Problem 54 - Thomas Calculus

Question

Exercises \(53-56\) give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes. $$ \frac{x^{2}}{16}-\frac{y^{2}}{9}=1, \quad \text { left } 2, \text { down } 1 $$

Step-by-Step Solution

Answer

New equation: \( \frac{(x+2)^2}{16} - \frac{(y+1)^2}{9} = 1 \). Center: \((-2, -1)\). Foci: \((-7, -1)\), \((3, -1)\). Vertices: \((-6, -1)\), \((2, -1)\). Asymptotes: \( y = \frac{3}{4}x + \frac{1}{2} \), \( y = -\frac{3}{4}x - \frac{5}{2} \).

Step 1: Understand the Transformation

The hyperbola given by \( \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 \) has its center at \((0, 0)\). We need to shift this hyperbola \(2\) units to the left and \(1\) unit down. This means the new center will be at \((-2, -1)\).

Step 2: Apply the Horizontal Shift

To shift the hyperbola \(2\) units to the left, replace \(x\) with \((x+2)\) in the equation. This adjusts the position of the hyperbola horizontally.

Step 3: Apply the Vertical Shift

To shift the hyperbola \(1\) unit down, replace \(y\) with \((y+1)\) in the equation. This will adjust the position of the hyperbola vertically.

Step 4: Write the Equation of the New Hyperbola

Substituting the transformations into the original equation, the new equation becomes: \[ \frac{(x+2)^{2}}{16} - \frac{(y+1)^{2}}{9} = 1 \]

Step 5: Identify the New Center

The new center of the hyperbola after the transformation is \((-2, -1)\).

Step 6: Determine the Vertices

The vertices of the original hyperbola are at \((-4, 0), (4, 0)\). After shifting, the new vertices are at \((-6, -1), (2, -1)\).

Step 7: Determine the Foci

The distance \( c \) from the center to each focus is found using the equation \( c^2 = a^2 + b^2 \). Here \( a^2 = 16 \) and \( b^2 = 9 \), hence \( c = \sqrt{25} = 5 \). The foci of the shifted hyperbola will be at \((-7, -1)\) and \((3, -1)\).

Step 8: Determine the Asymptotes

The asymptotes of the original hyperbola are given by the equations \( y = \frac{b}{a} x \) and \( y = -\frac{b}{a} x \). For our equation, that comes out to be \( y = \frac{3}{4} x \) and \( y = -\frac{3}{4} x \). After the shifts, these asymptotes become \( y + 1 = \frac{3}{4}(x + 2) \) and \( y + 1 = -\frac{3}{4}(x + 2) \), simplifying to \( y = \frac{3}{4}x + \frac{1}{2} \) and \( y = -\frac{3}{4}x - \frac{5}{2} \).

Key Concepts

Equation Transformation

When working with hyperbolas, transforming their equations by applying graphical shifts can alter their position on the coordinate plane. For example, the equation \( \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 \) represents a hyperbola centered at \((0, 0)\). However, if we need to shift this hyperbola, we apply changes directly to the variables within the equation.

To shift the hyperbola 2 units to the left, we replace \(x\) with \((x + 2)\). Similarly, to move it 1 unit down, we replace \(y\) with \((y + 1)\). This results in the new transformed equation:
\[ \frac{(x+2)^{2}}{16} - \frac{(y+1)^{2}}{9} = 1 \]
By understanding these operations, you can consistently manipulate the equation to reposition the hyperbola.

Center of Hyperbola

The center of the hyperbola greatly influences its position in space. Initially, our hyperbola given by \( \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 \) has a center at the origin, \((0, 0)\). Transformations involved shifting the graph to a new center based on given instructions.

Shifting the graph 2 units to the left and 1 unit down moves the center to \((-2, -1)\). Recognizing the transformed center is important, as it acts as a reference point for calculating other components such as vertices and foci. By clearly identifying shifts, the new center is accurately determined.

Vertices and Foci

Vertices and foci are critical points that define the shape and orientation of a hyperbola. For a hyperbola described by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the distance \(a\) from the center along the transverse axis defines its vertices. For the original equation with center at \((0,0)\), vertices are at \((-4, 0)\) and \((4, 0)\).

After shifting, the vertices move to \((-6, -1)\) and \((2, -1)\) to maintain the same relative distances from the new center.

• Vertices: Determine these by moving \(a\) units from the altered center \((-2, -1)\).
• Foci: For determining foci, use \(c^2 = a^2 + b^2\), which leads to \(c = 5\). Consequently, the foci shift to \((-7, -1)\) and \((3, -1)\).

The foci remain symmetric about the center, deeply influencing the hyperbola's appearance.

Graphical Shifts

Graphical shifts change the actual position of the hyperbola on the coordinate plane without altering its shape. Understanding how shifts affect the equation assists in accurately plotting the new graph.

After finding the standard equation of the hyperbola, graphical shifts are applied by altering terms inside the equation. It includes inside operation of replacing variables with expressions like \((x+2)\) and \((y+1)\) to manage horizontal and vertical movements.

• Horizontal Shifts: Achieved by changing \(x\) to \((x+h)\).
• Vertical Shifts: Managed by replacing \(y\) with \((y+k)\).

The hyperbola shifts accordingly, with the center, vertices, and foci moving to new positions. Shifts help in exploring spatial transformations which keep the analytical form intact but helps in targeting specific plane positions.