## 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.

1. anonymous

@amistre64

2. amistre64

well, what have you got so far?

3. 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.

4. amistre64

its mostly right ... we should make the formula more general, since the y parts are not always subtracted. $\pm\frac{(x-h)^2}{a^2}\mp\frac{(y-k)^2}{b^2}=1$

5. anonymous

ok so now what?

6. amistre64

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

7. amistre64

well, what do h and k represent to us?

8. anonymous

the center i believe

9. amistre64

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

10. anonymous

at the origin

11. 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?

12. anonymous

whether the hyperbola is horizontal or vertical? I think

13. amistre64

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

14. anonymous

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

15. 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?

16. amistre64

|dw:1437662925392:dw|

17. anonymous

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

18. 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

19. amistre64

|dw:1437663069638:dw|

20. amistre64

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

21. anonymous

Ok got it. So is bn 10 and an 8

22. amistre64

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

23. amistre64

correct

24. anonymous

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

25. anonymous

or can we write the equation now.

26. 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?

27. anonymous

yes

28. amistre64

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

29. anonymous

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

30. amistre64

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

31. anonymous

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

32. anonymous

is that right?

33. amistre64

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

34. anonymous

ok I will do it

35. anonymous

yep it's right! Thanks so much!

36. amistre64

youre welcome :)