\(\frac{dx}{dt} = -2a\sin 2t = -4a\sin t\cos t\) and \(\frac{dy}{dt} = 2\sqrt{2}a\cos t\).

\(m = \frac{dy}{dx} = \frac{2\sqrt{2}a\cos t}{-4a\sin t\cos t} = -\frac{1}{\sqrt{2}\sin t}\), so \(\sin t = -\frac{1}{\sqrt{2}m}\).

The point: \(y_0 = 2\sqrt{2}a\sin t = -\frac{2a}{m}\) and \(x_0 = a(1 - 2\sin^2 t) = a\left(1 - \frac{1}{m^2}\right)\).

Tangent: \(y = mx + (y_0 - mx_0) = mx - \frac{2a}{m} - am + \frac{a}{m} = mx - a\left(m + \frac{1}{m}\right)\)

Answer: (B)