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A rod $AB$ of length $15\,cm$ rests in between two coordinate axes in such a way that the end point A lies on $x-$ axis and end point $B$ lies on $y-$ axis. A point $P(x,\, y)$ is taken on the rod in such a way that $AP =6\, cm .$ Show that the locus of $P$ is an ellipse.
Solution
Let $AB$ be the rod making an angle $\theta $ with $OX$ as shown in Fig. and $P (x, \,y)$ the point on it such that $AP =6\, cm$
since $AB =15\,cm ,$ we have
$P B=9\, cm$
From $P$ draw $PQ$ and $PR$ perpendiculars on $y-$ axis and $x-$ axis, respectively.
From $\Delta PBQ$ , $\cos \theta=\frac{x}{9}$
From $\Delta PRA$, $\sin \theta=\frac{y}{6}$
since $\cos ^{2} \theta+\sin ^{2} \theta=1$
$\left(\frac{x}{9}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1$
or $\frac{x^{2}}{81}+\frac{y^{2}}{36}=1$
Thus the locus of $P$ is an cllipse.
