anonymous
  • anonymous
I WILL GIVE MEDAL AND FAN: Find the center, vertices, and foci of the ellipse with equation ((x^2/)/(100)) + ((Y^2)/(64))= 1
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@nincompoop
anonymous
  • anonymous
@ash2326
anonymous
  • anonymous
@ganeshie8

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UsukiDoll
  • UsukiDoll
standard ellipse equation For a wider-than-tall ellipse with center at (h, k), having vertices a units to either side of the center and foci c units to either side of the center, the ellipse equation is: \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\] For a taller-than-wide ellipse with center at (h, k), having vertices a units above and below the center and foci c units above and below the center, the ellipse equation is: \[\frac{(y-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1\] standard foci equation \[a^2-c^2=b^2\] standard center \[(h,k)\]
UsukiDoll
  • UsukiDoll
based on your equation \[\frac{x^2}{100}+\frac{y^2}{64}=1\] a > b there's a horizontal major axis
UsukiDoll
  • UsukiDoll
EXAMPLE: suppose we have the equation \[\frac{x^2}{4}+\frac{y^2}{1}=1\]
UsukiDoll
  • UsukiDoll
since a > b we have a horizontal major axis so we re-write this equation as where a = 2 and b = 1 \[\frac{x^2}{2^2}+\frac{y^2}{1^2}=1\]
UsukiDoll
  • UsukiDoll
before we start graphing, we want the center. Recall that the standard ellipse equation is \[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1 \] For this equation we just have \[x^2 \] and \[y^2 \] \[\frac{x^2-0}{2^2}+\frac{y^2-0}{1^2}=1\] which means that our center \[(h,k) \rightarrow (0,0)\]
UsukiDoll
  • UsukiDoll
|dw:1437650242909:dw|
UsukiDoll
  • UsukiDoll
for vertices we need (-a,0), (a,0), (-b,0),(0,b) since a = 2 and b = 1 our vertex points are (-2,0) (2,0) (-1,0)(0,1)
UsukiDoll
  • UsukiDoll
|dw:1437650445276:dw|
UsukiDoll
  • UsukiDoll
Foci : \[a^2-c^2=b^2 \] since a = 2 and b = 1 \[2^2-c^2=1^2 \] \[4-c^2=1 \] \[-c^2=1-4 \] \[-c^2=-3 \] \[c^2=3 \] \[\sqrt{c^2}=\sqrt{3} \] \[c=\sqrt{3} \]
anonymous
  • anonymous
lol nm i never saw that someone replied.
anonymous
  • anonymous
@phi can you actually help a little more on explaining this a bit better
phi
  • phi
I would hope that is enough info to answer the question
phi
  • phi
try reading through the posts. 1) can you figure out the center? 2) the vertices?
UsukiDoll
  • UsukiDoll
don't forget the foci
anonymous
  • anonymous
yeah hold on
UsukiDoll
  • UsukiDoll
alright. :)
anonymous
  • anonymous
the center is (0,0) right?
UsukiDoll
  • UsukiDoll
yes
anonymous
  • anonymous
for the foci: 10^2 - 8^2 = b^2 100 - 64 = b^2 36 = b^2 b = 6, so the foci is (-6,0) and (6,0)?
anonymous
  • anonymous
how's that?
UsukiDoll
  • UsukiDoll
errr. \[a^2-c^2=b^2 \] we are looking for c a = 10 and b = 6
anonymous
  • anonymous
oops, so: 100 - c^2 = 36 so c is 8?
UsukiDoll
  • UsukiDoll
but the vertices for b is (-6,0) (6,0)
UsukiDoll
  • UsukiDoll
we need vertices for a though (-a,0), (a,0)
UsukiDoll
  • UsukiDoll
yes c is 8
anonymous
  • anonymous
vertices for a = (-10,0) and (10,0)
UsukiDoll
  • UsukiDoll
yes. now we can use that information to graph. though...if we draw it with os tools it's gonna look ugly. I'll attach a file
UsukiDoll
  • UsukiDoll
anonymous
  • anonymous
ok so is this right? all together: center: (0,0) vertices: (-10,0); (10,0) foci: (-6,0); (6,0)
UsukiDoll
  • UsukiDoll
yeah
anonymous
  • anonymous
awesome thanks, help on another one?
