Problem 46 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: (3,0),(3,-6) asymptotes: \(y=x-6, y=-x\)

Step-by-Step Solution

Answer

The standard form of the equation of the hyperbola is \((y+3)^2/9 - (x-3)^2/9 = 1\).

Step 1: Identify the center

In this case, the center of the hyperbola is the midpoint of the vertices. To find the midpoint, one can use the formula \((x_1+x_2)/2, (y_1+y_2)/2\). Therefore, the center is \(h=(3+3)/2 = 3\) and \(k=(0+(-6))/2 = -3\).

Step 2: Calculate the a value

The value of 'a' is the distance from the center to either vertex. The x-coordinate, or h value, is the same for the center and vertices in our problem which indicates that the major axis is vertical, so 'a' is the absolute difference of their y-coordinate. a = \(|-3 - 0| = 3\).

Step 3: Calculate the b value using the asymptotes

From the equations of the asymptotes, we can observe that the slopes are +1 and -1, which are 'a/b' and '-a/b' for a vertically oriented hyperbola. Therefore, b = a in this case, so b = 3.

Step 4: Write the equation in standard form

Now that we have the center at (3, -3), a = 3 and b = 3, we can plug these into the equation for a vertical hyperbola: \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\). Substituting these values into the equation yields \((y+3)^2/3^2 - (x-3)^2/3^2 = 1\), and simplifying the equation gives - \((y+3)^2/9 - (x-3)^2/9 = 1\). This is the standard form of the requested hyperbola.

Key Concepts

Standard Form Equation

The standard form equation of a hyperbola is crucial as it provides a complete algebraic representation of the hyperbola. For a hyperbola centered at \((h, k)\), the general equation is:

• For a horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• For a vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

The terms \(a\) and \(b\) are important because they denote the distances that define the geometry of the hyperbola. In our exercise, the hyperbola is vertically oriented because the vertices share the same x-coordinate. Therefore, the standard form of our hyperbola is:\[ \frac{(y+3)^2}{9} - \frac{(x-3)^2}{9} = 1 \]This equation incorporates the center at \((3, -3)\) and the values of \(a = 3\) and \(b = 3\). The equations of the asymptotes confirm this orientation as well.

Vertices

The vertices of a hyperbola are the points where the hyperbola is closest or furthest apart on each axis. They are key to determining the center and orientation of the hyperbola. In our problem, the vertices were given as \((3,0)\) and \((3,-6)\).

The vertices help in finding the center. The midpoint formula was used to determine the center between these points.

• Calculate the center: \( \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) = (3, -3) \)

In this case, because the x-coordinates are the same for the vertices, the hyperbola is oriented vertically. This means the distance \(a\) along the y-axis is crucial for plotting hyperbola. Here, \(a = 3\), the absolute vertical distance from the center to each vertex.

Asymptotes

Asymptotes are lines that the hyperbola approaches but never actually reaches. They provide an essential guide to the slant of the hyperbola. Hyperbolas have two asymptotes. The slopes of these asymptotes indicate the values of \(a\) and \(b\).

For a hyperbola centered at \((h, k)\) with vertical orientation, the equations of the asymptotes are given by:

• \(y - k = \pm \frac{a}{b} (x - h)\)

In the exercise, the asymptotes provided were \(y = x - 6\) and \(y = -x\). These equations simplify to slopes of \(+1\) and \(-1\). These imply that \(\frac{a}{b} = 1\), leading to \(a = b\). Thus, with \(a = 3\), it was established that \(b = 3\) as well.

Center of Hyperbola

The center of a hyperbola is a fundamental point as it acts as a reference for the structure of the hyperbola. It is the midpoint between the vertices and determines the symmetry of the hyperbola. In our problem, calculating the center was the first step, using the formula \((\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\).

Given the vertices \((3,0)\) and \((3,-6)\), the center was found at:

• \((3, -3)\)

The center helps not only in defining the orientation — horizontal or vertical — of the hyperbola but also in determining the positioning of the entire curve in the coordinate plane. It will correlate directly with the asymptotes' equations and the vertex positions.