Chapter 11: Problem 3
Graph the following equations. $$ 5 x^{2}+6 x y+5 y^{2}-4 \sqrt{2} x+4 \sqrt{2} y=0 $$
Short Answer
Expert verified
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 by step solution
01
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 \(5x^2 + 6xy + 5y^2 - 4\sqrt{2}x + 4\sqrt{2}y = 0\).
02
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.
03
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.
04
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.
05
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 \(5((x+\frac{\sqrt{2}}{2})^2 + (y-\frac{\sqrt{2}}{2})^2) = 8\).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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
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.
- Identify its center, which shifts the ellipse from the origin to somewhere else on the graph.
- Determine the length of the major and minor axes, as these will indicate the vertex points of the ellipse.
- Trace the ellipse using these points, ensuring that the shape is symmetric about both axes.
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: \
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: \
- Start with the generic form: \(x^2 + bx\).
- Add and subtract \(\left(\frac{b}{2}\right)^2\) within the equation, allowing it to be expressed as \((x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2\).
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:
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.