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Given the slants of the asymptotes are +/- 3/5, a center of (2,-1), and its transverse axis is x. Write a possible equation.

a) (x-2)^2/9 - (y+1)^2/25 = 1
b) (x-2)^2/25 - (y+1)^2/9 = 1
c) (x-2)^2/3 - (y+1)^2/5 = 1
d) (x-2)^2/5 - (y+1)^2/3 = 1

1 Answer

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Final answer:

The correct equation of the hyperbola with the given asymptotes and center is (x-2)^2/25 - (y+1)^2/9 = 1, where the terms correspond to the squared values of a and b that are related to the slope of the asymptotes.

Step-by-step explanation:

The student is asking for the equation of a hyperbola with given asymptotes and center. Since the slants of the asymptotes are ± 3/5, we are looking for a hyperbola with a difference of squares in its equation. The denominators of the squared terms denote the squares of a and b, which relate to the asymptotes. Since the transverse axis is the x-axis, the equation will be of the form √(x-h)^2/a^2 - (y-k)^2/b^2 = 1 where (h,k) is the center of the hyperbola. Considering the slopes (slants) of the asymptotes are ±(a/b), we can find that a=3 and b=5, so we substitute these values into the equation, along with the center (2,-1).

Thus, the correct equation for the hyperbola is option (b): (x-2)^2/25 - (y+1)^2/9 = 1 since this corresponds to a^2=25 and b^2=9 and adheres to the relationship of the asymptotic slopes given (± 3/5).

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