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anonymous

  • one year ago

What are the foci of the following graph?

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  1. anonymous
    • one year ago
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    the asymptotes is 0/-20

  2. anonymous
    • one year ago
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    im lost

  3. anonymous
    • one year ago
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    well wait

  4. anonymous
    • one year ago
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    c2=0+20^2 c2=0+400 c=20

  5. Michele_Laino
    • one year ago
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    the general formula, is: \[\Large {c^2} = {a^2} + {b^2}\] if, the hyperbola is represented as by the subsequent equation: \[\Large \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\]

  6. anonymous
    • one year ago
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    x2+y2/20=1

  7. Michele_Laino
    • one year ago
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    from your graph, I see that: \[\Large 2a = 10\]

  8. anonymous
    • one year ago
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    x2/25+y2/20=1

  9. Michele_Laino
    • one year ago
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    and: \[\Large 2b = 4\]

  10. Michele_Laino
    • one year ago
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    2b is the height of the dashed rectangle

  11. Michele_Laino
    • one year ago
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    so: \[\Large a = 5,\quad b = 2\]

  12. anonymous
    • one year ago
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    x2/25+y2/4=1

  13. Michele_Laino
    • one year ago
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    I think it is: \[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]

  14. anonymous
    • one year ago
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    if it was an ellipses it be +

  15. Michele_Laino
    • one year ago
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    your graph is a hyperbola, not an ellipse

  16. anonymous
    • one year ago
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    ik, i was jsut saying to make sure i know the difference ellipse is + hyperbola is -

  17. anonymous
    • one year ago
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    right?

  18. Michele_Laino
    • one year ago
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    ok! right!

  19. anonymous
    • one year ago
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    okay. now i have another question that uses the same graph, can we just solve it here if thats okay?

  20. Michele_Laino
    • one year ago
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    ok!

  21. anonymous
    • one year ago
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    what is the equation of the graph?

  22. Michele_Laino
    • one year ago
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    I wrote that equation before: \[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]

  23. anonymous
    • one year ago
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    its the saem for the foci and equation?

  24. anonymous
    • one year ago
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    same*

  25. Michele_Laino
    • one year ago
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    the equation for the foci is: \[\Large {c^2} = {a^2} + {b^2} = 25 + 4 = ...\]

  26. Michele_Laino
    • one year ago
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    what is c?

  27. anonymous
    • one year ago
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    29

  28. Michele_Laino
    • one year ago
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    more precisely we have: \[\Large c = \pm \sqrt {29} \]

  29. anonymous
    • one year ago
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    okay i see i see

  30. Michele_Laino
    • one year ago
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    so the requested foci are the subsequent points: \[\Large \begin{gathered} {F_1} = \left( { - \sqrt {29} ,0} \right) \hfill \\ {F_2} = \left( {\sqrt {29} ,0} \right) \hfill \\ \end{gathered} \]

  31. anonymous
    • one year ago
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    okay. thank you. now ill start a new post bc i hav another question, or can we keep going here? its all up to u if u want more medals

  32. Michele_Laino
    • one year ago
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    please we have to make a traslation, since that equation above is referring to a hyperbola centrered at the origin of our system of coordinates

  33. anonymous
    • one year ago
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    okay. how do we do that?

  34. Michele_Laino
    • one year ago
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    from you graph, we can note that the center of our hyperbola is located at the center of the dashed rectangle, which is the point (2,1)

  35. anonymous
    • one year ago
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    okay.

  36. Michele_Laino
    • one year ago
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    oopss... (1,2) not (2,1)

  37. anonymous
    • one year ago
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    lol okay, what next?

  38. Michele_Laino
    • one year ago
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    in other words, the equation of our traslation, is: \[\Large \left\{ \begin{gathered} x = X + 1 \hfill \\ y = Y + 2 \hfill \\ \end{gathered} \right.\] where X,Y is the new coordinates system located at (1,2)

  39. anonymous
    • one year ago
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    x=2? y=4?

  40. Michele_Laino
    • one year ago
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    |dw:1439649900978:dw|

  41. anonymous
    • one year ago
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    okay

  42. Michele_Laino
    • one year ago
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    so the coordinates of the foci referred to the X,Y system are: \[\Large \begin{gathered} {F_1} = \left( { - \sqrt {29} ,0} \right) \hfill \\ {F_2} = \left( {\sqrt {29} ,0} \right) \hfill \\ \end{gathered} \] whereas the coordinates of the foci referred to the x,y system are: \[\Large \begin{gathered} {F_1} = \left( { - \sqrt {29} + 1,2} \right) \hfill \\ {F_2} = \left( {\sqrt {29} + 1,2} \right) \hfill \\ \end{gathered} \]

  43. Michele_Laino
    • one year ago
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    namely I have applied the traslation: \[\Large \left\{ \begin{gathered} x = X + 1 \hfill \\ y = Y + 2 \hfill \\ \end{gathered} \right.\]

  44. Michele_Laino
    • one year ago
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    for example, let's consider F1

  45. Michele_Laino
    • one year ago
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    we have: \[\Large X = - \sqrt {29} ,\quad Y = 0\] am I right?

  46. anonymous
    • one year ago
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    yes

  47. Michele_Laino
    • one year ago
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    ok! now I apply my traslation, so I get: \[\Large \left\{ \begin{gathered} x = X + 1 = - \sqrt {29} + 1 \hfill \\ y = Y + 2 = 0 + 2 = 2 \hfill \\ \end{gathered} \right.\]

  48. Michele_Laino
    • one year ago
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    similarly for F2

  49. anonymous
    • one year ago
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    okay so which will the anwser be? or are we not there yet.

  50. Michele_Laino
    • one year ago
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    your answer is: "The coordinates of the foci, of our hyperbola, are: \[\Large \begin{gathered} {F_1} = \left( { - \sqrt {29} + 1,2} \right) \hfill \\ {F_2} = \left( {\sqrt {29} + 1,2} \right) \hfill \\ \end{gathered} \]" that's all!

  51. anonymous
    • one year ago
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    so -squ29 +1,2 and squ29 +1,2

  52. Michele_Laino
    • one year ago
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    yes!

  53. anonymous
    • one year ago
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    thank you! now cn we move on or do u want me to start a new post?

  54. Michele_Laino
    • one year ago
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    is your question about the same exercise?

  55. anonymous
    • one year ago
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    i have like 3 more about hyperbola and then im done

  56. Michele_Laino
    • one year ago
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    if the hyperbola is different I think it is better if you open a new question

  57. anonymous
    • one year ago
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    okay.

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