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.

image of rod on axes with horizontal from y axis YC and vertical from x axis XCE drawn so that ED is horizontal. Angles X A C, Y C B and E C D are marked as alpha

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:

Image of a quarter of an ellipse centred on the origin intersecting the x and y axes at 9 and 4 respectively, and the line segment from point (6,4) to (9,6)