Problem 41 - Precalculus: Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices: \((\pm 1,0),\) asymptotes: \(y=\pm 5 x\)

Step-by-Step Solution

Answer

The equation is \( x^2 - \frac{y^2}{25} = 1 \).

Step 1: Identify the Standard Form of a Hyperbola

A hyperbola with horizontal transverse axis and center at the origin has the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, \( a \) is the distance from the center to each vertex.

Step 2: Determine the Value of 'a'

The vertices are at \((\pm 1, 0)\), which means \(a = 1\). Thus, \( a^2 = 1^2 = 1 \).

Step 3: Relate Asymptotes to 'b' and 'a'

The asymptotes for a hyperbola with a horizontal transverse axis follow the equation: \( y = \pm \frac{b}{a}x \). Given that the asymptotes are \( y = \pm 5x \), it follows that \( \frac{b}{a} = 5 \).

Step 4: Calculate 'b' Using the Asymptote Equation

Using \( \frac{b}{a} = 5 \) and knowing \( a = 1 \), we have \( b = 5 \). Hence, \( b^2 = 5^2 = 25 \).

Step 5: Write the Equation of the Hyperbola

Substitute \( a^2 = 1 \) and \( b^2 = 25 \) into the standard equation. Thus, the equation of the hyperbola is \( \frac{x^2}{1} - \frac{y^2}{25} = 1 \), which simplifies to \( x^2 - \frac{y^2}{25} = 1 \).

Key Concepts

Equation of a Hyperbola

A hyperbola is a type of conic section that is defined by its equation, which reflects its shape and orientation in the coordinate plane. For hyperbolas centered at the origin, the standard form of their equation is typically written as

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

The form we use depends on the axis along which the hyperbola opens. For a hyperbola with a horizontal transverse axis, like in our exercise, the formula utilizes the first form.
Here, \( a \) and \( b \) play crucial roles:

• \( a \) represents the distance from the center to each vertex along the transverse axis.
• \( b \) relates to the slope of the asymptotes and influences the steepness of the hyperbola branches.

By knowing the vertices’ position and the asymptote slopes, we can determine \( a \) and \( b \), and plug these values into the standard form to get the specific equation of our hyperbola.

Asymptotes

Asymptotes are imaginary lines that the branches of a hyperbola gradually approach, but never intersect. These lines create a framework within which the hyperbola is perfectly symmetric.
For a hyperbola with a horizontal transverse axis centered at the origin, the equations for the asymptotes are:

• \( y = \pm \frac{b}{a}x \)

This equation indicates that the slope of the asymptotes is \( \frac{b}{a} \), which dictates the angle at which the branches of the hyperbola open outwards.
In our case, the asymptotes \( y = \pm 5x \) suggest that the slope \( \frac{b}{a} \) is 5. This relationship between \( b \), \( a \) and the asymptotes helps in calculating \( b \) when \( a \) is known, thus fully defining the hyperbola's shape.

Vertices

Vertices are fundamental points on a hyperbola where the curve is closest to or farthest from the center. For hyperbolas centered at the origin, the vertices lie on the transverse axis.In the common standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) for a hyperbola with a horizontal transverse axis:

• The vertices are located at \( (+a, 0) \) and \( (-a, 0) \).

These points reveal the hyperbola's span across the transverse axis.
By using the given vertices \( (\pm 1, 0) \), you can immediately deduce that \( a = 1 \).
This information not only assists in finding the equation of the hyperbola but also helps understand how the curve positions itself within its bounding asymptotes.