Question

A doorway in the shape of an elliptical arch (a halfellipse) is 10 feet wide and 4 feet high at the center. A box 2 feet high is to be pushed through the doorway. How wide can the box be?

Short answer

The box can be 5√3 feet wide.

Step-by-step solution

Understand the ellipse equation

The equation of an ellipse centered at the origin with semi-major axis a and semi-minor axis b is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Since the arch is half of an ellipse, we still use the full equation for calculations.

Identify ellipse dimensions

For this problem, the semi-major axis \(a\) is half the width of the doorway, which is 5 feet, and the height at the center (the semi-minor axis \(b\)) is 4 feet. So, \( a = 5 \) and \( b = 4 \).

Write the ellipse equation

Substitute \(a = 5\) and \(b = 4\) into the ellipse equation: \( \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1 \). This simplifies to \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \).

Substitute box height into equation

We need the point on the ellipse corresponding to the box's height of 2 feet. Substitute \(y = 2\) into the ellipse equation: \( \frac{x^2}{25} + \frac{2^2}{16} = 1 \).

Solve for x

First simplify \(\frac{2^2}{16} = \frac{4}{16} = \frac{1}{4}\). The equation becomes \( \frac{x^2}{25} + \frac{1}{4} = 1 \). Solve for \(x^2\) by subtracting \(\frac{1}{4}\) from 1: \( \frac{x^2}{25} = \frac{3}{4} \). By cross-multiplying, we find \(x^2 = \frac{75}{4} \). Take the square root to find \(x = \sqrt{\frac{75}{4}} = \frac{\sqrt{75}}{2} = \frac{5\sqrt{3}}{2} \).

Determine the width of the box

The width corresponding to \(x\) is \(2x\), since the box is centered along the x-axis. Therefore, the width of the box is \(2 \times \frac{5\sqrt{3}}{2} = 5\sqrt{3} \) feet.

Key concepts

Ellipse Equation

The equation of an ellipse is fundamental to understanding its geometrical properties. An ellipse centered at the origin of a coordinate system is described by the equation: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where:

• \(a\) is the semi-major axis,
• \(b\) is the semi-minor axis.

This equation is important for tasks like determining the shape and size of elliptical objects, such as arches and orbits.
In our doorway problem, even though it's a half-ellipse, we use the full equation for determining where points on the ellipse lie. Calculating the intersection of specific heights with this curve helps solve practical problems like ensuring a box can fit through an arch without colliding with it.

Semi-major Axis

The semi-major axis of an ellipse is the longest radius that extends from the center to the ellipse's edge. It is crucial in determining the overall width of an ellipse's shape. For this specific problem, the doorway arch has a width of 10 feet, therefore, the semi-major axis is half of this width: \[ a = \frac{10}{2} = 5 \text{ feet} \] This axis is always aligned with the ellipse's major direction (the longest diameter). Knowing the semi-major axis helps define the horizontal scope a box could span while passing through the arch, without exceeding the limits of the ellipse.

Semi-minor Axis

The semi-minor axis represents the shortest radius that goes from the center to the edge of the ellipse. It defines the maximum height from the center of the ellipse (or in this case, the highest point of the elliptical arch).
In our doorway, the height is given as 4 feet, meaning: \[ b = 4 \text{ feet} \] This measurement is less than the semi-major axis, characterizing the "flattened" top of the ellipse compared to its width. Understanding the semi-minor axis aids in calculating the highest possible edges where an object, like a box, would not touch the top portion of the arch.

Geometry Problem Solving

Solving geometry problems involving ellipses often requires applying the ellipse equation to find specific points of interest. In this scenario, we need to know if a box can fit through an elliptical arch. By setting the height of the box as a fixed coordinate and substituting it into the ellipse equation, we can solve for the corresponding width.
Let's recap: the problem determined the height of the box (2 feet), so we substitute \( y = 2 \) into the equation: \[ \frac{x^2}{25} + \frac{2^2}{16} = 1 \] Solving this identifies the possible width of the box that can pass beneath without hitting the arch. By finding \( x \), we calculate the maximum width the box can be: \[ 2x = 5\sqrt{3} \text{ feet} \] This shows how algebra combined with geometric understanding can effectively solve practical design scenarios involving ellipses.