Problem 18 - Precalculus: Mathematical 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 \(x y\) -term. (c) Sketch the graph. $$21 x^{2}+10 \sqrt{3} x y+31 y^{2}=144$$

Step-by-Step Solution

Answer

(a) Ellipse; (b) Use a rotation of 75° to eliminate the xy-term. (c) Sketch an ellipse based on new axes.

Step 1: Identify the Discriminant

The general form of a conic section is \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). Here, \(A = 21\), \(B = 10\sqrt{3}\), and \(C = 31\). The discriminant \(\Delta\) for a conic section is given by \(\Delta = B^2 - 4AC\). Calculate \(\Delta\):\[ \Delta = (10\sqrt{3})^2 - 4 \cdot 21 \cdot 31 = 300 - 2604 = -2304 \]Since \(\Delta > 0\) indicates a hyperbola, \(\Delta = 0\) indicates a parabola, and \(\Delta < 0\) indicates an ellipse, in this case, \(\Delta < 0\), so the graph is an **ellipse**.

Step 2: Find the Rotation Angle

To eliminate the \(xy\)-term by rotating the axes, we use the formula \(\cot(2\theta) = \frac{A-C}{B}\). Here, \(A = 21\), \(B = 10\sqrt{3}\), and \(C = 31\). Calculate:\[ \cot(2\theta) = \frac{21 - 31}{10\sqrt{3}} = \frac{-10}{10\sqrt{3}} = \frac{-1}{\sqrt{3}} \]This angle corresponds to \(2\theta = 150^\circ\), so \(\theta = 75^\circ\). Compute \(\sin(\theta)\) and \(\cos(\theta)\) and use them for further calculations to rotate the axes.

Step 3: Use the Rotation Matrix

To eliminate the \(xy\)-term, use the rotation of the coordinate system by angle \(\theta\). The rotation formulas are \(x = x'\cos(\theta) - y'\sin(\theta)\) and \(y = x'\sin(\theta) + y'\cos(\theta)\). Substitute these into the original equation and simplify to get the new equation without the \(xy\)-term. This involves computing new coefficients using:\[ A' = A\cos^2(\theta) + B\cos(\theta)\sin(\theta) + C\sin^2(\theta) \] \[ C' = A\sin^2(\theta) - B\cos(\theta)\sin(\theta) + C\cos^2(\theta) \]

Step 4: Sketch the Graph

After transforming the equation, plot the ellipse using the major and minor axes derived from the canonical form. An ellipse can be plotted by identifying its center and using its axes length obtained from the transformed equation. The axes' length and position will adjust based on the transformation.

Key Concepts

Discriminant of Conic Sections

To determine the type of conic section represented by a given quadratic equation, we use the discriminant. The general form of a conic section is given by:

• \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\)

The discriminant \(\Delta\) for conic sections is calculated using the formula:

• \(\Delta = B^2 - 4AC\)

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By evaluating the discriminant:

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

In the given problem, after calculating \(\Delta\), we found it to be \(-2304\), meaning it's less than zero. Thus, the graph is an **ellipse**. Recognizing this is a key step in understanding the basic nature of conic sections.

Rotation of Axes

To eliminate the \(xy\)-term from a conic's equation, we rotate the coordinate axes. This simplifies the equation and makes it easier to identify the conic section. The formula used to find the rotation angle \(\theta\) is given by:

• \(\cot(2\theta) = \frac{A - C}{B}\)

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Once \(\theta\) is known, we use trigonometric identities to calculate \(\sin(\theta)\) and \(\cos(\theta)\). These are employed in the rotation matrix:

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

The coefficients of the new equations are computed by substituting these into the original equation, allowing us to eliminate the \(xy\)-term. This transformation is crucial for simplifying conic sections without direct symmetry along the axes.

Graphing Ellipses

Graphing an ellipse involves understanding its basic structure. An ellipse is defined by its major and minor axes, which are perpendicular.

• The longest axis is called the major axis.
• The shortest one is the minor axis.

In this context, after the axes have been rotated, what used to be the midpoint center of the ellipse remains unchanged, though the orientation of the axes will affect its graph.
Utilize the new equation from the rotation step to find these axes. The lengths are obtained from the coefficients \(A'\) and \(C'\) of the transformed equation. The graph can then be sketched by marking the center and drawing out the ellipse along the major and minor axes, adjusting for any changes introduced by the rotational transformation.
Especially important is ensuring the axes' lengths derived from the semi-major and semi-minor axes are correctly scaled. This gives a clear depiction of the ellipse's true orientation and size.