The equation of the hyperbola with its transverse axis is parallel to x-axis, and its centre is (–2, 1) the length of transverse axis is 10 and eccentricity 6/5 is

$$\begin{gathered} Given centre=(-2,1),e=\frac{6}{5} and 2a=10\Rightarrow a=5 \\ Since transverse axis parallel to x-axis then \\ {b}^2=a^2(e^2-1) \\ =25\left(\frac{11}{25}\right) \\ =11 \\ So,Hyperbola is \\ \frac{{(x+2)}^2}{25} -\frac{{(y-1)}^2}{11} =1 \end{gathered}$$