Question

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the path of a moving point P on the rod which is 3 cm from the end in contact with the x-axis.

Answer 1

Let AB be the rod and P(x, y) be a point on it such that

AP = 3 cm and PB = 9 cm.

From P, draw $PM\perp OX$ and $PN\perp OY.$

Let AM = p and BN = q.

Then $\Delta BNP$ and $\Delta PMA$ are similar.

$\therefore \frac{BN}{PM}=\frac{BP}{PA}\Rightarrow \frac{q}{y}=\frac{9}{3}\Rightarrow q=3y$

And, $\frac{MA}{NP}=\frac{PA}{BP}\Rightarrow \frac{p}{x}=\frac{3}{9}\Rightarrow p=\frac{1}{3}x.$

$\therefore OA=OM+MA=x+p=x+\frac{1}{3}x=\frac{4x}{3},$

and $OB=ON+BN=y+q=y+3y=4y.$

In $\Delta BOA,$

we have

$O{A}^2+O{B}^2=A{B}^2$

$\Rightarrow {\left(\frac{4x}{3}\right)}^2+(4y{)}^2=(12{)}^2$

$\Rightarrow \frac{16x^2}{9}+16y^2=144$

$\Rightarrow \frac{x^2}{81}+\frac{y^2}{9}=1$

Hence, the path of P is an ellipse whose equation is $\frac{x^2}{81}+\frac{y^2}{9}=1$