Anyone good with hyperbolas?
Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1

- anonymous

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- anonymous

\[\frac{ x^2 }{ 3^2 } -\frac{ y^2 }{ 4^2 } = 1\]

- anonymous

Yes this was the problem we were working on last night. (: @saseal

- anonymous

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## More answers

- anonymous

\[c=\sqrt{a^2+b^2}\]

- anonymous

Ok, so would it be the SqRt of 9+16? so the SqRt of 27?

- anonymous

yea

- anonymous

um no

- anonymous

how did you get 9+16=27?

- anonymous

Isn't a^2 = 9 and b^2 = 16?

- anonymous

9+16=25

- anonymous

Oh my bad. Stupid arithmetic problem. Ok, so c = 5?

- anonymous

yea

- anonymous

its +/- 5 actually because hyperbola goes both ways, just like vertices

- anonymous

Oh ok!

- anonymous

So is that for foci or vertices? @saseal

- anonymous

c is for foci and a is your vertices

- anonymous

Ok...so would that make the foci:
(+/-5, 0)
And vertices:
(+/-3, 0)
@saseal

- anonymous

@saseal Could you help me with a couple more?

- anonymous

yea

- anonymous

and ok ill help you with the questions

- anonymous

So the answer is correct with the foci and vertices? How do I know where to put the 0? Meaning, the zero always goes in the y values?

- anonymous

@saseal

- anonymous

|dw:1439048329264:dw|this is for \[\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1\]

- anonymous

|dw:1439048416352:dw|this is for \[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]

- anonymous

Ok... @saseal

- anonymous

the center of your hyperbola is deternined by the x^2 like \[\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1\] has center at origin while\[\frac{ (x-1)^2 }{ a^2 } - \frac{( y+1)^2 }{ b^2 } =1\] has a center of (1,-1)

- anonymous

Ok I get that

- anonymous

its actually determined by x^2 and y^2 not just x^2, got distacted dem parents

- anonymous

Oh ok, still makes sense. (:

- anonymous

So my answer would be correct? @saseal

- anonymous

yea

- anonymous

Ok so my next question is:
Find the vertices and foci of the hyperbola with equation quantity x plus one squared divided by sixteen minus the quantity of y plus five squared divided by nine = 1

- anonymous

\[\frac{ (x+1)^2 }{ 4^2 } + \frac{ (y+5)^2 }{ 3^2 } = 1\] where do you think is the center of the hyperbola?

- anonymous

I graphed it and got the center to be ( -1, -5 ) @saseal

- anonymous

you should be able to get the center without even graphing it

- anonymous

you see the pattern there with the (x+1)^2 and the (y+5)^2 and the center?

- anonymous

Oh ok. Yes. -1, -5

- anonymous

I get the center is (-1, -5), so now what? @saseal

- anonymous

you know wheres your a and b?

- anonymous

My a is 16, b is 9. So would that make c = 5?

- anonymous

a^2 is 16 not a.
a is 4, dont be mistaken.

- anonymous

Oh ok! So would the c still be 5?

- anonymous

yes

- anonymous

now where do you think the hyperbola will open out to? left/right or up/down

- anonymous

left/right

- anonymous

correct

- anonymous

|dw:1439050835472:dw|

- anonymous

where do you think the vertices are?

- anonymous

(-5, -5) and (3, 5) ?

- anonymous

there you have it

- anonymous

now we solve for c to get your foci

- anonymous

\[c=\sqrt{a^2+b^2}\]

- anonymous

c = +/- 5

- anonymous

correct, now put that into the coords just like you did for vertices

- anonymous

(-6, -5) and (4, -5) ?

- anonymous

yup

- anonymous

Ok, can you help me with one more? And check one?

- anonymous

sure

- anonymous

For:
Find an equation in standard form for the hyperbola with vertices at (0, Â±2) and foci at (0, Â±7).
Would the answer be:
y^2/4 - x^2/45 = 1

- anonymous

the x^2/45 part is wrong

- anonymous

you need to meddle a bit with the foci formula, its not so straightforward

- anonymous

Ok so how do I do that?

- anonymous

ok if c^2 = a^2 + b^2
then c^2 - a^2 = b^2

- anonymous

so for the x^2 would we subtract x^2/49?

- anonymous

no

- anonymous

foci is c which is 7 and your 4 is a^2

- anonymous

so 33?

- anonymous

\[7^2-2^2=?\]

- anonymous

so yea its 45 lol

- anonymous

i didnt square the stuff i got which is 6.708 thats why lol

- anonymous

Oh ok!

- anonymous

And the last problem on this lesson is:
Find an equation in standard form for the hyperbola with vertices at (0, Â±10) and asymptotes at
y = Â±five divided by six times x.

- anonymous

im used to putting em with the ^2 on

- anonymous

Oh ok

- anonymous

this is definitely a hyperbola opening up/down

- anonymous

|dw:1439051936537:dw|

- anonymous

I agree

- anonymous

\[\frac{ y^2 }{ 10^2 } - \frac{ x^2 }{ b^2 } = 1\]\[y=\pm \frac{ 5 }{ 6 }x\]

- anonymous

So...how could I get it into standard form equation?

- anonymous

and gradient is \[y=\frac{ b }{ a } x \]

- anonymous

\[\frac{ b }{ a } = \frac{ 5 }{ 6 } = \frac{ b }{ 10 }\]

- anonymous

Ok. So how can I solve for the b and a terms?

- anonymous

multiply through\[\frac{ 5 }{ 6 } = \frac{ 50 }{ 60 } = \frac{ 6b }{ 60 }\]\[6b=50\]\[b=8.333333333333\]

- anonymous

Ok that's what I thought.

- anonymous

lemme recheck

- anonymous

So would my answer be:
y^2/64 - x^2/100 ? Because I just checked and b should equal 8...

- anonymous

yup i made a mistake at the slope part b/a is for hyperbola opening left/right
b/a is for opening up/down

- anonymous

Ok so my final answer is
y^2/64 - x^2/100 = 1?

- anonymous

no lemme redo the last part

- anonymous

cause isn't a^2 = 10 and b^2 = 8 ?

- anonymous

i used the wrong formula for the asymtote lol

- anonymous

lemme explain to ya

- anonymous

|dw:1439052898040:dw|

- anonymous

|dw:1439052978322:dw|

- anonymous

see the difference?

- anonymous

Yes

- anonymous

One is going up/down, the other left/right

- anonymous

\[\frac{ a }{ b } = \frac{ 5 }{ 6 } = \frac{ 10 }{ b }\] so b is 12

- anonymous

Oh ok. Well I was just looking on open study, to see what others did, and someone else listed the answer as:
y^2/64 - x^2/100 = 1
But I understand also what you're saying.

- anonymous

same question?

- anonymous

yes

- anonymous

he prolly flipped the gradient like i did

- anonymous

Oh ok. So would the answer be:
y^2/144 - x^2/100 = 1?

- anonymous

\[\frac{ y^2 }{ 10^2 }-\frac{ x^2 }{ 12^2 } = 1\]

- anonymous

Oh ok! So simplified:
y^2/100 - x^2/144 = 1

- anonymous

Correct? @saseal

- anonymous

yea

- anonymous

Ok thanks!

- anonymous

@saseal could you help me again? so sorry!!!

- anonymous

ok

- anonymous

maybe just open a new one lol this one is getting really long

- anonymous

Ok lol

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