Find the locus of point R.

Question

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.

Answer 1

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: $$y=\frac{10}3\sqrt{1-\biggl(\frac{3x}5\biggr)^2},\quad\frac a3\le x\le 5.$$

Answer 2

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 $q=x/3$.

Answer 3

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?