The parabola $P$ has equation ${y^2} = 4ax$. The distinct points ${\text{U}}\left( {a{u^2},{\text{ }}2au} \right)$ and ${\text{V}}\left( {a{v^2},{\text{ }}2av} \right)$ lie on $P$, where $u,{\text{ }}v \ne 0$. Given that ${\rm{U\hat OV}}$ is a right angle, where ${\text{O}}$ denotes the origin,

(a)     show that $v =  - \frac{4}{\mu }$;

(b)     find expressions for the coordinates of ${\text{W}}$, the midpoint of $[{\text{UV}}]$, in terms of $a$ and $u$;

(c)     show that the locus of ${\text{W}}$, as $u$ varies, is the parabola ${P'}$ with equation ${y^2} = 2ax - 8{a^2}$;

(d)     determine the coordinates of the vertex of ${P'}$.

(a)     gradient of ${\text{OU}} = \frac{{2au}}{{a{u^2}}} = \frac{2}{u}$     A1

gradient of ${\text{OV}} = \frac{{2av}}{{a{v^2}}} = \frac{2}{v}$     A1

since the lines are perpendicular,

$\frac{2}{u} \times \frac{2}{v} =  - 1$     M1

so $v =  - \frac{4}{u}$     AG

[3 marks]

(b)     coordinates of ${\text{W}}$ are $\left( {\frac{{a({u^2} + {v^2})}}{2},{\text{ }}\frac{{2a(u + v)}}{2}} \right)$     M1

$= \left( {\frac{a}{2}\left( {{u^2} + \frac{{16}}{{{u^2}}}} \right),{\text{ }}a\left( {u - \frac{4}{u}} \right)} \right)$     A1

[2 marks]

(c)     putting

$x = \frac{a}{2}\left( {{u^2} + \frac{{16}}{{{u^2}}}} \right);{\text{ }}y = a\left( {u - \frac{4}{u}} \right)$     M1

it follows that

${y^2} = {a^2}\left( {{u^2} + \frac{{16}}{{{u^2}}} - 8} \right)$     A1

$= 2ax - 8{a^2}$     AG

Note: Accept verification.

[2 marks]

(d)     since ${y^2} = 2a(x - 4a)$     (M1)

the vertex is at $(4a,{\text{ }}0)$     A1

[2 marks]