Problem 42 - Precalculus with Limits

Question

In Exercises 35-46, find the standard form of the equation of the hyperbola with the given characteristics. Vertices: \((1, 2), (1, -2); \quad\) passes through the point: \((0, \sqrt{5})\)

Step-by-Step Solution

Answer

The standard form of the equation of the hyperbola is \(\frac{y^2}{4} - \frac{(x-1)^2}{5} = 1\).

Step 1: Identify the center and vertices

First, we identify the center of the hyperbola by averaging the x-coordinates and y-coordinates of the given vertices. The average of the x-coordinates \(1\) and \(1\) is \(1\). The average of the y-coordinates \(2\) and \(-2\) is \(0\). Therefore, the center of the hyperbola is at the point \((1,0)\). The distance from the center to each vertex is equal to \(a\), which is \(2\). So, \(a = 2\). The hyperbola is vertical because the vertices are aligned vertically.

Step 2: Identify the point (x,y)

The hyperbola passes through the point \((0, \sqrt{5})\). Substitute these values into the equation \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), where \(h\) and \(k\) are the coordinates of the center and \(a\) and \(b\) are the distances from the center to the vertices and foci, respectively. Plugging in the values, we get \(\frac{( \sqrt{5})^2}{2^2} - \frac{(0-1)^2}{b^2} = 1\). Solving for \(b\), we find that \(b = \sqrt{5}\).

Step 3: Write the equation

Substitute the values of \(h\), \(k\), \(a\), and \(b\) into the standard form equation for a hyperbola, \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\). This gives us \(\frac{(y-0)^2}{2^2} - \frac{(x-1)^2}{(\sqrt{5})^2} = 1\), which simplifies to \(\frac{y^2}{4} - \frac{(x-1)^2}{5} = 1\).

Key Concepts

Standard Form of Hyperbola

A hyperbola is a type of conic section that is formed by the intersection of a plane with two cones that face each other. The hyperbola has two branches that mirror each other. To understand the shape and orientation of a hyperbola, it's important to know its standard form equation. There are two standard forms based on the orientation of the hyperbola: horizontal and vertical. For a vertical hyperbola, like in our exercise, the equation takes the form:

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

The terms \((h, k)\) are the coordinates of the center of the hyperbola. In the above equation, the hyperbola opens vertically since the \((y-k)^2\) term comes first. The values \(a\) and \(b\) specify the distances from the center to the vertices and the foci, respectively. By substituting specific values into these equations, one can derive the equation of the hyperbola. Understanding this structure allows you to interpret key properties of the hyperbola, such as the direction it opens and its proportions.

Vertices of Hyperbola

The vertices of a hyperbola are crucial points that determine its basic shape and orientation. They are the nearest points on each branch of the hyperbola to its center. In this exercise, the vertices are given as \((1, 2)\) and \((1, -2)\). The x-coordinates are the same, meaning the vertices are vertically aligned, confirming a vertical hyperbola.
To find the exact position of these vertices from the center \((1, 0)\), we calculate the distance along the y-axis. This distance is denoted as \(a\), which is 2 in our case because each vertex is 2 units away from the center.
Knowing the vertices not only helps to sketch the hyperbola but also plays a vital role in the formulation of its equation. It helps to identify the values of \(a\), which are necessary to form the standard equation of the hyperbola.

Hyperbola Equation

The process of finding the equation of the hyperbola involves substituting the known values into its standard form. Initially, we identify the center \((h, k)\), which is \((1, 0)\) in this exercise.
Since the hyperbola is vertical, we use the form \(\frac{(y-0)^2}{2^2} - \frac{(x-1)^2}{b^2} = 1 \). The value of \(a\) is derived from the distance between the center and the vertices, hence \(a = 2\). Then, we use the given point \((0, \sqrt{5})\), which lies on the hyperbola, to find \(b\).
By substituting \(x = 0\) and \(y = \sqrt{5}\) into the equation, we solve for \(b\), finding \(b = \sqrt{5}\). Now that we have all the needed values, we can write the equation as:

• \(\frac{y^2}{4} - \frac{(x-1)^2}{5} = 1\)

This equation represents the specific hyperbola in question, defined by its vertices and the point through which it passes. Understanding each term and value in the hyperbola equation is key for interpreting and graphing the curve accurately.