Remove the xy-term by rotation of axes, reduce the resulting equation to standard form and trace the curve on the new axes: 13x^2-10xy+13y^2=72
tan(2θ) = -10/0
2θ = -pi/2
θ = -pi/4
x' = (1/√2)x + (1/√2)y
y' = (-1/√2)x + (1/√2)y
13((1/√2)x + (1/√2)y)^2 - 10((1/√2)x + (1/√2)y)((-1/√2)x + (1/√2)y) + 13((-1/√2)x + (1/√2)y)^2 = 72
8x^2 + 18y^2 = 72
x^2/9 + y^2/4 = 1
http://www.wolframalpha.com/input/?i=plot+13x%5E2-10xy%2B13y%5E2%3D72%2C+y%3Dx%2C+y%3D-x
To remove the xy-term by rotation of axes, we need to perform a rotation such that the new coordinate axes are aligned with the major and minor axes of the ellipse defined by the given equation.
The general equation for rotating the axes counterclockwise by an angle θ is given by:
x = x' * cos(θ) - y' * sin(θ)
y = x' * sin(θ) + y' * cos(θ)
To determine the angle of rotation, we can use the formula:
tan(2θ) = 2 * B / (A - C)
where A, B, and C are the coefficients of x^2, xy, and y^2 respectively.
In the given equation: 13x^2 - 10xy + 13y^2 = 72, A = 13, B = -10, and C = 13.
Plugging these values into the formula, we get:
tan(2θ) = 2 * (-10) / (13 - 13) = 2 * (-10) / 0
Since the denominator is zero, we cannot directly solve for θ using this formula. However, we can make an observation:
For equations in the form Ax^2 + Bxy + Cy^2 = D, when A = C, the axes are already aligned with the x and y axes. In this case, the equation is already in standard form.
Let's check if A = C for our equation:
A = 13, C = 13
Since A = C, the equation is already in standard form.
Next, let's find the values of x and y that trace the curve on the new axes. To do this, we can set both x and y to zero to find the coordinates of one point on the ellipse:
13(0)^2 - 10(0)(0) + 13(0)^2 = 0
Hence, one point on the ellipse is (0, 0).
By following these steps, we determine that the given equation is already in standard form and the curve on the new axes passes through the point (0, 0).
To remove the xy-term by rotation of axes, we need to find the angle of rotation that will eliminate the xy-term from the equation. Here's how you can do it:
1. Start with the equation: 13x^2 - 10xy + 13y^2 = 72.
2. The coefficient of the xy-term (-10) is not zero, indicating that the axes are not aligned with the principal axes. To align them, we need to rotate the coordinate system by an angle θ.
3. To find θ, we can use the formula: tan(2θ) = 2(B/(A-C)), where A, B, and C are the coefficients of x^2, xy, and y^2 terms, respectively.
In this case, A = 13, B = -10, and C = 13.
tan(2θ) = 2((-10)/(13-13)) = 2(10/0) = undefined.
4. Since the tan(2θ) is undefined, we cannot use the formula directly to find θ. However, we can determine the angle θ indirectly using another approach.
5. The equation of the ellipse can be written in matrix form as: [x, y] * [A B/2, B/2 C] * [x; y] = 72.
Here, the matrix [A B/2, B/2 C] represents the quadratic form of the equation.
6. Diagonalize the matrix [A B/2, B/2 C] to obtain its eigenvalues (λ1 and λ2) and the corresponding eigenvectors (v1 and v2).
7. Solve for the eigenvalues and eigenvectors of the matrix by solving the characteristic equation: |[A - λ, B/2; B/2, C - λ]| = 0, where |.| indicates the determinant.
[A - λ, B/2; B/2, C - λ] = [13 - λ, -5/2; -5/2, 13 - λ].
Setting the determinant equal to zero, we get: (13 - λ)(13 - λ) - (-5/2)(-5/2) = 0.
Simplifying, we have: λ^2 - 26λ + 169/4 = 0.
Solve this equation to find the eigenvalues λ1 and λ2. Let's say we get λ1 = 1/4 and λ2 = 169/4.
8. Find the corresponding eigenvectors v1 and v2 for each eigenvalue λ1 and λ2.
For λ1 = 1/4, solving the equation ([13 - λ, -5/2; -5/2, 13 - λ]) * [v1; v2] = 0, we get v1 = [-1; 2/5].
For λ2 = 169/4, solving the equation ([13 - λ, -5/2; -5/2, 13 - λ]) * [v1; v2] = 0, we get v2 = [2; 1/5].
9. The angle θ is given by the equation: tan(θ) = (2 * (B/2)) / (A - C), where B/2 represents the off-diagonal term in the matrix [A B/2, B/2 C].
In this case, B/2 = -5/2 and A - C = 0 (since A = C = 13).
tan(θ) = (2 * (-5/2)) / (0) = undefined.
10. Since tan(θ) is undefined, we conclude that the angle of rotation required to remove the xy-term is θ = 45 degrees or π/4 radians.
11. To reduce the resulting equation to standard form and trace the curve on the new axes, we substitute x = x' * cos(θ) - y' * sin(θ) and y = x' * sin(θ) + y' * cos(θ), where x' and y' are the new coordinates.
Applying the substitutions to the original equation, we get:
13(x' * cos(θ) - y' * sin(θ))^2 - 10(x' * cos(θ) - y' * sin(θ))(x' * sin(θ) + y' * cos(θ)) + 13(x' * sin(θ) + y' * cos(θ))^2 = 72.
Simplifying this equation gives the equation in the new coordinate system.
12. The equation obtained after simplification will be in the form: Ax'^2 + Cy'^2 = D, where A, C, and D are constants.
13. Plotting this equation on the new axes will give you the trace of the curve.
You can now follow these steps to remove the xy-term by rotating the axes, reduce the resulting equation to standard form, and trace the curve on the new axes.