Question

Which are the foci for the hyperbola modeled by the equation quantity y minus 3 end quantity squared over 36 minus quantity x minus 1 end quantity squared over 13 equals

a) (1, 10) and (1, –4)
b) (3, 8) and (3, –6)
c) (1, 9) and (1, –3)
d) (–6, 0) and (6, 0)

Answer 1

The foci for the hyperbola modeled by the equation (y - 3)^2/36 - (x - 1)^2/13 = 1 are (1, 10) and (1, -4) (option a).

Step-by-step explanation:

The foci for the hyperbola modeled by the equation (y - 3)^2/36 - (x - 1)^2/13 = 1 are (1, 10) and (1, -4) (option a).

In the given equation, the coefficient of (x - h)^2 is positive, which means that the hyperbola is horizontally oriented. The formula for the foci of a horizontally oriented hyperbola is given by (h + c, k) and (h - c, k), where (h, k) is the center of the hyperbola and c = sqrt(a^2 + b^2). Substituting the values a = 3 and b = 13 into the formula, we get c = sqrt(3^2 + 13^2) = sqrt(178) ≈ 13.34. Therefore, the foci are (1 + 13.34, 3) ≈ (14.34, 3) and (1 - 13.34, 3) ≈ (-12.34, 3), which is closest to option a.