Question

9x^2+4y^2+36x-24y+36=0
find vertex foci and center

Answer 1

This is an ellipse, no?

Answer 2

yes

Answer 3

Okay... group terms together, x's together, y's together
\[\large \color{red}{9x^2+36x}+\color{blue}{4y^2-24y}+36 = 0\]

Answer 4

\[\large \color{red}{9(x^2+4x)}+\color{blue}{4(y^2-6y)}+36 = 0\]

Answer 5

ok

Answer 6

You know how to complete the square?

Answer 7

yea

Answer 8

So, I want you to complete the square of the red (x) part

Answer 9

so you add 4 to 36 and 4 to 0

Answer 10

and multiply the 4 by 9

Answer 11

\[\large \color{red}{9(x^2+4x+4)}+\color{blue}{4(y^2-6y)}+36 = 0+\color{red}{(4\times9)}\]

Answer 12

You add the 4 within the parenthesis.

Answer 13

oh ok

Answer 14

what to add the blue (y) part?

Answer 15

same thing but you add3 times 4

Answer 16

you don't add 3

Answer 17

complete the square, remember?

Answer 18

ya i know after completing the square i got 3 so i add 3 within the parantheses and 3(9) after the equal sign

Answer 19

you take half of 6 and then SQUARE it.

Answer 20

ooops
i just added wrong

Answer 21

so... fix it :)

Answer 22

ok its 9

Answer 23

give me a minute to write my equation

Answer 24

sure.

Answer 25

9(x^{2}+4x+4) + 4(y^{2}-6y+9) +36 = 9(4)+ 4(9)

Answer 26

= 72

Answer 27

Nicely done :)
\[\large \color{red}{9(x^2+4x+4)}+\color{blue}{4(y^2-6y)}+36 = 0+\color{red}{(4\times9)}+\color{blue}{(9\times 4)}\]
\[\large \color{red}{9(x^2+4x+4)}+\color{blue}{4(y^2-6y+9)}+36 = \color{violet}{72}\]

Answer 28

9(x+2)^2 + 4 ( x-3 ) ^ 2 + 36 = 72

Answer 29

now divide by 72 on both sides

Answer 30

You're... good at this :D

\[\large \color{red}{9(x+2)^2}+\color{blue}{4(y-3)^2}+36 = 72\]

Answer 31

Not yet, you still have a pesky 36 on the left side :)

Answer 32

oh so you move it to the other side

Answer 33

yup :)

Answer 34

= 108

Answer 35

divide by 108 on both sides

Answer 36

I don't think so... you subtract 36 on the right side, don'cha? :)

Answer 37

ugh stupid mistakes

Answer 38

Don't worry, that's why I'm here :)

Answer 39

= 36

Answer 40

( x+2) ^2 div 4 + ( x-3)^2 div 9 = 1

Answer 41

now a = 3 and b = 2 so its a vertical ellipse

Answer 42

Yeah, it would seem so. :)
and no, it's a horizontal ellipse if that's the case :D

Answer 43

no its vertical

Answer 44

oh yes... it's b that's 3 and a that's 2...

Answer 45

so its vertical ?

Answer 46

yup

Answer 47

center = (-2,3)

Answer 48

\[\large \color{red}{\frac{(x+2)^2}{2^2}}+\color{blue}{\frac{(y-3)^2}{3^2}} = 1\]

Answer 49

yes i o put x on accident before

Answer 50

center = (-2,3)

Answer 51

no worries. correct with your centre.
Now your vertices?

Answer 52

(-2,6) and (-2,5)

Answer 53

nope...

Answer 54

c= \[2\sqrt{3}\]

Answer 55

why not
doesnt vertex = ( h,k+a)

Answer 56

Sure, also (h,k-a)

Answer 57

ya

Answer 58

so ( -2,6) and (-2, 5 )

Answer 59

How do you get (-2,5)?

Answer 60

oooh its supposed to be (-2,0)

Answer 61

much better.
Now your foci? You got what value for c?

Answer 62

\[2\sqrt{3}\]

Answer 63

and so your foci are at...?

Answer 64

Wait, no, I don't think so...

Answer 65

( 3,3+ c ) and

Answer 66

\[c = \sqrt{a^2-b^2}\]

Answer 67

oh ya its square nroot of 13 not 12
so c is equal to 3.6

Answer 68

c must not be bigger than a.
Or else your foci would be outside the ellipse.
I think you calculated
\[\huge \sqrt{a^2 + b^2}\]
when in fact
\[\huge \color{blue}{c=\sqrt{a^2-b^2}}\]

Answer 69

ok ok i got this foci = ( 3, 5.25 ) and ( 3, .75)

Answer 70

what did you get for c?

Answer 71

2.25

Answer 72

how? You're actually supposed to get something like 2.236

Answer 73

your right im blind

Answer 74

okay, well, as long as you got c correctly, your foci should be at
(h , k + c) and (h , k - c)
Do that, and you're done :)

Answer 75

ok now he gave us another question similiar so im gonna try doing that then come back to review just the answers

Answer 76

thank you so much for your time

Answer 77

You're welcome :)

Answer 78

are you still there ?