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anonymous

  • one year ago

Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±(5/4) x.

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  1. anonymous
    • one year ago
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    @amistre64

  2. amistre64
    • one year ago
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    well, what have you got so far?

  3. anonymous
    • one year ago
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    the formula: (x-h)^2/a^2 - (y-k)^2/b^2 = 1 and for asymptotes: y-k = +- (b/a) (x-h) Is that right? if so I am trying to figure out to plug in and solve.

  4. amistre64
    • one year ago
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    its mostly right ... we should make the formula more general, since the y parts are not always subtracted. \[\pm\frac{(x-h)^2}{a^2}\mp\frac{(y-k)^2}{b^2}=1\]

  5. anonymous
    • one year ago
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    ok so now what?

  6. amistre64
    • one year ago
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    also, your asymptotes are actually simplified versions of b/a so there might be a common factor to adjust with so that: \[y-k=\pm\frac{bn}{an}(x-h)\]

  7. amistre64
    • one year ago
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    well, what do h and k represent to us?

  8. anonymous
    • one year ago
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    the center i believe

  9. amistre64
    • one year ago
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    good, center is the point (h,k) and what would our center be for this?

  10. anonymous
    • one year ago
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    at the origin

  11. amistre64
    • one year ago
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    so 0,0 ... we should also determine if the x and y parts in our formula are + or - how do we know which part is subtracted from the other?

  12. anonymous
    • one year ago
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    whether the hyperbola is horizontal or vertical? I think

  13. amistre64
    • one year ago
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    youre doing good ... what would you say is the correct orientation for this one then?

  14. anonymous
    • one year ago
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    |dw:1437662838662:dw| I think is what it would look like, so vertical?

  15. amistre64
    • one year ago
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    vertical, or rather .. opens in the y direction, to y + and x is - \[\frac{y^2}{(bn)^2}-\frac{x^2}{(an)^2}=1\] \[\frac{bn}{an}=\frac54\] so, we only need to determine bn, an ... what are your thoughts for it?

  16. amistre64
    • one year ago
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    |dw:1437662925392:dw|

  17. anonymous
    • one year ago
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    I think b has to be 5 and an will be 2 but I am not sure. Can you explain this part?

  18. amistre64
    • one year ago
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    the asymptotes are best seen from a box drawn between the vertices ... b and a represent the vertices of an ellipse/hyperbola. hypers are just an insideout ellipse

  19. amistre64
    • one year ago
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    |dw:1437663069638:dw|

  20. amistre64
    • one year ago
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    we know that b parts are 10 from the center ... not 5

  21. anonymous
    • one year ago
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    Ok got it. So is bn 10 and an 8

  22. amistre64
    • one year ago
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    bn = 5 ------ an = 4 but bn = 10 10 = 5(2) -------- an = 4(2)

  23. amistre64
    • one year ago
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    correct

  24. anonymous
    • one year ago
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    so now how to we find a and b from an and bn.

  25. anonymous
    • one year ago
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    or can we write the equation now.

  26. amistre64
    • one year ago
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    well, your original asymptote equation was misleading y = b/a x is not exactly correct, since the slope (b/a) is expressed in simplest form \[\frac{5}{4}=\frac{10}{8}\] but b is not 5 and a is not 4 .. see the confusion that arises?

  27. anonymous
    • one year ago
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    yes

  28. amistre64
    • one year ago
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    so, i rewrote it as bn and an .. we want bn and an :)

  29. anonymous
    • one year ago
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    ok but how do I get the final equation of the hyperbola?

  30. amistre64
    • one year ago
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    \[\frac{y^2}{(bn)^2}-\frac{x^2}{(an)^2}=1\] we know bn and an ...

  31. anonymous
    • one year ago
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    oh ok thats easy so we get the equation: \[\frac{ y^2 }{ 100 }-\frac{ x^2 }{ 64} = 1\]

  32. anonymous
    • one year ago
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    is that right?

  33. amistre64
    • one year ago
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    i believe its good, we can dbl chk with the wolf

  34. anonymous
    • one year ago
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    ok I will do it

  35. anonymous
    • one year ago
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    yep it's right! Thanks so much!

  36. amistre64
    • one year ago
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    youre welcome :)

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