Problem 35 - Precalculus

Question

Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. Use a graphing utility to graph the hyperbola and its asymptotes. $$4 x^{2}-9 y^{2}=36$$

Step-by-Step Solution

Answer

The center of the hyperbola is at (0,0). The vertices are at (3,0) and (-3,0). The foci are at \((\sqrt{13}, 0)\) and \((- \sqrt{13}, 0)\). The equations of the asymptotes are \(y = \frac{2}{3}x\) and \(y = - \frac{2}{3}x\).

Step 1: Put the equation in standard form

Divide the given equation \(4 x^{2} - 9 y^{2} = 36\) by 36 to place it in standard form. This yields \((\frac{x}{3})^{2} - (\frac{y}{2})^{2} = 1\). From this equation, we can see that h=0, k=0 (the center of the hyperbola), a=3 and b=2.

Step 2: Determine the vertices and foci

The vertices of the hyperbola are of the form (a, 0) and (-a, 0). Substituting a=3, we get the vertices as (3, 0) and (-3, 0). The foci are at (c, 0) and (-c, 0). To find c, we can use the relationship \(a^{2} + b^{2} = c^{2}\). Solving for c, we find \(c= \sqrt{3^{2} + 2^{2}} = \sqrt{13}\). So the foci are at \((\sqrt{13}, 0)\) and \((- \sqrt{13}, 0)\).

Step 3: Find the equations of the asymptotes

The equations of the asymptotes for a hyperbola are given by the equations \(y = \pm\frac{b}{a}x)\). Substituting b=2 and a=3 into the equations, we get \(y = \pm\frac{2}{3}x\). Therefore, the equations of the asymptotes are \(y = \frac{2}{3}x\) and \(y = - \frac{2}{3}x\).

Step 4: Graph the hyperbola and its asymptotes

Using a graphing utility, graph the equation of the hyperbola along with the equations of its asymptotes to visualize the solution. The center of the hyperbola will be at the origin, with vertices along the x-axis at x=a and x=-a, and lines representing the asymptotes.

Key Concepts

Equations of Asymptotes

Asymptotes are lines that a graph of a curve approaches but never actually touches. For hyperbolas, these lines give a good indication of the general shape and direction of the hyperbola.
The standard form of a hyperbola's equation is important in identifying the equations of its asymptotes. For hyperbolas of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are straigth lines given by:

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

In our example, \( a = 3 \) and \( b = 2 \). Substituting into the formula, the asymptotes become \( y = \frac{2}{3}x \) and \( y = -\frac{2}{3}x \).
These straight lines run through the center of the hyperbola at (0,0) and extend indefinitely, providing boundary directions for the hyperbola.

Vertices of Hyperbola

The vertices of a hyperbola are crucial points where the hyperbola intersects its principal axis. In a standard form hyperbola equation of the type \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the hyperbola opens along the x-axis.
The vertices are located at \((a, 0)\) and \((-a, 0)\). In the example given, since \(a=3\), the vertices are at (3, 0) and (-3, 0).
This means the hyperbola stretches from the point on the x-axis at +3 to the point at -3, and does not rise above or fall below the x-axis.

Foci of Hyperbola

The foci are important points inside a hyperbola that help in defining its shape and properties. For hyperbolas, the distance between the foci determines how "stretched" the hyperbola is.
In general, the foci of a hyperbola given by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are located at (c, 0) and (-c, 0). Here, \(c\) is calculated using the formula \(c = \sqrt{a^2 + b^2}\).
For our specific example, where \(a=3\) and \(b=2\), we find \(c = \sqrt{3^2 + 2^2} = \sqrt{13}\). Thus, the foci are at \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\).
These points lie on the x-axis, outside of the vertices, showing where the two intersecting branches of the hyperbola curve towards and diverge from.