Given that the tangents at the points P and Q on the parabola \({y^2} = 4ax\) are perpendicular, find the locus of the midpoint of PQ.

EITHER

attempt to differentiate       (M1)

let \(y = 2at \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}t}} = 2a\) and \(x = a{t^2} \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}t}} = 2at\)     A1

hence \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}t}} \times \frac{{{\text{d}}t}}{{{\text{d}}x}} = \frac{{2a}}{{2at}} = \frac{1}{t}\)      A1

let P have coordinates \(\left( {at_1^2,\,2a{t_1}} \right)\) and Q have coordinates \(\left( {at_2^2,\,2a{t_2}} \right)\)      (M1)

therefore gradient of tangent at P is \(\frac{1}{{{t_1}}}\) and gradient of tangent at Q is \(\frac{1}{{{t_2}}}\)    A1

since these tangents are perpendicular \(\frac{1}{{{t_1}}} \times \frac{1}{{{t_2}}} =  - 1 \Rightarrow {t_1}{t_2} =  - 1\)    A1

mid-point of PQ is \(\left( {\frac{{a\left( {t_1^2 + t_2^2} \right)}}{2},\,\,a\left( {{t_1} + {t_2}} \right)} \right)\)      A1

\({y^2} = {a^2}\left( {t_1^2 + 2{t_1}{t_2} + t_2^2} \right)\)     M1

\({y^2} = {a^2}\left( {\frac{{2x}}{a} - 2} \right)\,\,\,\left( { \Rightarrow {y^2} = 2ax - 2{a^2}} \right)\)    A1

OR

attempt to differentiate       (M1)

\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4a\)      A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2a}}{y}\)

let coordinates of P be \(\left( {{x_1},\,{y_1}} \right)\) and the coordinates of Q be \(\left( {{x_2},\,{y_2}} \right)\)       (M1)

coordinates of midpoint of PQ are \(\left( {\frac{{{x_1} + {x_2}}}{2},\,\,\frac{{{y_1} + {y_2}}}{2}} \right)\)      M1

if the tangents are perpendicular \(\frac{{2a}}{{{y_1}}} \times \frac{{2a}}{{{y_2}}} =  - 1\)      A1

\( \Rightarrow {y_1}{y_2} =  - 4{a^2}\)

\(y_1^2 + y_2^2 = 4a\left( {{x_1} + {x_2}} \right)\)      A1

\(\frac{{y_1^2 + 2{y_1}{y_2} + y_2^2}}{4} = \frac{{4a\left( {{x_1} + {x_2}} \right) + 2{y_1}{y_2}}}{4}\)      M1

\(\left( {\frac{{{y_1} + {y_2}}}{2}} \right) = 2a\frac{{{x_1} + {x_2}}}{2} - \frac{{8{a^2}}}{4}\)      A1

hence equation of locus is \({y^2} = 2ax - 2{a^2}\)     A1

[9 marks]