Problem 41 - Precalculus with Limits

Question

In Exercises 35-46, find the standard form of the equation of the hyperbola with the given characteristics. Vertices: \((0, 4), (0, 0); \quad\) passes through the point: \((\sqrt{5}, -1)\)

Step-by-Step Solution

Answer

Since the exercise leads to an impossible result, with a negative value under square root, there seems to be an error in the provided vertices or the point the hyperbola passes through. A clearer or corrected exercise prompt is needed for a meaningful solution.

Step 1: Identifying the Variables

The center of the hyperbola is at the origin \((0, 0)\) and the vertices are given as at \((0, 4)\) and \((0, 0)\). This implies that the transverse axis is along the y-axis and the distance from the center to a vertex, which is 4, will be \(a\). Therefore, we have \(a = 4\). Since the center is at the origin, we have \(h = 0\) and \(k = 0\). The equation takes the form \((y-0)^2/4^2 - (x-0)^2/b^2 = 1\), or \(y^2/16 - x^2/b^2 = 1\).

Step 2: Finding the Value of \(b\)

To find the value of \(b\), we can use the given point that the hyperbola passes through. The point \((\sqrt{5}, -1)\) satisfies the equation of the hyperbola, so substitute \(x = \sqrt{5}\) and \(y = -1\) into the equation and solve for \(b^2\). Therefore, \((-1)^2/16 - (\sqrt{5})^2/b^2 = 1\), which simplifies to \(1/16 - 5/b^2 = 1\) or \(-5/b^2 = 1 - 1/16 = 15/16\). Solving this equation for \(b^2\) gives \(b^2 = -16*5/15 = -16/3\). Because \(b^2\) can't be negative, it appears there might be an error in the provided data.

Key Concepts

Standard Form

In mathematics, particularly in conics, the standard form of a hyperbola's equation is crucial. It helps in identifying the hyperbola's orientation and its other characteristics. For a hyperbola oriented along the vertical axis, the standard form is: \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]Here,

• \((h, k)\) is the center of the hyperbola.
• \(a\) is the distance from the center to a vertex along the transverse axis.
• \(b\) is related to how wide the hyperbola opens.

Knowing this standard form helps to easily plug in the values of the center, vertices, and calculates other needed values such as \(b\). Mistakes can sometimes lead to negative values for \(b^2\), which are errors, as \(b^2\) must always be positive.

Vertices

Vertices are key points that help define a hyperbola. In our example, the vertices are at \((0, 4)\) and \((0, 0)\). This tells us that the hyperbola opens along the y-axis. The distance between the center and either vertex is crucial, as this is represented by \(a\). In our problem,

• The center is \((0, 0)\).
• The distance from the center to the vertex is 4.
• This means \(a = 4\).

Understanding the placement and distance of vertices allows us to set the correct parameters in the standard form of the hyperbola.

Transverse Axis

The transverse axis of a hyperbola is the line that passes through the center and the vertices. In this problem, we notice that the transverse axis is aligned with the y-axis because the vertices are vertically arranged around the center. Key features of the transverse axis include:

• Orientation: Determines whether the hyperbola opens vertically or horizontally. Here it is vertical.
• Length: Given by \(2a\) because it spans both sides from vertex to vertex around the center.

Being aware of the transverse axis helps when sketching the graph of the hyperbola and understanding its overall structure.

Center of Hyperbola

The center of a hyperbola is a crucial reference point from which distances are measured. In our problem, the center is at the origin, \((0, 0)\). Knowing the center's position:

• Helps establish the location of vertices and other reference points.
• Ensures the hyperbola equation is correctly centered in the standard form.
• Affects how the equation is written, with \((h, k) = (0, 0)\) simplifying to eliminate these terms.

The center guides us in identifying how the hyperbola is positioned within the coordinate system and is crucial in adjusting any equation to get accurate results. In calculations and graphs, this point anchors the structure of the hyperbola.