Problem 39 - Algebra 2

Question

Write each equation in standard form. State whether the graph of the equation is a parabola, circle, ellipse, or hyperbola. Then graph the equation. $$ 6 x^{2}-24 x-5 y^{2}-10 y-11=0 $$

Step-by-Step Solution

Answer

The equation is a hyperbola, and its graph is centered at (2, -1).

Step 1: Rearrange the Equation

First, we need to rearrange the given equation by grouping the terms involving the same variables. Starting with the equation:\[ 6x^{2} - 24x - 5y^{2} - 10y - 11 = 0 \]Group the terms:\[ (6x^{2} - 24x) + (-5y^{2} - 10y) = 11 \]

Step 2: Complete the Square for x-terms

Consider the x-term group \(6x^{2} - 24x\). Factor out the 6:\[ 6(x^{2} - 4x) \]To complete the square inside the parentheses, take half of -4 (the coefficient of x), square it, and add and subtract it:\[ 6(x^{2} - 4x + 4 - 4) = 6((x-2)^{2} - 4) \]This simplifies to:\[ 6(x-2)^{2} - 24 \]

Step 3: Complete the Square for y-terms

Now, consider the y-term group \(-5y^{2} - 10y\). Factor out -5:\[ -5(y^{2} + 2y) \]To complete the square inside the parentheses, take half of 2 (the coefficient of y), square it, and add and subtract it:\[ -5(y^{2} + 2y + 1 - 1) = -5((y+1)^{2} - 1) \]This simplifies to:\[ -5(y+1)^{2} + 5 \]

Step 4: Substitute Back and Simplify

Substitute the completed squares back into the equation:\[ 6(x-2)^{2} - 24 - 5(y+1)^{2} + 5 = 11 \]Combine the constants on the right-hand side:\[ 6(x-2)^{2} - 5(y+1)^{2} = 30 \]

Step 5: Divide to Convert to Standard Form

To express the conic in standard form, divide the entire equation by 30:\[ \frac{6(x-2)^{2}}{30} - \frac{5(y+1)^{2}}{30} = 1 \]This simplifies to:\[ \frac{(x-2)^{2}}{5} - \frac{(y+1)^{2}}{6} = 1 \]This is the standard form of a hyperbola.

Step 6: Identify and Graph the Conic

The equation is in the standard form of a hyperbola with horizontal transverse axis:\[ \frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1 \]where \( h = 2 \), \( k = -1 \), \( a^{2} = 5 \), \( b^{2} = 6 \). Graph this hyperbola by plotting its center at (2, -1) and drawing the hyperbola along its axes. The vertices are at \((2 \pm \sqrt{5}, -1)\). The asymptotes slope through the center.

Key Concepts

Standard Form

In mathematics, the standard form for the equation of a conic section is a way to represent the equation that fully describes the conic's shape and position. Converting an equation to its standard form simplifies graphing and understanding the properties of the conic. Here is the breakdown for conic sections:

• For a circle, the standard form is \( (x-h)^2 + (y-k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius.
• For an ellipse, the standard form is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
• For a hyperbola, the standard form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).
• For a parabola, it is typically \( (y-k)^2 = 4p(x-h) \) or \( (x-h)^2 = 4p(y-k) \).

In this exercise, through the process of completing the square, we convert an equation into the standard form of a hyperbola.

Hyperbola

A hyperbola is one of the four types of conic sections, which also include circles, ellipses, and parabolas. A hyperbola is defined by its property that for any point on the hyperbola, the difference in distances to two fixed points (foci) is constant.

• The standard form of a hyperbola with its transverse axis parallel to the x-axis is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).
• If the transverse axis is along the y-axis, the form switches to \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).

For this problem, the equation simplifies to \( \frac{(x-2)^2}{5} - \frac{(y+1)^2}{6} = 1 \), meaning the hyperbola has a horizontal transverse axis, centered at \( (2, -1) \). Hyperbolas are characterized by two distinct branches.

Completing the Square

Completing the square is an algebraic technique used to transform a quadratic expression into a perfect square trinomial, making equations easier to solve or analyze. This method is instrumental in converting conic section equations to standard form.To complete the square:

• Focus on terms involving a single variable, such as \( ax^2 + bx \).
• Factor out the coefficient of \( x^2 \) if it’s not 1.
• Add and subtract the square of half the linear coefficient, ensuring the equation's balance.

In this exercise, completing the square was done for terms involving both \( x \) and \( y \), transforming the expressions into sum/difference of squares. This is crucial for expressing the hyperbola in standard form.

Graphing Equations

Graphing equations involves plotting them on a coordinate plane to visualize their shape and position. For conic sections, the standard form makes it easier to identify the key features needed for graphing, such as the center, vertices, and asymptotes.

• In hyperbolas, the center obtained from the standard form serves as a starting point for graphing.
• Vertices are determined from the terms \( a^2 \) and \( b^2 \), which control the spread of the hyperbola.
• Asymptotes, which are lines that the hyperbola approaches but never touches, can be calculated because they help in sketching the hyperbola's shape.

The given equation defines a hyperbola with a center at \( (2, -1) \), and it has asymptotes helping to shape its graph across the coordinate plane. Understanding each element from the standard form facilitates efficient and accurate graphing.