Review question

# What is the locus of these points fixed to a rod? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5754

## Solution

The ends $A$ and $B$ of a rod of fixed length move on the positive $x$ and $y$ axes respectively. $C$ is a point on $AB$, and $CD$ is perpendicular to $AB$, as shown in the figure. The lengths of $AC$, $BC$ and $CD$ are $a$, $b$ and $c$ respectively.

In the question the rod is moving attached to the axes, so one aspect which changes is the angle that it makes with the horizontal (angle $OAC$). We can then annotate our original diagram to help in writing our parametric equations.

Let $\alpha$ be the acute angle the rod makes with the $x$-axis. We can mark in other angles $\alpha$ on our diagram as shown.

It is possible to use a different variable as the parameter, for instance the $x$-coordinate of $C$. The final result would be the same.

1. Find parametric equations for the locus of $C$.
2. Find parametric equations for the locus of $D$.
Now using triangle $CYB$ and $CAX$ we can write the parametric equations for $C$. \begin{align*} x &= b \cos \alpha \\ y &= a \sin \alpha \\ \end{align*} To get the equations for $D$ we need to add sides from triangle $CED$. \begin{align} x &= b \cos \alpha+ c \sin \alpha \label{eq:1}\\ y &= a \sin \alpha+ c \cos \alpha \label{eq:2} \end{align}

1. If $c^2=ab$ show that the locus of $D$ consists of part of a straight line and find the coordinates of the extremities of this locus.

We are told to consider when $c^2=ab$. Using the parametric equations for $D$ with this information we can find a relationship between $x$ and $y$. Multiplying $\eqref{eq:1}$ by $a$ and $\eqref{eq:2}$ by $b$ we can create equations with a matching term so that subtraction eliminates them. $ax-cy=ab\cos \alpha - c^2 \cos \alpha$ Since we know that $c^2=ab$ this reduces to $ax-cy=0$ which is a straight line through the origin with gradient $\frac {a}{c}$.

If you used other ways of eliminating $\alpha$ from these equations you may have got $y=\frac{c}{b}x$ or $y=\sqrt{\frac{a}{b}}x$ or $y=\left ( \frac{c-a}{b-c} \right )x$. Can you show that these are all equivalent?

The extremities of the locus will be when $\alpha$ is either $0$ (the rod is lying flat) or $90 ^\circ$ (the rod is vertical) these are the points $(b,c)$ and $(c,a)$ respectively.
1. Sketch the loci of $C$ and $D$ for the case $a=4$, $b=9$, $c=6$.

The parametric equations for the locus of $C$ tell us this is (one quarter of) an ellipse with semi-minor axis $a$ and semi-major axis $b$. (If we don’t recognise this we can rearrange for $\sin \alpha$ and $\cos \alpha$ and use the Pythagorean identity to eliminate $\alpha$.)

We can now sketch these two loci: