Let $F_1$ and $F_2$ be two points where the flag posts are fixed on theground. The origin O is the mid-point of $F_1F_2$. ∴ $OF_1$ = $OF_2$ = $\frac{1}{2} F_1F_2$ = $\frac{1}{2} \times 8$ = 4 m. ∴ Coordinates of $F_1$ are (–4, 0) and $F_2$ are (4, 0) Let P(a, b) be any point on the track. ∴ $PF_1 + PF_2$ = 10 ∴ $\sqrt{(\alpha+4)^2+(\beta-0)^2}$ + $\beta\left( (\alpha-4)^2+(\beta-0)^2 \right)$ = 10 ⇒ $\sqrt{\alpha^2+16+8\alpha+\beta^2}$ = 10 - $\sqrt{\alpha^2+16-8\alpha+\beta^2}$ Squaring both sides, we get $\alpha^2 + \beta^2$ + 8α + 16 = 100 + $\alpha^2+\beta^2$ - 8α + 16 - 20 $\sqrt{\alpha^2+\beta^2-8\alpha+16}$ ⇒ 16α - 100 = - 20 $\sqrt{\alpha^2+\beta^2-8\alpha+16}$ Again squaring both sides, we get $(16\alpha-100)^2$ = $(-20\sqrt{\alpha^2+\beta^2-8\alpha+16})^2$ ⇒ $256a^2$ + 10000 – 3200a = 400($a^2 + b^2$ – 8a + 16) ⇒ $256a^2$ + 10000 – 3200a = $400a^2 + 400b^2$ – 3200a + 6400 ⇒ $144a^2 + 400b^2$ = 3600 ⇒ $\frac{144\alpha^2}{3600} + \frac{400\beta^2}{3600}$ = 1 ⇒ $\frac{\alpha^2}{25} + \frac{\beta^2}{9}$ = 1 Thus required equation of locus of point P is $\frac{x^2}{25} + \frac{y^2}{9}$ = 1.