anonymous
  • anonymous
What are the foci of the following graph?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
the asymptotes is 0/-20
anonymous
  • anonymous
im lost
anonymous
  • anonymous
well wait

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anonymous
  • anonymous
c2=0+20^2 c2=0+400 c=20
Michele_Laino
  • Michele_Laino
the general formula, is: \[\Large {c^2} = {a^2} + {b^2}\] if, the hyperbola is represented as by the subsequent equation: \[\Large \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\]
anonymous
  • anonymous
x2+y2/20=1
Michele_Laino
  • Michele_Laino
from your graph, I see that: \[\Large 2a = 10\]
anonymous
  • anonymous
x2/25+y2/20=1
Michele_Laino
  • Michele_Laino
and: \[\Large 2b = 4\]
Michele_Laino
  • Michele_Laino
2b is the height of the dashed rectangle
Michele_Laino
  • Michele_Laino
so: \[\Large a = 5,\quad b = 2\]
anonymous
  • anonymous
x2/25+y2/4=1
Michele_Laino
  • Michele_Laino
I think it is: \[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]
anonymous
  • anonymous
if it was an ellipses it be +
Michele_Laino
  • Michele_Laino
your graph is a hyperbola, not an ellipse
anonymous
  • anonymous
ik, i was jsut saying to make sure i know the difference ellipse is + hyperbola is -
anonymous
  • anonymous
right?
Michele_Laino
  • Michele_Laino
ok! right!
anonymous
  • anonymous
okay. now i have another question that uses the same graph, can we just solve it here if thats okay?
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
what is the equation of the graph?
Michele_Laino
  • Michele_Laino
I wrote that equation before: \[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]
anonymous
  • anonymous
its the saem for the foci and equation?
anonymous
  • anonymous
same*
Michele_Laino
  • Michele_Laino
the equation for the foci is: \[\Large {c^2} = {a^2} + {b^2} = 25 + 4 = ...\]
Michele_Laino
  • Michele_Laino
what is c?
anonymous
  • anonymous
29
Michele_Laino
  • Michele_Laino
more precisely we have: \[\Large c = \pm \sqrt {29} \]
anonymous
  • anonymous
okay i see i see
Michele_Laino
  • Michele_Laino
so 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
  • anonymous
okay. 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
  • Michele_Laino
please 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
  • anonymous
okay. how do we do that?
Michele_Laino
  • Michele_Laino
from 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)
anonymous
  • anonymous
okay.
Michele_Laino
  • Michele_Laino
oopss... (1,2) not (2,1)
anonymous
  • anonymous
lol okay, what next?
Michele_Laino
  • Michele_Laino
in 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)
anonymous
  • anonymous
x=2? y=4?
Michele_Laino
  • Michele_Laino
|dw:1439649900978:dw|
anonymous
  • anonymous
okay
Michele_Laino
  • Michele_Laino
so 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
  • Michele_Laino
namely I have applied the traslation: \[\Large \left\{ \begin{gathered} x = X + 1 \hfill \\ y = Y + 2 \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
for example, let's consider F1
Michele_Laino
  • Michele_Laino
we have: \[\Large X = - \sqrt {29} ,\quad Y = 0\] am I right?
anonymous
  • anonymous
yes
Michele_Laino
  • Michele_Laino
ok! 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
  • Michele_Laino
similarly for F2
anonymous
  • anonymous
okay so which will the anwser be? or are we not there yet.
Michele_Laino
  • Michele_Laino
your 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
  • anonymous
so -squ29 +1,2 and squ29 +1,2
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
thank you! now cn we move on or do u want me to start a new post?
Michele_Laino
  • Michele_Laino
is your question about the same exercise?
anonymous
  • anonymous
i have like 3 more about hyperbola and then im done
Michele_Laino
  • Michele_Laino
if the hyperbola is different I think it is better if you open a new question
anonymous
  • anonymous
okay.

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