UsukiDoll
  • UsukiDoll
sure
anonymous
  • anonymous
Find the center, vertices, and foci of the ellipse with equation 3x^2 + 6y^2 = 18
UsukiDoll
  • UsukiDoll
alright... we don't have the ellipse in standard form. so we need to divide 18 all over the equation
anonymous
  • anonymous
ok so...: (3/18)x^2 + (6/18)y^2 that looks weird
UsukiDoll
  • UsukiDoll
yeah but we need a 1 and dividing 18 all over the equation makes that =1 happen
anonymous
  • anonymous
oh so i did it right?
UsukiDoll
  • UsukiDoll
yeah
anonymous
  • anonymous
so now what
UsukiDoll
  • UsukiDoll
I'm just trying to think what's going on... I don't have perfect squares in here.
anonymous
  • anonymous
ok
UsukiDoll
  • UsukiDoll
find the center.. that's the easiest part >_<
anonymous
  • anonymous
actually not sure how, i know it is (h,k) but where is that in our equation
anonymous
  • anonymous
is it (0,0) again. i went on wolphram alpha and it simplifies the equation to \[\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1\]
UsukiDoll
  • UsukiDoll
If our equation was similar to \[7x^2 +16(y+2)^2=112\] and then divide 112 all over then our center would be (0,-2)
anonymous
  • anonymous
ok actually, looking at my answer choices, the center can only be (0,0) lol so moving on to the vertices.
UsukiDoll
  • UsukiDoll
but when we reduced that fraction from earlier \[\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1 \] there is no k and no h so center is (0,0) the vertices is next
anonymous
  • anonymous
ok
UsukiDoll
  • UsukiDoll
and I just figured it out in my noggin we have square roots this time
UsukiDoll
  • UsukiDoll
\[\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1 \] \[\frac{ x^2 }{ (\sqrt{6})^2 } + \frac{ y^2 }{ (\sqrt{3})^2 } = 1\]
anonymous
  • anonymous
so i got vertices: \[(-\sqrt6,0); (\sqrt6,0)\] foci: \[(-\sqrt3,0); (\sqrt3,0)\]
UsukiDoll
  • UsukiDoll
yes!
anonymous
  • anonymous
is that right?
anonymous
  • anonymous
yay!
anonymous
  • anonymous
one last one i have to work backwards on i thin. can you help?
UsukiDoll
  • UsukiDoll
I guess x.x ok y not
anonymous
  • anonymous
Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. No clue how to work this one. Sorry for being such a pain...
UsukiDoll
  • UsukiDoll
vertical major axis -> 18 minor axis of length -> 16
UsukiDoll
  • UsukiDoll
minor would be the foci and the major would be the vertices since we have vertical major axis, our equation should be in the form of \[\frac{(y-h)^2}{a^2}+\frac{(x-k)^2}{b^2}=1 \]
anonymous
  • anonymous
ok so 18 ^2 for a^2? and 16 ^2 for b^2
UsukiDoll
  • UsukiDoll
yeah.....
anonymous
  • anonymous
so how do I get the equation plug in these numbers so I get: \[(\frac{ x^2 }{ 18 })+(\frac{ y^2 }{ 16}) = 1\]is that right?
UsukiDoll
  • UsukiDoll
I haven't done this in a while, but these sites are confusing me... this vertical formula is switched \[\frac{(x-k)^2}{b^2}+\frac{(y-h)^2}{a^2}=1\]
UsukiDoll
  • UsukiDoll
our center is still (0,0) that's one good part
anonymous
  • anonymous
ok so we are going to have x^2 and y^2 by themselves on the top but what about the bottom.
UsukiDoll
  • UsukiDoll
vertical major axis -> 18 minor axis of length -> 16
UsukiDoll
  • UsukiDoll
a = length of semi-major axis b = length of semi-minor axis vertices: (h, k + a), (h, k - a) co-vertices: (h + b, k), (h - b, k) [endpoints of the minor axis]
UsukiDoll
  • UsukiDoll
now I'm confusing myself.. but .. a =18 and b = 16
UsukiDoll
  • UsukiDoll
yeah if the number under the y part is larger then it's going to be vertical
UsukiDoll
  • UsukiDoll
so a has to be 18 and b has to be 16
UsukiDoll
  • UsukiDoll
but ... we're still not done we have to write those numbers to the second power
UsukiDoll
  • UsukiDoll
\[\frac{(x-k)^2}{b^2}+\frac{(y-h)^2}{a^2}=1 \]
anonymous
  • anonymous
(x^2/16^2) + (y^2/18^2) = 1?
