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]