Ellipses help!

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Ellipses help!

Mathematics
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@ganeshie8 I'm not sure how to keep 1 on the right side while simplifying the left

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Other answers:

it's not ellipse
http://prntscr.com/7cbqdf :-)
Sorry, I meant hyperbola
so is it horizontal or vertical ?
do you know ?
Sorry I didn't get your responses until now! grr pc
horizontal?
\[\frac{ x^2 }{ \frac{ 1 }{ 36 } }-\frac{ y^2 }{ \frac{ 1 }{ 64 } }=1\]
a= 1/6 b=1/8 c=5/24
yeah?
it's hyperbola so its not depends on a(bigger number) if x comes first then it's horizontal and if y comes first then it's vertical \[x^2 - y^2 = 1 \]^^^ horizontal \[y^2 - x^2 =1\] vertical ^
ooo I didn't catch that in class. That is helpful ^-^ so yeah it's horizontal.
yep right
do you know if the a b and c are right?
gimme a sec
alright
ok :) I'm at school and going to drive home so brb in like 15 :)
oh okay
...so that was waay more than 15, sorry! back now.
alright to find c u have to use this formula \[\huge\rm c^2 = a^2 + b^2\] plug in a^2 and b^2 value solve for c
Earlier I posted what I got for a, b, and c :)
a= 1/6 b=1/8 c=5/24
ohh nvm i forgot
that's right now bec it's horizontal formula for focus \[\huge\rm (h \pm c ,k)\] (h,k) is the center point add c value in x coordinate
center is ( 0 ,0 ) in that question
so focus is = ?
(0, plus or minus 5/24)
x-coordinate not y it's horizontal so add h +/- c
\[\huge\rm (h \pm \color{red}{c} ,k)\] (h,k) is the center point h + c and h - c
alright seems like u r afk so for vertex \[\huge\rm (h \pm a ,k)\] and for asy \[\huge\rm y = k \pm \frac{ b }{ a }(x-h)\] (h.k) is the center
pm means \[\huge\rm (h + a , k) and ( h -a ,k)\]
did you get my msg?
wait what? focal point is... (0,a) then?
nope
wat msg ?
I sent you a message on here. Or i thought i did
(0,c) ?
what is the h?
(h , k) is the center which is (0,0) in this question did u read my comments ? :-)
Yeah, but if it's that then the foci are (0, pm 5/24)
nope add into x-coordinate not y \[\huge\rm (h \pm \color{red}{c} ,k)\] h + c and h - c
I don't understand that D: c - 5/24
* c = 5/24
what is h and k ?
(0,0)
yep right nice face! 0.0 alright so it should be like this \[(0 \pm \frac{ 5 }{ 24} , 0)\]
why is the k there? The previous problems I've done never have had that
what k ?
After the c :P
k = 0 y- coordinate now bec it's horizontal u have to add into x-coordinate not y so it's yep right nice face! 0.0 alright so it should be like this \[( \pm \frac{ 5 }{ 24} , 0)\]
so the final answer should be (0, pm5/24, 0)
vertex is (0, pm1/6, 0)
asy y=0 pm6/8(x-0)
\(\color{blue}{\text{Originally Posted by}}\) @Babynini so the final answer should be (0, pm5/24, 0) \(\color{blue}{\text{End of Quote}}\) why two zeros ? \[(0 \pm \frac{ 5 }{ 24} , 0)\] \[0 + \frac{ 5 }{ 24 } =? ?\] \[0-\frac{ 5 }{ 24 }=?\]
..oh xD foci: -5/24, 0 5/24, 0
yes right
asy y=pm6/8(x-0)
yep solve that reduce the fraction
or just pm6/8x
reduce the fraction
6/8= ?
xD 3/4
yep right so y = 3/4x
okies :)
done!?
it's all green checks! :) thanks so much
np :-)

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