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frx

  • 2 years ago

The equation for an ellipse is \[\frac{ x ^{2} }{ a ^{2} }+\frac{ y ^{2} }{ b ^{2}}=1\] and the foci is given by (c,0) where \[c ^{2}=a ^{2}-b ^{2}\] and my question is if b>a is the foci given by (0,c) where \[c ^{2}=a ^{2}-b ^{2}\]

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  1. asnaseer
    • 2 years ago
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    yes, if a>b then foci are at \((\pm c, 0)\) otherwise they are at \((0, \pm c)\). see here for more details: http://hotmath.com/hotmath_help/topics/ellipse.html

  2. henpen
    • 2 years ago
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    I suppose you could try to derive the equation as a locus from the two foci (c,0) and (-c,0).|dw:1354462354704:dw| \[\sqrt{y^2+(x-c)^2}+\sqrt{y^2+(x+c)^2}=k\]

  3. Chlorophyll
    • 2 years ago
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    It seems you're confused with the notation: Foci always belong to major axis, and major axis always named a!

  4. henpen
    • 2 years ago
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    \[y^2+(x-c)^2+y^2+(x+c)^2+\sqrt{y^2+(x-c)^2}\sqrt{y^2+(x+c)^2}=k^2\] \[y^2+(x-c)^2+y^2+(x+c)^2-k^2=-\sqrt{y^2+(x-c)^2}\sqrt{y^2+(x+c)^2}\] \[(y^2+(x-c)^2+y^2+(x+c)^2-k^2)^2=(y^2+(x-c)^2)(y^2+(x+c)^2)\]ans so on

  5. henpen
    • 2 years ago
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    *and

  6. frx
    • 2 years ago
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    |dw:1354462531860:dw|

  7. frx
    • 2 years ago
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    So thats the deal when a>b, right?

  8. frx
    • 2 years ago
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    |dw:1354462661891:dw| And that b>a

  9. frx
    • 2 years ago
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    oh wrong it should be rotated 90 degrees

  10. frx
    • 2 years ago
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    |dw:1354462770084:dw|

  11. asnaseer
    • 2 years ago
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    yes, 'a' always represents the "major axis"

  12. frx
    • 2 years ago
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    Ok great thank you guys!

  13. asnaseer
    • 2 years ago
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    yw :)

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