Problem 43 - Thomas Calculus

Question

You may wish to review Section 1.2 before solving Exercises \(39-56\) The hyperbola \(\left(x^{2} / 16\right)-\left(y^{2} / 9\right)=1\) is shifted 2 units to the right to generate the hyperbola $$ \frac{(x-2)^{2}}{16}-\frac{y^{2}}{9}=1 $$ $$ \begin{array}{l}{\text { a. Find the center, foci, vertices, and asymptotes of the new }} \\ {\text { hyperbola. }} \\ {\text { b. Plot the new center, foci, vertices, and asymptotes, and sketch }} \\ {\text { in the hyperbola. }}\end{array} $$

Step-by-Step Solution

Answer

a. Center: \( (2,0) \); vertices: \( (6,0), (-2,0) \); foci: \( (7,0), (-3,0) \); asymptotes: \( y = \frac{3}{4}(x-2), y = -\frac{3}{4}(x-2) \). b. Graph these elements and sketch the hyperbola.

Step 1: Reviewing the standard form

The standard form of a hyperbola with a horizontal transverse axis is given by \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Here, the center of the hyperbola is at point \((h, k)\), \(a\) is the distance from the center to each vertex along the x-axis, and \(b\) is the distance from the center to each co-vertex along the y-axis.

Step 2: Identify the center

Comparing the given hyperbola equation \( \frac{(x-2)^2}{16} - \frac{y^2}{9} = 1 \) with the standard form, we find \(h = 2\) and \(k = 0\). Thus, the center of the hyperbola is \((2, 0)\).

Step 3: Determine vertices

The vertices of the hyperbola are located along the transverse axis at distances of \(a\) from the center. Since \(a^2=16\), \(a=4\). Thus, the vertices are \((2+4, 0) = (6, 0)\) and \((2-4, 0) = (-2, 0)\).

Step 4: Locate the foci

The foci are found at distances \(c\) from the center, where \(c\) is given by \(c^2 = a^2 + b^2\). Here, \(c^2 = 16 + 9 = 25\), so \(c = 5\). Therefore, the foci are \((2+5, 0) = (7, 0)\) and \((2-5, 0) = (-3, 0)\).

Step 5: Find the asymptotes

The equations of the asymptotes for the hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) are \( y = k \pm \frac{b}{a}(x-h) \). Substituting in \( h = 2 \), \( k = 0 \), \( a = 4 \) and \( b = 3 \), the asymptotes are \( y = \frac{3}{4}(x - 2) \) and \( y = -\frac{3}{4}(x - 2) \).

Step 6: Sketch the graph

To sketch the hyperbola, plot the center \((2,0)\), vertices \((6,0)\) and \((-2,0)\), and foci \((7,0)\) and \((-3,0)\). Draw the asymptotes as slant lines passing through the center with slopes \(\frac{3}{4}\) and \(-\frac{3}{4}\). Finally, draw the hyperbola's two branches opening to the left and right, ensuring they approach the asymptotes as they extend outward.

Key Concepts

Hyperbola

A hyperbola is one of the four main types of conic sections, which are the curves formed by intersecting a plane with a cone. Unlike a circle or ellipse, a hyperbola consists of two separate branches. These branches resemble curved lines that mirror each other across center.
A hyperbola with a horizontal transverse axis is described by the standard form equation: \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] Here,

• \(h\) and \(k\) are the coordinates of the center,
• \(a\) and \(b\) are the distances from the center to the vertices and co-vertices, respectively.

The equation provided, \(\frac{(x-2)^2}{16} - \frac{y^2}{9} = 1\), suggests a hyperbola with its center shifted 2 units to the right from the origin, to the point \((2, 0)\). In this setup, the transverse axis lies horizontally along the x-axis.

Foci and Vertices

Finding the foci and vertices of a hyperbola is crucial for understanding its shape and behavior. The vertices of the hyperbola are the points where the hyperbola crosses the transverse axis. For the equation \(\frac{(x-2)^2}{16} - \frac{y^2}{9} = 1\),

• \(a^2 = 16\), which means \(a = 4\).

Thus, the vertices from the center \((2, 0)\) are at points \((6, 0)\) and \((-2, 0)\).
The foci are located further from the center than the vertices, providing another crucial indicator of the hyperbola's shape. To find the distance \(c\) to the foci from the center, use the relation: \[ c^2 = a^2 + b^2 \]Given \(b^2 = 9\) as well, we have \(c^2 = 25\), leading to \(c = 5\). Therefore, the foci are located at the coordinates \((7, 0)\) and \((-3, 0)\).
These markers help in graphing the hyperbola and understanding its dimensions.

Asymptotes

Asymptotes of a hyperbola are lines that the branches of the hyperbola approach but never touch. They provide a "guideline" for drawing the hyperbola accurately. These lines extend straight through the center and depict the general direction the hyperbola will open.
The formula for asymptotes of a horizontal hyperbola is: \[y = k \pm \frac{b}{a}(x-h)\]For our specific hyperbola \(\frac{(x-2)^2}{16} - \frac{y^2}{9} = 1\), we plug in \(h = 2\), \(k = 0\), \(a = 4\), and \(b = 3\) into this equation:

• The first asymptote: \(y = \frac{3}{4}(x - 2)\)
• The second asymptote: \(y = -\frac{3}{4}(x - 2)\)

These lines cross at the center and extend in opposite diagonal directions. When sketching the hyperbola, the branches should become increasingly close to these asymptotes but never intersect them. The steepness and direction of these asymptotes give insight into how 'wide' or 'narrow' the hyperbola will appear on a graph.