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anonymous

  • 5 years ago

Find the standard form of the equation of the hyperbola with the given characteristics. foci: (+-3,0) Asymptotes: y=+-2x

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  1. amistre64
    • 5 years ago
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    since the foci are located parallel to the x axis; we can make a note of that and say that they open to the left and right

  2. anonymous
    • 5 years ago
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    I know the origin is at (0,0) my equation should look like: y^2/a^2-x^2/b^2=1

  3. amistre64
    • 5 years ago
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    the center appears to already be at 0; otherwise youd have different 'x' parts for the foci

  4. anonymous
    • 5 years ago
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    jk it would be x^2/a^2-y^2/b^2=1

  5. amistre64
    • 5 years ago
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    the asymptote tells us what a and b are really; since it is in the form y = (b/a)x

  6. amistre64
    • 5 years ago
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    2 b -- x = --- x 1 a the y component is b and the x compnent is a

  7. anonymous
    • 5 years ago
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    Right and I know I need to find c^2=a^2+b^2 I know C is 3^2 right?

  8. amistre64
    • 5 years ago
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    x^2 y^2 --- - --- = 1 1^2 2^2

  9. amistre64
    • 5 years ago
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    a is only needed if you are looking to find a or b; we know both a and b from the asympotote

  10. anonymous
    • 5 years ago
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    ok but that can't be right....I think we may have a wrong....

  11. anonymous
    • 5 years ago
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    my options are x^2 y^2 --- - --- = 1 9 4 x^2 y^2 --- - --- = 1 9/5 36/5 x^2 y^2 --- - --- = 1 36/5 9/5

  12. anonymous
    • 5 years ago
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    x^2 y^2 --- - --- = 1 9 4 I would guess this one but I'm not sure

  13. amistre64
    • 5 years ago
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    lets recheck then :)

  14. amistre64
    • 5 years ago
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    we need to determine the "rectangle" of the hyperbola; to so that, lets write it as an ellipst

  15. anonymous
    • 5 years ago
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    uh.. ok?

  16. amistre64
    • 5 years ago
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    x^2 y^2 --- + ---- = 1 the vertices of this ellipse are the top and sides a^2 b^2 of your asympotoe information

  17. anonymous
    • 5 years ago
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    correct

  18. amistre64
    • 5 years ago
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    2x is the asympy.... this tells us the from the center, we go up 2 and over 1 to for the top right corner of the rectangle; and our ellipse will be 2 high and 1 left right?

  19. anonymous
    • 5 years ago
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    right

  20. amistre64
    • 5 years ago
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  21. anonymous
    • 5 years ago
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    ok I got it

  22. amistre64
    • 5 years ago
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    and it should end up looking like this inthe end

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  23. anonymous
    • 5 years ago
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    ok.... ugh this stupid math class their answer key never matches up to what I get lol

  24. amistre64
    • 5 years ago
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    our y compnent is 2^2 regardless right :)

  25. amistre64
    • 5 years ago
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    id go with your first option; dbl chk by making y=0 and seeing what you get for an x value.. maybe :)

  26. anonymous
    • 5 years ago
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    ok thanks

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