Problem 64 - Precalculus: Functions and Graphs

Question

Find an equation for the indicated part of the hyperbola. Lower halves of the branches of \(\frac{x^{2}}{9}-\frac{y^{2}}{49}=1\)

Step-by-Step Solution

Answer

The equation for the lower halves is \( y = -\sqrt{\frac{49x^{2}}{9} - 49} \).

Step 1: Identify the Given Equation

The given equation \( \frac{x^{2}}{9} - \frac{y^{2}}{49} = 1 \) represents a hyperbola. This hyperbola is centered at the origin \((0,0)\) since no \(x\) or \(y\) shifts are indicated, and the transverse axis is horizontal.

Step 2: Rewrite as a Function for y²

We start by isolating \(y^{2}\) on one side of the equation. From the equation \( \frac{x^{2}}{9} - \frac{y^{2}}{49} = 1 \), we can rearrange it to: \(-\frac{y^{2}}{49} = 1 - \frac{x^{2}}{9}\).

Step 3: Solve for y²

Multiply through by 49 to solve for \( y^2 \): \(-y^{2} = 49(1 - \frac{x^{2}}{9})\). Simplifying gives \(-y^{2} = 49 - \frac{49x^{2}}{9}\), or \(y^{2} = \frac{49x^{2}}{9} - 49\).

Step 4: Solve for y

Take the square root of both sides to solve for \( y \): \(y = \pm \sqrt{\frac{49x^{2}}{9} - 49}\).

Step 5: Choose the Lower Half

Since we are interested in the lower halves of the branches, we select the negative root: \(y = -\sqrt{\frac{49x^{2}}{9} - 49}\).

Key Concepts

Equation of a Hyperbola

A hyperbola is a type of conic section that looks like two mirrored curves or branches facing away from each other. The equation of a hyperbola with a horizontal transverse axis is generally given as \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). In this form:

• \( a^2 \) is associated with the \( x \)-term and represents the distance along the transverse axis.
• \( b^2 \) is linked to the \( y \)-term and relates to the distance along the conjugate axis.

The equation we started with, \( \frac{x^2}{9} - \frac{y^2}{49} = 1 \), is indeed a hyperbola, centered at the origin \((0,0)\), because it adheres to the standard form. Here, \( a^2 = 9 \) and \( b^2 = 49 \). Knowing these insights, one can see that the hyperbola extends along the x-axis by \( 3 \) units on either side of the center (because \( \sqrt{9} = 3 \)), and it opens up along the y-axis with imaginary branches determined by \( b \). This imaginary aspect shows up in practice as the width between the vertices.

Solving Hyperbola Equations

To hone our understanding of equations and their transformations, solving equations is a crucial skill. Let's break down the process we followed to solve for \( y \) in a hyperbola equation:1. ***Identify and Isolate Terms:*** Start by isolating the \( y^2 \) term in the given hyperbola equation. Here, the equation \( \frac{x^2}{9} - \frac{y^2}{49} = 1 \) was rewritten as \( -\frac{y^2}{49} = 1 - \frac{x^2}{9} \).2. ***Eliminate the Fractions:*** Multiply by the denominator on the \( y^2 \) side to make the term simpler. Thus, multiplying through by 49, this becomes \( -y^2 = 49 - \frac{49x^2}{9} \).3. ***Solve for \( y^2 \):*** Rearrange to express \( y^2 \) alone. This leads to \( y^2 = \frac{49x^2}{9} - 49 \).4. ***Extract the Square Root:*** Calculate the square root to find \( y \) - remember, this gives both positive and negative roots, \( y = \pm \sqrt{\frac{49x^2}{9} - 49} \).
When calculating square roots, always consider whether the context requires both roots or just one of them.

Functions in Precalculus

In precalculus, functions are fundamental, serving as a bridge to understanding equations in mathematics. A function can express relationships between two quantities, often defined by an equation that binds the dependent and independent variables. Let's explore some key ideas:- ***Definition:*** A function is a relation where every input (or \( x \)-value) is paired with exactly one output (or \( y \)-value). In our exploration of hyperbolas, the function form comes into play when solving for \( y \). - ***Graph Interpretation:*** When considering hyperbolas, the graph is split into branches. For our equation, \( y = -\sqrt{\frac{49x^2}{9} - 49} \), we specifically chose the negative root to focus on the lower half branches. - ***Contextual Use:*** Precalculus often uses functions for 'real world' scenarios—understanding motion, curve trajectories, and more. By mastering functions and their interpretations, one gains tools for modeling real-life situations.Through practicing and understanding functions, students can significantly enhance their mathematical skills, paving the way for calculus and beyond.