Problem 24 - Algebra and Trigonometry

Question

A polar equation of a conic is given. (a) Show that the conic is an ellipse, and sketch its graph. (b) Find the vertices and directrix, and indicate them on the graph. (c) Find the center of the ellipse and the lengths of the major and minor axes. $$ r=\frac{18}{4+3 \cos \theta} $$

Step-by-Step Solution

Answer

It's an ellipse with vertices at (10.29, 0) and (-10.29, 0), center at (0, 0), and axes 10.29 and 7.73 long.

Step 1: Identify the Conic Type

The given polar equation is of the form \( r = \frac{ed}{1 + e\cos\theta} \). By comparing, the equation \( r = \frac{18}{4 + 3\cos\theta} \) can be rewritten as \( r = \frac{18/4}{1 + \frac{3}{4}\cos\theta} \), implying \( e = \frac{3}{4} \). Since \( e < 1 \), the conic is an ellipse.

Step 2: Identify the Semi-Major Axis Length

The semi-major axis \( a \) can be found using the formula \( a = \frac{d}{1-e^2} \). Here, \( d = \frac{18}{4} = 4.5 \) and \( e = \frac{3}{4} \). Therefore, \( a = \frac{4.5}{1 - (\frac{3}{4})^2} = \frac{4.5}{1 - \frac{9}{16}} = \frac{4.5}{\frac{7}{16}} = \frac{4.5 \times 16}{7} \approx 10.29 \).

Step 3: Identify the Semi-Minor Axis Length

The semi-minor axis \( b \) can be found using the formula \( b = a\sqrt{1-e^2} \). Therefore, \( b = 10.29\sqrt{1-(\frac{3}{4})^2} = 10.29\sqrt{\frac{7}{16}} = 10.29\times\frac{\sqrt{7}}{4} \approx 7.73 \).

Step 4: Find the Vertices and Directrix

The vertices are at \( (a, 0) \) and \( (-a, 0) \), which gives \( (10.29, 0) \) and \( (-10.29, 0) \). The directrix is \( x = \pm\frac{d}{e} = \pm\frac{4.5}{\frac{3}{4}} = \pm 6 \).

Step 5: Identify the Center and Draw the Ellipse

The center of the ellipse is at the origin \((0,0)\). Sketch the graph by plotting the vertices at \((10.29,0)\) and \((-10.29,0)\), drawing the ellipse around these points while noting that the directrix is at \(x=\pm 6\).

Step 6: Summarize Findings

We found that the conic is an ellipse. The center is at \((0,0)\), vertices are \((10.29, 0)\) and \((-10.29, 0)\), and the lengths of the semi-major and semi-minor axes are approximately 10.29 and 7.73, respectively.

Key Concepts

Ellipse Identification

To determine if a polar equation represents an ellipse, it’s crucial to understand the role of eccentricity, denoted by \( e \). The eccentricity is a measure of how "stretched out" the conic section is. For an ellipse, the eccentricity satisfies the condition \( e < 1 \).

Given the equation \( r = \frac{18}{4 + 3\cos\theta} \), we can rewrite it in the standard conic polar form \( r = \frac{ed}{1 + e\cos\theta} \). The comparison leads to \( e = \frac{3}{4} \) (since it matches the coefficients comparable to \( e\cos\theta \)). Because \( e = \frac{3}{4} < 1 \), this equation indeed represents an ellipse.

Identifying an ellipse by its polar equation involves recognizing this fundamental property tied to its eccentricity.

Semi-Major and Semi-Minor Axes

The semi-major and semi-minor axes are important dimensions of an ellipse. These axes define the maximum and minimum radii, denoting how far the ellipse stretches.

To find the semi-major axis, \( a \), in polar form, we use the formula \( a = \frac{d}{1-e^2} \). For our problem:

• \( e = \frac{3}{4} \)
• \( d = 4.5 \)

Plug these into the formula to get \( a = \frac{4.5}{1-(\frac{3}{4})^2} \approx 10.29 \).

For the semi-minor axis, \( b \), we use \( b = a\sqrt{1-e^2} \). This calculation gives us \( b \approx 7.73 \).

These lengths tell us how wide and tall the ellipse stretches in relation to its center.

Vertices and Directrix

In an ellipse, the vertices represent the farthest points on its major axis and are key to its shape. They are located at \((a, 0)\) and \((-a, 0)\). Here, substituting our results, we find the vertices at \((10.29, 0)\) and \((-10.29, 0)\). These points help in sketching the ellipse and understanding its orientation.

The directrix is another essential feature of an ellipse, often serving as a reference line away from which distances are measured to define the ellipse. Using the formula for the location of the directrix, \( x = \pm\frac{d}{e} \), we find it at \( x = \pm 6 \) in our example.

By marking the vertices and determining the position of the directrix, you can more accurately plot and understand the geometric properties of the ellipse.