The given equation: 9x^2 - 16y^2 + 36x + 32y + 164 = 0

Identifying the conic:
To identify the conic, we can look at the coefficients of x^2 and y^2. In this equation, the coefficient of x^2 is 9 and the coefficient of y^2 is -16. Since the coefficients have opposite signs, we can conclude that the given equation represents a hyperbola.

Standard form of a hyperbola:
The standard form of a hyperbola with the center (h, k), horizontal transverse axis, and focal length c is given by:
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1, if the hyperbola opens horizontally
or
[(y - k)^2 / a^2] - [(x - h)^2 / b^2] = 1, if the hyperbola opens vertically

Comparing the given equation with the standard form:
9x^2 - 16y^2 + 36x + 32y + 164 = 0
Dividing the equation by 164 to simplify:
(x^2/16) - (y^2/9) + (2x/9) + (y/5) + 1 = 0

Comparing the simplified equation with the standard form, we can determine the values of a, b, h, and k:
a^2 = 16 => a = 4
b^2 = 9 => b = 3
2h/9 = 2 => h = 9/2 = 4.5
k/5 = 0 => k = 0

Therefore, the center of the hyperbola is (4.5, 0), and the values of a and b are 4 and 3 respectively.

Identifying the vertex:
The vertex of a hyperbola is the point where the hyperbola intersects its transverse axis. Since the transverse axis is horizontal in this case, the vertex is given by (h ± a, k). Therefore, the vertex of the hyperbola is (4.5 ± 4, 0), which can be written as (8.5, 0) and (0.5, 0).

Identifying the focus:
The distance between the center and the focus is given by c, where c^2 = a^2 + b^2. In this case, c^2 = 16 + 9 = 25, so c = 5. Since the transverse axis is horizontal, the foci are located at (h ± c, k). Therefore, the foci of the hyperbola are (4.5 ± 5, 0), which can be written as (9.5, 0) and (-0.5, 0).

Identifying the equation of the directrix:
The distance between the center and the directrix is given by d, where d^2 = a^2 + b^2. In this case, d^2 = 16 + 9 =