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anonymous

  • one year ago

What is the equation of the following graph?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Michele_Laino

  3. Michele_Laino
    • one year ago
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    that is a traslated hyperbola

  4. anonymous
    • one year ago
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    |dw:1439232785706:dw| I have to use this equation right?

  5. Michele_Laino
    • one year ago
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    the requested equation is an equation like this: \[\Large \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{b^2}}} = 1\]

  6. Michele_Laino
    • one year ago
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    from your graph we can see that: \[\Large 2a = 10\]

  7. anonymous
    • one year ago
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    wait, how do you know A is 2?

  8. anonymous
    • one year ago
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    isn't a "x" and b "y" ?

  9. Michele_Laino
    • one year ago
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    what is A?

  10. Michele_Laino
    • one year ago
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    referring to your drawing, we have: 2a=10, to understand that it is suffice to count the number of little squares

  11. Michele_Laino
    • one year ago
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    namely 2*a is the base of the dashed rectangle, whereas 2*b is the height of the dashed rectangle so 2*b=4

  12. Michele_Laino
    • one year ago
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    then we have: \[\Large \begin{gathered} \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{5^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{2^2}}} = 1 \hfill \\ \hfill \\ \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{4} = 1 \hfill \\ \end{gathered} \]

  13. Michele_Laino
    • one year ago
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    now, we establish what are x_0 and y_0

  14. anonymous
    • one year ago
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    im so confused ...

  15. Michele_Laino
    • one year ago
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    from your graph, I see that your hyperbola passes at point (6,2), so we have: \[\Large \frac{{{{\left( {6 - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {2 - {y_0}} \right)}^2}}}{4} = 1\]

  16. anonymous
    • one year ago
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    but how did you get the 25?

  17. Michele_Laino
    • one year ago
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    here is why:

  18. Michele_Laino
    • one year ago
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  19. anonymous
    • one year ago
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    Okay, I get that! Now, do I have to simplify that equation, or is it left just like that?

  20. Michele_Laino
    • one year ago
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    as I said before, we have to establish the value of x_0 and y_0

  21. anonymous
    • one year ago
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    and how do we do that?

  22. Michele_Laino
    • one year ago
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    as we can see, the center of our dashed rectangle is located at point (1,2), right?

  23. Michele_Laino
    • one year ago
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    do you see that point?

  24. anonymous
    • one year ago
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    right!

  25. Michele_Laino
    • one year ago
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    in other words our hyperbola has been traslated by 1 unit to right and by 2 units to up

  26. anonymous
    • one year ago
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    okay, and what is the next step now?

  27. Michele_Laino
    • one year ago
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    that means x_0=1 and y_0=2, so, substituting them intoi my equation above, we get: \[\Large \frac{{{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1\]

  28. Michele_Laino
    • one year ago
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    into*

  29. anonymous
    • one year ago
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    so where did they 6 & 2 go?

  30. Michele_Laino
    • one year ago
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    x=6 and y=2 are the coordinates of a point which belongs to our hyperbola, infact, if we substitute x=6 and y=2 into my last formula, we get: \[\large {\text{left side}} = \frac{{{{\left( {6 - 1} \right)}^2}}}{{25}} - \frac{{{{\left( {2 - 2} \right)}^2}}}{4} = \frac{{25}}{{25}} - 0 = 1 = {\text{right side}}\] namely I got an identity

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