Problem 46 - Precalculus with Limits

Question

In Exercises 35-46, find the standard form of the equation of the hyperbola with the given characteristics. Vertices: \((3, 0), (3, 4); \quad\) asymptotes: \(y=\frac{2}{3}x, y=4-\frac{2}{3}x\)

Step-by-Step Solution

Answer

The standard form equation of the given hyperbola is \(\frac{(y - 2)^2}{4} - \frac{(x - 3)^2}{16/9} = 1\)

Step 1: Determine the center of the hyperbola

The hyperbola's center lies midway between the given vertices, which can be calculated by taking the average of their coordinates. These vertices are \((3, 0), (3, 4)\), so the center is at \((3, 2)\) after taking the average.

Step 2: Calculate the values of 'a' and 'b'

'a' is the distance from the center to a vertex. Given vertices are at \((3, 0)\) and \((3, 4)\), and the center is \((3, 2)\). So, \(a = 2\). 'b' can be determined from the slope of the asymptotes, which is \(b/a\). Given asymptotes are \(y=\frac{2}{3}x\) and \(y=4-\frac{2}{3}x\), so the slope is \(\frac{2}{3}\), hence \(b = 4/3\).

Step 3: Determine the orientation of the hyperbola

Since the vertices given lie vertically in respect to the center, the orientation of the hyperbola is vertical.

Step 4: Write the standard form of the equation of the hyperbola

The standard form equation for a vertical hyperbola centered at \((h, k)\) is \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\). Substituting for \(h, k, a\) and \(b\), we get the standard form equation as \(\frac{(y - 2)^2}{4} - \frac{(x - 3)^2}{16/9} = 1\)

Key Concepts

Center of a Hyperbola

The center of a hyperbola is a crucial point as it acts as a reference for the entire hyperbola. It is essentially the midpoint between its two vertices.
In our example, the vertices are at \((3, 0)\) and \((3, 4)\). Calculating the center involves averaging the x-coordinates and y-coordinates of the vertices. For this hyperbola, the midpoints are:\[\left(\frac{3+3}{2}, \frac{0+4}{2}\right) = (3, 2)\]
Thus, the center is at \((3, 2)\). This kind of calculation helps in defining the position of the hyperbola in the coordinate system.
Knowing the center allows us to understand the symmetry of the hyperbola. Since hyperbolas are symmetric around both axes, this point of symmetry is vital in graphing the curve properly.

Asymptotes of a Hyperbola

Asymptotes are lines that a hyperbola approaches but never actually reaches. They provide guidance for the shape and orientation of the hyperbola.
In this example, the asymptotes given are \(y=\frac{2}{3}x\) and \(y=4-\frac{2}{3}x\). These equations are crucial in finding the slopes and ensuring that the hyperbola is anchored correctly.

• The slope of the asymptotes, \(\frac{2}{3}\), connects to the calculation of \(b/a\), where 'a' and 'b' are distances from the center to the vertices and co-vertices, respectively.
• Using the equation \(b/a = \text{slope of asymptote}\), if \(a=2\) (from vertex information), then \(b = 4/3\).

The intersection of the asymptotes is always at the center of the hyperbola. This feature helps in graphing because it offers precision in checking whether the drawn hyperbola matches the correct orientation defined by its asymptotes.

Vertices of a Hyperbola

The vertices of a hyperbola are the turning points, where the curve is closest or furthest from the center along the axis of symmetry.
For vertical hyperbolas, like the example given with two vertices at \((3,0)\) and \((3,4)\), the vertices lie along the y-axis when centered at \((3, 2)\). These points indicate the maximum distance from the center in the vertical direction.
Calculating 'a', the distance from the center to a vertex, is simple in this case as it's on a single axis. Here, \(a=2\) because each vertex is 2 units from \((3,2)\).
Vertices not only define the span of the hyperbola but help in writing the standard equation, which is important for graph drawings and interpretations. Knowing the position of the vertices particularly assists while setting up the equation for the hyperbola in its standard form.