Question

distance between directrices equal to 9 square root 2 / 2, latus rectum equal to 2/3. find the equation of ellipse, center at origin.

Answer 1

i am beginning to hate conic sections

Answer 2

me too..

Answer 3

but i have no choice..

Answer 4

it has been months for this
i think we can do this, but it will take me a minute or two

Answer 5

this would be last topic before our midterm exam.

Answer 6

okay thanks!! :)

Answer 7

still here
lets try this, hope i don't get lost

Answer 8

yes yes

Answer 9

damn wrong already. i thought the directrix was given, but it is not
back to the drawing board

Answer 10

oh maybe it is
one more piece of paper

Answer 11

when a question includes distance between directrices i got stuck up..

Answer 12

ok one more time
directrix is \(x=\frac{9\sqrt{2}}{4}\)

Answer 13

since your are given the distance, which is double that
then this means
\[\frac{a}{e}=\frac{9\sqrt{2}}{4}\] and so
\[a=\frac{9\sqrt{2}}{4}e\]

Answer 14

okay...

Answer 15

also
\[\frac{2b^2}{a}=\frac{2}{3}\] and so
\[\frac{b^2}{a}=\frac{1}{3}\] making \[b^2=\frac{a}{3}\]
and also \[b^2=a^2(1-e^2)\] together these give
\[\frac{a}{3}=a^2(1-e^2)\] so
\[\frac{1}{3}=a(1-e^2)\]

Answer 16

then since \(a=\frac{9\sqrt2}{4}\) the last equation becomes
\[\frac{1}{3}=\frac{9\sqrt2}{4}(1-e^2)\]

Answer 17

i can get now the real value of e then substitute it to the equation to get a. yeheeey

Answer 18

yeah but i am having a hell of a time solving it

Answer 19

oh actually i guess you just need \(e^2\)

Answer 20

yes yes

Answer 21

ok well i hope you are on the right path, because i gotta run
good luck!

Answer 22

THANK YOU SO MUCH!!