Problem 79 - A Graphical Approach to College Algebra

Question

Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. \(3 y^{2}+24 y-2 x^{2}+12 x+24=0\)

Step-by-Step Solution

Answer

The hyperbola is centered at (3, -4) with vertices (3, -4±√2).

Step 1: Group and Rearrange Terms

First, rearrange and group the terms involving the same variables together. The given equation is: \(3 y^{2} + 24 y - 2 x^{2} + 12 x + 24 = 0\). Rearrange them as \(-2x^{2} + 12x + 3y^{2} + 24y = -24\).

Step 2: Factor Out Constants

Factor out the constants in front of the quadratic terms. For the \(x\) terms: \(-2(x^{2} - 6x)\) and for the \(y\) terms: \(3(y^{2} + 8y)\). Now the equation looks like: \(-2(x^{2} - 6x) + 3(y^{2} + 8y) = -24\).

Step 3: Complete the Square

Complete the square for both the \(x\) group and the \(y\) group.1. For \(x\): the expression is \(x^{2} - 6x\). Take half of \(-6\), which is \(-3\), and square it to get \(9\). Add and subtract \(9\) inside the parenthesis: \(-2((x^{2} - 6x + 9) - 9)\).2. For \(y\): the expression is \(y^{2} + 8y\). Take half of \(8\), which is \(4\), and square it to get \(16\). Add and subtract \(16\) inside the parenthesis: \(3((y^{2} + 8y + 16) - 16)\).The equation becomes \(-2((x-3)^{2} - 9) + 3((y+4)^{2} - 16) = -24\).

Step 4: Simplify Equation

Distribute the constants and simplify the equation:\(-2(x-3)^{2} + 18 + 3(y+4)^{2} - 48 = -24\).Combine the constants:\(-2(x-3)^{2} + 3(y+4)^{2} = 6\).Divide the entire equation by \(6\) to get the standard form:\(\frac{(y+4)^{2}}{2} - \frac{(x-3)^{2}}{3} = 1\).

Step 5: Identify Center and Vertices

The equation \(\frac{(y+4)^{2}}{2} - \frac{(x-3)^{2}}{3} = 1\) is in the standard form of a vertical hyperbola centered at \((h, k) = (3, -4)\).The vertices are determined by the \(b^{2}\) value from the \(\frac{(y+4)^{2}}{2}\) term and are \((3, -4 + \sqrt{2})\) and \((3, -4 - \sqrt{2})\).

Key Concepts

Standard Form

The standard form of a hyperbola's equation is a way to express the equation in a simpler, more useful format, making it easier to identify the characteristics of the hyperbola. For hyperbolas centered at \((h, k)\), the standard form is given by:

• Vertical Hyperbola: \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]
• Horizontal Hyperbola: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]

In the problem, the equation of the hyperbola is transformed into the standard form \(\frac{(y+4)^2}{2} - \frac{(x-3)^2}{3} = 1.\)This indicates it is a vertical hyperbola centered at \((h, k) = (3, -4)\).Recognizing whether the equation describes a vertical or horizontal hyperbola comes down to the order and sign of the terms in the standard form.This form also allows easy identification of the hyperbola's other properties, like center, vertices, and axes.

Center of Hyperbola

The center of a hyperbola is a crucial element used to define its position on the coordinate plane. The center is denoted as \((h, k)\) and corresponds to the point around which the hyperbola is symmetrically located.From the transformed standard form of the equation \(\frac{(y+4)^2}{2} - \frac{(x-3)^2}{3} = 1\),we determine the center by examining the variables within the parentheses:

• The term \((x-3)^2\) suggests the x-coordinate of the center is \(h = 3.\)
• The term \((y+4)^2\) indicates the y-coordinate of the center is \(k = -4.\)

Thus, the center of the hyperbola is \((3, -4)\).Knowing the center helps in plotting the hyperbola, as it offers a reference point for placing the vertices and determining the axes of symmetry.

Vertices of Hyperbola

The vertices of a hyperbola are points that mark the turning points of the curves along the axes of the hyperbola. For a vertical hyperbola, these vertices lie along the line parallel to the y-axis.In the standard form equation \(\frac{(y+4)^2}{2} - \frac{(x-3)^2}{3} = 1,\)we determine the vertices by using the information from the \(\frac{(y+4)^2}{a^2}\) term where \(a^2 = 2.\)

• The distance from the center to each vertex along the y-axis is given by \(\pm a,\)where \(a = \sqrt{2}.\)
• Starting from the center \((3, -4),\) the vertices are found by moving \(\sqrt{2}\)units up and down along the y-axis:
  • The first vertex is at \((3, -4 + \sqrt{2})\).
  • The second vertex is at \((3, -4 - \sqrt{2})\).

The vertices provide key geometric components, offering insights into the extend and orientation of the hyperbola's curves.