Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation. \(\frac{(x+3)^{2}}{4}-\frac{(y+2)^{2}}{16}=1\)

Step-by-Step Solution

Answer

The hyperbola is centered at (-3, -2) with vertices at (-1,-2) and (-5,-2); it opens horizontally with asymptotes given by y + 2 = ±2(x + 3).

Step 1: Identify the Type of Conic Section

The equation given is of the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), which represents a hyperbola centered at \((h, k)\). Since the x-term is positive, we know that it is a horizontally opening hyperbola.

Step 2: Identify the Center of the Hyperbola

The standard form of the hyperbola equation is matched to \( \frac{(x+3)^2}{4} - \frac{(y+2)^2}{16} = 1 \), which shows that \( h = -3 \) and \( k = -2 \). Thus, the center of the hyperbola is \((-3, -2)\).

Step 3: Determine the Transverse and Conjugate Axes

For the hyperbola equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), \( a^2 = 4 \) and \( b^2 = 16 \). So, \( a = 2 \) and \( b = 4 \). The transverse axis is horizontal (since the x-term is positive), with a length of \( 2a = 4 \). The conjugate axis is vertical, with a length of \( 2b = 8 \).

Step 4: Locate the Vertices and Co-Vertices

The vertices are found on the transverse axis, which is horizontal in this case. They are located at \((h \pm a, k)\), giving the points \((-3 \pm 2, -2)\) or \((-1, -2)\) and \((-5, -2)\). The co-vertices are on the conjugate axis at \((h, k \pm b)\), which are \((-3, -2 \pm 4)\) or \((-3, 2)\) and \((-3, -6)\).

Step 5: Determine the Asymptotes

The asymptotes of a hyperbola centered at \((h, k)\) with horizontal transverse axis have the equations \( y - k = \pm \frac{b}{a} (x - h) \). Substituting \( a = 2 \), \( b = 4 \), \( h = -3 \), and \( k = -2 \), we get the asymptotes \( y + 2 = \pm 2(x + 3) \).

Step 6: Sketch the Graph

To sketch the hyperbola, plot the center at \((-3, -2)\), locate the vertices \((-1, -2)\) and \((-5, -2)\), and the co-vertices \((-3, 2)\) and \((-3, -6)\). Draw a rectangle that passes through these points with sides parallel to the axes. The diagonals of this rectangle are the asymptotes. Finally, sketch the hyperbola opening horizontally between the vertices.

Key Concepts

Conic Sections

Conic sections are important curves that emerge from slicing a cone with a plane. These intriguing shapes include circles, ellipses, parabolas, and hyperbolas. Imagine a cone made of clay. If you slice it in different ways, you'll get different shapes.
If you slice horizontally, you get a circle. Cut at an angle, and you find an ellipse. A more dramatic slash that touches the base results in a parabola. Finally, when the plane cuts through both nappes of the cone, you uncover a hyperbola.
Each one has its unique equation, but they all stem from similar principles, showcasing the elegance and symmetry inherent in mathematics. Understanding conic sections begins with recognizing their source and appreciating the beauty of these geometric forms.

Graphing Hyperbolas

Graphing hyperbolas involves identifying key components that define their shape on a cartesian plane. First, it's essential to spot the center of the hyperbola, a point from which the curve opens. This point is not on the hyperbola itself, but it helps in drawing the axes and asymptotes.
The hyperbola consists of two separate curves, pointing away from each other. These curves are defined by their transverse axis, along which the vertices are located, and the conjugate axis. The vertices are critical, marking where each curve arises.
Asymptotes are straight lines that the hyperbola approaches but never touches. They serve as guides when sketching the hyperbola. By drawing a rectangle using the vertices and co-vertices, the asymptotes form the diagonals, guiding the general shape of the hyperbola. Understanding these elements makes plotting a hyperbola an engaging puzzle, where each piece connects to reveal a striking geometric figure.

Equation of Hyperbolas

The equation of a hyperbola is fundamental in creating its graph and understanding its properties. A standard equation looks like \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] This form tells us a lot about the hyperbola's features.
Firstly, the \(h\) and \(k\) values in the equation represent the center of the hyperbola. The terms \(a^2\) and \(b^2\) guide us to calculate the lengths of the transverse and conjugate axes. These axes construct the frame on which the hyperbola sits.
Depending on the term that is positive in the standard form, the hyperbola will open horizontally or vertically. In cases where the \(x\)-term is positive, expect a horizontal opening, and with a positive \(y\)-term, the opening is vertical.
With this equation, we can also derive the slopes of the asymptotes and locate the vertices and co-vertices, laying down the groundwork for sketching the hyperbola. It’s like a roadmap to crafting this beautiful mathematical structure.