Problem 18
Question
Rotate the axes to eliminate the \(x y\) -term in the equation. Then write the equation in standard form. Sketch the graph of the resulting equation, showing both sets of axes. $$2 x^{2}+x y+2 y^{2}-8=0$$
Step-by-Step Solution
Verified Answer
The equation after rotating the axes (which was figured out to be a rotation by 0 degrees in this case) remained the same: \(2x^{2}+xy+2y^{2}=8\). This is now in standard form. A sketch of the graph would be a rough representation of this equation plotted on a coordinate axis, showing parabolic behaviour. However, to gain a more accurate visual representation, it's recommended to use graph plotting software.
1Step 1: Identifying the Rotation Angle Theta
To get rid of the xy-term, the axes need to be rotated by an angle \(\Theta\). The rotation of axes formula for quadratic equations will be used. The formula is \(\tan 2 \Theta = \frac{B}{A-C}\). For this problem, the equation is \(2x^{2}+xy+2y^{2}-8=0\), so A=2, B=1, and C=2. Substituting this into the rotation of axes formula results in \(\tan 2 \Theta = \frac{1}{2-2} = 0\), which implies that \(\Theta = 0^{\circ}\).
2Step 2: Rotating of Coordinates Axes
Since the rotation angle \(\Theta\) is \(0^{\circ}\), no rotation of the axes is required and the coordinates remain the same. Therefore, the equation \(2x^{2}+xy+2y^{2}-8=0\) remains the same.
3Step 3: Writing the Equation in Standard Form
The current equation is \(2x^{2}+xy+2y^{2}-8=0\). Bring 8 to the other side to have an equation = 0 for the standard form. The standard form of the equation is \(2x^{2}+xy+2y^{2} = 8\).
4Step 4: Sketching the Graph
To sketch the graph of the equation, first plot x and y for a few values that satisfy the equation \(2x^{2}+xy+2y^{2} = 8\). Visually represent the points plotted on the Cartesian plane, then connect the points to get the approximate shape of the graph. Note however, that the actual sketching might require more advanced graphical skills or software support.
Key Concepts
Eliminating xy-termStandard Form of Quadratic EquationGraph SketchingCoordinate Rotation
Eliminating xy-term
Understanding how to eliminate the xy-term from a quadratic equation involves a bit of geometry. In the context of conics, an equation with an xy-term suggests that the axes of the conic are not aligned with the coordinate axes, making it difficult to identify the type of conic section represented. By rotating the coordinate system, the xy-term can be removed, simplifying the equation.
Imagine spinning the entire coordinate grid around the origin: what you're essentially doing is aligning the new axes with the conic's axes. The rotation angle, \(\Theta\), at which the xy-term disappears, can be calculated using the formula \(\tan 2 \Theta = \frac{B}{A-C}\), where A, B, and C are coefficients from the quadratic equation \(Ax^2 + Bxy + Cy^2\). When B, the coefficient of the xy-term, is zero, the axes don't need any rotation—like in our exercise, where the calculated rotation angle turned out to be \(0^\circ\).
Imagine spinning the entire coordinate grid around the origin: what you're essentially doing is aligning the new axes with the conic's axes. The rotation angle, \(\Theta\), at which the xy-term disappears, can be calculated using the formula \(\tan 2 \Theta = \frac{B}{A-C}\), where A, B, and C are coefficients from the quadratic equation \(Ax^2 + Bxy + Cy^2\). When B, the coefficient of the xy-term, is zero, the axes don't need any rotation—like in our exercise, where the calculated rotation angle turned out to be \(0^\circ\).
Standard Form of Quadratic Equation
When studying quadratic equations, encountering them in their standard form is highly beneficial. The standard form of a quadratic equation in two variables x and y is \(Ax^2 + Cy^2 + Dx + Ey + F = 0\), where A, C, D, E, and F are constants, and surprisingly, B is absent because the xy-term has been eliminated.
This format allows us to quickly identify the nature of the graph we are dealing with—whether it's an ellipse, a circle, a parabola, or a hyperbola. It is much more straightforward to glean information about the conic section's orientation, vertex, and axes without the xy-term complicating matters. To get to the standard form, we often complete the square for both x and y, transfer the constant to the other side of the equation, or simply rearrange the terms for clarity, just as we moved the 8 in our example equation to right.
This format allows us to quickly identify the nature of the graph we are dealing with—whether it's an ellipse, a circle, a parabola, or a hyperbola. It is much more straightforward to glean information about the conic section's orientation, vertex, and axes without the xy-term complicating matters. To get to the standard form, we often complete the square for both x and y, transfer the constant to the other side of the equation, or simply rearrange the terms for clarity, just as we moved the 8 in our example equation to right.
Graph Sketching
Graph sketching is an essential tool for visualizing the behavior of equations in algebra and calculus. To sketch a quadratic equation's graph, such as the one in our exercise, you plot several points that satisfy the equation and then connect these points to form the conic section. With the standard form achieved, analyzing and sketching the graph becomes more manageable.
By understanding the coefficients of the standard form, one can predict whether the resulting graph will be an ellipse, circle, parabola, or a hyperbola, along with their various properties such as foci, vertices, and axes of symmetry. Graphing technology can be immensely helpful here, especially when dealing with complex equations, but a basic graph can be sketched by hand by plotting enough points to reveal the shape of the curve.
By understanding the coefficients of the standard form, one can predict whether the resulting graph will be an ellipse, circle, parabola, or a hyperbola, along with their various properties such as foci, vertices, and axes of symmetry. Graphing technology can be immensely helpful here, especially when dealing with complex equations, but a basic graph can be sketched by hand by plotting enough points to reveal the shape of the curve.
Coordinate Rotation
Coordinate rotation is a transformation technique in mathematics where the original coordinate system is rotated around the origin by a specific angle to produce a new coordinate system. This process is pivotal in simplifying equations and understanding the geometry of conics.
It is essentially a method of changing the perspective from where we observe our graphs. In the new system, points have new coordinates according to the rotation. The transformation equations for rotating coordinates by angle \(\Theta\) are \(x' = x\cos \Theta - y\sin \Theta\) and \(y' = x\sin \Theta + y\cos \Theta\), where \(x'\) and \(y'\) are the new coordinates. Rotation can unveil symmetry, reveal the most straightforward representation of a graph, and solve problems that seemed complex in the original coordinate frame, as we aimed to do with eliminating the xy-term in our problem, though it turned out no rotation was required in this instance.
It is essentially a method of changing the perspective from where we observe our graphs. In the new system, points have new coordinates according to the rotation. The transformation equations for rotating coordinates by angle \(\Theta\) are \(x' = x\cos \Theta - y\sin \Theta\) and \(y' = x\sin \Theta + y\cos \Theta\), where \(x'\) and \(y'\) are the new coordinates. Rotation can unveil symmetry, reveal the most straightforward representation of a graph, and solve problems that seemed complex in the original coordinate frame, as we aimed to do with eliminating the xy-term in our problem, though it turned out no rotation was required in this instance.
Other exercises in this chapter
Problem 18
(A) sketch the curve represented by the parametric equations (indicate the orientation of the curve) and (b) eliminate the parameter and write the resulting rec
View solution Problem 18
Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Focus: \(\left(-\frac{3}{2}, 0\right)\)
View solution Problem 18
Find the standard form of the equation of the ellipse with the given characteristics and center at the origin. Vertices: (0,±5)\(;\) passes through the point (4
View solution Problem 19
Identify the conic and sketch its graph. $$r=\frac{2}{2-\cos \theta}$$
View solution