6.4 Q-7

Question Statement

Find the equation of the parabola formed by the cables of a suspension bridge, where the span is $a$ meters and the vertical height of the supporting towers is $b$ meters.

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Background and Explanation

In this problem, we model the cables of a suspension bridge as a parabola. The parabola’s vertex is at the lowest point of the cable, and the supporting towers are symmetrically placed at the ends of the span. Key details:

1. The span is the horizontal distance between the two supporting towers, denoted as $a$.
2. The vertical height is the maximum height of the towers above the vertex of the parabola, denoted as $b$.
3. We use the standard equation of a parabola:

$$x^2 = 4Py$$

where $P$ represents the focal distance.

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Solution

Step 1: Choose the Coordinate System

• Place the vertex of the parabola at the origin $(0, 0)$.
• The parabola opens upward with the axis of symmetry along the $y$-axis.
• The two towers are located at $\left(\frac{a}{2}, b\right)$ and $\left(-\frac{a}{2}, b\right)$.

Step 2: Substitute the Point $\left(\frac{a}{2}, b\right)$ into the Parabola’s Equation

The equation of the parabola is:

$$x^2 = 4Py$$

Substitute $\left(\frac{a}{2}, b\right)$ into the equation:

$$\left(\frac{a}{2}\right)^2 = 4P \cdot b$$

Step 3: Solve for $P$

Simplify the equation:

$$\frac{a^2}{4} = 4Pb$$

Rearrange to solve for $P$:

$$P = \frac{a^2}{16b}$$

Step 4: Rewrite the Parabola’s Equation

Substitute $P = \frac{a^2}{16b}$ into the equation $x^2 = 4Py$:

$$x^2 = 4 \cdot \frac{a^2}{16b} \cdot y$$

Simplify:

$$x^2 = \frac{a^2}{4b} \cdot y$$

Final Equation of the Parabola

$$x^2 = \frac{a^2}{4b}y$$

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Key Formulas or Methods Used

1. Standard Equation of a Parabola:

$$x^2 = 4Py$$

2. Symmetry of a Parabola: The vertex is at the lowest point, and the parabola is symmetric about the $y$-axis.

3. Substitution: Use the known coordinates of a point on the parabola to determine the constant $P$.

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Summary of Steps

1. Place the vertex of the parabola at the origin $(0, 0)$.
2. Use the general equation $x^2 = 4Py$.
3. Substitute the coordinates of one tower $\left(\frac{a}{2}, b\right)$ into the equation.
4. Solve for $P$, the focal distance.
5. Rewrite the equation of the parabola with the derived value of $P$.