Question

Find the center of the ellipse. See attachment.

Answer 1

My first reaction is to complete the square, but the 5 in front of x^2 is throwing me off.

Answer 2

factor out the 5 then

Answer 3

\[ax^2+bx\]
\[a(x^2+\frac bax)\]

Answer 4

Divide everything by 5, then complete the square. Multiply by 5 later, if necessary...

Answer 5

\[a(x^2+\frac bax)\]
\[a(x^2+\frac bax+\frac{b^2}{2a^2}-\frac{b^2}{2a^2})\]
\[a(x^2+\frac bax+\frac{b^2}{2a^2})-a\frac{b^2}{2a^2}\]
\[a(x^2+\frac bax+\frac{b^2}{2a^2})-\frac{b^2}{2a}\]

Answer 6

lol, i forgot to square my 2

Answer 7

\[5\left( x+\frac{ 2 }{ 5 } \right)^{2}+(y ^{2}-\frac{ 2 }{25 }) = 1\]

Answer 8

Is that looking correct so far? Now I have to get the y part in proper form.

Answer 9

the y part is rather moot:

y^2 is already in "centered" form: (y+0)^2

the centered from of the x parts is (x+b/2a)^2
^^^

Answer 10

So \[5\left( x+\frac{ 2 }{ 10 } \right)^{2}\]? Why 2a, though?

Answer 11

i thought i went over that ...

\[ax^2+bx\to a(x^2+\frac bax)\]
complete the square

Answer 12

the center is only concerned with the simplification of the innards ... so the addon extra from teh completed square is just clutter

Answer 13

\[x^2+\frac bax\]
\[(x+\frac b{2a})^2-(...)\]
the rest of it is for determing axis lengths and whatnot

Answer 14

Sorry, I haven't been able to work on this until now. Thank you for your help with the midpoint! So now I have \[\left( x+\frac{ 1 }{ 5 } \right)^{2}-\frac{ 1 }{ 5 }+y ^{2}=1\] I have it centered at -1/5 and 0. Now, how do I determine the length of the major diameter?

Answer 15

Sorry, I have \[5\left( x+\frac{ 1 }{ 5 } \right)^{2}-\frac{ 1 }{ 5 }+y ^{2}=1\]

Answer 16

I don't think amistre64 is on anymore, so any other help would be greatly appreciated! :)

Answer 17

http://answers.yahoo.com/question/index?qid=20110720081710AADizRn.
This helped me work through it.

Answer 18

you construct it into a general form again
\[5\left( x+\frac{ 1 }{ 5 } \right)^{2}-\frac{ 1 }{ 5 }+y ^{2}=1\]
\[5\left( x+\frac{ 1 }{ 5 } \right)^{2}+y ^{2}=\frac65\]
\[\frac{5\left( x+\frac{ 1 }{ 5 } \right)^{2}}{6/5}+\frac{y ^{2}}{6/5}=1\]
\[\frac{( x+\frac{ 1 }{ 5 })^{2}}{6/25}+\frac{y ^{2}}{6/5}=1\]
the bottoms of the x and y parts tell us something about the axis lengths. the sqrts reveal half lengths; i would say that 6/5 is bigger than 6/25, therefore sqrt(6/5) is half the length of the major axis giving us a full length of 2sqrt(6/5)

the same concept can be applied to smaller one to determine the length of the minor axis