Problem 15 - Precalculus: Mathematics for Calculus

Question

An equation of a hyperbola is given. (a) Find the vertices, foci, and asymptotes of the hyperbola. (b) Determine the length of the transverse axis. (c) Sketch a graph of the hyperbola. $$x^{2}-y^{2}=1$$

Step-by-Step Solution

Answer

Vertices: \((\pm1, 0)\), Foci: \((\pm\sqrt{2}, 0)\), Asymptotes: \(y = \pm x\), Transverse axis: 2.

Step 1: Identify the Standard Form of the Hyperbola

The given equation is \( x^2 - y^2 = 1 \). This can be rewritten as \( \frac{x^2}{1} - \frac{y^2}{1} = 1 \). This fits the standard form of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) where \( a^2 = 1 \) and \( b^2 = 1 \).

Step 2: Determine the Vertices

For the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at \( (\pm a, 0) \). Since \( a^2 = 1 \), \( a = 1 \). Therefore, the vertices are \( (\pm1, 0) \).

Step 3: Calculate the Foci

The distance to each focus from the center is given by \( c = \sqrt{a^2 + b^2} \). Here, \( a = 1 \) and \( b = 1 \), so \( c = \sqrt{1 + 1} = \sqrt{2} \). Thus, the foci are \((\pm \sqrt{2}, 0)\).

Step 4: Find the Asymptotes

The asymptotes for the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by the lines \( y = \pm \frac{b}{a}x \). Here \( a = 1 \) and \( b = 1 \), so the asymptotes are \( y = \pm x \).

Step 5: Determine the Length of the Transverse Axis

The transverse axis of the hyperbola is the line segment that passes through the vertices. Its length is \( 2a \). Since \( a = 1 \), the length of the transverse axis is \( 2 \times 1 = 2 \).

Step 6: Sketch the Graph

To sketch the hyperbola, plot the vertices at \( (1, 0) \) and \( (-1, 0) \). Draw the asymptotes \( y = x \) and \( y = -x \) as dashed lines. The hyperbola will open to the left and right along the x-axis, approaching these asymptotes. The foci \( (\sqrt{2}, 0) \) and \( (-\sqrt{2}, 0) \) lie inside the arms of the hyperbola.

Key Concepts

Vertices of Hyperbola

In the study of hyperbolas, vertices are two significant points that indicate where the hyperbola intersects its transverse axis. For a hyperbola represented by the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are located at \((\pm a, 0)\). This means if you look at our specific hyperbola equation, \( x^2 - y^2 = 1 \), we can clearly identify that it matches the standard form and has \( a^2 = 1 \).
From this, we conclude that \( a = 1 \), leading us to the vertices at \((1, 0)\) and \((-1, 0)\). These vertices give us a valuable clue about how the hyperbola opens, in this case, along the x-axis.

Foci of Hyperbola

The foci of a hyperbola play an important role in defining its unique shape. They are located inside the arms of the hyperbola and are used to derive the equation itself. When given the typical form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are determined using the formula \( c = \sqrt{a^2 + b^2} \).
If we analyze the equation \( x^2 - y^2 = 1 \), it's clear that \( a = 1 \) and \( b = 1 \), suggesting \( c = \sqrt{1 + 1} = \sqrt{2} \). To find the exact coordinates, simply place \( c \) along the transverse axis: \((\pm \sqrt{2}, 0)\). Recognizing these points can aid in sketching the curve accurately and giving us an insight into how the hyperbola stretches out.

Asymptotes of Hyperbola

Asymptotes are straight lines that the hyperbola approaches but never actually meets. They provide a guiding framework around which the hyperbola is shaped and can help visually when sketching. The standard equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) tells us that the asymptotes will be of the form \( y = \pm \frac{b}{a}x \).
For \( x^2 - y^2 = 1 \), since \( b = 1 \) and \( a = 1 \), the asymptotes simplify to \( y = \pm x \).
This means the hyperbola will open asymptotically approaching these lines, with its paths curving closer and closer but never intersecting \( y = x \) and \( y = -x \). Understanding these asymptotes helps delineate the complete outer boundaries and behavior of the hyperbola, giving a clearer picture of its configuration.