What are the foci of the following graph?

- anonymous

What are the foci of the following graph?

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- anonymous

the asymptotes is 0/-20

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- anonymous

im lost

- anonymous

well wait

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## More answers

- anonymous

c2=0+20^2
c2=0+400
c=20

- 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

x2+y2/20=1

- Michele_Laino

from your graph, I see that:
\[\Large 2a = 10\]

- anonymous

x2/25+y2/20=1

- Michele_Laino

and:
\[\Large 2b = 4\]

- Michele_Laino

2b is the height of the dashed rectangle

- Michele_Laino

so:
\[\Large a = 5,\quad b = 2\]

- anonymous

x2/25+y2/4=1

- Michele_Laino

I think it is:
\[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]

- anonymous

if it was an ellipses it be +

- Michele_Laino

your graph is a hyperbola, not an ellipse

- anonymous

ik, i was jsut saying to make sure i know the difference
ellipse is +
hyperbola is -

- anonymous

right?

- Michele_Laino

ok! right!

- anonymous

okay. now i have another question that uses the same graph, can we just solve it here if thats okay?

- Michele_Laino

ok!

- anonymous

what is the equation of the graph?

- Michele_Laino

I wrote that equation before:
\[\Large \frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{4} = 1\]

- anonymous

its the saem for the foci and equation?

- anonymous

same*

- Michele_Laino

the equation for the foci is:
\[\Large {c^2} = {a^2} + {b^2} = 25 + 4 = ...\]

- Michele_Laino

what is c?

- anonymous

29

- Michele_Laino

more precisely we have:
\[\Large c = \pm \sqrt {29} \]

- anonymous

okay i see i see

- 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

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

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

okay. how do we do that?

- 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

okay.

- Michele_Laino

oopss... (1,2) not (2,1)

- anonymous

lol okay, what next?

- 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

x=2?
y=4?

- Michele_Laino

|dw:1439649900978:dw|

- anonymous

okay

- 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

namely I have applied the traslation:
\[\Large \left\{ \begin{gathered}
x = X + 1 \hfill \\
y = Y + 2 \hfill \\
\end{gathered} \right.\]

- Michele_Laino

for example, let's consider F1

- Michele_Laino

we have:
\[\Large X = - \sqrt {29} ,\quad Y = 0\]
am I right?

- anonymous

yes

- 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

similarly for F2

- anonymous

okay so which will the anwser be? or are we not there yet.

- 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

so -squ29 +1,2
and squ29 +1,2

- Michele_Laino

yes!

- anonymous

thank you! now cn we move on or do u want me to start a new post?

- Michele_Laino

is your question about the same exercise?

- anonymous

i have like 3 more about hyperbola and then im done

- Michele_Laino

if the hyperbola is different I think it is better if you open a new question

- anonymous

okay.

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