Question

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

A) (–2, –6 ±square root of 7)
B) (3, –6) and (–7, –6)
C) (–2, –1) and (–2, –11)
D) (-2 ±square root of 7 , –6)

How would I get the equation to the regular formula of ellipses? like x^2/a^2 + y^2/b^2 = 1.
A question from @Loujoelou , since his internet is currently slow.

Answer 1

@litchlani can u help with this one?

Answer 2

you make it look like
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] by completing the square

Answer 3

it won't look the way you wrote because the center is not \((0,0)\)

Answer 4

first group it up like this :
\[16(x^{2}+4x)+16(y^{2}+y)+244=0\]
then complete the square for the thing in the brackets

Answer 5

that wrong wait

Answer 6

? if its wrong please erase the comment, don't want to get confused between right and wrong

Answer 7

second one should be \[9(y^2+12y)\] sry

Answer 8

oh okay..

Answer 9

now u can complete the square for the things in the brackets , you know how to do it ?

Answer 10

yeah should be good thank you :)

Answer 11

thanks to both of you i should be able to find the foci by myself