Problem 17 - College Algebra with Corequisite Support

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. \(\frac{(x-2)^{2}}{49}-\frac{(y+7)^{2}}{49}=1\)

Step-by-Step Solution

Answer

Vertices: (9, -7) and (-5, -7). Foci: \((2 + 7\sqrt{2}, -7)\) and \((2 - 7\sqrt{2}, -7)\). Asymptotes: \(y = x - 9\), \(y = -x - 5\).

Step 1: Identify the Standard Form

The given equation is in the standard form for a hyperbola: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Here, \(h = 2\), \(k = -7\), \(a^2 = 49\) and \(b^2 = 49\). Hence, \(a = 7\) and \(b = 7\). Since the \(x^2\) term is positive, the transverse axis is horizontal.

Step 2: Find the Center

The center \((h, k)\) of the hyperbola is found from the equation. Thus, the center of this hyperbola is \((2, -7)\).

Step 3: Determine the Vertices

For a hyperbola with a horizontal transverse axis, the vertices are located at \((h \pm a, k)\). Here, \(h = 2\), \(a = 7\), and \(k = -7\). The vertices are \((2+7,-7)\) and \((2-7,-7)\), which results in \((9, -7)\) and \((-5, -7)\).

Step 4: Calculate the Foci

The foci for a hyperbola with a horizontal transverse axis are given by \((h \pm c, k)\), where \(c\) is calculated as \(c^2 = a^2 + b^2\). Since \(a = 7\) and \(b = 7\), it follows that \(c^2 = 49 + 49 = 98\). Therefore, \(c = \sqrt{98} = 7\sqrt{2}\). The coordinates for the foci are \((2 + 7\sqrt{2}, -7)\) and \((2 - 7\sqrt{2}, -7)\).

Step 5: Write the Equations of the Asymptotes

The equations for the asymptotes of a hyperbola with a horizontal transverse axis are given by \(y = k \pm \frac{b}{a}(x-h)\). Using \(h = 2\), \(k = -7\), \(a = 7\), and \(b = 7\), the equations become \(y = -7 \pm \frac{7}{7}(x - 2)\). This simplifies to the equations \(y = -7 + (x - 2)\) and \(y = -7 - (x - 2)\), which further simplify to \(y = x - 9\) and \(y = -x - 5\).

Key Concepts

Standard Form of a Hyperbola

The equation of a hyperbola in its standard form is crucial in understanding its graph and behavior. The standard form of a hyperbola with a horizontal transverse axis is

• \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)

Here,

• \((h, k)\) represents the center of the hyperbola.
• \(a^2\) is the square of the distance from the center to the vertices along the x-axis.
• \(b^2\) is the square of the distance from the center to the "conjugate vertices" along the y-axis.

In our example, the equation

• \(\frac{(x-2)^2}{49} - \frac{(y+7)^2}{49} = 1\)

shows that

• the center is at \((2, -7)\),
• \(a = 7\), and
• \(b = 7\).

This form helps us deduce other essential aspects like the vertices, foci, and asymptotes.

Vertices of a Hyperbola

The vertices of a hyperbola are key points that represent where the hyperbola intersects its transverse axis. For a hyperbola centered at \((h, k)\), if the transverse axis is horizontal, the vertices are at

• \((h \pm a, k)\).

For our equation, with

• \(h = 2\), \(k = -7\), and \(a = 7\),

we find the vertices by substituting:

• \((2 + 7, -7)\) and \((2 - 7, -7)\),

resulting in

• \((9, -7)\) and \((-5, -7)\).

These points provide critical structural limits of the hyperbola along its broadest path.

Foci of a Hyperbola

The foci of a hyperbola are two points located symmetrically around the center, lying along the transverse axis. They aid in defining the hyperbola's shape. To find the foci, calculate \(c\) using:

• \(c^2 = a^2 + b^2\).

Given

• \(a = 7\) and \(b = 7\),
• \(c^2 = 49 + 49 = 98\),

yielding

• \(c = \sqrt{98} = 7\sqrt{2}\).

Thus, the foci are at

• \((h \pm c, k)\), or \((2 \pm 7\sqrt{2}, -7)\),

meaning

• \((2 + 7\sqrt{2}, -7)\) and \((2 - 7\sqrt{2}, -7)\).

These foci determine the hyperbola's curvature.

Asymptotes of a Hyperbola

Asymptotes of a hyperbola are lines that the hyperbola approaches but never touches. They act as boundary lines guiding the hyperbola's path far from the center. For a hyperbola with a horizontal transverse axis, the equations for the asymptotes are

• \(y = k \pm \frac{b}{a}(x-h)\).

Given

• \(h = 2\), \(k = -7\), \(a = 7\), and \(b = 7\),

we find the equations to be:

• \(y = -7 \pm \frac{7}{7}(x - 2)\),

which simplifies to

• \(y = x - 9\) and \(y = -x - 5\).

These asymptotes help in understanding the overall direction and openness of the hyperbola's branches.