Problem 84 - 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. \(5 x^{2}+10 x-7 y^{2}+28 y=58\)

Step-by-Step Solution

Answer

Center: \((-1, 2)\), Vertices: \((-1 \pm \sqrt{7}, 2)\).

Step 1: Rearrange the Equation

Start by rearranging the terms in the equation to group the x-terms together and the y-terms together. The original equation is: \[ 5x^2 + 10x - 7y^2 + 28y = 58 \]Rearrange it as follows:\[ 5x^2 + 10x - 7y^2 + 28y = 58 \]

Step 2: Complete the Square for x-terms

To simplify the x-terms, factor out the coefficient of \(x^2\), which is 5, and complete the square:\[ 5(x^2 + 2x) = 5((x+1)^2 - 1) = 5(x+1)^2 - 5 \]

Step 3: Complete the Square for y-terms

For the y-terms, factor out -7 and complete the square:\[ -7(y^2 - 4y) = -7((y-2)^2 - 4) = -7(y-2)^2 + 28 \]

Step 4: Substitute Completed Squares Back

Substitute the completed squares from Steps 2 and 3 back into the equation:\[ 5(x+1)^2 - 5 - 7(y-2)^2 + 28 = 58 \]

Step 5: Simplify the Equation

Combine constants on the right side:\[ 5(x+1)^2 - 7(y-2)^2 = 58 + 5 - 28 \]\[ 5(x+1)^2 - 7(y-2)^2 = 35 \]

Step 6: Divide Through by 35

To express the equation in standard form, divide the whole equation by 35:\[ \frac{(x+1)^2}{7} - \frac{(y-2)^2}{5} = 1 \]

Step 7: Identify the Center and Vertices

The standard form for a hyperbola centered at \((h, k)\) is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]. From \(\frac{(x+1)^2}{7} - \frac{(y-2)^2}{5} = 1\), we identify:\(h = -1\), \(k = 2\), \(a^2=7\), and \(b^2=5\). The center is \((-1, 2)\). The vertices are at \((-1 \pm \sqrt{7}, 2)\).

Key Concepts

Equation of Hyperbola

To understand hyperbolas, it's important to start with the equation in standard form. The general equation for a hyperbola centered at \((h, k)\) is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). This equation gives us a detailed description of the hyperbola's shape and position in the coordinate plane.
- **\(a^2\)** and **\(b^2\)** are constants that define how stretched the hyperbola is along the x-axis and y-axis respectively.- The signs tell us how the hyperbola opens: '-' means the hyperbola opens along the x-direction, while a '+' after would open along the y-direction.
Transforming equations into this standard form helps easily identify properties of the hyperbola, such as its center and vertices.

Completing the Square

Completing the square is a crucial step in rewriting equations. It allows us to express quadratic expressions in a more manageable form.
- Begin by grouping x-terms and y-terms separately.- For the x-terms in the equation \(5x^2 + 10x\): 1. Factor out the coefficient of \(x^2\), which is 5. 2. Rewrite as: \(5(x^2 + 2x)\). 3. Completing the square inside gives \(5((x + 1)^2 - 1)\), leading to \(5(x + 1)^2 - 5\).- Similarly, for the y-terms \(-7y^2 + 28y\), factor out -7 and complete the square to get \(-7(y - 2)^2 + 28\).
Incorporate these steps to simplify and transform the equation to the standard form, making it easier to identify the center and vertices.

Center of Hyperbola

The center of a hyperbola is a key point that helps define its position. Once we transform the hyperbola into its standard form, identifying the center becomes straightforward.
- For an equation in the form \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), the center is \((h, k)\).- In our example, transforming the equation provides \((x + 1)^2\) and \((y - 2)^2\), leading to the center \((-1, 2)\).
The center acts as a reference point for describing other features of the hyperbola, such as its vertices and asymptotes.

Vertices of Hyperbola

Vertices are essential points on a hyperbola that indicate its extent along its transverse axis. From the standard form equation, calculating these becomes easier.
- The vertices are located at a distance \(a\) from the center along the axis of symmetry of the hyperbola.- In our example, with \(a^2 = 7\), we identify that \(a = \sqrt{7}\).- Thus, the vertices can be found at \((-1 \pm \sqrt{7}, 2)\), reflecting their positions along the x-axis given our equation.
Understanding the position of the vertices is essential in sketching the hyperbola and assessing its basic shape.