Question

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

Answer 1

@amistre64

Answer 2

well, what have you got so far?

Answer 3

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.

Answer 4

its mostly right ...

we should make the formula more general, since the y parts are not always subtracted.

\[\pm\frac{(x-h)^2}{a^2}\mp\frac{(y-k)^2}{b^2}=1\]

Answer 5

ok so now what?

Answer 6

also, your asymptotes are actually simplified versions of b/a so there might be a common factor to adjust with so that:
\[y-k=\pm\frac{bn}{an}(x-h)\]

Answer 7

well, what do h and k represent to us?

Answer 8

the center i believe

Answer 9

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

Answer 10

at the origin

Answer 11

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?

Answer 12

whether the hyperbola is horizontal or vertical? I think

Answer 13

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

Answer 14

I think is what it would look like, so vertical?

Answer 15

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?