Question

HELP PLEASE!!!! find the equation of ellipse. center @origin.
Passing through (2, square root 30 / 3.), distance between directrices 24 square root 7 / 7.

Answer 1

\[\frac{ x^{2} }{ a^{2} }+\frac{ y ^{2} }{ b^{2}}=1\]
can you identify this equation?

Answer 2

yes its ellipse

Answer 3

so plug in the values of (x,y) i.e \[(2,\sqrt{30}/3)\] in that equation

Answer 4

okayy..

Answer 5

now dist. of directrix from origin is a/e where e^2=1-(b/a)^2

Answer 6

so dist between 2 directrices will be 2a/e

Answer 7

yes and eccentricity not given...

Answer 8

so we find eccentricity in terms of a^2 and b^2

Answer 9

how do we get a? b?

Answer 10

okay let's visualize the problem first

Answer 11

you want the equation right?

Answer 12

so you want a, b

Answer 13

yes yes

Answer 14

so first lets use the co-ordinates given and plug it in the master eqn of ellipse

Answer 15

so you get a eqn with 2 unknown i.e a,b

Answer 16

you need 1 more to find the exact values for a,b

Answer 17

for that 2a/e = dist between directrices

Answer 18

2 coordinates?

Answer 19

itz co-ordinate not plural ..sorry

Answer 20

typo..

Answer 21

co-ordinate is (2, sqrt30/3) use it in ellipse eqn

Answer 22

eqn will be 4/a^2 +10/b^2 =1

Answer 23

yeah. thats what i did

Answer 24

now you answer me what's 2a/e?

Answer 25

distance between directrix and i dont know how to get e

Answer 26

\[e ^{2}=1-\ \frac{ b ^{2} }{ a ^{2}}\]

Answer 27

therefore plug in this eqn to find dist ..
i.e \[4a ^{2}/e ^{2}=dist ^{2}\]

Answer 28

where dist =

Answer 29

24 sqrt 7/7

Answer 30

4a^2 / e^2.. how did you come up with this??

Answer 31

look at the graph directix lies at a/e.. therefore dist between both of them is 2a/e

Answer 32

got it?

Answer 33

yey!

Answer 34

yippie!!

Answer 35

and.. how about getiing e?

Answer 36

anything else?

Answer 37

i already ut the dist on the equation and what's next?

Answer 38

already put the distance on the equation

Answer 39

e defined like that.. and for ellipse its the eqn written above

ingeneral e=distance of locus of point from focus/ perpendicular dist. of locus point from directrix

Answer 40

now you get a eqn in a^2 and b^2 again

Answer 41

right??

Answer 42

yes..

Answer 43

so you can now solve for 2 linear eqn with 2 variable .. right..??

note that you need to find a^2 and b^2 not a, b.. so i used the term linear eqn as you can put a^2=p and b^2=q

Answer 44

still confused?

Answer 45

yes..

Answer 46

at which part?

Answer 47

the e part.. dont know how to solve. what i know is the first equation would be the 4/a^2 +10/b^2 =1 and yeah stuck up no idea about the secodn equation..

Answer 48

suppose u have a eqn ... 2a/e=24 sqrt7/7

Answer 49

now u have a eqn of e in terms of a^2,b^2.. (the eqn just below the graph i drew)

Answer 50

now substitute this eqn of e in the dist eqn above..

Answer 51

square both sides.. u'll get eqn. in a^2 and b^2

Answer 52

and..

Answer 53

so now substitute a^2 by p and b^2 by q in both the obtained .. u get eqn in p,q ..

now from one eqn find p in terms of q ..
then substitute this in the second equation