What is the equation of the following graph?

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- anonymous

What is the equation of the following graph?

- jamiebookeater

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- anonymous

##### 1 Attachment

- anonymous

- Michele_Laino

that is a traslated hyperbola

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## More answers

- anonymous

|dw:1439232785706:dw| I have to use this equation right?

- 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

from your graph we can see that:
\[\Large 2a = 10\]

- anonymous

wait, how do you know A is 2?

- anonymous

isn't a "x" and b "y" ?

- Michele_Laino

what is A?

- 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

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

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

now, we establish what are x_0 and y_0

- anonymous

im so confused ...

- 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

but how did you get the 25?

- Michele_Laino

here is why:

- Michele_Laino

##### 1 Attachment

- anonymous

Okay, I get that! Now, do I have to simplify that equation, or is it left just like that?

- Michele_Laino

as I said before, we have to establish the value of x_0 and y_0

- anonymous

and how do we do that?

- Michele_Laino

as we can see, the center of our dashed rectangle is located at point (1,2), right?

- Michele_Laino

do you see that point?

- anonymous

right!

- Michele_Laino

in other words our hyperbola has been traslated by 1 unit to right and by 2 units to up

- anonymous

okay, and what is the next step now?

- 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

into*

- anonymous

so where did they 6 & 2 go?

- 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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