What are the foci of the ellipse given by the equation 16x^2 + 9y^2 + 64x + 108y + 244 = 0??

1 Answer
Sep 24, 2015

The foci are located at (-2, (-6 - sqrt(7))) and (-2, (-6 + sqrt(7)))

Explanation:

16 x^2 + 9 y^2 + 64 x + 108 y + 244 = 0

may be rearranged (for convenience, to group terms variously in x and y to see what is happening) to:

16 x^2 + 64 x + 9 y^2 + 108 y + 244 = 0

Noting ("completing the square")

16 x^2 + 64 x = (4 x + 8)^2 - 64 = 4^2 (x + 2)^2 - 64

and

9 y^2 + 108 y = (3 y + 18)^2 - 324 = 3^2(y + 6)^2 - 324

The equation may be rewritten

4^2 (x + 2)^2 - 64 + 3^2(y + 6)^2 - 324 + 244 = 0

That is

4^2 (x + 2)^2 + 3^2(y + 6)^2 - 144 = 0

Rearranging

4^2 (x + 2)^2 + 3^2(y + 6)^2 = 144

=> 4^2 (x + 2)^2 + 3^2(y + 6)^2 = 12^2

=> 4^2/12^2 (x + 2)^2 + 3^2/12^2(y + 6)^2 = 1

=> (4/12)^2 (x + 2)^2 + (3/12)^2(y + 6)^2 = 1

=> (1/3)^2 (x + 2)^2 + (1/4)^2(y + 6)^2 = 1

=> (x + 2)^2 / 3^2 + (y + 6)^2 / 4^2 = 1

From which it may be noted:

  • the respective hemiradii are 3 and 4 (the square roots of the denominators of the terms in x and y, conventionally denoted by b and a, respectively, where a conventionally denotes the major hemiradius (here, 4) and b typically denotes the minor hemiradius (here, 3)).
  • the ellipse is not in canonical position as the major axis is aligned with the y-axis rather than with the x-axis and the ellipse is not centred on the origin.
  • the centre of the ellipse is located at (-2, -6) (the negative of the constant terms added to x and y, respectively, to make "perfect" squares)

Denoting the absolute value of the distance of the foci from the centre of the ellipse by c, it is noted

c^2 = a^2 - b^2,

that is,

c = sqrt(a^2 - b^2)

For this ellipse,

c = sqrt(4^2 - 3^2) = sqrt(16 - 9) = sqrt(7)

So, the foci are located at

(-2, -6 - sqrt(7))

and

(-2, -6 + sqrt(7))