Babynini
  • Babynini
Ellipses help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
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Babynini
  • Babynini
1 Attachment
Babynini
  • Babynini
@ganeshie8 I'm not sure how to keep 1 on the right side while simplifying the left
Babynini
  • Babynini
@freckles ? :)

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

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