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anonymous
 one year ago
I WILL GIVE MEDAL AND FAN:
Find the center, vertices, and foci of the ellipse with equation ((x^2/)/(100)) + ((Y^2)/(64))= 1
anonymous
 one year ago
I WILL GIVE MEDAL AND FAN: Find the center, vertices, and foci of the ellipse with equation ((x^2/)/(100)) + ((Y^2)/(64))= 1

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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2standard ellipse equation For a widerthantall 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: \[\frac{(xh)^2}{a^2}+\frac{(yk)^2}{b^2}=1\] For a tallerthanwide 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: \[\frac{(yh)^2}{a^2}+\frac{(xk)^2}{b^2}=1\] standard foci equation \[a^2c^2=b^2\] standard center \[(h,k)\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2based on your equation \[\frac{x^2}{100}+\frac{y^2}{64}=1\] a > b there's a horizontal major axis

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2EXAMPLE: suppose we have the equation \[\frac{x^2}{4}+\frac{y^2}{1}=1\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2since a > b we have a horizontal major axis so we rewrite this equation as where a = 2 and b = 1 \[\frac{x^2}{2^2}+\frac{y^2}{1^2}=1\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2before we start graphing, we want the center. Recall that the standard ellipse equation is \[\frac{(xh)^2}{a^2}+\frac{(yk)^2}{b^2}=1 \] For this equation we just have \[x^2 \] and \[y^2 \] \[\frac{x^20}{2^2}+\frac{y^20}{1^2}=1\] which means that our center \[(h,k) \rightarrow (0,0)\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437650242909:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437650445276:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2Foci : \[a^2c^2=b^2 \] since a = 2 and b = 1 \[2^2c^2=1^2 \] \[4c^2=1 \] \[c^2=14 \] \[c^2=3 \] \[c^2=3 \] \[\sqrt{c^2}=\sqrt{3} \] \[c=\sqrt{3} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol nm i never saw that someone replied.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@phi can you actually help a little more on explaining this a bit better

phi
 one year ago
Best ResponseYou've already chosen the best response.0I would hope that is enough info to answer the question

phi
 one year ago
Best ResponseYou've already chosen the best response.0try reading through the posts. 1) can you figure out the center? 2) the vertices?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2don't forget the foci

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the center is (0,0) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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)?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2errr. \[a^2c^2=b^2 \] we are looking for c a = 10 and b = 6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, so: 100  c^2 = 36 so c is 8?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but the vertices for b is (6,0) (6,0)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2we need vertices for a though (a,0), (a,0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0vertices for a = (10,0) and (10,0)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yes. 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so is this right? all together: center: (0,0) vertices: (10,0); (10,0) foci: (6,0); (6,0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0awesome thanks, help on another one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the center, vertices, and foci of the ellipse with equation 3x^2 + 6y^2 = 18

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2alright... we don't have the ellipse in standard form. so we need to divide 18 all over the equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so...: (3/18)x^2 + (6/18)y^2 that looks weird

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yeah but we need a 1 and dividing 18 all over the equation makes that =1 happen

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so i did it right?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I'm just trying to think what's going on... I don't have perfect squares in here.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2find the center.. that's the easiest part >_<

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually not sure how, i know it is (h,k) but where is that in our equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is 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
 one year ago
Best ResponseYou've already chosen the best response.2If 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
 one year ago
Best ResponseYou've already chosen the best response.0ok actually, looking at my answer choices, the center can only be (0,0) lol so moving on to the vertices.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but 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

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2and I just figured it out in my noggin we have square roots this time

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{ x^2 }{ 6 } + \frac{ y^2 }{ 3 } = 1 \] \[\frac{ x^2 }{ (\sqrt{6})^2 } + \frac{ y^2 }{ (\sqrt{3})^2 } = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i got vertices: \[(\sqrt6,0); (\sqrt6,0)\] foci: \[(\sqrt3,0); (\sqrt3,0)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one last one i have to work backwards on i thin. can you help?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I guess x.x ok y not

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find 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
 one year ago
Best ResponseYou've already chosen the best response.2vertical major axis > 18 minor axis of length > 16

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2minor 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 \[\frac{(yh)^2}{a^2}+\frac{(xk)^2}{b^2}=1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so 18 ^2 for a^2? and 16 ^2 for b^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.2I haven't done this in a while, but these sites are confusing me... this vertical formula is switched \[\frac{(xk)^2}{b^2}+\frac{(yh)^2}{a^2}=1\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2our center is still (0,0) that's one good part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so we are going to have x^2 and y^2 by themselves on the top but what about the bottom.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2vertical major axis > 18 minor axis of length > 16

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2a = length of semimajor axis b = length of semiminor axis vertices: (h, k + a), (h, k  a) covertices: (h + b, k), (h  b, k) [endpoints of the minor axis]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2now I'm confusing myself.. but .. a =18 and b = 16

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yeah if the number under the y part is larger then it's going to be vertical

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so a has to be 18 and b has to be 16

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2but ... we're still not done we have to write those numbers to the second power

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{(xk)^2}{b^2}+\frac{(yh)^2}{a^2}=1 \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(x^2/16^2) + (y^2/18^2) = 1?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2yup I think you may have to expand like 16 x 16 and 18 x 18

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you want me to give the answer choices? doesn't look like any of them are working.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x^2 }{64 }+\frac{ y^2 }{ 81 } = 1\] \[\frac{ x^2 }{ 9 }+\frac{ y^2 }{ 8 } = 1\] \[\frac{ x^2 }{ 81 }+\frac{ y^2 }{ 64 } = 1\] \[\frac{ x^2 }{ 8 }+\frac{ y^2 }{ 9 } = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that took a while lol. so how do we get one of those.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2what the heck that doesn't make sense.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did we do anything wrong?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2on one choice that would mean a = 9, b = 8 and on the other a = 8 and b = 9 last choice won't work

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2second choice won't work either.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2did you copy the right numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's why i'm confused

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2then that would mean that you were given the wrong numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0here i will get phi back in here maybe: @phi

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2this doesn't match... I mean @twistnflip tell her about the last problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.0remember that a and b are the "semimajor" and "semiminor" axes. (comparable to the radius) in other words, divide the given axes lengths by 2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2*facepalm* missed that part.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2k... I see it now 18/2 = 9 16/2 = 8

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2and since it's vertical that means the denominator in the y portion has to be larger

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so i get the equation: \[\frac{ x^2 }{ 8} + \frac{ y^2 }{ 9 }\]?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Haha, 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
 one year ago
Best ResponseYou've already chosen the best response.2getting there. there's still one more step

phi
 one year ago
Best ResponseYou've already chosen the best response.0you want a^2 and b^2 in your equation (not just a and b)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437655407458:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2since a = 9 and that formula has a^2 9^2 = ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2umm @Astrophysics it's vertical

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\(\color{#0cbb34}{\text{Originally Posted by}}\) @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
 one year ago
Best ResponseYou've already chosen the best response.0I know, I just put that to show it's squared, didn't really pay attention to the vertical/ horizontal :P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.081 and 8 ^2 = 64 so i have \[\frac{ x^2 }{ 64} + \frac{ y^2 }{81 } = 1\]

phi
 one year ago
Best ResponseYou've already chosen the best response.0you want it too be harder?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0haha no thanks so much guys! you da bomb
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