Let $AB$ be the span of the cable $AB=40m$.Let $V$ be the vertex of the parabolic suspension and let it be the origin.The height of the lowest point is $5m$ above the road.
The pillars are $55m$ high and so the span $AB$ is $55-5=50m$ above the vertex.The vertical line through the vertex to the axis of the parabola and it bisects $AB$ at $(0,50)$.
The coordinates of $A$ and $B$ are $(-20,50)$ and $(20,50)$.
The equation of the parabola is of the form $x^2=4ay$
Substituting the coordinates $(20,50)$
The equation of the parabola is $x^2=8y$
Now $CD$ is the additional support which is at a height of $30m$ above the ground ,(i.e) 25m above $V$.
The coordinates of $C$ and $D$ are $(-x,25)$ and $(x,25)$.
Substituting $(x,25)$ we have
$x^2=200\Rightarrow x=10\sqrt 2$
Therefore $C$ and $D$ are the points $C(-10\sqrt 2,25),D(10\sqrt 2,25)$.
The length of the support is $CD=20\sqrt 2m$