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]