Problem 11 - Precalculus Mathematics 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, √26), (0, -√26). Asymptotes: y = ±(1/5)x.

Step 1: Identify Standard Form

The given equation is in the form \( y^2 - \frac{x^2}{25} = 1 \). This matches the standard form of a hyperbola centered at the origin: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), where \( a^2 = 1 \) and \( b^2 = 25 \).

Step 2: Find Vertices

For a hyperbola in this form, the vertices are located at \( (0, \pm a) \). Here, \( a^2 = 1 \), so \( a = 1 \). Thus, the vertices are \( (0, 1) \) and \( (0, -1) \).

Step 3: Calculate Foci

The distance to each focus \( c \) is found using \( c^2 = a^2 + b^2 \). Thus, \( c^2 = 1 + 25 = 26 \), so \( c = \sqrt{26} \). The foci are located at \( (0, \pm \sqrt{26}) \).

Step 4: Determine Asymptotes

For this form of hyperbola, the asymptotes are lines given by the equation \( y = \pm \frac{a}{b}x \). Here, \( a = 1 \) and \( b = 5 \), so the equations of the asymptotes are \( y = \pm \frac{1}{5}x \).

Step 5: Sketch the Hyperbola

Draw the coordinate axes. Plot the vertices at \( (0, 1) \) and \( (0, -1) \). Plot the foci at \( (0, \sqrt{26}) \) and \( (0, -\sqrt{26}) \). Draw the asymptotes: the lines \( y = \frac{1}{5}x \) and \( y = -\frac{1}{5}x \). Finally, sketch the hyperbola opening up and down, approaching but never touching the asymptotes.

Key Concepts

Vertices of a Hyperbola

Understanding the vertices of a hyperbola is essential in sketching its graph and grasping its overall shape. In a hyperbola centered at the origin, the vertices are positioned along the axis corresponding to the positive term in the equation. For the given hyperbola equation \( y^2 - \frac{x^2}{25} = 1 \), the vertices are aligned along the y-axis as the positive term is \( y^2 \).

• Standard for this form: The formula \( \left(0, \pm a\right) \) gives the position of the vertices.
• Calculation: Here, \( a^2 = 1 \), which gives \( a = 1 \).
• Resulting vertices: These are \( (0, 1) \) and \( (0, -1) \).

In the graph, the vertices mark key points where the hyperbola makes its sharpest turns. These points are at an equal distance of \( a \) from the center, forming a vertical line segment in this case. This characteristic helps to set the shape and opening direction of the hyperbola.

Foci of a Hyperbola

The foci of a hyperbola are pivotal in defining its geometry. They are not just arbitrary points but are crucial in determining the shape and spread of the hyperbola's branches. The foci lie on the same axis as the vertices but outside them, further from the center.

• Formula for calculation: The distance to each focus \( c \) is derived from \( c^2 = a^2 + b^2 \).
• Given values: For our hyperbola, \( a^2 = 1 \) and \( b^2 = 25 \).
• Calculating \( c \): This results in \( c^2 = 26 \), so \( c = \sqrt{26} \).
• Placing the foci: Positioned at \( (0, \pm \sqrt{26}) \).

These foci determine how "stretched" the hyperbola is. In particular, the closer together they are, the narrower the hyperbola. In this example, the relatively large value of \( \sqrt{26} \) compared to \( 1 \) suggests wider opening branches.

Asymptotes of a Hyperbola

Asymptotes serve as invisible guidelines toward which the branches of a hyperbola extend, without ever actually meeting them. They are linear equations that capture the diagonal trajectory of the hyperbola as it extends infinitely.

• Equation form: For our hyperbola, use the line equations \( y = \pm \frac{a}{b}x \).
• Specific calculation: With \( a = 1 \) and \( b = 5 \), the asymptotes are \( y = \pm \frac{1}{5}x \).
• Graph utilization: When graphing, these lines provide the avenues of approach for the hyperbola's branches.

These asymptotes essentially frame the hyperbola, outlining its limits in the plane. The branches of the hyperbola grow closer to these lines as they move away from the center but never actually touch them. This behavior is crucial for both understanding and sketching hyperbolas accurately.