anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • Nnesha
I have few minutes lets see if i can help u :=)
Nnesha
  • Nnesha
1st) is it vertical or horizontal ?

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anonymous
  • anonymous
Ok thank you! And for the first one it is horizontal. @Nnesha
Nnesha
  • Nnesha
yes right so formula for foci \[\huge\rm (h \pm c , k)\]
Nnesha
  • Nnesha
first of all you need to find c
Nnesha
  • Nnesha
use this equation \[\huge\rm c^2=a^2-b^2\]
anonymous
  • anonymous
Would c = SqRt656? @Nnesha
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
  • Nnesha
no what is a^2 and b^2 ?
Nnesha
  • Nnesha
\[\huge\rm \frac{ (x-h)^2 }{ a^2 }+\frac{ (y-k)^2 }{ b^2 }\] standard form of ellipse
anonymous
  • anonymous
Wouldn't a^2 = 256 and b^2 = 400?
anonymous
  • anonymous
Oh wait.
anonymous
  • anonymous
12 = c?
Nnesha
  • 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
  • Nnesha
yes !
Nnesha
  • Nnesha
so now what are the foci points ?
anonymous
  • anonymous
+/12, 0?
anonymous
  • anonymous
Forgot the ( ) sorry
Nnesha
  • Nnesha
\[(\pm 12,0)\] looks great should be plus minus sign
anonymous
  • anonymous
Yes that's what I meant, sorry.
anonymous
  • anonymous
So my answer would actually be A. Ok, thanks! Could you help me with the 2nd one?
anonymous
  • 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
  • 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
  • anonymous
? @Nnesha
Nnesha
  • Nnesha
if ellipse is vertical then bigger number supposed to be under x or y ?
anonymous
  • anonymous
Y?
Nnesha
  • Nnesha
yes right try to draw points on the paper
Nnesha
  • Nnesha
gtg sorry. :(
anonymous
  • anonymous
Ok :( Are you free at all tomorrow?
anonymous
  • anonymous
@saseal Could you help with the other two questions?
anonymous
  • anonymous
ok
anonymous
  • anonymous
Thank you! @saseal
anonymous
  • 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
  • anonymous
@saseal ^
anonymous
  • anonymous
first one is correct...2nd one is debatable because the question is not very specified
anonymous
  • anonymous
if they specifed the major axis is x and minor is y then you are correct
anonymous
  • anonymous
Hmm...how could I determine for sure?
anonymous
  • anonymous
hmm you are wrong...it says vertical major axis
anonymous
  • anonymous
Oh ok. So that would make the y axis the major axis?
anonymous
  • anonymous
Oh wait. would it be x^2/4 + y^2/9 = 1? @saseal
anonymous
  • anonymous
yea
anonymous
  • anonymous
Ok
anonymous
  • anonymous
Could you help me a couple more questions? I kind of am just getting this stuff. Lol. @saseal
anonymous
  • anonymous
ok
anonymous
  • 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
  • anonymous
\[\frac{ x^2 }{ 3^2 } - \frac{ y^2 }{ 4^2 } = 1\]
anonymous
  • anonymous
now it becomes y^2/4^2 - x^2/3^2
anonymous
  • anonymous
\[\frac{ y^2 }{ 4^2 } - \frac{ x^2 }{ 3^2 } =1\]
anonymous
  • anonymous
Ok so then what can I do?
anonymous
  • anonymous
you know where thid hyperbola opens open?
anonymous
  • anonymous
Not really like when it's graphed?
anonymous
  • anonymous
|dw:1439004618140:dw|
anonymous
  • anonymous
Ok so they open up where the two points are? Like the end points of the opening?
anonymous
  • anonymous
kinda...abit like parabola x2

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