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

1 Answer
Sep 24, 2015

Answer:

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))#