Problem 3 - College Trigonometry

Question

Graph the following equations. $$ 5 x^{2}+6 x y+5 y^{2}-4 \sqrt{2} x+4 \sqrt{2} y=0 $$

Step-by-Step Solution

Answer

The equation represents an ellipse centered at (\(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\)), rotated 45 degrees with respect to the x-axis.

Step 1: Identify the Form of the Equation

The given equation is a second-degree polynomial, specifically a \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). The equation is \(5x^2 + 6xy + 5y^2 - 4\sqrt{2}x + 4\sqrt{2}y = 0\).

Step 2: Coefficient Identification

Extract the coefficients from the equation: \(A = 5\), \(B = 6\), \(C = 5\), \(D = -4\sqrt{2}\), \(E = 4\sqrt{2}\), and \(F = 0\). This will help determine the type of conic section.

Step 3: Determine the Type of Conic Section

Compute the discriminant \(B^2 - 4AC\) using the known coefficients. Here, \(B^2 = 36\), \(4AC = 100\). So, the discriminant is \(B^2 - 4AC = 36 - 100 = -64\). A negative discriminant indicates that the conic is an ellipse.

Step 4: Complete the Square

To graph the equation, transform it by completing the square. Begin by rewriting the equation as \(5(x^2 + \frac{6}{5}xy + y^2) - 4\sqrt{2}x + 4\sqrt{2}y = 0\). Completing the square involves rearranging terms to form perfect squares.

Step 5: Rearrange and Simplify

Group terms to complete the square: \((x^2 + y^2) + \frac{6}{5}xy\). Use the identity: \((x + y)^2 = x^2 + 2xy + y^2\). Modify terms such that they can be expressed as \(5((x+\frac{\sqrt{2}}{2})^2 + (y-\frac{\sqrt{2}}{2})^2) = 8\).

Step 6: Graph the Ellipse

The equation after simplification is equivalent to an ellipse centered at \((\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) with axes aligned along lines x = y and x = -y, in the rotated coordinate system. Plot this ellipse with the calculated center and note the symmetric properties.

Key Concepts

Graphing Ellipses

Graphing ellipses can initially seem overwhelming, but with a structured approach, it becomes much more manageable. An ellipse is one of the conic sections, a special type of curve on a plane. It appears like an elongated circle or an oval shape. When we graph ellipses, the aim is to visualize how this specific mathematical equation translates into a geometric figure.

Ellipses are primarily defined by their standard form equation: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where

• \((h, k)\) are the coordinates of the ellipse's center.
• \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes respectively.

To graph an ellipse, you should:

1. Identify its center, which shifts the ellipse from the origin to somewhere else on the graph.
2. Determine the length of the major and minor axes, as these will indicate the vertex points of the ellipse.
3. Trace the ellipse using these points, ensuring that the shape is symmetric about both axes.

In the given problem, you will need to first simplify the equation, possibly by completing the square and then determining how the coefficients translate into the geometry of an ellipse. Plotting these characteristics will help create an accurate graph.

Complete the Square

Completing the square is a fundamental technique in algebra that makes complex equations easier to handle. Here, the main idea is to transform a quadratic equation into a form where you can more clearly discern certain geometric properties, such as the center or radius of a conic section.

The process involves rearranging and modifying the given equation. For example, given the terms \(x^2 + bx\), completing the square would involve constructing a perfect square trinomial: \

1. Start with the generic form: \(x^2 + bx\).
2. Add and subtract \(\left(\frac{b}{2}\right)^2\) within the equation, allowing it to be expressed as \((x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2\).

After applying this technique in the original equation, the resulting form will show clear perfect squares which make it easier to graph the geometry of the conic section. In our instance, it aids in isolating terms of an ellipse form. Completing the square realigns the equation into a recognizable ellipse-centered framework.

Discriminant of Quadratic Equations

The discriminant is a critical value in understanding the nature and properties of quadratic equations. It is particularly useful in identifying the type of conic section represented by an equation.

For a general conic equation of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the discriminant is calculated as:\[B^2 - 4AC\]The discriminant helps in classifying conic sections:

• If the discriminant is greater than zero, the conic is a hyperbola.
• If it equals zero, the equation represents a parabola.
• If it is less than zero, the equation indicates an ellipse.

In the equation at hand, the discriminant value is \(-64\), signifying that the conic section is an ellipse. This nifty tool helps decide the approach we take in further simplifying and graphing the equation by providing foresight into what geometric shape we should expect.