# A cable of a suspension bridge is in the form of a parabola whose span is $40mts$ . The road way is $5mts$ below the lowest point of the cable . if an extra support is provided across the cable $30mts$ above the ground level, Find the length of the support if the height of the pillars are $55mts$.

Toolbox:
• Standard parabolas :
• (i)$y^2=4ax$
• In this case the axis of symmetry is $x$-axis.
• The focus is $F(a,0)$ and the equation of the directrix is $x=-a$
• (ii)$y^2=-4ax$
• In this case the focus is $F(-a,0)$ and the equation of directrix is $x=a$
• (iii)$x^2=4ay$
• In this case the axis of symmetry is $y$-axis.
• The focus is $F(0,a)$ and the equation of directrix is $y=-a$
• (iv)$x^2=-4ay$
• In this case the focus is $F(0,-a)$ and the equation of directrix is $y=a$
• Shifting of origin from (0,0) to $(h,k)$ then if the new axes form $X-Y$ coordinate system then
• $x=X+a$
• $y=Y+k$
• The new coordinates of a point $P(x,y)$ become $X=x-a$,$Y=y-k$.
Step 1:
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 $x^2=4ay$
Step 2:
Substituting the coordinates $(20,50)$
$400=200a\Rightarrow a=2$
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 $(x,25)$.
Step 3:
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 $CD=20\sqrt 2m$