Problem 82 - 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. \(y^{2}+8 y-3 x^{2}+13=0\)

Step-by-Step Solution

Answer

Center: (0, -4); Vertices: (0, -4+√3) and (0, -4−√3)

Step 1: Group and Rearrange Terms

Start with the given equation \(y^{2}+8y-3x^{2}+13=0\). Group the \(y\) terms together and the \(x\) terms together, and move the constant to the other side of the equation:\[y^{2} + 8y = 3x^{2} - 13\]

Step 2: Complete the Square for y-Terms

To complete the square for the \(y\) terms, take half of the linear coefficient (8), square it, and add it inside the equation. Half of 8 is 4, and 4 squared is 16:\[y^{2} + 8y + 16 = 3x^{2} - 13 + 16\]. This simplifies to \((y+4)^2\) on the left side.

Step 3: Simplify and Rearrange to Standard Form

Substitute back and rearrange to form \((y+4)^2 = 3x^{2} + 3\). Then, divide through by 3 to isolate \(x^2\):\[\frac{(y+4)^2}{3} - x^2 = 1\]. This represents the standard form of a hyperbola \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) with \(h=0, k=-4, a^2=3,\) and \(b^2=3\).

Step 4: Identify the Center and Vertices

The center of the hyperbola is \((h, k)\) which in this case is \((0, -4)\). To find the vertices, use \((0, -4+a)\) and \((0, -4-a)\), where \(a=\sqrt{3}\):\((0, -4+\sqrt{3})\) and \((0, -4-\sqrt{3})\)

Key Concepts

Conic Sections

Conic sections are the curves obtained when a plane intersects a cone. These sections include circles, ellipses, parabolas, and hyperbolas, each defined by a unique set of equations and properties. Conic sections play a vital role in geometry and have practical applications in fields like astronomy and engineering.
An important property of conic sections is their ability to be represented in a specific form of algebraic equations. Each conic type has its own "signature" equation that distinguishes it from the others. For instance, hyperbolas, like the one in the original exercise, are defined by the equation \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).

• **Circles** have all points equidistant from a center.
• **Ellipses** are stretched circles with two focal points.
• **Parabolas** curve around a singular focus point.
• **Hyperbolas** have two symmetrical branches around two foci.

Understanding these forms helps in converting complex algebraic expressions into recognizable and manageable forms of conic sections.

Standard Form Equation

The standard form equation for a hyperbola is a specific arrangement that makes its properties and graph easier to understand. This form typically looks like \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) or \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Which format is used depends on whether the hyperbola opens upward and downward or left and right.
The numbers in the standard form give immediate insights into its key features:

• \((h, k)\) is the **center** of the hyperbola.
• \(a^2\) and \(b^2\) determine the distances from the center to the vertices and covertices.

The original exercise highlights rearranging and simplifying an equation to bring it into this form to directly read off properties like center and vertices, making analysis more intuitive. This process allows complex equations to become simple, useful tools for graphing and further problem-solving.

Completing the Square

Completing the square is a key technique used in algebra to transform quadratic equations into a form that reveals their characteristics instantly. It's essential in converting equations into standard form, especially for conic sections. This technique involves modifying a quadratic equation so that it becomes a perfect square trinomial.
In the original equation \(y^2 + 8y\), completing the square involves these steps:

• Take half of the coefficient of \(y\), which is 8, divide by 2 to get 4.
• Square it to get 16, then add and subtract this inside the equation.

This transforms \(y^2 + 8y + 16\) into a perfect square, \((y+4)^2\).
By repeating this for the other variable if needed and rearranging, equations can be simplified into easily understood and graphically useful forms. Mastery of completing the square not only aids in solving direct algebra problems but serve as a powerful tool in calculus and analytical geometry as well.