Problem 45 - Precalculus

Question

Find the standard form of the equation of the hyperbola with the given characteristics. Vertices: (0,2),(6,2) asymptotes: \(y=\frac{2}{3} x, y=4-\frac{2}{3} x\)

Step-by-Step Solution

Answer

The standard form of the given hyperbola is \( \frac{(x-3)^2}{9} - \frac{(y-2)^2}{20.25} = 1 \).

Step 1: Identify the center (h, k)

The center of the hyperbola is the midpoint of the two given vertices. Calculate the midpoint by averaging the x-values and the y-values of the vertices: \( h = \frac{0+6}{2} = 3, k = \frac{2+2}{2} = 2 \). So, (h, k) is (3, 2).

Step 2: Identify the distance a

'a' is the distance from the center to any vertex. This can be calculated by subtracting the x-coordinate of the center (h) from the x-coordinate of the vertex: \( a = |6-3| = 3\).

Step 3: Identify the distance b

b is related to the slope of the asymptote which is ±2/3 in this case. The value of 'b' can be calculated using the formula \(b = \frac{a}{slope} = \frac{3}{2/3} = 4.5 \)

Step 4: Write the equation of the hyperbola

Now plug in h, k, a, and b into the standard form of the hyperbola. The standard form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Plug in the values to get \( \frac{(x-3)^2}{3^2} - \frac{(y-2)^2}{4.5^2} = 1 \)

Key Concepts

Vertices Calculation

To determine the vertices of a hyperbola, you begin with the given coordinates. The vertices are essentially the points that define the 'ends' of the hyperbola's central rectangle. In this problem, we have the vertices at (0,2) and (6,2). To better understand their role, we calculate the midpoint of these vertices to find the center of the hyperbola. By averaging their coordinates, you find that the center is at (3,2). Therefore, the vertices help us ascertain both the spread of the hyperbola and its central location.

The distance from the center to each vertex is known as 'a'. This is simply calculated as the difference between the vertex's x-values (since the y-values are the same here):

• Distance a = |6 - 3| = 3

Understanding these steps is crucial for constructing the hyperbola's equation properly.

Asymptotes Determination

Asymptotes are the 'guidelines' that the hyperbola approaches but never truly touches. They are crucial for understanding the hyperbola's orientation and shape. For a hyperbola centered at (h, k), the equations of the asymptotes are:

• Slope positive: \[y = k + \frac{b}{a}(x - h)\]
• Slope negative: \[y = k - \frac{b}{a}(x - h)\]

Here, the given asymptotes are \(y = \frac{2}{3}x\) and \(y = 4 - \frac{2}{3}x.\)

The slope is \(\pm 2/3,\) which helps in determining 'b' using the formula: \(b = \frac{a}{\text{slope}}.\) This indicates that 'b' is \(b = \frac{3}{2/3} = 4.5.\)This calculation shows how asymptotes contribute to the understanding of 'b', further detailing the hyperbola's shape.

Center of Hyperbola

Finding the center of a hyperbola is essential as it provides a reference point for the entire structure. As identified earlier, the center is the midpoint between the vertices, and is crucial in formulating the standard equation. In our instance with vertices (0,2) and (6,2), the center comes out to be at (3,2).

• This is computed by averaging the x-coordinates: \(h = \frac{0+6}{2} = 3\)
• And the y-coordinates: \(k = \frac{2+2}{2} = 2\)

With this, we gain the center (h, k) = (3, 2), a vital starting point for both the vertices and the asymptotes.

The center is significant because it directly influences the equation form, as all transformations relate back to this point.

Standard Form of Hyperbola

The standard form of a hyperbola's equation allows us to precisely understand and plot the hyperbola's characteristics. For a hyperbola oriented horizontally, the equation is given by:

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

Here, 'h' and 'k' are the coordinates of the center. 'a' and 'b' are the distances determined from the vertices and the slopes of the asymptotes respectively. In our situation:

• Center: (3, 2)
• a = 3 (distance from center to vertex)
• b = 4.5 (calculated from the slope of asymptote)

Substitute these values into the standard form:

• \[\frac{(x-3)^2}{3^2} - \frac{(y-2)^2}{4.5^2} = 1\]

This form provides a comprehensive frame for plotting the hyperbola, detailing its curve, position, and orientation.