How do you derive the equation of an ellipse given five points ?

To derive the equation of an ellipse given five points, you need to solve a system of equations that arises from substituting the coordinates of the points into the general equation of the ellipse. The general equation of an ellipse can be written as:

\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]

Here are the steps to derive the equation:

1. **Substitute Each Point**: Substitute the coordinates of each of the five points \((x_i, y_i)\) into the general equation. This will give you a system of five linear equations with six unknowns: \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\).

    \[
    \begin{cases}
    A x_1^2 + B x_1 y_1 + C y_1^2 + D x_1 + E y_1 + F = 0 \\
    A x_2^2 + B x_2 y_2 + C y_2^2 + D x_2 + E y_2 + F = 0 \\
    A x_3^2 + B x_3 y_3 + C y_3^2 + D x_3 + E y_3 + F = 0 \\
    A x_4^2 + B x_4 y_4 + C y_4^2 + D x_4 + E y_4 + F = 0 \\
    A x_5^2 + B x_5 y_5 + C y_5^2 + D x_5 + E y_5 + F = 0 \\
    \end{cases}
    \]

2. **Form the Matrix Equation**: Write the system of equations in matrix form \(M \mathbf{p} = \mathbf{0}\), where \(M\) is the matrix of coefficients, \(\mathbf{p}\) is the vector of unknowns \([A, B, C, D, E, F]^T\), and \(\mathbf{0}\) is the zero vector.

    \[
    \begin{pmatrix}
    x_1^2 & x_1 y_1 & y_1^2 & x_1 & y_1 & 1 \\
    x_2^2 & x_2 y_2 & y_2^2 & x_2 & y_2 & 1 \\
    x_3^2 & x_3 y_3 & y_3^2 & x_3 & y_3 & 1 \\
    x_4^2 & x_4 y_4 & y_4^2 & x_4 & y_4 & 1 \\
    x_5^2 & x_5 y_5 & y_5^2 & x_5 & y_5 & 1 \\
    \end{pmatrix}
    \begin{pmatrix}
    A \\
    B \\
    C \\
    D \\
    E \\
    F \\
    \end{pmatrix}
    =
    \begin{pmatrix}
    0 \\
    0 \\
    0 \\
    0 \\
    0 \\
    \end{pmatrix}
    \]

3. **Solve the System**: Since the system is homogeneous (i.e., the right-hand side is zero), it always has the trivial solution \(\mathbf{p} = \mathbf{0}\). However, we are interested in the non-trivial solution. To find this, we solve for the vector \(\mathbf{p}\) that lies in the null space of the matrix \(M\).

4. **Apply Constraints**: The constraint that the ellipse is non-degenerate (i.e., a proper ellipse and not a degenerate conic) ensures that the matrix \(M\) has rank 5, which means the solution space for \(\mathbf{p}\) is one-dimensional. This gives us the coefficients \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) up to a multiplicative constant.

5. **Normalize the Equation**: Normalize the resulting vector \(\mathbf{p}\) so that one of the coefficients (typically \(F\)) is set to a specific value, usually \(1\) or \(-1\).

Would you like further details on any specific step or have any other questions?

Here are some questions you might consider asking:
1. What is the general form of the conic section equation?
2. How can you determine if a given conic section is an ellipse, parabola, or hyperbola?
3. What is the process of finding the foci of an ellipse?
4. How can you derive the equation of a hyperbola given five points?
5. What are some real-world applications of ellipses?

**Tip:** When solving systems of linear equations, it is often useful to use matrix methods like Gaussian elimination or finding the null space to simplify the process.

Learn how to derive the equation of an ellipse given five points using systems of equations and linear algebra. This guide covers the general equation of an ellipse and matrix methods to solve the system. Suitable for advanced high school and college students studying conic sections and linear algebra.

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