Problem 25 - A History of Mathematics: An Introduction

Question

Show that de Witt's equation $$ y^{2}+\frac{2 b x y}{a}+2 c y=\frac{f x^{2}}{a}+e x+d $$ represents a hyperbola. (Use the substitution \(z=y+\frac{b}{a} x+\) \(c\) and show that this substitution, when combined with a substitution of the form \(x^{\prime}=\beta x\), converts the original oblique \(x-y\) coordinate system into a new \(x^{\prime}-z\) coordinate system based on perpendicular axes.) Sketch the curve.

Step-by-Step Solution

Answer

Question: Show that de Witt's equation, which is given by $$ \left(y + \frac{b}{a}x + c\right)^2 + \frac{2bx}{a}\left(y + \frac{b}{a}x + c\right) + 2cy = \frac{fx^2}{a} + ex + d $$ represents a hyperbola when the given substitution (mentioned in the answer) is used. Also, sketch the curve. Answer: The given equation represents a hyperbola when the substitutions \(z = y + \frac{b}{a}x + c\) and \(x' = \beta x\) are used. After using these substitutions, the equation is converted into the standard form of a hyperbola: $$ \frac{x^{\prime 2}}{A^2} - \frac{z^2}{B^2} = 1 $$ where \(A^2 = \frac{a^2}{f}\) and \(B^2 = d - c^2\). To sketch the curve, find the vertices (\(\pm A\), \(0\)), the foci (\(\pm\sqrt{A^2 + B^2}\), \(0\)), and asymptotes given by the equations \(z=\pm\frac{B}{A}x^\prime\). The curve will be open in the positive and negative \(x^\prime\) direction because \(\frac{x^{\prime 2}}{A^2}\) has a positive coefficient.

Step 1: Apply the given substitution \(z=y+\frac{b}{a} x+c\)

Replace y with \(z - \frac{b}{a}x - c\) in de Witt's equation: $$ \left(z - \frac{b}{a}x - c\right)^2 + \frac{2bx}{a}\left(z - \frac{b}{a}x - c\right) + 2cy = \frac{fx^2}{a} + ex + d $$

Step 2: Simplify the equation

Expand and simplify the equation: $$ z^2 - 2z\left(\frac{b}{a}x + c\right) + \left(\frac{b}{a}x + c\right)^2 + 2bz - 2b^2x - 2bc + 2cz = \frac{fx^2}{a} + ex + d $$ We can rewrite this equation as: $$ z^2 - 2z\left(\frac{b}{a}x + c\right) + \frac{b^2}{a^2}x^2 + 2\frac{bc}{a}x + c^2 + 2bz - 2b^2x - 2bc + 2cz = \frac{fx^2}{a} + ex + d $$ Now, combine like terms: $$ \left(\frac{b^2}{a^2}-\frac{f}{a}+f\right)x^2 + \left(-2\frac{b}{a}+e\right)x + d - c^2 = z^2 - 2\left(\frac{b}{a}x + c\right)z + 2bz + 2cz $$

Step 3: Use the substitution \(x^\prime = \beta x\) and find the \(\beta\) value that makes the equation a hyperbola standard form

Replace \(x\) with \(\frac{x^\prime}{\beta}\) in the equation: $$ \left(\frac{b^2}{a^2}-\frac{f}{a}+f\right)\left(\frac{x^\prime}{\beta}\right)^2 + \left(-2\frac{b}{a}+e\right)\frac{x^\prime}{\beta} + d - c^2 = z^2 - 2\left(\frac{b}{a}\frac{x^\prime}{\beta} + c\right)z + 2b\frac{x^\prime}{\beta} + 2c\frac{x^\prime}{\beta} $$ We want the coefficient of \(z^2\) to be 1 and the coefficient of \({x^\prime}^2\) to be negative to represent a hyperbola. From the above equation, we get $$ \beta^2 = \frac{a^2}{f} $$

Step 4: Calculate the new coordinates using the substitution and make the equation represent the standard form

Now, we can write our equation in the standard form of a hyperbola: $$ \frac{a^2}{f}x^{\prime 2} - z^2 = d - c^2 $$ This equation is in the form: $$ \frac{x^{\prime 2}}{A^2} - \frac{z^2}{B^2} = 1 $$ where \(A^2 = \frac{a^2}{f}\) and \(B^2 = d - c^2\).

