Question

What are the foci of the following graph?

Answer 1

the asymptotes is 0/-20

Answer 2

im lost

Answer 3

well wait

Answer 4

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

Answer 5

the general formula, is:
\[\Large {c^2} = {a^2} + {b^2}\]
if, the hyperbola is represented as by the subsequent equation:
\[\Large \frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\]

Answer 6

x2+y2/20=1

Answer 7

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

Answer 8

x2/25+y2/20=1

Answer 9

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

Answer 10

2b is the height of the dashed rectangle

Answer 11

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

Answer 12

x2/25+y2/4=1

Answer 13

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

Answer 14

if it was an ellipses it be +

Answer 15

your graph is a hyperbola, not an ellipse

Answer 16

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

Answer 17

right?

Answer 18

ok! right!

Answer 19

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

Answer 20

ok!

Answer 21

what is the equation of the graph?

Answer 22

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

Answer 23

its the saem for the foci and equation?

Answer 24

same*

Answer 25

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

Answer 26

what is c?

Answer 27

29

Answer 28

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

Answer 29

okay i see i see

Answer 30

so the requested foci are the subsequent points:
\[\Large \begin{gathered}
{F_1} = \left( { - \sqrt {29} ,0} \right) \hfill \\
{F_2} = \left( {\sqrt {29} ,0} \right) \hfill \\
\end{gathered} \]

Answer 31

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

Answer 32

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

Answer 33

okay. how do we do that?

Answer 34

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)

Answer 35

okay.

Answer 36

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

Answer 37

lol okay, what next?

Answer 38

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)

Answer 39

x=2?
y=4?

Answer 40

okay

Answer 41

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:
\[\Large \begin{gathered}
{F_1} = \left( { - \sqrt {29} + 1,2} \right) \hfill \\
{F_2} = \left( {\sqrt {29} + 1,2} \right) \hfill \\
\end{gathered} \]

Answer 42

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

Answer 43

for example, let's consider F1

Answer 44

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

Answer 45

yes

Answer 46

ok! now I apply my traslation, so I get:
\[\Large \left\{ \begin{gathered}
x = X + 1 = - \sqrt {29} + 1 \hfill \\
y = Y + 2 = 0 + 2 = 2 \hfill \\
\end{gathered} \right.\]

Answer 47

similarly for F2

Answer 48

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

Answer 49

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!

Answer 50

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

Answer 51

yes!

Answer 52

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

Answer 53

is your question about the same exercise?

Answer 54

i have like 3 more about hyperbola and then im done

Answer 55

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

Answer 56

okay.