(a)   (i)     By first noting that \({\text{OP}} = x\sec \alpha \), show that \(x' = x\cos \theta  - y\sin \theta \) and find a similar expression for \(y'\).

       (ii)     Hence write down the \(2 \times 2\) matrix which represents the anticlockwise rotation about \({\text{O}}\) which takes \({\text{P}}\) to \({\text{P'}}\).

(b)     The ellipse \(E\) has equation \(5{x^2} + 5{y^2} - 6xy = 8\).

(i)     Show that if \(E\) is rotated clockwise about the origin through \(45^\circ\), its equation becomes \(\frac{{{x^2}}}{4} + {y^2} = 1\).

(ii)     Hence determine the coordinates of the foci of \(E\).

(a)     (i)     \(x' = x\sec \alpha \cos (\theta  + \alpha )\)     M1

\( = x\sec \alpha (\cos \theta \cos \alpha  - \sin \theta \sin \alpha )\)     A1

\( = x\cos \theta  - x\tan \alpha \sin \theta \)     A1

\( = x\cos \theta  - y\sin \theta \)     AG

\(y' = x\sec \alpha \sin (\theta  + \alpha )\)     M1

\( = x\sec \alpha (\sin \theta \cos \alpha  + \cos \theta \sin \alpha )\)     A1

\( = x\sin \theta  + x\tan \alpha \cos \theta \)

\( = x\sin \theta  + y\cos \theta \)     A1

(ii)     the matrix \(\left[ {\begin{array}{*{20}{c}}{\cos \theta }&{ - \sin \theta }\\{\sin \theta }&{\cos \theta }\end{array}} \right]\) represents the rotation     A1

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(b)     (i)     the above relationship can be written in the form

\(\left[ \begin{array}{l}x\\y\end{array} \right] = \left[ {\begin{array}{*{20}{c}}{\cos \theta }&{\sin \theta }\\{ - \sin \theta }&{\cos \theta }\end{array}} \right]\left[ \begin{array}{l}{x'}\\{y'}\end{array} \right]\)     M1

let \(\theta  =  - \frac{\pi }{4}\)

\(x = \frac{{x'}}{{\sqrt 2 }} - \frac{{y'}}{{\sqrt 2 }}\)     A1

\(y = \frac{{x'}}{{\sqrt 2 }} + \frac{{y'}}{{\sqrt 2 }}\)

substituting in the equation of the ellipse,

\(5{\left( {\frac{{x'}}{{\sqrt 2 }} - \frac{{y'}}{{\sqrt 2 }}} \right)^2} + 5{\left( {\frac{{x'}}{{\sqrt 2 }} + \frac{{y'}}{{\sqrt 2 }}} \right)^2} - 6\left( {\frac{{x'}}{{\sqrt 2 }} - \frac{{y'}}{{\sqrt 2 }}} \right)\left( {\frac{{x'}}{{\sqrt 2 }} + \frac{{y'}}{{\sqrt 2 }}} \right) = 8\)     M1

\(5\left( {\frac{{{{x'}^2}}}{2} + \frac{{{{y'}^2}}}{2} - x'y'} \right) + 5\left( {\frac{{{{x'}^2}}}{2} + \frac{{{{y'}^2}}}{2} + x'y'} \right) - 6\left( {\frac{{{{x'}^2}}}{2} - \frac{{{{y'}^2}}}{2}} \right) = 8\)     A1

leading to \(\frac{{{{x'}^2}}}{4} + {y'^2} = 1\)     AG

Note: Award M1A0M1A0 for using \(\theta  = \frac{\pi }{4}\) leading to \(\frac{{{{y'}^2}}}{4} + {x'^2} = 1\).

(ii)     in the usual notation, \(a = 2\), \(b = 1\)     (M1)

the coordinates of the foci of the rotated ellipse are \(\left( {\sqrt {3,} {\text{ 0}}} \right)\) and \(\left( { - \sqrt {3,} {\text{ 0}}} \right)\)     A1

the coordinates of the foci of \({\text{E}}\) are therefore \(\left( {\frac{{\sqrt 3 }}{{\sqrt 2 }},{\text{ }}\frac{{\sqrt 3 }}{{\sqrt 2 }}} \right)\) and \(\left( {\frac{{ - \sqrt 3 }}{{\sqrt 2 }},{\text{ }}\frac{{ - \sqrt 3 }}{{\sqrt 2 }}} \right)\)     A1

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