Problem 14 - Algebra 2

Question

Find the foci of each hyperbola. Then draw the graph. $$ \frac{y^{2}}{25}-\frac{x^{2}}{100}=1 $$

Step-by-Step Solution

Answer

The foci of the hyperbola are at (0, -√125) and (0, √125).

Step 1: Identify the standard form of hyperbola

The standard form for a vertical hyperbola is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Here, a^2 = 25 and b^2 = 100. So, a = 5 and b = 10.

Step 2: Find the foci

The distance from the center to the foci is given by \(\sqrt{a^2 + b^2}\). Doing the calculation we get \(\sqrt{25 + 100} = \sqrt{125}\). So the foci are at (0, ±\(\sqrt{125}\)). That gives us two points, (0, -√125) and (0, √125)

Step 3: Graph the hyperbola

Draw a vertical hyperbola at the center point (0,0) with foci at (0, -√125) and (0, √125).

Key Concepts

Standard Form

Understanding the standard form of a hyperbola is an essential starting point for solving related problems. When discussing hyperbolas, the standard form is used to describe its mathematical equation and characteristics. For a vertical hyperbola, the standard form is:

• \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

The "a" value is related to the semi-major axis, while "b" relates to the semi-minor axis. These values help to determine the shape and orientation of the hyperbola. Analyzing the given equation \( \frac{y^2}{25} - \frac{x^2}{100} = 1 \), we compare it to the standard form: here, \(a^2 = 25\) and \(b^2 = 100\). By extracting the square roots, we find \(a = 5\) and \(b = 10\). These numbers define the hyperbola's dimensions, with \(a\) identifying the distance from the center to the vertices along the y-axis and \(b\) representing the equivalent measurement along the x-axis.Don't forget that the hyperbola's eccentricity, a measure of how "stretched" it is, can be calculated using these values. The formula is \( e = \sqrt{1 + \frac{b^2}{a^2}} \). Knowing these basics helps us unravel more complex aspects of the hyperbola, like finding its foci and graphing.

Foci

The foci of a hyperbola are crucial in defining its shape and are always located inside the open ends of the figure. To find the foci, we use the formula:

• \( \sqrt{a^2 + b^2} \)

In our case, \( a^2 = 25 \) and \( b^2 = 100 \). Plugging these values into the formula gives us \( \sqrt{25 + 100} = \sqrt{125} \).For a hyperbola centered at the origin, such as in our example formula, the foci lie on the y-axis, because the hyperbola is vertical. This positions the foci at points (0, \(\sqrt{125}\)) and (0, -\(\sqrt{125}\)). It's important to visualize these foci as they dictate the curvature of the hyperbola and indicate how wide the branches of the hyperbola extend.Remember, as the foci move farther from the center, the hyperbola appears more "stretched." Therefore, the distance between the center and each focus is a key parameter in understanding the hyperbola’s graph and its geometric properties.

Graphing Hyperbolas

Graphing hyperbolas might seem challenging, but it's straightforward with the right approach. The process starts with identifying key features from the standard form equation: center, vertices, and foci. In the equation \( \frac{y^2}{25} - \frac{x^2}{100} = 1 \), we previously identified the center at (0, 0), with the vertices positioned \( a \) units from the center along the y-axis, at points (0, 5) and (0, -5).To graph:

• Sketch the asymptotes, which in a perfectly centered hyperbola, pass through its center. They intersect at angles determined by \(b/a\).
• Draw the vertices and foci on the coordinate plane.
• Using the vertices and asymptotes as guides, sketch the hyperbola's two branches opening upwards and downwards.

Ensure the branches approach but never touch the asymptotes. Keep in mind the distances calculated, such as the distance to the foci \( \sqrt{125} \), help perfect the placement and proportions.Visualizing hyperbolas dynamically enhances understanding of how these mathematical loci operate within coordinate geometry and fosters better comprehension of their equations.