powerangers69
hyperbola help



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powerangers69
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dw:1351156478191:dw

powerangers69
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how can you find b in \[\frac{x^2}{3^2}  \frac{y^2}{b^2} =1 \] v?

Jonask
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intercept

soty2013
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@Jonask is correct

powerangers69
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do i sub (4,0) in ?

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

powerangers69
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? what do i do what tiwtha

powerangers69
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whats interecept??

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

powerangers69
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can do intercept way??

estudier
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What is "intercept way"? What intercept?

powerangers69
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@Jonask ??

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

Jonask
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@estudier is correct

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

estudier
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Can u do this? It is not difficult.

powerangers69
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i think you can find by subbing in a point.. but i dunno which point

powerangers69
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yeah but i dont think i can remember ..

powerangers69
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can you find the corodinate of point directly above foci??

estudier
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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

powerangers69
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yeah but i dont wanna remember that if in test _

powerangers69
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they are related by a^2+b^2=c^2 as well right

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