(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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