Problem 22 - 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. $$ x^{2}-2 y^{2}=2 $$

Step-by-Step Solution

Answer

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

Step 1: Standard Form of Hyperbola Equation

First, rewrite the given hyperbola equation, \(x^2 - 2y^2 = 2\), in its standard form. Divide every term by 2 to isolate the equation to \(\frac{x^2}{2} - \frac{y^2}{1} = 1\). This shows the hyperbola is centered at the origin \((0,0)\) with the transverse axis along the x-axis.

Step 2: Identify Vertices

For the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices are \((\pm a, 0)\). Here, \(a^2 = 2\) so \(a = \sqrt{2}\). Thus, the vertices are \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\).

Step 3: Identify 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, \(a^2 = 2\) and \(b^2 = 1\), so \(c^2 = 3\) which yields \(c = \sqrt{3}\). Thus, the foci are \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).

Step 4: Equations of Asymptotes

The asymptotes of a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are \(y = \pm \frac{b}{a}x\). Substituting \(b = 1\) and \(a = \sqrt{2}\), the asymptote equations are \(y = \pm \frac{1}{\sqrt{2}}x\) or simplified, \(y = \pm \frac{\sqrt{2}}{2}x\).

Step 5: Sketch the Hyperbola

Draw the hyperbola with its center at the origin, the vertices at \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0)\), the foci at \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\), and the asymptotes \(y = \pm \frac{\sqrt{2}}{2}x\). The hyperbola opens horizontally.

Key Concepts

Hyperbola Equation

A hyperbola is a type of conic section that can be represented by a specific kind of equation. The standard form of a hyperbola equation is given by: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] This form indicates that the hyperbola opens horizontally, affecting how it is graphed and analyzed. In the example given, the equation \( x^2 - 2y^2 = 2 \) is first transformed into standard form by dividing all terms by 2, resulting in: \[ \frac{x^2}{2} - \frac{y^2}{1} = 1 \] This transformation helps to clearly identify the values of \(a^2\) and \(b^2\). Identifying these parameters is crucial because they allow us to find other important characteristics of the hyperbola such as vertices, foci, and asymptotes.

Vertices of Hyperbola

The vertices of a hyperbola are critical points where the hyperbola intersects its transverse axis. For a hyperbola in the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are located at: \(( \pm a, 0 )\) if the hyperbola is horizontal, or \((0, \pm a)\) if it's vertical. In our specific exercise, \(a^2 = 2\), giving \(a = \sqrt{2}\). Consequently, the vertices of this hyperbola are at:

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

The vertices not only indicate the orientation and width of the hyperbola but are also primary reference points for sketching the graph.

Foci of Hyperbola

The foci of a hyperbola locate the points from which distances are measured to determine points on the curve. For the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are positioned at: \(( \pm c, 0 )\) for a horizontal hyperbola. The value of \(c\) is defined by the relationship: \[ c^2 = a^2 + b^2 \] Plugging the known values \(a^2 = 2\) and \(b^2 = 1\) yields: \[ c^2 = 3 \Rightarrow c = \sqrt{3} \] Thus, the foci are calculated as:

• \( (\sqrt{3}, 0) \)
• \( (-\sqrt{3}, 0) \)

Understanding the location of the foci helps in explaining the unique properties of a hyperbola, such as its eccentricity.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never intersects. These lines provide a guide for sketching the hyperbola's shape. For the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equations of the asymptotes are given by: \[ y = \pm \frac{b}{a}x \] In our hyperbola, \(b = 1\) and \(a = \sqrt{2}\), so the asymptotes are computed as: \[ y = \pm \frac{1}{\sqrt{2}}x \] Which simplifies to: \[ y = \pm \frac{\sqrt{2}}{2}x \] These asymptotes effectively help shape the hyperbola's direction, showing that the hyperbola spreads wider as it moves away from its center, yet always nearing these lines without actually touching them.