Chapter 8: Problem 24
Find the lengths of the semi-axes of the ellipse $$ 73 x^{2}+72 x y+52 y^{2}=100 $$ and determine its orientation.
Short Answer
Expert verified
Eigenvalues provide semi-axis lengths, and orientation is determined by \( \theta \) found from \( \tan(2\theta) = \frac{24}{7} \).
Step by step solution
01
- Identify and write the equation
The given equation of the ellipse is \[73 x^{2} + 72 x y + 52 y^{2} = 100\]
02
- Convert to standard form
First convert the given equation into standard form by using the rotation of axes technique to eliminate the cross-term (xy term). This will involve expressing the given ellipse in rotated coordinates where there is no xy term. Assume the given form \[Ax^2 + Bxy + Cy^2 = 1\] where A = 73, B = 72, C = 52, and equate it to 1: \[Ax'^2 + Cy'^2 = 1\]
03
- Use rotation formulas
The angle of rotation \( \theta \) where the xy term vanishes is given by: \[ \cot(2\theta) = \frac{A - C}{B} = \frac{73 - 52}{72} = \frac{21}{72} = \frac{7}{24}\] Solve for \(\theta\) using \(\cot(2\theta) = \frac{7}{24}\). \(\tan(2\theta) = \frac{24}{7}\)
04
- Find Eigenvalues for semi-axes
To find the semi-major and semi-minor axis lengths, solve for the eigenvalues of the quadratic form represented by the matrix: \[ \begin{bmatrix} 73 & 36 \36 & 52 \end{bmatrix} \] Calculate the eigenvalues by solving the characteristic equation det(\(A - \lambda I\)) = 0: \[ (73 - \lambda)(52 - \lambda) - (36)^2 = \lambda^2 - 125\lambda + 1756 = 0 \] Solve this quadratic equation to get the eigenvalues \(\lambda_1\) and \(\lambda_2\).
05
- Determine the semi-axes lengths
Once the eigenvalues \(\lambda_1\) and \(\lambda_2\) are determined, the semi-major and semi-minor axis lengths can be calculated: \[ a = \sqrt{\frac{1}{\lambda_{min}}} \] \[ b = \sqrt{\frac{1}{\lambda_{max}}} \] Plug in the eigenvalues obtained to find the lengths of the semi-axes.
06
- Determine the orientation
The orientation of the ellipse in the xy-plane is given by the angle \(\theta\) used in the rotation which can be inferred from \(\tan(2\theta) = \frac{24}{7}\). Calculate \(\theta\) using \[ \theta = \frac{1}{2} \tan^{-1}(\frac{24}{7}) \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Semi-Major Axis
The semi-major axis of an ellipse is the longest radius of the ellipse. It extends from the center to the farthest point on the ellipse's boundary. To determine its length, you can use the eigenvalues obtained from solving the characteristic equation of the quadratic form of the ellipse.
For the given ellipse \(73 x^{2} + 72 x y + 52 y^{2} = 100\), the semi-major axis length \(a\) is calculated by \(\text{semi-major axis length} = a = \sqrt{\frac{1}{\lambda_{\text{min}}}}\), where \(\lambda_{\text{min}}\) is the smaller eigenvalue. Eigenvalues are derived from the matrix representation of the quadratic form and solving the characteristic equation.
If \(\lambda_1\) and \(\lambda_2\) are the eigenvalues, the semi-major axis is associated with \(\lambda_{\text{min}}\).
For the given ellipse \(73 x^{2} + 72 x y + 52 y^{2} = 100\), the semi-major axis length \(a\) is calculated by \(\text{semi-major axis length} = a = \sqrt{\frac{1}{\lambda_{\text{min}}}}\), where \(\lambda_{\text{min}}\) is the smaller eigenvalue. Eigenvalues are derived from the matrix representation of the quadratic form and solving the characteristic equation.
If \(\lambda_1\) and \(\lambda_2\) are the eigenvalues, the semi-major axis is associated with \(\lambda_{\text{min}}\).
Semi-Minor Axis
The semi-minor axis is the shortest radius of the ellipse, stretching from the center to the closest point on the ellipse. Finding this length involves a similar process to that of the semi-major axis but uses the larger eigenvalue.
