Problem 17 - 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. $$x^{2}+2 \sqrt{3} x y-y^{2}+2=0$$

Step-by-Step Solution

Answer

It's a hyperbola; rotate by \(30^\circ\) to eliminate \(xy\)-term; sketch based on transformed equation.

Step 1: Identify the coefficients

The given equation is \(x^2 + 2 \sqrt{3} xy - y^2 + 2 = 0\). Identify the coefficients: \(A = 1\), \(B = 2\sqrt{3}\), and \(C = -1\). These coefficients will help us determine the type of conic section and assist in the rotation of axes.

Step 2: Determine the type of conic section using the discriminant

The discriminant for a general second-degree equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) is \(B^2 - 4AC\). Plug in the coefficients: \((2\sqrt{3})^2 - 4(1)(-1) = 12 + 4 = 16\). Since the discriminant is positive, the graph is a hyperbola.

Step 3: Determine the angle of rotation

To eliminate the \(xy\)-term from the equation, we use the rotation of axes. The angle \(\theta\) is given by \(\tan(2\theta) = \frac{B}{A-C}\). Here, \(\tan(2\theta) = \frac{2\sqrt{3}}{2} = \sqrt{3}\). This implies that \(2\theta = \frac{\pi}{3}\), so \(\theta = \frac{\pi}{6}\).

Step 4: Perform the rotation to eliminate the xy-term

Using \(x = x'\cos(\theta) - y'\sin(\theta)\) and \(y = x'\sin(\theta) + y'\cos(\theta)\), with \(\theta = \frac{\pi}{6}\), substitute these into the original equation. Simplify the equation to find the expression without the \(xy\)-term. The rotation results in a form \(Ax'^2 + Cy'^2 + F = 0\) where the \(xy\)-term is eliminated.

Step 5: Sketch the graph of the hyperbola

After the rotation, the new equation can be graphed as a standard hyperbola. Identify the asymptotes and the center of the hyperbola. Sketch the hyperbola based on these features, considering the rotated axes orientation.

Key Concepts

Discriminant

In the world of conic sections, the discriminant helps us figure out if we are dealing with a parabola, ellipse, or hyperbola. The discriminant is applied to the general quadratic equation: \[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]The formula for the discriminant is \(B^2 - 4AC\). What you do is plug in your known coefficients from the equation into this formula. This simple calculation tells you a lot:

• If the discriminant is zero, the graph is a parabola.
• If it is negative, you are looking at an ellipse (or a circle if \(A = C\)).
• If it is positive, like in the exercise we have (where \((2\sqrt{3})^2 - 4(1)(-1) = 16\)), then the graph represents a hyperbola.

So, by solving this one part, we learn a lot about the shape of our graph, allowing us to proceed with more targeted methods for graphing and transformation.

Rotation of Axes

The rotation of axes is a very useful technique when dealing with conic sections, especially if there's an \(xy\)-term in the equation. This term can make it harder to analyze and graph the conic section directly. By rotating the axes, the goal is to "eliminate" that pesky \(xy\)-term, transforming it into a more standard form.To rotate the axes, we decide on an angle \(\theta\). This angle is chosen so that it simplifies the equation. You calculate this angle using:\[\tan(2\theta) = \frac{B}{A-C}\]In our example, \(\tan(2\theta) = \sqrt{3}\), meaning \(2\theta = \frac{\pi}{3}\), so \(\theta = \frac{\pi}{6}\).What we then do is substitute the rotated axes equations:

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

into the original equation. This transforms the equation into one without an \(xy\)-term, allowing us to sketch it more easily. It's like simplifying a complex operation in a calculator so that you can view the result from a clearer perspective.

Hyperbola

Understanding the hyperbola graph is easier when we recognize its basic properties. A hyperbola is a type of conic section defined by an equation that, depending on its orientation, either opens vertically or horizontally.With the rotation transforming our equation into a simpler hyperbolic form, we can examine it more clearly. A hyperbola's general features include things like foci, asymptotes, and vertices. These features come from the new, standard form of the equation, such as:\[ \frac{x'^2}{a^2} - \frac{y'^2}{b^2} = 1 \] Here are some key characteristics:

• The center of the hyperbola is where the axes of symmetry cross.
• The asymptotes act as guidelines showing how the branches of the hyperbola open.
• The vertices are points on the graph where each branch is closest to the center.

After completing the rotation of axes to remove the \(xy\)-term, sketching the hyperbola becomes an exercise in plotting these defining features accurately. Visualizing the hyperbola this way, especially after simplifying the rotation, helps consolidate our understanding of its structure on this rotated plane.