Question

A rod of length l slides between two perpendicular lines. Find the locus of the point on the rod which divides it in the ratio 1 : 2.

Answer 1

Let the two perpendicular lines be the coordinate axes. Let AB be a rod of length l and the coordinates of A and B be (a, 0) and (0, b) respectively.
As the rod AB slides, the values of a and b change. Let P(h, k) be a point on AB.



Here, BP:AP = 1:2 .

$$\begin{gathered} \therefore h=\frac{a+0}{3},k=\frac{0+2b}{3} \\ \Rightarrow a=3h,b=\frac{3k}{2} \end{gathered}$$ ... (1)

The length of the given rod is l.

$$\begin{gathered} \therefore AB=l \\ \Rightarrow \sqrt{a^2+b^2}=l \\ \Rightarrow a^2+b^2=l^2 \end{gathered}$$

Using equation (1), we get:

$$\begin{gathered} \Rightarrow 9h^2+{\left(\frac{3k}{2}\right)}^2=l^2 \\ \Rightarrow h^2+\frac{k^2}{4}=\frac{l^2}{9} \end{gathered}$$

Hence, the locus of (h, k) is $x^2+\frac{y^2}{4}=\frac{l^2}{9}$ .