Problem 41 - Precalculus Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Foci: \((\pm 5,0),\) length of transverse axis: 6

Step-by-Step Solution

Answer

The equation is \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \).

Step 1: Understand the Basic Form of the Hyperbola Equation

For hyperbolas with horizontal transverse axes, the standard form of the equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). In this case, the transverse axis is horizontal since the foci are positioned along the x-axis.

Step 2: Determine the Values of a and c

The length of the transverse axis is given as 6, so \( 2a = 6 \), thus \( a = 3 \). The foci are given at \( (+5,0) \) and \( (-5,0) \), so \( c = 5 \).

Step 3: Calculate b Using the Relationship Between a, b, and c

For a hyperbola, the relationship \( c^2 = a^2 + b^2 \) holds true. Substitute the known values: \( 5^2 = 3^2 + b^2 \). This gives \( 25 = 9 + b^2 \).

Step 4: Solve for b

Subtract 9 from 25 to get \( b^2 = 16 \). So, \( b = 4 \).

Step 5: Write the Equation of the Hyperbola

Now with \( a = 3 \), \( b = 4 \), and using the standard form for hyperbolas with a horizontal transverse axis: \( \frac{x^2}{3^2} - \frac{y^2}{4^2} = 1 \). Thus, the equation is \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \).

Key Concepts

Transverse Axis

In the context of hyperbolas, the transverse axis is a key feature. It is the line segment that passes through the center of the hyperbola and the two vertices. This axis is crucial because it helps to define the orientation and shape of the hyperbola. For a hyperbola aligned along the coordinate axes, the transverse axis can be either horizontal or vertical.

• If the transverse axis is horizontal, the foci and vertices are located at coordinates \( (\pm c, 0) \) and \( (\pm a, 0) \), respectively.
• If the transverse axis is vertical, the foci and vertices shift to \( (0, \pm c) \) and \( (0, \pm a) \).

In the exercise, the transverse axis is horizontal because the foci are given as \( (\pm 5, 0) \). The length of this axis is also given directly as 6, which means \( 2a = 6 \), and therefore \( a = 3 \). This tells us the hyperbola stretches 3 units left and right of the center.

Foci

The foci (plural for 'focus') are special points located along the transverse axis of a hyperbola. They have specific roles in defining the hyperbola's shape. The absolute beauty of the hyperbola lies in its property that the difference of the distances from any point on the hyperbola to the two foci is constant.

For our given problem, the foci are \( (\pm 5, 0) \), meaning the foci are positioned 5 units apart from the center of the hyperbola which is the origin for this case. The value for \( c \) (the distance from the center to each focus) is simply half the distance between the foci, hence \( c = 5 \).

• The foci are used to find the value of \( b \) in the hyperbola's equation by using the relationship \( c^2 = a^2 + b^2 \).

Utilizing this relationship, calculated \( b \) as 4 which further helps to complete the standard form of the hyperbola equation.

Standard Form of Hyperbola Equation

To write the equation of a hyperbola, the standard form comes into play. It varies depending on whether the transverse axis is horizontal or vertical. For a hyperbola with a horizontal transverse axis, the equation reads:
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]
This denotes that the x-axis is the direction of stretch, and the value \( a \) corresponds to the semi-transverse axis, while \( b \) is related to the semi-conjugate axis.

In our specific example:

• The value of \( a \) is calculated as 3, derived from the relation \( 2a = \text{transverse axis length} = 6 \).
• The foci provide \( c = 5 \) helping derive \( b \) using \( c^2 = a^2 + b^2 \).
• Finally, solving gives \( b = 4 \).

Plug these into the standard equation to get:
\[\frac{x^2}{9} - \frac{y^2}{16} = 1\]
This clearly describes every feature of the hyperbola, its orientation, and dimensions.