Let AB be the rod making an angle Ɵ with OX and P(x,y) be the point on it such that

AP = 3cm.

Then, PB = AB – AP = (12 – 3) cm = 9cm [AB = 12cm]

From P, draw PQ ⊥ OY and PR ⊥ OX.

In ΔPBQ, cos θ = \( \dfrac{x}{9}\)

Sin θ = \( \dfrac{y}{3}\)

we know that, sin^{2} θ +cos^{2} θ = 1,

So,

\( (\dfrac{y}{3})^2 + (\dfrac{x}{9})^2 = 1 \)or

\(\dfrac{ x^2}{81} + \dfrac{y^2}{9}= 1\)

Hence, the equation of the locus of point P on the rod is \( \dfrac{ x^2}{81} + \dfrac{y^2}{9}= 1\)

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