Question

Simple question - Need to know if I put this in the correct FORM

READ BELOW FOR EQUATION

Answer 1

\[x ^{2}+4y ^{2}+2y=6x+3y ^{2}-1\]
IS A CIRCLE
Vertex (3,-1) Radius (9)

Correct form:
\[(x-3)^{2}-\frac{ y ^{2} }{3}-\frac{ y }{ 6 }+\frac{ y ^{2} }{ 4 }=\frac{ -2 }{ 3 } \]

Is this right?

Answer 2

@satellite73 please take a look real quick if you can

Answer 3

Not really correct form. you need to get y to look like what x look like,

for example: \[(x-h)^2 + (y-k)^2\]

Answer 4

How do I do that? I've never done a problem like this before

Can I combine the y^2 together?

Answer 5

@Luigi0210

Answer 6

yeah you can.

Answer 7

\(-\dfrac{y^2}{3} + \dfrac{y^2}{4}\)

Answer 8

Doesn't the denominator have to be the same? 12?

Answer 9

@mathslover @phi @texaschic101 I need help on this please!!I have class in 40 min!!

Answer 10

Yes, the denominator has to be same in order to add :

LCM of 3 and 4 is 12

so, \(\dfrac{-4y^2 + 3y^2}{12}\)
= \(\dfrac{-y^2}{12}\)

Answer 11

so then, it would be \[(x-3)^{?}-\frac{ y ^{?} }{ 12 }+\frac{ y }{ 6 }=\frac{ -2 }{ }\]

Answer 12

* -2/3

Answer 13

based on the info
Vertex (3,-1) Radius (9)

the equation should transform to
\[ (x-3)^2 + (y+1)^2 = 81 \]

Answer 14

though I assume you mean center (not vertex)

Answer 15

Yes Center is (3,1)

Answer 16

(3,-1)******

So (x-3)^2+(y+1)^2=81 still is correcr?

Answer 17

I get a radius of 3 (not 9)

Answer 18

Well the given equation can be written as :

\(x^2 + y^2 + 2y -6x + 1 = 0 \)

\((x^2 - 6x) + (y^2 + 2y + 1) =0 \)
\((x^2 -6x + 9 -9) + (y+1)^2 = 0\)
\((x-3)^2 + (y+1)^2 = 9\)

Answer 19

\[ x ^{2}+4y ^{2}+2y=6x+3y ^{2}-1 \\
x^2 -6x + y^2 +2y = -1 \\
\]
complete the square for x by add (-6/2)^2 to both sides
complete the square for y by adding (2/2)^2 to both sides

Answer 20

But, I am not getting 81 in RHS... @phi - I think, the question is a little bit wrong, the radius should be 3..

Answer 21

yes, I think mighty figured out the wrong radius.

Answer 22

Yeah...

Answer 23

Yes, my fault. It's the SQUARE ROOT of 9, not 9.

So (x-3)^2+(y+1)^2=9 is correct

Answer 24

yeah

Answer 25

Man, thank you guys so much

Answer 26

Thanks to phi :)

Answer 27

yes. If you complete the square (see above for the details posted by math) you get that equation

Answer 28

mathslover, if you're still there I have a quick Q for ya!

Answer 29

Yes, sure! I am always here.

Answer 30

in the equation (x-2)^2 OVER 9 + (y-1)^2 OVER 4 =1

(an Ellipse)

is the center (2,1)?

Answer 31

@mathslover

Answer 32

Not studied Ellipse yet :( Very sorry.. @amistre64 may help you!

Answer 33

ellipses are just flattened circles

Answer 34

Nevermind, sorry I got it! It's (2,-1). I put (y-1) instead of (Y+1)

Answer 35

\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]
if a=b, its a circle, otherwise its an ellipse if one oof the terms iis negatiivee iits a hyperbola, if only one term is squared, it a parrabola

Answer 36

im thinking my keypad is getting old ...

Answer 37

Haha I know the feeling. OpenStudy is glitching up for me today.

Thanks for the help, I understand it now!

Answer 38

good luck :)