anonymous
  • anonymous
Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±(5/4) x.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
amistre64
  • amistre64
well, what have you got so far?
anonymous
  • anonymous
the formula: (x-h)^2/a^2 - (y-k)^2/b^2 = 1 and for asymptotes: y-k = +- (b/a) (x-h) Is that right? if so I am trying to figure out to plug in and solve.

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amistre64
  • amistre64
its mostly right ... we should make the formula more general, since the y parts are not always subtracted. \[\pm\frac{(x-h)^2}{a^2}\mp\frac{(y-k)^2}{b^2}=1\]
anonymous
  • anonymous
ok so now what?
amistre64
  • amistre64
also, your asymptotes are actually simplified versions of b/a so there might be a common factor to adjust with so that: \[y-k=\pm\frac{bn}{an}(x-h)\]
amistre64
  • amistre64
well, what do h and k represent to us?
anonymous
  • anonymous
the center i believe
amistre64
  • amistre64
good, center is the point (h,k) and what would our center be for this?
anonymous
  • anonymous
at the origin
amistre64
  • amistre64
so 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
  • anonymous
whether the hyperbola is horizontal or vertical? I think
amistre64
  • amistre64
youre doing good ... what would you say is the correct orientation for this one then?
anonymous
  • anonymous
|dw:1437662838662:dw| I think is what it would look like, so vertical?
amistre64
  • amistre64
vertical, 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
  • amistre64
|dw:1437662925392:dw|
anonymous
  • anonymous
I think b has to be 5 and an will be 2 but I am not sure. Can you explain this part?
amistre64
  • amistre64
the 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
  • amistre64
|dw:1437663069638:dw|
amistre64
  • amistre64
we know that b parts are 10 from the center ... not 5
anonymous
  • anonymous
Ok got it. So is bn 10 and an 8
amistre64
  • amistre64
bn = 5 ------ an = 4 but bn = 10 10 = 5(2) -------- an = 4(2)
amistre64
  • amistre64
correct
anonymous
  • anonymous
so now how to we find a and b from an and bn.
anonymous
  • anonymous
or can we write the equation now.
amistre64
  • amistre64
well, 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?
anonymous
  • anonymous
yes
amistre64
  • amistre64
so, i rewrote it as bn and an .. we want bn and an :)
anonymous
  • anonymous
ok but how do I get the final equation of the hyperbola?
amistre64
  • amistre64
\[\frac{y^2}{(bn)^2}-\frac{x^2}{(an)^2}=1\] we know bn and an ...
anonymous
  • anonymous
oh ok thats easy so we get the equation: \[\frac{ y^2 }{ 100 }-\frac{ x^2 }{ 64} = 1\]
anonymous
  • anonymous
is that right?
amistre64
  • amistre64
i believe its good, we can dbl chk with the wolf
anonymous
  • anonymous
ok I will do it
anonymous
  • anonymous
yep it's right! Thanks so much!
amistre64
  • amistre64
youre welcome :)

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