Let AB be a rod of length 12 cm and P(x, y) be any point on the rod such that PA = 3 cm and PB = 9 cm. Let AR = a and BQ = b Then ΔARP ~ ΔPQB ∴ $\frac{AR}{PQ}$ = $\frac{AP}{PB}$ ∴ $\frac{a}{x}$ = $\frac{3}{9}$ ⇒ 9a = 3x ⇒ a = $\frac{x}{3}$ and $\frac{BQ}{BP}$ = $\frac{PR}{PA}$ ∴ $\frac{b}{9}$ = $\frac{y}{3}$ ⇒ 3b = 9y ⇒ b = 3y Now, OA = OR + AR = x + a = x + $\frac{x}{3}$ = $\frac{4x}{3}$ OB = OQ + BQ = y + b = y + 3y = 4y In right angled ΔAOB, $AB^2$ = $OA^2$ + $OB^2$ ∴ $(120)^2$ = $\left(\frac{4x}{3}\right)^2 + (4y)^2$ ⇒ 144 = $\frac{16x^2}{9}+16y^2$ ⇒ $\frac{16x^2}{9 \times 144} + \frac{16y^2}{144}$ = 1 ⇒ $\frac{x^2}{81}+\frac{y^2}{9}$ = 1, which is the required locus of point P and which represents an ellipse.