Identity and graph, Hyperbola?

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Identity and graph, Hyperbola?

Mathematics
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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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Other answers:

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

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