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, sin2 θ +cos2 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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