Problem 44 - Thomas Calculus

Question

You may wish to review Section 1.2 before solving Exercises \(39-56\) The hyperbola \(\left(y^{2} / 4\right)-\left(x^{2} / 5\right)=1\) is shifted 2 units down to generate the hyperbola $$ \frac{(y+2)^{2}}{4}-\frac{x^{2}}{5}=1 $$ $$ \begin{array}{l}{\text { a. Find the center, foci, vertices, and asymptotes of the new }} \\ {\text { hyperbola. }} \\ {\text { b. Plot the new center, foci, vertices, and asymptotes, and sketch }} \\ {\text { in the hyperbola. }}\end{array} $$

Step-by-Step Solution

Answer

Center: (0, -2); Vertices: (0, 0), (0, -4); Foci: (0, 1), (0, -5); Asymptotes: \(y + 2 = \pm \frac{2}{\sqrt{5}}x\).

Step 1: Identify the Center

The given equation of the hyperbola is \( \frac{(y+2)^{2}}{4} - \frac{x^{2}}{5} = 1 \). This equation is in the form \( \frac{(y-k)^{2}}{a^2} - \frac{(x-h)^{2}}{b^2} = 1 \), which describes a vertical hyperbola. The center of this hyperbola is at \((h, k)\). In this equation, \(h = 0\) and \(k = -2\), so the center is at \((0, -2)\).

Step 2: Calculate the Vertices

For a vertical hyperbola \( \frac{(y-k)^{2}}{a^2} - \frac{(x-h)^{2}}{b^2} = 1 \), the vertices are found at \((h, k\pm a)\). Here, \(a^2 = 4\), so \(a = 2\). The vertices are at \((0, -2\pm 2)\), which means the vertices are at \((0, 0)\) and \((0, -4)\).

Step 3: Find the Foci

The distance from the center to each focus is given by \(c\), where \(c^2 = a^2 + b^2\). Here, \(a^2 = 4\) and \(b^2 = 5\), so \(c^2 = 9\), which gives \(c = 3\). The foci are located vertically from the center at \((h, k\pm c)\), or \((0, -2\pm 3)\). Thus, the foci are at \((0, 1)\) and \((0, -5)\).

Step 4: Determine the Asymptotes

The equations of the asymptotes for a hyperbola of form \( \frac{(y-k)^{2}}{a^2} - \frac{(x-h)^{2}}{b^2} = 1 \) are given by \(y - k = \pm \frac{a}{b}(x - h)\). For this hyperbola, \(a = 2\), \(b = \sqrt{5}\), \(h = 0\), and \(k = -2\). Thus the equations of the asymptotes are \(y + 2 = \pm \frac{2}{\sqrt{5}}x\).

Step 5: Plot and Sketch

To sketch the hyperbola, plot the center \((0, -2)\), the vertices \((0, 0)\) and \((0, -4)\), and the foci \((0, 1)\) and \((0, -5)\). Also, sketch the asymptotes given by \(y + 2 = \pm \frac{2}{\sqrt{5}}x\). Then, draw the hyperbola, ensuring it approaches these asymptotes as it extends away from the vertices.

Key Concepts

Conic Sections

Conic sections are curves generated by the intersection of a plane with a double-napped cone. They are fundamental to the study of geometry and include four main types: circles, ellipses, parabolas, and hyperbolas.

• A circle is formed when the plane cuts the cone parallel to the base.
• An ellipse results when the intersection is at an angle, but not steep enough to form a parabola.
• A parabola is created when the plane cuts parallel to a generatrix of the cone.
• A hyperbola appears when the plane intersects both nappes of the cone.

Hyperbolas, like the one described in the exercise, are two separate curves that mirror each other and have unique geometric properties. Understanding conic sections is crucial as they form the basis for many physical phenomena in physics and engineering.

Equation of a Hyperbola

A hyperbola's equation takes different forms depending on its orientation (horizontal or vertical). The general form of a vertical hyperbola, like the one in our exercise, is\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]

• In this equation, \( (h, k) \) represents the center of the hyperbola.
• \( a \) and \( b \) determine the shape and spread of the hyperbola.

The term \((y-k)^2\) appears before \((x-h)^2\), indicating it is a vertical hyperbola. If these terms were switched, it would be horizontal. The coefficients under the squared terms, \(a^2\) and \(b^2\), play a significant role in defining the hyperbola's structure, influencing the distance to the vertices and the asymptotic slopes.

Center of a Hyperbola

The center of a hyperbola is a crucial point that acts as the geometric midpoint around which the hyperbola is symmetrically arranged. It can be found directly from the equation.- In a hyperbola's standard form equation \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), the center is located at the coordinates \((h, k)\).
In this exercise, the hyperbola is described by the equation \( \frac{(y+2)^2}{4} - \frac{x^2}{5} = 1 \), which clearly shows that:

• The value for \(k\) is \(-2\), and the value for \(h\) is 0.
• Therefore, the center is at the point \((0, -2)\).

This point serves as a reference for locating other key elements, like vertices and foci, thereby playing a fundamental role in fully understanding a hyperbola's geometry.

Vertices of a Hyperbola

In a hyperbola, vertices are points where the hyperbola is closest to its center. They lie along the transverse axis, which corresponds to the direction where the equation has positive terms.

• For a vertical hyperbola, described by \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\), vertices are found at \((h, k \pm a)\).
• In this exercise, \(a^2 = 4\), so \(a = 2\).

The vertices of the hyperbola in the exercise are:\( (0, -2+2) = (0, 0) \) and \( (0, -2-2) = (0, -4) \).
Recognizing where the vertices are is key, as they identify the fundamental starting points from which the hyperbola stretches toward its asymptotes. Using these points is essential for graphing the hyperbola correctly.