Problem 19 - Precalculus with Limits

Question

In Exercises 13-26, rotate the axes to eliminate the \(xy\)-term in the equation. Then write the equation in standard form. Sketch the graph of the resulting equation, showing both sets of axes. \(5x^2 -6xy+5y^2- 12 = 0\)

Step-by-Step Solution

Answer

The equation represents a circle with a radius of \(\sqrt{\frac{12}{5}}\), centered at the origin. Sketching this on the graph shows a circle at the origin with a radius of \(\sqrt{\frac{12}{5}}\).

Step 1: Identifying the Rotation Angle

The formula to find the rotation angle (\(\theta\)) is \(\theta = \frac{1}{2} \arctan \frac{2B}{A-C}\), where \(A\), \(B\), and \(C\) are coefficients of the quadratic equation. In this case, \(A = 5\), \(B = -3\), and \(C = 5\). After substituting these into the rotation formula, you would get \(\theta = \frac{1}{2} \arctan \frac{2(-3)}{5-5} = 0\). This result indicates that no rotation is needed because the angle of rotation is zero.

Step 2: Rewriting the Equation into Standard Form

Since no rotation is needed as per the previous step, the original equation can be rewritten into standard form as follows: \(5x^2 -6xy+5y^2- 12 = 0\) becomes \(5x^2+5y^2 = 12\), by moving the \(12\) to the other side of the equation.

Step 3: Determining the Type of Conic

The standard form equation \(5x^2+5y^2 = 12\) specifies a circle, as it fits the standard form of a circle equation \(x^2 + y^2 = r^2\). The radius (\(r\)) of the circle can be found by dividing \(12\) by \(5\) and taking the square root, which yields \(\sqrt{\frac{12}{5}}\).

Step 4: Sketching the Conic

The equation describes a circle centered at the origin with a radius of \(\sqrt{\frac{12}{5}}\). Sketching the conic involves drawing a circle with the calculated radius at the origin. Because no rotation occurred, the original axes and the new axes coincide.

Key Concepts

Rotation of Axes

When faced with a quadratic equation containing an xy-term, such as in the exercise 5x^2 -6xy+5y^2- 12 = 0t, one strategy to simplify it is through the rotation of axes. This mathematical operation involves changing the orientation of the x and y-axes to eliminate the xy-term and simplify the equation. The angle \(\theta\)t for rotation is determined by the formula \(\theta = \frac{1}{2} \arctan \frac{2B}{A-C}\)nt, which hinges on the coefficients A, B, and C from the general form Ax^2 + Bxy + Cy^2nt.
Conveniently, in some cases like the one in our problem, the calculation might suggest that no rotation is necessary, as the angle \(\theta\)t is zero. This indicates that the conic section described by the equation is already aligned with the axes, and therefore, the xy-term does not need to be eliminated through rotation.

Standard Form of a Quadratic Equation

The 'standard form' of a quadratic equation is a way of writing the equation so that it's easily recognizable and can be used to identify the shape and position of the conic section it represents. For equations like 5x^2 -6xy+5y^2- 12 = 0t, the aim is to rearrange the equation to eliminate the xy-term and present it in a cleaner, more digestible format. The standard form highlights the key features of the conic, such as the center, size, and orientation. In our example, when noted that no rotation is needed, the standard form becomes 5x^2+5y^2 = 12t, which reveals the equation of a circle when the constants are isolated.

Conic Sections Identification

Identifying conic sections from their equations is crucial for understanding their geometric properties. Our example yields the standard form equation 5x^2+5y^2 = 12t, which fits the general pattern of a circle's equation, x^2 + y^2 = r^2t.
In conic sections identification, coefficients and the presence of terms inform us about the nature of the graph. When the coefficients of x^2t and y^2t are equal and positive, and there's no xy-term, as is the case after simplifying our given equation, the graphical representation would be a circle. By further examining the constants, one can determine the radius and location of the center of the circle.

Sketching Graphs of Conic Sections

Graphing the conic sections provides visual insight into their characteristics. The final step in our problem involves sketching the conic given by the equation in standard form. In this case, we were to sketch a circle with a radius of \(\sqrt{\frac{12}{5}}\)t, centered at the origin. Even though there was no need for rotation, and thus, no changes in the axes, sketching is still vital. It is the graphical representation that solidifies understanding by connecting algebraic equations with geometric figures. For the student, practicing the sketching of conic sections such as circles, ellipses, parabolas, and hyperbolas is important for mastery, as it aids in visualizing and interpreting their equations.