anonymous
  • anonymous
HELP!!!!!!!!!!!!!1 find the equation of ellipse. center @origin. 2. Passing through (2 square root 2, -2), distance between foci 4 square root 2.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i bet we can do this, probably take a bit
anonymous
  • anonymous
thank you thank you..
anonymous
  • 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 \(b\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
also since \(2\sqrt{2}>2\) we know it looks like this |dw:1378778894408:dw|
anonymous
  • 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
  • anonymous
|dw:1378779069012:dw|
anonymous
  • anonymous
how are we doing so far?
anonymous
  • anonymous
yes that's my graph
anonymous
  • anonymous
k good
anonymous
  • anonymous
already i screwed up, should be \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] with \(a>b\)
anonymous
  • 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
  • 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
  • anonymous
c is 2 square root 2.. ?
anonymous
  • anonymous
half of distance of foci..
anonymous
  • anonymous
yes that is c but of course we need \(a\) and \(b\) so we need some more information
anonymous
  • 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
  • anonymous
yes!
anonymous
  • anonymous
the distance between \((-2\sqrt2,0)\) and \((2\sqrt2,-2)\) is \(6\)
anonymous
  • anonymous
that i did with pencil and paper using the distance formula
anonymous
  • anonymous
but you should check it, because i may have screwed up
anonymous
  • anonymous
the answer is x^2 + 2y^2 = 16..
anonymous
  • anonymous
i am not there yet for sure
anonymous
  • anonymous
im sorry but no idea getting a and b..
anonymous
  • anonymous
hmm i am not getting that i am getting \(a=3\)
anonymous
  • anonymous
did you get 6 for the distance?
anonymous
  • anonymous
oh crap!! i forgot to add the 2
anonymous
  • 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
  • anonymous
yes 6
anonymous
  • 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
  • anonymous
|dw:1378780490412:dw|
anonymous
  • 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
  • anonymous
i bet there is an easier way to do it, but i don't know it
anonymous
  • anonymous
oh nvm it is obvious if the total distance is \(8\) then \(a=4\)
anonymous
  • anonymous
|dw:1378780695362:dw|
anonymous
  • 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
  • anonymous
thank you!!
anonymous
  • 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
  • anonymous
yw
anonymous
  • anonymous
@satellite73 can you pls help me?? i am sorry..
anonymous
  • anonymous
how about.. Passing through (2, - square root 2) latus rectum square root 3.
anonymous
  • anonymous
lol you always have these conic section problems that no one wants to do!
anonymous
  • anonymous
i dont know how to get the eccentricity given these conditions..
anonymous
  • anonymous
yes yes :(
anonymous
  • anonymous
crap, all i know is the latus rectum is \(\frac{2b^2}{a}\)
anonymous
  • anonymous
me too..
anonymous
  • 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
  • anonymous
only the point.. i mean (2, - square root 2) is present on my graph..
anonymous
  • 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
  • anonymous
i have an idea but it may be wrong
anonymous
  • anonymous
thank you so much, imso sorry for having thick face..
anonymous
  • anonymous
center of the ellipse is \((0,0)\) right?
anonymous
  • anonymous
yes yes. the answer is x^2 + 4y^2 = 12.. (stated on the book)
anonymous
  • anonymous
latus rectum goes through the foci, length is \(\sqrt3\)
anonymous
  • anonymous
|dw:1378782479308:dw|
anonymous
  • anonymous
that -2 is a typo, should be \(-c\)
anonymous
  • anonymous
use distance formula to solve for \(c\)
anonymous
  • anonymous
just like the other one?
anonymous
  • anonymous
what are the points?
anonymous
  • anonymous
(2 square root 2, 2) (-2 square root 2, -2)
anonymous
  • anonymous
?
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
  • anonymous
no i mean ( 2, - square root 2) (-2, - square root 2)
anonymous
  • anonymous
oh wait, you have two points? well i guess they would both be on it
anonymous
  • anonymous
after i solve for c? how to get a and b?
anonymous
  • anonymous
|dw:1378783214089:dw|
anonymous
  • anonymous
you can solve for \(c\) i think knowing that the sum of the distances are equal
anonymous
  • anonymous
let me try to write it with pencil and paper before i screw up here
anonymous
  • anonymous
yes..
anonymous
  • anonymous
oh. i really need to go.. 30 mins before my class starts.. :((
anonymous
  • anonymous
ok sorry i get \(c=3\) but i used worlfram good luck !
anonymous
  • 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
  • anonymous
thank you.. thank you. i need to close this, for me to ask one more.. no.. 6 questions.. aaaaaaaaaaah!!!!
anonymous
  • anonymous
before i log out..
anonymous
  • anonymous
ok close away

Looking for something else?

Not the answer you are looking for? Search for more explanations.