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 (see Examples \(3-5) .\) 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: Reorder and Group Terms

First, let's reorder the terms for convenience: \[ 16x^2 + 192x - 9y^2 + 90y - 495 = 0. \] We will group the x-terms and y-terms together: \[ (16x^2 + 192x) + (-9y^2 + 90y) = 495. \]

Step 2: Factor Out Coefficients of Quadratic Terms

Factor out the coefficients of the squared terms separately:\[ 16(x^2 + 12x) - 9(y^2 - 10y) = 495. \]

Step 3: Complete the Square for the x-Terms

To complete the square for the x-terms, find the value to add and subtract: The expression is \( x^2 + 12x \). Half of \( 12 \) is \( 6 \), and \( 6^2 = 36 \). Therefore, we rewrite:\[ 16(x^2 + 12x + 36 - 36). \] Which becomes: \[ 16((x+6)^2 - 36). \] When simplified, it becomes the following expansion: \[ 16(x+6)^2 - 576. \]

Step 4: Complete the Square for the y-Terms

Now, complete the square for the y-terms: The expression is \( y^2 - 10y \). Half of \( -10 \) is \( -5 \), and \( (-5)^2 = 25 \). Rewrite as:\[ -9(y^2 - 10y + 25 - 25). \]Which becomes: \[ -9((y-5)^2 - 25). \] When simplified, it becomes: \[ -9(y-5)^2 + 225. \]

Step 5: Substitute Completed Squares Back Into Equation

Substitute the completed squares back into the equation:\[ 16(x+6)^2 - 576 - 9(y-5)^2 + 225 = 495. \]

Step 6: Combine and Simplify Constants

Combine and simplify the constants:\[ 16(x+6)^2 - 9(y-5)^2 = 495 + 576 - 225. \] This simplifies to:\[ 16(x+6)^2 - 9(y-5)^2 = 846. \]

Step 7: Identify the Conic

Recognize the form \[ \frac{(x+6)^2}{52.875} - \frac{(y-5)^2}{94} = 1. \] This equation is in the standard form of a hyperbola, \[ \frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1. \] Therefore, the given equation represents a hyperbola.

Key Concepts

Completing the Square

Completing the square is a method used to simplify and solve quadratic equations. It's useful when dealing with conic sections, like circles, ellipses, and hyperbolas. Here's how it works:

When you have a quadratic expression like \(ax^2 + bx + c\), the goal is to transform it into a perfect square trinomial, which looks like \((x+p)^2 - q\). This form is easier to understand and manipulate.

To complete the square:

• Take the coefficient of the \(x\) term (\(b\)), divide it by 2, and then square the result (\(\left(\frac{b}{2}\right)^2\)).
• Add and subtract that squared number inside the expression.
• Factor the perfect square trinomial and simplify where possible.

In the original exercise, we applied this to both \(x\)-terms and \(y\)-terms to help identify the conic section represented by the equation. This process enabled the equation to be rewritten in a more recognizable form, suitable for identifying hyperbolas.

Equation of a Hyperbola

A hyperbola is one of the conic sections formed by intersecting a plane with a double cone. The standard equation for a hyperbola is expressed as \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). This formula shows a horizontally oriented hyperbola.

Each component in the equation has a specific function:

• \((h, k)\) represents the center of the hyperbola.
• \(a\) and \(b\) are real numbers that determine the shape and size of the hyperbola's axes.

For a vertical hyperbola, the \(x\) and \(y\) terms are switched, making the equation \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\).

In the exercise, after completing the square for \(x\) and \(y\), the equation resembled the standard hyperbola form. Recognizing these elements allows us to identify the conic section accurately as a hyperbola.

Quadratic Equations

Quadratic equations are second-degree polynomial equations that have the general form \(ax^2 + bx + c = 0\). These equations are foundational in algebra and are particularly important in the study of conic sections.

There are various methods to solve quadratic equations:

• Factoring, where you express the quadratic as a product of its roots.
• Completing the square, which rearranges the equation into a perfect square trinomial.
• Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), which is a straightforward solution method for any quadratic equation.

In our exercise, understanding how to manipulate quadratic equations is crucial as it is the first step to simplifying and identifying the type of conic section we are dealing with by completing the square.

Standard Form of Conics

The standard form of conics refers to the equations that represent different conic sections like circles, ellipses, hyperbolas, and parabolas. Converting conic equations into their standard form helps in identifying and analyzing their properties.

Each conic section has its own standard form:

• **Circle:** \((x-h)^2 + (y-k)^2 = r^2\)
• **Ellipse:** \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
• **Hyperbola:** \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\)
• **Parabola:** Standard form for a vertical parabola is \(y = ax^2 + bx + c\).
  

In the original exercise, after performing completing the square, the equation \(16(x+6)^2 - 9(y-5)^2 = 846\) was rewritten to fit the standard form of a hyperbola, aiding in easily identifying the conic as such.