For the ellipse\(73 x^{2} + 72 x y + 52 y^{2} = 100\), the semi-minor axis length \(b\) is given by \(\text{semi-minor axis length} = b = \sqrt{\frac{1}{\lambda_{\text{max}}}}\), where \(\lambda_{\text{max}}\) is the larger eigenvalue calculated from the characteristic equation. Solving the characteristic equation provides the eigenvalues which are then used to determine the length of both axes.
The semi-minor axis is associated with the larger eigenvalue, i.e., \(\lambda_{\text{max}}\).
For the ellipse\(73 x^{2} + 72 x y + 52 y^{2} = 100\), the semi-minor axis length \(b\) is given by \(\text{semi-minor axis length} = b = \sqrt{\frac{1}{\lambda_{\text{max}}}}\), where \(\lambda_{\text{max}}\) is the larger eigenvalue calculated from the characteristic equation. Solving the characteristic equation provides the eigenvalues which are then used to determine the length of both axes.
The semi-minor axis is associated with the larger eigenvalue, i.e., \(\lambda_{\text{max}}\).
Eigenvalues
Eigenvalues are critical in determining the lengths of the semi-axes of an ellipse. They are found by solving the characteristic equation derived from the quadratic form of the ellipse's equation.
In the context of the exercise \(73 x^{2} + 72 x y + 52 y^{2} = 100\), eigenvalues come from the matrix representation: \[ \begin{bmatrix} 73 & 36 \ 36 & 52 \ \end{bmatrix} \].
To find these values:
In the context of the exercise \(73 x^{2} + 72 x y + 52 y^{2} = 100\), eigenvalues come from the matrix representation: \[ \begin{bmatrix} 73 & 36 \ 36 & 52 \ \end{bmatrix} \].
To find these values:
- Set up the characteristic equation: \(\text{det}(A - \lambda I) = 0\).
- Solve \( (73 - \lambda)(52 - \lambda) - (36)^{2} = \lambda^{2} - 125\lambda + 1756 = 0\).
- The solutions to this quadratic equation are your eigenvalues \(\lambda_1\) and \(\lambda_2\).
Characteristic Equation
The characteristic equation is a polynomial that is used to find the eigenvalues of a matrix. For an ellipse, these eigenvalues help determine the lengths of the semi-major and semi-minor axes.
For the given ellipse equation \(73 x^{2} + 72 x y + 52 y^{2} = 100\), the transformation matrix is \[\begin{bmatrix} 73 & 36 \ 36 & 52 \ \end{bmatrix} \].
The characteristic equation for this matrix is \(\text{det}(A - \lambda I) = 0\). This translates to: \( (73 - \lambda)(52 - \lambda) - (36)^{2} = \lambda^{2} - 125\lambda + 1756 = 0\).
Solving this quadratic equation provides the eigenvalues, which are crucial for defining the ellipse's axes.
For the given ellipse equation \(73 x^{2} + 72 x y + 52 y^{2} = 100\), the transformation matrix is \[\begin{bmatrix} 73 & 36 \ 36 & 52 \ \end{bmatrix} \].
The characteristic equation for this matrix is \(\text{det}(A - \lambda I) = 0\). This translates to: \( (73 - \lambda)(52 - \lambda) - (36)^{2} = \lambda^{2} - 125\lambda + 1756 = 0\).
Solving this quadratic equation provides the eigenvalues, which are crucial for defining the ellipse's axes.
Rotation of Coordinate Axes
Sometimes, to simplify the equation of an ellipse, we perform a rotation of the coordinate axes to eliminate the cross-term (\( xy \) term). This makes it easier to identify key features of the ellipse like its axes lengths and orientation.
For the given ellipse \(73 x^{2} + 72 x y + 52 y^{2} = 100\), eliminate the \( xy \) term through rotation:
For the given ellipse \(73 x^{2} + 72 x y + 52 y^{2} = 100\), eliminate the \( xy \) term through rotation:
- Calculate the angle of rotation \( \theta \) using \( \cot(2\theta) = \frac{A - C}{B} \).
- For \( A = 73 \), \( C = 52 \), and \( B = 72 \), this results in \( \cot(2\theta) = \frac{21}{72} = \frac{7}{24} \).
- Solve \( \tan(2\theta) = \frac{24}{7} \) for \( \theta \).