Stitz-Zeager_College_Algebra_e-book

We set w t and from e 2 1 1 3 get y 2 t 250

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Unformatted text preview: right-hand side equal to 1. 9y 2 − x2 − 6x 9y 2 − 1 x2 + 6x 9y 2 − x2 + 6x + 9 9y 2 − (x + 3)2 y 2 (x + 3)2 1− 1 9 = = = = 10 10 10 − 1(9) 1 =1 Now that this equation is in the standard form of Equation 7.7, we see that x − h is x + 3 so h = −3, and y − k is y so k = 0. Hence, our hyperbola is centered at (−3, 0). We find that a2 = 1 so a = 1, and b2 = 1 so b = 1 . This means that we move 1 unit to the left and right of the 9 3 center and 1/3 units up and down from the center to arrive at points on the guide rectangle. Since the x2 term is being subtracted from the y 2 term, we know the branches of the hyperbola open upwards and downwards. This means the transverse axis lies along the vertical line x = −3 and the conjugate axis lies along the x-axis. Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the vertices are 1 of a unit above and below (−3, 0) at 3 440 Hooked on Conics 1 −3, 1 and −3, − 3 . To find the foci, we need c = 3 √ 1 9 a2 + b2 = √ lie on the transverse axis, we move √ −3, − 10 3 10 3 √ +1 = 10 3. S...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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