If you take the point on the bridge directly beneath the lowest point on the cable to be the origin, then the parabola has equation
[tex]y = ax^2+bx+10[/tex]
300 ft to either side of the origin, the parabola reaches a value of y = 90, so
[tex]a(300)^2+b(300)+10 = 90 \implies 4500a+15b = 4[/tex]
[tex]a(-300)^2+b(-300)+10 = 90 \implies 4500a-15b = 4[/tex]
Adding these together eliminates b and lets you solve for a :
[tex](4500a+15b)+(4500a-15b)=4+4 \\\\ 9000a = 8 \\\\ a = \dfrac1{1125}[/tex]
Solving for b gives
[tex]4500\left(\dfrac1{1125}\right)+15b=4 \\\\ 4+15b = 4 \\\\ 15b=0 \\\\ b=0[/tex]
So the parabola's equation is
[tex]y = \dfrac{x^2}{1125}+10[/tex]
150 ft away from the origin, the cable is at a height y of
[tex]y = \dfrac{150^2}{1125}+10 = \boxed{30}[/tex]
ft above the bridge.
