Parabola help :)

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Parabola help :)

Mathematics
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\[16x^2-128x-9y^2+400=0\] \[16(x^2-8x)-9y^2+400=0\] \[16(x^2-8x+16-16)-9y^2=-288\] \[16(x-4)^2-256-9y^2=-288\] \[16(x-4)^2-9y^2=-32\]
@jim_thompson5910 I know we just did one like this, but i'm not quite sure where to go next with 9y^2 because it has no match.

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you don't need to complete the square for y since it's already done think of -9y^2 as -9(y-0)^2
how did you go from step 2 to step 3?
factored out 16
idk where that -288 comes from
haha oh shoot, that was meant to be -400
I don't know where it came from either 0.0
wherever -288 is it's -400
ok
Yeah. So the thing i'm at right now is \[16(x-4)^2-9y^2=-144\]
er so the right side is negative. So i divide everything by negative yeah?
\[\frac{ (x-4)^2 }{ -9}+\frac{ y^2 }{ 16 }=1\]
hrm that seems wrong?
@ganeshie8 can you check this? :) it doesn't match up with the graph.
Looks good, just rewrite it as \[\large \frac{y^2}{16}-\frac{(x-4)^2}{9}=1\]
but that makes the center (-1,4)?
center is still (4, 0)
\[\large \frac{y^2}{16}-\frac{(x-4)^2}{9}=1\] is same as \[\large \frac{(y-0)^2}{16}-\frac{(x-4)^2}{9}=1\]
http://www.wolframalpha.com/input/?i=16x%5E2+%E2%88%92+9y%5E2+%E2%88%92+128x+%2B+400+%3D+0 that's where im confused because the center should be (4,4)
rearranging the equation wont change anything
(plus or minus)
oooh oh i see! that's the vertex, not center.
right?
Yes center is the midpoint of vertices
|dw:1433392386227:dw|
ahh, ok! I wasn't sure about that, so that's where I was confused. Thank you :)
a=4, b=3, c=5
now how do I find vertex, foci, asymptotes, dirtrex?
asy: a/b(x-h)+k so for mine it's \[\pm \frac{ 4 }{ 3 }(x-4)\]
(with a y = before that)
looks good |dw:1433393241031:dw|
scroll down to double check ur answers http://www.wolframalpha.com/input/?i=focus+16x%5E2+%E2%88%92+9y%5E2+%E2%88%92+128x+%2B+400+%3D+0
is this correct so far? :)
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(and DNE on the directrix as well)
looks gud
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thank you so so much! ;)
|dw:1433393976693:dw|
Excellent!!!

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