91Ó°ÊÓ

Let's begin by understanding the concept of the semi-major axis. An ellipse has two main axes: the semi-major axis and the semi-minor axis. The semi-major axis, denoted by \(a\), is the longest radius that spans from the center of the ellipse to its perimeter.
In the problem's context, the circular hole would maintain its diameter if the roof were flat. Here, the radius of the vent pipe is 4 inches. Since the semi-major axis is not affected by the roof's pitch, it stays as 4 inches. This is calculated as:
\[a = \text{radius} = 4\text{ inches}\]
So, even when the roof has a slope, the semi-major axis of the elliptical hole stays 4 inches.

Next, let's dive into understanding the semi-minor axis. The semi-minor axis, denoted by \(b\), is the shortest radius that runs perpendicular to the semi-major axis in an ellipse.
The pitch of the roof impacts this axis significantly. The pitch given is \(\frac{5}{4}\), which means for a 4-unit run, there's a 5-unit rise. Because of this, the semi-minor axis is calculated by adjusting the semi-major axis by the pitch ratio.
The formula to determine the semi-minor axis \(b\) is:
\[b = \frac{a}{\text{pitch}}\]
Substituting the given values:
\[b = \frac{4}{\frac{5}{4}} = \frac{4 \times 4}{5} = 3.2 \text{ inches}\]
This shows that while the semi-major axis (\(a\)) remains 4 inches, the semi-minor axis (\(b\)) adjusts to 3.2 inches due to the roof pitch.

The roof pitch is crucial in understanding how the elliptical hole needs to be cut. Roof pitch defines the steepness of a roof and is given as a ratio of rise over run.
In this case, the roof pitch is \(\frac{5}{4}\), indicating that for every 4 horizontal units, the vertical rise is 5 units.
This roof pitch transforms a circular cutout into an elliptical shape. It influences only the semi-minor axis of the ellipse. The semi-major axis remains constant at the pipe's radius.
Recognizing and accurately implementing the pitch ratio allows for precise adjustments to the dimensions for correct fitting without errors. This is achieved through the formula:
\[ \text{semi-minor axis } (b) = \frac{\text{semi-major axis } (a)}{\text{pitch}} \]

The radius of the vent pipe is the starting point for the entire calculation. A vent pipe's radius is half of its diameter. In this problem, the vent pipe's diameter is 8 inches, giving a radius of 4 inches.
This radius directly gives us the length of the semi-major axis of the elliptical hole in a sloped roof. For a flat roof, this radius would cut a perfect circle. On a sloped roof, the same hole becomes an ellipse.
The semi-major axis thus is:
\[a = 4 \text{ inches}\]
And the semi-minor axis is computed with respect to the roof pitch:
\[b = \frac{4}{\frac{5}{4}} = 3.2 \text{ inches}\]
Understanding the radius as the fundamental measurement helps streamline the calculations for the semi-major and semi-minor axes.