Babynini
  • Babynini
Identity and graph, Hyperbola?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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Babynini
  • Babynini
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Babynini
  • Babynini
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
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

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Babynini
  • Babynini
oo thanks.
Babynini
  • Babynini
ellipse?
jim_thompson5910
  • jim_thompson5910
yep it's an ellipse
Babynini
  • Babynini
fabulous, so...what next? I grouped stuff and factored out but i'm not sure how to finish the squares.
Babynini
  • Babynini
4(x^2-8x ) 25(y^2+6y )=-189
jim_thompson5910
  • jim_thompson5910
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
Babynini
  • Babynini
hmm ok yeah
Babynini
  • Babynini
so now we've got 4(x-4)^2 ?
Babynini
  • Babynini
and is the right side 25(x-3)^2
Babynini
  • Babynini
(sorry, by right I meant the y stuff)
jim_thompson5910
  • jim_thompson5910
you're forgetting about the -16 in the parenthesis
Babynini
  • Babynini
how do I write that?
jim_thompson5910
  • jim_thompson5910
\[\Large 4(x^2-8x + \underline{ \ \ \ \ \ \ \ \ } \ ) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4(x^2-8x+16-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4((x^2-8x+16)-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4((x-4)^2-16) + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\] \[\Large 4(x-4)^2-64 + 25(y^2+6y + \underline{ \ \ \ \ \ \ \ \ } \ )=-189\]
jim_thompson5910
  • jim_thompson5910
what goes in the blank for the y terms?
Babynini
  • Babynini
ahh ok. y^2-6y+9
Babynini
  • Babynini
25((x-3)^2-9)
Babynini
  • Babynini
25(x-3)^2-225 ?
jim_thompson5910
  • jim_thompson5910
yeah +9-9 goes in the blank on line 1 for the y terms
jim_thompson5910
  • jim_thompson5910
be careful it is NOT y^2 - 6y it's y^2 + 6y
Babynini
  • Babynini
sorry, my bad!
jim_thompson5910
  • jim_thompson5910
so you should have this \[\Large 4(x-4)^2-64 + 25(y+3)^2 - 225=-189\]
Babynini
  • Babynini
yeah. ok so since we're adding and doing stuff to the left side we need to do the same on the right, yeah?
jim_thompson5910
  • jim_thompson5910
no need
jim_thompson5910
  • jim_thompson5910
that's why I had +16 and -16 to balance things out
jim_thompson5910
  • jim_thompson5910
same for +9-9
Babynini
  • Babynini
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
jim_thompson5910
  • jim_thompson5910
well you could add things to both sides, but you'd have to be careful
Babynini
  • Babynini
Yep yep. So now what do we do?
Babynini
  • Babynini
add 64 and 225 to both sides?
jim_thompson5910
  • jim_thompson5910
yep
Babynini
  • Babynini
so 100 on the right side :P
Babynini
  • Babynini
then divide the whole thing by 100 so we get 0 on the right again?
jim_thompson5910
  • jim_thompson5910
you mean 1 on the right side
jim_thompson5910
  • jim_thompson5910
but yeah
Babynini
  • Babynini
\[\frac{ 4(x-4)^2 }{ 100 }+\frac{ 25(y+3)^2 }{ 100 }=1\]
Babynini
  • Babynini
haha yeah sorry.
jim_thompson5910
  • jim_thompson5910
and you can rewrite that into \[\Large \frac{(x-4)^2}{25} + \frac{(y+3)^2}{4} = 1\]
Babynini
  • Babynini
perfect :)
Babynini
  • Babynini
so now we get all the other stuff from there?
jim_thompson5910
  • jim_thompson5910
yeah you can find the center, foci, vertices, co-vertices, etc
Babynini
  • Babynini
a = 5 b = 4 c= sq29
jim_thompson5910
  • jim_thompson5910
how did you get c?
