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.

Solution: Let $AB$ is the rod with length $12$ cm. and$P\left(x,y\right)$ is any point such that,
$AP=3$ cm and $PB=12-3=9$ cm
we can draw perpendiculars from $P$ to X-axis and Y-axis at point $R$ and $Q$ respectively.
Then, $PQ=x$ and $PR=y$
Please refer to video forthe diagram.
Now, in $\Delta PBQ$,
$\frac{PQ}{PB}=\cos \theta$
$\Rightarrow \frac{x}{9}=\cos \theta \to \left(1\right)$
Now, in $\Delta APR$,
$\frac{PR}{PA}=\sin \theta$
$\Rightarrow \frac{y}{3}=\sin \theta \to \left(2\right)$
Now, squaring and adding (1) and (2),
$\frac{x^2}{81}+\frac{y^2}{9}={\cos }^2\theta +{\sin }^2\theta$
$\Rightarrow \frac{x^2}{81}+\frac{y^2}{9}=1$
,which is the required equation.