Ray emanating from the point (0, 6) is incident on the ellilpse 25x² + 16$y^2$= 1600 at the point P with ordinate 5. After reflection, ray cuts the y-axis at B. The length of PB is

$$\begin{gathered} Given ellipse \\ \frac{{\mathrm{x}}^2}{8^2}+\frac{{\mathrm{y}}^2}{{10}^2}=1 \end{gathered}$$
Coordinates of  foci are
$\left(0,\pm \sqrt{\left({10}^2-8^2\right)}\right)\text{ i.e. }(0,\pm 6)$
Let coordinates of P are  $(\mathrm{\lambda },5)$
$\therefore \frac{{\mathrm{\lambda }}^2}{8^2}+\frac{5^2}{{10}^2}=1\Rightarrow \mathrm{\lambda }=\pm 4\sqrt{3}$
$\therefore \mathrm{P}\equiv (\pm 4\sqrt{3},5)$(lie in I or II quadrants)
According to reflection property, a ray passing
through focus B'(0,6) will passing through B(0, –6)
(other focus). If P lies in I quadrant, then
$\mathrm{PB}=\sqrt{48+(5+6{)}^2}=\sqrt{169}=13$