What are the foci of the following graph?

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What are the foci of the following graph?

Mathematics
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the asymptotes is 0/-20
im lost
well wait

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Other answers:

c2=0+20^2 c2=0+400 c=20
the general formula, is: \[\Large {c^2} = {a^2} + {b^2}\] if, the hyperbola is represented as by the subsequent equation: \[\Large \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\]
x2+y2/20=1
from your graph, I see that: \[\Large 2a = 10\]
x2/25+y2/20=1
and: \[\Large 2b = 4\]
2b is the height of the dashed rectangle
so: \[\Large a = 5,\quad b = 2\]
x2/25+y2/4=1
I think it is: \[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]
if it was an ellipses it be +
your graph is a hyperbola, not an ellipse
ik, i was jsut saying to make sure i know the difference ellipse is + hyperbola is -
right?
ok! right!
okay. now i have another question that uses the same graph, can we just solve it here if thats okay?
ok!
what is the equation of the graph?
I wrote that equation before: \[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]
its the saem for the foci and equation?
same*
the equation for the foci is: \[\Large {c^2} = {a^2} + {b^2} = 25 + 4 = ...\]
what is c?
29
more precisely we have: \[\Large c = \pm \sqrt {29} \]
okay i see i see
so the requested foci are the subsequent points: \[\Large \begin{gathered} {F_1} = \left( { - \sqrt {29} ,0} \right) \hfill \\ {F_2} = \left( {\sqrt {29} ,0} \right) \hfill \\ \end{gathered} \]
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
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
okay. how do we do that?
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)
okay.
oopss... (1,2) not (2,1)
lol okay, what next?
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)
x=2? y=4?
|dw:1439649900978:dw|
okay
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: \[\Large \begin{gathered} {F_1} = \left( { - \sqrt {29} + 1,2} \right) \hfill \\ {F_2} = \left( {\sqrt {29} + 1,2} \right) \hfill \\ \end{gathered} \]
namely I have applied the traslation: \[\Large \left\{ \begin{gathered} x = X + 1 \hfill \\ y = Y + 2 \hfill \\ \end{gathered} \right.\]
for example, let's consider F1
we have: \[\Large X = - \sqrt {29} ,\quad Y = 0\] am I right?
yes
ok! now I apply my traslation, so I get: \[\Large \left\{ \begin{gathered} x = X + 1 = - \sqrt {29} + 1 \hfill \\ y = Y + 2 = 0 + 2 = 2 \hfill \\ \end{gathered} \right.\]
similarly for F2
okay so which will the anwser be? or are we not there yet.
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!
so -squ29 +1,2 and squ29 +1,2
yes!
thank you! now cn we move on or do u want me to start a new post?
is your question about the same exercise?
i have like 3 more about hyperbola and then im done
if the hyperbola is different I think it is better if you open a new question
okay.

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