - h)² / a²] + [(y - k)² / b²] = 1 where a > b

[(x - 4)² / 9] + [(y + 2)² / 4] = 1 <== equation for an ellipse

a² = 9

a = ± 3

b² = 4

b = ± 2

Find c.

c² = a² - b²

c² = 9 - 4

c² = 5

√c² = √5

c = ± √5

I think the foci are at (h, c) and (k, c)

(h, c) = (4, ± √5)

(k, c) = (-2, ± √5)

Answer: The foci are at (4, -√5), (4, √5), (-2, -√5), and (-2, √5)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

equation 2: x² + 4y² + 10x - 8y + 13 = 0

Group like terms.

(x² + 10x) + (4y² - 8y) + 13 = 0

Factor out numbers if possible. You want the squared term to have a coefficient of 1.

(x² + 10x) + 4(y² - 2y) + 13 = 0

Add blanks as placeholders.

(x² + 10x + ____) + 4(y² - 2y + ____) + 13 = 0 + ____ + 4(____)

Notice the coefficient on the RHS to account for what you factored out.

Take the coefficient of the x term: 10

Divide it by 2: 10 / 2 = 5

Square it: (5)² = 25

Add to two of the blanks. You want to put it with the blank with no coefficient.

(x² + 10x + 25) + 4(y² - 2y + ____) + 13 = 0 + 25 + 4(____)

Take the coefficient of the y term: -2

Divide it by 2: -2 / 2 = -1

Square it: (-1)² = 1

Add to two of the blanks. You want to put it with the 9 coefficient.

(x² + 10x + 25) + 4(y² - 2y + 1) + 13 = 0 + 25 + 4(1)

x² + 10x + 25 is the factored form of a perfect square binomial.

x² + 10x + 25 = (x + 5)²

Apply this to the equation.

(x + 5)² + 4(y² - 2y + 1) + 13 = 0 + 25 + 4(1)

y² - 2y + 1 is the factored form of a perfect square binomial.

y² - 2y + 1 = (y - 1)²

Apply this to the equation.

(x + 5)² + 4(y - 1)² + 13 = 0 + 25 + 4(1)

Simplify the rest.

(x + 5)² + 4(y - 1)² + 13 = 0 + 25 + 4(1)

(x + 5)² + 4(y - 1)² + 13 = 0 + 25 + 4

(x + 5)² + 4(y - 1)² + 13 = 29

(x + 5)² + 4(y - 1)² = 29 - 13

(x + 5)² + 4(y - 1)² = 16

Divide both sides by 16.

[(x + 5)² / 16] + [4(y - 1)² / 16] = 16 / 16

[(x + 5)² / 16] + [(y - 1)² / 4] = 1 <== equation for an ellipse

Remember that the general ellipse equation is:

[(x - h)² / a²] + [(y - k)² / b²] = 1 where a > b

[(x + 5)² / 16] + [(y - 1)² / 4] = 1 <== equation for an ellipse

a² = 16

a = ± 4

b² = 4

b = ± 2

Find c.

c² = a² - b²

c² = 16 - 4

c² = 12

√c² = √12

c = ± √12

c = ± √(4 * 3)

c = ± 2√3

I think the foci are at (h, c) and (k, c)

(h, c) = (-5, ± 2√3)

(k, c) = (1, ± 2√3)

