Given a rod of length 12 cm moves with its ends always touching the coordinate axes.

Again given a point P on the rod, which is 3 cm from the end in contact with the x-axis.

It is shown in the figure.

Here AP = 3 cm, AB = 12

Now BP = AB - AP

=> BP = 12 - 3

=> BP = 9 cm

Again from figure,

∠PAO = ∠BPO = θ    (since PQ || OA and are corresponding angles)

Now in ΔBPO,

cosθ = QP/BP

=> cosθ = x/9 .............1

Again in ΔPAr,

sinθ = PR/PA

=> sinθ = y/3 ........2

Now square equation 1 and 2 and then add them, we get

cos² θ + sin² θ = x² /81 + y² /9

=> x² /81 + y² /9 = 1      (since cos² θ + sin² θ = 1 )

So the equation of the locus of a point P is x² /81 + y² /9 = 1