A bar of length 20 units moves with its ends in two fixed perpendicular straight lines. P is a point on the Rod which is at a distance of 8 units from one end. The locus of P is an ellipse. Then its ecentricity is

Let $\overline{\mathrm{AB}}$ be the bar such that AB=20

Let P=(x, y) be point on AB such that PB=a and  AP=20-a

Now $\mathrm{cos\theta }=\frac{\mathrm{x}}{\mathrm{a}}$ and $\mathrm{sin\theta }=\frac{\mathrm{y}}{20-\mathrm{a}}$

 $$\begin{gathered} Since {\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=1 then \\ \Rightarrow \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{(20-\mathrm{a})}^2}=1 \end{gathered}$$

Given PB=8

$\Rightarrow {\mathrm{a}}^2=64$ and Let ${(20-\mathrm{a})}^2={\left(12\right)}^2=144$

$\therefore$ locus is $\frac{{\mathrm{x}}^2}{64}+\frac{{\mathrm{y}}^2}{144}=1(\mathrm{a}<\mathrm{b})$

Now $\mathrm{e}=\sqrt{\frac{144-64}{144}}=\sqrt{\frac{80}{144}}=\frac{\sqrt{5}}{3}$