Problem 46 - Calculus Early Transcendentals

Question

Find an equation of the following hyperbolas, assuming the center is at the origin. Sketch a graph labeling the vertices, foci, and asymptotes. Use a graphing utility to check your work. A hyperbola with vertices (±1,0) that passes through \(\left(\frac{5}{3}, 8\right)\)

Step-by-Step Solution

Answer

Answer: The equation of the hyperbola is \(x^2 - \frac{y^2}{144}=1\). The coordinates of the foci are \((\pm \sqrt{145}, 0)\). The equations of the asymptotes are \(y=\pm 12x\).

Step 1: Determine major axis (a) and vertices

Since the vertices are given as (±1, 0), it is a horizontal hyperbola with a major axis length of 2. Therefore, we find that the value of a is half the length of the major axis, which gives us \(a = 1\). The vertices of the hyperbola are \((\pm a,0) = (\pm 1,0)\). These are the points where the hyperbola intersects its major axis.

Step 2: Identify the given point on the hyperbola

The hyperbola passes through the point \(\left(\frac{5}{3}, 8\right)\). We will use this point to find the other key elements such as the minor axis and focal distance.

Step 3: Find the minor axis (b) and focal distance (c)

The equation for a horizontal hyperbola, with its center at the origin, is given by \(\frac{x^2}{a^2} - \frac{y^2}{b^2}=1\). Since the hyperbola passes through the point \(\left(\frac{5}{3}, 8\right)\), we can substitute the coordinates of this point into the equation to find b:$$\frac{\left(\frac{5}{3}\right)^2}{1^2} - \frac{8^2}{b^2}=1$$$$\Rightarrow \frac{25}{9} - \frac{64}{b^2}=1$$Now, solve for \(b^2\) by making \(b^2\) subject of formula: $$b^2 = \frac{64}{1 - \frac{25}{9}} = \frac{64}{\frac{4}{9}} = 144 \Rightarrow b = 12$$Now that we have the value of a and b, we can find the focal distance c using the relationship \(c^2 = a^2 + b^2\): $$c^2=1^2+12^2=1+144=145 \Rightarrow c=\sqrt{145}$$

Step 4: Write the equation of the hyperbola

Now that we found a, b, and c, we can write the equation of the hyperbola: $$\frac{x^2}{1^2} - \frac{y^2}{12^2}=1$$Which simplifies to:$$x^2 - \frac{y^2}{144}=1$$

Step 5: Find the coordinates of the foci and the equations of the asymptotes

The coordinates of the foci for a horizontal hyperbola are given by \((\pm c, 0)\), so the foci are at: $$(\pm \sqrt{145}, 0)$$The equations of the asymptotes for a horizontal hyperbola are given by: $$y = \pm\frac{b}{a}x$$For our hyperbola, the equations of the asymptotes are:$$y = \pm\frac{12}{1}x = \pm 12x$$

Step 6: Sketch the hyperbola

Using the information found above, the hyperbola can be sketched. Plot the vertices (±1,0), the foci \((\pm \sqrt{145}, 0)\), and the asymptotes \(y=\pm12x\). Make sure to label all these components on the graph. To check your work, use a graphing utility to graph the equation of the hyperbola \(x^2 - \frac{y^2}{144}=1\) and verify that it matches the sketch.

Key Concepts

Equation of a Hyperbola

A hyperbola is a type of conic section defined by its two symmetrical open curves. The general equation for a hyperbola centered at the origin with its transverse (major) axis along the x-axis is: \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\] Here, \(a\) and \(b\) are real numbers that determine the shape and size of the hyperbola.

• \(a\) is the semi-major axis length, representing the distance from the center to each vertex along the x-axis.
• \(b\) is the semi-minor axis length, perpendicular to \(a\).

This specific problem involves a horizontal hyperbola, indicated by the vertices \((\pm 1, 0)\). Thus, the equation takes the form: \[x^2 - \frac{y^2}{144} = 1\] Substituting values into the formula helps in sketching and understanding the hyperbola's behavior.

Vertices and Foci

Vertices and foci are essential features that describe a hyperbola's geometric and symmetrical properties. The vertices are the points on the hyperbola closest to the center. For the hyperbola given in this exercise, the vertices are located at \((\pm 1, 0)\). These positions align with the equation's "a" value, which is 1 in this case. This means that from the center, the hyperbola extends one unit to the left and right. Foci are points located along the major axis, further from the center than the vertices. They are key in defining the hyperbola's shape due to their relationship with the distance to any point on the hyperbola. The coordinates for the foci in this example are determined using:\[c^2 = a^2 + b^2\] Given \(a = 1\) and \(b = 12\), we find:\[c = \sqrt{145}\] This reveals the foci as \((\pm \sqrt{145}, 0)\), contributing to the hyperbola's full description.

Asymptotes

Asymptotes are imaginary lines that the hyperbola approaches but never intersects. They provide guidelines for sketching the hyperbola and understanding its extent in the coordinate plane.For a standard hyperbola centered at the origin, the equation of the asymptotes is crucial. It is given by:\[y = \pm \frac{b}{a} x\] In this exercise, where \(a = 1\) and \(b = 12\), the asymptotes are:\[y = \pm 12x\] This formula shows that, at a great distance from the center, the hyperbola's branches will become nearly indistinguishable from the lines \(y = 12x\) and \(y = -12x\). The closeness of the hyperbolic branches to their asymptotes explains the hyperbola's opening angle and overall shape.