Question

put the equation into standard form. find the center the lines which contain the major and minor axes, vertices the endpoints of the minor axis, foci, and eccentricity

Answer 1

\[9x^2+25y^2-54x-50y-119=0\]
so i put it in order
\[9x^2-54x+25y^2-50y=119\]
now having a hard time completing the square

Answer 2

@e.mccormick and @hartnn

Answer 3

if i factor out a 9 i get and 25 i get
\[9(x^2-6x+9)+25(y^2-10y+25)=119\]
is this right so far ?

Answer 4

Hmmm.... how did you get the y part?

Answer 5

oops should be a 2 not 25

Answer 6

damn here \[9(x^2-6x+9)+25(y^2-2y+1)=119\] ?

Answer 7

so end will end up being 225? then divide by 225 guess i just needed to talk it through

Answer 8

You also have to ballance what you add with something you subtract. I leave things in ( ) to make it a little clearer what I am doing with the sign, but it is optional.
\(9(x^2-6x+(-3)^2 - (-3)^2)+25(y^2-2y +(-1)^2-(-1)^2)=119\)

Answer 9

thanks i got it :) just guess i needed to talk it out haha

Answer 10

Yah, and do you know what that simplifies to?

Answer 11

its
\[\frac{ (x-3)^2 }{ 25 }+\frac{ (y-1)^2 }{ 9 }=1\]

Answer 12

The \(+(-3)^2\) becomes part of the square. The \(-(-3)^2\) gets evaluated. Same sort of thing on the y part.

\(9((x-3)^2 - 9)+25((y-1)^2-1)=119\)

I did not do the rest yet.

Answer 13

\((x-3)^2 - 81 +(y-1)^2-25=119\)

\((x-3)^2 +(y-1)^2=119+ 81+ 25\)

\((x-3)^2 +(y-1)^2=225\)

Yah, looks like it.

Answer 14

Oops. missed typing in the 9 and 25. LOL.

Answer 15

:)

Answer 16

I meant:
\(9(x-3)^2 - 81 +25(y-1)^2-25=119\)

\(9(x-3)^2 +25(y-1)^2=119+ 81+ 25\)

\(9(x-3)^2 +25(y-1)^2=225\)

=P

Answer 17

Yah, so: \(\dfrac{ (x-3)^2 }{ 5^2 }+\dfrac{ (y-1)^2 }{ 3^2 }=1\)