Question Statement

An architect's design for a building includes some large pillars with cross-sections in the shape of hyperbolas. The curve can be modeled by the equation $\frac{x^2}{16}-\frac{y^2}{9}=1$ where the units are in meters. If the pillars are $4.2\text{m}$ tall, find:

1. The width of the top of each pillar
2. The width at the middle of the pillar

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Background and Explanation

This problem involves a rectangular hyperbola centered at the origin with a horizontal transverse axis. The key insight is recognizing that the pillar's height extends symmetrically above and below the x-axis, so the "top" of the $4.2\text{m}$ pillar corresponds to $y=2.1\text{m}$.

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Solution

Part (a): Width at the Top of the Pillar

Since the pillar is $4.2\text{m}$ tall and the hyperbola is symmetric about the x-axis, the top edge is located at $y=\frac{4.2}{2}=2.1\text{m}$.

Let $w$ be the total width of the top. By symmetry, the right edge point $A$ has coordinates $\left(\frac{w}{2},2.1\right)$.

Since point $A$ lies on the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$, we substitute $x=\frac{w}{2}$ and $y=2.1$:

$\begin{array}{rlrl}& & \frac{{\left(\frac{w}{2}\right)}^2}{16}-\frac{(2.1{)}^2}{9}& =1\\ \Rightarrow & & \frac{w^2}{64}-\frac{4.41}{9}& =1\\ \Rightarrow & & \frac{w^2}{64}& =1+\frac{4.41}{9}\\ & & \frac{w^2}{64}& =1+0.49\\ & & & =1.49\\ \Rightarrow & & w^2& =(1.49)(64)\\ \Rightarrow & & w^2& =95.36\\ \Rightarrow & & w& =\sqrt{9.765}\text{m}\end{array}$

Therefore, the width of the top of the pillar is $\sqrt{9.765}\text{m}$ (approximately $9.77\text{m}$).

Part (b): Width at the Middle of the Pillar

At the middle of the pillar, the height is $y=0$ (at the center line of the hyperbola).

Let $w_1$ be the width at the middle. The coordinates of point $B$ (the right edge) are $\left(\frac{w_1}{2},0\right)$.

Substituting into the hyperbola equation:

$\begin{array}{ll}\text{Thus}& \frac{{\left(w_1^2\right)}^2}{16}-\frac{(0{)}^2}{9}=1\\ \Rightarrow & \frac{w_1^2}{64}=1\\ \Rightarrow & w_1^2=64\\ \Rightarrow & w_1=\sqrt{64}=8\\ & w_1=8\text{m}\end{array}$

Thus, the width at the middle of the pillar is $8\text{m}$.

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Key Formulas or Methods Used

• Standard hyperbola equation: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ (where $a^2=16$ and $b^2=9$)
• Coordinate substitution method: Substituting known $y$-values to solve for corresponding $x$-coordinates
• Symmetry principle: Total width equals $2x$ (distance from center line to edge)
• Algebraic isolation: Solving for $w$ by isolating the variable step-by-step

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Summary of Steps

1. Identify vertical position: Determine that the top of the $4.2\text{m}$ pillar is at $y=2.1\text{m}$ (half the total height).
2. Set up top width variable: Let width $=w$, so the edge point is $\left(\frac{w}{2},2.1\right)$.
3. Substitute and solve for top: Plug into $\frac{x^2}{16}-\frac{y^2}{9}=1$, simplify $\frac{4.41}{9}=0.49$, solve for $w^2=95.36$, then find $w$.
4. Identify middle position: Recognize the middle is at $y=0$ (the x-axis).
5. Set up middle width variable: Let width $=w_1$, so the edge point is $\left(\frac{w_1}{2},0\right)$.
6. Substitute and solve for middle: Plug $y=0$ into the hyperbola equation to get $\frac{w_1^2}{64}=1$, yielding $w_1=8\text{m}$.