UsukiDoll
  • UsukiDoll
yup I think you may have to expand like 16 x 16 and 18 x 18
anonymous
  • anonymous
do you want me to give the answer choices? doesn't look like any of them are working.
UsukiDoll
  • UsukiDoll
yeah what are they
anonymous
  • anonymous
\[\frac{ x^2 }{64 }+\frac{ y^2 }{ 81 } = 1\] \[\frac{ x^2 }{ 9 }+\frac{ y^2 }{ 8 } = 1\] \[\frac{ x^2 }{ 81 }+\frac{ y^2 }{ 64 } = 1\] \[\frac{ x^2 }{ 8 }+\frac{ y^2 }{ 9 } = 1\]
anonymous
  • anonymous
that took a while lol. so how do we get one of those.
UsukiDoll
  • UsukiDoll
what the heck that doesn't make sense.
anonymous
  • anonymous
did we do anything wrong?
UsukiDoll
  • UsukiDoll
on one choice that would mean a = 9, b = 8 and on the other a = 8 and b = 9 last choice won't work
UsukiDoll
  • UsukiDoll
second choice won't work either.
UsukiDoll
  • UsukiDoll
did you copy the right numbers?
anonymous
  • anonymous
yep
anonymous
  • anonymous
that's why i'm confused
UsukiDoll
  • UsukiDoll
then that would mean that you were given the wrong numbers
anonymous
  • anonymous
here i will get phi back in here maybe: @phi
UsukiDoll
  • UsukiDoll
@Astrophysics help!
Astrophysics
  • Astrophysics
What's up?
UsukiDoll
  • UsukiDoll
this doesn't match... I mean @twistnflip tell her about the last problem
anonymous
  • anonymous
ok so Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. But we have done the work and the answers don't match. I posted them with the equation editor a few comments ago. What do we do?
phi
  • phi
remember that a and b are the "semi-major" and "semi-minor" axes. (comparable to the radius) in other words, divide the given axes lengths by 2
UsukiDoll
  • UsukiDoll
*facepalm* missed that part.
phi
  • phi
thus you want a=9 and b=8
UsukiDoll
  • UsukiDoll
k... I see it now 18/2 = 9 16/2 = 8
UsukiDoll
  • UsukiDoll
and since it's vertical that means the denominator in the y portion has to be larger
anonymous
  • anonymous
so i get the equation: \[\frac{ x^2 }{ 8} + \frac{ y^2 }{ 9 }\]?
Astrophysics
  • Astrophysics
Haha, well here is a good reference on the standard form of an ellipse, and will help you from not making these kinds of mistakes :P |dw:1437655357843:dw|
UsukiDoll
  • UsukiDoll
getting there. there's still one more step
anonymous
  • anonymous
which is?
phi
  • phi
you want a^2 and b^2 in your equation (not just a and b)
Astrophysics
  • Astrophysics
|dw:1437655407458:dw|
UsukiDoll
  • UsukiDoll
since a = 9 and that formula has a^2 9^2 = ?
UsukiDoll
  • UsukiDoll
umm @Astrophysics it's vertical
UsukiDoll
  • UsukiDoll
\(\color{#0cbb34}{\text{Originally Posted by}}\) @twistnflip ok so Find an equation in standard form for the ellipse with the vertical major axis of length 18, and minor axis of length 16. But we have done the work and the answers don't match. I posted them with the equation editor a few comments ago. What do we do? \(\color{#0cbb34}{\text{End of Quote}}\) vertical major axis
Astrophysics
  • Astrophysics
I know, I just put that to show it's squared, didn't really pay attention to the vertical/ horizontal :P
anonymous
  • anonymous
81 and 8 ^2 = 64 so i have \[\frac{ x^2 }{ 64} + \frac{ y^2 }{81 } = 1\]
UsukiDoll
  • UsukiDoll
that's better.
anonymous
  • anonymous
so that's it?
phi
  • phi
you want it too be harder?
anonymous
  • anonymous
haha no thanks so much guys! you da bomb
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