Problem 17 - 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. \(\frac{(x-2)^{2}}{49}-\frac{(y+7)^{2}}{49}=1\)

Step-by-Step Solution

Answer

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

Step 1: Verify Standard Form

The given equation is \( \frac{(x-2)^{2}}{49}-\frac{(y+7)^{2}}{49}=1 \). The standard form for a hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) is satisfied as it compares directly. Therefore, the equation is already in standard form.

Step 2: Identify the Center

In the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center of the hyperbola is given by \((h, k)\). Here, \( h = 2 \) and \( k = -7 \). Thus, the center is \((2, -7)\).

Step 3: Determine Semi-major and Semi-minor Axes

The values \( a^2 = 49 \) and \( b^2 = 49 \) are given. Thus, \( a = 7 \) and \( b = 7 \). Since \( a = b \), the hyperbola is symmetric about the axes and both directions have equal importance. However, the equation form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) suggests a horizontally oriented hyperbola.

Step 4: Identify the Vertices

For a horizontal hyperbola, vertices are at \((h \pm a, k)\). So substituting the values, the vertices are \((2 + 7, -7) = (9, -7)\) and \((2 - 7, -7 ) = (-5, -7)\).

Step 5: Calculate the Foci

For a horizontal hyperbola, the foci are located at \((h \pm c, k)\) where \( c = \sqrt{a^2 + b^2} \). Hence, \( c = \sqrt{49 + 49} = \sqrt{98} = 7\sqrt{2} \). The foci are \((2 + 7\sqrt{2}, -7)\) and \((2 - 7\sqrt{2}, -7)\).

Step 6: Write Equations of Asymptotes

The equations of the asymptotes for a horizontal hyperbola are given by \( y = k \pm \frac{b}{a}(x-h) \). Substitution yields the asymptote equations \( y = -7 \pm \frac{7}{7}(x - 2) \), which simplifies to \( y = -7 \pm (x - 2) \). Therefore, the asymptotes are \( y = x - 9 \) and \( y = -x - 5 \).

Key Concepts

Standard Form of a Hyperbola

The standard form of a hyperbola helps us identify its key characteristics and position in the coordinate plane. Standard form comes in two typical equations:

• Horizontal hyperbola: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]
• Vertical hyperbola: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]

For the exercise at hand, the given equation is \[ \frac{(x-2)^{2}}{49} - \frac{(y+7)^{2}}{49} = 1 \].This matches the form of a horizontal hyperbola, where the \(x\) term is positive and comes first.
This recognition allows us to proceed with identifying other properties like the center, vertices, and asymptotes based on the equation in this specific standard form.

Center of a Hyperbola

The center of a hyperbola is a crucial part of understanding its placement on the graph. In a hyperbola's standard form, the center is denoted by the coordinates \((h, k)\).These are derived directly from the terms in the equation, specifically from the forms \((x-h)^2\) and \((y-k)^2\).
For our example \( \frac{(x-2)^{2}}{49} - \frac{(y+7)^{2}}{49} = 1 \), we identify \(h = 2\) and \(k = -7\).
Therefore, the center of this hyperbola is at \((2, -7)\).This point is the midpoint between the vertices and serves as a reference point for describing other elements like the foci and asymptotes.

Vertices and Foci of a Hyperbola

The vertices and foci are essential elements that describe the shape and extent of a hyperbola. For a horizontal hyperbola, like in our exercise, the vertices are located on the transverse axis. They can be calculated using the formula \((h \pm a, k)\).
Given \(a = 7\), the vertices are found at \((2 \pm 7, -7)\), which computes to \((9, -7)\) and \((-5, -7)\).
Foci for a horizontal hyperbola stretch beyond the vertices along the transverse axis. Their locations are determined by \((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2}\).
Plugging in the values \(c = \sqrt{98} = 7\sqrt{2}\), the foci become \((2 \pm 7\sqrt{2}, -7)\).This places them further out than the vertices, illustrating the hyperbola's "opening" along the x-direction.

Asymptotes of a Hyperbola

Asymptotes are lines that the hyperbola approaches but never touches. They help delineate the trajectory of the hyperbola's branches. In the case of a horizontal hyperbola, the asymptotes are described by the equations:

• \(y = k \pm \frac{b}{a}(x-h)\)

Substituting the values from our example, \(a = 7\), \(b = 7\), \(h = 2\), and \(k = -7\), the calculations simplify to \(y = -7 \pm (x-2)\).
These yield the asymptote lines: \(y = x - 9\) and \(y = -x - 5\).Asymptotes form an "X" pattern around the hyperbola's center, guiding its overall shape and ensuring that, as x or y extend to infinity, the hyperbola moves closer to, but never crosses, these lines.