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anonymous
 one year ago
Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ±(5/4) x.
anonymous
 one year ago
Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±(5/4) x.

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.2well, what have you got so far?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the formula: (xh)^2/a^2  (yk)^2/b^2 = 1 and for asymptotes: yk = + (b/a) (xh) Is that right? if so I am trying to figure out to plug in and solve.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2its mostly right ... we should make the formula more general, since the y parts are not always subtracted. \[\pm\frac{(xh)^2}{a^2}\mp\frac{(yk)^2}{b^2}=1\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2also, your asymptotes are actually simplified versions of b/a so there might be a common factor to adjust with so that: \[yk=\pm\frac{bn}{an}(xh)\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2well, what do h and k represent to us?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the center i believe

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2good, center is the point (h,k) and what would our center be for this?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2so 0,0 ... we should also determine if the x and y parts in our formula are + or  how do we know which part is subtracted from the other?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whether the hyperbola is horizontal or vertical? I think

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2youre doing good ... what would you say is the correct orientation for this one then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437662838662:dw I think is what it would look like, so vertical?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2vertical, or rather .. opens in the y direction, to y + and x is  \[\frac{y^2}{(bn)^2}\frac{x^2}{(an)^2}=1\] \[\frac{bn}{an}=\frac54\] so, we only need to determine bn, an ... what are your thoughts for it?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437662925392:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think b has to be 5 and an will be 2 but I am not sure. Can you explain this part?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2the asymptotes are best seen from a box drawn between the vertices ... b and a represent the vertices of an ellipse/hyperbola. hypers are just an insideout ellipse

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437663069638:dw

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2we know that b parts are 10 from the center ... not 5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok got it. So is bn 10 and an 8

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2bn = 5  an = 4 but bn = 10 10 = 5(2)  an = 4(2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now how to we find a and b from an and bn.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or can we write the equation now.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2well, your original asymptote equation was misleading y = b/a x is not exactly correct, since the slope (b/a) is expressed in simplest form \[\frac{5}{4}=\frac{10}{8}\] but b is not 5 and a is not 4 .. see the confusion that arises?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2so, i rewrote it as bn and an .. we want bn and an :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok but how do I get the final equation of the hyperbola?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{y^2}{(bn)^2}\frac{x^2}{(an)^2}=1\] we know bn and an ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok thats easy so we get the equation: \[\frac{ y^2 }{ 100 }\frac{ x^2 }{ 64} = 1\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.2i believe its good, we can dbl chk with the wolf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep it's right! Thanks so much!
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