Problem 5 - Algebra and Trigonometry

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. \(\frac{x^{2}}{4}-\frac{y^{2}}{16}=1\)

Step-by-Step Solution

Answer

Vertices: (2,0) and (-2,0); Foci: (\(2\sqrt{5}\), 0) and (-\(2\sqrt{5}\), 0); Asymptotes: \(y = 2x\) and \(y = -2x\).

Step 1: Identify the Standard Form of the Hyperbola

The given equation is \(\frac{x^{2}}{4}-\frac{y^{2}}{16}=1\), which is in the standard form of a horizontal hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, \(a^2 = 4\) and \(b^2 = 16\), so \(a = 2\) and \(b = 4\).

Step 2: Calculate the Vertices

For a horizontal hyperbola, the vertices are located at \((\pm a, 0)\). Therefore, substituting \(a = 2\), the vertices are \((2,0)\) and \((-2,0)\).

Step 3: Calculate the Foci

The foci of a hyperbola are located at \((\pm c, 0)\) where \(c\) is calculated as \(c^2 = a^2 + b^2\). Substituting \(a = 2\) and \(b = 4\) gives \(c^2 = 2^2 + 4^2 = 4 + 16 = 20\). Solving for \(c\) yields \(c = \sqrt{20} = 2\sqrt{5}\). Therefore, the foci are located at \((2\sqrt{5}, 0)\) and \((-2\sqrt{5}, 0)\).

Step 4: Determine the Asymptotes

The equations of the asymptotes for a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are \(y = \pm \frac{b}{a}x\). With \(a = 2\) and \(b = 4\), the asymptotes are \(y = \pm 2x\).

Step 5: Sketch the Graph

Begin by plotting the vertices \((2,0)\) and \((-2,0)\). Next, sketch the asymptotes \(y = 2x\) and \(y = -2x\), as they intersect at the center, which is \((0,0)\). Lastly, draw the hyperbola opening to the left and right along the x-axis and approaching the asymptotes.

Key Concepts

Vertices

When dealing with hyperbolas, vertices play a crucial role in understanding their shape and orientation. For a hyperbola expressed in the standard form

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

For our specific hyperbola \(\frac{x^{2}}{4}-\frac{y^{2}}{16}=1\), we identify it as a horizontal hyperbola since \(x^2\) comes first. Here's how you locate the vertices:

• Since \(a^2 = 4\), we find \(a = 2\).
• The vertices are symmetrically placed around the center of the hyperbola along the x-axis at \((\pm a, 0)\).

Therefore, the vertices for this hyperbola are located at \((2, 0)\) and \((-2, 0)\). These vertices help to indicate the direction the hyperbola opens, which in this case is horizontally along the x-axis.

Foci

The foci of a hyperbola are points that dictate its "sharpness" or "openness." Each hyperbola has two foci, and they're found using the distance formula:

• Square the semi-major axis, \(a^2\).
• Square the semi-minor axis, \(b^2\).
• Calculate \(c^2 = a^2 + b^2\) to find \(c\).

In our example, where \(a = 2\) and \(b = 4\), we compute:

• \(c^2 = 4 + 16 = 20\) so \(c = \sqrt{20} = 2\sqrt{5}\)

The foci are positioned symmetrically along the transverse axis at \((\pm c, 0)\). Thus, the coordinates are \((2\sqrt{5}, 0)\) and \((-2\sqrt{5}, 0)\). The presence of these points can help visualize the extent of the hyperbola's opening beyond the vertices.

Asymptotes

Asymptotes provide a framework for the shape of a hyperbola. They are diagonal lines that the arms of the hyperbola approach but never touch or cross. For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the asymptotes are given by:

• \(y = \pm \frac{b}{a}x\)

In our example with \(a = 2\) and \(b = 4\), the slope of the asymptotes is \(\pm \frac{b}{a} = \pm 2\). This results in the equations of the asymptotes:

• \(y = 2x\)
• \(y = -2x\)

These asymptotes intersect at the center of the hyperbola, which is the origin \((0,0)\) in this case, guiding us in sketching the overall shape. The arms of the hyperbola will always curve closer to these lines but will never intersect them, giving us an understanding of how wide the opening appears.