Given: \[ x \cos \theta = 3 \Rightarrow \cos \theta = \frac{3}{x} \Rightarrow \sec \theta = \frac{x}{3} \] Also, \[ 4 \tan \theta = y \Rightarrow \tan \theta = \frac{y}{4} \] We know: \[ \sec^2 \theta - \tan^2 \theta = 1 \Rightarrow \left(\frac{x}{3}\right)^2 - \left(\frac{y}{4}\right)^2 = 1 \Rightarrow \frac{x^2}{9} - \frac{y^2}{16} = 1 \] (Answer)