Problem 12 - Algebra and Trigonometry

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. \(x^{2}-y^{2}+4=0\)

Step-by-Step Solution

Answer

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

Step 1: Transform the Equation

The given equation is \(x^2 - y^2 + 4 = 0\). First, subtract 4 from both sides to get \(x^2 - y^2 = -4\). Now, rewrite it by dividing each term by 4 to get \(\frac{x^2}{4} - \frac{y^2}{4} = 1\). This is the standard form of a hyperbola.

Step 2: Identify Key Components

In the equation \(\frac{x^2}{4} - \frac{y^2}{4} = 1\), identify \(a^2\) and \(b^2\). Here \(a^2 = 4\) and \(b^2 = 4\). Therefore, \(a = 2\) and \(b = 2\). The transverse axis is horizontal as \(x^2\) term is positive.

Step 3: Find the Vertices

For a hyperbola of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), vertices are at \((\pm a, 0)\). So, the vertices are in this case \((\pm 2, 0)\) or \((-2, 0)\) and \((2, 0)\).

Step 4: Calculate the Foci

The foci of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by \((\pm c, 0)\) where \(c^2 = a^2 + b^2\). Here, \(c^2 = 4 + 4 = 8\), thus \(c = \sqrt{8} = 2\sqrt{2}\). Therefore, the foci are at \((\pm 2\sqrt{2}, 0)\).

Step 5: Determine the Asymptotes

For the given hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{b}{a}x\). Here, \(b = 2\) and \(a = 2\), hence the asymptotes are \(y = \pm x\).

Step 6: Sketch the Hyperbola

Using the vertices \((-2, 0)\) and \((2, 0)\), sketch the hyperbola opening to the left and right. Draw the asymptotes \(y = x\) and \(y = -x\) as guidelines for the branches of the hyperbola. The foci \((-2\sqrt{2}, 0)\) and \((2\sqrt{2}, 0)\) should be plotted for an accurate graph.

Key Concepts

Vertices of a Hyperbola

In the context of a hyperbola, the vertices are crucial as they help define the shape and orientation. Unlike an ellipse, which is closed, a hyperbola is open with two separate branches.

For the equation \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]the vertices are the points on the transverse axis where the hyperbola intersects. The transverse axis is the line segment that joins the vertices, and its length is always equal to \(2a\).

• The coordinates of the vertices are \((\pm a, 0)\) for a hyperbola centered at the origin, aligned horizontally.
• In the exercise given, \(a = 2\) since \(a^2 = 4\), which makes our vertices \((-2, 0)\) and \((2, 0)\).

These vertices define the boundary of the hyperbola and help set up the graphing process.

Remember, for vertical hyperbolas, the structure changes: the equation becomes \[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1,\]and the vertices are located at \((0, \pm a)\). This flexibility makes it essential to always reconsider the form of the hyperbola before identifying the vertices.

Foci of a Hyperbola

The foci are another defining feature fundamental to understanding and graphing hyperbolas. In any hyperbola, each branch gets closer and closer to its corresponding focus but never completely intersects it.

In the equation \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]the foci serve as a navigational point, residing further from the center than the vertices. The distance of the foci from the center, represented by \(c\), can be calculated using the formula:

\[c^2 = a^2 + b^2.\]

• For the given hyperbola, we calculated \(c^2\) as \(8\), making \(c = \sqrt{8} = 2\sqrt{2}\).
• Thus, the foci locations will be \((\pm 2\sqrt{2}, 0)\).

These foci are critical for constructing the graph of the hyperbola, emphasizing the direction in which the branches extend.

Just like vertices, if the hyperbola is vertically oriented, the foci positions transition to \((0, \pm c)\). Hence, always confirm the alignment of the main axis before determining the foci.

Asymptotes of a Hyperbola

Asymptotes are significant features that help in sketching the hyperbola. They act as invisible boundaries, indicating the direction in which each branch of the hyperbola travels.

For the equation \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]with a horizontal transverse axis, the asymptotes are represented by:

\[y = \pm \frac{b}{a}x.\]

For the given problem:

• We identify \(b = 2\) and \(a = 2\) from \(b^2 = 4\) and \(a^2 = 4\), so \(\frac{b}{a} = 1\).
• This results in the asymptotes being \(y = x\) and \(y = -x\).

These lines are crucial to sketching the hyperbola as they provide a guideline illustrating how the branches approach but never intersect these lines.

Understanding asymptotes also offers insight into handling vertical hyperbolas, for which the asymptotes remain the same, \(y = \pm \frac{b}{a}x\), but the application in the graph may differ between horizontal and vertical orientations.