Problem 43 - Thomas Calculus

Question

The hyperbola \(\left(x^{2} / 16\right)-\left(y^{2} / 9\right)=1\) is shifted 2 units to the right to generate the hyperbola $$\frac{(x-2)^{2}}{16}-\frac{y^{2}}{9}=1$$ a. Find the center, foci, vertices, and asymptotes of the new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

Step-by-Step Solution

Answer

The new hyperbola has center (2,0), vertices (-2,0) and (6,0), foci (-3,0) and (7,0), asymptotes: \(y = \pm \frac{3}{4}(x-2)\).

Step 1: Identify the New Center

The equation of the new hyperbola is \( \frac{(x-2)^2}{16} - \frac{y^2}{9} = 1 \). The hyperbola is translated 2 units to the right from its original center point. Therefore, the new center is located at \((h, k) = (2,0)\).

Step 2: Determine the Vertices

The hyperbola has a horizontal transverse axis because the \(x\) term comes first in the equation. The distance from the center to each vertex on the transverse axis is \(a = 4\), because \(a^2 = 16\). Thus, the vertices of the hyperbola are located at \((h \pm a, k) = (2 \pm 4, 0) = (-2, 0) \) and \((6, 0)\).

Step 3: Find the Foci

The distance from the center to each focus is \(c\), where \(c^2 = a^2 + b^2\). Here, \(b^2 = 9\) and \(a^2 = 16\), thus \(c^2 = 16 + 9 = 25\), so \(c = 5\). Therefore, the foci of the hyperbola are at \((h \pm c, k) = (2 \pm 5, 0) = (-3, 0)\) and \((7, 0)\).

Step 4: Calculate the Asymptotes

The equations for the asymptotes of a hyperbola with a horizontal transverse axis are given by \(y = \pm \frac{b}{a}(x - h)+ k\). In this case, \(b = 3, a = 4, h = 2,\) and \(k = 0\). So, the equations of the asymptotes are \(y = \pm \frac{3}{4}(x - 2)\).

Step 5: Sketch the Hyperbola

To sketch the hyperbola:- Plot the center at \((2, 0)\).- Place the vertices at \((-2, 0)\) and \((6, 0)\).- Mark the foci at \((-3, 0)\) and \((7, 0)\).- Draw the asymptotes using the lines \(y = \frac{3}{4}(x - 2)\) and \(y = -\frac{3}{4}(x - 2)\). - Sketch the hyperbola approaching but not crossing these asymptotes.

Key Concepts

Conic Sections

Conic sections are an essential part of geometry, derived from the intersection of a plane with a double-napped cone. These intersections can produce four main types of curves, namely circles, ellipses, parabolas, and hyperbolas. Each type has distinct equations and properties that define its shape, size, and orientation.

Here is a breakdown of these conic sections:

• Circle: A special type of ellipse, generated when the plane intersects the cone parallel to its base. It is symmetrical in all directions.
• Ellipse: Formed when a plane cuts through both nappes but not parallel to the base. It appears as an elongated circle.
• Parabola: Occurs when the plane is parallel to the slant of the cone. It has a distinct U-shape.
• Hyperbola: Results when the plane intersects both nappes. It consists of two separate curves mirroring each other.

In this context, we focus on hyperbolas, which have unique properties and formulas that differ significantly from other conic sections, such as the presence of asymptotes and two foci.

Center of Hyperbola

The center of a hyperbola is the midpoint between its foci and serves as the pivotal point from which the hyperbola is defined. In the standard form of a hyperbola equation, such as \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) (for a horizontal hyperbola) or \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \) (for a vertical hyperbola), the center is located at the point \((h, k)\).

For the given hyperbola equation \( \frac{(x-2)^2}{16} - \frac{y^2}{9} = 1 \), the center has been shifted from the origin to \((2, 0)\). This means the entire hyperbola moves, just like the center does, maintaining its shape and orientation. Understanding the placement of the center is crucial as it helps in plotting the hyperbola and determining other features like the vertices and asymptotes.

Vertices of Hyperbola

Vertices of a hyperbola are key points where the hyperbola intersects its transverse axis. For hyperbolas, this transverse axis is similar to the major axis in ellipses, determining the widest part of the curves. The vertices are a distance \(a\) from the center, where \(a\) is derived from the equation's denominator related to the \(x\)-term for horizontal hyperbolas.

For the equation \( \frac{(x-2)^2}{16} - \frac{y^2}{9} = 1 \), the vertices lie along the line \(y = 0\) at positions \(x = h \pm a\), giving us the coordinates \((-2, 0)\) and \((6, 0)\).

This is because the term \(a^2 = 16\) indicates \(a = 4\), thus elongating the hyperbola across the \(x\)-axis. Placing these vertices accurately helps in sketching the hyperbola's precise shape and orientation.

Asymptotes of Hyperbola

Asymptotes are straight lines that closely approach the curves of a hyperbola but never actually intersect with them. They provide a boundary that the hyperbola gets infinitely close to but never crosses. For a hyperbola, these lines help in determining its spread and directional flow, assisting in sketching its overall shape.

For horizontal hyperbolas like \( \frac{(x-2)^2}{16} - \frac{y^2}{9} = 1 \), the asymptotes are lines of the form \( y = \pm \frac{b}{a}(x - h) + k \). Here, \(b = 3\) and \(a = 4\), leading to the asymptote equations:

• \( y = \frac{3}{4}(x - 2) \)
• \( y = -\frac{3}{4}(x - 2) \)

Each asymptote forms an angle derived from the ratio \(\frac{b}{a}\), capturing the essence of the hyperbola's "openness." Marking these lines is essential; they offer guidance on how the hyperbola spreads on the graph as it extends towards these imaginary boundaries.