Question

What are the foci of the ellipse given by the equation 4x2 + 9y2 – 32x + 18y + 37 = 0?

Answer 1

You have to group all the like terms together and then I'm thinking you also have to compete the square, too.

Answer 2

Do this:

Answer 3

could you walk me through it please

Answer 4

\[4x ^{2}-32x+9y ^{2}+18y=-37\]That's the beginnings of the proper set-up for this ellipse. Any ellipse, for that matter! Yes, I will walk you through it. See what I have done thus far?

Answer 5

Now group the x terms in parenthesis and then the y terms in parenthesis. Like this (getting ready to complete the square).

Answer 6

\[(4x ^{2}-32x)+(9y ^{2}+18y)=-37\]In order to complete the square you have to have a coefficient of 1 in front of both the x^2 and the y^2 terms. So factor out what you need to in order to make the coefficients a 1:

Answer 7

\[4(x ^{2}-8x)+9(y ^{2}+2y)=-37\]Now it's time to complete the square.

Answer 8

Take half the x term, square it, then add it in. Like this: Half of the x term (8) is 4; square 4 and get 16; add 16 into the parenthesis like this:

Answer 9

\[4(x ^{2}-8x+16)+9(y ^{2}+2y)=-37\]Now do the same with the y's:

Answer 10

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

Answer 11

Since you added them on to the left side of the equation, you also have to add them onto the right side. This is tricky, though, because you have a 4 and a 9 on the outside of the parenthesis. That means that when you add the 16 to the right side, when you pull it out of the parenthesis, you also have to keep in mind that you are multiplying by the 4. So it would look like this: (this is just the x terms, ok? I'm going to leave out the y's for just a sec. One crazy thing at a time, is what I say!)

Answer 12

\[4(x ^{2}-8x+16)+...=-37+(4 * 16)\]I really hope that makes sense, cuz that's just the way it is!

Answer 13

Now do it with the y's as well. This is only the y's; then we will put it all together. I just don't think it will all fit into the equation editor; it's too long all at one time!

Answer 14

\[...+9(y ^{2}+2y+1)=-37+(4*16)+(9*1)\]

Answer 15

Now you can reduce those parenthesis, because they are now perfect square binomials:

Answer 16

\[4(x-4)^{2}+9(y+1)^{2}=-37 +64+9\]That right side equals\[4(x-4)^{2}+9(y+1)^{2}=36\]

Answer 17

We all know that an ellipse has to = 1, so divide the equation by 36 to get a 1 on the right side. That looks like this:

Answer 18

\[\frac{ 4(x-4)^{2} }{ 36 }+\frac{ 9(y+1)^{2} }{ 36 }=1\]Reduce between the 4 & 36 and the 9 & 36 to get this:

Answer 19

\[\frac{ (x-4)^{2} }{ 9 }+\frac{ (y+1)^{2} }{ 4 }=1\]That is your ellipse now. The rule with ellipses is that a is always greater than b, and whichever variable the a is under --the x or the y--is the major axis. Here the major axis is the x axis. a^2 = 9 and a = 3; b^2 = 4 and b = 2.

Answer 20

The equation for the foci is a^2 - b^2 = c^2. Or 9 - 4 = c^2 or c = sq rt of 5. Because this is an ellipse with the major vertices on the x axis, the foci are also on the x axis, which puts your sq rt 5 in the x position of your coordinates:

Answer 21

\[(+\sqrt{5},0)(-\sqrt{5},0)\]Those are the foci!!!! Hope you understood that ok. I'm sure it's not the first time you saw this! Take it slow and read each step completely. It's a long and tedious process, but it is the ONLY way to get the job done! Any other questions, just tag me or someone else who loves doing these as much as I do, and we'll get you taken care of!

Answer 22

(4, –1 ±square root of 5)

(4, –1 ±square root of 13)

(4 ±square root of 5 , –1)

(4 ±square root of 5 , 1)

Are the answer choices

Answer 23

Let me get back to you o that in just a bit.

Answer 24

I forgot to take into consideration the center is not at the origin. The center is at (4, -1). So that means take your foci of \[(\sqrt{5},0)\]and add the center's coordinates into it like this:

Answer 25

\[(\sqrt{5}+4,0-1)=(4+\sqrt{5},-1)\]and the other one, the one with the negative sq rt 5 is this:

Answer 26

\[(4-\sqrt{5},-1)\]THOSE are your foci!!!

Answer 27

That is choice c (or the third one down).

Answer 28

ok thank you, do you mind helping me with two more quaestions

Answer 29

ok

Answer 30

I can help you!