Problem 7 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$y^{2}-\frac{x^{2}}{25}=1$$

Step-by-Step Solution

Answer

Vertices: \((0, 1)\), \((0, -1)\); Foci: \((0, \sqrt{26})\), \((0, -\sqrt{26})\); Asymptotes: \(y = \pm \frac{1}{5} x\).

Step 1: Recognize the Hyperbola Form

The equation \(y^2 - \frac{x^2}{25} = 1\) is in the standard form of a hyperbola: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). This indicates a vertical hyperbola.

Step 2: Identify Parameters

In the equation \(y^2 - \frac{x^2}{25} = 1\), compare it with \(\frac{y^2}{1} - \frac{x^2}{25} = 1\). Here, \(a^2 = 1\) implies \(a = 1\), and \(b^2 = 25\) implies \(b = 5\).

Step 3: Calculate the Vertices

For a vertical hyperbola centered at the origin with equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are at \((0, \pm a)\). So the vertices are \((0, 1)\) and \((0, -1)\).

Step 4: Determine the Foci

To find the foci of the hyperbola, use the formula \(c^2 = a^2 + b^2\). Therefore, \(c^2 = 1 + 25 = 26\), which gives \(c = \sqrt{26}\). The foci are \((0, \pm \sqrt{26})\).

Step 5: Find the Asymptotes

The equations of the asymptotes for a vertical hyperbola are \(y = \pm \frac{a}{b} x\). Here, the asymptotes are \(y = \pm \frac{1}{5}x\).

Step 6: Sketch the Graph

Draw the vertical hyperbola centered at origin with vertices at \((0, 1)\) and \((0, -1)\). The foci are at \((0, \sqrt{26})\) and \((0, -\sqrt{26})\). Draw dashed lines for the asymptotes \(y = \frac{1}{5}x\) and \(y = -\frac{1}{5}x\) to complete the sketch.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are critical points that help define its shape and orientation. In this particular hyperbola, identified by the equation \(y^2 - \frac{x^2}{25} = 1\), we're dealing with a vertical hyperbola since the \(y^2\) term comes first. For vertical hyperbolas in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices lie on the y-axis and are located at \((0, \pm a)\).
To find these vertices, we identify \(a^2\) from the equation, which is \(1\). Solving for \(a\), we find \(a = 1\), hence the vertices are \((0, 1)\) and \((0, -1)\).

• Vertices give you important points to plot initially.
• The distance between them is directly correlated to the size and steepness of the hyperbola.

These vertices help provide a framework to begin sketching or further analyzing anything involving the hyperbola.

Foci of Hyperbola

The foci of a hyperbola are key points located along the axis of symmetry, further away from the center than the vertices. For the hyperbola \(y^2 - \frac{x^2}{25} = 1\), which is vertical, the foci alignment is along the y-axis. For hyperbolas, the foci are found using the formula \(c^2 = a^2 + b^2\).
In this case, \(a^2 = 1\) and \(b^2 = 25\) which leads us to \(c^2 = 1 + 25 = 26\). Solving for \(c\), we get \(c = \sqrt{26}\). Hence, the foci are located at \((0, \pm \sqrt{26})\).

• Foci are farther from the center than the vertices and add depth to the hyperbola's symmetry.
• Distance from the center can affect the "tightness" of the branches.

Having these foci points helps in understanding the degree of "spread" the hyperbola has.

Asymptotes of Hyperbola

Asymptotes are invisible lines that a hyperbola approaches but never touches. They help in defining the broad shape of the hyperbola as they set the directions of its branches. For the vertical hyperbola \(y^2 - \frac{x^2}{25} = 1\), the asymptotes are derived from the slopes represented in \(y = \pm \frac{a}{b} x\).
Here, \(a = 1\) and \(b = 5\), thus the equations become \(y = \pm \frac{1}{5}x\). These asymptotes guide the hyperbola's orientation in space and provide a clear limit on how it behaves as it extends.

• Asymptotes contribute to the hyperbola's end behavior.
• They are essential for sketching, especially when curves extend beyond plotted points.

By drawing these lines, you create a framework within which the hyperbola's curve can be precisely sketched or analyzed.