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A rod of length $12 \,cm$ moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $P$ on the rod, which is $3\, cm$ from the end in contact with the $x-$ axis.
$\frac{x^{2}}{81}+\frac{y^2} {9}=1$
$\frac{x^{2}}{81}+\frac{y^2} {9}=1$
$\frac{x^{2}}{81}+\frac{y^2} {9}=1$
$\frac{x^{2}}{81}+\frac{y^2} {9}=1$
Solution
Let $AB$ be the rod making an angle $\theta$ with $O X$ and $P ( x ,\, y )$ be the point on it such that $AP =3\,cm$
Then, $PB = AB – AP =(12-3)\, cm =9\, cm$ $[ AB =12 \,cm ]$
From $P$, draw $PQ \perp OY$ and $PR \perp OX$.
In $\Delta PBQ$ , $\cos \theta=\frac{ PQ }{ PB }=\frac{x}{9}$
In $\Delta PRA$ , $\sin \theta=\frac{ PR }{ PA }=\frac{y}{3}$
since, $\sin ^{2} \theta+\cos ^{2} \theta=1$
$\left(\frac{y}{3}\right)^{2}+\left(\frac{x}{9}\right)^{2}=1$
Or, $\frac{x^{2}}{81}+\frac{y^{2}}{9}=1$
Thus, the equation of the locus of point $P$ on the rod is $\frac{x^{2}}{81}+\frac{y^2} {9}=1$.
