Find the locus of point R. A ladder, PQ, 5m long leaning against a wall. Assuming the floor and the wall represent the x-axis and y-axis respectively. R is a point on PQ such that RQ is 2 times RP. Given that P(0, b) and Q(a, 0). Find the locus of point R when the ladder PQ slides down the wall.
 A: Hint: Let $P(0,p)$, $Q(q,0)$, and $P(x,y)$, hence $p^2+q^2=25$.  From the intercept theorem you'll be able to derive $p=3y/2$ and $q=x/3$.
A: From the ratio of RP:RQ given we can find the values of RP and RQ to be $\frac 53$ and $\frac{10}3$ respectively. Say at any moment the rod is leaning at angle $\theta$ with the X-axis. Now project RQ on the axes and notice that the coordinates of point R can be written as $\left(\frac 53\cos\theta,\frac{10}3\sin\theta\right)$. 
Does that parametrization look familiar?
A: Set $R=(x,y)$ and $Q=(0,t),\enspace 0\le t \le5$. By Thales' theorem, $\overrightarrow{PR}=\dfrac13\overrightarrow{PQ}$ implies $x=\dfrac13 t$. Furthermore, $RQ=\dfrac{10}3$, hence if $H$ is the projection of $R$ on the $x$-axis, Pythagoras' theorem in the right triangle $RHQ$ yields
$$y^2+4x^2=\frac{100}9,\quad \frac a3\le x\le 5,\enspace y\ge 0 $$
which is the equation of an elliptical arc.
As a function of $x$, we may write:
$$y=\frac{10}3\sqrt{1-\biggl(\frac{3x}5\biggr)^2},\quad\frac a3\le x\le 5.$$
