Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

the asymptotes is 0/-20

im lost

well wait

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

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

okay. how do we do that?

okay.

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

lol okay, what next?

x=2?
y=4?

|dw:1439649900978:dw|

okay

for example, let's consider F1

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

yes

similarly for F2

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

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.