Question

Find the eccentricity, directrices and focal width of the ellipse
2x^2=1-y^2

Answer 1

ok so 2x^2+y^2=1

Answer 2

what to do next?

Answer 3

@mathmate

Answer 4

i was trying to get it in the from x^2/a^2+y^2/b^2=1

Answer 5

To find a and b:
compare
\(x^2/a^2+y^2/b^2=1\)
with
\(x^2/(1/\sqrt 2)^2+y^2=1\)

Answer 6

ok y couldnt we just put x^2/(1/2)+y^2=1
or thats not correct?

Answer 7

we have a^2=(1/2)
so a=sqrt(1/2)=1/sqrt(2)=sqrt(2)/2

Answer 8

@blow_pop2000
Do you mind finding someone to chat with in the chat area, like what you're doing in the History section? Thank you.

Answer 9

we have a=sqrt(2)/2, b=1

Answer 10

ok so a=sq.rt 2/2 then a^2=2/4 or 1/2 ok

Answer 11

exactly, that's how you get back 2x^2+y^2=1

Answer 12

so eccentricity would be c/b
where c^2= b^2-a^2
c^2=1-1/2
c^2=1/2
c=sq.rt2/2
so eccentricity =(sq.rt 2/2)/1

Answer 13

right?

Answer 14

exactly! e=sqrt(2)/2

Answer 15

and c=sqrt(2)/2 as well.

Answer 16

So now you'll need the directrices.