Let $$\lambda x - 2y = \mu$$ be a tangent to the hyperbola $$a^2x^2 - y^2 = b^2$$. Then $$\left(\frac{\lambda}{a}\right)^2 - \left(\frac{\mu}{b}\right)^2$$ is equal to

The hyperbola is $$a^2x^2 - y^2 = b^2$$.

Rewrite in standard form by dividing by $$b^2$$:

$$ \frac{x^2}{(b/a)^2} - \frac{y^2}{b^2} = 1 $$

This is a hyperbola of the form $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$ where $$A = \frac{b}{a}$$ and $$B = b$$.

The tangent line is $$\lambda x - 2y = \mu$$, i.e., $$y = \frac{\lambda}{2}x - \frac{\mu}{2}$$.

This is of the form $$y = mx + c$$ where $$m = \frac{\lambda}{2}$$ and $$c = -\frac{\mu}{2}$$.

For a line $$y = mx + c$$ to be tangent to $$\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$$, the condition is:

$$ c^2 = A^2m^2 - B^2 $$

Substituting:

$$ \left(-\frac{\mu}{2}\right)^2 = \left(\frac{b}{a}\right)^2\left(\frac{\lambda}{2}\right)^2 - b^2 $$

$$ \frac{\mu^2}{4} = \frac{b^2\lambda^2}{4a^2} - b^2 $$

Multiply through by $$\frac{4}{b^2}$$:

$$ \frac{\mu^2}{b^2} = \frac{\lambda^2}{a^2} - 4 $$

Rearranging:

$$ \frac{\lambda^2}{a^2} - \frac{\mu^2}{b^2} = 4 $$

$$ \left(\frac{\lambda}{a}\right)^2 - \left(\frac{\mu}{b}\right)^2 = 4 $$

The answer is Option D: 4.