Problem 40 - Precalculus Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Foci: \((\pm 3,0),\) hyperbola passes through \((4,1)\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{x^2}{8} - \frac{y^2}{1} = 1\).

Step 1: Identify the Center and Foci

The foci of the hyperbola are given as \((\pm 3, 0)\), indicating the center of the hyperbola is at the origin \((0, 0)\), and the foci are aligned along the x-axis. Therefore, the transverse axis is horizontal.

Step 2: Use the Distance Between the Foci

The distance from the center to the foci \((c)\) is 3. Therefore, \(c = 3\). The relation \(c^2 = a^2 + b^2\) is used, where \(a\) relates to the semi-major axis and \(b\) to the semi-minor axis.

Step 3: Write the Standard Equation

Since the transverse axis is horizontal, the standard form of the hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

Step 4: Use the Point on the Hyperbola

Substitute the point \((4, 1)\) into the equation to find \(a^2\) and \(b^2\). Substitute \(x = 4\) and \(y = 1\) into the equation: \[ \frac{4^2}{a^2} - \frac{1^2}{b^2} = 1 \] Simplifying, we have \(\frac{16}{a^2} - \frac{1}{b^2} = 1\). This is one equation in two unknowns.

Step 5: Solve for a and b Using Known c

Using \(c^2 = a^2 + b^2\) with \(c=3\), we have \(9 = a^2 + b^2\). Now you need to solve the system of equations: \(\frac{16}{a^2} - \frac{1}{b^2} = 1\) and \(a^2 + b^2 = 9\).

Step 6: Express b in terms of a

From \(a^2 + b^2 = 9\), express \(b^2 = 9 - a^2\). Substitute \(b^2 = 9 - a^2\) into \(\frac{16}{a^2} - \frac{1}{b^2} = 1\) to solve for \(a^2\).

Step 7: Solve for a^2

Substitute \(b^2 = 9 - a^2\) into the equation \(\frac{16}{a^2} - \frac{1}{9-a^2} = 1\). Clear the fractions and solve: \[ 16(9 - a^2) - a^2 = a^2(9 - a^2) \]Simplifying and solving, find \(a^2 = 8, b^2 = 1\) as it satisfies both equations.

Step 8: Write the Final Equation of the Hyperbola

With \(a^2 = 8\) and \(b^2 = 1\), substitute into the standard form: \[ \frac{x^2}{8} - \frac{y^2}{1} = 1 \].This is the equation of the hyperbola.

Key Concepts

Foci

The foci of a hyperbola are two specific points located inside each of the two curves that make up the hyperbola. They play a crucial role in the definition and properties of the hyperbola. For a hyperbola centered at the origin, if the foci lie on the x-axis, they are symmetric around the center and can be represented as \((\pm c, 0)\). Here, \((\pm c, 0)\) indicates that foci are at a distance \(+c\) and \(-c\) from the center, which in this exercise are given as \((\pm 3, 0)\). Hence, the distance from the center of the hyperbola to each focus point is known as 'c' and is equal to 3.

• The foci help to determine the shape and orientation of the hyperbola.
• The distance between the two foci is always greater than the distance between the vertices.

Remember, the role of the foci in calculating hyperbolas is key because they ensure the accuracy and placement of the curves in relation to each other.

Equation of Hyperbola

To write the equation of a hyperbola, we need to understand its standard equation format. Hyperbolas have two standard forms depending on their orientation. They are derived based on the transverse axis:

• Horizontal transverse axis: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
• Vertical transverse axis: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

For the given exercise, the hyperbola has a horizontal transverse axis. Thus, the standard equation format \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is used. Here, \(a\) represents the distance from the center to a vertex along the transverse axis, and \(b\) the distance from the center to a point along the conjugate axis. The relationship between these and the foci is given by \(c^2 = a^2 + b^2\), linking them to the focal length \(c\). This ensures that the shape is correctly represented along its axes.

Transverse Axis

The transverse axis is the line segment that passes through the center of the hyperbola and its vertices. It is the axis that includes the foci. For a hyperbola centered at the origin, if the transverse axis is horizontal, it implies that the vertices and foci lie along the x-axis. This makes the hyperbola wider horizontally.

• The length of the transverse axis is \(2a\).
• The position of the transverse axis impacts the orientation of the hyperbola.

In this exercise, the given condition and the foci alignment indicate that the hyperbola has a horizontal transverse axis. The vertices, therefore, would lie along the line \(y = 0\), simplifying the analysis and construction of the hyperbola's equation.

Horizontal Hyperbola

A horizontal hyperbola is a hyperbola where the transverse axis is parallel to the x-axis. This orientation is essential as it determines the standard form of its equation. For a hyperbola such as the one provided here, the horizontal orientation implies:

• The center, denoted as \(h, k\), is at the origin, i.e., \(0,0\).
• The vertices are located at points \( (\pm a, 0) \).
• The foci are along the horizontal line, \((\pm c, 0)\).

Hence, with the hyperbola's center and consequently its transverse axis being along the x-axis, the equation simplifies to a form where \(x\) takes precedence. The wider stance of the hyperbola is reflected in its horizontally stretched nature. Knowing these aspects aids in not only plotting the hyperbola but also in predicting the behavior of its curves.