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

Step-by-Step Solution

Answer

The vertices are at (-5, -7) and (9, -7); foci at (2-7√2, -7) and (2+7√2, -7); asymptotes y = x-9 and y = -x-5.

Step 1: Standard Form Confirmation

The given equation \( \frac{(x-2)^2}{49} - \frac{(y+7)^2}{49} = 1 \) is already in the standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) for a horizontal hyperbola with center at \((h, k)\). Here, \(h=2\), \(k=-7\), \(a^2=49\), and \(b^2=49\). Thus, \(a = 7\) and \(b = 7\).

Step 2: Identify Vertices

For a hyperbola of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the vertices are located \(a\) units away from the center (\(h, k\)) along the x-axis. So, the vertices are at \((h-a, k)\) and \((h+a, k)\). Substituting the known values: \((2-7, -7) = (-5, -7)\) and \((2+7, -7) = (9, -7)\).

Step 3: Calculate Foci

The foci of the hyperbola are \(c\) units away from the center along the x-axis, where \(c^2 = a^2 + b^2\). Here, \(c^2 = 49 + 49 = 98\), so \(c = \sqrt{98} = 7\sqrt{2}\). Therefore, the foci are at \((h-c, k)\) and \((h+c, k)\), or \((2-7\sqrt{2}, -7)\) and \((2+7\sqrt{2}, -7)\).

Step 4: Write Asymptote Equations

For a hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the equations of the asymptotes are \( y = k \pm \frac{b}{a}(x-h) \). Substituting the known values gives: \( y = -7 \pm \frac{7}{7}(x-2) \), simplifying to the asymptote equations \( y = x - 9 \) and \( y = -x - 5 \).

Key Concepts

Hyperbola

A hyperbola is a fascinating type of conic section, which appears like two symmetrical open curves. These curves are formed when a plane intersects both halves of a double cone. Unlike ellipses or parabolas, hyperbolas consist of two separate branches. Analyzing this geometric shape helps us explore various natural phenomena and engineering applications.

The general form of a hyperbola's equation is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]for a horizontal hyperbola, or \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]for a vertical hyperbola. The hyperbola in this exercise is horizontal, denoted by the subtraction of the \( (y-k)^2 \)term, and its equation opens left and right.

• Center: The point \((h, k)\)is the center of the hyperbola. In our example, the center is located at \((2, -7)\).
• a and b: These parameters determine the spread and orientation of the hyperbola. Both are computed from the denominators of the equation and in this case, \( a = 7 \) and \( b = 7 \).

Vertices of a Hyperbola

The vertices of a hyperbola are vital as they mark the points on the graph where each branch of the hyperbola comes closest to its center.

For a horizontal hyperbola such as our example\[ \frac{(x-2)^2}{49} - \frac{(y+7)^2}{49} = 1 \]the vertices lie on the x-axis. They are located \(a\)units from the center in either direction. This means you add and subtract the value of \(a\)from the x-coordinate of the center:

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

The vertices are at:

• \((h-a, k) = (2-7, -7) = (-5, -7)\)
• \((h+a, k) = (2+7, -7) = (9, -7)\)

These points show us where each branch of the hyperbola begins its dramatic curve away from the center.

Foci of a Hyperbola

The foci of a hyperbola are two critical points located inside each branch of the hyperbola. These points are instrumental in the geometric concept of a hyperbola as they help define its shape. For a horizontal hyperbola, the foci are positioned \(c\)units from the center along the x-axis.

The formula to find \(c\)is derived from:\[ c^2 = a^2 + b^2 \]In this example, both \(a^2\)and \(b^2\) are \(49\).Plugging in these values:\[ c^2 = 49 + 49 = 98 \]Thus,\( c = \sqrt{98} = 7\sqrt{2} \)

The foci are then calculated as:

• \((h-c, k) = (2-7\sqrt{2}, -7)\)
• \((h+c, k) = (2+7\sqrt{2}, -7)\)

These positions help us understand how tightly or widely a hyperbola is curved.

Asymptotes of a Hyperbola

Asymptotes are crucial in sketching and understanding the hyperbola's behavior as they show the direction each branch will continue indefinitely without crossing. These lines approach the curve's branches and are pointers to its infinite nature.

For a hyperbola described by\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]the asymptotes are straight lines given by:
\( y = k \pm \frac{b}{a}(x-h) \)

In our exercise, substituting the center (\(h=2\), \(k=-7\)), \(a = 7\), \(b = 7\), the asymptote equations simplify to:

• y = (1)(x - 2) - 7 \(y = x - 9\)
• y = -(1)(x - 2) - 7 \(y = -x - 5\)

These lines serve as invisible guides to the hyperbola's extensive reach, helping us to conceptually anchor the shape and spread of its branches.