Anyone here good with ellipses? If so, please help!!
Find the center, vertices, and foci of the ellipse with equation x squared divided by four hundred plus y squared divided by two hundred and fifty six = 1
A. Center: (0, 0); Vertices: (-20, 0), (20, 0); Foci: (-12,0), (12, 0)
B. Center: (0, 0); Vertices: (-20, 0), (20, 0); Foci: (-16,0), (16, 0)
C. Center: (0, 0); Vertices: (0, -20), (0, 20); Foci: (0,-16), (0, 16)
D. Center: (0, 0); Vertices: (0, -20), (0, 20); Foci: (0,-12), (0, 12)
I know the answer is either B or C

- anonymous

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- anonymous

And also for
Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4.
I am pretty sure the answer is:
x^2/9 + y^2/4 = 1
Is that correct?

- Nnesha

I have few minutes lets see if i can help u :=)

- Nnesha

1st) is it vertical or horizontal ?

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## More answers

- anonymous

Ok thank you! And for the first one it is horizontal. @Nnesha

- Nnesha

yes right so formula for foci \[\huge\rm (h \pm c , k)\]

- Nnesha

first of all you need to find c

- Nnesha

use this equation \[\huge\rm c^2=a^2-b^2\]

- anonymous

Would c = SqRt656? @Nnesha

- Nnesha

remember (h,k) is the center point
it's horizontal so only x values would change y would stay same
no change in y values

- Nnesha

no what is a^2 and b^2 ?

- Nnesha

\[\huge\rm \frac{ (x-h)^2 }{ a^2 }+\frac{ (y-k)^2 }{ b^2 }\]
standard form of ellipse

- anonymous

Wouldn't a^2 = 256 and b^2 = 400?

- anonymous

Oh wait.

- anonymous

12 = c?

- Nnesha

okay for ellipse
a= bigger number
if a is under the x then it's horizontal
and if a is under y then it would be vertical

- Nnesha

yes !

- Nnesha

so now what are the foci points ?

- anonymous

+/12, 0?

- anonymous

Forgot the ( ) sorry

- Nnesha

\[(\pm 12,0)\] looks great
should be plus minus sign

- anonymous

Yes that's what I meant, sorry.

- anonymous

So my answer would actually be A. Ok, thanks! Could you help me with the 2nd one?

- anonymous

And also with this one:
The rectangular piece of property measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides.
I believe my answer would be:
x^2/25 + y^2/4 = 1
But I just wanted to make sure.

- Nnesha

\(\color{blue}{\text{Originally Posted by}}\) @jdoherty
And also for
Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4.
I am pretty sure the answer is:
x^2/9 + y^2/4 = 1
Is that correct?
\(\color{blue}{\text{End of Quote}}\)
*vertical*

- anonymous

? @Nnesha

- Nnesha

if ellipse is vertical then bigger number supposed to be under x or y ?

- anonymous

Y?

- Nnesha

yes right
try to draw points on the paper

- Nnesha

gtg sorry. :(

- anonymous

Ok :( Are you free at all tomorrow?

- anonymous

@saseal Could you help with the other two questions?

- anonymous

ok

- anonymous

Thank you! @saseal

- anonymous

1. The rectangular piece of property measures 10 mi by 4 mi. Find an equation for the ellipse if the path is to touch the center of the property line on all 4 sides.
I believe my answer would be:
x^2/25 + y^2/4 = 1
But I just wanted to make sure.
2. Find an equation in standard form for the ellipse with the vertical major axis of length 6, and minor axis of length 4.
I am pretty sure the answer is:
x^2/9 + y^2/4 = 1
Is that correct?

- anonymous

@saseal ^

- anonymous

first one is correct...2nd one is debatable because the question is not very specified

- anonymous

if they specifed the major axis is x and minor is y then you are correct

- anonymous

Hmm...how could I determine for sure?

- anonymous

hmm you are wrong...it says vertical major axis

- anonymous

Oh ok. So that would make the y axis the major axis?

- anonymous

Oh wait. would it be x^2/4 + y^2/9 = 1? @saseal

- anonymous

yea

- anonymous

Ok

- anonymous

Could you help me a couple more questions? I kind of am just getting this stuff. Lol. @saseal

- anonymous

ok

- anonymous

Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1
This is for Hyperbolas...

- anonymous

\[\frac{ x^2 }{ 3^2 } - \frac{ y^2 }{ 4^2 } = 1\]

- anonymous

now it becomes y^2/4^2 - x^2/3^2

- anonymous

\[\frac{ y^2 }{ 4^2 } - \frac{ x^2 }{ 3^2 } =1\]

- anonymous

Ok so then what can I do?

- anonymous

you know where thid hyperbola opens open?

- anonymous

Not really like when it's graphed?

- anonymous

|dw:1439004618140:dw|

- anonymous

Ok so they open up where the two points are? Like the end points of the opening?

- anonymous

kinda...abit like parabola x2

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