Question

Identity and graph, Hyperbola?

Answer 1

@jim_thompson5910

Answer 2

it's not a hyperbola

you can use something like geogebra (which is what I just used) or desmos to check

https://www.geogebra.org/

https://www.desmos.com/calculator

Answer 3

oo thanks.

Answer 4

ellipse?

Answer 5

yep it's an ellipse

Answer 6

fabulous, so...what next?
I grouped stuff and factored out but i'm not sure how to finish the squares.

Answer 7

4(x^2-8x ) 25(y^2+6y )=-189

Answer 8

the x coordinate is -8

cut that in half to get -4

then square it to get 16

you will add and subtract 16 inside the parenthesis

\[\Large 4(x^2-8x \ \ \ \ \ \ \ \ ) + 25(y^2+6y \ \ \ \ \ \ \ \ )=-189\]

\[\Large 4(x^2-8x {\color{red}{+16-16}} ) + 25(y^2+6y \ \ \ \ \ \ \ \ )=-189\]

the +16-16 is to make sure that the expression doesn't change (since we're effectively adding 0). Then notice how x^2-8x+16 factors to (x-4)^2

Answer 9

hmm ok yeah

Answer 10

so now we've got
4(x-4)^2 ?

Answer 11

and is the right side
25(x-3)^2

Answer 12

(sorry, by right I meant the y stuff)

Answer 13

you're forgetting about the -16 in the parenthesis

Answer 14

how do I write that?

Answer 15

\[\Large 4(x^2-8x + \underline{ \ \ \ \ \ \ \ \ } \ ) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\]

\[\Large 4(x^2-8x+16-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\]

\[\Large 4((x^2-8x+16)-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\]

\[\Large 4((x-4)^2-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\]

\[\Large 4(x-4)^2-64 + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\]

Answer 16

what goes in the blank for the y terms?

Answer 17

ahh ok.
y^2-6y+9

Answer 18

25((x-3)^2-9)

Answer 19

25(x-3)^2-225 ?

Answer 20

yeah +9-9 goes in the blank on line 1 for the y terms

Answer 21

be careful it is NOT y^2 - 6y

it's y^2 + 6y

Answer 22

sorry, my bad!

Answer 23

so you should have this

\[\Large 4(x-4)^2-64 + 25(y+3)^2 - 225=-189\]

Answer 24

yeah. ok so since we're adding and doing stuff to the left side we need to do the same on the right, yeah?

Answer 25

no need

Answer 26

that's why I had +16 and -16 to balance things out

Answer 27

same for +9-9

Answer 28

oh! that's fabulous :) I was wondering because in class I remember the +9 but and doing stuff to the left side also but I like this better haha

Answer 29

well you could add things to both sides, but you'd have to be careful

Answer 30

Yep yep.
So now what do we do?

Answer 31

add 64 and 225 to both sides?

Answer 32

yep

Answer 33

so 100 on the right side :P

Answer 34

then divide the whole thing by 100 so we get 0 on the right again?

Answer 35

you mean 1 on the right side

Answer 36

but yeah

Answer 37

\[\frac{ 4(x-4)^2 }{ 100 }+\frac{ 25(y+3)^2 }{ 100 }=1\]

Answer 38

haha yeah sorry.

Answer 39

and you can rewrite that into

\[\Large \frac{(x-4)^2}{25} + \frac{(y+3)^2}{4} = 1\]

Answer 40

perfect :)

Answer 41

so now we get all the other stuff from there?

Answer 42

yeah you can find the center, foci, vertices, co-vertices, etc

Answer 43

a = 5
b = 4
c= sq29

Answer 44

how did you get c?

Answer 45

oops b = 2

Answer 46

a^2+b^2=c^2

Answer 47

25+4=c^2

Answer 48

http://www.mathwarehouse.com/ellipse/images/formul-focus.gif

Answer 49

ou. then c = sqrt 21

Answer 50

yep

Answer 51

kks, and then the ellipse will be sideways yeah?

Answer 52

so foci:
\[(0,\pm \sqrt{21})\]

Answer 53

not quite

Answer 54

the foci will shift along with the center