Problem 28 - Intermediate Algebra

Question

Graph each hyperbola. See Example 3. $$ \frac{(y-2)^{2}}{4}-\frac{(x+1)^{2}}{1}=1 $$

Step-by-Step Solution

Answer

Center at (-1, 2), vertices at (-1, 4) and (-1, 0), asymptotes: \( y=2x+4 \) and \( y=-2x \).

Step 1: Identify the form of the hyperbola

The given equation is \( \frac{(y-2)^2}{4} - \frac{(x+1)^2}{1} = 1 \). This is in the standard form \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), indicating a vertical hyperbola.

Step 2: Determine the center of the hyperbola

For the equation \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \), the center is at \((h, k)\). Here, \( h = -1 \) and \( k = 2 \), so the center is at \( (-1, 2) \).

Step 3: Identify a and b values

From the equation \( \frac{(y-2)^2}{4} - \frac{(x+1)^2}{1} = 1 \), determine \( a^2 = 4 \) so \( a = 2 \), and \( b^2 = 1 \) so \( b = 1 \).

Step 4: Find the vertices

The vertices are found \( a \) units above and below the center. Thus, for this hyperbola, the vertices are located at \( (-1, 2+2) = (-1, 4) \) and \( (-1, 2-2) = (-1, 0) \).

Step 5: Calculate the slopes of the asymptotes

The slopes of the asymptotes for a vertical hyperbola are given by \( \pm \frac{a}{b} = \pm \frac{2}{1} = \pm 2 \).

Step 6: Write the equations of the asymptotes

Using the center \( (-1, 2) \) and the slopes \( m = \pm 2 \), the equations of the asymptotes are \( y - 2 = 2(x + 1) \) which simplifies to \( y = 2x + 4 \), and \( y - 2 = -2(x + 1) \) which simplifies to \( y = -2x \).

Step 7: Sketch the graph

Plot the center of the hyperbola at \( (-1, 2) \), then draw the asymptotes using their equations. Mark the vertices at \( (-1, 4) \) and \( (-1, 0) \). This will help outline the shape of the hyperbola, opening upwards and downwards through the vertices, approaching the asymptotes.

Key Concepts

Vertical Hyperbola

A vertical hyperbola is a specific type of hyperbola that opens up and down. This is determined by its standard equation form which is \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]In this equation:

• The center of the hyperbola is \((h, k)\)
• The variable \(a\) represents the distance from the center to the vertices along the y-axis, indicating an up-down orientation.
• If the equation was reversed to \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) it would signify a horizontal hyperbola.

To put it simply, if the \(y\) term leads the equation, it forms a vertical hyperbola. This concept is crucial for identifying the graph's orientation, ensuring the hyperbola is sketched correctly. Depending on the values of \(a\) and \(b\), the exact shape may vary, but the directional opening will always remain vertical.

Conic Sections

Conic sections are figures created by the intersection of a plane with a cone. They encompass different shapes:

• Circle
• Ellipse
• Parabola
• Hyperbola

Each of these shapes has a distinct definition and equation. Hyperbolas, like the vertical hyperbola in the exercise, occur when the plane intersects both nappes or sides of the cone, resulting in two mirror-image arcs.
Hyperbolas have different properties, such as two foci and two directrices, compared to the single focus and directrix of a parabola. Understanding conic sections helps in grasping concepts of geometry, especially when relating equations to graph shapes.
When analyzing a hyperbola, recognizing it as a conic section provides a mathematical framework for understanding its features and transformations.

Asymptotes

Asymptotes are lines that a curve approaches as it heads towards infinity. For hyperbolas, asymptotes are lines that help guide the shape and direction.
For a vertical hyperbola, the slopes of the asymptotes are determined by \(\pm \frac{a}{b} \) in the equation \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]Here,

• The slopes of the asymptotes become \(\pm\) the ratio of \(a\) to \(b\).
• In the provided problem, the calculated slopes are \(2\) and \(-2\).

Asymptotes provide a boundary for the approaching curve, but the curve will never actually intersect them, offering key insight when sketching the graph of the hyperbola. It's useful to draw the asymptotes first when graphing, as they set up a guideline for sketching the hyperbola's fins.

Vertices of Hyperbola

Vertices are pivotal points of a hyperbola that indicate where the curve turns and changes direction. For a vertical hyperbola, the vertices are determined by the value \(a\) in the standard equation:\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]

• The vertices lie along the \(y\)-axis, precisely \(a\) units above and below the center \((h, k)\).

In our exercise:

• With the center at \((-1, 2)\) and \(a = 2\), the vertices are positioned at \((-1, 4)\) and \((-1, 0)\).

These points are critical in accurately sketching the hyperbola, as they provide a visual cue for the curve's maximum and minimum reaches along the axis. Always plot the vertices after establishing the center, giving the framework necessary to shape the hyperbola accurately around its center.