HELP!!!!!!!!!!!!!1 find the equation of ellipse. center @origin.
2. Passing through (2 square root 2, -2), distance between foci 4 square root 2.

- anonymous

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- anonymous

i bet we can do this, probably take a bit

- anonymous

thank you thank you..

- anonymous

if i screw up we can fix it
ok first of all we know that since the center is at the origin it has to look like
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] so our goal is to find \(a\) and \(b\)

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- anonymous

also since \(2\sqrt{2}>2\) we know it looks like this |dw:1378778894408:dw|

- anonymous

so maybe we should rewrite it as
\[\frac{y^2}{a^2}+\frac{x^2}{b^2}=1\] as \(a>b\)
also since distance between foci \(4\sqrt2\) we know that the foci are at \((-2\sqrt{2},0)\) and \((-2\sqrt2,0)\)

- anonymous

|dw:1378779069012:dw|

- anonymous

how are we doing so far?

- anonymous

yes that's my graph

- anonymous

k good

- anonymous

already i screwed up, should be
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] with \(a>b\)

- anonymous

now by definition the ellipse is all point equidistant from the foci
so we can find what that distance is using the distance formula

- anonymous

let me write this down with pencil and paper, because there is no way for me to do it in my head and type it here

- anonymous

c is 2 square root 2.. ?

- anonymous

half of distance of foci..

- anonymous

yes that is c but of course we need \(a\) and \(b\) so we need some more information

- anonymous

ok correct me if i am wrong
the distance between \((2\sqrt{2},0)\) and \((2\sqrt2,-2)\) is \(2\)
i am sure i am not wrong about that

- anonymous

yes!

- anonymous

the distance between \((-2\sqrt2,0)\) and \((2\sqrt2,-2)\) is \(6\)

- anonymous

that i did with pencil and paper using the distance formula

- anonymous

but you should check it, because i may have screwed up

- anonymous

the answer is x^2 + 2y^2 = 16..

- anonymous

i am not there yet for sure

- anonymous

im sorry but no idea getting a and b..

- anonymous

hmm i am not getting that
i am getting \(a=3\)

- anonymous

did you get 6 for the distance?

- anonymous

oh crap!! i forgot to add the 2

- anonymous

distance between \((-2\sqrt2,0)\) and \((-2\sqrt2,-2)\) is\(6\)
distance between \((2\sqrt2,0)\) and \((2\sqrt2,-)\) is \(2\) total distance is \(8\)

- anonymous

yes 6

- anonymous

to get \(a\) note that the fixed distance is \(8\) so from \((-2\sqrt2,0)\) to \(a\) and back to \((2\sqrt2,0)\) is \(8\)

- anonymous

|dw:1378780490412:dw|

- anonymous

we can solve by \(4\sqrt{2}+2x=8\) so \(2x=8-4\sqrt2\) making \(x=4-2\sqrt{2}\) and so
\(a=2\sqrt{2}+4-2\sqrt{2}=4\)

- anonymous

i bet there is an easier way to do it, but i don't know it

- anonymous

oh nvm it is obvious
if the total distance is \(8\) then \(a=4\)

- anonymous

|dw:1378780695362:dw|

- anonymous

ok so now we have
\[\frac{x^2}{16}+\frac{y^2}{b^2}=1\] and we can find \(b\) since we know \(a=4,c=2\sqrt2\)

- anonymous

thank you!!

- anonymous

\(b^2=a^2-c^2\)
\[b^2=16-8\]
\[b^2=8\] gives
\[\frac{x^2}{16}+\frac{y^2}{8}=1\] multiply by \(16\) to get your answer
sorry it took so long

- anonymous

yw

- anonymous

@satellite73 can you pls help me?? i am sorry..

- anonymous

how about.. Passing through (2, - square root 2) latus rectum square root 3.

- anonymous

lol
you always have these conic section problems that no one wants to do!

- anonymous

i dont know how to get the eccentricity given these conditions..

- anonymous

yes yes :(

- anonymous

crap, all i know is the latus rectum is \(\frac{2b^2}{a}\)

- anonymous

me too..

- anonymous

ok i didn't leave i am googling
trying to connect the dots
obviously we need to use the point on the ellipse

- anonymous

only the point.. i mean (2, - square root 2) is present on my graph..

- anonymous

@satellite73 i really need to go, still have my class. but please pretty please help me.. i need your help. our midterm exam approaching and im pretty sure my professor will give this kind of exam..

- anonymous

i have an idea but it may be wrong

- anonymous

thank you so much, imso sorry for having thick face..

- anonymous

center of the ellipse is \((0,0)\) right?

- anonymous

yes yes. the answer is x^2 + 4y^2 = 12.. (stated on the book)

- anonymous

latus rectum goes through the foci, length is \(\sqrt3\)

- anonymous

|dw:1378782479308:dw|

- anonymous

that -2 is a typo, should be \(-c\)

- anonymous

use distance formula
to solve for \(c\)

- anonymous

just like the other one?

- anonymous

what are the points?

- anonymous

(2 square root 2, 2) (-2 square root 2, -2)

- anonymous

?

- anonymous

i am trying to do it with pencil and paper
from \((-c,0)\) to \((2,-\sqrt{2})\) is
\[\sqrt{(-c-2)^2+2}\]
from \((c,0)\) to \((2,-\sqrt2)\) is \(\sqrt{(c-2)^2+2}\)

- anonymous

no i mean ( 2, - square root 2) (-2, - square root 2)

- anonymous

oh wait, you have two points? well i guess they would both be on it

- anonymous

after i solve for c? how to get a and b?

- anonymous

|dw:1378783214089:dw|

- anonymous

you can solve for \(c\) i think knowing that the sum of the distances are equal

- anonymous

let me try to write it with pencil and paper before i screw up here

- anonymous

yes..

- anonymous

oh. i really need to go.. 30 mins before my class starts.. :((

- anonymous

ok sorry
i get \(c=3\) but i used worlfram
good luck !

- anonymous

i am sure there is a better way, but i do not know it
this seems to be a hard problem and i cannot find a similar one one the internet

- anonymous

thank you.. thank you. i need to close this, for me to ask one more.. no.. 6 questions.. aaaaaaaaaaah!!!!

- anonymous

before i log out..

- anonymous

ok close away

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