The equation of the hyperbola whose foci are f1(-5,1), f2 (1,1) and eccentricity 3/2 is:

**Solution:**

Given foci F_1(-5, 1) and F_2(1, 1) , eccentricity e = \frac{3}{2} .

1. Center (h, k) is midway between the foci:
h = \frac{-5 + 1}{2} = -2, \quad k = 1
So, the center is at (-2, 1) .

2. Distance between the foci is 6, so c = 3 .

3. Eccentricity e = \frac{c}{a} gives a = \frac{c}{e} = \frac{3}{\frac{3}{2}} = 2 .

4. Using c^2 = a^2 + b^2 , we find b^2 = 9 - 4 = 5 .

5. Thus, the equation of the hyperbola with a horizontal transverse axis is:
\frac{(x + 2)^2}{4} - \frac{(y - 1)^2}{5} = 1

\boxed{\frac{(x + 2)^2}{4} - \frac{(y - 1)^2}{5} = 1}