Problem 43 - Precalculus with Calculus Previews

Question

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Center \((2,4),\) one vertex \((2,5),\) one asymptote \(2 y\) \(-x-6=0\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Center and Vertex

The center of the hyperbola is given as \((2, 4)\), and one vertex is at \((2, 5)\). This suggests that the hyperbola is vertical since the vertex shares the x-coordinate (2) with the center.

Step 2: Determine the Distance 'a'

The distance from the center \((2, 4)\) to the vertex \((2, 5)\) is 1 unit in the y-direction. Therefore, \(a = 1\).

Step 3: Write the General Equation

The equation of a vertically oriented hyperbola centered at \((h, k)\) is \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\). Thus, the equation is initially \(\frac{(y - 4)^2}{1^2} - \frac{(x - 2)^2}{b^2} = 1\).

Step 4: Solve for the Slope of Asymptotes

The asymptotes of the hyperbola have slopes \(\pm \frac{a}{b}\). The given asymptote is \(2y - x - 6 = 0\), or rearranged, \(y = \frac{1}{2}x + 3\). The slope here is \(\frac{1}{2}\).

Step 5: Determine 'b' Using Slope

Since \(\frac{a}{b} = \frac{1}{2}\) and \(a = 1\), we have \(\frac{1}{b} = \frac{1}{2}\). Solving for \(b\), we get \(b = 2\).

Step 6: Form the Equation of the Hyperbola

Substitute \(a = 1\) and \(b = 2\) into the equation: \(\frac{(y - 4)^2}{1^2} - \frac{(x - 2)^2}{2^2} = 1\), which simplifies to \((y - 4)^2 - \frac{(x - 2)^2}{4} = 1\).

Key Concepts

Conic Sections

Conic sections are curves obtained by slicing a double cone with a plane. Depending on the angle and position of the slice, four distinct types of conic sections can be created:

• Circle
• Ellipse
• Parabola
• Hyperbola

Each conic section has unique properties and equations. The hyperbola is particularly interesting because it forms two separate curves called branches. These branches mirror each other and open in opposite directions.
Hyperbolas arise from slicing the cone at a steep angle, which is more vertical than the cone itself.
They have equations in a standard form resembling \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]for vertical hyperbolas, where \(a\) and \(b\) define important characteristics such as dimensions.

Asymptotes

Asymptotes are lines that the hyperbola approaches but never touches. These lines are crucial for defining the shape and orientation of the hyperbola.
Hyperbolas have two asymptotes which intersect at the center of the hyperbola. Depending on the orientation, either vertical or horizontal, the slopes of these asymptotes are defined by the terms \(\pm \frac{a}{b}\) or \(\pm \frac{b}{a}\), respectively.
In our problem, one of the asymptotes is defined by the equation \(2y - x - 6 = 0\), which simplifies to \(y = \frac{1}{2}x + 3\).
This indicates the slope is \(\frac{1}{2}\), helping determine the relationship between \(a\) and \(b\). It shows how the hyperbola will spread and provide clues for plotting it more accurately.

Vertices

Vertices are key points on the hyperbola. They mark the closest parts of the branches to the center and illustrate the hyperbola's open direction.
For a vertical hyperbola, the vertices lie along the y-axis or parallel to it. They are equidistant from the center, determined by the number \(a\), which defines how far they are from the center along the axis of symmetry.
In our exercise, the center is \( (2, 4) \), and a vertex is given at \((2, 5)\).
This means the distance \(a = 1\), which is calculated as the difference in the y-coordinates, \(5 - 4 = 1\). This finding directly supports developing the hyperbola equation.

Hyperbola Center

The center of a hyperbola is its midpoint between the branches. It is where the symmetry lines intersect each other.
Every hyperbola center is crucial in determining its equation and overall orientation. For this specific problem, the center is given as \( (2, 4) \), a fixed point that aligns with the coordinates of symmetry for the hyperbola's branches and asymptotes.
In the equation of the hyperbola, this center \( (h, k) \) becomes \( (2, 4) \), substituted into the formula, influencing where the branches and symmetry axes are.
Knowing the center helps in plotting and understanding how the hyperbola fits among other conic sections, complementing the defined asymptote equations and vertex locations.