Problem 21 - 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. $$\sqrt{3} x^{2}+3 x y=3$$

Step-by-Step Solution

Answer

The graph is a hyperbola, shown after rotating axes by 30° to eliminate the xy-term.

Step 1: Identify the Equation Type using the Discriminant

For the equation \[\sqrt{3}x^2 + 3xy = 3\]we first compare it with the general second-degree equation:\[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.\]Here, \(A = \sqrt{3}\), \(B = 3\), \(C = 0\), \(D = 0\), \(E = 0\), and \(F = -3\). The discriminant \(\Delta\) for conics is given by: \[\Delta = B^2 - 4AC.\] Substitute \(A\), \(B\), and \(C\) into the equation:\[\Delta = 3^2 - 4 \cdot \sqrt{3} \cdot 0 = 9.\]Since \(\Delta > 0\), the graph is a hyperbola.

Step 2: Rotation of Axes to Eliminate the xy-Term

To eliminate the \(xy\)-term, use the formula \(\tan(2\theta) = \frac{B}{A-C}\) to find the rotation angle. Substitute the values:\[\tan(2\theta) = \frac{3}{\sqrt{3} - 0} = \frac{3}{\sqrt{3}} = \sqrt{3}.\]The angle \(2\theta = 60^\circ\), hence, \(\theta = 30^\circ\).The transformation equations for rotation are:\[x = x'\cos\theta - y'\sin\theta\] \[y = x'\sin\theta + y'\cos\theta.\]With \(\theta = 30^\circ\), \(\cos\theta = \frac{\sqrt{3}}{2}\) and \(\sin\theta = \frac{1}{2}\). Substitute these into the equations to change variables, eliminating the \(xy\)-term.

Step 3: Simplifying Rotated Equation

Substituting from Step 2 transformations into the original equation, every instance of \(x\) and \(y\) is replaced with expressions involving \(x'\) and \(y'\). This results in new coefficients for an equation in \(x'\) and \(y'\). Solving will eliminate the \(x'y'\)-term, yielding an equation\(Ax'^2 + Cy'^2 + Dx' + Ey' + F = 0.\)Find the new values for \(A\) and \(C\), confirming the absence of an \(x'y'\) term.

Step 4: Sketching the Graph

The transformed equation can be converted into a more recognizable form, such as \[Ax'^2 - By'^2 = 1\].This confirms a hyperbola. Sketch the hyperbola with axes parallel to those adjusted by the rotation angle found in Step 2. The focus of the hyperbola will be along the major axis transposed through rotation.

Key Concepts

Discriminant

In mathematics, the discriminant is a useful tool for identifying the type of conic section represented by a second-degree polynomial. Conic sections include parabolas, ellipses, and hyperbolas.
To calculate the discriminant (\(\Delta\)) for a conic equation of the form \[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,\] the formula used is: \[\Delta = B^2 - 4AC.\]
The value of \(\Delta\) determines the nature of the conic:

• If \(\Delta = 0\), the conic is a parabola.
• If \(\Delta < 0\), it is an ellipse or a circle.
• If \(\Delta > 0\), it is a hyperbola.

In our case, with the equation \(\sqrt{3}x^2 + 3xy = 3\), the discriminant was calculated to be 9, confirming the graph is a hyperbola, as \(\Delta > 0\).

Rotation of Axes

The rotation of axes is a method used to simplify conic equations by removing the \(xy\)-term. This term makes equations complex, but fortunately, a simple rotation can often eliminate it.
To determine the angle of rotation (\(\theta\)), we use:\[\tan(2\theta) = \frac{B}{A-C}. \]After substitution, if \(\tan(2\theta) = \sqrt{3}\), then calculate \(2\theta\) using trig values, which gives us \(60^\circ\). So, \(\theta = 30^\circ\).
The coordinates are transformed using the following rotation equations:

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

This simplifies the equation by eliminating the \(xy\)-term, making it easier to work with and interpret.

Hyperbola Sketching

Sketching a hyperbola involves understanding its shape and orientation. With the equation rewritten from the rotated axes, such as \[Ax'^2 - By'^2 = 1,\] you can readily identify it as a hyperbola, characterized by its two distinct open curves.
To sketch, we need to identify the axes, vertices, and foci. Hyperbolas have two axes:

• The transverse axis, across which the vertices are located.
• The conjugate axis, perpendicular to the transverse.

With the rotation of axes previously conducted, ensure the hyperbola's orientation aligns with these new axes. The vertices will be equidistant from the center along the transverse axis. Use the standard form to guide you in plotting: the curves open away from each other on this axis. This understanding allows for accurate sketching, taking into account major transformations accomplished during rotation.