We are tasked to find the envelope of the system of lines defined by:

$\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1,$ where $\theta$ is the parameter.

Solution:

1. The equation of the system of lines can be rewritten as: $F(x, y, \theta) = \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta - 1 = 0.$

2. For the envelope of the system, the condition for the envelope is: $\frac{\partial F}{\partial \theta} = 0.$ Compute $\frac{\partial F}{\partial \theta}$: $\frac{\partial F}{\partial \theta} = -\frac{x}{a} \sin \theta + \frac{y}{b} \cos \theta.$

3. Solve $\frac{\partial F}{\partial \theta} = 0$: $-\frac{x}{a} \sin \theta + \frac{y}{b} \cos \theta = 0.$ Rearrange: $\frac{\sin \theta}{\cos \theta} = \frac{y/a}{x/b}.$ Simplify: $\tan \theta = \frac{by}{ax}.$

4. Substitute $\tan \theta = \frac{by}{ax}$ into the original equation: From $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we can find $\sin \theta$ and $\cos \theta$: $\sin \theta = \frac{by}{\sqrt{(ax)^2 + (by)^2}}, \quad \cos \theta = \frac{ax}{\sqrt{(ax)^2 + (by)^2}}.$ Substitute these into the original equation: $\frac{x}{a} \cdot \frac{ax}{\sqrt{(ax)^2 + (by)^2}} + \frac{y}{b} \cdot \frac{by}{\sqrt{(ax)^2 + (by)^2}} = 1.$ Simplify: $\frac{x^2}{\sqrt{(ax)^2 + (by)^2}} + \frac{y^2}{\sqrt{(ax)^2 + (by)^2}} = 1.$ Combine terms: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

Conclusion:

The envelope of the system of lines is an ellipse given by: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

Let me know if you'd like further clarification or detailed steps!

Follow-Up Questions:

1. What is the geometric meaning of an envelope in general?
2. How is the envelope different from a general family of curves?
3. Can this technique be extended to other parameterized line systems?
4. What changes if the parameter $\theta$ is replaced with another variable?
5. How can this method be applied in physics or engineering problems?

Tip: Always ensure you differentiate with respect to the parameter properly when finding envelopes.