91Ó°ÊÓ

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 \(\lambda_{\text{min}}\).

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., \(\lambda_{\text{max}}\).

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:

• 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 \(\lambda_2\).

These eigenvalues represent the squared reciprocals of the semi-axes lengths.

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.

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:

• 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 \( \theta \).

The angle \( \theta \) gives the orientation of the ellipse after rotation, simplifying the original equation into one without the \( xy \) term.