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

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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 ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.