Problem 32 - Precalculus: Functions and Graphs

Question

Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Foci \(F(\pm 34,0), \quad\) asymptotes \(y=\pm \frac{3}{5} x\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{x^2}{850} - \frac{y^2}{306} = 1\).

Step 1: Identifying the Parameters

Since the foci of the hyperbola are at \((\pm 34,0)\), the transverse axis is horizontal. This tells us that the standard form of the hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). The foci provide the value of \(c\), where \(c = 34\).

Step 2: Using the Asymptotes

The asymptotes of the hyperbola are given by the equations \(y = \pm \frac{3}{5}x\). Comparing this to \(y = \pm \frac{b}{a}x\), we find that \(\frac{b}{a} = \frac{3}{5}\).

Step 3: Finding Relationship Between \(a\), \(b\), and \(c\)

Recall the relationship \(c^2 = a^2 + b^2\) for hyperbolas. We've identified \(c = 34\), \(\frac{b}{a} = \frac{3}{5}\). Therefore, \(b = \frac{3}{5}a\).

Step 4: Substitute and Solve for \(a\)

Substitute \(b = \frac{3}{5}a\) into \(c^2 = a^2 + b^2\) to get \(34^2 = a^2 + \left(\frac{3}{5}a\right)^2\). Simplify this to \(1156 = a^2 + \frac{9}{25}a^2\).

Step 5: Simplifying the Expression

Combine terms: \(1156 = \left(1 + \frac{9}{25}\right)a^2\). This simplifies to \(1156 = \frac{34}{25}a^2\).

Step 6: Solving for \(a\)

Multiply both sides by \(\frac{25}{34}\) to isolate \(a^2\): \(a^2 = \frac{1156 \times 25}{34}\). Calculate to find \(a^2 = 850\).

Step 7: Calculate \(b^2\)

Using \(b = \frac{3}{5}a\), find \(b^2 = \left(\frac{3}{5}\right)^2 a^2 = \frac{9}{25} \times 850\). Compute to find \(b^2 = 306\).

Step 8: Write the Equation of the Hyperbola

Substitute \(a^2 = 850\) and \(b^2 = 306\) into the hyperbola equation to get \(\frac{x^2}{850} - \frac{y^2}{306} = 1\).

Key Concepts

Conic Sections

Conic sections are curves that result from the intersection of a plane and a double-napped cone. These curves include circles, ellipses, parabolas, and hyperbolas.
Each type of conic has distinct features and equations. In this context, we focus on hyperbolas. A hyperbola consists of two separate curves called branches, which mirror each other. The hyperbola forms when the plane cuts through both nappes of the cone. Among its remarkable geometric properties is the set of lines called asymptotes that guide the curvature of the hyperbola.
Understanding conic sections, especially hyperbolas, involves knowing how their equations represent their shapes and positions on the coordinate plane.

Transverse Axis

The transverse axis of a hyperbola is the line segment that passes through its center, connecting the vertices of the hyperbola. It dictates the orientation of the hyperbola.In many standard forms of hyperbolic equations, if the transverse axis is horizontal, the equation remains as:

• \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

When the transverse axis is vertical, the hyperbola can look like:

• \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)

In our exercise, the given foci indicate a horizontal transverse axis, guiding us to use the appropriate equation form.
This helps to frame the equation of the hyperbola accordingly.

Asymptotes

Asymptotes are lines that the branches of a hyperbola approach but never actually touch. They are fundamental to understanding the hyperbola's shape and direction.For hyperbolas, the equations of asymptotes emerge naturally from the equation:

• In horizontal form: \(y = \pm \frac{b}{a}x\)
• In vertical form: \(x = \pm \frac{a}{b}y\)

In this problem, we are provided with the asymptotes \(y = \pm \frac{3}{5}x\).
From these, we establish the ratio \(\frac{b}{a} = \frac{3}{5}\), which further helps in solving the hyperbola equation.

Foci of Hyperbola

The foci (plural for focus) of a hyperbola are two fixed points located along the transverse axis. As a defining characteristic, the difference in distances from any point on the hyperbola to these foci is constant.Understanding foci is essential for solving hyperbolic equations. Given foci provide a parameter \(c\), the distance from the center of the hyperbola to each focus.
In the original exercise, we have \(c = 34\), putting the foci at \((\pm 34, 0)\).This value of \(c\) plays a crucial role in determining the values of \(a\) and \(b\) using the relationship:

• \(c^2 = a^2 + b^2\)

which helps to derive the complete equation of the hyperbola.