Question

Ellipses help!

Answer 1

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

Answer 2

@freckles ? :)

Answer 3

it's not ellipse

Answer 4

http://prntscr.com/7cbqdf :-)

Answer 5

Sorry, I meant hyperbola

Answer 6

so is it horizontal or vertical ?

Answer 7

do you know ?

Answer 8

Sorry I didn't get your responses until now! grr pc

Answer 9

horizontal?

Answer 10

\[\frac{ x^2 }{ \frac{ 1 }{ 36 } }-\frac{ y^2 }{ \frac{ 1 }{ 64 } }=1\]

Answer 11

a= 1/6
b=1/8
c=5/24

Answer 12

yeah?

Answer 13

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 ^

Answer 14

ooo I didn't catch that in class. That is helpful ^-^ so yeah it's horizontal.

Answer 15

yep right

Answer 16

do you know if the a b and c are right?

Answer 17

gimme a sec

Answer 18

alright

Answer 19

ok :) I'm at school and going to drive home so brb in like 15 :)

Answer 20

oh okay

Answer 21

...so that was waay more than 15, sorry! back now.

Answer 22

@Nnesha

Answer 23

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

Answer 24

Earlier I posted what I got for a, b, and c :)

Answer 25

a= 1/6
b=1/8
c=5/24

Answer 26

ohh nvm i forgot

Answer 27

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

Answer 28

center is ( 0 ,0 ) in that question

Answer 29

so focus is = ?

Answer 30

(0, plus or minus 5/24)

Answer 31

x-coordinate not y
it's horizontal so add h +/- c

Answer 32

\[\huge\rm (h \pm \color{red}{c} ,k)\] (h,k) is the center point
h + c and h - c

Answer 33

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

Answer 34

pm means
\[\huge\rm (h + a , k) and ( h -a ,k)\]

Answer 35

did you get my msg?

Answer 36

wait what?
focal point is... (0,a) then?

Answer 37

nope

Answer 38

wat msg ?

Answer 39

I sent you a message on here. Or i thought i did

Answer 40

(0,c) ?

Answer 41

what is the h?

Answer 42

(h , k) is the center which is (0,0) in this question
did u read my comments ? :-)

Answer 43

Yeah, but if it's that then the foci are (0, pm 5/24)

Answer 44

nope add into x-coordinate not y \[\huge\rm (h \pm \color{red}{c} ,k)\] h + c and h - c

Answer 45

I don't understand that D: c - 5/24

Answer 46

* c = 5/24

Answer 47

what is h and k ?

Answer 48

(0,0)

Answer 49

yep right nice face! 0.0 alright so it should be like this \[(0 \pm \frac{ 5 }{ 24} , 0)\]

Answer 50

why is the k there? The previous problems I've done never have had that

Answer 51

what k ?

Answer 52

After the c :P

Answer 53

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 \[( \pm \frac{ 5 }{ 24} , 0)\]

Answer 54

so the final answer should be
(0, pm5/24, 0)

Answer 55

vertex is (0, pm1/6, 0)

Answer 56

asy y=0 pm6/8(x-0)

Answer 57

\(\color{blue}{\text{Originally Posted by}}\) @Babynini
so the final answer should be
(0, pm5/24, 0)
\(\color{blue}{\text{End of Quote}}\)
why two zeros ?
\[(0 \pm \frac{ 5 }{ 24} , 0)\]
\[0 + \frac{ 5 }{ 24 } =? ?\] \[0-\frac{ 5 }{ 24 }=?\]

Answer 58

..oh xD
foci:
-5/24, 0
5/24, 0

Answer 59

yes right

Answer 60

asy y=pm6/8(x-0)

Answer 61

yep solve that reduce the fraction

Answer 62

or just pm6/8x

Answer 63

reduce the fraction

Answer 64

6/8= ?

Answer 65

xD 3/4

Answer 66

yep right so y = 3/4x

Answer 67

okies :)

Answer 68

done!?

Answer 69

it's all green checks! :) thanks so much

Answer 70

np :-)