Question

convert each equation to standard form by completing the square on x and y. Then graph the ellipse and give the location of its foci.

4x^2+y^2+16x-6y-39=0

4x^2+25y^2-24x+100y+36=0

Answer 1

and also (x+2)^2/16 + (y-3)^2

graph each elipse and give the location of its foci

Answer 2

\[(x+2)^{2}\div16 + (y-3)^2=1\]

Answer 3

i nned help on this one^

Answer 4

4x^2+16x+y^2-6y-39=0
4(x^2+4x)+y^2-6y-39=0
4(x^2+4x+4-4)+(y^2-6y+9-9)-39=0
4(x+2)^2-16+(y-3)^2-9-39=0
4(x+2)^2+(y-3)^2=64
\[\frac{(x+2)^2}{16}+\frac{(y-3)^2}{64}=1\]

Answer 5

center is (-2,3)
vertical axis is the major axis
c^2=64-16=48
c=4sqrt(3)
focii(-2, 3 +/- 4sqrt(3))

Answer 6

i need steps how do youdo it

Answer 7

do what?

Answer 8

how did you transfer the equation

Answer 9

i showed all the steps, completed the square, put all the constants on the right, then divided by the constant

Answer 10

-16-9-39=-64, add 64 to both sides

Answer 11

nooo i need help on (x+2)^2/16 + (y-3)^2

Answer 12

where'd that one come from?

Answer 13

i need help on that one if you could

Answer 14

does it equal one?

Answer 15

my equation came from completing the square of the first equation in your question

Answer 16

yess the equation = 1