anonymous
  • anonymous
ellopse foci (0,+-2)(+-1,0)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
the ellipse must have a centre of (x,y) to get those foci
anonymous
  • anonymous
My possible answers are: A. x^2/4+y^2, B. x^2/5+y^2, C. x^2+y^2/5 or D. x^2+y^2/4
amistre64
  • amistre64
an ellipse only has 2 foci, youve posted 3 .... or at least thats the way this is being interpreted by others

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amistre64
  • amistre64
lol, youve posted 4, its sunday and i just cant count on the weekends
anonymous
  • anonymous
Its the co-ordinates of the 2 foci
amistre64
  • amistre64
the +- implies (0,a) and (0,-a)
anonymous
  • anonymous
Well +2 and -2 interpret to equal the same thing. The equation c^2=a^2-B^2 is supposed to be the correct way to go but the vertices of the vertical would be (0,a),(0,-A) and co-verties (b,0),(-b,0) which confuses me so much since being C to solve is all that seems to be offered... >.<
amistre64
  • amistre64
so, you have been given a set of vertices, not foci
amistre64
  • amistre64
the equation of abc on an ellipse is a^2 = b^2 + c^2
anonymous
  • anonymous
Exactly, I'm not trying to solve just to correctly format it
anonymous
  • anonymous
The foci would be c^2=A^2-B^2
anonymous
  • anonymous
Sorry no you're right lm thinking of solving for the foci
amistre64
  • amistre64
if all you are trying to do is pick the correct equation that has these verts .... (+-p,0) suggests that under x^2 needs to be (p)^2 (0,+-q) suggests that under y^2 needs to be (q)^2 and the foci, again, does not follow the pythag thrm set up
amistre64
  • amistre64
c^2 = a^2 - b^2 is correct, yes
anonymous
  • anonymous
So which one do you think it is? I am kind of thinking B
amistre64
  • amistre64
if you want to solve for the foci, just remember slope = y/x, which corresponds to b/a in the conventional setup of these things
amistre64
  • amistre64
use the information given:\[\frac {x^2}{M}+\frac {y^2}{N}=1\] when x=0, and y=+-2 \[\frac {0^2}{M}+\frac {(\pm2)^2}{N}=1\] what does N have to be? when x=+-1, and y=0 \[\frac {(\pm1)^2}{M}+\frac {0^2}{N}=1\] what does M have to be?
anonymous
  • anonymous
For the top one it would have to be 4 to even out to equal on and the bottom one 0, right?
anonymous
  • anonymous
equal 1*
amistre64
  • amistre64
4 and 1, yes :) so x^2 + y^2/4 = 1 is the equation that fits
anonymous
  • anonymous
Thank you!
amistre64
  • amistre64
good luck ;)

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