Problem 24 - The Calculus with Analytic Geometry

Question

In Exercises 7 through 28 , draw a sketch of the graph of the equation. $$ 3 x^{2}-13 x y-10 y^{2}=0 $$

Step-by-Step Solution

Answer

Identify the conic section as a hyperbola, then rewrite, diagonalize, and sketch the graph considering the diagonal form.

Step 1: Identify the type of conic section

The given equation is quadratic and has both \(x^2\) and \(y^2\) terms. To identify the type of conic section, check the discriminant (calculated as \(B^2 - 4AC\)). Rewrite the equation in standard form: \(Ax^2 + Bxy + Cy^2 = 0\). Here, \(A = 3\), \(B = -13\), and \(C = -10\). So, the discriminant is \(B^2 - 4AC = (-13)^2 - 4(3)(-10) = 169 + 120 = 289\). Since the discriminant is positive, the conic section is a hyperbola.

Step 2: Rewrite the equation in matrix form

Rewrite the quadratic equation \(3x^2 - 13xy - 10y^2 = 0\) in a matrix form using the quadratic form \([x \, y] \begin{bmatrix} A & B/2 \ B/2 & C \end{bmatrix} [x \, y]^T = 0\). Here, \([x \, y] \begin{bmatrix} 3 & -6.5 \ -6.5 & -10 \end{bmatrix} [x \, y]^T = 0\).

Step 3: Diagonalize the matrix

Find the eigenvalues and eigenvectors of the coefficient matrix to diagonalize it. This simplifies the equation, showing how the hyperbola is oriented. Let \(D\) be the diagonal matrix consisting of eigenvalues, and \(P\) the matrix of eigenvectors, we can then use \(P^{-1}AP = D\). This transformation rotates the axes so the conic can be graphed more easily.

Step 4: Sketch the graph

Using the diagonalized form, the equation of the hyperbola can be rewritten and its principal axes identified. Sketching the hyperbola, note that the asymptotes can be found from the transformed equation and direct the direction of the branches. For this case, since the hyperbola is rotated, consider how the rotation affects the standard hyperbola shape. Simplistically, trace the graph considering major features such as vertices, center, and asymptotes.

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a double-napped cone. The type of curve formed depends on the angle of the plane relative to the cone. There are four types of conic sections:

• Circle
• Ellipse
• Parabola
• Hyperbola

In our exercise, we encountered a hyperbola, identifiable by the positive discriminant.

Discriminant

The discriminant helps determine the type of conic section represented by the quadratic equation. For a general quadratic equation in the form \(Ax^2 + Bxy + Cy^2 = 0\), the discriminant (\(B^2 - 4AC\)) offers telltale signs:

• Negative Discriminant: Represents an ellipse or circle.
• Zero Discriminant: Represents a parabola.
• Positive Discriminant: Represents a hyperbola.

In our problem, the discriminant was positive, identifying the conic section as a hyperbola.

Matrix Diagonalization

Matrix diagonalization simplifies complex quadratic equations by changing the coordinate system. For the quadratic form \([x \, y] \begin{bmatrix} A & B/2 \ B/2 & C \ \, \, \, \, \, \, [x \, y]^T = 0\)\, diagonalization helps by finding eigenvalues and eigenvectors of the matrix. The matrix is denoted as \([3 & -6.5 \ -6.5 & -10]\). By diagonalizing, we transform this matrix to a simpler form, \(D\), making it easier to draw the hyperbola along its principal axes.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental in transforming a matrix to its diagonal form. The eigenvalues represent the scaling factors while eigenvectors define the new directions. Given the matrix \([3 & -6.5 \ -6.5 & -10]\), we solve the characteristic polynomial to find:

• Eigenvalues \(\lambda_1\) and \(\lambda_2\)
• Eigenvectors \(\mathbf{v_1}\) and \(\mathbf{v_2}\)

With these, we construct matrices \(P\) and \(D\):

• \(P\) consisting of eigenvectors
• \(D\) consisting of eigenvalues along its diagonal

Through \(P^{-1}AP = D\), the quadratic form is simplified.

Asymptotes and Vertices

Asymptotes and vertices are crucial for sketching hyperbolas. Asymptotes are imaginary lines guiding the shape and direction of the hyperbola, intersecting at the hyperbola's center. Vertices are the closest points on either branch. To find these for diagonalized matrices:

• Rewrite the equation in the standard hyperbola form after diagonalization.
• Identify asymptotes by setting the diagonal terms equal to zero.
• Locate vertices by determining the intersection points of the branches using the equation.

This method offers a visual blueprint for accurately drawing the hyperbola.