Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Name the conic or limiting form represented by the given equation. Usually you will need to use the process of completing the square. $$ 16 x^{2}-9 y^{2}+192 x+90 y-495=0 $$

Step-by-Step Solution

Answer

The given equation represents a hyperbola.

Step 1: Rearrange the Equation

First, we rearrange the given equation according to the terms to prepare them for completing the square: \[16x^2 + 192x - 9y^2 + 90y = 495\]

Step 2: Factor and Prepare to Complete the Square for x

For the \(x\)-terms, factor out 16 from \(x^2\) and \(x\) terms: \[16(x^2 + 12x)\]

Step 3: Complete the Square for x

To complete the square inside the \(x\)-terms:- Take half of 12, square it, which is 36, and add and subtract it inside the parenthesis: \[16(x^2 + 12x + 36 - 36)\]- Rewrite it as: \[16((x+6)^2 - 36) = 16(x+6)^2 - 576\]

Step 4: Factor and Prepare to Complete the Square for y

For the \(y\)-terms, factor out -9 from the \(y^2\) and \(y\) terms:\[-9(y^2 - 10y)\]

Step 5: Complete the Square for y

To complete the square inside the \(y\)-terms:- Take half of 10, square it, which is 25, and add and subtract it inside the parenthesis: \[-9(y^2 - 10y + 25 - 25)\]- Rewrite it as:\[-9((y-5)^2 - 25) = -9(y-5)^2 + 225\]

Step 6: Combine Completed Squares and Constants

Combine everything into the equation:\[16(x+6)^2 - 576 - 9(y-5)^2 + 225 = 495\]Simplify it to:\[16(x+6)^2 - 9(y-5)^2 = 846\]

Step 7: Divide by Constant to Standardize Equation

Divide the entire equation by 846 to get it into the standard form of a hyperbola:\[\frac{16(x+6)^2}{846} - \frac{9(y-5)^2}{846} = 1\]Simplify further:\[\frac{(x+6)^2}{52.875} - \frac{(y-5)^2}{94} = 1\]

Key Concepts

Completing the Square

Completing the square is a mathematical technique used to transform a quadratic equation into a perfect square trinomial. This is particularly useful for understanding the properties of conic sections, such as circles, ellipses, or hyperbolas.
To complete the square, you follow these steps:

• Identify the quadratic term and linear term in the expression like \(x^2 + bx\).
• Take half of the coefficient of the linear term (\(b\)), square it, and add and subtract this square within the equation.
• Reformulate the expression as a squared binomial: for example, \(x^2 + bx + \left(\frac{b}{2}\right)^2 = (x + \frac{b}{2})^2\).

In the given problem, we apply this technique twice, once for the \(x\)-terms and once for \(y\)-terms. This allows us to rewrite the original quadratic terms in a way that makes it easier to identify the conic section from the equation's structure.

Hyperbola

A hyperbola is a type of conic section that can be recognized by its distinct, two-part structure. It is typically expressed in the standard form of an equation:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]where \((h, k)\) is the center of the hyperbola, and \(a\) and \(b\) determine the distances from the center to vertices and co-vertices respectively.
Hyperbolas have several key properties:

• They consist of two separate curves called "branches" that mirror each other.
• The asymptotes are lines that the curves approach but never touch, offering the lines of symmetry.
• The foci are points located along the axis of each branch at a distance greater than the center to vertex distance.

In the problem above, converting the equation using completing the square leads to a format that reveals it as a hyperbola, due to the subtraction in the structure \((x+6)^2 - (y-5)^2\). This indicates the distinctive two-branches of this conic section.

Standard Form Equation

The standard form equation is essential in identifying and classifying conic sections. For hyperbolas, the standard form provides clear insights into the hyperbola's characteristics and orientation.
To convert a quadratic equation into a hyperbola's standard form:

• Ensure all terms are on one side and the equation is set to zero.
• Use completing the square for \(x\) and \(y\) terms separately to create perfect square trinomials.
• Divide through by the constant to make the equation equal 1, which is crucial for matching the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\).

For the given exercise, after completing the square and manipulating constants, our equation was transformed into this recognizable standard form of a hyperbola. This highlights the beauty of expressing a complex equation into a form revealing its geometric nature and properties.