202212131544 Solution to 1975-CE-AMATH-I-X

A rod $PQ$ of length $4$ units slides with $P$ on the $x$ -axis and $Q$ on the $y$ -axis. $R$ is the point on $PQ$ such that $PR:RQ=1:3$ . Find the equation of the locus of $R$ .

Roughwork.

Let $P=(a,0)$ and $Q=(0,b)$ where $a,b\in [0,4]$ and $a^2+b^2=(4)^2$ . The locus of $R(x,y)$ is

$\begin{aligned} &\quad\enspace \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{3}{4}a \\ \frac{1}{4}b \end{pmatrix} \\ &\Longrightarrow \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} \frac{4}{3}x \\ 4y \end{pmatrix} \\ &\Longrightarrow \bigg(\frac{4}{3}x\bigg)^2+ (4y)^2 = 16 \\ &\Longrightarrow \enspace R:\big\{ x^2+9y^2-9=0\big\} \\ \end{aligned}$

This problem is not to be attempted.

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