The point \((x,{\text{ }}y)\) is rotated through an anticlockwise angle \(\alpha \) about the origin to become the point \((X,{\text{ }}Y)\). Assume that the rotation can be represented by

\[\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right].\]

Show, by considering the images of the points \((1,{\text{ }}0)\) and \((0,{\text{ }}1)\) under this rotation that

\[\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right].\]

By expressing \((x,{\text{ }}y)\) in terms of \((X,{\text{ }}Y)\), determine the equation of the rotated hyperbola in terms of \(X\) and \(Y\).

Verify that the coefficient of \(XY\) in the equation is zero when \(\tan \alpha  = \frac{1}{2}\).

Determine the equation of the rotated hyperbola in this case, giving your answer in the form \(\frac{{{X^2}}}{{{A^2}}} - \frac{{{Y^2}}}{{{B^2}}} = 1\).

Hence find the coordinates of the foci of the hyperbola prior to rotation.

consider \(\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a \\ c \end{array}} \right]\)     (M1)

the image of \((1,{\text{ }}0)\) is \((\cos \alpha ,{\text{ }}\sin \alpha )\)     A1

therefore \(a = \cos \alpha ,{\text{ }}c = \sin \alpha \)     AG

consider \(\left[ {\begin{array}{*{20}{c}} a&b \\ c&d \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0 \\ 1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} b \\ d \end{array}} \right]\)

the image of \((0,{\text{ }}1)\) is \(( - \sin \alpha ,{\text{ }}\cos \alpha )\)     A1

therefore \(b =  - \sin \alpha ,{\text{ }}d = \cos \alpha \)     AG

[3 marks]

\(\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {\cos \alpha }&{ - \sin \alpha } \\ {\sin \alpha }&{\cos \alpha } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \end{array}} \right] \Rightarrow \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} X \\ Y \end{array}} \right]\)

or \(x = X\cos \alpha  + Y\sin \alpha ,{\text{ }}y =  - X\sin \alpha  + Y\cos \alpha \)     A1

substituting in the equation of the hyperbola,     M1

\({(X\cos \alpha  + Y\sin \alpha )^2} - 4(X\cos \alpha  + Y\sin \alpha )( - X\sin \alpha  + Y\cos \alpha )\)

\( - 2{( - X\sin \alpha  + Y\cos \alpha )^2} = 3\)     A1

\({X^2}({\cos ^2}\alpha  - 2{\sin ^2}\alpha  + 4\sin \alpha \cos \alpha ) + \)

\(XY(2\sin \alpha \cos \alpha  - 4{\cos ^2}\alpha  + 4{\sin ^2}\alpha  + 4\sin \alpha \cos \alpha ) + \)

\({Y^2}({\sin ^2}\alpha  - 2{\cos ^2}\alpha  - 4\sin \alpha \cos \alpha ) = 3\)

[??? marks]

when \(\tan \alpha  = \frac{1}{2},{\text{ }}\sin \alpha  = \frac{1}{{\sqrt 5 }}\) and \(\cos \alpha  = \frac{2}{{\sqrt 5 }}\)     A1

the \(XY{\text{ term}} = 6\sin \alpha \cos \alpha  - 4{\cos ^2}\alpha  + 4{\sin ^2}\alpha \)     M1

\( = 6 \times \frac{1}{{\sqrt 5 }} \times \frac{2}{{\sqrt 5 }} - 4 \times \frac{4}{5} + 4 \times \frac{1}{5}\left( {\frac{{12}}{5} - \frac{{16}}{5} + \frac{4}{5}} \right)\)     A1

\( = 0\)     AG

[??? marks]

the equation of the rotated hyperbola is

\(2{X^2} - 3{Y^2} = 3\)     M1A1

\(\frac{{{X^2}}}{{{{\left( {\sqrt {\frac{3}{2}} } \right)}^2}}} - \frac{{{Y^2}}}{{{{(1)}^2}}} = 1\)     A1

\(\left( {{\text{accept }}\frac{{{X^2}}}{{\frac{3}{2}}} - \frac{{{Y^2}}}{1} = 1} \right)\)

[??? marks]

the coordinates of the foci of the rotated hyperbola

are \(\left( { \pm \sqrt {\frac{3}{2} + 1} ,{\text{ }}0} \right) = \left( { \pm \sqrt {\frac{5}{2}} ,{\text{ }}0} \right)\)     M1A1

the coordinates of the foci prior to rotation were given by

\(\left[ {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt 5 }}}&{\frac{1}{{\sqrt 5 }}} \\ { - \frac{1}{{\sqrt 5 }}}&{\frac{2}{{\sqrt 5 }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { \pm \sqrt {\frac{5}{2}} } \\ 0 \end{array}} \right]\)

M1A1

\(\left[ {\begin{array}{*{20}{c}} { \pm \sqrt 2 } \\ { \mp \frac{1}{{\sqrt 2 }}} \end{array}} \right]\)     A1

[??? marks]