Problem 42 - Contemporary Precalculus

Question

Use the information given in Special Topics 10.3. A and summarized in the endpapers at the beginning of this book to find a parameterization of the conic section whose rectangular equation is given. Confirm your answer by graphing. $$\frac{(x+5)^{2}}{25}-\frac{(y-3)^{2}}{50}=1$$

Step-by-Step Solution

Answer

Question: Find a parameterization of the conic section given by the rectangular equation: $$\frac{(x+5)^{2}}{25} - \frac{(y-3)^{2}}{50} = 1$$ Answer: The parameterization for the given conic section is given by the equations: $$x(t) = -5 + 5\cdot\cosh(t)\\ y(t) = 3 + \sqrt{50}\cdot\sinh(t)$$

Step 1: Identify h, k, a, and b

In our given equation: $$\frac{(x+5)^{2}}{25} - \frac{(y-3)^{2}}{50} = 1$$ We can identify h, k, a, and b by matching: $$h = -5, k = 3, a^2 = 25, b^2 = 50$$ Therefore: $$a = 5, b = \sqrt{50}$$

Step 2: Create parameterization using h, k, a, and b

Now using the parameterization formula for a hyperbola: $$x(t) = h + a\cdot\cosh(t)\\ y(t) = k + b\cdot\sinh(t)$$ Insert the values of a, b, h, and k, and we have the new parameterized equations: $$x(t) = -5 + 5\cdot\cosh(t)\\ y(t) = 3 + \sqrt{50}\cdot\sinh(t)$$

Step 3: Confirmation by graphing

To confirm our answer, we can graph the hyperbola using the parameterized equation and compare it with the original one. If both graphs are the same hyperbola, then we have found the correct parameterization. In most graphing tools (like Desmos, or graphing calculator software), enter the following: Parameterized equation: 1. \(x(t) = -5 + 5\cdot\cosh(t)\) 2. \(y(t) = 3 + \sqrt{50}\cdot\sinh(t)\) Original equation: 1. \(\frac{(x+5)^{2}}{25} - \frac{(y-3)^{2}}{50} = 1\) Comparing the graphs will show that they indeed represent the same hyperbola, confirming our parameterization is correct.

Key Concepts

Hyperbola Equation

Understanding the hyperbola equation is integral for students delving into the world of conic sections. A hyperbola, by definition, is the set of all points \( (x, y) \) such that the absolute difference of the distances to two fixed points (foci) is constant. The standard equation for a hyperbola centered at the point \( (h, k) \) with axes parallel to the coordinate axes can be expressed as:

\[\frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1\]or\[\frac{(y - k)^{2}}{b^{2}} - \frac{(x - h)^{2}}{a^{2}} = 1\]
depending on whether it's horizontally or vertically oriented. Here, \( a \) and \( b \) are the distances from the center to the vertices and co-vertices, respectively. In our exercise, the given equation:\[\frac{(x+5)^{2}}{25}-\frac{(y-3)^{2}}{50}=1\]is identified as a horizontal hyperbola due to its subtraction form, where \( a^2 \) is associated with the \( x \) term.

Rectangular Equation

The rectangular equation of a conic section refers to its expression in the standard Cartesian coordinate plane, where the coordinates of any point on the curve are denoted by \( x \) and \( y \) variables. This form is advantageous for analyzing the properties of the conic such as center, axes, asymptotes, and directrices. In our problem, the rectangular equation:\[\frac{(x+5)^{2}}{25}-\frac{(y-3)^{2}}{50}=1\]exhibits symmetry about the line \( x=-5 \) while being centered at \( (-5,3) \), where the shifts in \( x \) and \( y \) are denoted by \( h=-5 \) and \( k=3 \), respectively. The values \( a \) and \( b \) derived from the coefficients of \( x^2 \) and \( y^2 \) terms, respectively, are crucial for graphing and parameterizing the hyperbola.

Graphing Hyperbolas

Graphing hyperbolas can initially seem complex, but understanding their fundamental components simplifies the process. Start by plotting the center at the point \( (h, k) \). Then, draw the axes of the hyperbola corresponding to the distances \( a \) and \( b \) along the \( x \) and \( y \) axes from the center. From these, you can sketch the asymptotes as lines passing through the center at angles indicative of the slopes given by \( \pm \frac{b}{a} \) for the horizontal hyperbola or \( \pm \frac{a}{b} \) for the vertical hyperbola.

The actual curve of the hyperbola will approximate these lines but never touch them, opening along the transverse axis which, in the exercise provided, is the horizontal axis defined by \( a \) due to the larger value under the \( x \) term, which suggests the distance to the vertices is greater along the \( x \) axis. You can further refine your graph by locating the foci using the relationship \( c^2 = a^2 + b^2 \) to find the distance \( c \) from the center to each focus.

Polar Coordinates

Polar coordinates offer an alternative to the Cartesian system for defining positions of points. While the rectangular equation uses \( x \) and \( y \) coordinates, the polar coordinate system specifies location with a radius \( r \) and an angle \( \theta \), measured from the polar axis (equivalent to the positive \( x \) axis in Cartesian coordinates). Conversion between Cartesian and polar is done using the equations \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \).

This system is particularly useful for graphing equations like conic sections that have a clear radial symmetry or are centered at points other than the origin. For hyperbolas, parameterization can often simplify plotting since it can directly relate angle measures to points on the curve. Although hyperbolas are not as directly associated with polar coordinates as circles or ellipses, understanding how to move between coordinate systems is a valuable tool for students in higher mathematics.