Problem 23 - Algebra 2

Question

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ x^{2}-y^{2}=4 $$

Step-by-Step Solution

Answer

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

Step 1: Recognize the Form of the Equation

The equation given is already in the form for a standard hyperbola: \( x^2 - y^2 = 4 \). This can be rearranged to \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \), showing that it is a hyperbola centered at the origin \((0,0)\) with a horizontal transverse axis.

Step 2: Identify the Values of a and b

From the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we identify that \( a^2 = 4 \) and \( b^2 = 4 \). Therefore, \( a = 2 \) and \( b = 2 \).

Step 3: Calculate the Distance to the Foci

For a hyperbola in this form, the foci are located at \(( \pm c, 0)\), where \( c = \sqrt{a^2 + b^2} \). Substituting \( a^2 = 4 \) and \( b^2 = 4 \), we have \( c = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \). So the foci are \(( \pm 2\sqrt{2}, 0)\).

Step 4: Determine the Vertices

The vertices of the hyperbola are at \(( \pm a, 0) \). Since \( a = 2 \), the vertices are at \(( -2, 0) \) and \( (2, 0) \).

Step 5: Write the Equations of the Asymptotes

The slopes of the asymptotes for this hyperbola are \( \pm \frac{b}{a} \). Since \( a = 2 \) and \( b = 2 \), the slopes are \( \pm 1 \). Therefore, the equations of the asymptotes are \( y = x \) and \( y = -x \).

Step 6: Graph the Hyperbola

To sketch the hyperbola, plot the vertices \((-2, 0)\) and \((2, 0)\), and draw the asymptotes \(y=x\) and \(y=-x\), which intersect at the center (0,0). The hyperbola will approach but never intersect the asymptotes.

Key Concepts

Asymptotes

Asymptotes are crucial lines that provide guidance on the behavior of a hyperbola. They act as imaginary boundaries that the hyperbola gets infinitely close to but never actually touches or crosses. For the hyperbola given by the equation, \[ x^2 - y^2 = 4 \] after writing it in standard form \[ \frac{x^2}{4} - \frac{y^2}{4} = 1, \] you can deduce the equations for the asymptotes. Since the asymptotes of a hyperbola of this type have slopes determined by the formula \( \pm \frac{b}{a} \), where both \( a \) and \( b \) are equal to 2, the slopes are \( \pm 1 \). Therefore, the asymptote equations are

• \( y = x \)
• \( y = -x \)

Understanding asymptotes helps predict the hyperbola's direction and gives insight into its overall shape. They intersect at the hyperbola's center, which, in this case, is the origin (0,0).

Vertices

Vertices are key points that define the stretch and direction of a hyperbola. For a hyperbola centered on the origin, like \( x^2 - y^2 = 4 \), the vertices are located directly along the transverse axis. In standard hyperbolic equations of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices will be at \( (\pm a, 0) \) for horizontally oriented hyperbolas. Here, since \( a = 2 \), the vertices are at

• \((-2, 0)\)
• \((2, 0)\)

These vertices are pivotal in sketching the hyperbola accurately, as they represent its widest points along the horizontal axis.

Foci

Foci, or focal points, are special points inside each branch of the hyperbola. They are critical in defining the shape of a hyperbola. Unlike ellipses, hyperbolas have foci located outside the vertices. For our hyperbola \[ x^2 - y^2 = 4, \] the foci are calculated using the formula \( c = \sqrt{a^2 + b^2} \), where both \( a^2 \) and \( b^2 \) equal 4. Thus, \( c = \sqrt{8} = 2\sqrt{2} \). Therefore, the coordinates of the foci are:

• \( (2\sqrt{2}, 0) \)
• \( (-2\sqrt{2}, 0) \)

These points are situated along the same axis as the vertices and help describe the hyperbola's steepness and direction.

Equations

The equation of a hyperbola is the foundation for understanding its geometry. Starting with \[ x^2 - y^2 = 4, \] it was rearranged to fit the standard hyperbola form \[ \frac{x^2}{4} - \frac{y^2}{4} = 1. \] This rearrangement confirms the hyperbola's center is at (0,0) and that it has a horizontal transverse axis. This standard form makes it possible to identify elements such as

• \( a^2 = 4 \), implying \( a = 2 \)
• \( b^2 = 4 \), implying \( b = 2 \)

With these values, you can calculate further characteristics like the foci and asymptotes. The equation is essential, as it holds all necessary information for analyzing the hyperbola's properties and for graphing it successfully.