Problem 20 - 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. $$21 x^{2}+10 \sqrt{3} x y+31 y^{2}=144$$

Step-by-Step Solution

Answer

The equation represents an ellipse; rotate axes by 165° to remove the \(xy\)-term, then sketch the ellipse.

Step 1: Identify the equation form

The given equation is similar to the general form for conic sections: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0. \] In this case, we have \( A = 21 \), \( B = 10\sqrt{3} \), \( C = 31 \), and the other coefficients are zero.

Step 2: Determine the discriminant

The discriminant \( \Delta \) for conic sections is given by \( \Delta = B^2 - 4AC \). Substitute the values to calculate \( \Delta \): \[ \Delta = (10\sqrt{3})^2 - 4(21)(31) = 300 - 2604 = -2304. \] Since \( \Delta < 0 \), the graph of the equation is an ellipse.

Step 3: Find the rotation angle

To eliminate the \(xy\)-term, we use the rotation formula \( \cot(2\theta) = \frac{A-C}{B} \). Substituting the values, \( \cot(2\theta) = \frac{21-31}{10\sqrt{3}} = \frac{-10}{10\sqrt{3}} = \frac{-1}{\sqrt{3}} \). Solve for \( \theta \) using \( \cot(2\theta) = \frac{-1}{\sqrt{3}} \), which gives \( 2\theta = 150^\circ \) or \( -30^\circ \) (selecting appropriate angle gives \( \theta = -15^\circ \) or \( 165^\circ \)).

Step 4: Rotate the axes

Using \( \theta = 165^\circ \) for transformation: \( x = x'\cos\theta - y'\sin\theta \) and \( y = x'\sin\theta + y'\cos\theta \). Substitute \( \cos 165^\circ = -\cos 15^\circ \) and \( \sin 165^\circ = \sin 15^\circ \). Use trigonometric identities to update the equation.

Step 5: Simplify the new equation

After substitution and simplification, the new equation in terms of \( x' \) and \( y' \) will not have the \( x'y' \)-term, simplifying it closer to the standard form of an ellipse. Determine the new \( A', C' \) values to show it is an ellipse.

Step 6: Sketch the graph

The graph is an ellipse aligned with the new axes. Plot the ellipse using the transformed equation and mark axes after rotation based on calculated angle.

Key Concepts

Discriminant in Conics

The discriminant in conic sections is a crucial tool for identifying the type of conic represented by a quadratic equation. Specifically, for conic sections, the discriminant is given by the formula:\[\Delta = B^2 - 4AC\]where \( A, B, \) and \( C \) are coefficients from the general equation \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \). By evaluating \( \Delta \), we can determine the nature of the conic:

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

In our exercise, after calculating \( \Delta \), we find it to be \(-2304\), which is less than zero. This negative value confirms that the equation describes an ellipse.

Rotation of Axes

To simplify the graphing of conics, particularly when an \( xy \)-term is present in a conic equation, we often use a technique called rotation of axes. This method eliminates the \( xy \)-term by aligning the axes with the conic's principal axes.The angle of rotation \( \theta \) can be determined using the formula:\[\cot(2\theta) = \frac{A-C}{B}\]Once \( \theta \) is found, the equations for the new rotated axes, \( x' \) and \( y' \), become:

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

Selecting \( \theta = 165^\circ \) in our exercise, corresponding trigonometric values are substituted, and the transformation simplifies the original equation to one without the \( xy \)-term. This transformation helps in identifying the conic as an ellipse more efficiently.

Graphing Ellipses

An ellipse is a smooth, closed curve that resembles an elongated circle. Graphing one requires knowledge of its standard form equation, which is derived from the original equation after the rotation of axes eliminates any \( xy \)-terms.For an ellipse, the standard equation is generally written as:\[\frac{(x' - h')^2}{a^2} + \frac{(y' - k')^2}{b^2} = 1\]where \((h', k')\) is the center, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.When sketching the graph of an ellipse:

• First, identify the center using any \((h', k')\) shifts present.
• Determine \(a\) and \(b\) to understand the dilation and orientation of the ellipse.
• Sketch using these dimensions, ensuring to keep the aspect ratio consistent with \(a\) and \(b\).

For the ellipse in this exercise, plotting involves axis transformations and realigning the graph along the new axes to represent the ellipse correctly.