Question

Can anyone help me find the center, vertices, and foci of the ellipse?
4. x^2+4y^2+8x-48=0

Answer 1

Well of course. ^_^
Now, it might help if we have a goal in mind, and that goal is to play with this equation so that it looks something like this:
\[\Large \frac{(x-\color{red}h)^2}{\color{blue}a^2}+\frac{(y-\color{green}k)^2}{\color{orange}b^2}= 1\]
You follow?

Answer 2

yes.

Answer 3

Okay, first, bring everything with no variable next to it to the right-side. Can you do that?

Answer 4

yes .

Answer 5

Then please do it, and tell me what you get.

Answer 6

x^2+8x+4y^2=48

Answer 7

Okay good :)
\[\Large x^2 + 8x + 4y^2 = 48\]
Now, the y-part has nothing to worry about for now, what about the x-part?
You need to do what's called "completing the square"
heard of it?

Answer 8

yes.

Answer 9

Then can you complete the square here on this part
\[\Large \color{blue}{x^2 + 8x} + 4y^2 = 48\]

Answer 10

there is nothing to complete because the x value is already 1

Answer 11

No, I mean, for example, \[\Large z^2 + 6z\]
to complete the square, you just have to take the coefficient of the variable without a square (the 6z)
so that coefficient is 6.
Get half of it, and then square it.


Half of 6 is 3, and the square of 3 is 9.

Now add and subtract that to the expression
\[\Large z^2 + 6z \color{red}{+9 - 9}\]
and you'll notice that this part
\[\Large \color{blue}{z^2 +6z +9 }-9\] is now a perfect square:\[\Large \color{blue}{(z+3)^2}-9\]

Answer 12

\[\Large \color{blue}{x^2 + 8x} + 4y^2 = 48\]
do the same here.

Answer 13

i guess
x^2+8x+16-16
= (x+4)-16

Answer 14

That's correct, except you're missing an exponent :P
No matter...\[\Large \color{blue}{(x+4)^2-16} + 4y^2 = 48\]

Answer 15

Now, there's that -16, mind bringing that to the right-side too? It has no variable after all...

Answer 16

(x+4)^2+4y=64

Answer 17

This time your y is missing an exponent :P
But again, no matter,
\[\Large (x+4)^2 + 4y^2 = \color{blue}{64}\]

Answer 18

Ready for the next step?
Remember that we want it to look like this:
\[\Large \frac{(x-\color{red}h)^2}{\color{blue}a^2}+\frac{(y-\color{green}k)^2}{\color{orange}b^2}= 1\]so we want the right-side to be just 1.
How do we do that?

Answer 19

I guess by dividing it all by 64 ? ?

Answer 20

Yup.\[\Large \frac{(x+4)^2}{64}+ \frac{4y^2}{64}=1\]

simplify?

Answer 21

(x+4)^2/64+y^2/16=1 ? correct ?

Answer 22

Yup. Now all you have to do is identify your h,k, a and b.

Answer 23

Here, a reference:
\[\Large \frac{(x-\color{red}h)^2}{\color{blue}a^2}+\frac{(y-\color{green}k)^2}{\color{orange}b^2}= 1\]

Answer 24

center is (-4,0)
a=8 b=4
the major axis is x
the vertices is (4,0) && (-12,0)
&& the foci is (-4,0 +(-)4sqrt3 . correct ?

Answer 25

Impressive :D

Answer 26

can you help me with (x+6)^2/12+(y-4)^2/16=1 -- i cant figure out the 12.

Answer 27

sorry about that, numbers aren't always going to be pretty... you'll just have to make do with the fact that \(\large a = \sqrt{12}\)
no other way, sorry :)

Answer 28

ok , thanks !

Answer 29

no problem :)