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ok so 2x^2+y^2=1
what to do next?
i was trying to get it in the from x^2/a^2+y^2/b^2=1
To find a and b: compare \(x^2/a^2+y^2/b^2=1\) with \(x^2/(1/\sqrt 2)^2+y^2=1\)
ok y couldnt we just put x^2/(1/2)+y^2=1 or thats not correct?
we have a^2=(1/2) so a=sqrt(1/2)=1/sqrt(2)=sqrt(2)/2
@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.
we have a=sqrt(2)/2, b=1
ok so a=sq.rt 2/2 then a^2=2/4 or 1/2 ok
exactly, that's how you get back 2x^2+y^2=1
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
and c=sqrt(2)/2 as well.
So now you'll need the directrices. |dw:1446778893722:dw|
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
Yep, y=+/- sqrt(2) = b/e
so how do i write the focus points
For some reason, OS is very sluggish.
So the foci would be (0, \(\pm c)\) since the major axis is vertical.
i mean like does the ellipse always go through the x and y axis .. can it shift like above the axes
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
ok and the focal width would be ? idk how to do this 1
It would be 2c.
o 2*sq.rt2/2 =sq.rt2 wow almost everything is sq.rt 2 lol thanks for your help again
You're welcome! :)
how would u represent the focal width though ?
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.
See for example, http://www.mathopenref.com/ellipse.html which has no mention of the term.