Rotation matrix to construct canonical form of a conicC:9x2+4xy+6y2−10=0.I've found C is a non-degenerate ellipses (computing the cubic and the quadratic invariant), and then I've studied the characteristic polynomial&nbsp;p(t)=det9-t226-t&nbsp;&nbsp;The eigenvalue are&nbsp;t1=5,t2=10, with associated eigenvectors&nbsp;(−1,2),(2,1). Thus I construct the rotation matrix R by putting in columns the normalized eigenvectors (taking care that&nbsp;det(R)=1):R=1512-21&nbsp;&nbsp;Then&nbsp;(x,y)t=R(x′,y′)t, and after some computations I find the canonical form12x′2+45y′2=1.

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Rotation matrix to construct canonical form of a conicC:9x2+4xy+6y2−10=0.I&#039;ve found C is a non-degenerate ellipses (computing the cubic and the quadratic invariant), and then I&#039;ve studied the characteristic polynomial&amp;nbsp;p(t)=det9-t226-t&amp;nbsp;&amp;nbsp;The eigenvalue are&amp;nbsp;t1=5,t2=10, with associated eigenvectors&amp;nbsp;(−1,2),(2,1). Thus I construct the rotation matrix R by putting in columns the normalized eigenvectors (taking care that&amp;nbsp;det(R)=1):R=1512-21&amp;nbsp;&amp;nbsp;Then&amp;nbsp;(x,y)t=R(x′,y′)t, and after some computations I find the canonical form12x′2+45y′2=1.

lernarfnincln6g · Accepted Answer

Explanation:If q(x,y)=9x2+4xy+6y2−10and x′=15(x+2y) and y′=15(−2x+y),then q(x′,y′)=5x2+10y2−10. So, q(x′,y′)=0⇔12x2+y2=1.

Rotation matrix to construct canonical form of a

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