Question

A ladder of length $15$ m moves with its ends always touching the vertical wall and the horizontal floor. Determine the equation of the locus of a point. $P$ on the ladder, which is $6$ m from the end of the ladder in contact with the floor.

Answer 1

Let $AB$ be the ladder and $P(x_1,y_1)$ be a point on the ladder such that $AP=6$ m.
Draw $PD$ perpendicular to $X$-axis and $PC$ perpendicular to $y$-axis.
Clearly the triangles $ADP$ and $PCB$ are similar.
$\therefore \frac{PC}{DA}=\frac{PB}{AP}=\frac{BC}{PD}$
(i.e.,) $\frac{x_1}{DA}=\frac{9}{6}=\frac{BC}{y_1}$
$\Rightarrow DA=\frac{6x_1}{9}=\frac{2x_1}{3}=\frac{5}{3}{x}_1$
$\Rightarrow OB=OC+BC=y_1+\frac{3y_1}{2}=\frac{5}{2}{y}_1$
But ${OA}^2+{OB}^2={AB}^2\Rightarrow \frac{25}{9}{x}_1^2+\frac{25}{4}{y}_1^2=225$
$\Rightarrow \frac{x_1^2}{9}+\frac{y_1^2}{4}=9$
$\therefore$ the locus of $(x_1,y_1)$ is $\frac{x^2}{81}+\frac{y^2}{36}=1$, which is an ellipse.