Short Answer

In Exercise 40 through 43, consider the problem of fitting a conic through $m$ given points $P_{1} (x_{1}, y_{1}), . ......, P_{m} (x_{m}, y_{m})$ in the plane; see Exercise 53 through 62 in section 1.2. Recall that a conic is a curve in $ℝ^{2}$ that can be described by an equation of the form , $f (x, y) = c_{1} + c_{2} x + c_{3} y + c_{4} x^{2} + c_{5} x y + c_{6} y^{2} = 0$ where at least one of the coefficients is non zero.

40. Explain why fitting a conic through the points $P_{1} (x_{1}, y_{1}), . ......, P_{m} (x_{m}, y_{m})$ amounts to finding the kernel of an $m \times 6$ matrix $A$ . Give the entries of the row of $A$ .

Note that a one-dimensional subspace of the kernel of defines a unique conic, since the equations $f (x, y) = 0$ and $k f (x, y) = 0$ describe the same conic.

Thus, the entries for $i t h$ row are $[\begin{matrix} 1 & x_{i} & y_{i} & x_{i}^{2} & x_{i} y_{i} & y_{i}^{2} \end{matrix}]$ .

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TABLE OF CONTENTS

Step 1: Given information

A conic is a curve in $ℝ^{2}$ that can be described by an equation of the form , $f (x, y) = c_{1} + c_{2} x + c_{3} y + c_{4} x^{2} + c_{5} x y + c_{6} y^{2} = 0$ where at least one of the coefficients $c_{i}$ is non-zero.

Step 2:Explanation

To fit a conic through the points $P_{1} (x_{1}, y_{1}), ....., P_{m} (x_{m}, y_{m})$ is equivalent to find the kernel of an $m x_{6}$ of matrix and also these are solution to the equation.

$f (x, y) = c_{1} + c_{2} x + c_{3} y + c_{4} x^{2} + c_{5} x y + c_{6} y^{2} = 0$

Thus, the $i t h$ row have entries as follows:

$[\begin{matrix} 1 & x_{i} & y_{i} & x_{i}^{2} & x_{i} y_{i} & y_{i}^{2} \end{matrix}]$

Hence, the entries for $i t h$ row are $[\begin{matrix} 1 & x_{i} & y_{i} & x_{i}^{2} & x_{i} y_{i} & y_{i}^{2} \end{matrix}]$ .

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