Problem 8 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

Vertices at \((-\sqrt{2}, 0)\), \((\sqrt{2}, 0)\); Foci at \((-\sqrt{3}, 0)\), \((\sqrt{3}, 0)\); Asymptotes \(y = \pm \frac{1}{\sqrt{2}}x\).

Step 1: Identify the Standard Form

The given equation \(\frac{x^2}{2} - y^2 = 1\) is already in the form of a hyperbola. The standard form of a hyperbola with a horizontal transverse axis is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). In this case, \(a^2 = 2\) and \(b^2 = 1\), so \(a = \sqrt{2}\) and \(b = 1\).

Step 2: Determine the Vertices

For a hyperbola with a horizontal transverse axis, the vertices are located at \((\pm a, 0)\). Using \(a = \sqrt{2}\), the vertices are \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\).

Step 3: Determine the Foci

The distance from the center to each focus is \(c\), where \(c^2 = a^2 + b^2\). Substitute \(a^2 = 2\) and \(b^2 = 1\) to find \(c^2 = 3\), so \(c = \sqrt{3}\). Thus, the foci are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).

Step 4: Determine the Asymptotes

The asymptotes of a hyperbola in this form are given by the equations \(y = \pm \frac{b}{a}x\). Since \(a = \sqrt{2}\) and \(b = 1\), the equations are \(y = \pm \frac{1}{\sqrt{2}}x\).

Step 5: Sketch the Graph

To sketch the graph, plot the vertices \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\) and the foci \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\). Draw the asymptotes through the center at \((0,0)\) with slopes \(\pm \frac{1}{\sqrt{2}}\). The hyperbola opens horizontally and approaches the asymptotes.

Key Concepts

Vertices of a Hyperbola

In a hyperbola, the vertices are crucial points that provide vital information about its shape and orientation. For a hyperbola with a horizontal transverse axis, like the one given by the equation \(\frac{x^2}{2} - y^2 = 1\), the vertices are located along the x-axis. Here, the standard form equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) helps us determine the value of \(a\).

• In our example, \(a^2\) is 2, thus \(a = \sqrt{2}\).
• The vertices are located at \((\pm a, 0)\).

This means you will find the vertices at \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\). These points mark the closest points of each branch of the hyperbola to the center.
Remember, the distance \(\sqrt{2}\) measures how far each vertex is from the center at (0,0). This placement dictates the breadth of the hyperbola along its transverse axis.

Foci of a Hyperbola

Foci are special points inside each branch of a hyperbola that define its shape. The foci are always further away from the center than the vertices.
In our hyperbola's equation \(\frac{x^2}{2} - y^2 = 1\), we find the distance \(c\) from the center to the foci using the formula \(c^2 = a^2 + b^2\).

• Given \(a^2 = 2\) and \(b^2 = 1\), then \(c^2 = 2 + 1 = 3\), thus \(c = \sqrt{3}\).
• The coordinates of the foci are \((\pm c, 0)\).

This results in foci at \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).
Their position inside each branch means that as any point on the hyperbola moves, the sum of distances from these foci remains constant. This property creates the signature two-beam shape of a hyperbola.

Asymptotes of a Hyperbola

Asymptotes are invisible lines that the branches of a hyperbola approach as they extend to infinity. They give us an idea of the hyperbola's spread and direction.
In our equation \(\frac{x^2}{2} - y^2 = 1\), the asymptotes are described by the equations \(y = \pm \frac{b}{a}x\).

• With \(a = \sqrt{2}\) and \(b = 1\), the slopes of the asymptotes are \(\pm \frac{1}{\sqrt{2}}\).

This means the asymptotes are the lines \(y = \frac{1}{\sqrt{2}}x\) and \(y = -\frac{1}{\sqrt{2}}x\).
They intersect at the center of the hyperbola, \((0,0)\), and the branches of the hyperbola will align along these paths as they go out towards infinity. The trend of the branches becoming closer to the asymptotes demonstrates the highly open shape of hyperbolas compared to ellipses.