Problem 28 - Algebra and Trigonometry

Question

A polar equation of a conic is given. (a) Show that the conic is a hyperbola, and sketch its graph. (b) Find the vertices anddirectrix, and indicate them on the graph. (c) Find the center of the hyperbola, and sketch the asymptotes. $$ r=\frac{6}{2+7 \cos \theta} $$

Step-by-Step Solution

Answer

The conic is a hyperbola with vertices at \(\frac{2}{3}\) and \(\frac{6}{-5}\), centered around the pole and asymptotes passing through the origin.

Step 1: Understand the Polar Equation of Conic

The given polar equation is in the form \( r = \frac{ed}{1 + e\cos\theta} \) compared to the form \( r = \frac{ed}{1 + e\cos\theta} \) where \( e \) is the eccentricity and \( d \) is the semi-latus rectum of the conic. Since the denominator has the form \( 1 + e\cos\theta \), here \( e = 7 \) and \( d = 3 \). For conics, if \( e > 1 \), then the conic is a hyperbola.

Step 2: Conclusion on Conic Type

Given \( e = 7 \), which is greater than 1, we can confirm the conic described by the polar equation is a hyperbola.

Step 3: Identify the Vertices and Directrix

For a hyperbola, the vertices in polar form occur where \( \theta = 0 \) or \( \theta = \pi \). Substitute \( \theta = 0 \) into the polar equation to find the vertex:\[ r = \frac{6}{2 + 7 \cdot 1} = \frac{6}{9} = \frac{2}{3}. \] Substitute \( \theta = \pi \) into the polar equation to find the other vertex:\[ r = \frac{6}{2 - 7 \cdot 1} = \frac{6}{-5}. \]These points on the pole represent the vertices. The distance to the directrix (not easily visualizable in polar form) is \( \frac{d}{e} = \frac{3}{7} \).

Step 4: Sketch the Graph of the Hyperbola

Plotting these values will create a hyperbola, with vertices at \( \frac{2}{3} \) and \( \frac{6}{-5} \). The directrix is a line perpendicular to the major axis, offset at \( \frac{3}{7} \) from the foci. On the polar grid, sketch will be indicative based on these effects.

Step 5: Find the Center of the Hyperbola

Remember, the hyperbola does not have a center in the same sense as Cartesian hyperbolas since its structure is defined relative to the directrix and focus, here roughly centered around the origin in polar form.

Step 6: Sketch the Asymptotes of the Hyperbola

In polar form, asymptotes are not as straightforward as Cartesian forms. However, generally, as \( \theta \to \frac{\pi}{2} \pm n\pi \), the asymptotes pass through the origin, outward along radial lines intersecting the polar vertices. Draw continuous lines extending through these points, adjusted for the polar origin. These are the asymptotes.

Key Concepts

Hyperbola

A hyperbola is a type of conic section that can be defined using a polar equation. In polar coordinates, the equation takes the form \( r = \frac{ed}{1 + e\cos\theta} \) or \( r = \frac{ed}{1 + e\sin\theta} \). The parameter \( e \) stands for eccentricity, which tells us about the shape of the conic section. If \( e > 1 \), the conic is a hyperbola. Hyperbolas have two separate curves called branches.
These branches mirror each other and are divided by a central figure. In polar form, these branches are not as symmetrical as in Cartesian coordinates but spread out from the origin, creating the typical hyperbolic shape.
When graphing a hyperbola in polar coordinates, it’s important to notice how the hyperbola appears visually, depending on the polar angle \( \theta \). Understanding the equation helps in visualizing its structure and how it gets wider or narrower with changes in \( \theta \).

Vertices and Directrix

Vertices of a hyperbola in polar form refer to the closest and farthest points from the pole (origin) along the main axis. They can be found by substituting \( \theta = 0 \) or \( \theta = \pi \) into the polar equation. In our exercise, the vertices were found at \( r = \frac{2}{3} \) and \( r = \frac{6}{-5} \). These represent the distances at specific polar angles and give the hyperbola its characteristic shape.
The directrix of a conic section is a specific imaginary line used in defining the conic mathematically. For hyperbolas in polar coordinates, it is a line perpendicular to the axis of symmetry and equidistant from the vertices in polar form. Its distance from the pole is given by \( \frac{d}{e} \), providing crucial information on the hyperbola’s dimension. In our exercise's equation, this distance is \( \frac{3}{7} \). Understanding both vertices and directrices aids in plotting and visualizing the complete graph of the hyperbola.

Conic Sections

Conic sections are curves obtained by intersecting a plane with a cone. They include ellipses, parabolas, circles, and hyperbolas, depending on the angle and position of the intersecting plane. The equation of a conic in polar form is usually given with parameters that describe its specific type and dimensions.
Each conic section has unique properties:

• Circle: Eccentricity is zero (looks like a circle).
• Ellipse: Eccentricity is between 0 and 1.
• Parabola: Eccentricity equals 1 (U-shaped curve).
• Hyperbola: Eccentricity greater than 1 (two open curves).

For students, understanding these definitions is key to distinguishing between different conics and identifying them in both Cartesian and polar forms. Grasping conic sections in polar coordinates adds depth to their geometric understanding and is crucial for advanced applications, like satellite orbits.

Eccentricity

Eccentricity is a crucial parameter in defining all conics, describing how "stretched" they are compared to a circle. In a polar equation, it’s found directly as \( e \), such as \( e = 7 \) in our hyperbola example, indicating a hyperbola due to \( e > 1 \). For any conic section:

• \( e = 0 \): It's a circle, perfectly round.
• \( 0 < e < 1 \): It's an ellipse, more oblong.
• \( e = 1 \): It's a parabola, with a symmetric open curve.
• \( e > 1 \): It's a hyperbola, with two divergent branches.

Eccentricity impacts how the conic’s shape behaves on a graph. In polar equations, higher eccentricity in hyperbolas illustrates how wide their opening becomes relative to the origin and axis, drawing the graph into varied shapes. Therefore, mastering eccentricity allows students to unlock the behavior of these fascinating curves and predict their appearances under different mathematical conditions.