Problem 20 - Algebra and Trigonometry

Question

15–22 (a) Find the eccentricity and identify the conic. (b) Sketch the conic and label the vertices. $$r=\frac{5}{2-3 \sin \theta}$$

Step-by-Step Solution

Answer

The conic is a hyperbola with eccentricity 3. Vertex: (2.5, \( \frac{\pi}{2} \)) on the polar plot.

Step 1: Identify the Conic Section Formula Type

The given polar equation is in the form \( r = \frac{ed}{1 - e\sin \theta} \). This form indicates a conic section where the denominator's coefficient of \( \sin \theta \) determines \( e \) (eccentricity) and the direction of the directrix.

Step 2: Determine the Eccentricity

In the given equation, \( r = \frac{5}{2 - 3 \sin \theta} \), compare it to \( r = \frac{ed}{1 - e \sin \theta} \). The eccentricity \( e = 3 \) appears as the coefficient of \( \sin \theta \). Since \( e > 1 \), the conic is a hyperbola.

Step 3: Find the Directrix

The numerator \( ed = 5 \) gives \( d = \frac{5}{3} \). For a hyperbola when \( \sin \theta \) is involved, the directrix is parallel to the horizontal axis, positioned at \( y = -\frac{d}{1-e} = -\frac{5}{3-1} = -\frac{5}{2} \) units from the pole.

Step 4: Locate the Vertices

To find the vertices of the hyperbola in polar form, set \( \theta = 0 \) and \( \theta = \pi \):- For \( \theta = \frac{\pi}{2} \), \( r = \frac{5}{2} = 2.5 \) (vertex).- For \( \theta = \frac{3\pi}{2} \), no real vertex since the expression does not yield a valid radius.

Step 5: Sketch the Hyperbola

Draw the polar plot placing the pole (origin) and mark the vertex found at \( r = 2.5 \) with \( \theta = \frac{\pi}{2} \). Since \( e > 1 \), the conic is a hyperbola opening vertically, symmetrical to the directrix line found earlier. Mark the focus at the origin.

Key Concepts

Eccentricity

Eccentricity is a key characteristic of a conic section, which helps to distinguish its type. In polar coordinates, eccentricity, denoted as \( e \), is usually identified as the coefficient in front of the trigonometric function in the denominator of the conic's equation.

The general form for conic sections is \( r = \frac{ed}{1 - e\sin \theta} \) or \( r = \frac{ed}{1 - e\cos \theta} \). Here, \( e \) is the eccentricity.

- If \( e = 0 \), the conic is a circle.
- If \( 0 < e < 1 \), it's an ellipse.
- If \( e = 1 \), it's a parabola.
- If \( e > 1 \), the conic is a hyperbola.

For this exercise, we notice \( e = 3 \), which clearly indicates a hyperbola due to its value being greater than 1.

Polar Coordinates

Polar coordinates give us a unique way of expressing the position of points where each point is determined by a distance from the origin and an angle from a fixed direction. This method of location is particularly useful in describing curves like conic sections.

In comparison to Cartesian coordinates that use \( (x, y) \), polar coordinates use \( (r, \theta) \), where:

• \( r \) is the radius or the distance from the origin to the point
• \( \theta \) is the angle between the positive x-axis and the line connecting the origin to the point

By understanding how polar coordinates construct curves, you can graph equations like \( r = \frac{5}{2-3 \sin \theta} \) and identify essential features like vertices or directrices.

Hyperbola

A hyperbola is one of the types of conic sections characterized by an eccentricity greater than 1. Its equation in polar coordinates often involves a trigonometric function in the denominator.

Key properties of a hyperbola include two separate curves and a characteristic of opening either horizontally or vertically depending on the equation's structure and \( \theta \). In our case, the hyperbola is defined by \( r = \frac{5}{2 - 3 \sin \theta} \), indicating that the hyperbola opens vertically.

You'll find the hyperbola opens along the line of directrix which can be determined using the equation's structure. The major features include vertices and foci, critical points in understanding the hyperbola's shape and orientation.

Vertices of the Conic

Vertices in a conic section, such as a hyperbola, are the points where the curve takes a maximum or minimum distance from the center or focus. This exercise requires you to locate these vertices on a hyperbola in polar coordinates.

To find the vertices for a hyperbola like ours \( r = \frac{5}{2-3\sin\theta} \), you often set specific angles such as \( \theta = 0 \) and \( \theta = \pi \) for simplicity.

• When \( \theta = \frac{\pi}{2} \): \( r = 2.5 \), which gives us one of the vertices.
• Checking other values for \( \theta \) can confirm the location or absence of additional vertices.

These points are crucial for sketching and understanding the broader structure of the conic.