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Meehan98

  • one year ago

Identify an equation in standard form for an ellipse with its center at the origin, a vertex at (0, 12), and a focus at (0, 8).

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  1. Meehan98
    • one year ago
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    It appears that it's a horizontal ellipse so I know that the equation will be \[\frac{ x ^{2} }{ a ^{2} } + \frac{ y ^{2} }{ b ^{2} }=1\]

  2. anonymous
    • one year ago
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    That is right.

  3. Meehan98
    • one year ago
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    I just don't know how to find the other vertex in order to find b^2.

  4. anonymous
    • one year ago
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    Actually isn't it a verticle ellipse.

  5. anonymous
    • one year ago
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    Because the vertex (furtherest point) is 12 up from the origin

  6. Meehan98
    • one year ago
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    Oh yes, that's right!

  7. anonymous
    • one year ago
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    So the semi major axis is 12, in the y direction

  8. anonymous
    • one year ago
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    Therefore the equation is (y^2)/(12^2) + (x^2)/(b^2) = 1

  9. anonymous
    • one year ago
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    Now for the b value I think you use the rule that for any point on the ellipse, the sum of the distance from both foci is 2a (a is the semi major axis)

  10. anonymous
    • one year ago
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    Ill draw a graph to show

  11. Meehan98
    • one year ago
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    So, you use the constant sum of an ellipse formula: d=PF1 + PF2

  12. anonymous
    • one year ago
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    |dw:1441973292765:dw| in the picture i took the point to be the semi minor axis. Since the ellipse is symmertrical (both foci same distance) the distance from that point to each foci is a (semi major axis)

  13. anonymous
    • one year ago
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    as they are same length and both add to 2a

  14. anonymous
    • one year ago
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    Does that make sense?

  15. anonymous
    • one year ago
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    Are you there?

  16. Meehan98
    • one year ago
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    I'm sorry, just taking me little bit to understand. I'm used to strictly using formulas to get the solution.Okay, how do you know what point to pass through on the x-axis?

  17. anonymous
    • one year ago
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    What do you mean exactly?

  18. anonymous
    • one year ago
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    Btw you gotta understand really whats happening with everything in maths it makes things so much easier, and theres less things to remember.

  19. Meehan98
    • one year ago
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  20. Meehan98
    • one year ago
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    This is the problem. I know that the denominator for y^2 is 144, but I don't know the denominator for x^2 because I'm not given thenother focus.

  21. Meehan98
    • one year ago
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    Okay, I got it now! I used c^2=a^2-b^2 to get b^2= 80. So, my answer would be: (y^2/144) + (x^2/80)=1. Thanks for your help!

  22. anonymous
    • one year ago
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    Both foci are in line with the centre and are the same distance from the centre. Therefore if you know one you can find the other. If you take the point on the ellipse closest to the centre, the semi minor axis (the b value, last value we need) its not closer to one focus than the other. If the sum of the distances from any point to each foci is 2a, taking the closest point I just talked about, the distance from each focus to it is a (both distances are equal and must sum to 2a). Using pythagoras theorem where f is the distance from centre to focus and a is the semi major axis and b is the semi minor axis, (f^2) + (b^2) = (a^2) we know f=8 and a=12 therefore (8^2) + (b^2) = (12^2) 64 + (b^2) = 144 b^2 = 80 therefore (x^2)/(80) + (y^2)/(144) = 1

  23. anonymous
    • one year ago
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    Oh i just read yours lol, yes you did eveything right, good job :)

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