Stitz-Zeager_College_Algebra_e-book

For example when solving 2 above it is clear after we

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Unformatted text preview: ence, our hyperbola is centered at (2, 0). We see that a2 = 4 so a = 2, and b2 = 25 so b = 5. This means we move 2 units to the left and right of the center and 5 units up and down from the center to arrive at points on the guide rectangle. The asymptotes pass through the center of the hyperbola as well as the corners of the rectangle. This yields the following set up. y 7 6 5 4 3 2 1 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6 −7 Since the y 2 term is being subtracted from the x2 term, we know that the branches of the hyperbola open to the left and right. This means that the transverse axis lies along the x-axis. Hence, the conjugate axis lies along the vertical line x = 2. Since the vertices of the hyperbola are where the hyperbola intersects the transverse axis, we get that the√ vertices are√ units to the left and right of 2 √ (2, 0) at (0, 0) and (4, 0). To ﬁnd the foci, we need c = a2 + b2 = 4 + 25 = 29. Since the foci √ √ lie on the transverse axis, we move √ units to t...
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