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nick1337 · Accepted Answer

The given quadratic form x12+24x1x2−6x22=5 represents a conic section. To determine the type of conic section, we can analyze the discriminant of the quadratic equation.The general form of a quadratic equation in two variables is Ax12+Bx1x2+Cx22+Dx1+Ex2+F=0. Comparing it to the given equation, we have:A=1, B=24, C=−6, D=0, E=0, F=−5.The discriminant of the quadratic form is given by B2−4AC. Let&#039;s calculate it:B2−4AC=(24)2−4(1)(−6)=576+24=600.Since the discriminant is positive, B2−4AC&amp;gt;0, the conic section represents an ellipse.To transform the conic section to its principal axes, we need to perform a change of variables. Let y1 and y2 be the new coordinate variables.To express xT=[x1x2] in terms of the new coordinate vector yT=[y1y2], we can use the transformation matrix P:P=[abcd]We need to find the values of a, b, c, and d that diagonalize the quadratic form. We can find these values by solving the eigenvalue problem:A·P=P·Λwhere Λ is a diagonal matrix containing the eigenvalues.In our case, A=[AB/2B/2C]=[11212−6].Solving the eigenvalue problem, we find the eigenvalues λ1=18 and λ2=−23. The corresponding eigenvectors are [−41] and [31], respectively.We can choose one of these eigenvectors as a column of the transformation matrix P. Let&#039;s choose [−41] as the first column.Normalizing this vector, we get [−4/171/17].Now, we can express xT in terms of yT using the transformation:xT=P·yTSubstituting the values of P and yT, we have:[x1x2]=[−4/17b1/17d]·[y1y2]Simplifying this expression, we get:x1=−417y1+by2x2=117y1+dy2

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