Q.The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and $100$ m long is supported by vertical wires attached to the cable, the longest wire being $30$ m and the shortest being $6$ m. Find the length of a supporting wire attached to the roadway $18$ m from the middle.

Concept: Parabola Standard Form — the cable takes the shape $x^2 = 4ay$, with vertex at the lowest point.

Step 1 – Set up the coordinate system.

Place the vertex (lowest point of the cable) at $(0,0)$. The roadway is horizontal, $100$ m long, so the ends are at $x = \pm 50$ m. The shortest wire (at the vertex) is $6$ m, so the cable height at the vertex is $6$ m above the roadway. The longest wires at the ends are $30$ m, so the cable height at $x = \pm 50$ is $30$ m.

Step 2 – Find the parabola equation.

Let the parabola be $x^2 = 4ay$. At $x = 50$, the height of the cable above the vertex is $y = 30 - 6 = 24$ m.

Substitute: $50^2 = 4a(24) \implies 2500 = 96a \implies a = \frac{2500}{96} = \frac{625}{24}$.

Step 3 – Find the wire length at $x = 18$ m from the middle.

At $x = 18$, the cable height above the vertex is $y = \frac{x^2}{4a} = \frac{324}{4 \cdot \frac{625}{24}} = \frac{324 \cdot 24}{2500} = \frac{7776}{2500} = 3.1104$ m.

The total wire length = cable height above roadway = $y + 6 = 3.1104 + 6 = 9.1104$ m.

The problem describes a suspension bridge where the cable hangs in a parabola. The roadway is horizontal, 100 m long, and vertical wires connect the cable to the roadway. The longest wire (at the ends) is 30 m, the shortest (at the middle) is 6 m. We need the length of the wire 18 m from the middle.

The key insight: the parabola is symmetric about the midpoint. The shortest wire is at the vertex (lowest point of the cable), and the longest wires are at the ends. So we set up a coordinate system with the vertex at the origin, and the parabola opening upward.

1. Set up the coordinate system

Place the vertex of the parabola at $(0, 0)$. Since the shortest wire is 6 m, the vertex is 6 m above the roadway. But careful: the wire length is the vertical distance from the roadway to the cable. If the roadway is horizontal, we can think of the cable’s height above the roadway. Let the roadway be the x-axis, and the cable’s height above it be $y$. Then at the vertex, $y = 6$ (shortest wire). But if we set the vertex at $(0,0)$, then the roadway is 6 m below it. That’s fine — we just need consistent coordinates.

Better: Let the roadway be the line $y = 0$. The cable is above it. The shortest wire (at the middle) is 6 m, so the vertex of the parabola is at $(0, 6)$. The longest wire (at the ends) is 30 m, so at $x = \pm 50$ (half of 100 m), the cable height is 30.

So the parabola has vertex at $(0, 6)$ and opens upward. Its equation in standard form is:

Here $h = 0$, $k = 6$, so:

2. Find the value of $a$ using the endpoint condition

At $x = 50$, $y = 30$. Substitute:

So the parabola equation is:

3. Find the wire length 18 m from the middle

We need $y$ when $x = 18$:

Multiply both sides by 6:

4. Simplify

So $y = 9.1104$ m.

Thus the wire length is approximately $9.11$ m.