The 4' pole A is set into the ground at point A and the 6' pole B is 13' away at point B. The cable is fixed at point P a feet away from A. The 19' cable is attached to the tops of the poles, forming two right-angled triangles.
Using Pythagoras: 4^2+a^2=x^2, where x is the length of cable between P and the top of pole A. Also, 6^2+(13-a)^2=(19-x)^2. So we have x^2-a^2=16 and 36+169-26a+a^2=361-38x+x^2 and x^2-a^2=36+169-26a-361+38x.
x^2-a^2=16=38x-26a-156, so x=(16+156+26a)/38=(172+26a)/38=(86+13a)/19.
Substituting x=(86+13a)/19 in x^2-a^2=16, we have (86+13a)^2/361-a^2=16, which is (86+13a)^2-361a^2=16*361. So, (86+13a-19a)(86+13a+19a)=5776 and (86-6a)(86+32a)=5776.
7396+2236a-192a^2-5776=0⇒1620+2236a-192a^2=0 or 192a^2-2236a-1620=0. This is a quadratic equation:
48a^2-559a-405=0. From this a=12.3301' approx. The negative solution doesn't apply. So the cable needs to be fixed to the ground 12.33' from the 4' pole.
(The length of cable x=12.9627' from the top of the 4' pole to the ground.)