anonymous
  • anonymous
Find the eccentricity, directrices and focal width of the ellipse 2x^2=1-y^2
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
ok so 2x^2+y^2=1
anonymous
  • anonymous
what to do next?
anonymous
  • anonymous
@mathmate

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anonymous
  • anonymous
i was trying to get it in the from x^2/a^2+y^2/b^2=1
mathmate
  • mathmate
To find a and b: compare \(x^2/a^2+y^2/b^2=1\) with \(x^2/(1/\sqrt 2)^2+y^2=1\)
anonymous
  • anonymous
ok y couldnt we just put x^2/(1/2)+y^2=1 or thats not correct?
mathmate
  • mathmate
we have a^2=(1/2) so a=sqrt(1/2)=1/sqrt(2)=sqrt(2)/2
mathmate
  • mathmate
@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.
mathmate
  • mathmate
|dw:1446778123425:dw|
mathmate
  • mathmate
we have a=sqrt(2)/2, b=1
anonymous
  • anonymous
ok so a=sq.rt 2/2 then a^2=2/4 or 1/2 ok
mathmate
  • mathmate
exactly, that's how you get back 2x^2+y^2=1
anonymous
  • anonymous
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
anonymous
  • anonymous
right?
mathmate
  • mathmate
exactly! e=sqrt(2)/2
mathmate
  • mathmate
and c=sqrt(2)/2 as well.
mathmate
  • mathmate
So now you'll need the directrices. |dw:1446778893722:dw|
anonymous
  • anonymous
ok so the directrices y=+b^2/c and y=-b^2/c so y=1/sq.rt2/2 which gives y=sq.rt 2 and y=-sq.rt 2
anonymous
  • anonymous
right?
mathmate
  • mathmate
Yep, y=+/- sqrt(2) = b/e
anonymous
  • anonymous
so how do i write the focus points
mathmate
  • mathmate
For some reason, OS is very sluggish.
mathmate
  • mathmate
So the foci would be (0, \(\pm c)\) since the major axis is vertical.
anonymous
  • anonymous
i mean like does the ellipse always go through the x and y axis .. can it shift like above the axes
mathmate
  • mathmate
Well yes, it could. It's the same translation process as you did for parabolas. If the centre is shifted to say, x1, y1,then the equation is (x-x1)^2/a^2 + (y-y1)^2/b^2 = 1
anonymous
  • anonymous
ok and the focal width would be ? idk how to do this 1
mathmate
  • mathmate
It would be 2c.
anonymous
  • anonymous
o 2*sq.rt2/2 =sq.rt2 wow almost everything is sq.rt 2 lol thanks for your help again
mathmate
  • mathmate
You're welcome! :)
anonymous
  • anonymous
how would u represent the focal width though ?
mathmate
  • mathmate
It's the distance between the two foci. I was trying to look up an official definition, but could not. The distance 2c between the two foci is what i think is logical.
mathmate
  • mathmate
See for example, http://www.mathopenref.com/ellipse.html which has no mention of the term.

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