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anonymous
 one year ago
What is the equation of the following graph?
anonymous
 one year ago
What is the equation of the following graph?

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that is a traslated hyperbola

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439232785706:dw I have to use this equation right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1from your graph we can see that: \[\Large 2a = 10\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, how do you know A is 2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0isn't a "x" and b "y" ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1referring to your drawing, we have: 2a=10, to understand that it is suffice to count the number of little squares

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1namely 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
 one year ago
Best ResponseYou've already chosen the best response.1then 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
 one year ago
Best ResponseYou've already chosen the best response.1now, we establish what are x_0 and y_0

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1from 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
 one year ago
Best ResponseYou've already chosen the best response.0but how did you get the 25?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I get that! Now, do I have to simplify that equation, or is it left just like that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1as I said before, we have to establish the value of x_0 and y_0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and how do we do that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1as we can see, the center of our dashed rectangle is located at point (1,2), right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1do you see that point?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in other words our hyperbola has been traslated by 1 unit to right and by 2 units to up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, and what is the next step now?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that means x_0=1 and y_0=2, so, substituting them intoi my equation above, we get: \[\Large \frac{{{{\left( {x  1} \right)}^2}}}{{25}}  \frac{{{{\left( {y  2} \right)}^2}}}{4} = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so where did they 6 & 2 go?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1x=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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