The line $$12x\cos\theta + 5y\sin\theta = 60$$ is tangent to which of the following curves?

We are given the straight line

$$12x\cos\theta \;+\; 5y\sin\theta \;=\; 60.$$

Our task is to decide which one of the four given curves has this line as a tangent (for some value of the parameter $$\theta$$). To see this clearly we first put the equation of the line in a form that can be compared with the standard tangent form of conic sections.

Dividing every term by $$60$$ we get

$$\frac{12x\cos\theta}{60} \;+\; \frac{5y\sin\theta}{60} \;=\; 1.$$

Reducing the fractions step by step,

$$\frac{12}{60} \;=\; \frac{1}{5}, \qquad \frac{5}{60} \;=\; \frac{1}{12},$$

so the equation becomes

$$\frac{x\cos\theta}{5} \;+\; \frac{y\sin\theta}{12} \;=\; 1.$$

Now we recall the standard result for an ellipse. For the ellipse whose equation is

$$\frac{x^{2}}{a^{2}} \;+\; \frac{y^{2}}{b^{2}} \;=\; 1,$$

the equation of a tangent at the point $$\bigl(a\cos\phi,\;b\sin\phi\bigr)$$ on the ellipse is

$$\frac{x\cos\phi}{a} \;+\; \frac{y\sin\phi}{b} \;=\; 1.$$

Comparing our rearranged line

$$\frac{x\cos\theta}{5} \;+\; \frac{y\sin\theta}{12} \;=\; 1$$

term by term with the general tangent form

$$\frac{x\cos\phi}{a} \;+\; \frac{y\sin\phi}{b} \;=\; 1,$$

we see an exact match when we identify

$$a \;=\; 5, \qquad b \;=\; 12.$$

Hence the curve whose tangents are described by the given family of lines is the ellipse

$$\frac{x^{2}}{5^{2}} \;+\; \frac{y^{2}}{12^{2}} \;=\; 1,$$

that is,

$$\frac{x^{2}}{25} \;+\; \frac{y^{2}}{144} \;=\; 1.$$

To place this ellipse in the exact form used in the options, we clear denominators by multiplying every term by the least common multiple $$25 \times 144 = 3600$$:

$$144x^{2} \;+\; 25y^{2} \;=\; 3600.$$

This matches Option B exactly.

Hence, the correct answer is Option B.