anonymous
  • anonymous
Anyone good with hyperbolas? Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1
Mathematics
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anonymous
  • anonymous
Anyone good with hyperbolas? Find the vertices and foci of the hyperbola with equation x squared over nine minus y squared over sixteen = 1
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\frac{ x^2 }{ 3^2 } -\frac{ y^2 }{ 4^2 } = 1\]
anonymous
  • anonymous
Yes this was the problem we were working on last night. (: @saseal
anonymous
  • anonymous
|dw:1439046272960:dw|

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anonymous
  • anonymous
\[c=\sqrt{a^2+b^2}\]
anonymous
  • anonymous
Ok, so would it be the SqRt of 9+16? so the SqRt of 27?
anonymous
  • anonymous
yea
anonymous
  • anonymous
um no
anonymous
  • anonymous
how did you get 9+16=27?
anonymous
  • anonymous
Isn't a^2 = 9 and b^2 = 16?
anonymous
  • anonymous
9+16=25
anonymous
  • anonymous
Oh my bad. Stupid arithmetic problem. Ok, so c = 5?
anonymous
  • anonymous
yea
anonymous
  • anonymous
its +/- 5 actually because hyperbola goes both ways, just like vertices
anonymous
  • anonymous
Oh ok!
anonymous
  • anonymous
So is that for foci or vertices? @saseal
anonymous
  • anonymous
c is for foci and a is your vertices
anonymous
  • anonymous
Ok...so would that make the foci: (+/-5, 0) And vertices: (+/-3, 0) @saseal
anonymous
  • anonymous
@saseal Could you help me with a couple more?
anonymous
  • anonymous
yea
anonymous
  • anonymous
and ok ill help you with the questions
anonymous
  • 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
  • anonymous
anonymous
  • anonymous
|dw:1439048329264:dw|this is for \[\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 } = 1\]
anonymous
  • anonymous
|dw:1439048416352:dw|this is for \[\frac{ y^2 }{ a^2 } - \frac{ x^2 }{ b^2 } = 1\]
anonymous
  • anonymous
Ok... @saseal
anonymous
  • 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
  • anonymous
Ok I get that
anonymous
  • anonymous
its actually determined by x^2 and y^2 not just x^2, got distacted dem parents
anonymous
  • anonymous
Oh ok, still makes sense. (:
anonymous
  • anonymous
So my answer would be correct? @saseal
anonymous
  • anonymous
yea
anonymous
  • 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
  • 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
  • anonymous
I graphed it and got the center to be ( -1, -5 ) @saseal
anonymous
  • anonymous
you should be able to get the center without even graphing it
anonymous
  • anonymous
you see the pattern there with the (x+1)^2 and the (y+5)^2 and the center?
anonymous
  • anonymous
Oh ok. Yes. -1, -5
anonymous
  • anonymous
I get the center is (-1, -5), so now what? @saseal
anonymous
  • anonymous
you know wheres your a and b?
anonymous
  • anonymous
My a is 16, b is 9. So would that make c = 5?
anonymous
  • anonymous
a^2 is 16 not a. a is 4, dont be mistaken.
anonymous
  • anonymous
Oh ok! So would the c still be 5?
anonymous
  • anonymous
yes
anonymous
  • anonymous
now where do you think the hyperbola will open out to? left/right or up/down
anonymous
  • anonymous
left/right
anonymous
  • anonymous
correct
anonymous
  • anonymous
|dw:1439050835472:dw|
anonymous
  • anonymous
where do you think the vertices are?
anonymous
  • anonymous
(-5, -5) and (3, 5) ?
anonymous
  • anonymous
there you have it
anonymous
  • anonymous
now we solve for c to get your foci
anonymous
  • anonymous
\[c=\sqrt{a^2+b^2}\]
anonymous
  • anonymous
c = +/- 5
anonymous
  • anonymous
correct, now put that into the coords just like you did for vertices
anonymous
  • anonymous
(-6, -5) and (4, -5) ?
anonymous
  • anonymous
yup
anonymous
  • anonymous
Ok, can you help me with one more? And check one?
