Problem 21 - Algebra and Trigonometry

Question

\(15-28=(a)\) 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. $$ 21 x^{2}+10 \sqrt{3} x y+31 y^{2}=144 $$

Step-by-Step Solution

Answer

(a) The graph is an ellipse; (b) Use a 60° rotation to eliminate xy-term; (c) Sketch an ellipse.

Step 1: Identify the type of conic section using the discriminant

The equation of a conic section is given by the general second-degree equation: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]In this problem, we have: \( A = 21 \), \( B = 10\sqrt{3} \), \( C = 31 \).The discriminant \( \Delta \) is used to determine the type of conic and is given by: \[ \Delta = B^2 - 4AC \].Calculate \( \Delta \):\[ \Delta = (10\sqrt{3})^2 - 4(21)(31) \].\[ \Delta = 300 - 2604 = -2304 \].Since \( \Delta < 0 \), the equation represents an ellipse.

Step 2: Use a rotation of axes to eliminate the xy-term

To eliminate the \( xy \)-term, we need to find the angle \( \theta \) for rotation using the formula: \[ \tan(2\theta) = \frac{B}{A-C} \].Substitute the given values:\[ \tan(2\theta) = \frac{10\sqrt{3}}{21-31} = \frac{10\sqrt{3}}{-10} = -\sqrt{3} \].The angle \( 2\theta \) corresponding to \( -\sqrt{3} \) is \( 120^\circ \), which gives \( \theta = 60^\circ \).Use the formulas for rotation: \[ x = x'\cos(\theta) - y'\sin(\theta) \]\[ y = x'\sin(\theta) + y'\cos(\theta) \]Substitute \( \theta = 60^\circ \) to express \( x \) and \( y \) in terms of \( x' \) and \( y' \).

Step 3: Sketch the graph of the ellipse

After rotating the axes, rewrite the equation using the new coordinates \( x' \) and \( y' \). The equation simplifies, allowing for easier identification of standard form and principal axes.Identify the center, major, and minor axes of the ellipse from the simplified equation. Since the original equation was determined to form an ellipse, draw a standard ellipse using these axes centered at the origin, scaled proportionally according to the coefficients.

Key Concepts

Discriminant in Conic Sections

The discriminant is a special value that helps us identify the type of conic section represented by a given second-degree polynomial equation with terms up to the second degree. You will often see the general form as: \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \). In the context of conic sections, the discriminant \( \Delta \) can be calculated from: \[ \Delta = B^2 - 4AC \].For different conic sections, \( \Delta \) behaves differently:

• 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 example, with \( A = 21 \), \( B = 10\sqrt{3} \), and \( C = 31 \), we found that \( \Delta = -2304 \). Since \( \Delta < 0 \), it reveals that the equation represents an ellipse, a closed curve, where the sum of distances from any point on the curve to two fixed points (foci) is constant.

Ellipse Identification

Ellipses are one of the most intriguing shapes in conic sections, distinguishable by their oval shape which can have varying lengths for its principal axes. When identifying an ellipse from its equation, the squared terms \( x^2 \) and \( y^2 \) both have positive coefficients, and \( \Delta < 0 \) confirms it further. An ellipse can be oriented in various directions, but it is most simply understood in its standard form where its axes are aligned with the coordinate axes. To identify an ellipse's key features from its equation, one must:

• Convert the equation into the standard form: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \((h, k)\) is the center.
• Determine the lengths of the semi-major axis \( a \) and semi-minor axis \( b \).
• Identify the foci, which lie along the major axis, via the relation \( c^2 = a^2 - b^2 \).

These attributes allow you to sketch the ellipse accurately, identifying its width, height, and orientation.

Rotation of Axes in Conics

When a conic equation includes an \( xy \) term, it suggests that the conic is rotated with respect to the standard coordinate axes. To simplify the equation and complete an accurate sketch, a rotation transformation can eliminate this \( xy \) term. The rotation involves finding an angle \( \theta \) using:\[ \tan(2\theta) = \frac{B}{A-C} \]By substituting the necessary values, you can find the degree of rotation required. In our exercise, we computed \( \tan(2\theta) = -\sqrt{3} \), which results in \( \theta = 60^\circ \).The coordinates are transformed using:

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

This transformation aligns the major and minor axes of the ellipse with the new \( x' \) and \( y' \) axes, simplifying the equation further. It is a pivotal step to make sense of the equation geometrically and to draw the conic accurately by examining its central and major axes alignment.