Problem 2 - Precalculus: Mathematical for Calculus

Question

Write a polar equation of a conic that has its focus at the origin and satisfies the given conditions. Hyperbola, eccentricity \(\frac{4}{3},\) directrix \(x=-3\)

Step-by-Step Solution

Answer

The polar equation is \( r = \frac{12}{3 - 4\cos(\theta)} \).

Step 1: Identify the General Form of the Polar Equation

For a conic with a focus at the origin, the polar equation can generally be written as \( r = \frac{ed}{1 + e\cos(\theta)} \) if the directrix is vertical or \( r = \frac{ed}{1 + e\sin(\theta)} \) if the directrix is horizontal. In this problem, the directrix is vertical, hence we use \( r = \frac{ed}{1 + e\cos(\theta)} \).

Step 2: Insert Eccentricity into the Formula

The given eccentricity \( e = \frac{4}{3} \), which indicates that the conic is a hyperbola (since \( e > 1 \)). Substitute this value into the formula: \( r = \frac{\frac{4}{3}d}{1 + \frac{4}{3}\cos(\theta)} \).

Step 3: Determine the Appropriate Sign in the Formula

Since the directrix is \( x = -3 \), the conic opens to the left. This modifies the polar equation to \( r = \frac{ed}{1 - e\cos(\theta)} \) instead of an addition sign.

Step 4: Substitute the Directrix into the Formula

Convert the directrix equation into its distance. The absolute distance from the origin to the directrix is \( |d| = 3 \). Thus, substitute \( d = 3 \) into the equation to get \( r = \frac{4 \times 3}{3 - 4\cos(\theta)} \).

Step 5: Simplify the Formula

Simplify the equation \( r = \frac{12}{3 - 4\cos(\theta)} \) to get the final polar equation for this hyperbola.

Key Concepts

Understanding Conic Sections

Conic sections are curves obtained by slicing a cone with a plane. They form several distinct shapes depending on the angle and position of the slice. These shapes include circles, ellipses, parabolas, and hyperbolas.

• Circle: Created when the plane cuts the cone perpendicular to its axis.
• Ellipse: Occurs when the angle of the slice is less than that required for a parabola but less than 90 degrees.
• Parabola: Produced when the plane is parallel to a generating line of the cone.
• Hyperbola: Forms when the plane cuts through both halves of the cone, causing two separate curves.

Understanding conic sections is essential to solving problems in polar coordinates, where the focus can lie at the origin and the directrix forms a boundary for the setup of the equation.

The Hyperbola

A hyperbola consists of two separate curves called branches. These branches open in opposite directions and have distinct focal points commonly referred to as foci. For hyperbolas, the eccentricity is always greater than 1, distinguishing them from other conic sections like parabolas (eccentricity = 1) and ellipses (eccentricity < 1).
The standard form of the hyperbola in rectangular coordinates can be transformed into polar coordinates, especially when the focus is at the origin. This unique representation helps in understanding the nature of hyperbolas, particularly their symmetry and components like axes and vertices.

Eccentricity: Key to Conic Identity

Eccentricity is a measure that determines the shape of a conic section. It is denoted by the letter \(e\) in mathematics.

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

In our example, the eccentricity \(e = \frac{4}{3}\), which is greater than 1, confirms that the conic is indeed a hyperbola. This focuses the solution process, aligning the formula to reflect the hyperbolic nature using this particular \(e\) value in the polar equation framework.

Role of the Directrix in Polar Equations

In the context of polar equations for conics, the directrix is a fixed reference line. This reference line helps define the conic's relationship to the pole (or origin in polar coordinates). The directrix's position significantly influences the polar equation’s structure.
When the directrix is vertical, the standard form of a polar equation becomes \( r = \frac{ed}{1 \pm e\cos(\theta)} \), where \(d\) is the distance from the pole to the directrix. This configuration helps achieve the desired shape and orientation of the conic.
In the given exercise, the directrix is represented by the line \(x = -3\). Being a vertical line, and knowing it influences the conic to open to the left, the equation requires a subtraction sign: \( r = \frac{ed}{1 - e\cos(\theta)} \). Moreover, the absolute distance \(|d|\) is calculated as 3, ensuring precision in setting up the polar equation accordingly.