## powerangers69 Group Title hyperbola help one year ago one year ago

1. powerangers69

|dw:1351156478191:dw|

2. powerangers69

how can you find b in $\frac{x^2}{3^2} - \frac{y^2}{b^2} =1$ v?

intercept

4. soty2013

5. powerangers69

do i sub (4,0) in ?

6. estudier

The foci are (plus/minus a*e,0) where the eccentricity e = sqrt(1+ (b^2/a^2))

7. powerangers69

? what do i do what tiwtha

8. powerangers69

whats interecept??

9. estudier

foci, 4 = 4/e -< e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2

10. powerangers69

can do intercept way??

11. estudier

What is "intercept way"? What intercept?

12. powerangers69

the intercept only works in an ellipse i thought you drew an ellipse

@estudier is correct

15. estudier

S/B foci, 4 = 3/e -< e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2

16. estudier

Can u do this? It is not difficult.

17. powerangers69

i think you can find by subbing in a point.. but i dunno which point

18. powerangers69

yeah but i dont think i can remember ..

19. powerangers69

can you find the corodinate of point directly above foci??

20. estudier

foci, 4 = 3/e -> e = 4/3 = sqrt( 1 + b^2/3^2)) Solve for b^2 So 4/3 = sqt(1 + b^2/3^2) -> b^2 = 12/9

21. powerangers69

yeah but i dont wanna remember that if in test -_-

22. powerangers69

they are related by a^2+b^2=c^2 as well right

23. estudier

U can say c^2 = a^2 -b^2 (same thing)

24. estudier

Actually, I made mistake in figures: 4/3 = sqrt(1 +b^2/3^2) -> 16/9 = 1 + b^2/3^2 -> 3^2 * 16/9 = 3^2 + b^2 -> b^2 = 3^2 (16/9 -1) = 3^2 *7/9 = 7 You can check it here http://www.wolframalpha.com/input/?i=%28x^2%2F3^2%29++++-+y^2%2F7+%3D+1