Step 1:

The point $P(x,y)$ moves such that the difference of its distances between the points $F(5,0)$ and $F'(-5,0)$ is $8$.

Therefore it traces a hyperbola with length of the transverse axis $2a=8\Rightarrow a=4$

The foci $FF'$ centre of the hyperbola is the midpoint of $FF'$ (i.e)$(0,0)$

Step 2:

The transverse axis contains the points $F$ and $F'$.Therefore it is the $x$-axis.

$FF'=2ae$

Therefore $2ae=10$

$\Rightarrow ae=5$

$e=\large\frac{5}{a}=\frac{5}{4}$

Step 3:

The semi conjugate axes ' b' is given by

$b=a\sqrt{e^2-1}=4\sqrt{\large\frac{25}{16}-\normalsize 1}$

Step 4:

The equation of the hyperbola is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$

$\large\frac{x^2}{16}-\frac{y^2}{9}$$=1$

(i.e)$9x^2-16y^2=144$