Question

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

Answer 1

i bet we can do this, probably take a bit

Answer 2

thank you thank you..

Answer 3

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 \(b\)

Answer 4

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

Answer 5

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 \((-2\sqrt2,0)\)

Answer 6

how are we doing so far?

Answer 7

yes that's my graph

Answer 8

k good

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

c is 2 square root 2.. ?

Answer 13

half of distance of foci..

Answer 14

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

Answer 15

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

Answer 16

yes!

Answer 17

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

Answer 18

that i did with pencil and paper using the distance formula

Answer 19

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

Answer 20

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

Answer 21

i am not there yet for sure

Answer 22

im sorry but no idea getting a and b..

Answer 23

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

Answer 24

did you get 6 for the distance?

Answer 25

oh crap!! i forgot to add the 2

Answer 26

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

Answer 27

yes 6

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

thank you!!

Answer 34

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

Answer 35

yw

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

yes yes :(

Answer 41

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

Answer 42

me too..