Answer: The foci are at (-5, -2√3), (-5, 2√3), (1, -2√3), and (1, 2√3).

~~~~~~~~~~~~~~~~~~~~~~~~~~

Note: My recollection on how to find the foci is very vague, so I'm not sure about those, but I know the completing the square parts are correct.

Group like terms.

(4x² - 32x) + (9y² + 36y) + 64 = 0

Factor out numbers if possible. You want the squared term to have a coefficient of 1.

4(x² - 8x) + 9(y² + 4y) + 64 = 0

Add blanks as placeholders.

4(x² - 8x + ______) + 9(y² + 4y + ______) + 64 = 4(______) + 9(______)

Notice the coefficients on the RHS to account for what you factored out.

Take the coefficient of the x term: -8

Divide it by 2: -8 / 2 = -4

Square it: (-4)² = 16

Add to two of the blanks. You want to put it with the 4 coefficient.

4(x² - 8x + 16) + 9(y² + 4y + ______) + 64 = 4(16) + 9(______)

Take the coefficient of the y term: 4

Divide it by 2: 4 / 2 = 2

Square it: (2)² = 4

Add to two of the blanks. You want to put it with the 9 coefficient.

4(x² - 8x + 16) + 9(y² + 4y + 4) + 64 = 4(16) + 9(4)

x² - 8x + 16 is the factored form of a perfect square binomial.

x² - 8x + 16 = (x - 4)²

Apply this to the equation.

4(x - 4)² + 9(y² + 4y + 4) + 64 = 4(16) + 9(4)

y² + 4y + 4 is the factored form of a perfect square binomial.

y² + 4y + 4 = (y + 2)²

Apply this to the equation.

4(x - 4)² + 9(y + 2)² + 64 = 4(16) + 9(4)

Simplify the rest.

4(x - 4)² + 9(y + 2)² + 64 = 4(16) + 9(4)

4(x - 4)² + 9(y + 2)² + 64 = 64 + 36

4(x - 4)² + 9(y + 2)² + 64 = 100

4(x - 4)² + 9(y + 2)² = 100 - 64

4(x - 4)² + 9(y + 2)² = 36

Divide both sides by 36.

[4(x - 4)² / 36] + [9(y + 2)² / 36] = 36 / 36

[(x - 4)² / 9] + [(y + 2)² / 4] = 1 <== equation for an ellipse

Remember that the general ellipse equation is:

[(x - h)² / a²] + [(y - k)² / b²] = 1 where a > b

[(x - 4)² / 9] + [(y + 2)² / 4] = 1 <== equation for an ellipse

a² = 9

a = ± 3

b² = 4

b = ± 2

Find c.

c² = a² - b²

c² = 9 - 4

c² = 5

√c² = √5

c = ± √5

I think the foci are at (h, c) and (k, c)

(h, c) = (4, ± √5)

(k, c) = (-2, ± √5)

Answer: The foci are at (4, -√5), (4, √5), (-2, -√5), and (-2, √5)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

equation 2: x² + 4y² + 10x - 8y + 13 = 0

Group like terms.

(x² + 10x) + (4y² - 8y) + 13 = 0

Factor out numbers if possible. You want the squared term to have a coefficient of 1.

(x² + 10x) + 4(y² - 2y) + 13 = 0

Add blanks as placeholders.

(x² + 10x + ____) + 4(y² - 2y + ____) + 13 = 0 + ____ + 4(____)

Notice the coefficient on the RHS to account for what you factored out.

Take the coefficient of the x term: 10

Divide it by 2: 10 / 2 = 5

Square it: (5)² = 25

Add to two of the blanks. You want to put it with the blank with no coefficient.

(x² + 10x + 25) + 4(y² - 2y + ____) + 13 = 0 + 25 + 4(____)

Take the coefficient of the y term: -2

Divide it by 2: -2 / 2 = -1

Square it: (-1)² = 1

Add to two of the blanks. You want to put it with the 9 coefficient.

(x² + 10x + 25) + 4(y² - 2y + 1) + 13 = 0 + 25 + 4(1)

x² + 10x + 25 is the factored form of a perfect square binomial.

x² + 10x + 25 = (x + 5)²

Apply this to the equation.

(x + 5)² + 4(y² - 2y + 1) + 13 = 0 + 25 + 4(1)

y² - 2y + 1 is the factored form of a perfect square binomial.

y² - 2y + 1 = (y - 1)²

Apply this to the equation.

(x + 5)² + 4(y - 1)² + 13 = 0 + 25 + 4(1)

Simplify the rest.

(x + 5)² + 4(y - 1)² + 13 = 0 + 25 + 4(1)

(x + 5)² + 4(y - 1)² + 13 = 0 + 25 + 4

(x + 5)² + 4(y - 1)² + 13 = 29

(x + 5)² + 4(y - 1)² = 29 - 13

(x + 5)² + 4(y - 1)² = 16

Divide both sides by 16.

[(x + 5)² / 16] + [4(y - 1)² / 16] = 16 / 16

[(x + 5)² / 16] + [(y - 1)² / 4] = 1 <== equation for an ellipse

Remember that the general ellipse equation is:

[(x - h)² / a²] + [(y - k)² / b²] = 1 where a > b

[(x + 5)² / 16] + [(y - 1)² / 4] = 1 <== equation for an ellipse

a² = 16

a = ± 4

b² = 4

b = ± 2

Find c.

c² = a² - b²

c² = 16 - 4

c² = 12

√c² = √12

c = ± √12

c = ± √(4 * 3)

c = ± 2√3

I think the foci are at (h, c) and (k, c)

(h, c) = (-5, ± 2√3)

(k, c) = (1, ± 2√3)

Answer: The foci are at (-5, -2√3), (-5, 2√3), (1, -2√3), and (1, 2√3).

~~~~~~~~~~~~~~~~~~~~~~~~~~

Note: My recollection on how to find the foci is very vague, so I'm not sure about those, but I know the completing the square parts are correct.

### Source:

For more info on ellipses:

http://www.analyzemath.com/EllipseProble...

http://www.analyzemath.com/EllipseProble...

convert to standard form by completing the square ; then find the location of the foci of the ellipse?

equation 1: 4x^2+9y^2-32x+36y+64=0

equation 2: x^2+4y^2+10x-8y+13=0

equation 2: x^2+4y^2+10x-8y+13=0

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