Babynini
  • Babynini
oops b = 2
Babynini
  • Babynini
a^2+b^2=c^2
Babynini
  • Babynini
25+4=c^2
jim_thompson5910
  • jim_thompson5910
http://www.mathwarehouse.com/ellipse/images/formul-focus.gif
Babynini
  • Babynini
ou. then c = sqrt 21
jim_thompson5910
  • jim_thompson5910
yep
Babynini
  • Babynini
kks, and then the ellipse will be sideways yeah?
Babynini
  • Babynini
so foci: \[(0,\pm \sqrt{21})\]
jim_thompson5910
  • jim_thompson5910
not quite
jim_thompson5910
  • jim_thompson5910
the foci will shift along with the center
jim_thompson5910
  • jim_thompson5910
|dw:1433386754426:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1433386763450:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1433386776494:dw|
Babynini
  • Babynini
so we've got to find the center first.
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
in this case, the major axis is horizontal, so the foci + center all lay on the same horizontal level. Only the x values will change
jim_thompson5910
  • jim_thompson5910
oh nearly forgot the vertices |dw:1433387027602:dw|
Babynini
  • Babynini
wouldn't want to forget those! :P
Babynini
  • Babynini
ok. that makes sense.
jim_thompson5910
  • jim_thompson5910
for vertical ellipses (where b > a) \[\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] we have this schematic |dw:1433387213931:dw|
Babynini
  • Babynini
this one isn't horizontal? o.0
jim_thompson5910
  • jim_thompson5910
no I'm saying IF it were vertical, then you'd have what you see above
Babynini
  • Babynini
ooh right right
jim_thompson5910
  • jim_thompson5910
the ellipse you have is horizontal, like this |dw:1433387651593:dw|
Babynini
  • Babynini
so how do we find the center? :)
jim_thompson5910
  • jim_thompson5910
Compare \[\Large \frac{(x-4)^2}{25} + \frac{(y+3)^2}{4} = 1\] with \[\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]
jim_thompson5910
  • jim_thompson5910
The center of \[\Large \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] is (h,k)
Babynini
  • Babynini
so (4,-3)
jim_thompson5910
  • jim_thompson5910
yes
Babynini
  • Babynini
so foci : (4- sq21, -3) , (4sq21, -3)
Babynini
  • Babynini
?
jim_thompson5910
  • jim_thompson5910
I think you meant to say \[\Large \left(4 {\color{red}{ \textbf{+}}}\sqrt{21},-3\right)\] for the second focus
Babynini
  • Babynini
yeah
jim_thompson5910
  • jim_thompson5910
you have the correct foci
Babynini
  • Babynini
haha yay! Ok, vertex: (+/-5, -3) ?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
did you mean \[\Large \left(4 \pm 5, -3\right)\]??
Babynini
  • Babynini
ai ai ai yes sorry!
Babynini
  • Babynini
(9,-3) (-1,-3)
jim_thompson5910
  • jim_thompson5910
both correct
Babynini
  • Babynini
phew.
Babynini
  • Babynini
Length of major axis: 2a? minor axis 2b?
jim_thompson5910
  • jim_thompson5910
yes correct
jim_thompson5910
  • jim_thompson5910
only because a > b
Babynini
  • Babynini
so major: 50 minor: 8
jim_thompson5910
  • jim_thompson5910
50?
Babynini
  • Babynini
lol I did (a^2)2 major = 10
jim_thompson5910
  • jim_thompson5910
major is 10, yep
jim_thompson5910
  • jim_thompson5910
minor isn't 8
Babynini
  • Babynini
4
jim_thompson5910
  • jim_thompson5910
yep minor = 4
Babynini
  • Babynini
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Babynini
  • Babynini
all good? :)
jim_thompson5910
  • jim_thompson5910
everything is perfect
Babynini
  • Babynini
thanks so much! a thousand medals for your efforts :)
jim_thompson5910
  • jim_thompson5910
you're welcome

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