# Ellipse help, Finding foci?

I need help figuring out the focal points for the following equations:

A. 4x^2 + 9y^2 − 8x + 54y + 49 = 0

B. 16x^2 + 32y^2 + 64x − 256y + 64 = 0

C. 9x^2 + 25y^2 + 90x + 350y + 1225 = 0
asked Apr 30, 2013

A.

4x2 + 9y2 - 8x + 54y + 49 = 0

(2x)2 -2(2x)(2) + 4 + (3y)2 + 2(3y)(9) + 81 = 36

(2x - 2)2 + (3y + 9)2  = 36

4(x - 1)2 + 9(y + 3)2 = 36

(x - 1)2 / 9 + (y + 3)2 / 4 = 1

The ellipse of the center = (h , k) = (1 , -3)

a2 = 9 and b2 = 4

If a > b then c2 = a2 - b2 = 9 - 4 = 5

c = √5

The foci's are (h + c , k) and (h , k + c) = (1 + √5 , -3) and (1 , -3 + √5).

answered Apr 30, 2013

Foci: (h + c , k ), (h - c, k )

Focal points of

B.

16x2 + 32y2+ 64x - 256y + 64 = 0

(4x)2 + 2(4x)(8) + (8)^2 + (4√2 y)2 - 2(4√2 y)(8√2) + 128 = 128

(4x + 8)2 + (4√2 y - 8√2)2  = 128

16(x + 2)2 + 32(y - 2)2 = 128

(x + 2)2 / 8 + (y - 2)2 / 4 = 1

The ellipse of the center = (h , k) = (-2 , 2)

a2 = 8 and b2 = 4

If a > b then c2 = a2 - b2 = 8 - 4 = 4

c = 2

The foci's are (h + c , k) and (h , k + c) = (-2 + 2 , 2) and (-2 , 2 + 2) = (0 , 2) and (-2 , 4).

answered Apr 30, 2013

Foci: (h + c , k ), (h - c, k )

Focal points of are (2, 4), (-6, 4).

C.

B.

9x2 + 25y2+ 90x + 350y + 1225 = 0

(3x)2 + 2(3x)(15) + (15)^2 + (5 y)2 + 2(5 y)(35) + (35)^2 = 225

(3x + 15)2 + (5y + 35)2  = 225

9(x + 5)2 + 25(y + 7)2 = 225

(x + 5)2 / 25 + (y + 7)2 / 9 = 1

The ellipse of the center = (h , k) = (-5 , -7)

a2 = 25 and b2 = 9

If a > b then c2 = a2 - b2 = 25 - 9 = 16

c = 4

The foci's are (h + c , k) and (h , k + c) = (-5 + 4 , -7) and (-5 , -7 + 4) = (-1 , -7) and (-5 , -3).

answered Apr 30, 2013

Foci: (h + c , k ), (h - c, k )

Focal points of are (-1, -7), (-9, -7) .

A) The ellipse equation

To change the expressions (x 2- 2x ) and (y 2 - 4y ) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)²to each side of the equation.

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 36 to set it equal to 1.

Compare it to standard form of ellipse

a 2 > b 2

If the larger denominator is under the "x" term, then the ellipse is horizontal.

center (h, k ) = (1, -3)

a  = length of semi-major axis = 3

= length of semi-minor axis = 2.

c is the distance from the center to each focus.

Foci: (h + c , k ), (h - c, k )

Focal points of are .

answered Jun 5, 2014

B) The ellipse equation

To change the expressions (x 2+ 4x ) and (y 2 - 8y ) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)²to each side of the equation.

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 512 to set it equal to 1

Compare it to standard form of ellipse

a 2 > b 2

If the larger denominator is under the "x " term, then the ellipse is horizontal.

center (h, k ) = (-2, 4)

a 2 = 32, b 2 = 16

is the distance from the center to each focus.

Foci: (h + c , k ), (h - c, k )

Focal points of are .

answered Jun 5, 2014

c) The ellipse equation

To change the expressions (x 2+ 10x ) and (y 2 + 14y ) into a perfect square trinomial,

add (half the x coefficient)² and add (half the y coefficient)²to each side of the equation.

The standard form for an ellipse is in a form = 1, So divide both sides of equation by 225 to set it equal to 1

Compare it to standard form of ellipse

a 2 > b 2

If the larger denominator is under the "x " term, then the ellipse is horizontal.

center (h, k ) = (-5, -7)

a 2 = 25, b 2 = 9

is the distance from the center to each focus.

Foci: (h + c , k ), (h - c, k )

Focal points of are

answered Jun 5, 2014