Step-by-step Solution:

CONCEPT:
If \( x = f(t) \), \( y = g(t) \), where \( t \) is a parameter, then \[ \frac{dy}{dx} = \frac{g'(t)}{f'(t)} \] The slope of the tangent to a curve \( y = f(x) \) at any point is given by: \[ m = \frac{dy}{dx} \] The equation of a tangent to a curve \( y = f(x) \) is given by: \[ y - y_1 = m (x - x_1) \] CALCULATION:
The parametric equations of the curve are: \[ x = a \cos 2t, \quad y = 2\sqrt{2} \, a \sin t \] Since the slope of the tangent to a curve \( y = f(x) \) at any point is given by \( m = \frac{dy}{dx} \), we calculate: \[ m = \frac{dy}{dx} = \frac{-\sqrt{2}}{2 \times \sin t} \] Solving for \( \sin t \): \[ \sin t = \frac{-\sqrt{2}}{2m} \] Using the given equations: \[ x = a \cos 2t, \quad y = 2\sqrt{2} \, a \sin t \] Substituting \( \sin t \) in \( x \) and \( y \), we get: \[ y = \frac{-2a}{m}, \quad x = a \left( 1 - \frac{1}{m^2} \right) \] Since the equation of a tangent to the curve \( y = f(x) \) is: \[ y - y_1 = m (x - x_1) \] We use the point \( \left( a \left( 1 - \frac{1}{m^2} \right), \frac{-2a}{m} \right) \) to get: \[ y + \frac{2a}{m} = m \left[ x - a \left( 1 - \frac{1}{m^2} \right) \right] \] Simplifying further, we obtain: \[ y = mx - a \left( m + \frac{1}{m} \right) \] Thus, option B is the correct answer.