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

##### 1 Answer

The foci are located at

#### Explanation:

may be rearranged (for convenience, to group terms variously in

Noting ("completing the square")

and

The equation may be rewritten

That is

Rearranging

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

that is,

For this ellipse,

So, the foci are located at

and