Question

Eliminate the cross-product term by a suitable rotation of axes and then, if necessary, translate axes (complete the squares) to put the equation in standard form. Finally, graph the equation showing the rotated axes. \(-\frac{1}{2} x^{2}+7 x y-\frac{1}{2} y^{2}-6 \sqrt{2} x-6 \sqrt{2} y=0\)

Short answer

After eliminating the cross-term and transformation, the equation represents two intersecting lines, resulting in a degenerate parabola.

Step-by-step solution

Identify Cross-Product Term

The given equation is \(-\frac{1}{2}x^2 + 7xy - \frac{1}{2}y^2 - 6\sqrt{2}x - 6\sqrt{2}y = 0\). Here, the cross-product term is \(7xy\).

Find Angle of Rotation

To eliminate the cross-product term, we find the angle \(\theta\) such that \( \tan(2\theta) = \frac{B}{A-C} \). Here, \(A = -\frac{1}{2}, B = 7, C = -\frac{1}{2}\). Thus, \( \tan(2\theta) = \frac{7}{0} = \infty \). Therefore, \(2\theta = \frac{\pi}{2} + k\pi\) and \(\theta = \frac{\pi}{4} + k\frac{\pi}{2}\). We choose \(\theta = \frac{\pi}{4}\) for rotation.

Apply Rotation Formulas

Substitute the rotation of axes formulas: \(x = x'\cos(\theta) - y'\sin(\theta)\) and \(y = x'\sin(\theta) + y'\cos(\theta)\) where \(\theta = \frac{\pi}{4}\) into the original equation. Simplifying, we substitute: \(x = \frac{x' - y'}{\sqrt{2}}\) and \(y = \frac{x' + y'}{\sqrt{2}}\).

Simplify the Equation

Substitute the values for \(x\) and \(y\) into the given equation and simplify. The terms involving \(x'y'\) should cancel out, leaving us with terms only involving \((x')^2\) and \((y')^2\). After simplifying, you obtain the rotated equation: \(-8(x')^2 + (y')^2 - 24(x')\sqrt{2} = 0\).

Complete the Square

Focus on the term \(-24(x')\sqrt{2}\). Complete the square to rewrite the equation. Factor the quadratic term and integrate it to form a perfect square: \(-8((x') + 1.5\sqrt{2})^2 + (y')^2 = 0\).

Translate Axes

Since we have completed the square with \((x') + 1.5\sqrt{2}\), translate the axes such that \(X = x' + 1.5\sqrt{2}\), which yields the equation: \(-(8)X^2 + (y')^2 = 0\).

Determine Standard Form

The equation \(-8X^2 + (Y')^2 = 0\) is now in the standard form of a conic section, specifically a parabolic form since the equation reflects a degenerate case (intersecting lines).

Graph the Equation

Plot the graph using the new rotated and translated axes. Since it's a degenerate parabola, the graph depicts two intersecting lines rather than a single curve.

Key concepts

Rotation of Axes

The rotation of axes is a powerful technique in algebra used to simplify equations of conic sections by eliminating cross-product terms. Imagine having an equation in the form of a conic section, such as ellipses, parabolas, or hyperbolas, but with the added complication of a term like \(xy\) (cross-product term). This term can make it difficult to understand or graph the equation in its standard form.

• To rotate the axes, we need to find an angle \( \theta \) which helps remove the cross-product term.
• The formula \( \tan(2\theta) = \frac{B}{A-C} \) is used, where \(A\), \(B\), and \(C\) are coefficients from the general quadratic equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\).
• By choosing the appropriate \(\theta\), often \(\frac{\pi}{4}\) in certain alignments, we replace \(x\) and \(y\) with new coordinates \(x'\) and \(y'\), calculated using \((x = x'\cos(\theta) - y'\sin(\theta))\) and \((y = x'\sin(\theta) + y'\cos(\theta))\).

This transformation helps to erase the \(xy\) term and offers a simpler, more interpretable equation format.

Completing the Square

Completing the square is a useful algebraic method to rewrite quadratic expressions so that they reveal the nature of the conic section more explicitly. It’s typically used after rotation, where we expect terms like \( (x')^2 \) or \( (y')^2 \) to remain in the equation. By fiddling with these terms, we can simplify the expression into a perfect square format, which often corresponds to the standard form of the conic section.Here’s how you can complete the square:

• Look for terms with variables like \((x')\) or \((y')\), especially the linear ones such as \(-24(x')\).
• Factor out any coefficients, if necessary, and adjust the terms inside the parentheses to make it a perfect square, like \(-8((x') + 1.5\sqrt{2})^2 \).
• This may also shift the graph, which usually leads to translating the axes – a simple adjustment involving the introduction of \(X = x' + 1.5\sqrt{2}\), thus aligning your equation in a neat standard parabolic or circular form.

This technique aids in visualizing the conic shapes better and positions you to graph them accurately.

Graphing Conics

Graphing conic sections involves sketching shapes like circles, ellipses, parabolas, or hyperbolas. After transforming the equation through rotation and completing the square, you arrive at a standard form which is much easier to graph.Once in standard form, plotting becomes straightforward:

• The equation \(-8X^2 + (y')^2 = 0\) represents a degenerate conic, which is a parabolic form where the 'curve' manifests as intersecting lines on the graph.
• Identifying that these are intersecting lines involves understanding that for each \(X\), there's a corresponding \(y'\), resulting in linear rulings rather than a closed shape.
• The graph thus vividly depicts the breakdown of the conic into simpler linear elements, highlighting the elegant geometry behind conic sections.

This practical approach turns what could be complex shapes into simple graphical nuances that beautifully illustrated how algebra translates into geometry.