Problem 35 - Precalculus Mathematics for Calculus

Question

Show by eliminating the parameter \(\theta\) that the following parametric equations represent a hyperbola: $$x=a \tan \theta \quad y=b \sec \theta$$

Step-by-Step Solution

Answer

The parametric equations represent a hyperbola: \\\( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \\\).

Step 1: Express \\( \tan \theta \\\) in terms of \\( x \\\)

We are given that \( x = a \tan \theta \). To eliminate the parameter \( \theta \), first solve for \( \tan \theta \) in terms of \( x \): \( \tan \theta = \frac{x}{a} \).

Step 2: Express \\( \sec \theta \\\) using \\( y \\\)

From the given parametric equation \( y = b \sec \theta \), solve for \( \sec \theta \): \( \sec \theta = \frac{y}{b} \).

Step 3: Use the identity \\( \sec^2 \theta = 1 + \tan^2 \theta \\\)

Utilize the Pythagorean identity \( \sec^2 \theta = 1 + \tan^2 \theta \) to relate the expressions for \( \sec \theta \) and \( \tan \theta \). Substitute \( \tan \theta = \frac{x}{a} \) to get \( \sec^2 \theta = 1 + \frac{x^2}{a^2} \).

Step 4: Substitute for \\( \sec^2 \theta \\\) in terms of \\( y \\\)

From Step 2, \( \sec^2 \theta = \left(\frac{y}{b}\right)^2 = \frac{y^2}{b^2} \).

Step 5: Equate the expressions for \\( \sec^2 \theta \\\)

Equating the two expressions from Steps 3 and 4 gives us: \(\frac{y^2}{b^2} = 1 + \frac{x^2}{a^2}\)

Step 6: Rearrange into the standard form of a hyperbola

Rearrange \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \). This is the standard equation of a hyperbola centered at the origin. Thus, the given parametric equations represent a hyperbola.

Key Concepts

Eliminating the Parameter

When working with parametric equations, the concept of eliminating the parameter is used to convert the parametric form into a more familiar Cartesian form. This process involves finding a relationship between the variables by removing the parameter. In our exercise, we have the parametric equations:

• For x: \( x = a \tan \theta \)
• For y: \( y = b \sec \theta \)

To eliminate \( \theta \), solve each equation for \( \tan \theta \) and \( \sec \theta \) respectively:

• \( \tan \theta = \frac{x}{a} \)
• \( \sec \theta = \frac{y}{b} \)

These expressions allow us to link \( x \) and \( y \) directly using a known trigonometric identity, which we will explore in the next section. By establishing this link, we eliminate the parameter \( \theta \) effectively, enabling us to work with a standard form equation.

Hyperbola

A hyperbola is a type of conic section, and you can recognize its equation in the form:\[ \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \]This specific equation describes a hyperbola centered at the origin, with transverse and conjugate axes of lengths determined by \(a\) and \(b\). In the exercise, by eliminating the parameter \( \theta \), we transform the parametric equations into this familiar hyperbolic form. The resulting expression \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \) shows that both variables in the parametric equations come together to describe a hyperbola.Understanding this shows how geometry and algebra connect through parametric and Cartesian formats. This helps define the paths of motion in physics or graphically represent relationships in algebra. Hyperbolas have practical applications in real-world scenarios, such as satellite paths or navigation systems, emphasizing their significance.

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry:\[ \sec^2\theta = 1 + \tan^2\theta \]This identity is key to simplifying the parametric equations into a single equation. After expressing \( \tan \theta \) and \( \sec \theta \) in terms of \( x \) and \( y \), substitute these back into the Pythagorean identity:

• \( \sec^2 \theta = 1 + \frac{x^2}{a^2} \)
• \( \sec^2 \theta = \frac{y^2}{b^2} \)

By equating the two expressions for \( \sec^2 \theta \), we link the Cartesian coordinates:\[ \frac{y^2}{b^2} = 1 + \frac{x^2}{a^2} \]Rearranging this gives the standard hyperbola equation, confirming our transformation. The Pythagorean identity simplifies these steps, illustrating its power in connecting trigonometric relationships and algebraic expressions. It functions as a tool to bridge parametric definitions with their Cartesian counterparts, revealing hidden insights in mathematical structures.