Question

What are the foci of the ellipse given by the equation 4x^(2)-32x+9y^(2)+18y=37?

Answer 1

Yay i love these equations....not alright so you have to solve the equation first? do you know how to do that

Answer 2

no

Answer 3

alright do you know the equation for an ellipse

Answer 4

heres a reference i've made for these types of equations

Answer 5

first what you want to do is try to create the equation of an ellipse.

\[(x-h)^2/a^2 + (y-k)^2/b^2\]

Answer 6

=1

Answer 7

you have to do the same steps i did for prev problem

Answer 8

Thats what I am stuck on, idk how to put it into a equation

and dumbcow i didn't understand your way

Answer 9

so first you have your numbers grouped but you can divide out a common factor of 4 from x and 9 from y getting

\[4(x^2-8x)+9(y^2+2y)=37\]

Answer 10

theres really no other way to do it though to get it in standard form

Answer 11

now in order to get x-h^2 and (y-k)^2 you can obviously see that you'll need to complete the square inorder to factor it

Answer 12

I forget how to complete the square

Answer 13

sooooo

\[4(x^2-8x+16)+9(y^2+2y+1)=37\]

however whatever you add to one side you must add to the other and you have to take account for the factored out integers so 4*16 and 9*1

you'll get

\[4(x-8x+16)+9(y^2+2x+1)= 37+(16*4)+(9*1) \]

Answer 14

bandgeek
are you sure its not supposed to be equal to -37

Answer 15

yeah sorry, it equals -37

Answer 16

when completing the square you take your base x or your second term (i'm not 100% what they call it when they teach this) and you divide by 2 and square the number

Answer 17

so for the first one : -8x/2 = -4^2 = 16

Answer 18

now when you factor this it will be a perfect square and you can now factor it to (x-4)^2

Answer 19

so far so good?

Answer 20

let m know when you get back or what not?