Problem 9 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The vertices are \((\pm 1, 0)\); foci are \((\pm \sqrt{2}, 0)\); asymptotes are \(y = \pm x\).

Step 1: Identify the Standard Form of a Hyperbola

The given equation is \(x^2 - y^2 = 1\). Compare this with the standard form of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, \(a^2 = 1\) and \(b^2 = 1\), hence, \(a = 1\) and \(b = 1\).

Step 2: Determine the Center

For the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the center of the hyperbola \((h, k)\) is determined from the lack of terms \(x-h\) and \(y-k\). Here, the center is at the origin \((0, 0)\).

Step 3: Find the Vertices

The vertices of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are at \((h \pm a, k)\). With \(a = 1\) and center \((0, 0)\), the vertices are \((\pm 1, 0)\).

Step 4: Find the Foci

The foci of the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are located at \((h \pm c, k)\) where \(c = \sqrt{a^2 + b^2}\). Calculating \(c = \sqrt{1 + 1} = \sqrt{2}\), the foci are at \((\pm \sqrt{2}, 0)\).

Step 5: Determine the Asymptotes

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

Step 6: Sketch the Hyperbola

Sketching involves plotting the center \((0, 0)\), the vertices \((\pm 1, 0)\), and the foci \((\pm \sqrt{2}, 0)\). The asymptotes, \(y = x\) and \(y = -x\), should also be drawn as diagonal lines through the origin forming a 'X'. Sketch the hyperbola approaching these asymptotes.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are crucial points. They provide the tips of the hyperbola's "open ends." In the standard form, \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are positioned at \((h \pm a, k)\). Here, \(h\) and \(k\) represent the center of the hyperbola.

Given our equation \(x^2 - y^2 = 1\), which can be rewritten as \(\frac{x^2}{1} - \frac{y^2}{1} = 1\), we find that both \(a^2\) and \(b^2\) are 1. Thus, \(a = 1\) and the center is at origin \( (0, 0) \).

• This means the vertices are at \((0 \pm 1, 0)\), simplifying to \((1, 0)\) and \((-1, 0)\).

These vertices help define the shape and size of the hyperbola. In our case, they lie along the x-axis.

Foci of Hyperbola

The foci of a hyperbola play a similar role to the vertices, marking points inside each "branch" of the hyperbola, pulling the curve outward. For identifying the foci, you need a number \(c\) determined by the formula \(c = \sqrt{a^2 + b^2}\).

For the equation \(x^2 - y^2 = 1\), we have \(a^2 = 1\) and \(b^2 = 1\). By plugging into the formula, \(c = \sqrt{1 + 1} = \sqrt{2}\). The foci, based on the formula, are situated at \((h \pm c, k)\).

• Thus, the foci positions are \((0 \pm \sqrt{2}, 0)\), simplifying to \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\).

These points might not be visible in simple graphs, but they are vital when understanding the hyperbola's properties and its stretched shape along the x-axis.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never actually touches. They provide a "frame" for the hyperbola, indicating the direction in which the branches of the hyperbola extend.

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

• Using our values, \(a = 1\) and \(b = 1\), the equations simplify to \(y = \pm x\).

These lines pass through the center of the hyperbola at the origin \( (0, 0) \) and extend diagonally. When sketching, ensure that these asymptotes form a cross-like "X" to guide the shape of the hyperbola branches.