Problem 50 - Precalculus: Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Foci: \((0, \pm 1),\) length of transverse axis: 1

Step-by-Step Solution

Answer

The equation of the hyperbola is \(4y^2 - \frac{4}{3}x^2 = 1\).

Step 1: Identify the Form of the Equation

The foci of the hyperbola are given as \((0, \pm 1)\). This indicates that the hyperbola is vertical, meaning its equation will be in the form: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

Step 2: Recognize the Given Components

We know the foci are at \((0, \pm c)\), which gives \(c = 1\). Additionally, the length of the transverse axis is given as 1, so \(2a = 1\), meaning \(a = \frac{1}{2}\).

Step 3: Calculate the Value of b²

For hyperbolas, the relationship between \(a\), \(b\), and \(c\) is given by the equation:\[ c^2 = a^2 + b^2 \]Plug in the known values:\[ 1^2 = \left(\frac{1}{2}\right)^2 + b^2 \]This simplifies to:\[ 1 = \frac{1}{4} + b^2 \]Subtract \(\frac{1}{4}\) from both sides:\[ b^2 = 1 - \frac{1}{4} = \frac{3}{4} \]

Step 4: Form the Equation of the Hyperbola

With \(a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) and \(b^2 = \frac{3}{4}\), substitute these values into the standard form of the equation for a vertical hyperbola:\[ \frac{y^2}{\frac{1}{4}} - \frac{x^2}{\frac{3}{4}} = 1 \]This can be rewritten as:\[ 4y^2 - \frac{4}{3}x^2 = 1 \]

Key Concepts

Foci of Hyperbola

One of the defining features of a hyperbola is its foci. The foci are two distinct points located along the axis of symmetry of the hyperbola, around which the shape is defined. For the given problem, the foci of the hyperbola are at

• y = \( (0, \pm 1) \),this means that the hyperbola opens upward and downward, making it a vertical hyperbola.

Knowing the foci allows us to determine the value of \( c \), which is the distance from the center of the hyperbola to either focus. For this problem, \( c = 1 \). This means that each focus is 1 unit away from the center. The center of our hyperbola is at the origin, \( (0,0) \), since the coordinates are symmetrical around zero.

Transverse Axis

The transverse axis of a hyperbola is the line segment that passes through both foci and is aligned with the direction in which the hyperbola opens. For a vertical hyperbola, like the one described in the exercise, this axis is vertical. The length of the transverse axis is given as 1.This length is directly related to the parameter \( a \), which represents half of the transverse axis's length. Therefore, in this problem:

• \( 2a = 1 \)
• \( a = \frac{1}{2} \)

These simple relationships are derived from the geometric properties of the hyperbola. Knowing \( a \) helps us construct the equation of the hyperbola by determining the scale of the transverse axis compared to the distance of the foci from the center.

Hyperbola Formula

To form the equation of a vertical hyperbola, we use the general formula: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]The provided conditions allow us to determine the values of \( a^2 \) and \( b^2 \). From the solution steps:

• \( a^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
• Using the relationship \( c^2 = a^2 + b^2 \), we find
  • \( 1 = \frac{1}{4} + b^2 \)
  • \( b^2 = \frac{3}{4} \)

Substitute these values back into the formula to get:\[ \frac{y^2}{\frac{1}{4}} - \frac{x^2}{\frac{3}{4}} = 1 \]Simplifying, we multiply through by 4 to clear the denominators and get an equivalent, more straightforward form:\[ 4y^2 - \frac{4}{3}x^2 = 1 \]This represents our vertical hyperbola's equation based on the provided foci and transverse axis length. This equation tells us how the hyperbola's arms open and stretch in relation to the coordinate axes.