Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$x^{2}-2 y^{2}=8$$

Step-by-Step Solution

Answer

The vertices are \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\), foci are \((2\sqrt{3}, 0)\) and \((-2\sqrt{3}, 0)\), and the asymptotes are \(y = \pm \frac{\sqrt{2}}{2}x\).

Step 1: Write the standard form of the hyperbola equation

The given equation is \(x^2 - 2y^2 = 8\). To convert it into its standard form, divide the whole equation by \(8\):\[ \frac{x^2}{8} - \frac{2y^2}{8} = 1 \]Thus, the equation becomes:\[ \frac{x^2}{8} - \frac{y^2}{4} = 1 \]

Step 2: Compare to the standard hyperbola form

The standard form of a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Comparing this with \(\frac{x^2}{8} - \frac{y^2}{4} = 1\), we identify \(a^2 = 8\) and \(b^2 = 4\). Thus, \(a = \sqrt{8}\) and \(b = 2\).

Step 3: Determine the vertices

The vertices of the hyperbola are \((a, 0)\) and \((-a, 0)\) for a hyperbola that opens along the x-axis. Calculating, we have:\(a = \sqrt{8} = 2\sqrt{2}\)Thus, the vertices are \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\).

Step 4: Calculate the foci

The foci are located at \((c, 0)\) and \((-c, 0)\). To find \(c\), use the relation \(c^2 = a^2 + b^2\):\[ c^2 = 8 + 4 = 12 \]\[ c = \sqrt{12} = 2\sqrt{3} \]So, the foci are \((2\sqrt{3}, 0)\) and \((-2\sqrt{3}, 0)\).

Step 5: Determine the equations of the asymptotes

The equations of the asymptotes for a hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are:\[ y = \pm \frac{b}{a}x \]Substituting the values:\[ y = \pm \frac{2}{\sqrt{8}}x = \pm \frac{1}{\sqrt{2}}x \]\[ y = \pm \frac{\sqrt{2}}{2}x \]

Step 6: Sketch the graph

Draw the hyperbola opening along the x-axis with vertices at \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\). Draw the foci at \((2\sqrt{3}, 0)\) and \((-2\sqrt{3}, 0)\). Draw asymptotes with equations \(y = \frac{\sqrt{2}}{2}x\) and \(y = -\frac{\sqrt{2}}{2}x\). The hyperbola will approach these asymptotes but never touch them.

Key Concepts

Vertices of a Hyperbola

In the world of conic sections, a hyperbola is defined by its distinctive, open curves. Central to understanding these curves are the vertices of the hyperbola. The vertices are important points where the hyperbola intersects its principal axis. For a hyperbola like ours, which has its transverse axis along the x-axis, the vertices are located at

• \((a, 0)\)
• \((-a, 0)\)

To find the vertices from the equation \(x^2 - 2y^2 = 8\), we first rewrite it in its standard form: \(\frac{x^2}{8} - \frac{y^2}{4} = 1\). Here, \(a^2 = 8\), thus \(a = \sqrt{8} = 2\sqrt{2}\). Therefore, the vertices are located at the points \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\), lying symmetrically around the origin along the x-axis.

Foci of a Hyperbola

The foci of a hyperbola are two crucial points that define its shape and orientation. These points are always situated along the transverse axis, beyond the vertices. In the case of hyperbolas, the distances from any point on the hyperbola to each of the foci have a constant difference.

To determine the foci for our hyperbola, we use the relationship \(c^2 = a^2 + b^2\). Given \(a^2 = 8\) and \(b^2 = 4\), we find \(c^2 = 12\), giving us \(c = \sqrt{12} = 2\sqrt{3}\). Therefore, the foci for this hyperbola are located at

• \((2\sqrt{3}, 0)\)
• \((-2\sqrt{3}, 0)\)

These points are even farther out from the center, reinforcing the elongation of the hyperbola along the x-axis.

Asymptotes of a Hyperbola

Asymptotes play a vital role in understanding hyperbolas, as they represent the lines that the hyperbola approaches but never intersects. They're key when graphing a hyperbola because they shape its direction and extent.

For a hyperbola given by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{b}{a}x\). From our exercise, with \(a = \sqrt{8}\) and \(b = 2\), we derive the equations of the asymptotes as follows:

• \(y = \pm \frac{2}{\sqrt{8}}x = \pm \frac{1}{\sqrt{2}}x\)
• \(y = \pm \frac{\sqrt{2}}{2}x\)

These asymptotes function as guidelines the hyperbola hugs but doesn't touch, ever expanding outwards along these slanted lines.

Standard Form of a Hyperbola

The standard form of a hyperbola serves as a foundation for understanding and solving problems related to these conic sections. It's from this form that we derive critical information, such as vertices, foci, and asymptotes.

For hyperbolas with a horizontal transverse axis, the standard equation is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Our given equation \(x^2 - 2y^2 = 8\) can be transformed into this standard form by dividing through by 8, resulting in \(\frac{x^2}{8} - \frac{y^2}{4} = 1\).

By aligning the equation with the standard form, we identify \(a^2 = 8\) and \(b^2 = 4\). This step is primary because it allows us to calculate all other critical points and characteristics of the hyperbola, making understanding and graphing much more intuitive.

Graphing Hyperbolas

Graphing hyperbolas can be an enriching experience because you get to see the blend of mathematical logic and visual art. The shape of a hyperbola is dictated by its vertices, foci, and asymptotes.

To sketch the hyperbola given by \(\frac{x^2}{8} - \frac{y^2}{4} = 1\):

• Start by plotting the center at the origin \((0,0)\).
• Mark the vertices \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\).
• Plot the foci \((2\sqrt{3}, 0)\) and \((-2\sqrt{3}, 0)\).
• Draw the asymptotes \(y = \pm \frac{\sqrt{2}}{2}x\), creating slanting lines through the origin.

Finally, sketch the two branches of the hyperbola opening along the x-axis, approaching these asymptotes. The graph illustrates how the hyperbola grows closer to the asymptotes, never touching them, marking out the classic shape of a hyperbola.