Question

Sketch the graph of the following hyperbolas. Specify the coordinates of the vertices and foci, and find the equations of the asymptotes. Use a graphing utility to check your work. $$\frac{x^{2}}{3}-\frac{y^{2}}{5}=1$$

Short answer

Question: Sketch the graph of the hyperbola given by the equation $\frac{x^2}{3}-\frac{y^2}{5}=1$, and find its vertices, foci, and the equations of its asymptotes. Answer: The hyperbola has vertices at $(\pm\sqrt{3}, 0)$, foci at $(\pm\sqrt{8}, 0)$, and its asymptotes have the equations $y = \pm\frac{\sqrt{5}}{\sqrt{3}}x$. The graph is a horizontal hyperbola with its center at the origin.

Step-by-step solution

Identify the type of hyperbola

The general equation for a hyperbola can be written as: $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} =\pm 1$$ Comparing this with the given equation: $$\frac{x^{2}}{3}-\frac{y^{2}}{5}=1$$ We see that the center of the hyperbola is \((h,k)=(0,0)\). Since the x-term comes first with a positive sign, this is a horizontal hyperbola.

Find the coordinates of the vertices

For a horizontal hyperbola, the vertices are located at \((\pm a,0)\), where \(a^2=3\). Therefore, \(a=\sqrt{3}\). The coordinates of the vertices are \((\pm\sqrt{3},0)\).

Find the coordinates of the foci

For a horizontal hyperbola, the foci are located at \((\pm c,0)\), where \(c^2 = a^2 + b^2\). In this case, \(a^2=3\) and \(b^2=5\), so \(c^2 = 8\). Therefore, \(c=\sqrt{8}\). The coordinates of the foci are \((\pm\sqrt{8},0)\).

Find the equations of the asymptotes

For a horizontal hyperbola, the asymptotes have the equation: $$y = \pm\frac{b}{a}(x-h) + k$$ Since the center of the hyperbola is at the origin \((0,0)\), the equations of the asymptotes for this hyperbola are: $$y = \pm\frac{\sqrt{5}}{\sqrt{3}}x$$

Sketch the graph of the hyperbola

To sketch the graph, we first plot the center, vertices, and foci on the coordinate plane. Then, we sketch the asymptotes using their equations. Finally, we draw the graph of the hyperbola using the asymptotes as a guide. Finally, use a graphing utility to check the sketch of the hyperbola. The graph should match the sketched graph of the hyperbola, with the correct vertices, foci, and asymptotes shown.

Key concepts

Vertices

In a hyperbola, the vertices are significant points that help define its shape. They are the points where the hyperbola intersects its transverse axis. For a horizontal hyperbola like the one given here, the vertices are located at

• \((+a, 0)\)
• \((-a, 0)\)

In this equation,\(\frac{x^2}{3} - \frac{y^2}{5} = 1\), we identify \(a^2 = 3\), which leads us to \(a = \sqrt{3}\). Thus, the vertices of the hyperbola are located at the coordinates

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

To plot these points on a graph, simply move \(\sqrt{3}\) units to the right and left along the x-axis from the center point Understanding the location of the vertices is crucial for graphing the hyperbola accurately.

Foci

The foci of a hyperbola are internal reference points that help define the curve's width and orientation. They lie along the transverse axis, just like the vertices but are placed further away from the center. For a hyperbola of the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the foci coordinates are given by \((\pm c, 0)\) for a horizontal hyperbola. Here, the formula \(c^2 = a^2 + b^2\) helps us to calculate the distance to the foci. Given \(a^2 = 3\) and \(b^2 = 5\), we compute

• \(c^2 = 3 + 5\)
• \(c = \sqrt{8}\)

Therefore, the foci are located at

• \((\sqrt{8}, 0)\)
• \((-\sqrt{8}, 0)\)

Understanding these coordinates helps in sketching an accurate hyperbola and provides clarity on how it 'opens' on a graph.

Equations of Asymptotes

Asymptotes are imaginary lines that a hyperbola approaches but never touches. They are critical in determining the hyperbola's shape and direction.For a horizontal hyperbola, the equations of the asymptotes take the form:\[ y = \pm \frac{b}{a}(x-h) + k \]Since our hyperbola is centered at \((0, 0)\) (the origin), the equation simplifies to\[ y = \pm \frac{\sqrt{5}}{\sqrt{3}}x \]The slope of these lines is derived from the values of \(b\) and \(a\) (i.e., \(\frac{b}{a}\)). This gives us two lines with slopes \(\pm \frac{\sqrt{5}}{\sqrt{3}}\), meaning one ascends and the other descends symmetrically from the origin.These asymptotes act as boundary lines for the hyperbola, guiding its path and indicating how it stretches infinitely outward.

Graphing Utility

A graphing utility is a valuable tool for visualizing complex mathematical objects like hyperbolas. By using a graphing calculator or software, you can quickly generate a precise graph that confirms your manual calculations.To use a graphing utility, enter the hyperbola's equation \(\frac{x^2}{3} - \frac{y^2}{5} = 1\). The tool will display the hyperbola, highlighting its features such as vertices, foci, and asymptotes.Here are some tips when using a graphing utility:

• Check if the window shows all relevant points like vertices and foci.
• Make sure to adjust the viewing window for a clear view of the hyperbola's entire curve.

The utility provides a quick verification method, ensuring your graph aligns with computed values. This helps catch mistakes and strengthens your comprehension of the hyperbola's geometric structure.