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

- anonymous

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- anonymous

@amistre64

- amistre64

well, what have you got so far?

- 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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## More answers

- 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

ok so now what?

- 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

well, what do h and k represent to us?

- anonymous

the center i believe

- amistre64

good, center is the point (h,k) and what would our center be for this?

- anonymous

at the origin

- 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

whether the hyperbola is horizontal or vertical? I think

- amistre64

youre doing good ... what would you say is the correct orientation for this one then?

- anonymous

|dw:1437662838662:dw|
I think is what it would look like, so vertical?

- 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

|dw:1437662925392:dw|

- anonymous

I think b has to be 5 and an will be 2 but I am not sure. Can you explain this part?

- 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

|dw:1437663069638:dw|

- amistre64

we know that b parts are 10 from the center ... not 5

- anonymous

Ok got it. So is bn 10 and an 8

- amistre64

bn = 5
------
an = 4
but bn = 10
10 = 5(2)
--------
an = 4(2)

- amistre64

correct

- anonymous

so now how to we find a and b from an and bn.

- anonymous

or can we write the equation now.

- 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

yes

- amistre64

so, i rewrote it as bn and an .. we want bn and an :)

- anonymous

ok but how do I get the final equation of the hyperbola?

- amistre64

\[\frac{y^2}{(bn)^2}-\frac{x^2}{(an)^2}=1\]
we know bn and an ...

- anonymous

oh ok thats easy so we get the equation:
\[\frac{ y^2 }{ 100 }-\frac{ x^2 }{ 64} = 1\]

- anonymous

is that right?

- amistre64

i believe its good, we can dbl chk with the wolf

- anonymous

ok I will do it

- anonymous

yep it's right! Thanks so much!

- amistre64

youre welcome :)

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