Question

What are the foci of the hyperbola given by the equation 16x2 – 9y2 + 64x – 72y – 224 = 0 ?

Answer 1

i put (–3, 4) and (7, 4)

Answer 2

What did you get when you completed the square in x and in y?

Answer 3

16x²+64x+1024-9y²-72y+1296=224

Answer 4

That was a little rearranging. How about the Completing the Square part? It should end up like this:

\(16(x-Something)^{2} - 9(y-something)^{2} = Constant^{2} \)

Answer 5

BTW - You added 1024 and 1296 to one side of the equation. Does this make sense to you?

Answer 6

oh yes i forgot that part.
16x²+64x+1024-9y²-72y+1296=224+1024+1296
16(x²+4x+64)-9(y²-8y+144)=2544

Answer 7

wait did i do something wrong ?

Answer 8

\(16x^{2} + 64x = 16(x^{2} + 4x) = 16(x^{2} + 4x + 4 - 4) = 16(x^{2} + 4x + 4) - 16(4)\) =

\(16(x^{2} + 4x + 4) - 64 = 16(x+2)^{2} - 64\)

Not sure how you managed "64" inside the parentheses.

Answer 9

okk so whats the next step to get the answer ?

Answer 10

Do the same thing for 'y'. You seem to have the right idea. Just be a little more careful.

Answer 11

9(y-2)²-144 ?

Answer 12

Not quite.

\(-9y^{2} - 72y =\)
\(-9(y^{2} + 8y) =\)
\(-9(y^{2} + 8y + 16 - 16) =\)
\(-9(y^{2} + 8y + 16) - (-9)(16)\)
\(-9(y+4)^{2} + 144\)

You are SO CLOSE!! Keep working on it. You'll get it.

Answer 13

ooppss :( . Thankyouu.(: And the next step is ?

Answer 14

Almost done. Just put it all together.

Answer 15

16(x+2)² - 64 9(y+4)²+144

Answer 16

16(x+2)² - 64 + 9(y+4)²+144
like this ?

Answer 17

\( = 224 + 64 - 144\)

Answer 18

144

Answer 19

Sorry, I was a little confused. Result should be \(16(x+2)^{2} + 9(y+4)^{2} = 144\)

Now what?

Answer 20

umm 16+9 ?

Answer 21

What is that? Divide by 144 and see if anything nice happens.

Answer 22

16+9=25 but it doesnt go into 144 evenly

Answer 23

I'm still wondering why you did that.

\(\dfrac{(x+2)^2}{9} - \dfrac{(y+4)^{2}}{16} = 1\)

or

\(\dfrac{(x+2)^2}{3^{2}} - \dfrac{(y+4)^{2}}{4^{2}} = 1\)

Answer 24

wait so , ohh nvm okk. so now what do i do ?

Answer 25

Once you find this wonderful form, you just start picking off information.

Center: ??
a = ??
Location of vertices: ??
b = ??
Location of fake vertices: ??
Equations of asymptotes: ??
c = ??
Location of Foci: ??

Start filling in the blanks.

Answer 26

0.o. Ughh idk what im doing. i give up :(

Answer 27

Just one thing at a time.

Center: (-2,-4) -- Do you see where this comes from?

Answer 28

yhes from the parenthesis . just the opposite.

Answer 29

The standard form is x-h and y-k in there. So x + 2 = x - (-2) and there is the -2.

Now 'a' is half the length of the major axis.

a = 3 -- Do you see where that comes from?

Answer 30

yes i get it. so b i y-(-4) wich is 5 ?

Answer 31

or maybe not .

Answer 32

How did 4 magically turn into 5? The center is (-2,-4)

a = 3 (half the length of the major axis)
b = 4 (half the length of the phantom axis)

Those are from the denominators.

Answer 33

becausee you made x - (-2) =3
so i thought y-(-4) = 5 . oohh ok nvm.

Answer 34

hello ?

Answer 35

I do not understand what you are doing.

x + 2 leads to h = -2
y + 4 leads to k = -4

This means the center is at (-2,-4). This has nothing to do with a or b or the c, yet to come.

Do we have the center, yet?