Problem 22 - College Algebra

Question

For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. $$ -4 x^{2}+40 x+25 y^{2}-100 y+100=0 $$

Step-by-Step Solution

Answer

The hyperbola is \(\frac{(x-5)^2}{25} - \frac{(y-2)^2}{4} = 1\) with vertices \((0,2)\) and \((10,2)\), and foci \((5 \pm \sqrt{29}, 2)\). Asymptotes: \(y-2=\pm\frac{2}{5}(x-5)\).

Step 1: Start by Rearranging the Equation

Our given equation is \(-4x^2 + 40x + 25y^2 - 100y + 100 = 0\). We'll want to rearrange it to group the \(x\)-terms and \(y\)-terms separately: \(-4x^2 + 40x + 25y^2 - 100y = -100\).

Step 2: Complete the Square for x-terms

For the \(x\)-terms, take \(-4x^2 + 40x\), factor out \(-4\) to get: \(-4(x^2 - 10x)\). To complete the square, we take half of \(-10\), which is \(-5\), and square it to get \(25\). Thus, \(-4(x^2 - 10x)\) becomes \(-4((x - 5)^2 - 25)\).

Step 3: Complete the Square for y-terms

For the \(y\)-terms, take \(25(y^2 - 4y)\), factor out \(25\) to get: \(25(y^2 - 4y)\). To complete the square, we take half of \(-4\), which is \(-2\), and square it to get \(4\). Thus, \(25(y^2 - 4y)\) becomes \(25((y - 2)^2 - 4)\).

Step 4: Substitute Completed Squares into the Equation

Substitute back into the rearranged equation: \(-4((x-5)^2 - 25) + 25((y-2)^2 - 4) = -100\). Simplify to get: \(-4(x-5)^2 + 100 + 25(y-2)^2 - 100 = -100\). This simplifies to \(-4(x-5)^2 + 25(y-2)^2 = -100\).

Step 5: Divide to Achieve Standard Form

Divide the entire equation by \(-100\) to achieve the hyperbola's standard form: \(\frac{(x-5)^2}{25} - \frac{(y-2)^2}{4} = 1\). This is the equation of the hyperbola in standard form.

Step 6: Identify Vertices and Foci

For \(\frac{(x-5)^2}{25} - \frac{(y-2)^2}{4} = 1\), we have a horizontal hyperbola. The center is \((5, 2)\). The term \(a^2 = 25\) gives \(a = 5\), so the vertices are \((5 \pm 5, 2)\) or \((0, 2)\) and \((10, 2)\). The term \(b^2 = 4\) gives \(b = 2\). The foci are found using \(c^2 = a^2 + b^2 = 29\), so \(c = \sqrt{29}\). The foci are \((5 \pm \sqrt{29}, 2)\).

Step 7: Write Equations of Asymptotes

For a horizontal hyperbola \(\frac{(x-5)^2}{25} - \frac{(y-2)^2}{4} = 1\), the slopes of the asymptotes are \(\pm \frac{b}{a} = \pm \frac{2}{5}\). Therefore, the equations of the asymptotes are \(y - 2 = \pm \frac{2}{5}(x - 5)\).

Key Concepts

Standard Form

The standard form of a hyperbola is a way to write its equation that clearly shows its orientation and components. A hyperbola has two forms based on whether it is horizontal or vertical. For a horizontal hyperbola, the standard form is given by: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] Where

• \((h, k)\) is the center of the hyperbola
• \(a\) is the distance from the center to the vertices along the x-axis
• \(b\) is the distance from the center to the vertices along the y-axis

To convert the given general equation into the standard form, first rearrange and group the terms by their variables. Then, complete the square for each grouped variable, and adjust the equation so that it equals 1. As detailed in the solution, our equation simplifies to: \[ \frac{(x-5)^2}{25} - \frac{(y-2)^2}{4} = 1 \] indicating that it's a horizontal hyperbola centered at \((5, 2)\) with \(a^2 = 25\) and \(b^2 = 4\). This form is crucial for easily identifying other features of the hyperbola, including its vertices and foci.

Vertices and Foci

Vertices and foci are key features of a hyperbola that define its shape and scale. The vertices are the closest points on each branch of the hyperbola. From the standard form equation, once you have identified \(a\) and \(b\), you can easily find these features. For our equation:

• We calculated \(a = 5\) (since \(a^2 = 25\)).
• This indicates that our vertices are located \(5\) units away from the center \((5, 2)\) along the x-axis, leading us to the vertices \((0, 2)\) and \((10, 2)\).
• For the foci, use \(c^2 = a^2 + b^2\). Here, \(c^2 = 25 + 4 = 29\), so \(c = \sqrt{29}\).
• The foci are \(5 \pm \sqrt{29}\) units away from the center along the x-axis, at \((5 + \sqrt{29}, 2)\) and \((5 - \sqrt{29}, 2)\).

These points help visualize the hyperbola's structure and can be pivotal in graphing its shape accurately.

Asymptotes

Asymptotes in a hyperbola are crucial lines that the hyperbola approaches but never touches. They indicate the general direction that the 'arms' of the hyperbola point and are especially useful in graphing because they help sketch the hyperbola's shape. For a hyperbola in standard form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the equations of the asymptotes are derived based on its orientation. For a horizontal hyperbola, the asymptotes are calculated using the slope \(\pm \frac{b}{a}\). In our example, with \(a^2 = 25\) and \(b^2 = 4\), this calculation provides:

• \(a = 5\)
• \(b = 2\)
• Thus, the slope is \(\pm \frac{2}{5}\)

This leads to the asymptote equations: \[y - 2 = \pm \frac{2}{5}(x - 5)\] These asymptotes help frame the hyperbola within its coordinate plane, offering a reference for graphing and observing its boundaries without intersection.