Problem 32 - Precalculus: Functions and Graphs

Question

Find a polar equation of the conic with focus at the pole that has the given eccentricity and equation of directrix. $$e=\frac{3}{4}, \quad r \sin \theta=5$$

Step-by-Step Solution

Answer

The polar equation is \( r = \frac{15}{4 + 3\sin\theta} \).

Step 1: Understanding Polar Equations of Conics

A polar equation of a conic section with a focus at the pole is generally given by the formula \( r = \frac{ed}{1 - e\sin\theta} \) or \( r = \frac{ed}{1 + e\sin\theta} \), where \( e \) is the eccentricity and \( d \) is the distance from the pole to the directrix. Here, \( e = \frac{3}{4} \). Since the directrix equation given is \( r \sin \theta = 5 \), the distance \( d \) is 5.

Step 2: Identify the Conic Type

Since the eccentricity \( e = \frac{3}{4} \) is less than 1, we are dealing with an ellipse. For ellipses, the polar equation can be structured using \( r = \frac{ed}{1 + e\sin\theta} \) because the directrix equation \( r\sin\theta = 5 \) implies sin in the denominator.

Step 3: Plug Values into the Equation

Now that we have the formula \( r = \frac{ed}{1 + e\sin\theta} \), substitute \( e = \frac{3}{4} \) and \( d = 5 \) into the formula, yielding \( r = \frac{\frac{3}{4} \cdot 5}{1 + \frac{3}{4}\sin\theta} \).

Step 4: Simplify the Expression

Simplifying \( r = \frac{\frac{3}{4} \cdot 5}{1 + \frac{3}{4}\sin\theta} \) gives \( r = \frac{\frac{15}{4}}{1 + \frac{3}{4}\sin\theta} \). Further simplification leads to \( r = \frac{15}{4 + 3\sin\theta} \).

Key Concepts

Conic Sections

Conic sections are fascinating geometric shapes that arise when a plane intersects a cone. These shapes include circles, ellipses, parabolas, and hyperbolas. In polar coordinates, these sections can be described with equations that are often more intuitive compared to their Cartesian counterparts.

When portraying these conics in polar form, the pole usually represents the focus, and the radii, denoted as \( r \), change according to the angle \( \theta \). The essential aspect of these sections is dictated by the eccentricity \( e \), which differentiates whether the conic is a circle (\( e = 0 \)), an ellipse (\( 0 < e < 1 \)), a parabola (\( e = 1 \)), or a hyperbola (\( e > 1 \)).

Understanding conic sections in polar coordinates can simplify certain problems in geometry and physics, particularly those involving radial symmetry.

Eccentricity

Eccentricity \( e \) is a crucial parameter in determining the nature of a conic section. Think of it as a measure that tells us how much a conic deviates from being circular.

• If \( e = 0 \), the conic is a perfect circle, as there is no deviation from circularity.
• If \( 0 < e < 1 \), the conic is an ellipse, representing a slightly stretched circle.
• If \( e = 1 \), we have a parabola; a unique case where the conic extends infinitely in one direction.
• If \( e > 1 \), the conic becomes a hyperbola, appearing as two mirrored, open curves.

The eccentricity directly influences the polar equation format. For instance, in our exercise, the eccentricity \( e = \frac{3}{4} \) indicates an elliptical shape.

The eccentricity not only determines the shape but also affects how the equation of the conic is structured, especially in relation to the directrix.

Directrix

The directrix is a critical line used to define conics, offering another way to describe these shapes besides using focus and eccentricity. In polar coordinates, the directrix helps give a specific position to the conic relative to the pole.

Depending on the sign in the equation, the directrix can either attract or repel the conic section. In our particular exercise, the directrix is given by \( r \sin \theta = 5 \). This implies a distance of 5 from the pole along the line perpendicular to the \( \theta = 0 \) direction.

Understanding this line’s equation is crucial for properly setting the polar equation of the conic. It aids in knowing how much the conic will be stretched or compressed relative to the pole, ensuring accurate graphing and analysis of the curve.