Question

Find the vertices, the foci, the lengths \(a\) and \(b\) of the semiaxes, and the slope of the asymptotes for each hyperbola. Graph some. $$x^{2}-4 y^{2}=16$$

Short answer

Vertices: \((-4, 0)\), \((4, 0)\); Foci: \((-2\sqrt{5}, 0)\), \((2\sqrt{5}, 0)\); Lengths of semiaxes: \(a = 4\), \(b = 2\); Slope of the asymptotes: \(\pm\frac{1}{2}\).

Step-by-step solution

Identify the hyperbola equation

Begin by comparing the given equation with the standard form of the hyperbola equation, which is either \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for a horizontal orientation, or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for a vertical orientation. In this case, the given equation \(x^2 - 4y^2 = 16\) resembles the first form after dividing by 16.

Determine the orientation and standard form

Divide the entire equation by 16 to translate it into standard form, yielding \(\frac{x^2}{16} - \frac{y^2}{4} = 1\). Since \(x^2\) leads, this hyperbola opens horizontally.

Find the lengths of the semiaxes \(a\) and \(b\)

The lengths of the semiaxes can be read directly from the standard form; \(a^2 = 16\) and \(b^2 = 4\). Thus \(a = 4\) and \(b = 2\).

Determine the vertices

The vertices of a horizontally oriented hyperbola are \((\pm a, 0)\). In this case, the vertices are at \((\pm 4, 0)\) or \((-4, 0)\) and \((4, 0)\).

Find the foci

The foci are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\). Calculate \(c = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\). The foci are at \((-2\sqrt{5}, 0)\) and \((2\sqrt{5}, 0)\).

Calculate the slope of the asymptotes

For a hyperbola with the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes have a slope of \(\pm \frac{b}{a}\). Here, the asymptotes will have slopes of \(\pm \frac{2}{4} = \pm \frac{1}{2}\).

Graph the hyperbola

Draw the rectangle with vertices at \((\pm a, \pm b)\), which here are \((\pm 4, \pm 2)\). Sketch the asymptotes that pass through the corners of this rectangle and then draw the hyperbola's branches opening outward towards the vertices and approaching the asymptotes at infinity.

Key concepts

Standard Form of a Hyperbola

Understanding the standard form of a hyperbola is crucial for analyzing its properties and sketching its graph. The standard form is given by either \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for a hyperbola that opens left and right (horizontally), or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for a hyperbola that opens up and down (vertically). In these equations, \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively.

To solve an exercise like \(x^2 - 4y^2 = 16\), the first step involves rewriting the equation into the standard form. This is done by dividing each term by 16, resulting in the standard form \(\frac{x^2}{16} - \frac{y^2}{4} = 1\). Here, \(a^2 = 16\) and \(b^2 = 4\), leading to \(a = 4\) and \(b = 2\), which are fundamental in determining other components of the hyperbola.

Hyperbola Vertices and Foci

The vertices and foci are specific points that help define the shape and location of a hyperbola. For a hyperbola that opens horizontally like \(\frac{x^2}{16} - \frac{y^2}{4} = 1\), the vertices are found along the x-axis at \(\pm a, 0\), which corresponds to \(\pm 4, 0\) for our example. These are the points where the hyperbola intersects its transverse axis, which is the axis connecting the two branches of the hyperbola.

The foci (plural for focus) are points located inside each branch of the hyperbola, along the transverse axis, and they are defined as \(\pm c, 0\) for a horizontal hyperbola, where \(c = \sqrt{a^2 + b^2}\). In our exercise, the foci are calculated to be at \(\pm 2\sqrt{5}, 0\), illustrating that they are further from the center than the vertices, given that \(c\) is always greater than \(a\). The locations of the vertices and foci are key components in plotting the hyperbola on a coordinate system.

Asymptotes of a Hyperbola

Asymptotes of a hyperbola are straight lines that the hyperbola approaches but never touches as it extends towards infinity. These lines provide the skeleton for graphing a hyperbola, as they pass through the corners of a rectangle constructed around the hyperbola's center. For the hyperbola in our example, the asymptotes have equations of the form \(y = \pm \frac{b}{a}x\).

The slopes of the asymptotes are determined by the ratio of \(b\) and \(a\). In \(\frac{x^2}{16} - \frac{y^2}{4} = 1\), we calculate the slopes as \(\pm \frac{2}{4} = \pm \frac{1}{2}\). This tells us that for any point on the asymptote lines, moving 2 units horizontally from the center will result in a 1 unit vertical change. When graphing, these lines help to ensure that the hyperbola's branches will approach them correctly as they open outward to infinity, providing a visual boundary for the hyperbola's shape.