Question

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

Answer 1

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))`