**Unformatted text preview: **he left and right of (2, 0) to arrive at (2 − 29, 0)
29
(approximately (−3.39, 0)) and (2 + 29, 0) (approximately (7.39, 0)). To determine the equations
of the asymptotes, recall that the asymptotes go through the center of the hyperbola, (2, 0), as well
b
as the corners of guide rectangle, so they have slopes of ± a = ± 5 . Using the point-slope equation
2
of a line, Equation 2.2, yields 438 Hooked on Conics 5
y = ± (x − 2) + 0,
2
so we get y = 5 x − 5 and y = − 5 x + 5. Putting it all together, we get
2
2
y
7
6
5
4
3
2
1
−3 −2 −1
−1 1 2 3 4 5 6 7 x −2
−3
−4
−5
−6
−7 Example 7.5.2. Find the equation of the hyperbola with asymptotes y = ±2x and vertices (±5, 0).
Solution. Plotting the data given to us, we have
y
5 −5 5 x −5 This graph not only tells us that the branches of the hyperbola open to the left and to the right,
it also tells us that the center is (0, 0). Hence, our standard form is
x2 y 2
− 2 = 1.
a2
b 7.5 Hyperbolas 439 Since the vertices are (±5, 0), we have a = 5 so a2 = 25. In order to determine b2 , we recall th...

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