Given:
The vertices are at (-2, 0) and (2, 0). The conjugate axis's length is 6.
The two vertices have the same y coordinate, therefore, the hyperbola is the horizontal transverse axis type. The standard equation for the horizontal transverse axis type is
\(\frac{(x - h)^2}{a^2} - ((y - k)^2)/(b^2) = 1\)
We know that the general form for the vertices of this type is:
\((h + a, k) and (h - a, k)\)
comparing with vertices
\(k = 0\)

\(h + a = 2\)

\(h - a = -2\) Solve \(h = 0, a = 2\) The length of the conjugate axis is equal to 2b \(2b = 6\)

\(b = 3\) Standard form \(\frac{(x - 0)^2}{2^2} + \frac{(y - 0^2)}{3^2} = 1\)

\(\frac{x^2}{4} + \frac{y^2}{9} = 1\) General form \(9x^2 + 4y^2 = 36\)

\(9x^2 + 4y^2 - 36 = 0\)

\(h + a = 2\)

\(h - a = -2\) Solve \(h = 0, a = 2\) The length of the conjugate axis is equal to 2b \(2b = 6\)

\(b = 3\) Standard form \(\frac{(x - 0)^2}{2^2} + \frac{(y - 0^2)}{3^2} = 1\)

\(\frac{x^2}{4} + \frac{y^2}{9} = 1\) General form \(9x^2 + 4y^2 = 36\)

\(9x^2 + 4y^2 - 36 = 0\)