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is that correct? I feel like I did something wrong xD

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doesn't look correct
Yeah I figured.
scroll down http://www.wolframalpha.com/input/?i=focus+20x%5E2%2B4y%5E2%3D5
hrrm but I want to know the process of how to get there ( so i learn it better) :P
(x^2/20)+(y^2/4)=5
a^2=20 a=2sq5
b^2=4 b=2
\[20x^2+4y^2 = 5\] start by dividing both sides by \(5\)
oooh oh I forgot to get one on the right side
\[\frac{ 20x^2 }{ 5 }+\frac{ 4y^2 }{ 5 }=0\]
\[20x^2+4y^2 = 5\] dividing both sides by \(5\) gives \[\frac{20x^2}{5}+\frac{4y^2}{5} = \frac{5}{5}\] which is same as \[\frac{x^2}{1/4} + \frac{y^2}{5/4}=1\]
right
\(\frac{5}{5}\) equals \(1\) not \(0\)
aii yeah.
\[\frac{x^2}{1/4} + \frac{y^2}{5/4}=1\] \(a^2 = 5/4\) \(b^2 = 1/4\)
Wouldn't it be 4/5? not 5/4?
No remember an ellipse as following \[\frac{ x^2 }{ a^2 } + \frac{ y^2 }{ b^2 }=1\]
...then it'd be a^2 = 1/4 b^2 = 5/4 ? o.o
which ever is big is \(a\)
oou.
\[a \ge b > 0 \]
big : major : \(a\) small : minor : \(b\)
Yeah yeah, I gotcha now :)
5/4 is greater than 1/4 so the thing under \(y\) is \(a\) and consequently the ellipse stretches along \(y\) axis (vertical)
yeah
\(a^2 = 5/4\) \(b^2 = 1/4\) next use the chart for vertical ellipse |dw:1433224590567:dw|
It's best if you graph it, then everything becomes obvious say we have \[\frac{ x^2 }{ 9 }+\frac{ y^2 }{ 5 }=1\] |dw:1433224525998:dw| such a bad drawing but notice i square root 9 and 5 \[a^2 = 9 \implies \sqrt{9} = 5~~~~~b^2 = 5 \implies \sqrt{5}\]
Just an easy example to show you whats going on
hrm yeah. I think this one just messed me up because it's turned the other way.
Yeah ganeshie put up a nice chart :P
I can't see the chart o.o it looks fuzzy.
Try http://puu.sh/i9kBs/ee49513aac.png
ah, there it is :)
ok so for major and minor axis usually it's 2a for major 2b for minor same here?
Yeah that sounds good
soo it'd be 2a = 5/4 2b=1/4 ?? because they don't like those answers
Remember square :P
a^2 = 5/4
so 25/16
and 1/16
whats 25/16 ?
Major axis.
No, \[a = \frac{ \sqrt{5} }{ 2 }\] 2 will just get cancelled
-.- what? a = 5/4 a^2 = 25/16
who said a=5/4
\[a^2 = 5/4\]
oh yeah.
GOSH DARNIT
so yeah a = sq5/2
Yes
so 2a = ?
It's a game of substitution, what is a?
sq5/2
2(sq5/2)
Right, so now multiply that by 2
Good and we get?
(2sq5)/2
oh sq 5
Yes :) I actually put that up earlier
haha thanks for putting up with me xP i'm incorrigible. meeh.
and then 2b =
It's cool, ganeshie has taught me how to be patient :P
Yall are amazing lol I hate it when people i'm trying to help don't get it. So I offer a thousand apologies xD
\[b^2 = 1/4\] right? so do same as above
= 1?
Looks good
yay! for the first time in forever, I get an answer right :D
thanks so much.
Np :) sorry to butt in like that ganeshie, I just love ellipses.
^ xD
I have unlimited piles of ellipses on my homework *cough cough
though your hate for me might just over ride your love for ellipses
I have one emotion, and that's batman.
You...are a true sir.
he is a true super hero
length major axis : \(\sqrt{5}\) length of minor axis : \(1\) what about others
other what now?
Focus, vertex etc
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Ah nicely done :)
thanks to you guys :)

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