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anonymous
 one year ago
What are the foci of the following graph?
anonymous
 one year ago
What are the foci of the following graph?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the asymptotes is 0/20

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0c2=0+20^2 c2=0+400 c=20

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the general formula, is: \[\Large {c^2} = {a^2} + {b^2}\] if, the hyperbola is represented as by the subsequent equation: \[\Large \frac{{{x^2}}}{{{a^2}}}  \frac{{{y^2}}}{{{b^2}}} = 1\]

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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2and: \[\Large 2b = 4\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.22b is the height of the dashed rectangle

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so: \[\Large a = 5,\quad b = 2\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I think it is: \[\Large \frac{{{x^2}}}{{25}}  \frac{{{y^2}}}{4} = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if it was an ellipses it be +

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2your graph is a hyperbola, not an ellipse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ik, i was jsut saying to make sure i know the difference ellipse is + hyperbola is 

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay. now i have another question that uses the same graph, can we just solve it here if thats okay?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the equation of the graph?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I wrote that equation before: \[\Large \frac{{{x^2}}}{{25}}  \frac{{{y^2}}}{4} = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its the saem for the foci and equation?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2the equation for the foci is: \[\Large {c^2} = {a^2} + {b^2} = 25 + 4 = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2more precisely we have: \[\Large c = \pm \sqrt {29} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so the requested foci are the subsequent points: \[\Large \begin{gathered} {F_1} = \left( {  \sqrt {29} ,0} \right) \hfill \\ {F_2} = \left( {\sqrt {29} ,0} \right) \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay. 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2please we have to make a traslation, since that equation above is referring to a hyperbola centrered at the origin of our system of coordinates

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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2from 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)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oopss... (1,2) not (2,1)

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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2in 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)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2dw:1439649900978:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so 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: \[\Large \begin{gathered} {F_1} = \left( {  \sqrt {29} + 1,2} \right) \hfill \\ {F_2} = \left( {\sqrt {29} + 1,2} \right) \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2namely I have applied the traslation: \[\Large \left\{ \begin{gathered} x = X + 1 \hfill \\ y = Y + 2 \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for example, let's consider F1

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have: \[\Large X =  \sqrt {29} ,\quad Y = 0\] am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! now I apply my traslation, so I get: \[\Large \left\{ \begin{gathered} x = X + 1 =  \sqrt {29} + 1 \hfill \\ y = Y + 2 = 0 + 2 = 2 \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2similarly for F2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so which will the anwser be? or are we not there yet.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2your 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!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so squ29 +1,2 and squ29 +1,2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you! now cn we move on or do u want me to start a new post?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2is your question about the same exercise?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have like 3 more about hyperbola and then im done

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2if the hyperbola is different I think it is better if you open a new question
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