Step 5: Sketch the curve

To sketch the hyperbola, we need to find the vertices (\(\pm A\), \(0\)), the foci (in this case, since the axes are perpendicular, the foci can be found at coordinates (\(\pm\sqrt{A^2 + B^2}\), \(0\))), and asymptotes which are given by the equations \(z=\pm\frac{B}{A}x^\prime\). The curve will be open in the positive and negative \(x^\prime\) direction because \(\frac{x^{\prime 2}}{A^2}\) has a positive coefficient.

Key Concepts

Hyperbola

A hyperbola is one of the four well-known conic sections, the others being the circle, ellipse, and parabola. It is formed by the intersection of a double cone and a plane that cuts both halves of the cone. This creates two separate, open curves.

In general, a hyperbola is defined by the equation in the form \[ \frac{x^2}{A^2} - \frac{y^2}{B^2} = 1 \]where \( A \) and \( B \) are real numbers that determine the hyperbola's size and orientation. In this exercise, we have an equation that is transformed into this general form to prove it represents a hyperbola.

Hyperbolas have the following features:

• Vertices: The points on the hyperbola closest to or farthest from the center, respectively.
• Foci: Two fixed points located inside each curve which define its shape and size.
• Asymptotes: Lines that the hyperbola approaches but never quite reaches, showing the hyperbola's "direction" or "orientation." These are given by \( y = \pm \frac{B}{A}x \).

The main task involves algebraically manipulating the given equation to confirm these features align with a hyperbola.

Coordinate Transformation

Coordinate Transformation is a method used to simplify complex equations by moving from one coordinate system to another. In this exercise, we employ this method to convert an oblique coordinate system into a standard rectangular form using two separate transformations.

The first transformation is:\[z = y + \frac{b}{a}x + c\]This substitution helps transform the curve defined in the \(x-y\) plane into one involving the new variable \(z\), which simplifies further analysis.

The second transformation involves:\[x^\prime = \beta x\]By changing \(x\) to \(x^\prime\), we eliminate obliqueness and clarify the hyperbola's orientation. In the exercise, the specific value of \(\beta\) is determined such that the transformed equation fits the standard form of a hyperbola.

Through these substitutions, the problem transitions from a more complex oblique system to a simpler perpendicular coordinate system, where the standard form of conic sections is more easily identified.

Algebraic Manipulation

In mathematics, algebraic manipulation refers to the rearrangement and simplification of algebraic expressions to make equations easier to work with or solve. This concept is essential to transforming the given equation into the standard form of a hyperbola.

The initial step involves substituting and expanding the equation:\[\left(z - \frac{b}{a}x - c\right)^2 + \frac{2bx}{a}\left(z - \frac{b}{a}x - c\right) + 2cy = \frac{fx^2}{a} + ex + d\]By substituting \(y\) with the expression defined by \(z\), the equation becomes a function of \(z\) and \(x\). Careful expansion and recombination of terms then follow.

The strategy used here involves:

• Expanding and collecting like terms: This helps in identifying components such as squares and constants, which must form parts of the hyperbola's standard equation.
• Applying transformations: replaces variables to shift from an oblique to a simpler standard form.

Once the equation is simplified, by adjusting \(\beta \), it can mirror the hyperbola's equation with specific ratios (\[\frac{x^\prime 2}{A^2}\] and \[\frac{z^2}{B^2}\] ). Such algebraic manipulation is crucial for recognizing the underlying symmetry and ensures that the overall geometric relationship remains intact, confirming the hyperbolic nature of the equation.