SOLUTION: My Textbook gives a sample hyperbola equation (x^2 / 9) - (y^2 / 16) = 1 It then has y = ±(4/3)√x^2 - 9 I would like someone to explain to me step by step how my

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: My Textbook gives a sample hyperbola equation (x^2 / 9) - (y^2 / 16) = 1 It then has y = ±(4/3)√x^2 - 9 I would like someone to explain to me step by step how my       Log On


   

Question 273332: My Textbook gives a sample hyperbola equation
(x^2 / 9) - (y^2 / 16) = 1
It then has
y = ±(4/3)√x^2 - 9
I would like someone to explain to me step by step how my textbook got to this answer or is it that i am missing out on some detail.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
My Textbook gives a sample hyperbola equation
(x^2 / 9) - (y^2 / 16) = 1
It then has
y = ±(4/3)√x^2 - 9
=======================
(x^2/9) - (y^2/16) = 1
Multiply thru by 144 to get:
16x^2 - 9y^2 = 144
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Rearrange:
9y^2 = 16x^2-144
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Divide thru by 9 to get:
y^2 = (16/9)x^2 - 16
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y = +/- sqrt(16/9)x^2 - 16)
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Factor out the 16
---
y = +/-4sqrt((1/9)x^2 - 1)
y = +/-4sqrt[(1/9)(x^2 - 9)]
---
Bring out the (1/9)
---
y = +/-(4/3)sqrt(x^2 - 9)
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Cheers,
Stan H.