Problem 17 - Precalculus: Mathematical 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 \(x y\) -term. (c) Sketch the graph. $$x^{2}+2 \sqrt{3} x y-y^{2}+2=0$$

Step-by-Step Solution

Answer

The graph is a hyperbola, with rotation eliminating the xy-term.

Step 1: Identify the discriminant

To classify the conic section (parabola, ellipse, or hyperbola) based on its equation, we need to use the discriminant. For a general second degree equation of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the discriminant \(\Delta\) is given by \(\Delta = B^2 - 4AC\). In this equation, \(A = 1\), \(B = 2\sqrt{3}\), and \(C = -1\), so the discriminant \(\Delta = (2\sqrt{3})^2 - 4(1)(-1)\). Calculate this value.

Step 2: Calculate the discriminant

Calculate \((2\sqrt{3})^2\), which is \(12\), and \(-4 \times 1 \times -1\), which is \(4\). Therefore, the discriminant \(\Delta = 12 + 4 = 16\).

Step 3: Analyze the discriminant

Since \(\Delta = 16\) is greater than 0, the equation represents a hyperbola. For conic sections: if \(\Delta > 0\), it's a hyperbola; if \(\Delta = 0\), it's a parabola; and if \(\Delta < 0\), it's an ellipse.

Step 4: Use rotation to eliminate the \(xy\)-term

To eliminate the \(xy\)-term, we use the rotation of axes formulae. Let \(x = x'\cos(\theta) - y'\sin(\theta)\) and \(y = x'\sin(\theta) + y'\cos(\theta)\). The angle \(\theta\) is found using \(\tan(2\theta) = \frac{B}{A-C}\). Here, \(\tan(2\theta) = \frac{2\sqrt{3}}{1 + 1} = \sqrt{3}\), so \(2\theta = 60^\circ\) or \(\theta = 30^\circ\).

Step 5: Substitute new coordinates

Substitute \(\theta = 30^\circ\) into the rotation equations: \(x = x'\sqrt{3}/2 - y'/2\) and \(y = x'/2 + y'\sqrt{3}/2\). Replace these into the original equation \(x^2 + 2\sqrt{3}xy - y^2 + 2 = 0\) and simplify to remove the \(xy\) term. The goal is to express the equation in terms of \(x'\) and \(y'\) without the \(x'y'\) term.

Step 6: Simplify the rotated equation

After substitution and simplification, you'll have a new equation in the form \(Ax'^2 + By'^2 + F' = 0\), where the term involving \(x'y'\) is gone. This step typically involves algebraic simplification, which matches conventional expectations of rotating a hyperbola's axes.

Step 7: Sketch the graph

With the new equation without the \(xy\)-term, determine the appearance of the hyperbola. Identify the center, vertices, and asymptotes from the standard form of the hyperbola after rotation. Use these features to sketch the hyperbola, showing its characteristic "two branches" shape. For a standard hyperbola, the center is at the origin and the asymptotes help visualize its axes.

Key Concepts

Discriminant

The discriminant is a powerful tool for identifying the nature of conic sections in a quadratic equation of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). It helps determine whether the graph is a parabola, ellipse, or hyperbola. Here's how you use the discriminant:

• Calculate it using the formula \(\Delta = B^2 - 4AC\).
• If \(\Delta > 0\), the graph is a hyperbola.
• If \(\Delta = 0\), the graph is a parabola.
• If \(\Delta < 0\), the graph is an ellipse.

In our problem, the given quadratic equation is \(x^2+2 \sqrt{3} x y-y^2+2=0\) where \(A = 1\), \(B = 2\sqrt{3}\), and \(C = -1\). Calculating the discriminant, you find:\[\Delta = (2\sqrt{3})^2 - 4 \times 1 \times (-1) = 12 + 4 = 16\]Since \(\Delta = 16\), which is greater than zero, the equation represents a hyperbola.

Rotation of Axes

When dealing with conic sections, especially those with an \(xy\)-term, the rotation of axes can simplify the equation. This method removes the \(xy\)-term, making it easier to identify and sketch the conic.

• Start by finding the angle \(\theta\) using \(\tan(2\theta) = \frac{B}{A-C}\).
• For the problem equation, \(\tan(2\theta) = \frac{2\sqrt{3}}{1+1} = \sqrt{3}\), leading to \(2\theta = 60^\circ\) and \(\theta = 30^\circ\).

Next, use the new variables \(x'\) and \(y'\) in terms of the original coordinates to facilitate rotation:

• \(x = x'\cos(\theta) - y'\sin(\theta)\)
• \(y = x'\sin(\theta) + y'\cos(\theta)\)

Substituting \(\theta = 30^\circ\), we have:

• \(x = x'\cdot\sqrt{3}/2 - y'/2\)
• \(y = x'/2 + y'\sqrt{3}/2\)

Insert these into the original equation. Simplifying this expression, you should eliminate the \(xy\) term, leaving a cleaner conic section equation. The rotated axes transform the hyperbola into a more straightforward form to graph.

Hyperbola

A hyperbola is a type of conic section that appears as two mirrored, open curves, known for its intriguing properties and forms.Characteristics of a hyperbola include:

• It consists of two distinct branches.
• The standard equation of a hyperbola is \(\frac{x'^2}{a^2} - \frac{y'^2}{b^2} = 1\) or \(\frac{y'^2}{b^2} - \frac{x'^2}{a^2} = 1\), depending on its orientation (horizontal or vertical).
• It has asymptotes, lines that the branches approach but do not meet.

Finding the characteristics of a hyperbola:

• Identify the center, typically at the origin after rotation.
• Determine the vertices along the transverse axis, calculated using \(\pm a\) or \(\pm b\), depending on the axis of symmetry.
• The asymptotes can be found using the slopes \(\pm \frac{b}{a}\) or \(\pm \frac{a}{b}\).

For the given problem, once the equation is free of the \(xy\)-term after the axis rotation, graphing becomes more intuitive. You use the asymptotes and vertices to sketch the hyperbola's broad branches, symmetry, and bounds. This transforms the conic's mathematical complexity into a visually approachable graph.