Problem 33 - Precalculus Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Foci: \((0, \pm 2),\) vertices: \((0, \pm 1)\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(y^2 - \frac{x^2}{3} = 1\).

Step 1: Understand the Form of the Hyperbola

The foci and vertices use coordinates of the form (0, ±𝑐) and (0, ±𝑎). This tells us that the hyperbola is vertical and its center is at the origin. Thus, the equation of the hyperbola is of the form: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

Step 2: Identify 'a' and 'c'

Given the vertices are (0, ±1), we know that \(a = 1\). With foci at (0, ±2), the value of \(c\) is 2. Remember this relationship: \[ c^2 = a^2 + b^2 \]

Step 3: Solve for 'b'

We know \(c^2 = a^2 + b^2\). Substituting the known values, we have: \[ 2^2 = 1^2 + b^2 \] which simplifies to \[ 4 = 1 + b^2 \]. Solve for \(b^2\) by subtracting 1 from both sides: \[ 3 = b^2 \].

Step 4: Write the Equation of the Hyperbola

Now we have all the pieces. The equation of the hyperbola is: \[ \frac{y^2}{1^2} - \frac{x^2}{3} = 1 \] This simplifies to: \[ \frac{y^2}{1} - \frac{x^2}{3} = 1 \] or simply \[ y^2 - \frac{x^2}{3} = 1 \].

Key Concepts

Foci of a Hyperbola

In a hyperbola, the foci are two fixed points used to define the curve. For a vertical hyperbola, like the one described in our problem, the foci lie along the y-axis since the coordinates are given as \(0, \pm c\). In simpler terms, they are positioned symmetrically around the center, which is at the origin \(0, 0\).
For this example, the foci are at \(0, \pm 2\). This means they are positioned two units above and two units below the center. The distance from the center to each focus is represented by the variable \(c\).
The role of the foci is crucial, as it helps in shaping the hyperbola. The greater the distance between the foci, the "wider" the hyperbola appears. In our specific problem, knowing the distance to the foci also allows us to find other parameters like \(b^2\) using the relationship \(c^2 = a^2 + b^2\).

Vertices of a Hyperbola

Vertices are the points where the hyperbola intersects its principal axis. In a vertical hyperbola, the vertices lie on the y-axis. In our provided problem, the vertices are located at \(0, \pm 1\).
This tells us that the hyperbola reaches its closest and farthest points from the center along this vertical direction. The distance from the center to a vertex is denoted by \(a\), which equals 1 in this case. The vertices help in determining the shape and size of the hyperbola.
To sum up:

• Vertices are always along the primary axis.
• They help determine the transverse axis' length, which is \(2a\).

Vertical Hyperbola

In the context of hyperbolas, the orientation is key. A vertical hyperbola opens up and down along the y-axis. This is opposed to a horizontal hyperbola that opens left and right along the x-axis.
Our problem reveals a vertical hyperbola because:

• The coordinates of both the foci and vertices are centered around the y-axis ( \(0, \pm c\) and \(0, \pm a\)).
• The equation of the hyperbola takes the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

Remember, the distinguishing feature of a vertical hyperbola lies in those coordinates centering on the y-axis and the positioning of terms within its equation.

Hyperbola Equation Form

The general form of a hyperbola's equation is crucial because it helps identify its key components quickly. For a vertical hyperbola, the typical equation is:
\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]
Here is how to interpret this:

• \(a\) represents the distance from the center to each vertex on the y-axis.
• \(b\) is derived from the relationship \(c^2 = a^2 + b^2\), represented as the distance related to the co-vertices.

In this particular problem, with known values for \(a\) and \(c\), we solve for \(b^2\) using:
\[ 2^2 = 1^2 + b^2\]When simplified, it leads to \(b^2 = 3\). This gives us the full equation \(\frac{y^2}{1} - \frac{x^2}{3} = 1\), which succinctly describes this specific vertical hyperbola.