anonymous
  • anonymous
sure
anonymous
  • anonymous
For: Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7). Would the answer be: y^2/4 - x^2/45 = 1
anonymous
  • anonymous
the x^2/45 part is wrong
anonymous
  • anonymous
you need to meddle a bit with the foci formula, its not so straightforward
anonymous
  • anonymous
Ok so how do I do that?
anonymous
  • anonymous
ok if c^2 = a^2 + b^2 then c^2 - a^2 = b^2
anonymous
  • anonymous
so for the x^2 would we subtract x^2/49?
anonymous
  • anonymous
no
anonymous
  • anonymous
foci is c which is 7 and your 4 is a^2
anonymous
  • anonymous
so 33?
anonymous
  • anonymous
\[7^2-2^2=?\]
anonymous
  • anonymous
so yea its 45 lol
anonymous
  • anonymous
i didnt square the stuff i got which is 6.708 thats why lol
anonymous
  • anonymous
Oh ok!
anonymous
  • anonymous
And the last problem on this lesson is: Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±five divided by six times x.
anonymous
  • anonymous
im used to putting em with the ^2 on
anonymous
  • anonymous
Oh ok
anonymous
  • anonymous
this is definitely a hyperbola opening up/down
anonymous
  • anonymous
|dw:1439051936537:dw|
anonymous
  • anonymous
I agree
anonymous
  • anonymous
\[\frac{ y^2 }{ 10^2 } - \frac{ x^2 }{ b^2 } = 1\]\[y=\pm \frac{ 5 }{ 6 }x\]
anonymous
  • anonymous
So...how could I get it into standard form equation?
anonymous
  • anonymous
and gradient is \[y=\frac{ b }{ a } x \]
anonymous
  • anonymous
\[\frac{ b }{ a } = \frac{ 5 }{ 6 } = \frac{ b }{ 10 }\]
anonymous
  • anonymous
Ok. So how can I solve for the b and a terms?
anonymous
  • anonymous
multiply through\[\frac{ 5 }{ 6 } = \frac{ 50 }{ 60 } = \frac{ 6b }{ 60 }\]\[6b=50\]\[b=8.333333333333\]
anonymous
  • anonymous
Ok that's what I thought.
anonymous
  • anonymous
lemme recheck
anonymous
  • anonymous
So would my answer be: y^2/64 - x^2/100 ? Because I just checked and b should equal 8...
anonymous
  • 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
  • anonymous
Ok so my final answer is y^2/64 - x^2/100 = 1?
anonymous
  • anonymous
no lemme redo the last part
anonymous
  • anonymous
cause isn't a^2 = 10 and b^2 = 8 ?
anonymous
  • anonymous
i used the wrong formula for the asymtote lol
anonymous
  • anonymous
lemme explain to ya
anonymous
  • anonymous
|dw:1439052898040:dw|
anonymous
  • anonymous
|dw:1439052978322:dw|
anonymous
  • anonymous
see the difference?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
One is going up/down, the other left/right
anonymous
  • anonymous
\[\frac{ a }{ b } = \frac{ 5 }{ 6 } = \frac{ 10 }{ b }\] so b is 12
anonymous
  • 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
  • anonymous
same question?
anonymous
  • anonymous
yes
anonymous
  • anonymous
he prolly flipped the gradient like i did
anonymous
  • anonymous
Oh ok. So would the answer be: y^2/144 - x^2/100 = 1?
anonymous
  • anonymous
\[\frac{ y^2 }{ 10^2 }-\frac{ x^2 }{ 12^2 } = 1\]
anonymous
  • anonymous
Oh ok! So simplified: y^2/100 - x^2/144 = 1
anonymous
  • anonymous
Correct? @saseal
anonymous
  • anonymous
yea
anonymous
  • anonymous
Ok thanks!
anonymous
  • anonymous
@saseal could you help me again? so sorry!!!
anonymous
  • anonymous
ok
anonymous
  • anonymous
maybe just open a new one lol this one is getting really long
anonymous
  • anonymous
Ok lol

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