Question

what is the graph of:
x^2+4y^2-2y=8

Answer 1

but how do i fighure out the math

Answer 2

@xozombie

Answer 3

how do I figure out the math

Answer 4

Thats an ellipse graph.

Answer 5

but how did you figure that out without puting it in the calc

Answer 6

complete the square and you have
\[x^2+4(y-\frac{ 1 }{ 4 })^2=8+\frac{ 1 }{ 4 }=\frac{ 17 }{ 2 }\]
divide by 17/2
\[\frac{ x^2 }{ 17/2 }+\frac{ (y-1/4)^2 }{ 17/8 }=1\]
so its ellipse centered at (0,1/4)

Answer 7

ok, that's good

Answer 8

huh

Answer 9

im confused already

Answer 10

@amistre64 can you help this user?

Answer 11

thanks @Ⓐ Ashleyisakitty (Ambassador)

Answer 12

thanks @Ashleyisakitty

Answer 13

please help

Answer 14

thanks anyway

Answer 15

no...just a failing senior running out of time

Answer 16

for me it is

Answer 17

not in foster care

Answer 18

hello do you think you can help me

Answer 19

do you know how to complete the sqaure of \[x^2+4y^2-2y=8 \]

Answer 20

no

Answer 21

do i need to bring over the 8

Answer 22

lets forget the x^2 for know
so we want to deal with
\[4y^2-2y=8\]

Answer 23

ok

Answer 24

the rule for completing the square is that the co-efficient of y^2 as to be 1
so we have to factor the 4

Answer 25

is that ok

Answer 26

which is 2

Answer 27

yep

Answer 28

or 2y

Answer 29

huh what is that

Answer 30

\[4(y^2-\frac{y}{2})=8

Answer 31

sry i meant this ^^

Answer 32

\[4(y^2-\frac{y}{2})=8\]

Answer 33

again huh

Answer 34

im sorry i dont think ive ever seen this in my life

Answer 35

im sorr im probably asking too many questions...go on...

Answer 36

sry about that i was typing false

Answer 37

its ok

Answer 38

if we factor we have\[4(y^2-\frac{ y }{ 2 })=8\]
factoring here is like dividing each term by 4
except 8 ofcourse

Answer 39

oh ok

Answer 40

almost...

Answer 41

where did all the work go

Answer 42

@Jonask are you still there

Answer 43

hello

Answer 44

another mistake persisted so i had to re start
\[4(y^2-\frac{y}{2})=8 \]
\[4(\color{red}{y^2-\frac{y}{2}+\frac{1}{16}}+-\frac{1}{16})=8\]
\[4(\color{red}{(y-\frac{1}{4})^2}-\frac{1}{16}=8\]
\[4\color{red}{(y-\frac{1}{4})^2}-4\frac{1}{16}=8\]
\[\color{red}{(y-\frac{1}{4})^2}=8+\frac{1}{4}\]
\[\color{red}{(y-\frac{1}{4})^2}=\frac{33}{4}\]
original
\[\large x^2-4(y-\frac{ 1 }{ 2 })^2=\frac{ 33 }{ 8 } \implies \frac{ x^2 }{ \frac{ 33 }{ 8 } }+\frac{ (y-\frac{ 1 }{ 2 })^2 }{ \frac{ 33 }{ 32 } }=1\]

Answer 45

thats not the original... its a positive 4

Answer 46

yes you are right

Answer 47

soooooo

Answer 48

and also i forgot the 4s on the last two red steps

Answer 49

anyway the answer is correct just change that sign

Answer 50

sorry...my pc is slow...i might have to ask some one in person...school ends at 3:43 and I have to retake my exam

Answer 51

thanks!!!!!!!!!!!!!!!!!!!!!!!

Answer 52

wait before you go i have an easier way for you

Answer 53

for any conic section of the form
\[Ax^2+Bxy+Cy^2+Dx+Ey+F=0\]
consider
\[\Delta=B^2-4AC\]
if
\[\Delta \le 0 \]
ellipse
\[\Delta \ge 0 \]
hyperbola
\[\Delta = 0 \]
parabola

Answer 54

in our case
\[B=0,A=1,C=4 \implies \Delta=0^2-4*1*4 \le 0 \implies\text{ellipse}\]

Answer 55

@studyfreak_101

Answer 56

Sorry but guess what... I GRADUATED!!!!!!!!!!!!!!!!!!!!!

Answer 57

thats great news,well done