The foci of a hyperbola are S and S' with SS'=10. A point P on the hyperbola satisfies PS⊥ PS' and PS· PS'=18. If the transverse axis of the…

Since $\angle SPS'=90^\circ$, Pythagoras gives $PS^2+PS'^2=SS'^2=100$. For any point on a hyperbola $|PS-PS'|=2a$. Then $(PS-PS')^2=PS^2+PS'^2-2\,PS\cdot PS'=100-36=64$, so $2a=8$ and $a=4$. Consistency check: $c=5$, so $b^2=c^2-a^2=9$ and the hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1$. Solving this with the circle $x^2+y^2=25$ having $SS'$ as diameter (the condition for a right angle at $P$) gives real points $P$, and for them $PS\cdot PS'=18$, equal to $b^2\cdot2$ as expected from the area $b^2\cot45^\circ=9$ of the right triangle. Therefore $a=4$. Hence the length of the transverse axis is $8$ and the required value is $a=4$, a single digit as the answer format requires.