Problem 42 - College Algebra

Question

For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. $$ 16 x^{2}+64 x-4 y^{2}-8 y-4=0 $$

Step-by-Step Solution

Answer

The vertices are at (0, -1) and (-4, -1), and the foci are at \((-2+2\sqrt{5}, -1)\) and \((-2-2\sqrt{5}, -1)\).

Step 1: Simplify the Equation

First, rearrange the given equation to look like a standard form of a hyperbola. Start by combining similar terms and moving the constant to the other side of the equation:\[16x^2 + 64x - 4y^2 - 8y = 4\]

Step 2: Complete the Square for the x-term

Factor the coefficient of the \(x^2\) term from the first two terms in the equation:\[16(x^2 + 4x) - 4y^2 - 8y = 4\]Now, complete the square for the expression \(x^2 + 4x\):\[x^2 + 4x = (x+2)^2 - 4\]Substitute back:\[16((x+2)^2 - 4) - 4y^2 - 8y = 4\]Distribute 16:\[16(x+2)^2 - 64 - 4y^2 - 8y = 4\]Then, move \(-64\) to the other side:\[16(x+2)^2 - 4y^2 - 8y = 68\]

Step 3: Complete the Square for the y-term

Now factor \( -4 \) from the \( y \) terms:\[-4(y^2 + 2y) = -4((y+1)^2 - 1)\]Replace it in the equation:\[16(x+2)^2 - 4((y+1)^2 - 1) = 68\]Expand:\[16(x+2)^2 - 4(y+1)^2 + 4 = 68\]And simplify the right side:\[16(x+2)^2 - 4(y+1)^2 = 64\]

Step 4: Divide to Obtain the Standard Form

Divide the entire equation by 64 to express it as a standard form of a hyperbola:\[\frac{(x+2)^2}{4} - \frac{(y+1)^2}{16} = 1\]This matches the standard form of a horizontal hyperbola \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), with \(h = -2\), \(k = -1\), \(a = 2\), and \(b = 4\).

Step 5: Identify Vertices and Foci of the Hyperbola

For this hyperbola, the center is \((-2, -1)\).Vertices: Since \(a = 2\), the vertices are located \(a\) units left and right of the center along the x-axis: \[(-2+2, -1) \Rightarrow (0, -1) \] and \[(-2-2, -1) \Rightarrow (-4, -1)\].Foci: The distance to the foci is \(c\), where \(c = \sqrt{a^2 + b^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\).The foci are located \(c\) units from the center \((-2\sqrt{5}, -1)\) left and right:\[(-2+2\sqrt{5}, -1)\] and \[(-2-2\sqrt{5}, -1)\].

Key Concepts

Vertices of Hyperbola

To find the vertices of a hyperbola, it's important to understand that they are key points which lie along the major axis of the hyperbola. For a horizontal hyperbola, the vertices will be located along the x-axis, surrounding the center of the hyperbola.

If your hyperbola is in the standard form, \[\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1,\]then the vertices can be found using the center coordinates, \((h, k)\), and a value, \(a\).

Here are the steps to find them:

• Identify the center, \((h, k)\). For our equation, this is \((-2, -1)\).
• The distance \(a\) is found from the denominator under the \(x\) component, which in this exercise is \(2\).
• Add and subtract \(a\) from the x-coordinate of the center to get the vertices; therefore, vertices \((h + a, k)\) and \((h - a, k)\).

The vertices of this hyperbola are \((0, -1)\) and \((-4, -1)\). They are points where the hyperbola intersects its major axis.

These points are crucial as they define the 'width' across the hyperbola's opening in its principal direction.

Foci of Hyperbola

The foci of a hyperbola are two points located inside each 'branch' of the hyperbola, and they are a bit more auxiliary than vertices. They help in understanding the 'spread' of the hyperbola.

The foci of a hyperbola can be found for its standard form as follows:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,\]where the foci are found using the formula:\[c = \sqrt{a^2 + b^2}.\]

Steps to find the foci:

• Calculate \(c\), the distance from the center to each focus, using \(a\) and \(b\) from the standard form.
• In this solution, \(a = 2\) and \(b = 4\), leading to \(c = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\).
• Foci points are calculated as \((h + c, k)\) and \((h - c, k)\), which gives us specific coordinates for the foci on the hyperbola.

The foci's coordinates are \((-2 + 2\sqrt{5}, -1)\) and \((-2 - 2\sqrt{5}, -1)\). These are beyond the vertices, making the foci key indicators in how open or spread the hyperbola is.

Standard Form of a Hyperbola

The standard form of a hyperbola is crucial as it allows us to easily determine and define its properties, such as the center, vertices, and foci. It categorizes the hyperbola as either horizontal or vertical based on its orientation.

The general standard form for a horizontal hyperbola is:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.\]

Key Components

• \((h, k)\) are the coordinates of the center.
• \(a\) is the distance from the center to each vertex along the x-axis.
• \(b\) helps determine the steepness or 'opening' extent of the hyperbola.
• The leading positive term \((x-h)^2\) signifies that it's a horizontal hyperbola.

For the hyperbola in the exercise, the equation was transformed effectively into:\[\frac{(x+2)^2}{4} - \frac{(y+1)^2}{16} = 1.\] This form confirms it’s a horizontal hyperbola centered at \((-2, -1)\) with vertices and foci identified from the previous sections.

Understanding the transformation to this form through simplification is essential as it allows clear visualization and interpretation of the hyperbola's main attributes.