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anonymous
 3 years ago
hyperbola help
anonymous
 3 years ago
hyperbola help

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351156478191:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how can you find b in \[\frac{x^2}{3^2}  \frac{y^2}{b^2} =1 \] v?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The foci are (plus/minus a*e,0) where the eccentricity e = sqrt(1+ (b^2/a^2))

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0? what do i do what tiwtha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0foci, 4 = 4/e < e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can do intercept way??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is "intercept way"? What intercept?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the intercept only works in an ellipse i thought you drew an ellipse

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0S/B foci, 4 = 3/e < e = 1 = sqrt( 1 + b^2/3^2)) Solve for b^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can u do this? It is not difficult.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think you can find by subbing in a point.. but i dunno which point

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah but i dont think i can remember ..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you find the corodinate of point directly above foci??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0foci, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah but i dont wanna remember that if in test _

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0they are related by a^2+b^2=c^2 as well right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0U can say c^2 = a^2 b^2 (same thing)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually, 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
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