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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ x^2 }{ 3^2 } \frac{ y^2 }{ 4^2 } = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes this was the problem we were working on last night. (: @saseal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439046272960:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[c=\sqrt{a^2+b^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so would it be the SqRt of 9+16? so the SqRt of 27?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you get 9+16=27?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Isn't a^2 = 9 and b^2 = 16?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh my bad. Stupid arithmetic problem. Ok, so c = 5?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its +/ 5 actually because hyperbola goes both ways, just like vertices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So is that for foci or vertices? @saseal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0c is for foci and a is your vertices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok...so would that make the foci: (+/5, 0) And vertices: (+/3, 0) @saseal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@saseal Could you help me with a couple more?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and ok ill help you with the questions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439048329264:dwthis is for \[\frac{ x^2 }{ a^2 }  \frac{ y^2 }{ b^2 } = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439048416352:dwthis is for \[\frac{ y^2 }{ a^2 }  \frac{ x^2 }{ b^2 } = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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\[\frac{ (x1)^2 }{ a^2 }  \frac{( y+1)^2 }{ b^2 } =1\] has a center of (1,1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its actually determined by x^2 and y^2 not just x^2, got distacted dem parents

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok, still makes sense. (:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So my answer would be correct? @saseal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ (x+1)^2 }{ 4^2 } + \frac{ (y+5)^2 }{ 3^2 } = 1\] where do you think is the center of the hyperbola?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I graphed it and got the center to be ( 1, 5 ) @saseal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you should be able to get the center without even graphing it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you see the pattern there with the (x+1)^2 and the (y+5)^2 and the center?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get the center is (1, 5), so now what? @saseal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you know wheres your a and b?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My a is 16, b is 9. So would that make c = 5?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a^2 is 16 not a. a is 4, dont be mistaken.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok! So would the c still be 5?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now where do you think the hyperbola will open out to? left/right or up/down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439050835472:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where do you think the vertices are?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(5, 5) and (3, 5) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we solve for c to get your foci

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[c=\sqrt{a^2+b^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct, now put that into the coords just like you did for vertices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(6, 5) and (4, 5) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok, can you help me with one more? And check one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For: 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
 one year ago
Best ResponseYou've already chosen the best response.0the x^2/45 part is wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you need to meddle a bit with the foci formula, its not so straightforward

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so how do I do that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok if c^2 = a^2 + b^2 then c^2  a^2 = b^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so for the x^2 would we subtract x^2/49?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0foci is c which is 7 and your 4 is a^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i didnt square the stuff i got which is 6.708 thats why lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And 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
 one year ago
Best ResponseYou've already chosen the best response.0im used to putting em with the ^2 on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is definitely a hyperbola opening up/down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439051936537:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ y^2 }{ 10^2 }  \frac{ x^2 }{ b^2 } = 1\]\[y=\pm \frac{ 5 }{ 6 }x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So...how could I get it into standard form equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and gradient is \[y=\frac{ b }{ a } x \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ b }{ a } = \frac{ 5 }{ 6 } = \frac{ b }{ 10 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. So how can I solve for the b and a terms?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0multiply through\[\frac{ 5 }{ 6 } = \frac{ 50 }{ 60 } = \frac{ 6b }{ 60 }\]\[6b=50\]\[b=8.333333333333\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok that's what I thought.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So would my answer be: y^2/64  x^2/100 ? Because I just checked and b should equal 8...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yup i made a mistake at the slope part b/a is for hyperbola opening left/right b/a is for opening up/down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so my final answer is y^2/64  x^2/100 = 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no lemme redo the last part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cause isn't a^2 = 10 and b^2 = 8 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i used the wrong formula for the asymtote lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439052898040:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439052978322:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One is going up/down, the other left/right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ a }{ b } = \frac{ 5 }{ 6 } = \frac{ 10 }{ b }\] so b is 12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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
 one year ago
Best ResponseYou've already chosen the best response.0he prolly flipped the gradient like i did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok. So would the answer be: y^2/144  x^2/100 = 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ y^2 }{ 10^2 }\frac{ x^2 }{ 12^2 } = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok! So simplified: y^2/100  x^2/144 = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@saseal could you help me again? so sorry!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe just open a new one lol this one is getting really long
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