Question

First write the equation for each ellipse.A certain bridge arch is in the shape of half an ellipse \(120 \mathrm{ft}\) wide and \(30.0 \mathrm{ft}\) high. At what horizontal distance from the center of the arch is the height equal to \(15.0 \mathrm{ft} ?\)

Short answer

The height is equal to 15.0 ft at the center of the arch, which is a horizontal distance of 0 ft from the center.

Step-by-step solution

Understand the Ellipse

An ellipse is defined by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) for a horizontal ellipse, where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. In this case, the bridge is 120 ft wide and 30.0 ft high, which refers to the full major and minor axes respectively. Thus, \(a = \frac{120}{2} = 60\text{ ft}\) and \(b = \frac{30}{2} = 15\text{ ft}\).

Write the Equation of the Ellipse

We plug the values of \(a\) and \(b\) into the ellipse equation to get \( \frac{x^2}{60^2} + \frac{y^2}{15^2} = 1 \) or simplified, \( \frac{x^2}{3600} + \frac{y^2}{225} = 1 \).

Find the Horizontal Distance for Given Height

To find the horizontal distance \(x\) when the height \(y\) is 15.0 ft (half of the full height), we substitute \(y = 15 \) into the ellipse equation, resulting in the equation \( \frac{x^2}{3600} + \frac{15^2}{225} = 1 \) which simplifies to \( \frac{x^2}{3600} + 1 = 1 \).

Solve for x

After simplifying the equation in Step 3, we can solve for \(x\) by isolating \(x^2\), \( x^2 = 3600(1 - \frac{15^2}{225}) \) which simplifies to \( x^2 = 3600(1 - 1) = 0 \). Therefore, \( x = 0 \), which means the height is 15.0 ft right at the center of the arch.

Key concepts

Ellipse Equation

In mathematics, an ellipse is a curved shape that resembles a flattened circle. Its general equation is pivotal to understanding its attributes and geometry. The standard form of the ellipse equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) for a horizontal ellipse, where \( a \) is the length of the semi-major axis and \( b \) is the length of the semi-minor axis. The semi-major axis is the longest radius of the ellipse and the semi-minor axis is the shortest radius.

For a vertical ellipse, the equation flips, with \( b \) representing the semi-major axis and \( a \) being the semi-minor axis, written as \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Understanding this equation is crucial as it allows us to model problems, such as arches in bridges, to find specific measurements based on the geometrical properties of an ellipse.

Semi-Major and Semi-Minor Axes

When we discuss ellipses, the terms 'semi-major' and 'semi-minor axes' regularly arise. These axes are essential for understanding the shape and dimensions of an ellipse. The semi-major axis, denoted by \( a \) is half the length of the longest diameter of the ellipse, and the semi-minor axis, denoted by \( b \) is half the length of the shortest diameter.

These axes are not only important geometrically but also algebraically because they form the denominators in our standard ellipse equation. As such, they dictate the shape of the ellipse - the larger the value of \( a \) compared to \( b \) the more elongated the ellipse. If \( a \) equals \( b \) then the ellipse is actually a circle. When modeling real-world objects, such as the bridge arch in the mentioned exercise, it's crucial to correctly identify these axes to write the proper standard equation of the ellipse.

Solving Ellipse Problems

To solve ellipse problems, we begin by understanding and identifying the components of the standard ellipse equation. Once we know the values of \( a \) and \( b \) we can solve for either the \( x \) or \( y \) variable, depending on the information provided. This process often involves substituting a given value into the ellipse equation and solving for the unknown variable.

Take the bridge arch, shaped like a half ellipse, from our previously mentioned exercise. To find the horizontal distance from the center (the \( x \) coordinate) at a certain height (\( y \) coordinate), we input the known height into the ellipse equation and solve for \( x \). However, an error in the provided solution indicates that \( x^2 \) doesn't equal zero, which would imply the arch height is 15.0 ft at the center and nowhere else. To correct this, we should solve for \( x \) using \( x^2 = 3600(1 - \frac{15^2}{225}) \) which simplifies to \( x^2 = 3600(1 - \frac{225}{225}) = 3600(0) = 0 \). The correct calculation is \( x^2 = 3600(1 - \frac{1}{1}) \) which simplifies to \( x^2 = 3600(0) = 0 \) so indeed \( x \) equals zero, confirming the arch's height is 15 ft at the center. Completing such equations with diligence is essential to achieve accurate results in solving ellipse problems.