Problem 34 - Algebra and Trigonometry

Question

Find an equation for the hyperbola that satisfies the given conditions. Foci: \(( \pm 6,0),\) vertices: \(( \pm 2,0)\)

Step-by-Step Solution

Answer

Equation: \(\frac{x^2}{4} - \frac{y^2}{32} = 1\).

Step 1: Identify the Center

Since the vertices are at (±2, 0) \((\displaystyle \pm 2,0)\) and the hyperbola opens horizontally, the center of the hyperbola is at \((0, 0)\).

Step 2: Determine the Transverse Axis and \(a\)

The transverse axis is horizontal, and its length is the distance between the vertices, giving us \(2a = 4\). Thus, \(a = 2\).

Step 3: Determine the Distance to the Foci and \(c\)

The foci are at \((\pm 6, 0)\). The distance from the center to each focus, which is described by \(c\), is 6.

Step 4: Use the Relationship Between \(a\), \(b\), and \(c\)

The relationship between the distances is given by: \[ c^2 = a^2 + b^2 \].Substituting for \(c = 6\) and \(a = 2\):\[ 6^2 = 2^2 + b^2 \] \[ 36 = 4 + b^2 \] Subtracting 4 from both sides gives:\[ b^2 = 32 \]. Thus, \(b = \sqrt{32}\).

Step 5: Write the Equation for the Hyperbola

The standard form of the equation for a horizontally opening hyperbola centered at \((0, 0)\) is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]. We have \(a^2 = 4\) and \(b^2 = 32\), hence the equation is: \[ \frac{x^2}{4} - \frac{y^2}{32} = 1 \].

Key Concepts

Transverse Axis

The transverse axis of a hyperbola is a fundamental concept. It helps in understanding its structure and orientation. For our exercise, the transverse axis is horizontal because the hyperbola opens horizontally, indicated by the symmetry along the x-axis.
When given the vertices at points \(\pm 2, 0\), we can determine the length of this axis. This is simply the distance between these two vertices, which is 4. The transverse axis runs through the hyperbola's center, which is at the midpoint of the vertices in this case, \(0, 0\).
This axis plays an integral role in defining the parameter \(a\), which describes how far each vertex is from the center.

• The transverse axis defines the orientation of the hyperbola (horizontal or vertical).
• The parameter \(a\) represents half of its length.

The length of the transverse axis here is 4, leading to \(a = 2\). This is vital for writing the hyperbola's equation.

Distance to the Foci

In hyperbolas, the foci are two crucial points that determine its shape and properties. The foci in our problem are located at \(\pm 6, 0\). These points are also aligned horizontally.
The distance from the hyperbola's center (0, 0) to each focus is denoted by the parameter \(c\). For our specific case, \(c\) is 6, which means each focus is 6 units away from the center along the x-axis.
Understanding the foci's role helps in grasping how 'stretched' or 'open' the hyperbola is. The larger the distance between the foci, the more open the curve.

• The foci and their distance are key to the hyperbola's distinct shape.
• The parameter \(c\) represents this distance from the center to a focus.

This distance aids in composing the hyperbola's equation, especially in relation to other parameters like \(a\) and \(b\).

Relationship Between a, b, c

The parameters \(a, b,\) and \(c\) are interconnected in defining the hyperbola's geometry. The relationship is captured in the equation: \[ c^2 = a^2 + b^2 \]. This formula perfectly links these parameters and aids in finding unknown values when some are given.
In our exercise, \(a\) is 2, making \(a^2 = 4\). The distance to the foci, \(c\), is 6, giving \(c^2 = 36\). By substituting these known values into the equation, we solve for \(b\) as follows:
\[ 36 = 4 + b^2 \]
Solving this yields \(b^2 = 32\), hence, \(b = \sqrt{32}\).

• The equation \(c^2 = a^2 + b^2\) brings together the key parameters.
• This relationship ensures precise understanding of hyperbola formation.

Thus, these parameters not only define the hyperbola's characteristics but are vital for accurate equation formation.