Problem 16 - Precalculus: Mathematics for Calculus

Question

(a) Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola. (b) Use a rotation of axes to eliminate the xy-term. (c) Sketch the graph. $$x y+4=0$$

Step-by-Step Solution

Answer

The graph is a hyperbola; use rotation to derive \\(x'^2 - y'^2 = -8\\). Sketched as a hyperbola.

Step 1: Identify the Equation Type

The given equation is \(xy + 4 = 0\). To determine the type of conic section, we'll first write this in the standard second-degree form: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). For \(xy + 4 = 0\), we see \(A = 0, B = 1, C = 0, D = 0, E = 0, F = 4\). The discriminant \( ext{\Delta}\) is given by \( ext{\Delta} = B^2 - 4AC\). Substitute the values to find \( ext{\Delta}\).

Step 2: Calculate the Discriminant

Substitute \(A = 0\), \(B = 1\), \(C = 0\) into the discriminant formula: \(\text{\Delta} = B^2 - 4AC = 1^2 - 4(0)(0) = 1\). Since \(\text{\Delta} > 0\), this indicates a hyperbola.

Step 3: Find Rotation Angle to Eliminate xy-Term

For hyperbolas and other conic sections with xy-terms, we use a rotation to eliminate the xy-term. The angle \(\theta\) is found using \(\cot(2\theta) = \frac{A-C}{B} = \frac{0-0}{1} = 0\), leading to \(2\theta = 90^\circ\) or \(270^\circ\). Therefore, \(\theta = 45^\circ\) or \(135^\circ\).

Step 4: Apply the Rotation of Axes

Using the rotation formulas, substitute \(x = x'\cos\theta - y'\sin\theta\), \(y = x'\sin\theta + y'\cos\theta\), and simplify. For \(\theta = 45^\circ\), the rotation equations are \(x' = \frac{x-y}{\sqrt{2}}\), \(y' = \frac{x+y}{\sqrt{2}}\). Substitute these into the original equation and simplify.

Step 5: Derive Transformed Equation

Substitute \(x = \frac{x'-y'}{\sqrt{2}}\) and \(y = \frac{x'+y'}{\sqrt{2}}\) into \(xy + 4 = 0\). After simplification, we have \(x'^2 - y'^2 = -8\), which is a standard form of a hyperbola.

Step 6: Sketch the Graph

Now that we have the simpler form \(x'^2 - y'^2 = -8\), we identify this as a hyperbola. Sketching involves recognizing the symmetry about the axes because the equation in standard form suggests hyperbolas open along x-axis and y-axis. Since it derived from the rotated coordinates, adjust the standard hyperbola graph to align with the axes rotated by \(45^\circ\).

Key Concepts

Discriminant

In the world of conic sections, the discriminant plays a crucial role in identifying the type of conic section represented by a quadratic equation. Conic sections include parabolas, ellipses, and hyperbolas.
To find the discriminant (\(\Delta\)) of a conic, we use the standard quadratic form: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. \]

• The discriminant formula is:\[ \Delta = B^2 - 4AC. \]
• If \(\Delta = 0\), the equation represents a parabola.
• If \(\Delta > 0\), it signifies a hyperbola.
• If \(\Delta < 0\), the graph is an ellipse.

In the exercise, with the equation \(xy + 4 = 0\), the coefficients are \(A = 0\), \(B = 1\), \(C = 0\), so we substitute these into the formula to get \(\Delta = 1^2 - 4(0)(0) = 1\). Since \(\Delta > 0\), this indicates that the graph is a hyperbola.

Hyperbola

A hyperbola is one of the fascinating shapes found in the family of conic sections. It consists of two symmetrical, open curves. The equation given in the exercise: \( xy + 4 = 0 \), once simplified, represents a hyperbola.
Hyperbolas have distinct features:

• They are formed when the plane cuts through both nappe (or cones) of the double cone.
• The equation of a hyperbola in its standard form is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( - \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
• The hyperbola has two branches, each opening indefinitely.
• Its centering symmetry is usually around the origin or another specific point after translation.
• It has asymptotes that guide its shape but the graph does not actually touch these lines.

For this exercise, after eliminating the \(xy\)-term through rotation, the hyperbola's standard form helps us interpret and sketch it accurately.

Rotation of Axes

The rotation of axes is a useful technique when dealing with conic sections that contain an \(xy\)-term. This term can make the conic section difficult to interpret and graph, necessitating a transformation. By rotating the axes, we eliminate the \(xy\)-term and simplify the equation for better understanding.
Here's how rotation works:

• To find the rotation angle \(\theta\), use \(\cot(2\theta) = \frac{A-C}{B}\).
• For our exercise where \(A = 0\), \(C = 0\), \(B = 1\), the solution gives \(2\theta = 90^\circ\), making \(\theta = 45^\circ\).
• Substitute into rotation formulas: \(x = x'\cos\theta - y'\sin\theta\) and \(y = x'\sin\theta + y'\cos\theta\).
• With \(\theta = 45^\circ\), we use: \(x' = \frac{x-y}{\sqrt{2}}\) and \(y' = \frac{x+y}{\sqrt{2}}\).

This transformation makes the original equation \(xy + 4 = 0\) easier to analyze, converting it to \(x'^2 - y'^2 = -8\), a recognizable form of a hyperbola.