anonymous
  • anonymous
What is the equation of the following graph?
Mathematics
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anonymous
  • anonymous
What is the equation of the following graph?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
that is a traslated hyperbola

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anonymous
  • anonymous
|dw:1439232785706:dw| I have to use this equation right?
Michele_Laino
  • Michele_Laino
the requested equation is an equation like this: \[\Large \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{a^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{b^2}}} = 1\]
Michele_Laino
  • Michele_Laino
from your graph we can see that: \[\Large 2a = 10\]
anonymous
  • anonymous
wait, how do you know A is 2?
anonymous
  • anonymous
isn't a "x" and b "y" ?
Michele_Laino
  • Michele_Laino
what is A?
Michele_Laino
  • Michele_Laino
referring to your drawing, we have: 2a=10, to understand that it is suffice to count the number of little squares
Michele_Laino
  • Michele_Laino
namely 2*a is the base of the dashed rectangle, whereas 2*b is the height of the dashed rectangle so 2*b=4
Michele_Laino
  • Michele_Laino
then we have: \[\Large \begin{gathered} \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{{5^2}}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{{{2^2}}} = 1 \hfill \\ \hfill \\ \frac{{{{\left( {x - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {y - {y_0}} \right)}^2}}}{4} = 1 \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now, we establish what are x_0 and y_0
anonymous
  • anonymous
im so confused ...
Michele_Laino
  • Michele_Laino
from your graph, I see that your hyperbola passes at point (6,2), so we have: \[\Large \frac{{{{\left( {6 - {x_0}} \right)}^2}}}{{25}} - \frac{{{{\left( {2 - {y_0}} \right)}^2}}}{4} = 1\]
anonymous
  • anonymous
but how did you get the 25?
Michele_Laino
  • Michele_Laino
here is why:
Michele_Laino
  • Michele_Laino
1 Attachment
anonymous
  • anonymous
Okay, I get that! Now, do I have to simplify that equation, or is it left just like that?
Michele_Laino
  • Michele_Laino
as I said before, we have to establish the value of x_0 and y_0
anonymous
  • anonymous
and how do we do that?
Michele_Laino
  • Michele_Laino
as we can see, the center of our dashed rectangle is located at point (1,2), right?
Michele_Laino
  • Michele_Laino
do you see that point?
anonymous
  • anonymous
right!
Michele_Laino
  • Michele_Laino
in other words our hyperbola has been traslated by 1 unit to right and by 2 units to up
anonymous
  • anonymous
okay, and what is the next step now?
Michele_Laino
  • Michele_Laino
that means x_0=1 and y_0=2, so, substituting them intoi my equation above, we get: \[\Large \frac{{{{\left( {x - 1} \right)}^2}}}{{25}} - \frac{{{{\left( {y - 2} \right)}^2}}}{4} = 1\]
Michele_Laino
  • Michele_Laino
into*
anonymous
  • anonymous
so where did they 6 & 2 go?
Michele_Laino
  • Michele_Laino
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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