Question

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

Answer 1

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?

Answer 2

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

Answer 3

1st) is it vertical or horizontal ?

Answer 4

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

Answer 5

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

Answer 6

first of all you need to find c

Answer 7

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

Answer 8

Would c = SqRt656? @Nnesha

Answer 9

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

Answer 10

no what is a^2 and b^2 ?

Answer 11

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

Answer 12

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

Answer 13

Oh wait.

Answer 14

12 = c?

Answer 15

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

Answer 16

yes !

Answer 17

so now what are the foci points ?

Answer 18

+/12, 0?

Answer 19

Forgot the ( ) sorry

Answer 20

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

Answer 21

Yes that's what I meant, sorry.

Answer 22

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

Answer 23

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.

Answer 24

\(\color{blue}{\text{Originally Posted by}}\) @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*

Answer 25

? @Nnesha

Answer 26

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

Answer 27

Y?

Answer 28

yes right
try to draw points on the paper

Answer 29

gtg sorry. :(

Answer 30

Ok :( Are you free at all tomorrow?

Answer 31

@saseal Could you help with the other two questions?

Answer 32

ok

Answer 33

Thank you! @saseal

Answer 34

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?

Answer 35

@saseal ^

Answer 36

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

Answer 37

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

Answer 38

Hmm...how could I determine for sure?

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

yea

Answer 43

Ok

Answer 44

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

Answer 45

ok

Answer 46

Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1

This is for Hyperbolas...

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

Ok so then what can I do?

Answer 51

you know where thid hyperbola opens open?

Answer 52

Not really like when it's graphed?