Problem 11 - Algebra and Trigonometry

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); Asymptotes: y = ±(1/5)x.

Step 1: Recognize the Type of Conic

From the equation \(y^2 - \frac{x^2}{25} = 1\), we identify it as a hyperbola because it is in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). This represents a vertical hyperbola.

Step 2: Identify a and b

The standard form of a vertical hyperbola is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Comparing, we get \(a^2 = 1\), so \(a = 1\), and \(b^2 = 25\), so \(b = 5\).

Step 3: Find the Vertices

For a vertical hyperbola centered at the origin \((0,0)\), the vertices are located at \((0, \pm a)\). Therefore, the vertices are \((0, 1)\) and \((0, -1)\).

Step 4: Calculate the Foci

The formula for the foci of a vertical hyperbola is \((0, \pm c)\), where \(c = \sqrt{a^2 + b^2}\). Calculate \(c = \sqrt{1 + 25} = \sqrt{26}\). Hence, the foci are \((0, \pm \sqrt{26})\).

Step 5: Determine the Asymptotes

For vertical hyperbolas, the equations of the asymptotes are \(y = \pm \frac{a}{b}x\). Given \(a = 1\) and \(b = 5\), the asymptotes are \(y = \pm \frac{1}{5}x\).

Step 6: Sketch the Graph

Plot the vertices at \((0, 1)\) and \((0, -1)\), and mark the foci near \((0, \pm \sqrt{26})\). Draw the asymptotes as lines passing through the origin with slopes \(\pm \frac{1}{5}\). Sketch the hyperbola opening upwards and downwards, approaching the asymptotes but never crossing them.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are key points where the curve intersects its principal axis. For a vertical hyperbola centered at the origin, the equation is typically written as \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). The vertices are located along the y-axis at the coordinates \((0, \pm a)\).

• In our example, \(a^2 = 1\), giving \(a = 1\).
• This means the vertices are at \((0, 1)\) and \((0, -1)\).

Knowing the vertices helps us understand the shape and orientation of the hyperbola. They are crucial for sketching its graph accurately.

Foci of Hyperbola

The foci of a hyperbola are two special points that are used to define and construct the hyperbola. These points lie along the same axis as the vertices. For a hyperbola centered at the origin with its principal axis along the y-axis, the foci are found using the formula \((0, \pm c)\), where \(c = \sqrt{a^2 + b^2}\).

• With \(a = 1\) and \(b = 5\), we calculate \(c = \sqrt{1 + 25} = \sqrt{26}\).
• The foci are at \((0, \pm \sqrt{26})\).

The foci give insight into the curvature of the hyperbola, indicating how "wide" or "narrow" it opens.

Asymptotes of Hyperbola

The asymptotes of a hyperbola are straight lines that the curve approaches as it extends to infinity. They help define the ultimate direction and opening of the hyperbola, though the hyperbola never actually intersects these lines. For a vertical hyperbola, the asymptotes have equations \(y = \pm \frac{a}{b}x\).

• In our case, \(a = 1\) and \(b = 5\), giving asymptotes \(y = \pm \frac{1}{5}x\).

These asymptotic lines are essential in accurately sketching the hyperbola, as they provide a guideline within which the branches of the hyperbola will operate.

Conic Sections

Hyperbolas are one among the set of conic sections, geometric shapes that are created by cutting a cone with a plane. The principal conic sections include:

• Circle
• Ellipse
• Parabola
• Hyperbola

Hyperbolas appear when the plane cuts through both the top and bottom halves of the cone. Unlike circles and ellipses which are closed curves, hyperbolas are open and consist of two branches extending to infinity, reflecting the diverse nature of conic sections. Understanding these shapes enriches the study of geometry and many real-world phenomena.

Graphing Hyperbolas

Graphing a hyperbola involves plotting and drawing the curve based on its key features: vertices, foci, and asymptotes. The equation \(y^2 - \frac{x^2}{25} = 1\) describes a vertical hyperbola, which means it will open upwards and downwards.

• First, plot the vertices, \((0, 1)\) and \((0, -1)\), on the coordinate plane.
• Next, indicate the foci positions, \((0, \pm \sqrt{26})\), as they help define the curve's stretch.
• Draw the asymptotes, \(y = \pm \frac{1}{5}x\), through the origin to guide the shape.
• Finally, sketch the hyperbola, making sure it hugs the asymptotes without crossing them.

By understanding these components, graphing hyperbolas becomes systematic, allowing one to visualize complex mathematical concepts with clarity.