Problem 25 - College Algebra

Question

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. \(4 x^{2}+24 x-25 y^{2}+200 y-464=0\)

Step-by-Step Solution

Answer

Vertices: (2, 4), (-8, 4); Foci: (-3±√29, 4); Asymptotes: y = 4 ± (2/5)(x + 3).

Step 1: Move Constants to One Side

Let's start by moving the constant to the right-hand side of the equation. The given equation is: \[4x^2 + 24x - 25y^2 + 200y - 464 = 0\]Add 464 to both sides:\[4x^2 + 24x - 25y^2 + 200y = 464\]

Step 2: Group x and y Terms

We need to group the terms involving \(x\) and \(y\) separately to complete the square.\[\begin{align*} \text{x terms: } & 4(x^2 + 6x) ewline \text{y terms: } & -25(y^2 - 8y) \end{align*}\]

Step 3: Complete the Square for x Terms

Complete the square for the \(x\) terms.For the expression \(x^2 + 6x\), add and subtract \(9\) inside the parenthesis:\[x^2 + 6x = (x + 3)^2 - 9\]Multiply through by 4:\[4(x^2 + 6x) = 4((x + 3)^2 - 9) = 4(x + 3)^2 - 36\]

Step 4: Complete the Square for y Terms

Complete the square for the \(y\) terms.For the expression \(y^2 - 8y\), add and subtract \(16\) inside the parenthesis:\[y^2 - 8y = (y - 4)^2 - 16\]Multiply through by -25:\[-25(y^2 - 8y) = -25((y - 4)^2 - 16) = -25(y - 4)^2 + 400\]

Step 5: Substitute Completed Squares into Equation

Substitute the completed square expressions back into the equation:\[4(x + 3)^2 - 36 - 25(y - 4)^2 + 400 = 464\]Simplify by combining constants:\[4(x + 3)^2 - 25(y - 4)^2 = 100\]

Step 6: Divide by 100 to Standardize Equation

Divide the entire equation by 100 to express it in the standard form of a hyperbola:\[\frac{4(x + 3)^2}{100} - \frac{25(y - 4)^2}{100} = 1\]Simplify the fractions:\[\frac{(x + 3)^2}{25} - \frac{(y - 4)^2}{4} = 1\]This is the standard form equation of the hyperbola.

Step 7: Identify Vertices

The standard form equation is \(\frac{(x + 3)^2}{25} - \frac{(y - 4)^2}{4} = 1\). This is a horizontal hyperbola centered at \((-3, 4)\).The distance between the vertices is determined by \(2a\), where \(a^2 = 25\), hence \(a = 5\).Vertices are at \((-3 + 5, 4) = (2, 4)\) and \((-3 - 5, 4) = (-8, 4)\).

Step 8: Identify Foci

For a hyperbola, the relationship \(c^2 = a^2 + b^2\) holds, where \(a^2 = 25\) and \(b^2 = 4\).Calculate \(c^2 = 25 + 4 = 29\), so \(c = \sqrt{29}\).The foci are at \((-3 + \sqrt{29}, 4)\) and \((-3 - \sqrt{29}, 4)\).

Step 9: Write Equations of Asymptotes

The asymptotes of a hyperbola in the form \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\) are given by the equations:\[ y - k = \pm \frac{b}{a}(x - h) \]For our hyperbola, \(h = -3\), \(k = 4\), \(a = 5\), and \(b = 2\).Thus, the asymptotes are:\[y - 4 = \pm \frac{2}{5}(x + 3)\]

Key Concepts

Standard Form Equation

A hyperbola is a type of conic section that can be represented in a particular way using what's called a standard form equation. For a hyperbola, the standard form is crucial in understanding its geometric properties and how it behaves in a coordinate plane. It generally looks like this:

• For horizontal hyperbolas: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• For vertical hyperbolas: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

This form provides clarity on key features such as the center, vertices, and foci. In our solved exercise, the hyperbola was rearranged and simplified to the standard form as \( \frac{(x + 3)^2}{25} - \frac{(y - 4)^2}{4} = 1 \). Here, the center of the hyperbola is found at the point (-3, 4), reflecting the values of \( h \) and \( k \) in the equation.

Vertices

The vertices of a hyperbola are key points that show its extent along the axis of symmetry. They are the points where the hyperbola makes its sharpest turns. The distance between the vertices is denoted as \(2a\), with each vertex located at a distance \(a\) from the center. For the hyperbola in the original problem, \(a^2 = 25\), giving \(a = 5\).
The vertices for a horizontal hyperbola such as this one are calculated by moving left and right from the center (-3, 4):

• Move 5 units to the right to get (2, 4).
• Move 5 units to the left to get (-8, 4).

So, the vertices are located at (2, 4) and (-8, 4). Understanding vertices helps to graph the hyperbola's shape accurately.

Foci

The foci (or focal points) of a hyperbola are essential in understanding its eccentricity or nature of the curve. Located along the same axis as the vertices, the foci are a measure of how "open" the hyperbola is.
The distance from the center to each focus is given by \(c\), calculated through the relationship \(c^2 = a^2 + b^2\). For the problem at hand, since \(a^2 = 25\) and \(b^2 = 4\), the calculation yields:

• \(c^2 = 25 + 4 = 29\)
• \(c = \sqrt{29}\)

This means the foci are at

• \((-3 + \sqrt{29}, 4)\)
• \((-3 - \sqrt{29}, 4)\)

These points help us understand how the lines forming the hyperbola would appear relative to the curve itself.

Asymptotes

Asymptotes of a hyperbola are straight lines that provide insight into the direction the branches of the hyperbola will follow as they extend towards infinity. These lines are not part of the hyperbola itself but act as boundary guides. For a hyperbola centered at (\(h, k\)) and in standard form, the equations for asymptotes are:

• \( y - k = \pm \frac{b}{a}(x - h) \)

Applying this to the given hyperbola centered at (-3, 4) with \(a = 5\) and \(b = 2\), the equations of the asymptotes become:

• \(y - 4 = \frac{2}{5}(x + 3)\)
• \(y - 4 = -\frac{2}{5}(x + 3)\)

These lines help predict the overall shape and orientation of the hyperbola on the coordinate plane.